Algebraic Differential Calculus [Web draft ed.]

Table of contents :
Title: Algebraic Differential Calculus
Contents
Chapter I. Kaehler Differentials
1_1 Derivations Tangent Spaces and Vector Fields
1_2 The Module of Kaehler Differentials Tangent Bundles
1_3 Differential Modules of Field Extensions
1_4 Differential Modules of Local Rings
1_5 Differential Modules of Affine Algebras and their Localizations
1_6 Smooth Algebras
1_7 Differential Modules of Complete Intersections
1_8 Existence of Universally Finite Derivations
1_9 The Kaehler Differents (Jacobian Ideals) of an Algebra
Introduction to the bibliography I
Chapter I I. Differential Operators
2_1 Basic Properties of Differential Operators
2_2 Universal Differential Operators
2_3 Functorial Properties
2_4 Differential Operators of Smooth Algebras
Introduction to the bibliography I I
Chapter I I I. Differential Forms
3_1 Differential Algebras
3_2 The Universal Differential Algebra. Universal Extension of Differential Algebras
3_3 Functorial Properties of the Universal Extension
Introduction to the bibliography I I I
Chapter IV. Connections
4_1 Connections on Modules
4_2 Curvature and Torsion of Connections
4_3 Riemannian Algebras
4_4 The Levi-Civita Connection
4_5 Curvature of Riemannian Algebras
Solutions to some of the problems
Appendices
A Commutative Algebras
B Dimension Formulas in Algebras of Finite Type. Quasifinite and Equidimensional Algebras
C Complete Intersections
D The Fitting Ideals of a Module
Bibliography

Citation preview

Algebraic Dierential Calculus

Contents Chap.I. Kahler Dierentials

1. Derivations, Tangent Spaces and Vector Fields x 2. The Module of K ahler Dierentials. Tangent Bundles. x 3. Dierential Modules of Field Extensions x 4. Dierential Modules of Local Rings x 5. Dierential Modules of Ane Algebras and their Localizations x 6. Smooth Algebras x 7. Dierential Modules of Complete Intersections x 8. Existence of Universally Finite Derivations x 9. The K ahler Dierents (Jacobian Ideals) of an Algebra Introduction to the Bibliography I

1 14 34 49 64 84 94 101 113 120

1. Basic Properties of Dierential Operators x 2. Universal Dierential Operators x 3. Functorial Properties x 4. Dierential Operators of Smooth Algebras Introduction to the Bibliography II

123 131 142 153 163

1. Dierential Algebras x 2. Universal Extension of a Dierential Algebra x 3. Description of Universal Extensions in Special Cases Introduction to the Bibliography III

164 175 192 200

x

Chap.II. Dierential Operators x

Chap.III. Dierential Forms x

Chap.IV. Connections

1. Connections on Modules x 2. Curvature and Torsion of Connections x 3. Riemannian Algebras x 4. The Levi-Civit a Connection x 5. Curvature of Riemannian Algebras Solutions to Some of the Problems

x

Appendices

A. Commutative Algebras B. Dimension Formulas in Algebras of Finite Type Quasinite and Equidimensional Algebras C. Complete Intersections D. The Fitting Ideals of a Module

Bibliography

I

201 215 224 236 247 258 260 264 277 287 296

Chapter I. Kahler Dierentials

x 1. Derivations, Tangent Spaces and Vector Fields In this section we shall discuss the notion of derivation. Some examples of derivations are given, and some basic rules about derivations are proved. Then we introduce tangent spaces at the points of the spectrum of a ring and vector elds on the spectrum. These concepts are analogous to the corresponding notions for dierential manifolds, though none of the notions is a specialization of the other.

Let R0 be a ring, R an R0 {algebra and M an R {module. M can be regarded as an R0 {module via the structure homomorphism R0 ! R . 1.1. Definition. A

derivation of R=R0 into M is a mapping d : R ! M having the

following properties: a) d ist R0 {linear. b) (Product formula) For all a b 2 R

d(ab) = adb + bda Such derivations will also be called R0 {derivations. In case R0 = Z derivations of R= Z will simply be called derivations of R (absolute derivations). Since R itself is an R {module, we may consider in particular R0 {derivations d : R ! R . Examples: 1.2. Examples from analysis.

Let U  R n be a non-empty open set, R := E (U ) the R {algebra of C 1 {functions f : U ! R . Then the partial dierential operator

@ : E (U ) ! E (U ) @Xi

@f ) (f 7! @X i

is a derivation of E (U )= R . For P 2 U let R := EP be the R {algebra of germs of C 1 {functions at P . We have an induced mapping @ :E !E P @X P i

@f ] of its partial which sends the germ f ] of a C 1 {function f at P to the germ @X i derivative and which is an R {derivation.

1

R may be regarded as an EP {module via the homomorphism EP Then @f (P ) i : EP ! R  i f ] := @X

! R , f ] ! f (P ).

i

is also an R {derivation. Similar examples occur in complex analysis: If R := O(U ) is the C {algebra of holomorphic functions on a non-empty open set U  C , then the derivative gives a C {derivation of R into R . 1.3. Trivial derivations.

For any ring R and any R {module M

d: R ! M

(dr = 0 for all r 2 R )

is a derivation, which is called trivial. 1.4. Formal partial derivative in polynomial and power series algebras.

Let R := R0 X1  : : :  Xn ]] be the R0P{algebra of formal power series in indeterminates X1  : : :  Xn . For a power series F = 1:::n X11 : : : Xnn (1:::n 2 R0 ) the formal @F is dened by partial derivative @X i (1)

@F := P   1 i ;1 : : : X n i  1 :::n X1 : : : Xi n @Xi

The mapping

@ : R ! R (F 7! @F ) @Xi @Xi is an R0 {derivation, as is easily checked. Let fdX1  : : :  dXng be a set of indeterminates and let M := RdX1  RdXn be the free R {module on the basis fdX1  : : :  dXn g , the module of "formal dierentials". Then the mapping @F  dX +  + @F  dX ) d : R ! M (dF = @X 1 n @X 1

n

is an R0 {derivation of R into M . The element dF is called the formal dierential of F. If F 2 R0 X1 : : :  Xn ], the polynomial ring in X1  : : :  Xn over R0 , then we have n @F 2 R0 X1  : : :  Xn ] and dF 2 L obviously @X R0 X1 : : :  Xn]dXi . Hence @X@ i and d i i=1 induce R0 {derivations on the polynomial ring. @F for any F 2 R0 X ]]. Observe that In the case n = 1 we write F 0 (X ) instead of @X formal integration in R0 X ]] is only possible if Q  R0 . 2

L

P

1.5. The Euler derivation of a graded ring.

Let R = Rn be a graded ring. Then R0 is a subring of R . For r 2 R let r = rn n2 Z n2 Z be the decomposition into homogeneous elements rn 2 Rn . The mapping

d: R ! R

(r 7!

P nrn)

n2 Z

is an R0 {derivation, as is easily checked. It is called the Euler derivation of R . 1.6. The universal derivation.

As is well-known, the kernel I of the mapping

 : R R0 R ! R

(a  b 7! a  b)

is generated by the elements r  1 ; 1  r (r 2 R). The ring R R0 R has two R {module structures, given by the ring homomorphisms R ! R R0 R (a 7! a  1) and R ! R R0 R (a 7! 1  a). They induce two R {module structures on I . From those, we get induced R {module structures of M := I=I 2 , which however coincide: For r 2 R and a  1 ; 1  a + I 2 2 I=I 2 we have r  1  (a  1 ; 1  a + I 2) = 1  r  (a  1 ; 1  a + I 2), since (r  1 ; 1  r)  (a  1 ; 1  a) 2 I 2 . Dene a mapping d : R ! M (a 7! a  1 ; 1  a + I 2) Then clearly d is R0 {linear. Moreover, for a b 2 R

d(ab) = ab  1 ; 1  ab + I 2 = (a  1)  (b  1 ; 1  b + I 2) + (1  b)  (a  1 ; 1  a + I 2) = (a  1)  (b  1 ; 1  b + I 2) + (b  1)  (a  1 ; 1  a + I 2) = adb + bda

The special importance of this derivation will be explained in the next section (see 2.3). 1.7. Rules about derivations.

If R0 is an algebra over a ring R00 , then any R0 {algebra (any R0 {module) is also an R00 {algebra (an R00 {module). Clearly any R0 {derivation is also an R00 {derivation. In particular any R0 {derivation is a Z {derivation.

! M be an R {derivation of R into an R {module M . Then: a) d(1) = 0, hence d(n  1) = 0 for all n 2 Z . Pn b) d(x  : : :  xn) = x  : : :  x^i  : : :  xn dxi for all x  : : :  xn 2 R , in particular i 1.8. Let d : R 1

0

=1

1

1

dxn = nxn;1dx

for all x 2 R

3

c) P := ker d := f 2 R j d = 0g is a subring of R , and d is a derivation of R=P . The image of R0 in R is contained in P . d) For any ideal I of R and any n > 0 we have

d(I n)  I n;1  M If R and M are endowed with the I {topology, then d is continuous. e) Quotient formula. If s 2 R is a unit, then for any r 2 R

d(rs;1 ) = (s;1 )2 (sdr ; rds)

in particular

ds;1 = ;(s;1 )2ds

f) Suppose N  R0 is multiplicatively closed and R0 ! R induces a ring homomorphism (R0 )N ! R . Then d is also a derivation of R=(R0 )N . g) Derivation of polynomials. Let f 2 R0 X1 : : :  Xn] be a polynomial and let x1  : : :  xn 2 R . Then

d(f (x1  : : :  xn )) =

Pn

@f (x  : : :  x )dx 1 n i i=1 @Xi

In particular, if g1 : : :  gn 2 R0 Y1 : : :  Ym ]] are power series in indeterminates Y1  : : :  Ym and h := f (g1  : : :  gn), then n @f @h = P (g1  : : :  gn)  @gi @Yk i=1 @Xi @Yk

(k = 1 : : :  m)

More generally for f1  : : :  ft 2 R0 X1 : : :  Xn] and hi := fi (g1  : : :  gn) (i = 1 : : :  t) we have the chain rule for the Jacobian matrices @ (h1  : : :  ht ) = @ (f1  : : :  ft ) (g  : : :  g )  @ (g1  : : :  gn) n @ (Y  : : :  Y ) @ (Y1 : : :  Ym ) @ (X1  : : :  Xn ) 1 1 m h) Derivation of determinants. Let  = det(aik ) be the determinant of an n n matrix (aik ) with coecients aik 2 R . Then

d =

Pn ikdaik

ik=1

where ik := (;1)i+k Aik and Aik is the minor of (aik ) obtained by deleting the i {th row and k {th column. 4





 Z {linear, we obtain d(n  1) = n  d(1) = 0 for all n 2 Z . b) The rst formula is obtained from the product formula by induction, the second follows, if we set x =  = xn = x . c) Clearly, P is a subgroup of (R +). The product formula shows that P is a subring and that d is P {linear. For r 2 R we have d(r  1) = r d(1) = 0, hence P contains the image of R in R . d) The elements of I n are nite sums of products x  : : :  xn with xi 2 I (i = 1 : : :  n). The assertion follows from b). e) From 0 = d(1) = d(s  s; ) = sds; + s; ds we obtain ds; = ;(s; ) ds . The quotient

Proof: a) We have d(1) = d(1 1) = 1 d(1) + 1 d(1), hence d(1) = 0. Since d is 1

0

0

0

0

0

1

1

1

1

1

1 2

formula follows now by another application of the product formula. f) By e) the image of (R0)N in R is contained in ker d . So c) can be applied. g) Let P f = 1:::n X11 : : : Xnn (1 :::n 2 R0) P Since d is R0 {linear, we have df (x1  : : :  xn) = 1:::n d(x1 1 : : : xnn ) and from b) we Pn obtain d(x1 1 : : : xnn ) = i x1 1 : : : xi i ;1 : : : xnn dxi . Hence i=1

df (x1  : : :  xn ) =

Pn (P i ::: x : : : xi ; : : : xn )dxi = Pn @f (x  : : : xn)dxi

i=1

1

n 1

i 1

1

n

i=1 @Xi

1

Observe that this proof does not work for power series. Thus the proof does not cover the chain rule for power series f1 : : :  ft 2 R0 X1 : : :  Xn]], though this chain rule is true (with a more complicated proof) in case hi := fi (g1  : : :  gn) is dened (i = 1 : : :  t). h) If  = det(Xik ) is a determinant with indeterminates Xik (i k = 1 : : :  n) as coef@  = ik , as is seen by expanding the determinant with respect to the cients, then @X ik i {the row. The formula of h) follows from g). Examples of algebras with only trivial derivations are given in the following two propositions. In these cases one cannot expect much information from the theory of derivations. 1.9. Proposition. Let K=K0 be an algebraic eld extension and let K 0 be the subeld

of all separable elements of K=K0 . Then K 0  ker d for any K0 {derivation d : K ! M . In particular, if K=K0 is separably algebraic, any K0 {derivation d : K ! M is trivial.

2 K 0 let f 2 K X ] be the minimal polynomial of0 x over K . Since f 6 0. Then is separable, we have an equation f = (X ; x)  g with g 2 K X ], g(x) = 0 f (x) = g(x) = 6 0 by the product formula for derivations. By 1.8g) Proof: For x

hence dx = 0 and x 2 ker d .

0

0

0 = d(0) = d(f (x)) = f 0 (x)dx 5

1.10.Proposition. Let R be a ring whose characteristic is a prime number p . Then

Rp  ker d for any derivation d : R ! M . In particular, if K is a perfect eld of characteristic p > 0, any derivation d : K ! M is trivial.

2 R pwe have drp = prp; dr = 0. If K is a perfect eld of characteristic p > 0, then K = K , hence da = 0 for any a 2 K . Proof: For r

1

For an algebra R=R0 and an R {module M let DerR0 (R M ) denote the set of all R0 {derivations d : R ! M . Clearly, for d d0 2 DerR0 (R M ) and r 2 R we have

d + d0 2 DerR0 (R M ) and rd 2 DerR0 (R M ) hence DerR0 (R M ) is an R {module. If I  R is an ideal with IM = 0, then DerR0 (R M ) is even an R=I {module. 1.11.Examples: a) Let X be an n {dimensional C 1 {manifold, P

2 X , and let EP

be

the R {algebra of germs of C 1 {functions at P . Then TP (X ) := Der R (EP  R ) is called the tangent space of X at P . It can be shown that TP (X ) is an n {dimensional

R {vectorspace, and that the derivations d : EP ! R are in natural one-to-one correspondence with the "tangent vectors" dened in a more geometric way (tangent vectors of curves passing through P ), see BJ], x 2. b) An algebraic version of this concept is the following. Let X := Spec R , and let x 2 X be the point corresponding to the prime ideal p of R . Let k(x) be the residue eld of the local ring Ox = Rp at x . Further let X0 := Spec R0 and  : X ! X0 the morphism corresponding to the structure homomorphism R0 ! R . The k(x){vector space TX=X0 (x) := DerR0 (Ox k(x)) is called the tangent space of X=X0 at x . Its elements are called tangent vectors of X=X0 at x . Later we will try to determine the dimension of TX=X0 (x) (see 4.18c). For R=R0 and M as above let N be another R {module. For each d 2 DerR0 (R M ) and each ` 2 HomR(M N ) the composition ` d is an R0 { derivation from R into N , as denition 1.1 immediately shows. Thus we obtain an R {linear map DerR0 (R `): DerR0 (R M ) ! DerR0 (R N )

(d 7! ` d)

which makes DerR0 (R M ) a covariant functor of M . 1.12.Proposition. DerR0 (R M ), as a functor of M , is left exact: If 0

is an exact sequence of R {modules, then DerR0 (R)

DerR0 (R )

! M ;! N ;! P

0 ! DerR0 (R M ) ;;;;;;;;! DerR0 (R N ) ;;;;;;;;! DerR0 (R P ) is also exact. 6

Proof: Set  := DerR0 (R ),   := DerR0 (R  ). It is clear that  is injective

and im   ker   . Conversely, if   (d) =  d = 0 for d 2 DerR0 (R N ), then dR  ker  = im  and hence d 2 im  . Suppose now that S=R0 is another algebra and ' : S ! R is an R0 {algebra homomorphism. Then for each R0 {derivation d : R ! M the composition d ' is an R0 {derivation from S into M , if M is considered as an S {module via ' . In fact, it suces to show the product formula: For s s0 2 S we have

d('(ss0 )) = d('(s)  '(s0 )) = '(s)  d'(s0 ) + '(s0 )  d'(s) = s  d'(s0 ) + s0  d'(s)

Therefore we obtain an S {linear map

DerR0 (' M ): DerR0 (R M ) ! DerR0 (S M )

(d 7! d ')

1.13.Proposition. The sequence DerR0 ('M )

j

0 ! DerS (R M ) ;! DerR0 (R M ) ;;;;;;;;! DerR0 (S M )

is exact. (Any d 2 DerS (R M ) is also an R0 {derivation from R into M , and j is dened to be the corresponding inclusion mapping).

 ker ' . Conversely, if we have '(d) = d ' = 0 for d 2 DerR (R M ), then d is an S {derivation and hence d 2 im j . 1.14. Examples: a) Let X and Y be C 1 {manifolds and F : X ! Y a C 1 {mapping. Then F induces for any P 2 X an R { algebra homomorphism 'P : EF P ! EP ( g] 7! g F ])

Proof: Set ' := DerR0 (' M ). It is clear that im j 0

( )

where g] denotes the germ of a function g . The induced mapping

Der R ('P  R ): Der R (EP  R ) ! Der R (EF (P )  R )

maps the tangent space TP (X ) = Der R (EP  R ) at P into the tangent space TF (P ) (Y ) = Der R (EF (P )  R ) at F (P ). In analysis this mapping is denoted TP (F ) or dFP , and is called the dierential of F at P (or the linear approximation of F at P). b) Similarly let X := Spec R , Y := Spec S , and let F : X ! Y be the map induced by ' . For x 2 X set y := F (x). Further let  : X ! X0 and  : Y ! X0 be induced by the structure maps of R=R0 and S=R0 . The homomorphism ' induces a local homomorphism 'x : Oy ! Ox of R0 {algebras and an injection ` : k(y) ! k(x) of their residue elds. Thus we have linear maps DerR0 ('x k(x)) : DerR0 (Ox k(x)) ! DerR0 (Oy  k(x)) 7

and

DerR0 (Oy  `) : DerR0 (Oy  k(y)) ! DerR0 (Oy  k(x))

If ` is bijective, i.e. if Oy and Ox have the same residue eld, then DerR0 ('x k(x)) is denoted by TF=X0 (x). Thus we have a k(x){linear map

TF=X0 (x) : TX=X0 (x) ! TY=X0 (y) and by 1.13 a canonical exact sequence relating the tangent spaces at x and y TF=X0 (x)

0 ! TX=Y (x) ! TX=X0 (x) ;;;;;;! TY=X0 (y) Consider now the canonical epimorphism : R ! R=I of R onto a residue class algebra R=I and let M be an R=I {module. For d 2 DerR0 (R M ) the restriction d jI : I ! M is an R {linear map with d jI (I 2) = 0. It induces an R=I {linear map I=I 2 ! M . Therefore we have an R {linear map

 : DerR0 (R M ) ! HomR=I (I=I 2 M ) (d 7! d jI ) 1.15.Proposition. The following sequence of R {modules is exact: DerR0 ( M )

0 ! DerR0 (R=I M ) ;;;;;;;;! DerR0 (R M ) ;! HomR=I (I=I 2 M )

Proof: Since DerR (R=I M ) = 0 it follows from 1.13 that  := DerR0 (  M ) is injective.

If (d) = 0 for d 2 DerR0 (R M ), then d jI = 0 and an R0 {derivation R=I induced, hence d 2 im  . Clearly any d 2 im  belongs to ker  . 1.16.Example: For Z := Spec R=I the map F : Z

!M

is

! X induced by is injective, and

for x 2 Z the local rings OZx = Ox=I Ox and Ox have the same residue eld k(x). By 1.15 we have an exact sequence of k(x){vector spaces 0 ! TZ=X0 (x) ! TX=X0 (x) ! HomOZx (I Ox=I 2Ox k(x)) In particular TZ=X0 (x) may be regarded as a subspace of TX=X0 (x), and by Nakayama dimk(x) TX=X0 (x)=TZ=X0 (x) (I Ox) 8

1.17.Proposition. For an R0 {derivation d : R

!M

and a multiplicatively closed set N  R there is exactly one R0 {derivation dN : RN ! MN such that the canonical diagram

R

d

M

RN

dN

MN

;

(1) is commutative. The derivation dN is given by the quotient formula

dN ( r ) = dr ;2 rd

(r 2 R  2 N )

Further

DerR0 (R M ) ! DerR0 (RN  MN ) (d 7! dN ) is R {linear. It induces an RN {linear map  : DerR0 (R M )N ! DerR0 (RN  MN ). Proof: That dN has to satisfy the formula above follows from the quotient formula 1.8e).

One checks easily that by the formula a well-dened map dN : RN ! MN is given which is an R0 {derivation and makes the diagram (1) commutative. The last statement of the proposition is clear. See 2.8 for conditions under which  is injective or bijective. The R {module DerR0 (R R) is called the derivation module of the algebra R=R0 . It is also denoted by DerR0 R .

E

1.18.Examples: a) Let R := (X ) be the algebra of C 1 {functions on a dierential

manifold X and R0 := R . Then the elements V 2 DerRE (X ) are called vector elds on X . For P 2 X the canonical homomorphism P : E (X ) ! EP associating with each function f its germ f ] at P is surjective. This corresponds to localization in algebra. Each R {derivation V : E (X ) ! E (X ) induces a unique derivation VP : EP ! EP such that P V = VP P . This is analogous to 1.17. Let "P : EP ! R be the evaluation at P . Then V (P ) := "P VP 2 DerR(EP  R) = TP X is the tangent vector at P associated with V . Moreover for V W 2 DerRE (X ) we have V (P ) = W (P ) for all P 2 X if and only if V = W . Thus the derivations (vector elds) V 2 DerRE (X ) can be identied with the sets fV (P )gP 2X of their tangent vectors. b) Similarly for R=R0 resp. X=X0 as above DerR0 R is also called the module of vector elds on X with respect to X0 . For V 2 DerR0 R and x 2 X the image V (x) of V by the composed map DerR0 R ;;;! DerR0 Ox ! DerR0 (Ox k(x)) = TX=X0 (x) 1:17

9

is called the tangent vector associated with V at x . If R is a reduced ring, then for V W 2 DerR0 R we have V = W if and only if V (x) = W (x) for all x 2 X . In fact, it suces to show V = 0 if V (x) = 0 for all x 2 X . This condition implies that R is mapped by V into p Rp \ R = p for all prime ideals p of R , hence to 0 when R is reduced.

c) Let R = R0 X1 : : :  Xn] be a polynomial algebra. Then formula 1.7a) shows that DerR0 R is generated as an R {module by the partial derivatives @X@ 1  : : :  @X@ n . They form Pn in fact a basis of DerR0 R since they are linearly independent: If fi @X@ i = 0 (fi 2 R),

Pn then fi @X

i=1

= 0 (k = 1 : : :  n), i.e. fk = 0 (k = 1 : : :  n). Thus we have i=1 @X n @ DerR0 R = R @X i i=1 n and the vector elds V = fi @X@ i of R=R0 can be identied with the i=1 \functions" (f1  : : :  fn ). k i

M

P

n {tuples of

In analysis vector elds are \integrated". A corresponding algebraic problem is to nd for V as above polynomials '1 : : :  'n 2 R0 t] which solve the system of dierential equations '0i = fi ('1 : : :  'n) (i = 1 : : :  n) In classical terminology a non-constant polynomial f 2 R0 X1 : : :  Xn] in the kernel of V Pn @f = 0) is called a \rst integral" of V or of the dierential equation (i.e. such that fi @X i i=1 corresponding to V . 10

Assume again that R=R0 is an arbitrary algebra. In addition to being an R {module DerR0 R has also the structure of a Lie algebra. For V V 0 2 DerR0 R dene the bracket operation (or Lie product) by V V 0 ] := V V 0 ; V 0 V

Clearly V V 0 ] is an R0 {linear map. For a b 2 R

V V 0 ](ab) = V (V 0 (ab)) ; V 0 (V (ab)) = V (aV 0 b + bV 0 a) ; V 0 (aV b + bV a) = aV V 0 b + V aV 0 b + bV V 0 a + V bV 0 a ; aV 0 V b ; V 0 aV b ; bV 0 V a ; V 0 bV a = a V V 0 ](b) + b V V 0 ](a)

hence V V 0 ] 2 DerR0 R . One checks easily that the bracket  ] is R0 {bilinear, V V ] = 0 for any V and that the Jacobian relation

2 DerR R , 0

V V 0  V 00 ]] + V 0  V 00  V ]] + V 00  V V 0 ]] = 0

is satised for V V 0  V 00 2 DerR0 R , hence DerR0 R is a Lie algebra. 1.19.Proposition. Let N

 R be a multiplicatively closed subset.

map  : DerR0 R ! DerR0 RN described in 1.17 we have

For the canonical

( V V 0 ]) = (V ) (V 0 )] for all V V 0 2 DerR0 R

i.e.  is a homomorphism of Lie algebras. This follows by an easy calculation with fractions. Exercises:

P

1) Let R0 be a ring, (1 : : :  n) 2 Z n , d 2 Z . A polynomial F = a1 :::n X11 : : : Xnn 2 R0 X1 : : :  Xn] is called quasihomogeneous of weight (1  : : :  n) and degree d , if Pn a1:::n = 0 whenever ii 6= d . i=1 a) For such polynomials the Euler relation Pn iXi @F = d  F @Xi i=1

Pn

is satised, i.e. iXi @X@ i is the Euler derivation of the graded R0 {algebra i=1 R0 X1 : : :  Xn] b) Conversely, if Q  R0 and F satises Euler's relation, then F is quasihomogeneous of weight (1  : : :  n) and degree d . 11

2) Let R be a ring, F 2 R , and p 2 Spec (R). We say that F has order m at p , if F 2 p m Rp , F 2= p m+1Rp . Suppose R0 is a ring with Q  R0 and R := R0 X1 : : :  Xn]. Assume p 2 Spec (R) contains a polynomial F of degree m whose order at p is also m . Then p contains a polynomial of degree 1. (Hint: Consider the partial derivatives of F ). 3) Let R be a ring whose characteristic is a prime number p . If d : R ! R is a derivation, then so is dp := d  d ( p factors). Moreover for a 2 R the following formula of Hochschild holds (ad)p = ap  dp + (ad)p;1 (a)  d (see No1 ], No2 ] for generalizations). 4) Let R be a ring and  : R ! R a derivation. Let N  R X ] be multiplicatively closed and let  2 R X ]N . There is precisely one derivation  : R X ]N ! R X ]N with (X ) =  such that the diagram R R X ]N 





R

R X ]N

is commutative. 5) Let R be a ring, d : R ! M a derivation into an R {module M , I There is a derivation d : R=I ! M=RdI + IM given by the formula d(r + I ) = dr + RdI + IM (r 2 R) such that the diagram R R=I

 R an ideal.



d

d

M M=RdI + IM is commutative ( RdI is the submodule of M generated by all dx for x 2 I ). 6) Let L=K be a separably algebraic eld extension and  : K ! K a derivation. Then there is exactly one derivation d : L ! L with djK =  . Hint: Using Zorn's Lemma one can reduce to the case of a simple extension L = K x]. Let f 2 K X ] be the minimal polynomial of x over K . There is a K {epimorphism K X ](f ) ! L with kernel fK X ](f ) . Choose  2 K X ](f ) such that for the derivation  : K X ](f ) ! K X ](f ) of exercise 4) the condition  (fK X ](f ))  fK X ](f ) is satised. Now apply exercise 5). In the following exercises R=R0 is an algebra and g := DerR0 R . An ideal I  R is called g {invariant or a dierential ideal (dierentially closed ideal) if dI  I for all d 2 g . 12

7) a) Extension ideals I0R of ideals I0  R0 are g {invariant. b) If R = R0 X1  : : :  Xn ] is a polynomial algebra and Q  R0 , then the g {invariant ideals of R are exactly the extension ideals I0 R . 8) Sums, product, intersections and colon ideals of g {invariant ideals are g {invariant. Each ideal of R contains a unique largest g {invariant ideal. If I  R is a g {invariant ideal each d 2 DerR0 R induces an R0 {derivation d 2 DerR0 (R=I ) (r + I 7! dr + I ). 9) For p 2 Spec R let g p := DerR0 Rp . a) If p Rp is g p {invariant, then p is g {invariant. b) If Q  R and I is a g {invariant ideal of R , then all its minimal prime divisors are g {invariant. In particular, all minimal primes of R are g {invariant. 10) (see SS5 ],1.3) Suppose R is a noetherian ring and Q  R . If I is a g {invariant ideal of R , then all elements of Ass (R=I ) are g {invariant, and I has a reduced primary decomposition I = q 1 \  \ q r with g {invariant primary ideals q i (i = 1 : : :  r). In particular the non-embedded primary components of I and the radical of I are g {invariant. For more about g {invariant ideals, see H1 ], H2 ], Ku16 ], Mal1 ]- Mal3], SS5 ], S2 ], Si] and x 2, ex.4) and 5), x 5, ex.5)-6), x 6, ex.3).

13

x 2. The Module of Kahler Dierentials. Tangent Bundles This module is an important invariant of commutative algebras. Its dual is the derivation module of the algebra, and the spectrum of its symmetric algebra is the tangent bundle of the spectrum of the algebra. In this section we study some basic properties of the dierential module. Later a closer investigation of what it \knows" about algebras will follow. Kahler dierentials are the algebraic analogue of dierential 1{forms on dierentiable manifolds.

Let R=R0 be a commutative algebra and M an R {module. We rst relate the R0 {derivations d : R ! M to R0 {algebra homomorphisms of R into the algebra R n M of \dual numbers". Recall that this algebra is R  M with the multiplication given by (r1  m1 )  (r2  m2) := (r1 r2  r1 m2 + r2 m1) (ri 2 R mi 2 M ) i : R ! R n M (r 7! (r 0)) is an injective and p : R n M ! R ((r m) 7! r) a surjective R0 {homomorphism with p  i = idR . 2.1.Proposition.

a) For any R0 {derivation d : R ! M the map d~: R ! R n M (r 7! (r dr)) is an R0 {algebra-homomorphism with p  d~ = idR . b) For any R0 {algebra-homomorphism h : R ! R n M with p  h = idR there is a unique R0 {derivation d : R ! M with h = d~. Proof: a) Obviously d~ is R0 {linear and p  d~ = idR . Moreover, for a b 2 R we have ~  db ~ d~(ab) = (ab d(ab)) = (ab adb + bda) = (a da)  (b db) = da thus d~ is a R0 {algebra-homomorphism. b) Given h , set h(a) = (a a ) with a 2 M for any a 2 R . Let d : R ! M be the mapping a 7! a . Clearly d is R0 {linear. From h(ab) = h(a)  h(b) = (a a )(b b ) = (ab ab + ba ) we see that d is a derivation and h = d~. 0

0

0

0

0

0

0

! M is called universal, if the following condition is satised: For any R {derivation  : R ! N of R into an R {module N there is one and only one R {linear mapping ` : M ! N with  = `  d . In other words: If one knows a universal derivation d : R ! M of R=R , then any other derivation  of R=R can be written as a "specialization" `  d of d with a unique linear mapping ` . As usual, if d : R ! M and d : R ! M are universal derivations of R=R , then there is a unique R {linear map ` : M ! M such that d = `  d , and ` is an isomorphism. 2.2.Definition. An R0 {derivation d : R 0

0

0

1

1

2

1

2

2

0

2

1

So we can say that there is exactly one universal derivation of R=R0 up to canonical isomorphism. 14

2.3.Theorem. For any algebra R=R0 there exists a universal derivation. In fact, the

derivation described in example 1.6 is universal.

! N be an arbitrary derivation of ~ R=R0 and  : R ! R n N the R0 {homomorphism associated with it according to 2.1: We have ~(a) = (a a) for each a 2 R . By the universal property of R R0 R the R0 {homomorphisms ~ and i : R ! R n N (b 7! (b 0)) induce an R0 {homomorphism h : R R0 R ! R n N with h(a  b) = ~(a)  i(b) = (a a)  (b 0) = (ab ba) for all a b 2 R . For x = r  1 ; 1  r 2 I (r 2 R) we obtain h(x) = (r r) ; (r 0) = (0 r), and hence h(I 2) = 0. Therefore h induces an R {linear mapping ` : I=I 2 ! N with Proof: We shall use the notations of 1.6. Let  : R

`(r  1 ; 1  r + I 2) = `(dr) = r for all r 2 R Since the R {module I=I 2 is generated by fdrgr R , we obtain  = `  d , and there can be only one such mapping ` . Hence d is universal. 2

2.4.Definition. If d : R

! M is a universal derivation of R=R , then the R {module M 0

(which is unique up to canonical isomorphism) is denoted by R=R0 and is called the module of (Kahler) dierentials of R=R0 . The universal derivation of R=R0 is sometimes denoted by dR=R0 . For x 2 R the element dR=R0 x is called the dierential of x . The elements of 1R=R0 are also called 1-forms of R=R0 . The question, which relations exist between the properties of an algebra R=R0 and its module of dierentials 1R=R0 , is a major theme of this chapter. 1

2.5.Remarks:

a) As was shown in the proof of 2.3 we have a canonical isomorphism 1R=R0

= I=I 2

where I is the kernel of R R0 R ! R (a  b 7! ab). b) For a derivation  : R ! N let RR denote the submodule of N generated by frgr R . n P Thus RR is the set of all sums ri ri (ri  ri 2 R n 2 N ). If d : R ! 1R=R0 is i=1 the universal derivation of R=R0 , we have 1R=R0 = RdR , because I is generated by fr  1 ; 1  rgr R . c) It is clear by the denition 2.2 that 1R=R0 = 1R=R0 if R0 is the image of R0 in R . Moreover, if N  R0 is multiplicatively closed, and the structure homomorphism R0 ! R induces a ring homomorphism (R0 )N ! R , then 1R=R0 = 1R=(R0 )N , since the images of the elements of (R0 )N in R have derivation 0 by the quotient formula. Thus, if R = K is a eld, then 1K=R0 = 1K=K0 where K0 is the quotient eld of the image of R0 in K . 2

0

0

2

0

15

0

L

d) Suppose R = Rg is a graded ring over an abelian group (G +) with neutral g G element 0 such that the image of the structure homomorphism R0 ! R is contained in L 0 g a R . Then R R0 R is G {graded by (R R0 R) = R R0 Rb , and the kernel I of a+b=g R R0 R ! R is a homogeneous ideal. It follows that 1R=R0 = I=I 2 carries a canonical structure of a G {graded module over the graded ring R . For homogeneous x 2 R the dierential dR=R0 x is homogeneous of the same degree, and this determines the grading of 1R=R0 uniquely. 2

We can "compute" 1R=R0 already in some special cases. 2.6.Examples: a) Let K=K0 be a separably algebraic eld extension. Then 1K=K0 = 0,

because any derivation of K=K0 is trivial, as was shown in 1.9. b) Similarly, for a perfect eld K of characteristic > 0, we have 1K= Z = 0 by 1.10. c) Let R0 be a ring and R := R0fX g ] the polynomial algebra over R0 on a family of indeterminates X ( 2 ). Form the free R {module on the family of indeterminates fdXg  L M := RdX 2

2



2



and dene for each F 2 R the "formal dierential" as in 1.4 by

dF =

(1)

P @F dX   @X 2

It is easy to see that d is a derivation of R=R0 . We shall show that it is also P universal. @F X by In fact, if  : R ! N is an arbitrary derivation of R=R0 , then F = @X    1.8g). The R {linear mapping ` : M ! N with `(dX ) = X satises the equation `  d =  , and there can be only one such mapping ` , since fdXg  is a basis of M . We have shown that L 1R0 X ]=R0 = R0fXg]dX 2

2

f



g

2



and that the universal derivation dR0 X ]=R0 is given by (1). n n @F L P d) For R = R0 X1 : : :  Xn ]] consider the derivation d : R ! RdXi , F 7! @X dXi i=1 i=1 i of example 1.4. Although this example is very similar to that of the polynomial ring, d is not always universal, as will be seen later (3.5b). e) Dierentials (1{forms) on C {manifolds can be dened in various ways. The sheaf theoretic construction is similar to that of Kahler dierentials. The following process makes no explicit use of sheaf theory. Let U be an open set of a C {manifold X . For P 2 U let (TP X ) be the R {dual of the tangent space TP X , called the cotangent f

g

1

1



16

space at P . Then T (U ) := Q (TP X ) is an E (U ){module. For a function f 2 E (U ) 



P U

let dU f 2 T (U ) be the family f(dU f )P gP U with 2



2

(dU f )P (V (P )) = V (P )(fP )

(2)

for each V (P ) 2 TP X = DerR(EP  R) where fP is the germQ of f at P . Let 1 (X )  T (X ) be the set of all maps : X ! (TP X ) with P 

for all P 2 X such that is locally of the form

Pt

P X



2

2 (TP X )



gi dU fi with gi fi 2 E (U ), that is for Pt each P there is an open neighbourhood U of P in X such that Q = (gidU fi)Q for i=1 all Q 2 U . Then 1(X ) is a module over E (X ) called the module of dierential 1 {forms on X . The map dX : E (X ) ! 1(X ) dened as in (2) is an R {derivation. Except for trivial cases it is not the universal Lderivation of E (X )=R but of course a specialization thereo. If n 1 X = R , then  (X ) = E (X )  dX Xi with the coordinate functions Xi (i = 1 : : :  n), i=1 n @f P d X. and dX is the map given by dX f = @X i X i i=1

i=1

2.7.Proposition. For an R {module M the canonical R {linear mapping

HomR(1R=R0  M ) ! DerR0 (R M )

(` 7! `  dR=R0 )

is an isomorphism. In particular, there is a canonical isomorphism DerR0 R

= HomR (1R=R0  R)

which identies the derivation module of R=R0 canonically with the dual of the di erential module of R=R0 . The rst statement of the proposition is only a reformulation of the universal property of dR=R0 . (In a dierent language: 1R=R0 "represents" the functor DerR0 (R ;)). If R is a G {graded ring as in remark 2.5d), and 1R=R0 is nitely generated, then L for each G {graded R {module M = M a the derivation module DerR0 (R M ) = a G HomR(1R=R0  M ) is G {graded too. A derivation d : R ! M is homogeneous of degree g , if dRa  M g+a for all a 2 G . In particular DerR0 R is G {graded. Examples of G {graded rings for general abelian groups are the group rings R0G], see exercises 6) and 7). For G = Zn the G {graded rings and modules are basic in combinatorics and computer algebra, and for G = Z the graded rings and modules are of fundamental importance in projective algebraic geometry. 2

17

Proposition 2.7 also implies that there is an R {bilinear map

h  i : R=R DerR R ! R given by h! V i = `(!) for ! 2 R=R , V 2 DerR R , where ` is the linear form on R=R with `  dR=R = V . The R {linear map DerR R ! HomR (R=R  R) (V 7! h  V i) is bijective, while : R=R ! HomR (DerR R R) (! 7! h! i) in general is not. 1

1

0

1

0

0

1

0

0

1

0

0

0

0

0

The dual HomR (DerR0 R R) = (DerR0 R) of the derivation module, that is the bidual (1R=R0 ) of 1R=R0 , is sometimes called the module of Zariski dierentials. There is an R0 {derivation d : R ! (1R=R0 ) such that dr for r 2 R sends V 2 DerR0 R to hdr V i = V (r). The map is the canonical homomorphism into the bidual of 1R=R0 . Hence the following statements are equivalent: a) 1R=R0 is a reexive R {module. b) The modules of Zariski and of Kahler dierentials are canonically isomorphic. c) The canonical scalar product h  i : 1R=R0 DerR0 R ! R is non-degenerate (i.e. both induced linear maps are isomorphisms). In general 1R=R0 contains more information than (1R=R0 ) and is easier to handle. Zariski dierentials were introduced in Zariski's Harvard lecture notes 1957-58 about algebraic surfaces, see also ZF]. More about their properties can be found in Kn], Pl4 ] and Pl5 ]. 







Let S (1R=R0 ) denote the symmetric algebra of 1R=R0 . By its universal property the R {homomorphisms S (1R=R0 ) ! R are in natural one-to-one correspondence with the R {linear maps ` : 1R=R0 ! R , i.e. with the derivations d 2 DerR0 R . Remark: Let X := Spec R , X0 := Spec R0 . Then TX=X0 := Spec S (1R=R0 ) is called

1 the tangent bundle of X=X0 . The canonical map R ! S (R=R ) induces a morphism 0 pr : TX=X0 ! X called the projection of the tangent bundle onto the basis X . The derivations V 2 DerR0 R (vector elds on X ) are in one-to-one correspondence with the morphisms V : X ! TX=X0 such that pr  V = idX i.e. with the sections of pr : TX=X0 ! X . Here we use the fact that X {morphisms X ! TX=X0 are in one-to-one correspondence with the R {homomorphisms S (1R=R0 ) ! R . For example if R = R0t] is a polynomial algebra in a variable t , then S(1R=R0 )= R0t dt] is a polynomial algebra in two variables t dt . The vector eld @t@ on Spec R0t] = A1R0 corresponds to the R {homomorphism S(1R=R0 ) ! R with dt 7! 1. It denes the section @ 1

2 @t : AR0 ! TA1R0 =X0 = AR0 . The above denition of the tangent bundle is the one of EGA IV, 16.5. It is analogous to the corresponding denition of dierential topology. For a dierent notion and its relation to the present one, see SUV].

18

2.8.Proposition. (Localisation of the di erential module). If under the assumptions of

1.17 the derivation d : R ! M is universal, so is dN : RN ! MN (see 1.17 for the denition of dN ). Hence 1RN =R0

= (1R=R0 )N If 1R=R0 is nitely generated, then there is a canonical injection DerR0 (R M )N ! DerR0 (RN  MN )

for each R {module M which is an isomorphism of RN {modules in case 1R=R0 is nitely presentable. Then in particular DerR0 RN is generated as an RN {module by the image of DerR0 R in DerR0 RN .

! L be an R {derivation into an RN {module L . By the universal property of d there is an R {linear map h : M ! L such that the diagram

Proof: Let  : RN

0

d

R ;

M h

RN



L

is commutative. h induces an RN {linear map H : MN ! L with  = H  dN . Since MN = (RdR)N = RN dN RN there can be only one such map H . So we have proved the rst part of the proposition. As for the second we have DerR0 (RN  MN )

= HomRN (1RN =R0  MN )

= HomRN ((1R=R0 )N  MN )

= HomR(1R=R0  M )N = DerR0 (R M )N

where it was used in the third step that 1R=R0 is nitely presentable (Bourbaki, Alg. Comm. Chap.II, Prop. 19). Suppose 1R=R0 is nitely presentable and K is the full ring of quotions of R . Then the canonical map HomR (1R=R0  R) ! HomK (1K=R0  K ) is injective, hence the derivation module DerR0 R is torsion free. If R is noetherian and 1R=R0 is nitely generated, then DerR0 R is even reexive being the dual of 1R=R0 . 2.9.Examples:

a) Let R = R0X1 : : :  Xn] be a polynomial algebra over R0 . By 2.6c) 1R=R0 =

n L

i=1

19

RdXi

where the universal derivation dR=R0 is given by (1). The partial derivatives @X@ i (i = 1 : : :  n) form the dual basis to (dX1  : : :  dXn ), hence we see again (1.18c) that DerR0 R =

(3)

If N  R is multiplicatively closed, then 1RN =R0 =

n L i=1

@ R @X

i

n L RN dXi

i=1

n L

where the universal derivation dRN =R0 : RN ! RN dXi satises the quotient rule. i=1 Moreover n L @ DerR0 RN = RN @X i=1

i

@ @Xi denoting partial dierentiation on the quotient ring RN . By (3) the vector elds V n P on Spec R = AnR0 are of the form V = fi @X@ i (fi R). They can be identied with i=1 the n {tupels of \functions" (f1  : : :  fn) Rn . For x A nR0 the tangent space at x

2

2

2

T A nR0 =X0 (x) = Homk(x)(1 x=R0 = m x1 x=R0  k(x)) O

O

where X0 = Spec R0 is an n {dimensional k(x){vector space:

  n L @ T A nR0 =X0 (x) = k(x) @X i x i=1

Here ( @X@ i )x : Ox ! k(x) denotes the partial dierentiation composed with the canonical n P epimorphism Ox ! k(x). Given = ( 1  : : :  n) 2 k(x)n the tangent vector i( @X@ i )x i=1 is sometimes called the \directional derivative" with direction

. n P For a vector eld V = fi @X@ i (fi 2 R) on A nR0 the vector corresponding to V at x i=1 is  @  n P V (x) = fi (x)  @X i x

i=1

fi (x) denoting the residue class of fi in k(x). b) For an ideal I = (f1  : : :  fm ) in R let Z := Spec R=I . Then for z 2 Z the tangent space TZ=X0 (z) is by the exact sequence (2) of x 1 the kernel of the canonical map  : TAn=X0 (z) ! Hom

(I OAnz =I 2OAnz  k(z))

Zz

O

20

induced by restricting derivations d : OAnz ! k(z) to I OAnz . Thus for a tangent vector n P d = ai ( @X@ i )z (ai 2 k(z)) we have (d) = 0 if and only if i=1

dfj =

n @fj P ai (z) = 0 i=1 @Xi

(j = 1 : : :  m)

In other words: If we identify T A nR0 =X0 (z) with k(z)n , then TZ=X0 (z) is identied with the solution set of the system of linear equations n @fj P (z)  Yi = 0 i=1 @Xi

(j = 1 : : :  m)

In particular, if I = (f ) is the ideal of a \hypersurface" Z , then

TZ=X0 (z) = f(a1  : : :  an ) 2 k(z)n j

n P @f (z) = 0g ai @X i i=1

n P

@f (z )  Yi = 0 (the \tangent hyperplane"). is the hyperplane with the equation @X i=1 i c) Let R=R0 be an arbitrary algebra such that 1R=R0 is nitely generated. Let X :=Spec R , X0 :=Spec R0 . Then

TX=X0 (x)=DerR0 (Ox k(x))

= Hom x (1 x=R0  k(x))

= Homk(x)(1 x=R0 = m x1 x=R0  k(x)) O

O

O

O

has the same k(x){dimension as 1 x=R0 = mx 1 x=R0 , which by Nakayama's lemma is O

O

(1 x =R0 ) = ((1R=R0 )p ) O

where p is the prime ideal of R corresponding to x . It follows that dimk(x) TX=X0 (x) is an upper-semicontinuous function of x . Since symmetric algebras are compatible with localisation and formation of residues in the base ring we have by 2.8 canonical isomorphisms of k(x){ algebras (S (1R=R0 ))p = p (S ((1R=R0 ))p

= S (1Rp =R0 = p 1Rp =R0 )

= S (1 x=R0 = mx1 x=R0 ) O

O

This implies that the ber at x of the projection pr : TX=X0 ! X of the tangent bundle TX=X0 onto its basis is Spec S (1 x=R0 = m x1 x=R0 ), an a!ne space over k(x) of dimension (1 x=R0 ). By the universal property of the symmetric algebra the k(x){homomorphisms S (1 x=R0 = m x1 x=R0 ) ! k(x) are in one-to-one correspondence with the elements of O

O

O

O

O

Homk(x)(1 x=R0 = mx1 x=R0  k(x)) = TX=X0 (x) O

O

21

Let prx : Spec S (1 x=R0 = m x1 x=R0 ) ! Spec k(x) be the morphism induced by the structure homomorphism of the symmetric algebra. Its sections tx are in one-to-one correspondence with the tangent vectors of X=X0 at x . In other words: The tangent vectors at x can be identied with the k(x){rational points of O

O

pr 1(x) = Spec S (1 x=R0 = m x1 x=R0 ) ;

O

O

S

This generalizes the formula T (X ) = TP (X ) for the tangent bundle of a dierential P X manifold X . d) Let K be a eld and L = K (fX g) a purely transcendental extension eld of K . Since L is the quotient eld of the polynomial ring K fXg] we have by 2.6 and 2.8 

2

1L=K =

L LdX







2

the universal derivation dL=K being the extension of (1) by the quotient formula.

A family fag  of elements a 2 R is called a dierential basis of R=R0 if fdag  is a basis of 1R=R0 as R {module. Of course a dierential basis need not exist, even if 1R=R0 is a free R {module. It does when 1R=R0 is a nite free module over a local ring R , by Nakayama's lemma. If R=R0 has a nite dierential basis fa1  : : :  an g , then 2

2

n M DerR0 R = R @a@ i i=1

where f @a@ 1  : : :  @a@n g denotes the dual basis of fda1 : : :  dang . Thus @a@ i is the unique R0 {derivation of R satisfying

@ (a ) =  (k = 1 : : :  n) ik @ai k This generalizes partial dierentiation in polynomial algebras. Clearly the Lie product  @a@ i  @a@ j ] = 0 for i j = 1 : : :  n . Let  : R0 ! R be the structure homomorphism of the algebra R=R0 and let R =R0 be another algebra with structure homomorphism  . Suppose a commutative diagram 0

0

R 

(4)

h



R0

R

0



0



of ring homomorphisms is given. 22

R0 0

0

2.10.Proposition. (Functorial property of the di erential module). There is exactly one

R {linear map ` : 1R=R0 ! 1R =R0 such that the diagram 0

0

h

R 

R

0

dR =R0

dR=R0

0

0

1R=R0 ` 1R =R0 0

0

is commutative. ` induces an R {linear map R R 1R=R0 ! 1R =R0 . 0

0

0

0



Proof: The map dR =R0 h is an R0 {derivation of R . The existence and uniqueness of 0

0

1 . ` follows from the universal property of R=R 0

2.11.Example: Under the assumptions of 2.10 let X=X0 and X =X0 be the schemes 0

corresponding to the algebra R=R0 resp. R =R0 and f : X Then ` induces a homogeneous homomorphism 0

0

0

0

! X the map induced by h .

S (1R=R0 ) ! S (1R =R0 ) 0

0

which coincides with h on the elements of degree 0. Hence there is a functorial morphism Tf : TX =X0 ! TX=X0 such that 0

0

TX =X0 0

0



Tf

pr

X

TX=X0 pr

f

0

X

commutes. Tf is called the derivative of f .

Suppose S=R0 is another algebra and h : R ! S an R0 {homomorphism. Then we have a commutative diagram (4) and by 2.10 two S {linear maps

S R 1R=R0 ! 1S=R0 and 1S=R0 ! 1S=R 2.12.Theorem. (First fundamental exact sequence). The following sequence of S {modules

is exact:



S R 1R=R0 ;! 1S=R0 ;! 1S=R ! 0

23

Proof: We use the well-known fact that a sequence of R {modules and linear maps '

'

1 2 M1 ;;! M2 ;;! M3

is exact, if for each R {module M the sequence '

'

2 1 HomR (M3  M ) ;;! HomR (M2  M ) ;;! HomR (M1  M ) 



is exact (with 'i := HomR ('i M )). In our situation we have a canonical commutative diagram (with an S {module M ) 

0 

HomS (1S=R  M )

HomS (1S=R0  M )

o

0

HomS (S R 1R=R0  M )

o

DerR(S M )

DerR0 (S M )

o

DerR0 (R M )

where we have used the canonical isomorphism

HomS (S R 1R=R0  M )

= HomR(1R=R0  M )

As the row on the bottom of the diagram is exact by 1.13 the claim follows. Since im is the S {submodule of 1S=R0 generated by fdS=R0 r j r 2 Rg we obtain the formula (5)

1S=R

= 1S=R0 =hfdS=R0 r j r 2 Rgi

which allows to \compute" 1S=R once 1S=R0 is \known". For an ideal I  R the restriction of dR=R0 to I induces a map dR=R

0 I ;;;;! 1R=R0 ! 1R=R0 =I 1R=R0

which is R {linear and whose kernel contains I 2 . Therefore an R=I {linear map

 : I=I 2 ! 1R=R0 =I 1R=R0

is induced: (x + I 2) = dR=R0 x + I 1R=R0 for each x 2 I . Further the functorial map 1R=R0 ! 1R=I=R0 induces an R=I {linear map

 : 1R=R0 =I 1R=R0 ! 1R=I=R0

(dR=R0 r + I 1R=R0 7! dR=I=R0 (r + I )) 24

2.13.Theorem. (Second fundamental exact sequence). The following sequence of R=I -

modules is exact:

I=I 2 ;! 1R=R0 =I 1R=R0 ;! 1R=I=R0 ! 0 

Proof: For any R=I {module M we have a canonical commutative diagram

0 !

HomR=I (1R=I=R0  M ) HomR(1R=R0  M )  HomR=I (I=I 2 M ) 



o

o

o

0 DerR0 (R=I M ) DerR0 (R M ) HomR=I (I=I 2 M ) where the row on the bottom is exact by 1.15. From this the exactness of the sequence of the theorem follows. The R=I {module I=I 2 is called the conormal module of I or of the embedding Spec R=I ! Spec R . It is an important question in which cases the map  is injective. The image of  is the R=I -submodule RdR=R0 I + I 1R=R0 =I 1R=R0  1R=R0 =I 1R=R0 Observe that I 1R=R0  RdR=R0 I since for a 2 I and r 2 R adR=R0 r = dR=R0 (ar) ; rdR=R0 a 2 RdR=R0 I Therefore we have by the Noether isomorphism theorem (6) 1R=I=R0

= 1R=R0 =RdR=R0 I where the universal derivation dR=I=R0 is induced by the composition of dR=R0 with the canonical epimorphism 1R=R0 ! 1R=R0 =RdR=R0 I . If R and S are graded rings over an abelian group G and h : R ! S is a homomorphism of graded rings, then the functorial map 1R=R0 ! 1S=R0 is homogeneous of degree 0. The maps of the rst fundamental exact sequence are homogeneous of degree 0, and so are those of the second fundamental exact sequence in case I is a homogeneous ideal of R . A similar remark can be made about the derivation functor and the exact sequences (1.13, 1.14) connected with it, provided the corresponding dierential modules are nitely generated. The rule (6) is useful for the computation of the dierential module of residue class rings. For example one has the following general recipe: Suppose the algebra R=R0 has a presentation R = R0 fXg ]N =I with a polynomial algebra R0fX g], a multiplicatively closed set N  R0fXg], and an ideal I  R0 fXg]N . Let x denote the residue class of X in R and for f 2 R0fXg]N @f in R . let @x@f be the image of @X  2

25

2.14.Corollary. (Presentation of the di erential module by generators and relations).

Choose a system ff g where U 

2



of generators of I . Then 1R=R0

=

L





2

RdX=U

L RdX is the submodule generated by the elements 





2

df :=

f g

P @f  dX   @x 2

( 2 ")

Proof: Set P := R0  X ]N . Then by 2.6c) and 2.8

L

1P=R0 =





2

and by formula (6) above

P  dX

1R=R0

= 1P=R0 =I 1P=R0 + PdI

=

L 



2

RdX =PdI

PdI denoting the image of PdI . Since I is generated by ff g  it follows that PdI is q.e.d. generated, as an R {module, by df ( 2 "), hence PdI = U , 1 It follows from 2.14 that the symmetric algebra of R=R0 has a presentation P @f S(1R=R0 )

= RfYg ]=(f @x  Yg  )   Moreover, if in the situation of 2.14 the set  is nite, then L @ DerR0 (P R)

= HomP (1P=R0  R) = R  @x    2

2

2

2

where @x@ is the composition of @X@  with the canonical epimorphism P exact sequence 

0 ! DerR0 R ;! DerR0 (P R) ;! HomR(I=I 2 R)

! R.

By the

the derivation module DerR0 R is identied with ker  , that is with the submodule of DerR0 (P R) consisting of all Pr @ (r 2 R)    @x 2

such that

P   2

@f = 0 r @x

for all  2 "



26

2.15.Example: (Hypersurface algebras). Let

R = R0X1 : : :  Xn]=(f ) = R0x1  : : :  xn ] Then

n @f n L @f dX +  + @f dX i and S(1 )= RY  : : :  Y ]=( P  Yk ) RdXi =h @x 1 n 1 n R=R 0 @xn 1 i=1 k=1 @xk hence the tangent bundle TX=X0 of X := Spec R is the subscheme of the a!ne space n P AnR = Spec RX1 : : :  Xn] dened by the principal ideal ( @x@fk  Yk ). For the module of k=1 vector elds on X we have n P P @f = 0g DerR0 R = f ri @x@ j ri 2 R ri @x i i i=1 This module can be identied with the submodule of Rn consisting of all (r1  : : :  rn ) with n @f P n @xi  ri = 0. Given V = (r1  : : :  rn ) 2 R we then have for x 2 X

1R=R0 =

i=1

V (x) = (r1 (x) : : :  rn(x)) if we identify TAnR0 =X0 (x) with k(x)n as in 2.9a). Thus V (x) is the directional derivative dened by the vector (r1 (x) : : :  rn (x)). Suppose R is a local ring with maximal ideal m , and let L be the residue eld of R . Then we have by 2.13 an exact sequence of L {vector spaces 1 m = m 2 ! R=R = m 1R=R0 ! 1L=R0 ! 0 0 and we obtain by Nakayama's lemma the following estimate for the minimal number of generators of the dierential module (and hence also of the dimension of the tangent space Tm (Spec R)): If 1R=R0 is nitely generated, then

(1R=R0 ) dimL 1L=R0 + ( m ) This formula becomes useful as soon as we know more about dierential modules of elds, see x 3. Let R=R0 and S=R0 be algebras. Then S R0 1R=R0  R R0 1S=R0 is in a natural way an R R0 S {module. Moreover the map d : R R0 S ! S R0 1R=R0  R R0 1S=R0 given by the formula d(r  s) = s  dR=R0 r + r  dS=R0 s (r 2 R s 2 S ) is an R0 {derivation as is easily checked. 27

  R R

2.16.Theorem. d is the universal derivation of R R0 S=R0 , hence

1R





= S R

R0 S=R0

1R=R0

0

1S=R0

be an R0 {derivation into an R R0 S {module M . Let iR : R ! R R0 S and iS : S ! R R0 S be the canonical homomorphisms. Then   iR : R ! M is an R0 {derivation, so we have an R {linear map hR : 1R=R0 ! M with hR  dR=R0 =   iR . It extends to an R R0 S {linear map H1 : S R0 1R=R0 ! M with H1 (s  dR=R0 r) = (1  s)(r  1) for r 2 R , s 2 S . Similarly we have an R R0 S {linear map H2 : R R0 1S=R0 ! M and hence an R R0 S {linear map Proof: Let  : R R0 S

!M

0

H : S R0 1R=R0  R R0 1S=R0 ! M

By the construction of H we have for r 2 R , s 2 S

H (d(r  s)) = H (s  dR=R0 r + r  dS=R0 s) = (1  s)hR (dR=R0 r) + (r  1)hS (dS=R0 r) = (1  s)(r  1) + (r  1)(1  s) = (r  s)

That there can be only one R R0 S {linear map H with H  d =  is clear. 2.17.Corollary. (Base change formula)

1R The universal derivation dR

R0 S=R



dR

R0 S=R



R0 S=R



= R R

0

1S=R0

satises the formula

(r  s) = r  dS=R0 s

(r 2 R s 2 S )

Proof: Follows from 2.16 and the rst fundamental exact sequence.

f g

2.18.Example: Let R=R0 be an algebra and P := R X ] a polynomial algebra over

R . Write P = R R0 S with S := R0fXg]. Then by 2.16

1P=R0

= S R0 1R=R0  R R0 1S=R0

= P R 1R=R0  L PdX

For a polynomial f = given by the formula

df =

Pa P

1 s X11 

 Xss



(a1 s 

X11  Xss  dR=R0 a1 28



2

2 R) the dierential in P=R

s +



1

P @f dX   @X 2

0

is

P

We set f := X11  Xss  dR=R0 a1 s in the future (dierentiation of the coe!cients of a polynomial while the variables are kept xed). Suppose now that an algebra S=R is given which has a presentation 

S = RfXg]N =I with a multiplicatively closed set N  P := RfX g] and an ideal I  PN . Let x be the image of X in S and choose a system ff g  of generators of the ideal I . Let @f @f f (fx g) the image of the f dened above in @x denote the image of @X in S and L 1 1 S R R=R0 . Let U  S R R=R0  SdX be the S {submodule generated by the   elements P @f f (fx g) + @x dX    L 2.19.Theorem. 1S=R0

= (S R 1R=R0  SdX)=U . The universal derivation dS=R0   f is given as follows: For s 2 S choose  2 PN (f 2 P  2 N ) with image s . Then dS=R0 s is the image of 1  (f + P @f dX ) ; f  ( + P @ dX )  @X   2 @X  L in S R 1R=R0  SdX=U . (7)

2

2

2

2





2

Proof: Follows from 2.18 and 2.14.

The formula of 2.19 is used to compute the dierential module 1S=R0 when 1R=R0 is already known. 2.20.Example: (Dierential modules of symmetricL algebras). Let M be an R {module

and S := S(M ) its symmetric algebra. Write M = RX =U as a homomorphic image  P  of a free module and choose a system of generators f a Xg of the relation module  U . Then P S = RfXg ]=(f a Xg ) 2

2

hence and therefore

1S=R =

L 



2



P SdX=hf adX gi 

1S=R

= S R M where this isomorphism is induced by dS=R jM : M ! 1S=R . Here the rst fundamental exact sequence has the form (8)

S R 1R=R0 ! 1S=R0 ! S R M ! 0 29

Let S+ be the ideal generated by the elements of positive degree of S , hence S=S+ = R . The composition of functorial maps 1R=R0 ! 1S=R0 =S+  1S=R0 ! 1R=R0 is the identity, therefore 1R=R0 is an R {direct summand of 1S=R0 =S+ 1S=R0 . Considering the exact sequence (8) mod S+ we obtain an isomorphism of R {modules 1S=R0 =S+  1S=R0

= 1R=R0  M

(9)

Another useful general statement is a formula for the dierential module of a nite direct product of algebras. Let Si=R0 (i = 1 : : :  n) be algebras and let S := S1  Sn denote their direct product. Then the canonical projections S ! Si induce S {linear maps i : 1S=R0 ! 1Si =R0 and an S {linear map

 : 1S=R0 ! 1S1 =R0  1Sn=R

((!) = (1 (!) : : :  n(!)))

2.21.Theorem.  is an isomorphism. For s = (s1  : : :  sn )

the di erential dS=R0 s is given by

2 S , si 2 Si

(i = 1 : : :  n)

dS=R0 s = (dS1 =R0 s1  : : :  dSn =R0 sn) Proof: Let  : S

! M be an R {derivation into an S {module M and let 1 = e +  + en (ei = ei  ei ej = 0 (i = 6 j )) 0

1

2

be the decomposition of 1 into orthogonal idempotents corresponding to the product n P decomposition of S . From 0 = (1) = ei and 2ei ei = ei we conclude easily i=1 that ei = 0 (i = 1 : : :  n). Therefore  induces R0 {derivations i : ei S ! ei M with n P  = i . Since ei S

= Si we have a unique Si {linear map hi : 1Si =R0 ! ei M with i=1 i = hi  dSi =R0 and an S {linear map

h = (h1 : : :  hn): 1S1 =R0  1Sn =R ! e1M  enM = M

with  = h  (dS1 =R0  : : :  dSn =R0 ). It is clear that any such map h must be hi on 1Si =R , hence it is unique as well, and (dS1 =R0  : : :  dSn =R0 ) is the universal derivation of S=R0 . Let R=R0 be an algebra for which 1R=R0 is nitely presented, hence by 2.8 we have that DerR0 RN = (DerR0 R)N . Then clearly the local-global-principle for vector elds holds: 30

2.22.Proposition. For f1  : : :  fn

2 R with R = (f  : : :  fn) and V  V 2 DerR R the 1

following assertions are equivalent: a) (V1 )fi = (V2 )fi for i = 1 : : :  n . b) (V1)p = (V2 )p for all p 2 Spec R . c) V1 = V2 .

1

2

0

Similarly vector elds can be pasted as follows: Let Vi 2 DerR0 Rfi (i = 1 : : :  n) be given such that (Vi )fj = (Vj )fi for i j = 1 : : :  n . Then there is precisely one vector eld V 2 DerR0 R with Vi = Vfi for i = 1 : : :  n . Exercises:

1) Let K be a eld and A a subalgebra of the polynomial algebra K X ]. Let T (1A=K ) P denote the torsion of 1A=K . Show that 1A=K =T (1A=K ) is isomorphic to Af , the f A A {submodule of K X ] generated by the derivatives f of the f 2 A . L R be a positively graded ring. Assume that R := L R is generated 2) Let R = n + n n>0 n N by homogeneous elements whose degrees are units of R0 (for example R+ = (R1 )). a) The linear map ` : 1R=R0 ! R+ induced by the Euler derivation (1.5) is surjective. b) Assume R+ contains a non-zerodivisor of R and 1Q(R)=R0

= Q(R). Then ` induces an exact sequence (called the Euler sequence) 0

2

0

2

0 ! T (1R=R0 ) ! 1R=R0 ;! R+ ! 0 `

where T (1R=R0 ) is the torsion of 1R=R0 . 3) Under the assumptions of exercise 2b) let R+1 := fx 2 Q(R)jxR+  Rg . Then R+1

= HomR(R+  R). Dualize the Euler sequence in order to obtain an isomorphism ;

;

R+1

= DerR0 R (x 7! x) where  is the Euler derivation. 4) Let R=R0 be an algebra, I  R an ideal and R := R=I . The following conditions are equivalent: a) RdR=R0 I  I 1R=R0 . b) The canonical map  : I=I 2 ! 1R=R0 =I 1R=R0 is zero. 1 is bijective. c) The canonical map 1R=R0 =I 1R=R0 ! R=R Moreover the conditions a)-c) imply d) I is g {invariant (see x 1, exercises 9-12) and are equivalent to d) in case 1R=R0 is a projective R {module. ;

31

5) Let R=R0 be an algebra, g := DerR0 R , p 2 Spec R , g p := DerR0 Rp . If 1R=R0 is a nitely presented R {module, then the following conditions are equivalent: a) p is g {invariant. b) p Rp is g p {invariant. L 6) Let R0 be a ring, (G +) an abelian group, and R := R0 G] = R0xg be the group g G algebra of G over R0 . Set M := R  Z G . a) The R0 {linear map d : R ! M with dxg = xg  g for all g 2 G is a derivation. b) d is universal, i.e. 1R=R0 = R Z G . c) For any R {module N there is an isomorphism DerR0 (R0G] N )

= Hom Z (G N ) which is functorial in N . 7) (Abelian groups with isomorphic group rings are isomorphic). Under the assumptions of exercise 6) consider the ring-epimorphism P P " : ZG] ! Z ( ng xg 7! ng ) Any abelian group G can be made a Z G]{module via " . Let G and H be abelian groups such that there is an isomorphism of Z {algebras Z G]

= Z H ]. a) Show for any abelian group G there is an isomorphism Hom Z (G G )

= Hom Z (H G ) which is functorial in G . b) Conclude that G

= H. For generalizations, see Ku6 ]. 8) Under the assumptions of Thm.2.12 the canonical map S R 1R=R0 ! 1S=R0 is an isomorphism if and only if each R0 {derivation  : R ! M into an S {module M can be uniquely extended to a derivation d : S ! M . 9) (Universal extensions of derivations). Let R=R0 and S=R be algebras,  : R ! M an R0 {derivation with M = RR . An R0 {derivation D : S ! N into an S {module N is called an extension of  to S , if there is an R {linear map h : M ! N such that the following diagram commutes R S 2

0

0

0

0

0

%$"#



D

M h N D is called a universal extension of  to S , if for any extension D : S ! N of  there is a unique S {linear map ` : N ! N with D = `  D . 0

0

32

0

0

a) Show that a universal extension of  always exists. This can be done by rst showing b) and c) below and proving a transitive law for universal extensions. L b) Let S = RfX g ] be a polynomial algebra over R and N := S R N  SdX .   P For f = r1 n X11  : : :  Xnn (r1 n 2 R) let Df be given by the formula 2

2





Df =

P X 1  : : :  X n  r 1

1 n +

n



P @f dX  @X

Then D is a universal extension of  to S . c) Let S = R=I be a homomorphic image of R and D : S ! M=IM the derivation induced by  , i.e. (r + I ) = r + IM for r 2 R . Then D is a universal extension of  to S . (For more about (universal) extensions of derivations, see Chap.III, 2.23 and B3 ].) n P

10)Let R0 be a ring with Q  R0 and R = R0 X1 : : :  Xn]. Then for ! = fi dXi i=1 @f @f j 1 i 2 R=R0 there exists an F 2 R with dF = ! if and only if @Xj = @Xi (i j = 1 : : :  n). 11)Let R=R0 be an algebra whose module of dierentials 1R=R0 is nitely generated and projective. For x 2 X := Spec R , X0 := Spec R0 , let t 2 TX=X0 (x) be a tangent vector at x . Show that there is a neighbourhood D(f ) of x in X (f 2 R) and a vector eld V 2 DerR0 Rf with V (x) = t . 12)Let R=R0 be an algebra such that 1R=R0 is nitelyQpresentable and R is reduced. Let X := Spec R , X0 := Spec R0 and let ftxgx X 2 TX=X0 (x) be given such that for x X each x 2 X there is a neighbourhood D(f ) of x and a vector eld Vf 2 DerR0 Rf with Vf (y) = ty for each y 2 D(f ). Show that there exists a unique vector eld V on X=X0 with V (x) = tx for all x 2 X . 13)(Vector elds on S 1 ). Let K be a eld of characteristic 6= 2, and let R = K X Y ]=(X 2 + Y 2 ; 1) = K x y], where x y are the images of X Y in R . a) Show that 1R=K is a free R {module with basis fydx ; xdyg , but R=K has no dierential basis. @ ; x @ ) with f 2 R , b) Show that the vector elds on Spec R have the form f  (y @x @y @  @ 2 DerK R are as in 2.15. where @x @y 2

2

33

x 3.

Dierential Modules of Field Extensions

The purpose of this section is to nd out which informations are contained in the dierential modules of eld extensions. Since they are vector spaces we are interested in their dimensions and their bases. Of course any result about the dimension of the dierential module of a eld extension implies a similar result for the derivation module, which is the dual of the dierential module. In this section we obtain the rst applications of the theory of dierential modules: Some results about separability and the structure of eld extensions.

A eld extension L=K is called nitely generated, or an algebraic function eld, if elements x1  : : :  xn 2 L exist with L = K (x1  : : :  xn ). A transcendence basis fx g2 of a eld extension L=K is called separating, if L=K (fxg2) is separable (algebraic). An algebraic function eld is called separable, if it has a separating transcendence basis. The denition of separability for an arbitrary eld extension will be given later. Consider rst the fundamental exact sequence (2.12) 



S R 1R=R0 ;! 1S=R0 ;! 1S=R ! 0

(1)

For many questions one has to study the kernel of  , which we denote by TR0 (S=R). It is sometimes called the \imperfection module" of S=R with respect to R0 . If S 0=R is another algebra and  : S ! S 0 an R {homomorphism we have a canonical commutative diagram with exact rows 0

 ;

TR0 (S=R)

1 S R R=R 0



1S=R0



1S=R

0

0

1S =R0

0

1S =R

0

id

TR0 (S 0 =R) S 0 R 1R=R0   id induces an S {linear map 0

0

0

TR0 (): TR0 (S=R) ! TR0 (S 0 =R) which makes T a functor of S . If R is a eld, and  is injective, then   id and TR0 () are injective as well, hence TR0 (S=R) can be identied with its image in TR0 (S 0 =R). The following proposition allows the reduction of some questions about arbitrary eld extensions to algebraic function elds. 3.1.Proposition. Let L=K and K=K0 be eld extensions. Then

S

TK0 (L=K ) = TK0 (Z=K ) Z

where Z runs over all intermediate elds of L=K for which Z=K is nitely generated. 34

Proof: Let fx g2 be a system of generators of the eld extension L=K and fX g2

a family of indeterminates. Then L=K has a presentation as in x 2, (7) and by 2.19 there is an exact sequence 0 ! U ! L K 1K=K0 

L LdX ! L=K 1

2

0

!0

where U is the L {vector space generated by certain elements P  dX F (fx g) + @F  @x with F 2 K fXg]. It is easy to see that TK0 (L=K ) = (L K 1K=K0 ) \ U . Hence each z 2 TK0 (L=K ) can be written

z=

(2)

Pn i  mi

i=1

and

z=

(3)

(i 2 L mi 2 1K=K0 )

P  (F (fxg) + P @F  dX)

@x In (3) only nitely many  are dierent from zero, say 1  : : :  r . Let X1  : : :  Xs be the indeterminates which actually occur in Fj (j = 1 : : :  r). Set 



Z := K (1  : : :  n  1  : : :  r  x1  : : :  xs ) The relations (2) and (3) then hold in Z K 1K=K0 

Ls ZdXi and therefore z 2 TK (Z=K ).

i=1

0

3.2.Corollary. If L=K is separably algebraic, then TK0 (L=K ) = 0, and we have a

canonical isomorphism

L K 1K=K0 = 1L=K0

Proof: By (1) the second statement follows from the rst since 1L=K = 0 by 2.6a). In

order to prove the rst statement we may assume by 3.1 that L=K is nite. Then by the theorem of the primitive element L=K has a presentation L = K X ]=(f ) with a separable polynomial f 2 K X ]. If x is the image of X in L we have f 0 (x) 6= 0. In the exact sequence 0 ! U ! L K 1K=K0  LdX ! 1L=K0 ! 0 the vector space U is generated by f (x) + f 0 (x)dX . Hence

TK0 (L=K ) = L K 1K=K0 \ U = 0 35

L

3.3.Corollary. Suppose L=K has a separating transcendence basis fx g2 . Then

a) 1L=K = Ldx , i.e. fx g2 is a dierential basis of L=K . 2 b) dimL 1L=K < 1 if and only if trdeg(L=K ) < 1 . c) 1L=K = 0 if and only if L=K is algebraic.

L

Proof: Z := K (fx g2 ) is isomorphic over K to the quotient eld of K fX g2 ]. By

2.9d) we have 1Z=K = Zdx . Since L=Z is separably algebraic there is an isomorphism 2 L Z 1Z=K = 1L=K by 3.2, which proves a). The assertions b) and c) are then clear. In particular we see that for an algebraic function eld L=K which is separable dimL 1L=K = trdeg(L=K ) moreover for any separating transcendence basis fx1  : : :  xt g of L=K the dierentials fdx1  : : :  dxt g form a basis of 1L=K . This result will be strengthened later (3.10). 3.4.Corollary. Suppose K has characteristic 0. Then a family fx g2 of elements of

L is a transcendence basis of L=K if and only if it is a dierential basis of L=K .

Proof: It remains to show that fx g2 is a transcendence basis of L=K , if fdx g2

is a basis of 1L=K . From 3.3c) we obtain that K is algebraic over K (fx g2). If fx g2 were algebraically dependent over K , a nite subfamily fx1  : : :  xm g would be algebraically dependent over K . With Z := K (x1  : : :  xm ) we had dimZ 1Z=K < m by 3.3a) since trdeg(Z=K ) < m . But then the dierentials dxi of the xi in 1Z=K are linearly dependent over Z , hence so are their images in 1L=K , a contradiction. 3.5.Examples:

a) For an extension eld L of Q we have 1L=Q = 0 if and only if L is a (possibly innite) algebraic number eld. b) Let K be a eld of characteristic 0, R := K X ]], and L := Q(R) = K ((X )). It is known that trdeg(L=K ) = 1 , hence dimL 1L=K = 1 . Consequently 1R=K cannot be nitely generated. This implies that the natural K {derivation d : R ! RdX , f 7! f 0 dX is not universal. c) Let k be a eld of characteristic p > 0, t an indeterminate, and K := k(t). The polynomial f := X p + tY p + t 2 K X Y ] is irreducible, and R = K X Y ]=(f ) is a domain. The eld L := Q(R) has transcendence degree 1 over K . 36

Let x y be the images of X and Y in R . Then by 2.14 @f dY i = Rdx  Rdy

1R=K = RdX  RdY=h @f dX + @x @y @f = @f = 0. It follows that 1 since @X L=K = Ldx  Ldy is a vector space of dimension 2, @Y hence L=K cannot be separable. For a eld K of characteristic zero all what can be said about 1L=K is already contained in 3.4. In the following we assume that Char K =: p > 0. In addition to the notion of separating transcendence basis the notion of p {basis is of importance in this case. Suppose S=R is an algebra where S is of prime characteristic p . A system fxg2 of generators of the algebra S=RS p] is called a p {generating set of S=R . For such systems the elements

x11  : : :  xnn (n 2 N  0 i p ; 1) form a system of generators of S as an RS p]{module. A family fxg2 of elements x 2 S is said to be p {independent over R if the elements (4) are linearly independent over RS p], and it is called a p {basis of S=R , if the elements (4) form a basis of S=RS p]. This is equivalent with the fact that the kernel of the RS p]{homomorphism ' : RS p]fXg2] ! S (X 7! x ) is generated by the polynomials Xp ; xp ( 2 ). (4)

3.6.Proposition. Let fx g2 be a p {generating set of S=R . Then fdx g2 is a

system of generators of 1S=R . Moreover fxg2 is a p {basis of S=R if and only if it is a dierential basis of S=R . Proof: For any derivation d of S=R we have d(RS p ]) = 0, hence 1S=R = 1S=RS p ] .

The rst assertion of 3.6 is now clear. If fxg2 is a p {basis of S=R , using

S = RS p]fXg]=(fXp ; xp g) L SdX = L Sdx , since for f := Xp ; xp we have =

and 2.14 we obtain 1S=R 2 2 @f = 0 for all   2 . @x Conversely assume fdxg2 is a basis of 1S=R and f an element of the kernel I of the map ' dened above. Since Xp ; xPp 2 I we may assume degX f p ; 1 for alle  2 . Since f (fx g) = 0, we also have @x@f  dx = 0, hence @x@f = 0 for all  2  since  fdx g is linearly independent over S . By induction we see that all partial derivatives of f are in I . In S fXg2] the polynomial f can be written

f=

P

0i p;1

a1n (X1 ; x1 )1    (Xn ; xn )n 37

with a1 n 2 S and certain 1  : : :  n 2 . Then 1 ++n 1!  : : :  n!  a1 n = (@x @)1    (@xf )n = 0 for all ( 1 : : :  n) 1 n

Since i p;1 this yields a1n = 0, and f = 0. Hence I is generated by fXp ;xpg2 , and fxg is a p {basis of S=R . The proposition implies that the cardinality of a p {basis of S=R is independent of the chosen p {basis since the corresponding fact is true for bases of free modules. This cardinality is called the p {degree of S/R and is denoted degp(S=R). By Zorn's lemma it is easy to see that any eld extension L=K has a p {basis and hence a p {degree. More generally each p {independent family of L=K can be extended to a p {basis of L=K . For eld extension 3.6 can be strengthened in the following way: 3.7.Proposition. Let L=K be a eld extension and fx g2 a family of elements of L .

a) fxg2 is a p {basis of L=K if and only if it is a dierential basis of L=K . b) dimL 1L=K = degp(L=K ). c) 1L=K = 0 if and only if L = K Lp]. Each p {generating set of L=K contains a p {basis as a subset.

Proof: b) and c) follow from 3.6. For a) we have only to show that if fdx g is a

basis of 1L=K then fx g2 is a p {basis of L=K . Given such a basis we have for L0 := K Lp]fxg2]

1L=L = 1L=K =hfdx gi = 0 hence L = L0 Lp] by c). But since Lp  L0 this implies that L = L0 , and fxg is a p {generating set of L=K . By 3.6 it is then even a p {basis. The last statement of the proposition is clear, since from each generating set of the vector space 1L=K a basis can be selected. 0

3.8.Corollary. If L=K has a separating transcendence basis fx g , then fx g is also a

p {basis and degp(L=K ) = trdeg(L=K ). This follows from 3.3 and 3.7a), and is our rst result in which dierentials are only used in the proof. For algebraic function elds more precise statements can be made. 3.9.Proposition. For an algebraic function eld L=K the following assertions are equi-

valent: a) 1L=K = 0. b) L=K is separably algebraic.

38

Proof: b) ! a) was already shown in 2.6a).

a) ! b). If 1L=K = 0, then L = K Lp] = K Lp2 ] =    = K Lpe ] for all e > 0 by 3.7c). Let fx1  : : :  xt g be a transcendence basis of L=K , and let L0 the set of all elements of L that are separable over K (x1  : : :  xt ).e Then L=L0 is nite and purely inseparable: There are elements y1  : : :  yr 2e L with yeip 2 L0e for some e 2 N (i = 1e  : : :  r) such e that L = L0 y1 : : :  yr ]. But Lp = L0p y1p  : : :  yrp ]  L0 , hence L = K Lp ] = L0 . Thus L=K (x1  : : :  xt ) is separably algebraic. By 3.3 we have t = trdeg(L=K ) = dimL 1L=K = 0, hence L=K is separably algebraic. 3.10.Theorem. Let L=K be an algebraic function eld. Then

a) degp(L=K )  trdeg(L=K ). b) From each p {basis of L=K a transcendence basis of L=K can be selected. c) degp(L=K ) = trdeg(L=K ) if and only if L=K is separable. d) If L=K is separable, for elements x1  : : :  xt 2 L the following statements are equivalent:  ) fx1  : : :  xt g is a p {basis of L=K . ) fx1 : : :  xt g is a separating transcendence basis of L=K .  ) fdx1  : : :  dxt g is a basis of 1L=K . Proof: Let fx1  : : :  x g be a p {basis of L=K , and set L0 := K (x1  : : :  x ). Then

1L=L = 1L=K =hdx1  : : :  dx i = 0, hence L=L0 is separably algebraic by 3.9, and we have trdeg(L=K ) = trdeg(L0 =K ). Since x1 : : :  x generate L0 =K a transcendence basis can be selected from them which is then also a transcendence basis of L=K . This proves b) and a). If degp(L=K ) = trdeg(L=K ), then fx1 : : :  x g has to be already a transcendence basis of L=K , hence L=K is separable and c) is true (3.8). As for d) suppose now that L=K is separable. We have just seen that ) ! ). But ) !  ) follows from 3.3 and  ) ! ) from 3.7a). Thus the proof is complete. 0

3.11.Corollary. Let L=K be an algebraic function eld, t := trdeg(L=K ), and set

 := degp(L=K ). a) If L=K is separable, then L=K can be generated by t + 1 elements. b) If  > t , then L=K can be generated by  (and not by fewer) elements. Proof: a) is a consequence of the theorem of the primitive element.

b) Let fx1  : : :  x g be a p {basis of L=K . Select a transcendence basis from it, say fx ;t+1  : : :  x g . Let L0 := K (x ;t+1  : : :  x ), and denote by L00 the eld of all elements of L which are separable over L0 . We have Lpe  L00 for some e 2 N . Then

L = K Lp]x1 : : :  x ] = L0 Lp]x1  : : :  x ;t ] = L00 Lp]x1  : : :  x ;t ] e = L00 Lp ]x1 : : :  x ;t ] = L00 x1  : : :  x ;t ] 39

By the theorem of the primitive element L00 x1] can be generated over L0 by one element x . Hence we obtain L = K (x ;t+1  : : :  x )x x2  : : :  x ;t ] = K (x x2  : : :  x ), and L=K has  generators. Since dimL 1L=K =  it is clear that  ; 1 elements will not do. For arbitrary eld extensions L=K and K=K0 we wish to study the vector space TK0 (L=K ) more closely. Let Z be an intermediate eld of L=K . By the denition of T we have a commutative diagram with exact rows and columns 0 0 !"

TK0 (L=Z )

TK (L=Z )

0

L Z TK0 (Z=K )

L Z (Z K 1K=K0 )

L Z 1Z=K0

L Z 1Z=K

0

0

TK0 (L=K )

L K 1K=K0

1L=K0

1L=K

0

1L=Z

1L=Z

0 0 which induces the following commtutative diagram with exact rows and columns 0 0 0 #$%&'()*+,-./0123456789:
0. Set insepi (L=K ) := insep(K (Lp )=K ) for all i 2 N . a) finsepi(L=K )gi2N is a monotone decreasing sequence and there exists an r 2 N such that insepi (L=K ) = 0 for all i  r . Pi=0 insepi(L=K ), and let Z be an intermediate eld of L=K such that b) Let c := 1 Z=K is separable and L=Z algebraic. Then L : Z ]  pc . 47

8) Under the assumption as in 7) let r be the smallest number such that K (Lpr )=K is separable. A transcendence basis fx1  : : :  xt g of L=K ris called distinguished if r r p p fx1  : : :  xt g is a separating transcendence basis of K (Lp )=K . a) Show the existence of a distinguished transcendence basis. b) Given a distinguished transcendence basis fx1  : : :  xt g of L=K let Z denote the separable algebraic closure of K (x1  : : :  xt ) in L . ) Show that K (Z pi ) = K (Lpi ) for all i  r . Pi=0 insepi(L=K ). ) Conclude that L : Z ] = pc with c = 1 For more precise results about distinguished transcendence bases, see KrH]. 9) Let L=K be a eld extension where Char K =: p > 0. The following assertions are equivalent: a) L = K Kp Lp . b) Every p {basis of K=K p is a p {basis of L=Lp . c) The canonical map L K 1K=Kp ! 1L=Lp is bijective. d) L=K is separable and 1L=K = 0. e) Every (absolute) derivation d : K ! K into an L {vector space M has a unique extension to a derivation D : L ! M . These are the \formally etale" eld extensions in the sense of EGA], IV. x 19. 10)Let L=K be a separable eld extension where Char K =: p > 0. a) If B  L is a set of p {independent elements over K , then B is algebraically independent over K (M1 ], Thm. 1). b) There is an intermediate eld Z of L=K such that Z=K is purely transzendental and L=Z separable with 1Z=K = 0.

48

x 4. Dierential Modules of Local Rings The structure of the dierential module of a local ring is of course much more complicated than that of a eld. We wish to determine rst its minimal number of generators which is also the dimension of the tangent space at the maximal ideal. Then we shall study what consequences it has when the dierential module of a local ring vanishes.

Let (R m ) be a local ring with residue eld K := R= m and p := Char K . Let d denote the universal derivation of R= Z . By 2.13 there is a canonical exact sequence of K {vector spaces m = m 2 ;! 1R= Z = m 1R= Z

(1)



;! 1K= Z ! 0

where (x + m 2 ) = dx + m 1R= Z for all x 2 m . We have pR  m , and for each r 2 R we obtain d(pr) = pdr 2 m 1R= Z . Therefore  induces a K {linear map

 : m = m 2 + pR ! 1R= Z = m 1R= Z 0

with  (x + m 2 + pR) = dx + m 1R= Z . 0

4.1.Theorem. The canonical sequence 

0

0 ! m= m + pR ;;! 1R= Z = m 1R= Z ;! 1K= Z ! 0

(2)

2

is exact. Proof: It remains to be shown that  is injective. For R := R= m 2 +pR , m := m = m 2 +pR we have 1R= Z = 1R= Z =Rd( m 2 + pR) by x 2(6) and, since d( m 2 + pR) = d m 2 + pdR  m 1R= Z , we obtain 0

1 1 1R= Z = m 1R= Z = 1R= Z =Rd( m 2 + pR)= m 1R= Z =Rd( m 2 + pR) = R= Z = m R= Z

 can be identied with the analogous map 0

1 1 1  : m ! R= Z = m R= Z ((x) = d x + m R= Z )

where d denotes the universal derivation of R= Z . It is enough to show that  is injective. In other words: It suces to prove the theorem if m 2 + pR = 0 in the original local ring R . Then R is a complete and separated local ring having the same characteristic as its residue eld K . By Cohen's structure theorem (Matsumura M1 ], Thm. 60), which is 49

needed here only in a mild form ( m 2 = 0), there is a eld K  R , which is mapped isomorphically onto K by the canonical epimorphism R ! K . Let fx g  be a basis of the K {vector space m . Since m 2 = 0, we have 0

2

R=K  0

L





2

K x  xx = 0 for   2  0

Therefore the kernel I of the canonical K {homomorphism 0

K fXg ] ! R 0

(X 7! x )

2

is generated by all products XX (  2 ). By 2.19 we obtain

L 1R= Z  = (R K 1K = Z  RdX)=U   0

0

2

L

where U is the submodule of RdX generated by the elements xdX + x dX   (  2 ). Hence L 1R= Z = m 1R= Z  = 1K= Z  KdX 2

n P





2

n P

and  is identied with the mapping given by ixi 7! idXi (i 2 K ). Clearly i=1 i=1 this mapping is injective. 4.2.Corollary. Suppose that both m and 1R= Z are nitely generated. Then

( 1R= Z ) = edim R=pR + dimK 1K= Z Proof: By Nakayama's lemma

( 1R= Z ) = dimK ( 1R= Z = m 1R= Z ) = dimK ( m = m 2 + pR) + dimK 1K= Z The ring R~ := R=pR is local with maximal ideal m~ := m =pR and residue eld K . Since ~ = m~ 2 we see that dimK ( m = m 2 + pR) = dimK m~ = m~ 2 m = m 2 + pR = m =pR= m 2 + pR=pR = m = edim R~ . The term dimK 1K= Z is known by the results of x 3:

 Trdeg(K=Q ) if p = 0 dimK K= Z = degp(K=K p) if p > 0 1

In fact 4.2 and Nakayama's lemma yield the following more precise statement. 50

4.3.Corollary. Under the assumptions of 4.2 let x1  : : :  xm 2 m be elements whose residue classes in m=pR form a minimal system of generators of that ideal in R=pR ,

and let y1 : : :  yt 2 R be elements whose residue classes yi in K form a transcendence basis of K=Q , in case p = 0, and a p {basis of K=K p , in case p > 0. Then fdx1  : : :  dxm  dy1 : : :  dyt g is a minimal system of generators of 1R= Z . Now let (S n ) be a local ring with residue eld L . Let R be an arbitrary ring and : R ! S a ring homomorphism. We want to study 1S=R . p := 1 ( n ) is a prime ideal of R , and induces a ring homomorphism : R p ! S with ( p Rp )  n . By 2.5c) we have 1S=R = 1S=Rp , and therefore we may assume that R is a local ring with maximal ideal m and ( m )  n . In other words: S=R is a local algebra and K := R= m may be regarded as a subeld of L . ;

0

0

4.4.Theorem. Under these assumptions there is an exact sequence of L {vector spaces

T ! T Z (L=K ) ! n = n2 + m S ! 1S=R = n 1S=R ! 1L=K ! 0 0

where T is the kernel of the canonical homomorphism 0

L K 1R= Z = m 1R= Z ! 1S= Z = n 1S= Z Proof: Let p := Char K . In the commutative diagram ;  

0

0

T

TZ (L=K )

0

0

L K m = m 2 + pR

L K 1R=Z = m 1R=Z

L K 1K=Z

0

0

n = n 2 + pS

1S=Z = n 1S=Z

1L=Z

0

n = n 2 + mS

1S=R = n 1S=R

1L=K

0

0 0 0 the rst two rows are exact by 4.1. Clearly the left hand column is exact. The exactness of the last two columns follows from the denition of T and T Z (L=K ). The snake lemma gives us now the exact sequence of the theorem. 0

51

Since TZ (L=K ) = 0 when L=K is separable (3.15) we obtain 4.5.Corollary. If L=K is separable, then the sequence

0 ! n = n 2 + m S ! 1S=R = n 1S=R ! 1L=K ! 0 is exact. 4.6.Example: Let L=K be an algebraic function eld with trdeg(L=K ) = 1 where K

is a perfect eld. Let R  L be a discrete valuation ring with K  R and Q(R) = L . Then R is a localization of an ane K -algebra, hence 1R=K is nitely generated. The residue eld k of R is separably algebraic over K , hence 1k=K = 0. It follows from 4.5 that 1R=K = Rdt with a prime element t of R . Since L R 1R=K = 1L=K is an L {vector space of dimension 1 the R {module 1R=K is free and can be identied with its image in 1L=K . A dierential ! 2 1L=K is called regular at R if ! 2 1R=K , i.e. if ! = fdt with some f 2 R . It is called globally regular if it is in 1R=K for all discrete valuation rings R as above. It can be shown that the globally regular dierentials form a nite dimensional vector space over the algebraic closure K of K in L . Its dimension is the genus of the function eld L=K . If L is the eld of rational functions of an irreducible smooth projective algebraic curve C dened over an algebraically closed eld K , then the discrete valuation rings R as above are the local rings OP of C at its closed points P . In this situation the element ! 2 1L=K are called rational dierentials on the curve C and ! is called regular at P if ! 2 1 P =K . The genus of the curve is that of its function eld, i.e. the K {dimension of the vector space of globally regular dierentials. For irreducible curves with singularities the notion of regular dierential at a singular point P is more complicated than the above. See Ha], Chap. IV and Se2 ], Chap. II and IV for more about the role of Kahler dierentials in the theory of algebraic curves. O

Let us now consider local algebras S=R where L=K is not necessarily separable. 4.7.Lemma. Let (R m ) be a local ring and p > 0 an integer. Then each subring R  R with Rp  R is also local, and m := m \ R is its maximal ideal. We show that the elements of R n m are units in R . An element x 2 R n m is 0

0

0

0

0

0

0

0

0

certainly a unit in R , so xy = 1 for some y 2 R . But then xxp 1 yp = 1 and xp 1 yp 2 R hence x is also a unit of R . ;

;

0

0

4.8.Theorem. Suppose that under the conditions of 4.4 K is a eld of characteristic p > 0 and let m := n \ RS p]. Then there is a canonical exact sequence 0

0 ! n = n 2 + m S ! 1S=R = n 1S=R ! 1L=K ! 0 0

52

Proof: By the lemma R := RS p ] is a local ring with maximal ideal m . Clearly 0

0

K Lp] =: K is the residue eld of R . For the universal derivation d of S=R we have d(RS p])  p 1S=R  n 1S=R and in particular d m  n 1S=R . Hence the canonical map n = n 2 ! 1S=R = n 1S=R induces a map n = n 2 + m S ! 1S=R = n 1S=R with the same image. Moreover 0

0

0

0

1S=R = n 1S=R = 1S=R = n 1S=R 0

and

0

1L=K = 1L=KLp ] = 1L=K

0

By 4.4 we have an exact sequence

T ! T Z (L=K ) ! n = n 2 + m S ! 1S=R = n 1S=R ! 1L=K ! 0 0

0

0

0

0

0

where T is the kernel of the canonical map L K 1R = Z = m 1R = Z ! 1S= Z = n 1S= Z . We shall show that T ! T Z (L=K ) is surjective, which then will give us the desired exact sequence. Let fxg  be a p {basis of L=K , fyg  a system of representatives of fxg in S and R resp. K the universal derivation of R = Z (of K = Z ). Then by 2.19 0

0

0

0

0

0

2

0

0

0

2

0

0

0

L

1L= Z = L K 1K = Z =hfK (xp )g i  Ldx   and T Z (L=K ) is the subspace of L K 1K = Z spanned by f1 K (xp )g  . On the other hand yp 2 R and R (yp ) is contained in the kernel of the canonical map 1R = Z ! 1S= Z = n 1S= Z . If  denotes the image of R (yp ) in 1R = Z = m R1 = Z , then 1  2 T and 1 K (xp ) is the image of 1  in T (L=K ), q.e.d. 0

0

0

2

2

0

0

0

0

0

0

0

0

0

2

0

0

0

0

0

0

Under the assumptions of 4.4 let S=R be essentially of nite type, hence L=K is an algebraic function eld. If Char K = p > 0 let m be dened as in 4.8. In this case we call the number 0

insep(S=R) := degp(L=K ) ; Trdeg(L=K ) ; (edim S= m S ; edim S= m S ) 0

the inseparability of S=R . We set insep(S=R) = 0 when Char K = 0. 4.9.Lemma. We always have insep(S=R)  0, with equality when L=K is separable. Proof: It suces to consider the case Char K = p > 0. By the exact sequences of 4.4

and 4.7 we obtain an exact sequence

TZ (L=K ) ! n = n 2 + m S ! n = n 2 + m S ! 0 0

Since dimL TZ (L=K ) = degp(L=K ) ; Trdeg(L=K ) by 3.12b) it follows that edim S= mS edim S= m S + degp(L=K ) ; Trdeg(L=K ) 0

53

which implies insep(S=R)  0. If L=K is separable, then degp(L=K ) = Trdeg(L=K ) by 3.10c), and comparing the exact sequences of 4.5 and 4.8 we see that edim S= m S = edim S= m S , hence insep(S=R) = 0. For a local algebra S=R as above we call the number (S=R) := edim S= m S ; dim S= m S the regularity defect of S=R . Remember that for a noetherian local ring R we have always that edim R  dim R , and that such a ring is called regular if edim R = dim R . An arbitrary noetherian ring is called regular, if all its localizations are so. 0

4.10.Proposition. Let S=R be essentially of nite type. Then ( 1S=R ) = (S=R) + dim S= mS + Trdeg(L=K ) + insep(S=R) Proof: If L=K is separable this follows from 3.3 applying Nakayama's lemma to the

exact sequence of 4.5. In the case Char K = p > 0 we use instead the exact sequence of 4.8 and the fact that dimL 1L=K = degp(L=K ). 4.11.Proposition. Let S=R be an algebra which is essentially of nite type. Then the

following conditions are equivalent: a) 1S=R = 0. b) For all P 2 Spec S we have (with p := P \ R ) P SP = p SP and k( P)=k( p ) is separably algebraic. b') The conditions of b) hold for all P 2 Max (S ).

Proof: By the local-global principle we may assume that S=R is a local algebra. Let n be the maximal ideal of S , m the maximal ideal of R , L := S= n , and K := R= m .

a) ! b'). If 1S=R = 0, then the terms on the right hand side of 4.10 vanish. Since L=K is nitely generated, L=K is separably algebraic, moreover we see that edim S= mS = 0, in other words n = m S . b') ! a). Under the assumptions of b') we have 1L=K = 0 and n = n2 + m S = 0, hence 1S=R = n 1S=R = 0 by 4.5. Thus 1S=R = 0 by Nakayama's lemma, since 1S=R is nitely generated. 4.12.Corollary. Let K be a eld and A=K an algebra which is essentially of nite

type. Then the following are equivalent: a) 1A=K = 0. b) A  = L1    Lm where Li=K (i = 1 : : :  m) is a nite separable eld extension. c) For each eld extension L=K the L {algebra L K A is nite dimensional (as L {vector space) and reduced. d) If K is the algebraic closure of K , then K K A is a nite direct product of copies of K. 54

Proof: b)

! a). By 2.20 1L1



= 1L1=K    1Lm =K Lm =K 

and by 3.9 we have 1Li =K = 0, hence 1A=K = 0. a) ! b). Condition a) and 4.11 imply that P AP = 0 for each P 2 Max (A), consequently Max (A) = Min (A). Since A is noetherian Min (A) is a nite set, say f P 1 : : :  P m g . From P i AP i = 0 (i = 1 : : :  m) we obtain P 1 \    \ P m = 0 and by the Chinese remainder theorem A = A= P 1    A= P m The assumption 1A=K = 0 implies that 1A=P i =K = 0. Therefore A= P i is a nite and separable extension eld of K (i = 1 : : :  m) by 3.9. a) ! c). By 2.17 we have 1L K A=L  = L K 1A=K = 0 

hence L K A is a nite direct product of nite separable extension eld of L by b), in particular L K A is reduced and a nite L {algebra. c) ! d) is trivial. q.e.d. d) ! a). We have K K 1A=K  = 1K K A=K = 0 by 2.20, hence 1A=K = 0, Remember that for an algebra S=R a prime P 2 Spec S is called unramied over R (or S=R unramied at P ), if P SP = p SP where p := P \ R , and k( P )=k( p ) is separably algebraic. Otherwise S=R is called ramied at P . The algebra S=R is called etale at P , if S=R is unramied at P and SP =Rp is at. S=R is called unramied (etale), if S=R is unramied (etale) at all P 2 Spec S . Clearly, if N  R is a multiplicatively closed set and S = RN , then S=R is etale. In the situation of 4.11, the algebra 1 S=R is unramied if and only if S=R = 0. The set VS=R := f P 2 Spec S j S=R is ramied at P g is called the ramication locus of S=R . The etale locus of S=R is the set of all P 2 Spec S at which S=R is etale. 

4.13.Corollary. Under the assumptions of 4.11 we have

VS=R = Supp ( 1S=R ) In particular VS=R is a closed subset of Spec S . If R is noetherian, the etale locus of S=R is open in Spec S . By 4.11 a prime ideal P 2 Spec S is unramied over R if and only if 1SP =Rp =( 1S=R )P = 0, hence the formula of 4.13 holds. The etale locus is the intersection of Spec S n VS=R 55

with the set of all P 2 Spec S at which S=R is at. If R is noetherian, this set is open in Spec S (Matsumura M1 ], Thm. 53). Since Supp ( 1S=R ) is the set of all P 2 Spec S with P  Ann( 1S=R ) we may write, under the assumptions of 4.11

VS=R = f P 2 Spec S j P  Ann( 1S=R )g Proposition 4.11 allows easy proofs of some statements about unramied and etale extensions. 4.14.Corollary. Let Q R , so is Q .

 P be elements of Spec S . If P is unramied (etale) over

In fact if ( 1S=R )P = 0, then ( 1S=R )Q = 0. Moreover atness of SP =R implies atness of SQ =R .

4.15.Corollary. Assume that the algebra T=S is essentially of nite type.

a) If T=R is unramied, so is T=S . b) If T=S and S=R are unramied (etale), so is T=R . Proof: Use the exact sequence

T S 1S=R ! 1T=R ! 1T=S ! 0 and 4.11. 4.16.Corollary. a) Let R =R be an arbitrary algebra. If S=R is unramied (etale), so 0

is R R S=R . The converse is true, if R =R is faithfully at. b) Let S=R and T=R be algebras that are essentially of nite type. If S=R and T=R are unramied (etale), so is S R T=R . The converse is true if S=R and T=R are faithfully at. 0

0

0

For a) use the formula 1R R S=R = R R 1S=R and for b) the formula 1S RT=R = T R 1S=R  S R 1T=R (see 2.16). 0

0

0



For an algebra S=R a prime p 2 Spec R is called unramied in S , if all P 2 Spec S lying over p are unramied over R , otherwise p is called ramied in S . Let vS=R denote the set of all p 2 Spec R that are ramied in S . 4.17.Corollary. Under the assumptions of 4.11 suppose that Spec S

closed map. Then vS=R is a closed subset of Spec R .

In fact vS=R is the image of the closed set VS=R in Spec R . 56

! Spec R is a

4.18.Examples: a) Let K=Q be a nite eld extension (i.e. K is an algebraic number

eld) and let R be the integral closure of Z in K (the ring of integers of K ). Then there are only nitely many prime numbers p such that (p) is ramied in R . In fact, vR= Z is a closed subset of Spec ( Z ) by 4.17. The set vR= Z cannot be all of Spec ( Z ) since 1K=Q = K R 1R= Z = Q Z 1R= Z = 0 hence the annihilator of 1R= Z as a Z {module is 6= 0. It follows that vR= Z is a nite set. b) Let S=R be a nite ring extension where R and S are Dedekind domains. Assume that L := Q(S ) is separable over K := Q(R). As in a) we have 1L=K = L S 1S=R = 0, and hence 1S=R is a torsion module. VS=R is a nite set, and so is vS=R . For each p 2 Spec R the extension ideal p S has a unique representation p S = P e11  : : :  P err

as a power product of primes P i 2 Spec S . The number ei is called the ramication index of P i over R . Clearly P i is ramied over R if and only if ei > 1 or k( Pi )=k( p ) is inseparable. By the Chinese remainder theorem 1S=R =

Q

P VS=R 2

1SP =Rp ( p := P \ R)

Since SP is a discrete valuation ring, hence a principal ideal domain, it follows from 4.10 that P ( 1SP =Rp ) 1, if k( P)=k( p ) is separable, and P ( 1SP =Rp ) 1+degp(k( P )=k( p )) otherwise. Let eP be the ramication index of P over R and assume that k( P )=k( p) is separable. Then 1SP =Rp  = SP = P"P SP for some "P 2 N . We wish to compare the integers eP and "P . With SP := SP = pSP = SP = P eP SP we have 1SP =k(p )  = 1SP =Rp = p 1SP =Rp  = SP = P Min (eP "P )SP

By Cohen's structure theorem

SP = k( P)t]=(teP )

and therefore 1SP =k(p ) = SP dt=heP  eP 1 dti  = SP =(eP  eP 1 ) where  is the image of t in SP (a generator of the maximal ideal of SP ). Let p := Char k( P ). Comparing the two expressions for S1 P =k(p) we obtain:  ) If eP 6 0 mod p then "P = eP ; 1 and hence ;

;

1SP =Rp  = SP = P eP 1SP ;

57

 ) If eP  0 mod p , then "P  eP . P is called tamely ramied over R , if e P > 1, e P 6 0 mod p , and k ( P )=k ( p ) is separable. The example shows that 1S=R not only \knows" which primes of S are ramied over R but sometimes \discovers" the ramication index of tamely ramied primes too. For more information about modules of dierentials of discrete valuation rings, see BK]. c) Under the assumptions of 4.11 let X := Spec S , Y := Spec R , and let f : X ! Y be the morphism corresponding to the structure map R ! S . For x 2 X and y := f (x) let Ox resp. Oy be the local rings of X at x (of Y at y ), let m x resp. m y be their maximal ideals and k(x) resp. k(y) their residue elds. Call x(X=Y ) := (Ox=Oy ) the regularity defect of X=Y at x . Since ( 1 x= y ) is the dimension of the tangent space TX=Y (x) we obtain from 4.10 the formula O

O

dimk(x) TX=Y (x) = x(X=Y ) + dim Ox= my Ox + Trdegk(x)=k(y) + insepOx=Oy We say that X=Y is unramied at x , if S=R is so at the prime P corresponding to x . This is the case if and only if TX=Y (x) = 0. The following structure theorem for unramied extensions generalizes the theorem of the primitive element for nite separable eld extensions. 4.19.Theorem. Let S=R be a local algebra where (R m ) is noetherian and S=R is

essentially nite. a) If S=R is unramied, then there is a prime n 2 Spec (RX ]) with n \ R = m , a monic polynomial f 2 RX ] with f 2= n , and an R {epimorphism  : RX ]n =(f ) ! S , which is an isomorphism, if S=R is etale. b) If in addition R and S have the same residue eld, there is an  as in a) where n = ( m  X ) and f (0) 2 m . 0

Proof: That S=R is essentially nite means that S = T P where T=R is a nite subalgebra of S=R and P 2 Spec T . If S=R is unramied, then L := TP = P TP = TP = m TP is a nite and separable extension eld of K := R= m . Therefore we can nd a  2 L n f0g

with L = K ]. Let P 1  : : :  P r be the maximal ideals of the semilocal ring T (all lying over m ). By the Chinese remainder theorem

T= mT = T1    Tr

(3)

with Ti := TP i = m TP i (i = 1 : : :  r). Assume P 1 = P . Then T1 = L , and we can choose an x 2 T with image  in T1 and image 0 in Ti for i = 2 : : :  r . Write T := Rx], P := P \ T . Since x is contained in all maximal ideals of T but P we conclude that P is the only prime of T lying over P , hence S = T P = T P and 0

0

0

0

58

0

TP =TP is nite. But TP =TP is unramied and both rings have the same residue eld L . By Nakayama S = TP . After replacing T by T we may assume T = Rx]. Then T= mT = K X ]=(f ) with a monic polynomial f , and the decomposition (3) corresponds to the factorization of f into powers of irreducible polynomials. Since L is one of the factors of T= mT , the minimal polynomial g of  over K is one of the factors of f , and it is not a multiple factor. Therefore f = g  ' with ' 2 K X ], and g does not divide ' . Let n be the degree of f . By Nakayama T = R + Rx + : : : Rxn 1 , and there is a monic polynomial f 2 RX ] with f (x) = 0 and image f in K X ]. Let n be the inverse image of P by the R {epimorphism 0

0

0

0

0

0

0

0

0

;

0 : RX ] ! Rx]

(X 7! x)

0 induces an R {epimorphism  : RX ]n =(f ) ! S and a K {epimorphism  : K X ]n =(f ) ! L where n is the image of n in K X ]. Clearly  is an isomorphism. Assume f 2 n . Then f () = 0. But f () = g ()  '() since g() = 0. We have '() 6= 0 since g does not divide ' , and g () 6= 0, because  is separable over K . Thus we arrived at a contradiction, so f 2= n . Suppose now that S=R is etale and let I be the kernel of  . The exact sequence 0

0

0

0

0

0

0 ! I ! RX ]n =(f ) ;! S ! 0 remains exact after tensorization with K over R , because S is a at R {module. We obtain an exact sequence 0 ! I= mI ! K X ]n =(f ) ;! L ! 0 and, as we already know,  is an isomorphism. We conclude by Nakayama that I = 0. In case L = K there is a 2 R with image  in L . The R {isomorphism RY ] ! RX ] (Y 7! X ; ) induces an R {epimorphism

~ : RY ]n~ =(f~) ! S where n~ is the inverse image of n in RY ] and f~ the polynomial with f~(Y ) = f (Y + ). Again ~ is an isomorphism, if S=R is etale. Since x ; 2 P we have Y 2 n~ , and thus ~n = ( m  Y ). Moreover f~(0) = f ( ) 2 m since f () = 0, and f~ 2= n~ since f 2= n . The proof of the theorem is now complete. 0

0

4.20.Remark: Let S=R be a local algebra as in 4.19, and assume S=R is essentially of nite type and unramied. Since S= mS is nite over R= m one can conclude from Zariski's

Main Theorem that S=R is essentially nite. Therefore theorem 4.19 also holds under the weaker assumption that S=R be essentially of nite type. 59

4.21.Corollary. Let R=R0 be an algebra and S=R an etale algebra which is essentially

of nite type. Then

1S=R0  = S R 1R=R0 If R=R0 is essentially of nite type and R0 noetherian, then DerR0 S  = S R DerR0 R Proof: Consider the exact sequence

0 ! TR0 (S=R) ! S R 1R=R0 ! 1S=R0 ! 1S=R ! 0 Since S=R is unramied we have 1S=R = 0 by 4.11, and it remains to show that TR0 (S=R) = 0. By the theorem S = RX ]n =(f ) with a monic f 2 RX ] where f (x) is a unit in S if x denotes the image of X in S . It follows that 0

1S=R0 = S R 1R=R0  SdX=hf (x) + f (x)dX i 0

f (x) being a unit of S the canonical map S R 1R=R0 ! 1S=R0 is necessarily injective, hence TR0 (S=R) = 0. When R=R0 is essentially of nite type and R0 noetherian, then 1R=R0 is of nite presentation. The formula HomS (S R 1R=R0  S )  = S R HomR ( 1R=R0  R) implies the statement about the derivation modules. 0

4.22.Corollary. Under the assumptions of 4.21 for any derivation  : R

exist a unique derivation # : S ! S with #jR =  (a \prolongation" of  ).

! R there

Proof: We have 1S=Z = S R 1R=Z . Any S {linear form L : 1S=Z ! S is given as L = idS ` with a unique linear form ` : 1R=Z ! R which is a restriction of L . From this

the claim follows. In particular any derivation of a eld K has a unique prolongation to any separable algebraic extension eld (by 4.22 and Zorn's lemma). Our last observation about unramied algebras is the proposition following the next lemma. 4.23. Lemma. In each algebra R=R0 there is a (unique) maximal subalgebra R =R0 with 1R =R0 = 0. 0

0

Proof: Let R =R0 be the union of all subalgebras T=R0 with 1T=R0 = 0. Any x 2 R 0

0

is in such a T. By the functorial map 1T=R0 ! 1R =R0 it follows that d1R =R0 x = 0. Since 1R =R0 is generated by the d1R =R0 x with x 2 R we have R1 =R0 = 0. 0

0

0

0

60

0

0

4.24. Proposition. Let R=R0 be a nite algebra where R0 is a noetherian ring. Then

there is a (unique) maximal unramied subalgebra R =R0 of R=R0 . 0

Proof: Let R =R0 be the maximal subalgebra with 1R =R0 = 0. Since R0 is noetherian 0

0

and R=R0 nite, so is R =R0 . Hence R =R0 is unramied by 4.11. If R  R is unramied over R0 , then 1R =R0 = 0, and R  R . 0

0

00

00

00

0

We close this section with the following criterion for the existence of a p {basis which follows from 4.11. 4.25. Proposition. Let R be a local ring of prime characteristic p and R0 ! R a ring

homomorphism such that R=R0Rp] is nite. Then the following conditions are equivalent: a) R=R0 has a p {basis. b) 1R=R0 is a free R {module. Proof: a)

! b) was shown in 3.6.

b) ! a). R=R0 has a basis of the form fdx1  : : :  dx g with x1  : : :  x 2 R $ this follows from 1R=R0 = RdR and Nakayama's lemma. Let R := R0Rp]x1  : : :  x ]. By 4.6 this is a local ring and since 1R=R = 0 the algebra R=R is unramied. The residue class eld of R is separable over that of R but also purely inseparable, since Rp  R , hence the two residue class elds are equal. Let m be the maximal ideal of R and m that of R . Then m = m R and moreover R= m R = R = m , consequently R = R + m R . Since R=R is nite Nakayama's lemma implies that R = R and hence fx1 : : :  x g is a p {basis of R=R0 , by 3.6, q.e.d. 1

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

Let R be a ring of prime characteristic p > 0 and let S=R be an algebra. Suppose S and RS p] are noetherian rings. By a theorem of Fogarty (Fo], Prop. 1) 1S=R is a nitely generated S {module if and only if S is a nitely generated RS p]{module (The proposition is stated without the assumption that RS p] be noetherian but the proof covers only the case when this is supposed). In particular, if S is a noetherian ring of prime characteristic p , then 1S= Z is a nitely generated S {module if and only if S is nitely generated as an S p {module. This was proved rst by Brezuleanu-Radu BR3 ], 3.4. Exercises:

p

1) For a squarefree integer a form the quadratic number eld K := Q( a). Let R be the integral closure of Z in K (the ring of integers of K ). p a) If a 6 1 mod 4, then R = Z  Z a , if a  1 mod 4, then R = Z  Z  1 2 a . b) Determine the prime ideals (p) of Z which are ramied in R . p ;

61

2) For a noetherian ring R of prime characteristic p the following conditions are equivalent: a) R is a direct product of nitely many perfect elds of characteristic p . b) 1R=Rp = 0. c) R = Rp . 3) Let S=R be a local algebra, where S is a nite R {module. a) If ( 1S=R ) 1, then S = Rx] with some x 2 S . (Hint: We may assume that R is a eld (Nakayama). Using 4.12 one can construct an x 2 S with 1S=R = Sdx whose image in the residue eld L of S is a primitive element of L=R ). b) If S=R is etale, then there is an R {isomorphism S  = RX ]=(f ), where f is a monic polynomial and f (x) is a unit of S . 4) Under the hypothesis of exercise 3) assume that ( 1S=R ) = r . Then S = Rx1  : : :  xr ] with suitable elements x1  : : :  xr 2 S . Conclude that ( 1S=R )  edim S= mS , where m is the maximal ideal of R . 5) Let S=R be an algebra that is essentially of nite type and unramied. Then the kernel of the map S R S ! S (a b 7! ab) is generated by an idempotent element of S R S . 6) Let R=R0 be an algebra where R0 is a noetherian ring and R a nitely generated R0 {module. Then there is a unique maximal subalgebra R =R0 of R=R0 which is unramied. 7) Let R=R0 and S=R be algebras and I an ideal of S with I 2 = 0. Let : R ! S be the structure homomorphism and " : S ! S=I the canonical epimorphism. a) For each R0 {homomorphism  : R ! S for which S 0

0





(4)

R

S=I



S is commutative the induced map on the algebra of dual numbers (see x 2) ~ : R n I ! S

(r i) 7! (r) + i

is an R0 {homomorphism. b) The R0 {derivations  : R ! I are in natural one-to-one correspondence with the R0 {homomorphisms  : R ! S which make (4) commutative (the "liftings" of "  ). 62

8) An algebra R=R0 is called formally unramied (formally smooth, formally etale) if for any algebra S=R0 and any ideal I of S with I 2 = 0 the canonical mapping alg alg Homalg R0 (R S ) ! HomR0 (R S=I ) is injective (resp. surjective, bijective). Here HomR0 denotes the set of all R0 {algebra homomorphisms. Show that the following conditions are equivalent: a) R=R0 is formally unramied. b) 1R=R0 = 0 (Use exercise 7) and 2.5a). 9) Let K be a eld and (R m ) a local K {algebra with residue eld L . Assume that m 2 = (0) and R has a Cohen subeld which is an extension eld of K . We x one such subeld and denote it by L , so that K  L  R , R = L  m . Let CR=K be the set of all Cohen subelds of R which are extension elds of K . a) There is a natural one-to-one correspondence between CR=K and the set of K { homomorphisms  : L ! R . b) For each such  and a 2 L set d(a) := (a) ; a . Then d 2 DerK (L m ) and there is a natural one-to-one correspondence CR=K ! DerK (L m ). c) R has a unique Cohen subeld which is an extension eld of K if and only if 1L=K = 0. 10)For a eld K of characteristic p > 0 let f1  : : :  fn 2 K X1  : : :  Xn ] be polynomin ) is a unit in K X  : : :  X ]. Show that als whose Jacobian determinant @@((Xf11:::f 1 n :::X n) p p K X1 : : :  Xn] = K f1  : : :  fn  X1  : : :  Xn].

63

x 5.

Dierential Modules of Ane Algebras and their Localizations

The results of the last section will now be applied to ane algebras and their localizations, in particular to the coordinate rings and local rings of ane algebraic varieties. The main goal of this section is to prove dierential criteria for regularity and smoothness.

In the sequel let A always be an ane algebra over a eld K , and set X := Spec A . Let TX=K be the tangent bundle Spec S( 1A=K ). We shall use some facts about the Krull dimension dim A of A and of its localizations which we recall here. If A is a domain with Q(A) =: L , then dim A = Trdeg(L=K )

(1)

In the general case, for p 2 X , we call dimp A := dim Ap + dim A= p = dim Ap + Trdeg(k( p )=K ) the dimension of A (or of X ) at p and we write also dimp X for this number. Clearly dimp A  dim A is the supremum of the length of all prime ideal chains in A of which p is a member. The ring A (or the scheme X ) is called equidimensional, if and only if dimp A is independent of p 2 X . If Ap is a domain with Q(Ap ) = L , then dimp A = dim Ap + Trdeg(k( p )=K ) = Trdeg(L=K )

(2)

If p  p 2 X with p  p are given, and Ap is equidimensional, then 0

(3)

0

dim Ap = dim Ap + dim Ap = p Ap 0

0

A point p of X such that Ap is regular is called a regular point of X , otherwise a singular point or singularity of X . We write (Ap ) = edim Ap ; dim Ap for the regularity defect of Ap =K . We wish to characterize the local rings Ap which are regular in terms of their dierential modules. If k( p)=K is separable, then by 4.5 and (2) we have (4) ( 1Ap =K ) = edim Ap + Trdeg(k( p )=K ) = edim Ap ; dim Ap + dimp A = (Ap ) + dimp A 5.1.Theorem. (Regularity criterion for characteristic zero). If Char K = 0, then for each p 2 X the following conditions are equivalent:

a) Ap is a regular local ring. b) 1Ap =K is a free Ap {module. c) dimk(p) TX=K ( p) = dimp A . d) ( 1Ap =K )  dimp A .

64

Proof: a) ! b). With L := Q(A p ) we have by (2) and (4) above

( 1Ap =K ) = Trdeg(L=K ) = dimL 1L=K = dimL (L Ap 1Ap =K ) Hence 1Ap =K is a free Ap {module of rank dimp A by the following lemma: 5.2.Lemma. Let M be a nitely generated module over an integral domain R with quo-

tient eld L . If (M ) = dimL (L R M ), then M is a free R {module. Indeed: Let fm1 : : :  mr g a system of generators of M with r := (M ). Then the set f1  m1 : : :  1  mr g generates L R M and is a basis of that vector space. Hence fm1  : : :  mr g is linearly independent over R .

b) ! c). We rst show that the ring R := Ap is \generically reduced", that is, Rq is a eld (or equivalently q Rq = 0) for each q 2 Min (R). With M := q Rq we have an exact sequence (4.5) 0 ! M = M 2 ! 1Rq =K = M 1Rq =K ! 1k(q)=K ! 0 For each x 2 M n M 2 it follows from Nakayama's lemma that dx is part of a minimal system of generators of 1Rq =K , hence part of a basis of 1Rq =K = Rq R 1R=K . Suppose that M 6= (0), and choose x 2 M n M 2 . Since M is the only prime ideal of Rq we know that x is nilpotent: x = 0, x 1 6= 0 for some  > 1. Then dx = x 1dx = 0, but x 1 6= 0 since Char K = 0. This contradicts the fact that dx is part of a basis. Hence M = (0), and R is generically reduced. Choose q 2 Min (R) with dim R= q = dim R = dim Ap . Then Rq is a eld, and ;

;

;

( 1Ap =K ) = dimRq ( 1Rq =K ) = Trdeg(Rq =K ) = dim R= q + Trdeg(k( p )=K ) = dim Ap + Trdeg(k( p)=K ) = dimp A In particular dimk(p) TX=K ( p ) = dimk(p) Homk(p)( 1Ap =K = p 1Ap =K  k( p)) = dimp A . c) ! d) is clear by Nakayama's lemma. d) ! a) By (4) the assumption d) implies that edim Ap  dim Ap . The opposite inequality is always true, so Ap is a regular local ring, q.e.d. 5.3.Corollary. a) A is generically reduced (resp. reduced) if and only if 1A q =K is a free Aq {module for each q 2 Min (A) (each q 2 Ass (A)).

b) A is regular if and only if 1A=K is a projective A {module. Theorem 5.1 is not true if Char K 6= 0. A regularity criterion in this case requires the notion of admissible eld which was discussed in x 3 for algebraic function elds and which will be generalized here. 65

Assume now that Char K =: p > 0. 5.4.Definition. A subeld K0  K with K p  K0 , K : K0 ] < 1 is called admissible for p 2 X if

( 1Ap =K0 ) = (Ap ) + dimp A + degp(K=K0 )

(5)

K0 is called admissible for A if it is so for each p 2 X . 5.5.Remarks: a) If A is a domain and p = (0), then K0 is admissible for p if and only

if it is for Q(A)=K in the sense of dention 3.19 (see 3.20a)). b) If K : K p] < 1 , then K0 = K p is admissible for any ane K {algebra A . Indeed, for p 2 X we have by 4.1 and 3.20

( 1Ap =Kp ) = edim Ap + dimk(p) 1k(p)=Kp = edim Ap + Trdeg(k( p )=K ) + degp(K=K p) = edim Ap ; dim Ap + dimp A + degp(K=K p ) = (Ap ) + dimp A + degp(K=K p ) c) If k( p)=K is separable, then K0 = K is admissible for p by (4). Thus the existence of admissible elds is only a problem if inseparable residue elds occur. 5.6.Proposition. For a ne K {algebras A1  : : :  As there is a subeld K0  K which is

admissible for all Ai (i = 1 : : :  s).

Proof: We treat rst the case s = 1, A := A1 . Choose a presentation

A = K X1 : : :  Xn]=(F1  : : :  Fm ) = K x1  : : :  xn ] and a p {basis fyg (6)



2

of K=K p . Then by 2.19

L A  (1  y ) Ln AdX =U

1A=Kp

=



2





where U is generated by

Fj (x1  : : :  xn ) +

Pn @Fj dX

k=1 @xk

k

i=1

(j = 1 : : :  m)

There is a nite subset 0   such that

U

i

L A  (1  y) Ln AdXi



i=1

0

2

66

Set K0 := K pfyg  0 ]. Then fyg 0 is a p {basis of K=K0 . We shall show that K0 is admissible for A . Since we can take for 0 any nite subset of  containing 0 it is clear that an admissible eld exists simultaneously for nitely many ane K {algebras. For p 2 Spec A we have exact sequences (4.1) 2

2

n



0 ! p Ap = p 2Ap ;! 1Ap =Kp = p 1Ap =Kp ;! 1k(p)=Kp ! 0 and



0 ! TKp (k( p )=K ) ;! k( p ) K 1K=Kp ! 1k(p)=Kp ! 1k(p)=K ! 0

By (6)

1A=Kp =

where

E

=



L A  (1  y) E

2

n0

L A  (1  y ) Ln AdXi =U = 1 A=K i=1



0

2

0

Since p Ap = p2 Ap and TKp (k( p )=K ) are nite-dimensional k( p){vector spaces, we may assume after enlarging 0 by nitely many elements, if necessary, that im   Ep = pEp L and im  k( p)  (1  y ). Denote the new set 0 by 0( p ) and replace K0 by  0 K0p := K pfyg 0 (p)]. If we can show that K0p is admissible for p , then so is K0 , since degp(K=K0p ) = degp(K=K0 ) + degp(K0 =K0p ). By the choice of K0p the sequences 2

n

0 ! p Ap = p2 Ap ! 1Ap =K0p = p 1Ap =K0p ! 1k(p)=K0p ! 0 and

0 ! TKp (k( p )=K ) ! k( p) K 1K=K0p ! 1k(p)=K0p ! 1k(p)=K ! 0 are exact. Now we obtain using 3.12b)

( 1Ap =K0p ) = edim Ap + dimk(p) 1k(p)=K0p = edim Ap + dimk(p) 1k(p)=K + dimK 1K=K0p ; dimk(p) TKp (k( p)=K ) = edim Ap + Trdeg(k( p)=K ) + degp(K=K0p ) = edim Ap ; dim Ap + dimp A + degp(K=K0 ) = (Ap ) + dimp A + degp(K=K0p )

q.e.d. 67

5.7.Theorem. (Regularity criterion for characteristic p> 0). Assume that Char K =: p> 0

and that Ap is generically reduced. Let K0  K be an admissible subeld for A . Then the following assertions are equivalent: a) Ap is a regular local ring. b) 1Ap =K0 is a free Ap {module. c) 1Ap = Z is a free Ap {module. d) Ap has a p {basis over (Ap )p . Under these conditions rank 1Ap =K0 = dim Ap + Trdeg(k( p )=K ) + degp(K=K0 ) = dimp A + degp(K=K0 ) Proof: a) $ b) can be shown as in the proof of 5.1. Here the proof is easier since we

already assume Ap to be generically reduced. If Ap is regular with quotient eld L , then

( 1Ap =K0 ) = dim Ap + Trdeg(k( p)=K ) + degp(K=K0 ) = Trdeg(L=K ) + degp(K=K0 ) = dimL 1L=K0 and 1Ap =K0 is a free Ap {module by 5.2. Conversely, if 1Ap =K0 is free, set R := Ap and choose q 2 Min (R) with dim R= q = dim R . Then Rq is a eld and from edim Ap + Trdeg(k( p )=K ) + degp(K=K0 ) = ( 1Ap =K0 ) = dimRq 1Rq =K0 = Trdeg(Rq =K ) + degp(K=K0 ) we obtain edim Ap = Trdeg(Rq =K ) ; Trdeg(k( p )=K ) = dim R = dim Ap a) $ c). The construction of admissible subelds in the proof of 5.6 shows that there is an admissible subeld K0  K for A such that 1A= Z

= 1A=K0 A K0 1K0 = Z Here A K0 1K0 = Z is a free A {module and therefore 1Ap = Z is a free Ap {module if and only if 1Ap =K0 is one. This is equivalent with Ap being regular (a) $ b)). It remains to be shown: If Ap is regular, then Ap has a p {basis over (Ap )p . Choose K0 as above, set L := Q(Ap ), and let fxg  be a p {basis of K0 =K p . Since K0 is linearly disjoint to Lp over K p (by 3.22) fxg  is also a p {basis of K0(Ap )p] over (Ap )p . Since 1Ap =K0 is a free Ap {module there is a p {basis fy1  : : :  yt g of Ap over K0 (Ap )p ] by 4.16. Then fxg   fy1 : : :  yt g is a p {basis of Ap over (Ap )p . 2

2

2

68

5.8.Corollary. The following conditions are equivalent:

a) A is regular. b) If K0  K is an admissible subeld for A , then 1A=K0 is a projective A {module. c) 1A= Z is a projective A {module. As for c) observe that 1A= Z is a direct sum of a nite and a free A {module. In 5.7 and 5.8 the assumption that A be generically reduced cannot be omitted: If K is a perfect eld of characteristic p > 0 and A = K X ]=(X p), then 1A=K = AdX is free but A is not regular. We can give now further applications of the theory of dierential modules. 5.9.Proposition. For p  p 2 X with p  p , let A p be equidimensional. Then 0

0

(Ap )  (Ap ) 0

In particular, if Ap is regular, so is Ap . 0

Proof: If Char K := p > 0 let K0 be an admissible subeld for A , if Char K = 0 let

K0 := K . Then by 4.5 and denition 5.4

( 1Ap =K0 ) = edim Ap + Trdeg(k( p )=K ) + dimK 1K=K0 and

( 1Ap =K0 ) = edim Ap + Trdeg(k( p )=K ) + dimK 1K=K0 Since ( 1Ap =K0 )  ( 1Ap =K0 ) we obtain using (2) and (3) 0

0

0

0

edim Ap ; edim Ap  Trdeg(k( p )=K ) ; Trdeg(k( p)=K ) = dim Ap = p Ap = dim Ap ; dim Ap 0

0

0

0

hence (Ap )  (Ap ). q.e.d. It is known that the inequality of 5.9 holds for any noetherian ring R , of course with a dierent proof (see Le]). For a nitely generated module M over a noetherian ring R of nite Krull dimension let fM := p Max fdim R= p +  p (M )g Spec R 0

2

be the Forster-number of M . By a theorem of Forster For] M can be generated by fM elements. Huneke and Rossi HR] have shown that fM is the Krull dimension of the symmetric algebra S(M ), see B.11. 69

Suppose A is an ane algebra and K0  K a subeld which is admissible for A when Char K = p > 0, else K0 = K . Let X := Spec A . Then by (4) and (5)

f1A=K0 = Max f (A p ) ; dim A p + 2 dimp X g + r p X 2

with r := dimK 1K=K0 . Considering p 2 X with dimp A = dim A we obtain

f1A=K0  2 dim A + r with equality in case A is regular. If A is equidimensional, then

f1A=K0 = 2 dim A + r + Max f (A p ) ; dim A p g p X 2

hence f1A=K0 = 2 dim A + r if and only if (Ap )  dim Ap for all p 2 X , i.e. if the singularities of X have \small" regularity defect. 5.10.Corollary. If A is equidimensional and  (A p )  dim A p for all p 2 X , then 1A=K0 can be generated by 2 dim A + r elements. For a noetherian ring R Reg(R) := f p 2 Spec (R) j Rp is regularg is called the regular locus of R and Sing(R) := Spec (R) n Reg(R) the singular locus (singular set of Spec (R)). The regularity criteria allow the following applications. 5.11.Proposition. (Closedness of the singular locus). For an a ne algebra A over a eld K the set Sing(A) is closed in Spec (A). Proof: If Char K = 0, then Reg(A) is the set of all p 2 Spec (A) for which ( 1A=K ) p is free (5.1). This is well-known to be an open subset of Spec (A). If Char K =: p > 0, the proof is more complicated, since 5.7 is weaker than 5.1. If A is reduced we can argue as above. In the general case assume that Ass (A) = f p1  : : :  p tg . If p 2 Spec (A) contains two dierent elements of Ass (A), then Ap cannot be a domain, nor can it be regular. Assume p contains only one element of Ass (A), say p1 , which is necessarily in Min (A). Then p 1Ap = 0 if and only if p 1Ap 1 = 0. In this case p 2 Reg(A) if and only if p 2 Reg(A= p 1) where p := p = p1 . We see that Sing (A) is the union of V ( p i ) \ V ( p j ) (i 6= j ), of V ( p i) for p i 2 Ass (A) with p iAp i 6= 0, and of 'i(Sing (A= p i)) for p i 2 Ass (A) with pi Ap i = 0 where 'i : Spec (A= p i ) ! Spec (A) is the canonical map. Since Sing(A= p i) is closed in Spec (A= p i) we conclude that Sing (A) is closed too,

q.e.d.

70

5.1 and 5.7 are closely related to the Jacobian criterion for regularity, which we shall derive now. Choose a presentation of A=K (7)

A = K X1 : : :  Xn]=(F1  : : :  Fm ) = K x1  : : :  xn ]

where xi is the residue of Xi in A . Take p 2 Spec (A) and let i denote the image of xi in @Fk @Fk k k( p). Moreover @F @xi is the image of @Xi in A , and @i its image in k ( p ). Set K0 := K , if Char K = 0, and choose K0 to be admissible for A , if Char K > 0. Let  denote the universal derivation of K=K0 , and let 1K=K0 = Kt1    Ktr (t1  : : :  tr 2 K ). From (7) we then get the following presentation of the dierential module 1Ap =K0 = Ap K 1K=K0

(8)

where U is generated by

Fk (x1  : : :  xn) + The rst summand is of the form

Fk (x1  : : :  xn) =

Ln A dX =U p i i=1

Pn @Fk dX (k = 1 : : :  m) @xi i

i=1

Pr 'kj  (1  tj )

j =1

with 'kj 2 Ap

k We denote the image of 'kj in k( p) by @F @tj and call 0 @F1  : : :  @F1  @F1  : : :  @F1 1 @1 @n @t1 @tr B . . . .. C .. .. J ( p ) := @ .. . A @Fm  : : :  @Fm  @Fm  : : :  @Fm @1 @n @t1 @tr

the Jacobian matrix at the point p associated with the presentation (7) (and the k choice of K0 ). Its coecients are elements of k( p). Observe that the @F @tj do not occur if K is a perfect eld. 5.12.Theorem. (Jacobian criterion for regularity). For each p 2 X the following holds: a) rank(J ( p ))  n ; dimp A . b) p 2 Reg(A) if and only if rank(J ( p )) = n ; dimp A . Proof: We have ( 1A p =K0 ) = edim A p +Trdeg(k ( p )=K )+ r , see 5.4. On the other hand,

from (8) we see that

1Ap =K0 = p 1Ap =K0

=

Lr k( p)   Ln k( p)dXi =U j i=1

j =1

71

where j is the image of 1  tj (mod p ) and U is generated by

Pr @Fk  + Pn @Fk dX (k = 1 : : :  m) @tj j @i i i=1

j =1

Therefore we have

( 1Ap =K0 ) = n + r ; rank(J ( p ))

and we obtain rank(J ( p )) = n + r ; (edim Ap + Trdeg(k( p)=K ) + r) = n ; (edim Ap ; dim Ap ) ; (dim Ap + Trdeg(k( p)=K )) = n ; (Ap ) ; dimp A We conclude that in the general case rank(J ( p ))  n ; dimp A and that the equality sign holds if and only if Ap is regular, q.e.d. The Jacobian criterion is particularly useful to decide regularity in explicit examples while 5.1 and 5.7 are used in general considerations. For example they can be applied to study the behavior of regularity under base change. 5.13.Lemma. Let L=K be an arbitrary eld extension and K =K a separable eld ex0

tension. Then K K L is a reduced ring. 0

Proof: We may assume that K =K is nitely generated. Then K =K has a separating 0

0

transcendence basis fx1 : : :  xt g . Since

K K L = K K(x1 ::: xt) (K (x1  : : :  xt ) K L)  K K(x1 ::: xt) L(x1  : : :  xt ) 0

0

0

it suces to consider the case that K =K is a nite separable eld extension: Then K = K X ]=(f ) with a separable polynomial f 2 K X ]. But then K K L

= LX ]=(f ). As a polynomial of LX ] the separable polynomial f has no multiple factors. By the Chinese remainder theorem K K L is therefore the direct product of (separable) extension elds of L , hence reduced. The lemma generalizes as follows. 0

0

0

0

5.14.Proposition. Let A be an a ne algebra over a eld K and K =K a separable eld extension. For p 2 Spec (A) let Ap be regular. Then K K Ap is a regular ring, too. 0

0

Proof: If Char K = 0 we may use 5.1. The module 1K K A p =K = K K 1A p =K is 0

0

0

free over K K Ap because 1Ap =K is a free Ap {module. Therefore K K Ap is regular. Now let Char K =: p > 0. Since K K Ap is a reduced ring by 5.13 we may use 5.7. Let fxg  be a p {basis of K=K p . Since K p and K are linearly disjoint over K p (3.24) 0

0

0

2

0

72

is also part of a p {basis of K =K p . Moreover K K Ap = K Kp Ap =I where I is the ideal generated by fx  1 ; 1  xg  . Let d denote the universal derivation of K = Z and d the universal derivation of Ap = Z . Then it follows easily from 2.16 that

fx g

0



2

0

0

0

0

2

0

1K

K Ap = Z

0

= Ap K 1K = Z K K A1 p = Z =U 0

0

where U is generated by f1  d x ; 1  dxg  . Since the elements d x ( 2 ) form part of a basis of 1K = Z the elements 1  d x ; 1  dx form also part of a basis of the free K K Ap {module Ap K 1K = Z K K 1Ap = Z . Therefore 1K K Ap = Z is a free K K Ap {module, and K K Ap is regular by 5.7. 0

0

2

0

0

0

0

0

0

0

0

If K =K is an inseparable eld extension, then 5.13 is not true in general. For an example, take an imperfect eld K , choose  2 K with  2= K p , and set A := K X ]=(X p ; ). Then A is a eld and therefore regular, but A K A contains the nonzero nilpotent element x  1 ; 1  x where x is the residue class of X in A . 0

Remember that a noetherian local ring R containing a eld K is called geometrically regular over K , if K K R is regular for each nite eld extension K =K . Let A be an ane algebra over a eld K and X := Spec A . We say that X or A is smooth at p 2 X , if the local ring A p is geometrically regular over K . If Char K = 0 or more generally K is a perfect eld, then a regular local ring Ap is always geometrically regular over K by 5.14. For arbitrary ground elds K any localization of a polynomial algebra K X1 : : :  Xn] is geometrically regular over K . If A is an ane algebra over a eld K , and Ap is geometrically regular over K for some p 2 Spec A , then Aq is geometrically regular for each q 2 Spec A with q  p . This follows easily from the regularity criteria. 0

0

5.15.Lemma. Let A be an a ne algebra over a eld K and R := A p for some p 2 Spec (A). Let L=K be an algebraic eld extension. a) For M 2 Max (L K R) let S := (L K R)M . Then dim S = dim R and Trdeg(k( M )=L) = Trdeg(k( m )=K ) where m := p Ap .

b) If R is geometrically regular over K , then L K R is regular.

Proof: a) L K R is integral and faithfully at over R . Therefore M \ R = m and dim S = dim R . Moreover S= MS is algebraic over R= m , hence the statement about

transcendence degrees follows. b) For M 2 Max (LK R) there is a nite eld extension K =K with K  L such that M is already generated by the elements of M := M \(K K R). The ring S := (K K R)M is regular by assumption and we have dim R = dim S = dim S by a). Since M S = M S the ring S is regular as well. 0

0

0

0

0

73

0

0

0

0

5.16.Theorem. (Smoothness criterion for a ne algebras). For an a ne K {algebra A let X := Spec A and p 2 X . Let K be the algebraic closure of K . The following assertions

are equivalent: a) A is smooth at p . b) K K Ap is regular. c) 1Ap =K is a free Ap {module of rank dimp A . d) dimk(p) TX=K ( p ) = dimp A . e) ( 1Ap =K )  dimp A . f) K K Ap is regular for any eld extension K =K . 0

0

Proof:

a) ! b). For M 2 Max (K K Ap ) the local ring S := (K K Ap )M is regular by 5.15b). b) ! c). By 5.1 and 5.7 1S=K = S Ap 1Ap =K is a free S {module of rank dim S + Trdeg(L=K ) where L is the residue eld of S . By 5.15a) dim S = dim Ap and Trdeg(L=K ) = Trdeg(k( p )=K ). Since S=Ap is faithfully at 1Ap =K is a free Ap {module of rank dim Ap + Trdeg(k( p )=K ) = dimp A . The assertions c) ! d) ! e) are clear. e) ! a). Let K =K be a nite eld extension, M 2 Max (K K Ap ), and let S := (K K Ap )M . Assumption e) and 5.15 imply that 0

0

0

( 1S=K )  p ( 1A=K )  dim S + Trdeg(L=K ) 0

0

where L is the residue eld of S . We have S = (K K A)P with P 2 Spec (K K A), P \ A = p , and ( 1(K K A)P =K )  dimP (K K A) 0

0

0

0

0

Therefore it suces to show that condition e) implies regularity of Ap . Write A = K X1 : : :  Xn ]=I and let q be the inverse image of p in P := K X1 : : :  Xn]. The exact sequence Iq =Iq2 ! 1Pq =K =I 1Pq =K ! 1Ap =K ! 0 induces, after tensorization with k( q ) = k( p), the exact sequence of k( p){vector spaces 



Iq = q Iq ;! 1Pq =K = q 1Pq =K ;! 1Ap =K = p 1Ap =K ! 0 By the assumption e) it is clear that ker has at least the dimension n ; dimp A = dim Pq ; dim Ap . 74

Let a1  : : :  a 2 I be elements whose images in 1Pq =K = q 1Pq =K form a basis of ker , and let I := (a1  : : :  a )  Pq . Then in the canonical commutative diagram 

I =qI 

;





1Pq =K = q 1Pq =K



q Pq = q2 Pq

the mapping  induced by  is injective, and therefore the images of a1 : : :  a in q P q = q 2 P q are linearly independent over k ( p ), hence form part of a minimal system of generators of q Pq . It is known that in this case Pq =I is a regular local ring of dimension dim Pq ;  . Moreover Pq =I is a domain, and Ap is a homomorphic image of Pq =I . As was shown above dim Pq =I  dim Ap . This is only possible if Pq =I = Ap , hence Ap is regular. e) ! f) Let P 2 Spec (K K A) be a prime ideal lying over p . We use the wellknown fact that dimp A = dimP (K K A). Since 











0

0

( 1(K

K A) P =K 0 )  ( A p =K )  dimp A = dim P (K 1

0

0

K A)

and since e) $ a) was already proved it follows that (K K A)P is regular. As f) ! b) is obvious the proof is now complete. 0

5.17.Corollary. a) The set of all p 2 X at which X is smooth is open. b) If X is smooth at all p 2 X , then 1A=K can be generated by 2 dim A elements. Proof: a) Similarly as in the proof of 5.11 we can assume that A is a domain, hence equidimensional. By 5.16e) we then have to consider the set of all p 2 Spec A such that ( 1Ap =K )  dim A . It is open in Spec A . (It is also the set of all p for which 1Ap =K is

free of rank dim A ). b) The Forster number f1A=K is 2 dim A .

5.18.Corollary. If A p is regular and k ( p )=K separable, then A p is geometrically re-

gular over K .

Proof: By 4.5 condition 5.16e) is satised. 5.19.Corollary. For an algebraic function eld L=K the following conditions are equi-

valent: a) L is geometrically regular over K . b) For any algebraic eld extension K =K the ring K K L is reduced. c) L=K is separable. 0

0

75

Proof: Write L = A p with an ane domain A over K and p = (0). Then 5.16c) is

equivalent with the statement that dimL 1L=K = Trdeg(L=K ), which is also equivalent with L=K being separable, hence a) $ c). For a nite eld extension K =K the ring K K L is a direct product of zerodimensional local rings. It is regular if and only if these local rings are elds, which is also equivalent with K K L being reduced. a) $ b) follows now easily. 0

0

0

5.20.Corollary. (Jacobian criterion for smoothness). Given p 2 X and a presenta@Fk k tion A = K X1 : : :  Xn ]=(F1 : : :  Fm ) let @F @i denote the image of @Xi in k ( p ) and

 

k J ( p) = @F @i k=1 ::: m . Then the following conditions are equivalent: i=1 ::: n a) X is smooth at p . b) rank(J ( p)) = n ; dimp A .

Starting with 5.16 one has simply to copy the arguments in the proof of 5.12 to obtain this criterion. It indicates that in geometry the notion of geometric regularity is more natural than that of regularity. The following theorem will show that a ring Ap which is geometrically regular over K is \close" to being a localization of a polynomial algebra over K . 5.21.Theorem. (Structure theorem for geometrically regular local rings). Let A be an equidimensional a ne algebra over a eld K with dim A = n . For p 2 Spec A the

following conditions are equivalent: a) Ap is geometrically regular over K . b) There is a noetherian normalization K X1  : : :  Xn ]  A such that Ap =K X1  : : :  Xn ]q ( q := p \ K X1  : : :  Xn ]) is etale. If dim Ap > 0, these conditions are also equivalent with b') There exists a noetherian normalization K X1 : : :  Xn]  A such that the algebra Ap =K X1  : : :  Xn ]q is etale and k( p ) = k( q). Condition b') implies in particular that Ap and K X1 : : :  Xn]q have the same completion and that fX1 : : :  Xng is a di erential basis of Ap =K . Moreover

Ap

= K X1 : : :  Xn ]q Y ](q Y )=(F ) where F 2 K X1  : : :  Xn ]q Y ] is monic in Y , F 2= ( q  Y ), and F (0) 2 q K X1  : : :  Xn ]q . 0

Proof: b) ! a). Since A p is unramied over B := K X1  : : :  Xn ] q we have 1A p =B = 0

by 4.8. From the exact sequence

Ap B 1B=K ! 1Ap =K ! 1Ap =B ! 0 76

we obtain p ( 1A=K )  n = dim A , and 5.15 implies that Ap is geometrically regular over K . We now assume that a) is true. Set d := dim Ap and t := Trdeg(k( p )=K )! then n = d + t . Let A denote the image of A in k( p ) and a the image of a for each a 2 A . Choose a noetherian normalization K Y1 : : :  Yn]  A with p \ K Y1 : : :  Yn] = (Yt +1 : : :  Yn) for some t  n . Then k( p) is nite over K (Y 1  : : :  Y t ), and fY 1 : : :  Y t g is a transcendence basis of k( p )=K , hence t = t . We rst show that after a suitable change of the normalization we can assume fdY 1 : : :  dY tg to be part of a basis of 1k(p)=K . If Char K = 0, this is already the case. So we assume that Char K =: p > 0. Choose z1  : : :  zm 2 A such that fdz1  : : :  dz m g is a basis of 1k(p)=K . Observe that m  t . Let z1r + f1z1r 1 +    + fr = 0 (fi 2 K Y1 : : :  Yn ])e be an equation of integral dependence of z1 over K Y1 : : :  Yn]. Set X1 := z1 ; Y1p with some e > 0 and replace z1 by e X1 + Y1p in the above equation. If e is chosen large enough, we see that Y1 is integral over K X1 Y2 : : :  Yn], hence A is integral over that ring. Moreover dX 1 = dz1 . Similarly we can replace Y2 : : :  Yt in the noetherian normalization of A by elements X2 : : :  Xt with dX i = dzi (i = 2 : : :  t). Then K X1  : : :  Xt  Yt+1 : : :  Yn]  A is a noetherian normalization with p \ K X1  : : :  Xt  Yt+1 : : :  Yn] = (Yt+1  : : :  Yn), and fdX 1  : : :  dX tg is part of a basis of 1k(p)=K . It follows that fdX1  : : :  dXt g is part of a basis of 1Ap =K , and 1k(p)=K(X 1 ::: X t) is generated by n ; t = d elements. If d = 0, then Ap = k( p) is a separable extension eld of K . In fact, in this case 1 k(p)=K(X 1 ::: Xt) = 0, hence k( p) is separably algebraic over K (X 1  : : :  X t). We have shown that condition b) is satised in case d = 0. We now assume d > 0 and try to prove the stronger assertion b'). Since 1k(p)=K(X 1 ::: Xt ) is generated by d elements we know by 3.11 that the eld extension k( p )=K (X 1  : : :  X t) can be generated by d elements. As k( p) is generated over K (X 1 : : :  X t) by A , it is possible to nd elements t1 : : :  td 2 A such that k( p)=K (X 1 : : :  X t ) is generated by t1 : : :  td . In the next step we show that such elements exist so that fdX1 : : :  dXt dt1  : : :  dtd g is a basis of 1Ap =K . Suppose that fdX 1  : : :  dX t  dt1  : : :  dt g is a basis of 1k( p )=K (  d). Looking at the exact sequence 0

0

0

0

0

;

p A p = p 2 A p ! 1A p =K = p 1A p =K ! 1k( p )=K ! 0

we see that elements u +1 : : :  ud 2 p can be found such that the set of dierentials fdX1  : : :  dXt  dt1  : : :  dt  du +1  : : :  dud g is a basis of 1A p =K . If  < d , write

dt +1 =

Pt a dX + P b dt + Pd c du i i i i i i i=1 i=1

i= +1

77

(ai  bi  ci 2 Ap )

We can assume that c +1 is a unit of Ap because, if necessary, we can replace t +1 by t +1 + u +1 without disturbing the properties of t +1 . But then fdX1  : : :  dXt  dt1  : : :  dt +1  du +2  : : :  dudg is also a basis of 1A p =K and, repeating the process, we end up with a basis fdX1  : : :  dXt  dt1 : : :  dtdg of 1Ap =K . We now change the Noetherian normalization K X1  : : :  Xt  Yt+1 : : :  Yn] again so that fdX1  : : :  dXt  dYt+1  : : :  dYn g becomes a basis of 1A p =K and fX 1  : : :  X t  Y t+1  : : :  Y n g a system of generators of k( p )=K . Let tr1 + f1 tr1 1 + : : :  +fr = 0 (fi 2 K X1 : : :  Xt  Yt+1 : : :  Yn]) ;

be an equation of integral dependence of t1 over K X1 : : :  Xt  Yt+1 : : :  Yn]. Set Xt+1 := t1 ; Ytq+1 with a suciently large q > 1 such that Yt+1 (and hence A ) becomes integral over K X1 : : :  Xt+1 Yt+2 : : :  Yn]. Since Yt+1 2 p we have X t+1 = t1 and since dXt+1 = dt1 ; qYtq+11 dYt+1 we see that in the above basis of 1Ap =K we can replace dt1 by dXt+1 . Repeating this process with t2 and Yt+2 etc. we end up with a noetherian normalization K X1 : : :  Xn]  A such that fdX1 : : :  dXng is a basis of 1Ap =K and X 1  : : :  X n a system of generators of k( p)=K . With q := p \ K X1 : : :  Xn ] we have 1Ap =K X1 ::: Xn]q = 0, hence Ap is unramied over K X1 : : :  Xn]q . Clearly k( p) = k( q ) by construction. Since Ap is regular there is only one p 0 2 Min (A) with p 0  p . As A= p 0 is integral over the image of K X1 : : :  Xn] in A= p 0 and dim A= p 0 = n we have p 0 \ K X1  : : :  Xn ] = (0). The composed map K X1 : : :  Xn] ! A= p0 ! (A= p 0 )p = Ap is injective and hence K X1 : : :  Xn ]p ! Aq is injective as well. Moreover both rings have the same dimension. From p Ap = q Ap and k( p) = k( q) it follows that both rings also have the same completion and, in particular, that Ap =K X1 : : :  Xn ]q is at. The last statement of the theorem follows from 4.16. This completes the proof of the theorem. ;

5.22.Corollary. Let A=K be an a ne algebra where A is a domain and L := Q(A) is

separable over K . Then there exists a noetherian normalization K X1 : : :  Xn]  A of A such that L=K (X1  : : :  Xn ) is separably algebraic. Proof: Apply the theorem to the zero ideal of A .

Let us consider now the tangent bundle of ane algebraic K {schemes X = Spec A . If Char K = 0, let K0 := K , if Char K = p > 0, choose a subeld K0  K which is admissible for A , and let ft1 : : :  tr g be a p {basis of K=K0 . Remember (x 2) that if A = K X1 : : :  Xn]=(f1  : : :  fm ), i.e. if X is the closed subscheme of AnK dened by the ideal (f1  : : :  fm ), then the tangent bundle TX=K0 = Spec S( 1A=K0 ) can be described as the closed subscheme of A2Kn+r = Spec K X1 : : :  Xn Y1  : : :  Yn T1  : : :  Tr ] which is given 78

Pn

Pr

@fi  Yk + @fi  T` gi=1 ::: m ) where @fi is obtained by dierentiation by the ideal (ffi  @X @t` `=1 @t` k=1 k of the coecients of fi with respect to t` . Here of course r = 0 if K is a perfect eld. We wish to exhibit some of the properties of that subscheme, and we consider at rst its dimension.

5.23.Theorem. a) We have

dim TX=K0 = f1A=K0 = Max f (A p ) ; dim A p + 2 dimp Ag + r  2 dim X + r p X 2

b) In case X is regular we have dim TX=K0 = 2 dim X + r , and in case X is smooth dim TX=K = 2 dim X . c) When X is equidimensional, then dim TX=K0 =2 dim X + r if and only if (Ap )  dim Ap for all singularities p of X . Proof: a) follows from B.11 and what was said in connection with 5.10 about the Forster

number of the module of dierentials. b) If X is regular, i.e. (Ap ) = 0 for all p 2 X , then by a) dim TX=K0 = 2 dim X + r . In case X is smooth, then 1A=K is a projective A {module with p ( 1A=K ) = dimp X for p 2 X by 5.16, and dim TX=K = 2 dim X follows. c) When X is equidimensional we have (9) dim TX=K0 = 2 dim X + r + Max f (A p ) ; dim A p g p X 2

hence dim TX=K0 = 2 dim X + r if and only if the maximum is zero. Then (Ap )  dim Ap for all singularities p of X . Conversely, if this holds, then the minimal primes of A are regular, and the maximum disappears. 5.24.Corollary. Let X be reduced and equidimensional. Set S := S( 1A=K0 ) and for p 2 X let t( p ) be the kernel of S ! S p = p S p . Then the following conditions are equivalent:

a) TX=K0 is equidimensional. b) (Ap )  dim Ap for all p 2 X , and the equality sign holds in this relation exactly for those p 2 X for which t( p ) 2 Min S . Proof: a) ! b). Applying formula (9) with p 2 Min A yields dim TX=K0 = 2 dim X + r ,

and from 5.23c) we obtain assertion b). b) ! a). From (9) we have dim S=t( p) = 2 dim X + r for all p 2 X with t( p) 2 Min S , hence TX=K0 is equidimensional. The question for which p 2 X the primes t( p ) are minimal in S = S( 1A=K0 ) is not easy to answer. It is related to questions about the torsion of symmetric powers of the module of dierentials. 79

5.25.Example: Let X be reduced and irreducible, and assume X has exactly one singularity m . For p 2 X with p 6= m , since 1Ap =K0 is a free Ap {module, Sp = S( 1Ap =K0 ) is a polynomial algebra over Ap , hence t( p ) = p Sp \ S . Therefore t(0)  t( p) for all p 6= m . Further t( m ) = mS . If (Ap ) ; dim Ap > 0, then TX=K0 is not equidimensional. If (Am ) < dim Am , then dim S=t( m) < dim S=t(0). Equidimensionality of TX=K0 is then equivalent with t(0)  m S , hence with

torsion(Si ( 1A=K0 ))  m Si( 1A=K0 ) for all i 2 N+ For more about the components of symmetric algebras, see HR]. Now let us investigate regularity and smoothness of the tangent bundle. 5.26.Lemma. Let (R m ) be a noetherian local ring, M a nite R {module, S := S(M ) L and M := m S+ where S+ := Si . Then SM is regular if and only if R is regular

and M a free R {module.

i>0

Proof: We have

M=M2

= m = m 2 S+= m S+ + M 2

= m = m 2 M= m M

If SM is regular each minimal system of generators of M forms part of a regular system of parameters in SM . Then R = SM =S+  SM is a regular local ring. Moreover (S+)M =(S+)2M

= M is a free R {module, since (S+)M is generated by an SM {regular sequence. Conversely, if R is regular and M free, then it is clear that SM is regular. 5.27.Theorem. a) TX=K0 is regular if and only if X is.

b) TX=K is smooth if and only if X is. Proof: Let S := S( 1A=K0 ).

a) When TX=K0 is regular, then for each m 2 Max A the ring SM with M := m S+ is regular, and hence Am is regular by 5.26. When X is regular, then 1A=K0 is a projective A {module and the regularity of TX=K0 follows. b) When TX=K is smooth, then X is regular, and 1A=K is projective by 5.26. For m and M as in a) we have dim SM = dim Am + ( 1Am =K ) 80

and by formula (9) of x 2

( 1SM =K ) = 2( 1Am =K ) thus ( 1Am =K ) = dim Am , which implies smoothness of X . The converse statement of b) is clear. Finally we are interested in criteria for the irreducibility of TX=K0 . They follow from considerations of Simis-Vasconcelos SV] about symmetric algebras in general. 5.28.Lemma. Let R be a noetherian domain with dim R = dim R p + dim R= p for all p 2 Spec R , and let M be a nite R {module. For S := S(M ) the following assertions

are equivalent: a) Spec S is irreducible. b) S is equidimensional and for all p 2 Spec R with p 6= (0) we have

(Mp ) ; rank M < dim Rp Proof: Let K be the quotient eld of R .

a) ! b). If Spec S is irreducible, then t(0) := ker (M ! K R M ) is the only minimal prime ideal of S . The formula dim S=t( p) = dim R= p + (Mp )

( p 2 Spec R)

(appendix B, formula (7)) implies that for p 6= (0) dim S=t( p ) ; dim S = dim R= p ; dim R + (Mp ) ; rank M < 0 hence

(Mp ) ; rank M < dim Rp b) ! a). Assuming this condition for p 6= (0) we obtain dim S=t( p) < dim S . By equidimensionality t(0) is then the only minimal prime ideal of S , hence Spec S is irreducible. When we apply the lemma to S := S( 1A=K0 ) with A and K0 as before we obtain: 5.29.Theorem. Let X be reduced and irreducible. The following are equivalent:

a) TX=K0 is irreducible. b) TX=K0 is equidimensional and (Ap ) < dim Ap for all singularities p of X . Moreover TX=K0 is reduced and irreducible if and only if Si ( 1A=K0 ) is a torsion free A {module for all i 2 N+ . 81

Proof: In the present situation ( 1A p =K0 ) ; rank 1A=K0 = edim A p ; dim A p =  (A p ) for p 2 X . If p 6= (0) is regular, then clearly (Ap ) < dim Ap . Therefore the condition

p (X ) < dim Ap has to be required only for singularities of X . The last assertion of the theorem follows from B.12b).

5.30.Corollary. If dim X = 1 the following assertions are equivalent:

a) TX=K0 is irreducible. b) X is regular. c) Si( 1A=K0 ) is torsion free for i 2 N+ . d) TX=K0 is reduced and irreducible.

Proof: a) ! b). Since dim X = 1 the condition  (A p ) < dim A p can only be satised for regular p 2 X .

b) ! c) is clear since 1A=K0 is a projective A {module. c) ! d). Condition c) implies that t(0) = (0) is the only minimal prime of S( 1A=K0 ).

5.31.Remark: If K is a perfect eld Berger' problem B4 ] asks whether for 5.30c) the

weaker condition that 1A=K be torsion free implies regularity. For a survey about positive answers to this question, see the report B9 ]. Exercises:

1) The ane R {algebra A := R X Y Z ]=(X 2 + Y 2 + Z 2 ; 1) is regular. Use a result from analysis to show that 1A= R and Der R (A) are not free A {modules (see Sa], Prop.10 for the case of an arbitrary ground eld instead of R ). 2) Let K be a perfect eld of characteristic p > 0 and A := K X Y Z ]=(Z p + XY ). a) Determine Sing(A). b) Show that DerK (A) is a free A {module. 3) (R.Waldi) Let K be a eld with Char K 6= 2. Then A := K X Y ]=(X 2 + Y 2 ; 1) is a one-dimensional regular ane domain, and 1A=K is a free A {module generated by ydx ; xdy where x and y are the images of X Y in A . Show that 1A=K is generated by an exact dierential da (a 2 A) if and only if ;1 is a square in K . 4) Let R=R0 be an algebra and suppose R is local and 1R=R0 is nitely generated. Show that the following assertions are equivalent: a) The module of Zariski dierentials DerR0 (R) is a free R {module. b) DerR0 (R) is a free R {module. 

82

5) Let A be an equidimensional ane algebra over a eld K . Then for all i 2 N Sing i (A) := f p 2 Spec Aj(Ap ) > ig is a closed subset of Spec A . 6) Let A be an ane K {algebra and g := DerK A . a) A prime p 2 Spec A for which Ap is regular and k( p )=K is separable is g {invariant if and only if p 2 Min A . b) If Char K = p > 0 and Ap is geometrically regular over K , then p is g {invariant if and only if p Ap = q Ap where q := p \ K Ap]. c) If Char K = 0 and A is regular, then the radicals of g {invariant ideals of A are the intersections of minimal prime ideals of A . 7) Under the assumptions of 5.23 let A be reduced, and let S := S( 1A=K ). Then for p 2 Spec A the A p {module S p is at if and only if p 2 Reg A . 8) Let K be a eld, P = K X1 : : :  Xn] a polynomial algebra and I  P an ideal with radI = I . Let I (2) denote the second symbolic power of I , i.e. the preimage of I 2PS in P where S is the complement of the union of all minimal prime divisors of I . Set A := P=I . a) I (2)=I 2 is the torsion of the conormal module I=I 2 as an A {module. b) Assume that for each minimal prime divisor p of I the residue eld k( p) is separable over K . Then there is a natural exact sequence

O ! I=I (2) ! 1P=K =I 1P=K ! 1A=K ! 0 9) Let K be a perfect eld and A  B integral ane K {algebras. Assume A is regular. Then the following conditions are equivalent: a) Any  2 DerK A can be extended to a derivation  2 DerK B . b) Q(B)=Q(A) is a separable eld extension, and the image of B A 1A=K in 1B=K is a direct summand.

83

x 6. Smooth Algebras The goal of this section is to generalize the dierential criteria for smoothness of ane algebras to the case of algebras over a noetherian base ring. Moreover the structure theorems for etale algebras and geometrically regular ane algebras will be generalized.

We shall consider algebras S=R which are essentially of nite type, where R is a noetherian ring. Let X := Spec R , Y := Spec S and ' : Y ! X the map induced by the structure homomorphism R ! S .

2

6.1.Definition. S=R is called smooth at P Y if the following conditions are satised: a) SP =Rp is at ( p := P R ). b) SP = p SP is geometrically regular over k( p). The algebra S=R (or the morphism ' : Y X ) is called smooth if it is so at all P Y .

\

!

2

For example, each polynomal algebra S = R X1  : : :  Xn ] is smooth over R . If the algebra S=R is etale at P 2 Spec S , then it is also smooth at P by 5.19, the ring SP = pSP being a nite separable extension eld of k( p ). If R = K is a eld, smoothness of S at P is equivalent with SP being geometrically regular over K . In particular, if P 2 Min S , then S=K is smooth at P if and only if S P is a separable algebraic function eld over K . In the general case SP = pSP is the local ring of the ber ';1( p ) at P , a localization of an ane k( p ){algebra. Hence the dierential criteria for geometric regularity of x 5 can be applied. Consider a presentation SP = PN =IN where P := R X1  : : :  Xn] is a polynomial algebra, I  P an ideal and N 2 Spec P . Then (1)

SP = pSP = (PN = p PN )=J

where J is the image of IN in PN = pPN . Using the generalized principal ideal theorem of Krull we obtain (2)

(IN )  (J )  h(J ) = dim PN = pPN ; dim SP = pSP

In case R  S are integral domains with quotient elds K and L respectively one has the dimension formula B.1 (3)

dim SP + Trdeg(k( P )=k( p ))  dim Rp + Trdeg(L=K )

where the equality sign holds if Rp is universally catenary or S is a polynomial algebra over R . The formulas (2) and (3) play an essential role in what follows. 84

6.2.Theorem. (Smoothness criterion). With the notations of (1) let S=R be of nite

type. Then the following conditions are equivalent: a) S=R is smooth at P . b) SP =Rp is at and (1SP =R )  dimP (Sp = pSp ). b') SP =Rp is at and dimk(P) TY=X ( P )  dimP (Sp = pSp ). c) SP =Rp is at and 1SP =R is a free SP {module of rank dimP (Sp = p Sp ). d) The canonical sequence of SP {modules 0 ! IN =IN2 ! 1PN =R =I 1PN =R ! 1SP =R ! 0

is split-exact. If these conditions are satised, then IN is generated by a PN {regular sequence of length dim PN = pPN ; dim SP = pSP = n ; dimP Sp = pSp . If Char k( p ) = 0, then the conditions a)-d) are also equivalent with e) SP =Rp is at and 1SP =R is a free SP {module. Proof: We have

1SP =p SP =k(p ) = 1SP =R = p1SP =R Therefore by Nakayama's lemma (4)

(1SP =R ) = (1SP =p SP =k(p ))

Oberserve that Sp = pSp is an ane k( p){algebra. By theorem 5.16 the number (4) is  dimP (Sp =pSp ) if and only if SP = pSP is geometrically regular over k(p). Thus a) $ b) is proved. Clearly b') is equivalent with b). If SP =Rp is at, then the exact sequence 0 ! IN ! PN ! SP ! 0 induces by tensorization with k( p) over Rp an exact sequence 0 ! IN = p IN ! PN = pPN ! SP = pSP ! 0 hence J = IN = pIN . If we assume b), then SP = p SP is regular, as was shown above. Therefore IN = p IN is generated by elements that form part of a regular system of parameters of the regular local ring PN = pPN . This sequence has length

 := (IN = pIN ) = dim PN = p PN ; dim SP = pSP We shall apply now the following fact: 85

!

6.3.Lemma. Let R and P be noetherian local rings and R P a at local homomorphism. For an ideal a of R set R := R= a and P := P= aP . Given a sequence

t = (t1 : : :  th ) of elements ti 2 P let ti be the image of ti in P (i = 1 : : :  h) and t := (t1 : : :  th). Then the following conditions are equivalent: a) t is a P {regular sequence and P=(t) is a at R {module. b) t is a P {regular sequence and P=(t) is a at R {module. The lemma is an easy consequence of the local criterion of atness ( M3 ], 20.F). By 6.3 the ideal IN is generated by a PN {regular sequence of length  , hence IN =IN2 is a free SP {module of rank  . In the exact sequence IN =IN2 ! 1PN =R =I 1PN =R ! 1SP =R ! 0 the SP {module 1PN =R =I 1PN =R is free of rank n and by assumption b)

(1SP =R )  dimP (Sp = p Sp ) = dim SP = pSP + Trdeg(k( P )=k( p )) = n ; 

Let 1  : : :   be a basis of IN =IN2 and f!1 : : :  ! g a minimal set of generators of 1SP =R , so = P (1S=R ). The images of 1 : : :   together with a system of representatives of !1 : : :  ! in 1PN =R =I 1PN =R generate this module. Since  +  n these elements must then even be a basis. It follows that = n ;  and that f!1 : : :  ! g is a basis of 1SP =R . Moreover the sequence 0 ! IN =IN2 ! 1PN =R =I 1PN =R ! 1SP =R ! 0 is split exact. Thus we see that b) implies both c) and d). Since c) ! b) is trivial, it remains to be shown that d) ! b). If d) holds, then both IN =IN2 and 1SP =R are free SP {modules. Using (2) we see that (5)

(1SP =R )  n ; dim PN = p PN + dim SP = p SP = dim SP = pSP + Trdeg(k( P )=k( p )) = dimP (Sp = pSp )

and from (4) and 5.15 we conclude that SP = p SP is geometrically regular over k( p) and that the equality sign holds in (5). Then IN is generated by dim PN = p PN ; dim SP = pSP elements. Their images in PN = pPN form part of a regular system of parameters and hence a regular sequence. By the lemma 6.3 we conclude that SP =Rp ist at, so c) is proved. If Char k( p) = 0 and e) holds, then 1SP =p SP =k(p ) is a free SP = pSP {module. By 5.1 the local ring SP = p SP is regular. Since we are in characteristic 0 it is geometrically regular over k( p). Thus we have shown that e) implies a), q.e.d. 86

6.4.Corollary. a) If S=R is smooth at P , then the algebra S P =R p has a di erential basis a1  : : :  at with t = dimP Sp = pSp . b) The set of all P Spec S at which S=R is smooth is open in Spec S .

f

g

2

Proof: a) follows from 6.2c) and Nakayama's lemma.

b) Consider a presentation S = P=I where P is a polynomial algebra over R and I  P an ideal. By 6.2 the algebra S=R is smooth at P if and only if the canonical sequence 0 ! I=I 2 ! 1P=R =I 1P=R ! 1S=R ! 0

becomes split-exact after localization at P . It is known that the set of these prime ideals is open in Spec S .

2

6.5.Transitivity. Let T=S be another algebra of nite type. For Q Spec T let P := Q S and p := Q R . Assume that T=S is smooth at Q . Then T=R is smooth at Q if and only if S=R is at P . In particular if T=S is smooth, then T=R is smooth if

\

\

and only if S=R is.

Proof: Since T Q =S P is at we have that T Q =R p is at if and only if S P =R p is.

Moreover by the dimension formula for at extensions ( M3 ], Thm.19) dim TQ = p TQ = dim SP = pSP + dim TQ = P TQ

The additivity of the transcendence degree of the residue elds implies dimQ Tp = p Tp = dimP Sp = p Sp + dimQ TP = P TP

(6)

Write T = P=I with P = S X1 : : :  Xn ] and an ideal I  P . Let N be the preimage of Q in P . Then TQ = PN =IN , and according to 6.2 the ideal IN is generated by a PN {regular sequence ff1 : : :  fm g with fi 2 P (i = 1 : : :  m) where

m = n ; dimQ TP = P TP

With analogous notations as in 2.19 we have TQ =SP 1

where U = hf

Pn @x@f dXkgi

k=1

i k

=1:::m

n M = TQ dXk =U k=1

i and

TQ =Rp = TQ SP SP =Rp 1

1

87

n M  TQ dXk=V k=1

where V = hffi +

fP \

Pn @x@f dXk gi i k

=1:::m

i . Here TQ =SP 1

L

is free of rank dimQ TP = P TP ,

k=1 n n @f i dX forms part of a basis of TQ dXk . But this implies that hence k i =1 :::m @x k=1 k=1 k V TQ SP 1SP =Rp = 0 . It follows that 1SP =Rp is a free SP {module if and only if 1TQ =Rp is a free TQ {module. Then

g

hi

rank 1TQ =Rp = rank 1SP =Rp + dimQ TP = PTP and (6) implies that 1TQ =Rp has rank dimQ Tp = pTp if and only if 1SP =Rp has rank dimP Sp = pSp . We conclude from 6.2 that TQ =Rp is smooth if and only if SP =Rp is. are algebraically independent over R  S and the algebra S=R X1  : : :  Xn ] is nitely generated and etale, then S=R is smooth.

6.6.Example: If X1  : : :  Xn

2S

6.7.Base change. Let R0 =R be an algebra where R0 is a noetherian ring. Assume S=R is of nite type and set S 0 := R0 R S . If S=R is smooth at P Spec S , then S 0=R0 is smooth at any Q Spec S 0 lying over P . In particular, if S=R is smooth, so is S 0=R0 . 0 is a localization of Proof: Set q := Q R0 and p := P R = Q R . Since S Q R0q Rp SP and since atness is compatible with base change, we conclude that S 0 =R0q

2





\

2

\

is at. Moreover

\

Q

Sq0 = q Sq0 = k( q ) k(p) Sp = pSp

hence

dimQ (Sq0 = q Sq0 ) = dimP (Sp = pSp ) Since 1SQ =R = SQ0 SP 1SP =R (2.17), we have (1SQ =R )  (1SP =R ) = dimP (Sp = pSp ) = dimQ (Sq0 = q Sq0 ) 0

0

0

0

and 6.2 implies that S 0 =R0 is smooth at Q ,

q.e.d.

6.8.Tensor product. Let S=R and T=R be algebras of nite type where S=R is smooth.

Then S R T=R is smooth if and only if T=R is.



Proof: By base change S R T is smooth over T . Now apply 6.5.

Remember (B.21) that an algebra S=R of nite type is equidimensional of dimension d if each minimal prime of S contract to a minimal prime of R and dimP Sp = p Sp = d for all P 2 Spec S  p := P \ R The following smoothness criterion is a direct consequence of 6.2. 88

6.9.Corollary. Let S=R be of nite type and equidimensional of dimension d . Then

the following conditions are equivalent: a) S=R is smooth. b) S=R is at and 1S=R is a projective S {module of rank d . If we drop the assumption of equidimensionality but assume that Q  R , then a) is also equivalent with b') S=R is at and 1S=R is a projective S {module. Let S=R be an algebra of nite type, I  S an ideal and S := S=I . For P 2 Spec S let N 2 Spec S be its preimage. The smoothness criterion 6.2 generalizes as follows. 6.10.Theorem. Assume S N =R is smooth. Then the following statements are equivalent:

a) S P =R is smooth. b) The canonical sequence of S P {modules

0 ! IN =IN2 ! 1SN =R =I 1SN =R ! 1S P =R ! 0

is split-exact. c) The canonical sequence

0 ! DerRS P ! DerR (SN  S P ) ! HomS P (IN =IN2  S P ) ! 0

is split-exact. Moreover, if a)-c) are satised, then IN is generated by a regular sequence of length dim SN = p SN ; dim S P = pS P where p := P \ R = N \ R . Proof: Choose a presentation S = P=J where P = R X1  : : :  Xn ] is a polynomial

algebra over R and J an ideal of P . Let I be the preimage of I in P , hence I = I=J and S = P=I . Denote by Q the preimage of N in P . The canonical sequence 0 ! J=J \ I 2 ! I=I 2 ! I=I 2 ! 0

is exact and the canonical diagram

J=J \ I 2 ;

I=I 2 is commutative, hence we have

S S J=J 2 J=J \ I 2 = S S J=J 2 89

a) $ b) Consider the commutative diagram with exact rows and columns

  

0

0

K1

K2

0

JQ =JQ \ I 2Q

I Q =I 2Q

IN =IN2

0

0

S P SN JQ =JQ2

1PQ =R =I 1PQ =R

1SN =R =I 1SN =R

0

1S P =R

S1 P =R

0

0

where the (split-)exactness of the middle row follows from 6.2. By the snake lemma K1 = K2 . Therefore the equivalence of a) and b) follows from 6.2 b) ! c) Dualize the sequence in b) and observe that HomS P (1SN =R =I 1SN =R  S P ) = HomSN (1SN =R  S P ) = DerR(SN  S P )

c) ! b) Since DerR (SN  S P ) is nitely generated and free so are the S P {modules DerRS P and HomS P (IN =IN2  S P ). We recover the sequence in b) by dualizing the sequence in c). The last statement of the theorem follows from lemma 6.3. The structure theorems 4.19 and 5.21 can be generalized as follows. 6.11.Theorem. Let S=R be an algebra of nite type where R is noetherian. For P Spec S , p := P R the following assertions are equivalent: a) S=R is smooth at P . b) There exists a quasinormalization R T1 : : :  Td] Sf in the neighbourhood of P

2

\

!

(B.22) such that Sf =R T1 : : :  Td] is etale. c) The ring SP has a presentation of the form SP = R T1 : : :  Td Y ]N =(F ) where @F 2= N . N 2 Spec R T1  : : :  Td  Y ] and F 2 R T1  : : :  Td  Y ] is monic in Y with @Y 90

!

Proof: a) b) Choose a presentation S = R X1  : : :  Xn ]=I and set P := R X1  : : :  Xn ].

By 6.2 the canonical sequence

0 ! I=I 2 ! 1P=R =I 1P=R ! 1S=R ! 0 becomes split-exact after localization at P , and it is so after localization at a suitable f 2 S n P . Then Sf S I=I 2 is a projective Sf {module, and we can choose f so that Sf S I=I 2 is projective of rank r := rank SP S I=I 2 = n ; rank1SP =R = n ; dimPSf (Sf )p = p(Sf )p . It follows then that dimQ (Sf )p = p (Sf )q = n ; r =: d for each Q 2 Spec Sf , q := Q \ R , i.e. Sf =R is equidimensional of dimension d at P (B.21). For simplicity let us write S for Sf and P for P Sf . We have seen that Sp = p Sp is an equidimensional ane k( p){algebra of dimension d , and it is geometrically regular over k( p) by the smoothness assumption. By the structure theorem 5.21 there exists a noetherian normalization k( p ) 1 : : :  d]  Sp = pSp such that Sp =pSp is etale over k( p) 1 : : :  d] at P Sp = pSp . We may assume that 1 : : :  d are the images of T1  : : :  Td 2 S by the canonical map S ! Sp = p Sp . Then by B.24 the noetherian normalization can be lifted, i.e. there is a g 2 S n P such that R T1  : : :  Td ] ! Sg denes a quasinormalization of S=R in the neighbourhood of P . By assumption Sg =R is smooth at P Sg and hence also at. As S p = p S p is at over k( p) 1  : : :  d] at P Sp = p Sp it follows from B.33 that Sg =R T1  : : :  Td ] is at at P Sg . Since S p = p S p is unramied over k ( p ) 1  : : :  d ] at P Sf = p Sf so is Sg =R T1  : : :  Td ] at P Sg . The etale locus of an algebra of nite type is open, so we may assume that Sg =R T1 : : :  Td] is etale. Thus we have proved that a) ! b). b) ! c). This follows from the structure theorem for etale extension 4.19. @F 2= N we see c) ! a). Clearly SP =R is at as F is a monic polynomial in Y . Since @Y that 1SP =R is a free SP {module of rank d = dimP Sp = pSp . Therefore S=R is smooth at P by 6.2. In terms of the envelloping algebra S e := S R S of S=R one can give the following smoothness criterion.

2 Spec S let Q be thee preimage of P in Se and p := P \ R . be the kernel of the canonical map S ! S (a b 7! ab). The following

6.12.Proposition. For P

Let IS=R conditions are equivalent: a) S=R is smooth at P . b) SP =Rp is at, and there are elements a1 : : :  an 2 S such that the set of tensors f1 a1 ; a1 1 : : :  1 an ; an 1g forms an (Se)Q {regular sequence which generates (IS=R )Q . 91

!

7!

Proof: Consider (S e ) Q as an S P {module by S S e (a a 1). a) b) When S=R is smooth at P , then S e=S is smooth at Q by 6.7. It follows that S e=R is smooth at Q by transitivity. By 6.11 the kernel (IS=R )Q of the canonical epimorphism

!

(S e)Q ! SP is generated by a regular sequence. Since it is also generated by elements 1 a ; a 1 with a 2 S it follows from Nakayama's lemma that there is a regular sequence f1 a1 ; a1 1 : : :  1 an ; an 1g which generates (IS=R)Q . b) ! a) If b) is satised, then 1SP =Rp = ISP =Rp =IS2P =Rp = (IS=R )Q =(IS=R )2Q is a free SP {module of rank n . Since (S e)Q =(IS=R )Q = SP is at over Rp the kernel of the canonical map (S e)Q = p(S e)Q ! SP = p SP is generated by a regular sequence of length n , hence n = dim(S e)Q = p (S e)Q ; dim SP = pSP

But (S e )Q = p (S e)Q = (SP = pSP k(p) SP = pSP )Q with the prime ideal Q corresponding to Q . Using the dimension formula for at local homomorphisms ( M3 ],Thm.19) we conclude that dim(S e )Q = p (S e)Q = dim SP = pSP + dim(k( P ) k(p) SP = p SP )Q  dim SP =pSP + dimP (Sp =pSp )

hence n  dimP (Sp = pSp ). Therefore 6.2 implies that SP =Rp is smooth,

f

q.e.d.

g

6.13.Remark: In the situation of 6.12b) the proof shows that a1  : : :  an is a dierential basis of SP =Rp with n = dimP (Sp = pSp ). Conversely, if SP =Rp is at and a1  : : :  an a dierential basis of SP =Rp with n = dimP (Sp = pSp ), then the set of ten-

f

g

sors fa1 1 ; 1 a1  : : :  an 1 ; 1 an g is an SQe {regular sequence which generates (IS=R )Q . If S=R is a smooth algebra of nite type, then since IS=R is locally generated by a regular sequence grIS=R (S e) is locally at each P 2 Spec S the symmetric algebra of the free SP {module 1SP =R , hence we have grIS=R (S e) = S(1S=R )

Under the assumptions of 6.1 let TX=Y be the tangent bundle of X=Y . 6.14.Proposition. Let S=R be of nite type. Then the following conditions are equiva-

lent: a) X=Y is smooth. b) TX=Y =Y is smooth. Moreover if a) or b) holds and X=Y is equidimensional of dimension d (e.i. S=R is so), then TX=Y =Y is equidimensional of dimension 2d . 92

!

Proof: a) b) If X=Y is smooth, then 1S=R is a projective S {module. The localization of the symmetric algebra  := S(1S=R ) at P X is a polynomial algebra over SP . If is

2

therefore clear that TX=Y =Y is smooth. Moreover if X=Y is equidimensional of dimension d , then 1S=R is projective of rank d by 6.9, and it follows that TX=Y =Y is equidimensional of dimension 2d . b) ! a) If TX=Y =Y is smooth, then lemma 5.26 implies that for each P 2 X the local ring SP is regular and 1SP =R is a free SP {module. Again P is a polynomial algebra over SP . By 6.5 the smoothness of TX=Y =Y implies the smoothness of X=Y . Exercises:

1) Under the assumptions of 6.1 let S=R be at and Spec (S ) ! Spec (R) a closed surjective map. Then the set of all p 2 Spec (R) for which Sp = p Sp is smooth over k( p ) is open in Spec (R). 2) Let S=R be a at algebra of nite type, where (R m ) is a complete noetherian local ring. Suppose there are only nitely many prime ideals of S= mS at which S= mS is not smooth over R= m (\the special ber S= mS has only isolated singularities"). Then the set of all p 2 Spec (R) for which Sp = p Sp is smooth over k( p ) is open in Spec (R). (Hint: With an ideal I of S let V (I ) be the set of all P 2 Spec (S ) at which S=R is not smooth. Then S= mS + I is a nite dimensional vector space over R= m . Conclude that S=I is nite over R . Consider the image of V (I ) in Spec (R)). 3) Let S=R be a smooth algebra with Q  R , and set g := DerR S . The g {invariant prime ideals of S are the minimal prime divisors of the extension ideals p S with p 2 Spec R (Hint: x 5, exercise 6).

93

x 7. Dierential Modules of Complete Intersections The relevant facts about complete intersections are contained in appendix C. Here we shall study their dierentials. Complete intersection algebras can sometimes be characterized in terms of the projective dimension of their dierential module. Clearly smooth algebras of nite type are locally complete intersections.

If R is a ring and M is an R {module, pdR (M ) will always denote the projective dimension of M . Suppose S=R is an algebra which is essentially of nite type. Choose a presentation S = R X1  : : :  Xn]N =I with a multiplicatively closed set N of R X1 : : :  Xn]. Write P := R X1 : : :  Xn ]N . 7.1.Lemma. If pdR (S )
p  t each such product contains a factor (1  ai ; ai  1) = 1  api ; api  1 = 0, m hence I m+1 = (0) and R R0 Rpe ] M = PR=R e (M ). 0 Rp ] m Since (` dR=R0 Rpe ] )(x) = `(1  x) = D(x) for each x 2 M we obtain that

D 2 Di mR=R0 Rpe ] (M N ) Di mR=R0 (M N ) 136

q.e.d.

S 2.10.Corollary. Di 1 R=R0 (M N ) = e2N HomR0 Rpe ] (M N ). e Proof: By 1.8c) Di nR=R0 (M N ) consists of R0 Rp ]{linear maps if pe > n . Conversely

HomR0 Rpe ] (M N ) Di m R=R0 (M N ) for suciently large m by 2.9.

For example, if K is a eld of characteristic p > 0 and A a nite-dimensional local K {algebra, then by Cohen's structure theorem there is a unique extension eld L of K in A which is mapped onto the separable hull of K in the residue eld of A. We then have e p K A ] = L for suciently large e 2 N , hence Di K1 (A) = EndL (A). 2.11.Remark: Let R=R0 be an algebra, M an R {module. There is a canonical exact

sequence of R {modules

n ! P n;1 ! 0 0 ! I n=I n+1 ! PR=R R=R0 0 0

which splits in case n = 1, since PR=R = R . Moreover there are canonical exact sequences 0

n (M ) ! P n;1 (M ) ! 0 (n  2) I n=I n+1 R M ! PR=R R=R0 0

and

1 0 ! 1R=R0 R M ! PR=R (M ) ! M ! 0 0

n . Then the associated graded ring is Let J be the image of I in PR=R 0 n = R  I=I 2      I n =I n+1 grJ PR=R 0

2.12.Definition. We say that R=R0

any R {modules M and N

has only trivial dierential operators if for

0 Di 1 R=R0 (M N ) = Di R=R0 (M N ) = HomR (M N )

2.13.Proposition. The following assertions are equivalent:

a) R=R0 has only trivial dierential operators. n = R for all n 2 N . b) PR=R 0 n +1 b') I = I for alle n 2 N . c) 1R=R0 = 0. If R=R0 is essentially of nite type these conditions are also equivalent with d) R=R0 is unramied. 137

n (M )

n R M by 2.3b) and since by the universal property Proof: Since PR=R = PR=R 0 0 n R M N ) for all n 2 N it is clear that a),b) and b') are Di nR=R0 (M N )

= HomR (PR=R 0 equivalent. Moreover 1R=R0 = I=I 2 = 0 if and only if b') holds. The equivalence of c) and d) in case R=R0 is essentially of nite type was shown in I.4.8.

2.14.Corollary. Separable algebraic eld extensions have only trivial dierential ope-

rators. An algebraic function eld L=K has only trivial dierential operators if and only if L=K is separably algebraic. The dierential operators of algebraic function elds L=K with Char K = p > 0 are given by the formula 2.10. Separable algebraic function elds are smooth algebras whose dierential operators will be studied in x 4.

2.15.Proposition. Let R=R0 be an algebra and n

generated R {module if and only if 1R=R0 is so.

n 2 N+ . Then PR=R is a nitely 0

n is nitely generated over R . Then P 1 1 Proof: Assume n  2 and PR=R R=R0 = RR=R0 , 0

n , is nitely generated too, and so is 1 . being a homomorphic image of PR=R R=R0 0 Conversely, if 1R=R0 = I=I 2 is a nitely generated R {module, so is I=I n+1 , as follows n

by an easy induction, hence PR=R = R  I=I n+1 is a nitely generated R {module. 0 Suppose now that R is a local ring with maximal ideal m , and R is essentially of nite type over R0 . We are interested in the minimal number of generators of the R {module n , i.e. in the number PR=R 0 n ) = dimR= m P n = m P n R=R0 (n) := (PR=R R=R0 R=R0 0 n When PR=R is a projective R {module this number is also the minimal number of gene0 rators of the R {module Di nR0 (R). We call

R=R0 (X ) :=

1 X

n=0

R=R0 (n)  X n 2 ZX ]

the dierential Hilbert series of R=R0 . In order to compute this series we may assume that R0 is a local ring too, whose maximal ideal is m 0 := m \ R . Then with R := R= m 0R , m := m = m 0 R we have (1)

n = m P n = P n = mP n PR=R R=R0 0 R=K R=K

where K := R0= m 0 . Therefore it suces to consider a local algebra (R m ) which is essentially of nite type over a eld K . Let L := R= m . Then (2)

n+1 n = mP n

PR=K R=K = (L K R) M = M (L K R) M

138



where M is the kernel of L K R ! L K L ;! L , i.e. the image of IR=K in L K R . Set RL := (L K R)M and m L := M RL . 1 P Then R=K (X ) = dimL (RL = mnL+1)  X n is the ordinary Hilbert series (of order 1) n=0 of the local ring RL . Assume now that L=K is a separable eld extension of transcendence degree t . The local homomorphism R ! RL (r 7! 1  r) is at, and its ber

RL = mRL

= (L K R)M =(1  m )(L K R)M

= (L K L)M (where M is the image of M in L K L ) is a regular local ring of dimension t , since L=K was separable of transcendence degree t . It follows that the associated graded ring of RL is isomorphic to the polynomial ring (grm R)X1  : : :  Xt ], from which the statement 1 (0) R=K (X ) = (1 ; X  H (X ) R t ; 1 ) 1 P

with HR(0)(X ) := dimK ( m n= m n+1)  X n easily follows. Putting everything together we n=0 have shown: 2.16.Proposition. Let R=R0 be a local algebra which is essentially of nite type. Let L be the residue eld of R and K the residue eld of R0 , and assume that L=K is separable of transcendence degree t . Set R := R= m 0 R , m := m = m 0R where m resp. m 0 are the maximal ideals of R resp. R0 . Then 1 X 1 1 (0) R=R0 (X ) = (1 ; X )t+1  HR (X ) = (1 ; X )t dimK ( m n= m n+1)  X n n=0

If L=K is not separable the situation is more complicated, see Ku18 ] for details in this case. The invariants introduced by H. Kraft KrH] for inseparable algebraic function elds play then an essential role. Exercises: 1) Let (R m ) be a local noetherian algebra over a eld K such that m  = (0). Show

that Di nK (R) = EndK (R) for all n   ; 1. 2) Let R = R0X1 : : :  Xt ] be a polynomial algebra over a ring R0 . The dierential opePt rator E := Xk @X@ k is called Euler-Operator of R=R0 . For  = (1 : : :  t) 2 Nt k=1 with jj  n show the formula (n ; jj)!1!    t!D1 t = (1 ; E )(2 ; E )  : : :  (n ; jj ; E ) 139

@ jj

@X11    @Xtt

3) Let R=R0 be an algebra, M and N R {modules. Then HomR0 (M N ) is a (left) R R0 R {module whose scalar multiplication is given as follows: For ` 2 HomR0 (M N ) and r  s 2 R R0 R (r s 2 R) (r  s)  ` := r ` s Show that Di 1 R=R0 (M N ) is the set of all ` 2 HomR0 (M N ) annihilated by some power of IR=R0 , i.e. in terms of local cohomology 0 Di 1 R=R0 (M N ) = HIR=R0 (HomR0 (M N )) 4) Let K be a eld of characteristic p > 0 and L = K x] a purely inseparable aln over L is given by gebraic extension eld of degree pe . Then a basis of PL=K fdnL=K xi j i = 0 : : :  mg where m := Min fn + 1 peg . 5) Let D = fDn gn2N be a Hasse-Schmidt derivation of an algebra R=R0 into a ring A . It is called universal, if for any Hasse-Schmidt derivation  = fngn2N of R=R0 into a ring B there is exactly one ring homomorphism h : A ! B such that n = h Dn for all n 2 N . Show the existence of universal Hasse-Schmidt derivations. (Hint: Do this rst for polynomial algebras over R0 and then for homomorphic images of polynomial algebras.) 6) Let D = fDngn2N be the universal Hasse-Schmidt derivation of an algebra R=R0 into a ring A . Write D(R=R0 ) := A . 1 L a) D(R=R0 ) is a graded R {algebra: D(R=R0 ) = Dk (R=R0 ) where Dk (R=R0 ) k=0 is the R {module generated by the elements D 1 a1  : : :  D r ar with a1  : : :  ar 2 R , 1 : : :  r 2 N+ , 1 +    + r = k . b) D1 (R=R0 )

= 1R=R0 . 7) Let D = fDngn2N be the universal Hasse-Schmidt derivation of an algebra R=R0 into D(R=R0 ). Dene for each n 2 N an R {linear map fn : R R0 R ! D(R=R0 ) by n P fn(a  b) = a Di (b). i=0 a) For z = (1  a1 ; a1  1)  : : :  (1  at ; at  1) show that  0 for t > 0 fn(z) = D1 a1  : : :  Dn an for t = n hence fn induces an R {linear map fn : I n=I n+1 ! Dn(R=R0 ), where I := IR=R . n b) Let hn : PR=R ! D(R=R0 ) be the R {linear map with Dn = hn dnR=R0 . Show 0 that the induced map hn : I n=I n+1 ! D(R=R0 ) equals fn . c) Show that by the maps hn (n 2 N) a homomorphism of graded R {algebra grI (R R0 R) ! D(R=R0 ) is induced. (See O1 ] for more details about this homomorphism.) 140

8) Under the assumptions of 2.7a) let Q R0 . Then there is an R0 {isomorphism Di 1 = R0fX1 : : :  Xt  U1 : : :  Ut g=I R0 (R)

@ ) (Ui 7! @X i

where R0fX1  : : :  Xt  U1  : : :  Ut g is the free R0 {algebra on X1  : : :  Xt  U1  : : :  Ut (non-commutative polynomial algebra) and I is the two-sided ideal generated by the elements

Xi Xk ; Xk Xi Ui Uk ; Uk Ui  Xi Uk ; Uk Xi ; ik (i k = 1 : : :  t) : The substitution Xi 7! Ui Ui 7! ;Xi (i = 1 : : :  t) induces an R0 {automorphism of Di 1 R0 (R), called formal Fourier transform. 9) Under the assumption of example 2.7b) show a) If m 6= 0, then there exist a dierential operator D of R=K with D( m ) = R . b) If Char K = 0, then d( m ) m for each d 2 DerK R , and if Char K = p > 0, then ( m) m for any component  of a Hasse{Schmidt derivation of R=K .

141

x 3. Functorial Properties This section is devoted to studying the behaviour of the algebra of principal parts under some of the most important algebraic operations such as localization, formation of residue class algebras, tensor products, base change and direct limits.

a) Localization Let R=R0 be an algebra, M an R {module and S  R a multiplicatively closed set. n (M ) 3.1.Proposition. The functorial map PR=R 0

phism

! PRnS =R0 (MS ) induces an isomor-

n (M )S ;! P n PR=R e RS =R0 (MS ) 0

of PRnS =R0 {modules. The universal dierential operators dnRS =R0 MS and dnR=R0 M are related by the formula

dnRS =R0 MS ( ms ) = sn1+1



X n i=0

(;1)n;i

n +1  i n n ; i i  s  dR=R0 M (s m)

for m 2 M , s 2 S . In particular dnRS =R0 MS ( m1 ) =

dnR=R0M (m) 1

n )S Proof: We rst show that the canonical map (PR=R 0

.

! PRnS =R0 is an isomorphism of

RS {algebras. In fact with S  S := fa  b j a b 2 S g we obtain

n+1 )S S = (P n )S S PRnS =R0 = RS R0 RS =IRn+1 = (R R0 R=IR=R R=R0 S =R0 0 n )S 1 is For s 2 S we have 1  s = s  1 ; (s  1 ; 1  s). The image of s  1 in (PR=R 0 a unit and the image of s  1 ; 1  s is nilpotent. It follows that the image of 1  s in n )S 1 is also a unit, hence (PR=R 0 n )S S = (P n )S 1 = (P n )S (PR=R R=R0 R=R0 0

Further n )S R MS = (P n R M )S = P n (M )S PRnS =R0 (MS ) = PRnS =R0 RS MS = (PR=R S R=R0 R=R0 0

which proves the rst assertion of the proposition. The second is a special case of formula 1.8d). 142

3.2.Corollary. Let N and P be R {modules. For each D 2 Di nR=R0 (M N ) there is

exactly one dierential operator DS 2 Di nRS =R0 (MS  NS ) such that the diagram

M

D

N

MS

DS

NS

;

commutes. DS is given by the formula

X n + 1  n m 1 n ; i i n ; i DS ( s ) = sn+1  (;1)  s  D(s m) i i=0

for m 2 M , s 2 S . Moreover for D0 2 Di mR=R0 (N P ) (D0  D)S = DS0  DS n (M ) N ) and if  dnR=R0 M with an ` 2 HomR (PR=R 0 `S 2 HomRS (PRnS =R0 (MS ) NS ) is the unique extension of ` , then DS := `S  dnRS =R0 MS

Proof: If we write D = `

clearly satis es the conditions of the corollary. The uniqueness of the DS implies the last formula of 3.2.

n (M ) is a nitely presented R {module. Then the R { 3.3.Corollary. Assume PR=R 0

linear map Di nR=R0 (M N ) ! Di nRS =R0 (MS  NS ) (D 7! DS ) induces an isomorphism Di nR=R0 (M N )S ;e! Di nRS =R0 (MS  NS )

This follows since Hom commutes with localization. We call DS the localization of D with respect to S . As usual set DS =: Df for S = ff 0  f 1  : : :  g , f 2 R and DS =: Dp for S = R n p , p 2 Spec R . n (M ) is a nitely presentable R {module and f1  : : :  ft 2 R 3.4.Corollary. Assume PR=R 0

are elements with R = (f1  : : :  ft ). a) (Local-global principle for dierential operators) For D1  D2 2 Di nR=R0 (M N ) the following assertions are equivalent:  ) (D1 )fi = (D2 )fi for i = 1 : : :  t .  ) (D1 )p = (D2 )p for all p 2 Spec R .  ) D1 = D2 . 143

b) (Pasting of dierential operators) Let Di 2 Di nRfi =R0 (Mfi  Nfi ) (i = 1 : : :  t) be given such that (Di )fj = (Dj )fi for i j = 1 : : :  t . Then there is a unique dierential operator D 2 Di nR=R0 (M N ) such that Di = Dfi (i = 1 : : :  t). n If R=R0 is essentially of nite type and R0 noetherian, then PR=R is certainly a nitely 0 n n presentable R {module, hence Di R0 (R) = HomR(PR=R0  R) is a reexive R {module, in particular it is torsion free. It follows that Di 1 R0 (R) is also a reexive R {module. b) Ground ring extension Let T=R be an algebra, M and N two T {modules. Clearly Di nT=R (M N ) is the T {submodule of Di nT=R0 (M N ) consisting of all dierential operators which are R { 1 linear. In particular Di 1 R (T )  Di R0 (T ) is the T {subalgebra of R {linear dierential operators. 3.5.Proposition. The structural map R

modules

n { ! T induces an exact sequence of PT=R 0

(n)  P n (M ) ! P n (M ) ! P n (M ) ! 0 IR=R R T=R0 T=R0 T=R 0

Proof: We have a canonical isomorphism T R T

= T R0 T=(fr  1 ; 1  rgr2R ), hence ( n ) n n n n n PT=R = PT=R0 =(fdT=R0 r ; rgr2R ) = PT=R0 =IR=R0 PT=R which gives the exact sequence 0 (n)  P n ! P n ! P n ! 0 IR=R R T=R0 T=R0 T=R 0

Now tensor with M . In particular there is an exact sequence (1)

(n)  P n ! I (n) ! I (n) ! 0 IR=R R T=R0 T=R0 T=R 0

(n) = 0, and If T=R has only trivial dierential operators, then IT=R

(2)

n = T  I (n) = T  T  I (n) PT=R T=R0 R=R0 0

(n) is the T {submodule of P n (n) where T  IR=R T=R0 generated by the image of IR=R0 . In case 0 (n) = 0 and P n 2 n , hence IR=R 1R=R0 = 0 we have IR=R0 = IR=R T=R0 = PT=R . It follows that 0 0 Di nR0 (T ) = Di nR(T ) for all n and

1 Di 1 R0 (T ) = Di R (T )

144

This holds in particular when R=R0 is the maximal subalgebra of T=R0 with 1R=R0 = 0 (see I. 4.23). For example, if R0 is noetherian, T=R0 nite and R=R0 the maximal unrami ed subal1 gebra of T=R0 (see I.4.24), then Di 1 R0 (eT ) = Di eR+1(T ). If in addition T contains a eld of characteristic p > 0, then S := R0T p ] = R0tp ] = R0S p] for suciently large e, as T is a noetherian R0 {module. Then 1S=R = 0 and therefore S  R . From 2.10 we infer Di 1 R0 (T ) = EndR (T ) c) Residue class algebras Let R=R0 be an algebra, M an R {module, I  R an ideal. Set R := R=I and M := M=IM . n (M ) ! P n (M ) induces an isomorphism 3.6.Proposition. The functorial map PR=R 0 R=R0 n {modules of PR=R 0

n (M )=(I dn I )  P n (M ) ;! P n (M ) PR=R e R=R0 R=R0 R=R0 0 For m 2 M with residue class m 2 M we have n (M ) dnR=R0 M (m) = dnR=R0 M (m) + (I dnR=R0 I )  PR=R 0 Proof: It suces to consider M = R . We have R R0 R = R R0 R=I R0 R + R R0 I hence n = R R R=I n+1 = R R R=I n+1 + I R R + R R I PR=R 0 0 0 0 R=R0 R=R0 0 n =(I  P n + (dn I )  P n ) = PR=R R=R0 R=R0 R=R0 0 In the following let N be an R {module, set N := N=IN and DIn (M N ) := fD 2 Di nR=R0 (M N )jD(IM )  IN g This is an R {submodule of Di nR=R0 (M N ) and we have a wellde ned map n (M N )  : DIn(M N ) ! Di R=R 0 where D := (D) is given by D(m + IM ) = D(m) + IN The product formula 1.1b) for D immediately implies the product formula for D . The R {module I  Di nR=R0 (M N ) is contained in DIn (M N ), therefore we obtain (I  Di nR=R0 (M N )) = 0, hence there is a canonical homomorphism of R=I {modules  : DIn(M N )=I  Di nR=R0 (M N ) ! Di nR=R0 (M N ) n (M ) is a nitely generated projective R {module. Then 3.7.Proposition. Suppose PR=R 0  is an isomorphism.

145

Proof: a)  is surjective. Let D

2 Di nR=R0 (M N ) be given and let " : M ! M=IM

be the canonical epimorphism. Then

n (M ) N ) D  " 2 Di nR=R0 (M N ) = HomR(PR=R 0 n (M ) is projective the operator D  " can be lifted: D  " = "0  D with some Since PR=R 0 n (M ) N ) where "0 : N ! N is the canonical epimorphism. For a 2 I , D 2 HomR (PR=R 0 m 2 M we have "0 (D(am)) = D("(am)) = D(0) = 0, hence D(am) 2 I  N and D 2 DIn(M N ). Clearly (D) = D . b)  is injective. Suppose (D) = 0 for some D 2 DIn (M N ), i.e. D(M )  IN . Write n (M ) IN ). Since P n (M ) is a nitely generated D = `  dnR=R0 M with ` 2 HomR(PR=R R=R0 0 n (M ) N ) and projective R {module we have ` 2 I  HomR (PR=R D 2 I  Di nR=R0 (M N ). 0 This implies that  is injective.

1 3.8.Corollary. Let DI1 (R) := fD 2 Di 1 R0 (R)jD(I )  I g . Then DI (R) is a subring of Di 1 (R) and I  Di 1 (R) is a two-sided ideal in D1 (R). The canonical map R0

R0

I

1  : DI1(R)=I Di 1 R0 (R) ! Di R0 (R)

is a ring homomorphism and an isomorphism provided PRn0 (R) is a nite projective R { module for all n 2 N . Proof: For D1  D2 2 DI1 (R) clearly D1  D2 2 DI1 (R) and (D1  D2 ) = (D1 )  (D2 ). Further for D 2 I Di 1 (R) we have D1  D D  D2 2 D1 (R). The claim follows from R0

I

the proposition. Observe that  is an isomorphism if R = R=I is a presentation of R as a homomorphic image of a polynomial algebra R = R0 X1 : : :  Xt] whose dierential operators are well understood (see 2.7a). d) Tensor products Let R=R0 and S=R0 be two algebra, T := R R0 S . Then the canonical isomorphism T R0 T ;! e (R R0 R) R0 (S R0 S ) 0 0 (r  s)  (r  s ) 7! (r  r0 )  (s  s0 ) maps IT=R0 onto the ideal IR=R0 R0 S e + Re R0 IS=R0 , the kernel of the canoncial map (R R0 R) R0 (S R0 S ) ! R R0 S . Hence (IT=R0 )n+1 is mapped onto n X n +1 e e n +1 k n+1;k IR=R0 R0 S + R R0 IS=R0 + IR=R  IS=R 0 0 k=1

and we have 146

3.9.Proposition. There is a canonical isomorphism n

n PT=R P n R0 PS=R 0 = R=R0 0

X n

(n) )k  (I (n) )n+1;k (IR=R S=R0 0

k=1

of T {algebras. Further dnT=R0 is the map induced by n R P n dnR=R0  dnS=R0 : R R0 S ! PR=R 0 S=R 0

modulo

n (n) k P (n) )n+1;k . (IR=R0 )  (IS=R 0

k=1

Proof: Since n+1 n R P n = Re =I n+1 R S e=I n+1

e e n+1 e e PR=R 0 S=R0 R=R0 0 S=R0 = R R0 S =IR=R0  S + R  IS=R0 0

the rst statement follows from Noether's isomorphism theorem. We have dnR=R0 (r  s) = +1 n+1 which by (3) corresponds to (1  r)  (1  s) + nP k  I n+1;k . (1  1)  (r  s) + IT=R IR=R S=R0 0 k=0 From this the second statement of the proposition follows easily. 3.10.Corollary. Let D 2 Di kR0 (R) and D0 2 Di `R0 (S ) where k + ` n . Then

D R0 D0 : R R0 S ! R R0 S is a dierential operator of T=R0 of order n . n ! R and h0 : P n ! S be the linear maps corresponding to D Proof: Let h : PR=R S=R0 0 resp. D0 according to the universal property 2.5. Then D R D0 is the composition 0

dnR=R0 dnS=R0

hh

n R P n ;;;! R R S R R0 S ;;;;;;;;;! PR=R 0 S=R0 0 0 0

(n) )k+1 ) = 0 and h0 ((I (n) )e+1 ) = 0. Therefore We have h((IR=R S=R0 0

(h  h0)(

n X

=1

(n) )  (I (n) )n+1;) = 0 (IR=R S=R 0

due to the assumption that k + ` n . This implies that the map D  D0 factors through n PT=R and hence is a dierential operator of T=R0 of order n . 0 147

3.11.Example: Let T = RfXg2 ] be a polynomial algebra over R . For D 2 Di nR0 (R)

de ne D~ : T ! T by applying D to the coecients of the polynomials:

D~ (a1 t X11    Xtt ) = D(a1 t )X11    Xtt Then D~ 2 Di nR0 (T ). e) Base change Let R=R0 and S=R0 be algebras, M an R {module. 3.12.Proposition. There is a canonical isomorphism of S R0 R {algebras n PSnR0 R=S ;e! S R0 PR=R 0

where dnSR0 R=S corresponds to idS dnS=R . Proof: Consider the canonical isomorphism of S R0 R {modules (S R0 R) S (S R0 M ) ;e! S R0 (R R0 M ) sending (s  r)  (t  m) to st  (r  m) for s t 2 S r 2 R m 2 M . In case M=R it is an isomorphism of S  R {algebras which maps ISR0 R=S onto S R0 IR=R0 , hence an isomorphism as required in the proposition is induced. It sends

dnSR0 RSR0 M (s  m) = 1S  1R  (s  m) + ISn+1R0 R=S (S R0 R S S R0 M ) for s 2 S r 2 R m 2 M to n+1 (R R M ) (idS dnR=R0 M )(s  m) = s  (1R  m) + S  IR=R 0 0

which proves the assertion about universal dierential operators. If N is another R {module we have an isomorphism of S R0 R {modules n HomSR0 R (S R0 PR=R  S R0 N ) ;e! HomSR0 R (PSnR0 RSR0 M  S R0 N ) 0 M

and a canonical S R0 R {linear map n n  N ) ! HomSR0 R (S R0 PR=R  S R0 N ) S R0 HomR(PR=R 0 M 0 M n which is an isomorphism in case S is at over R0 and PR=R a nitely presentable 0 M R {module.

148

n 3.13 Proposition. Suppose S=R0 is at and PR=R a nitely presentable R {module. 0 M

Then there is a canonical isomorphism of S R0 R {modules

S R0 Di nR=R0 (M N ) ;e! Di nSR0 R(S R0 M S R0 N ) Further there is a canonical isomorphism of S R0 R {algebras 1 S R0 Di 1 R0 (R) ;e! Di S (S R0 R)

It sends 1  D to idS D for any dierential operator D of R=R0 . The last assertion follows from the rst if we take M=N=R and form the union over all n 2 N. f) Direct limits Let fR ' g be a direct system of R0 {algebras R over a directed set . This means that  is partially ordered, and for  0 2  there exists 00 2  with 00 , 0 00 . For  0 2  with 0 there is an R0 {homomorphism ' : R ! R such that ' = idR and '  ' = ' for 0 00 . Let R := ; lim ! R be the direct limit of  fR ' g . By the functorial property of P n there are ring homomorphisms  : PRn=R0 ! PRn =R0 ( 0 ) which are compatible with the universal dierential operators, hence a morphism of direct systems fdnR=R0 g : fRg ! fPRn =R0 g 0

0

0

0

00

0

00

0

0

0

is induced. It de nes a ring homomorphism n dnR=R0 : R ! lim ;! PR=R0 n n n 3.13.Proposition. We have lim ;! PR =R0 = PR=R0 , and dR=R0 is the universal dierential

operator of R=R0 .

Proof: The tensor product of algebras is compatible with direct limits, i.e.

R R0 R

= lim ;! R R0 R n+1 is the union of the images of the ideals I n+1 Under this isomorphism the ideal IR=R R =R0 0 ( 2 ). From this the statement of the proposition easily follows.

149

3.14.Example: Let R = R0 fX g2M ] be a polynomial algebra over R0 in an arbitrary family fX g2M of indeterminates. Then R is the union (direct limit) of the system of

all polynomial algebras R0X1  : : :  Xt ] in nite subsets f1     t g  M . Let N(M ) be the setPof all functions  : M ! N ( 7!  ) with only nitely many  6= 0, and set jj =  . Then 2M

L Ru n = SPn PR=R R0 X1 :::Xt ]=R0 = 0 2N(M ) jjn

Q

with u = u where u is as in 2.4d). Here the dierential operator dnR=R0 acts on 2M polynomials of R0X1  : : :  Xt ] as dnR0 X1 :::Xt ]=R0 does. Exercices:

1) Let K=K0 be a eld extension where Char K0 = p > 0. Further let L=K be a simple extension of the form L = K X ]=(X pe ; a) where a 2 K0 K pe ]. Show that n = dimL PL=K 0

e ;1ng minfpP

i=0

n;i dimK PK=K 0

2) (Dierential operators of direct products). Let Ri=R0 (i = 1 : : :  t) be algebras and R := R1      Rt their direct product. The map n ! Pn n PR=R R1 =R0      PRt =R 0 with dnR=R0 (r1  : : :  rt ) 7! (dnR1=R0 r1  : : :  dnRt =R0 rt ) for (r1  : : :  rt ) 2 R , ri 2 Ri is an isomorphism of R {algebras. Further the map Di nR0 (R1 )      Di nR0 (Rt ) ! Di nR0 (R) which sends (D1  : : :  Dt ) with Di 2 Di nR0 (Ri ) to D with D(r1  : : :  rt ) = (D1 (r1 ) : : :  Dt (rt )) is an isomorphism of R {modules. It induces an isomorphism of R {algebras 1 1 Di 1 R0 (R1 )      Di R0 (Rt ) ;e! Di R0 (R) 3) Let A be a nite dimensional algebra over a eld K and A = A1      Ah its decomposition as a direct product of its localizations Ai at the maximal ideals of A . Let Li be the extension eld of K in Ai which is mapped onto the separable hull of K in the residue eld of Ai (i=1, : : : ,h) by the canonical epimorphism. Then there is a canonical isomorphism Di 1 = EndL1 (A1 )      EndLh (Ah ) K (A)

. 150

4) (Dierential operators of symmetric algebras) Let R be a ring, M an R {module and 1 S(M ) = L Si(M ) its symmetric algebra. Determine PSn(M )=R and Di 1 R S(M ). i=0

5) (Dierential group, L operators of group algebras) Let R be an ring, G an abelian 1 RG] = Rxg its group algebra over R . Determine PR (RG]) and Di R (RG]). g2G

151

x 4. Dierential Operators of Smooth Algebras The algebra of principal parts and the dierential operators of (local) smooth algebras R=R0 can be described rather explicitely. In case Q R0 we shall see that the algebra of dierential operators is the Weyl-algebra RDerR0 R], and we shall determine its two-sided ideals.

Let R0 be a noetherian ring and R=R0 an algebra which is essentially of nite type. The following theorem generalizes example 2.4d).

f

g

4.1.Theorem. Assume R=R0 is a smooth algebra and a1  : : :  at is a dierential basis

of

R=R0 .

Then the R {homomorphism n ' : RU1  : : :  Ut ] ! PR=R (Ui 7! dnR=R0 (ai ) ; ai ) 0 induces an isomorphism of R {algebras n RU1  : : :  Ut ]=(U1  : : :  Ut )n+1 ; e! PR=R 0

In particular with ui := dnR=R0 (ai ) ; ai (i = 1 : : :  t) we have n PR=R 0

=

M

jjn

1 Ru 1

n PR=R is a free R {module of rank 0



 ut

t

t+n n



 u1 1

 ut

t

= 0 for j j > n

. The elements

(dnR=R0 a1)1  : : :  (dnR=R0 at )t with n . form also an R {basis of PR=R 0

jj  n

P 2 Spec R let Q denote the preimage of P in Re by the canonical map Re ! R (ab 7! ab), and let p := P \R0 . By 3.1 we have PRnP =(R0 ) p = (Re ) Q =(IR=R0 )nQ+1 Proof: For

and

2 ) = (IR=R0 )Q =(IR=R0 )2Q 1RP =(R0 )p = (IR=R0 =IR=R 0 P 2 Since the images of 1  ai ; ai  1 (i = 1 : : :  t) in (IR=R0 =IR=R ) form a basis of this 0 P R P {module they are a minimal set of generators of (IR=R0 ) Q and moreover an (Re ) Q { regular sequence (see I.6.13). Therefore the localization 'P of ' induces a surjective R P {homomorphism gr('P ) : gr(U ) RP U1  : : :  Ut ] ! grI (n) (PRnP =(R0 )p )

P

R =(R0 )

p

with kernel (U1  : : :  Ut)n+1 . But then 'P is surjective, too, with kernel (U1  : : :  Ut)n+1 . The rst assertion of the theorem follows from the local-global principle, and the remaining statements are clear. 153

4.2.Example:

Let L=K be a separable algebraic function eld. Then L=K is a smooth algebra and any separating transcendence basis fa1 : : :  at g of L=K is a dierential basis. Therefore the n . statements of theorem 4.1 are true for PL=K n ! R be the R {linear form with Under the assumptions of 4.1 let `1t : PR=R 0 `1 t

(u1 1



ut t ) =

 1 if i = i (i = 1 : : :  t)

Let D1 t := `1 t dnR=R0 . Then since

0 otherwise

n  R) Di nR0 (R) = HomR0 (PR=R 0

we have

Di nR0 (R) =

and Di 1 R0 (R) =

(1)

where D1 t satises the rule (2)



D1 t (a11

M jjn

M 2Nt

 

1 at t ) = 1



R D1 t



R D1 t

 ::: 

 

t a1 ;1 1 t

 at ; t

t

In fact D1 t (a11  at t ) = `1t ((dn a1 )1  ((dn at )t ) = `1t ((u1 + a1 )1  (ut + at )t ) = `1t =

  1

1

X  





 

t a1 ;1 1 t  1

t a11 ;1 at t ;t t 1



at t ;t u1 1



ut t





is uniquely determined by (2) since f(dn a1)1  (dn at )t j ( 1 : : :  t ) 2 Ntg is a n basis of PR=R over R . It follows that each D 2 Di nR0 (R) is uniquely determined by its 0 action on the monomials a11  : : :  at t with j j  n . In fact, if we write D1 t

D

=

X

jjn



1 t D1 t

154

(1t 2 R)

then D(a01  a0t ) = 00 , and by induction along the lexicographic order of the (1  : : :  t ) with j j  n one concludes that all coecients 1t are uniquely determined by the values D(a11  at t ) with j j  n . For example (3)

@a1 1

@



t = (1 )! @a t

 (t)!  D  1

t

(1  : : :  t 2 Nt)

for the same reason. The operators D0k0 are called the "divided powers" of the partial derivatives @a@k . For each f 2 R we have dnR=R0 f

(4)

=

X

1 D1  t (f )u 1

4.3.Iteration rule: For (1  : : :  t ), (1  : : :  t ) D1 t

t

2 Nt

 +    +   = 1 1  t t D

D 

t

1

 ut

t

1

Proof: Apply both sides of the equation to the elements a11

use the uniqueness assertion.

4.4.Product rule: For (1  : : :  t )



D1 t (f g ) =

Proof: Choose n

2 Nt and f g 2 R P

i +i =i (i=1:::t)

 jj so that D 

is a ring homomorphism



t

1



=

j jnj jn

 at

t

(( 1  t) 2 Nt ) and



D1 t (f ) D1 t (g )

= `1t dnR=R0 as above. Then since dnR=R0

dnR=R0 (f g ) = dnR=R0 (f ) dnR=R0 (g )

X

1 +1t +t

D1 t (f )D1 t (g )u1 1 +1

 ut



t+ t

and the product formula follows. 4.5.Commutation rule: For (1  : : :  t ), (1  : : :  t ) 2 Nt and f 2 R

X  1 + 1   t + t  D1 t (f D1 t ) =  t D1t (f )  D1+ 1t+ t

1 i + i =i

In particular: D1t f =

P

i + i =i



D1 t (f ) D 1  t .

155

2 R by the product rule X D  (f D  (g )) =

Proof: For g

1

t

t

1

X  1 + 1   =

i + i =i

1



(D1 t (f ) D 1 t (D1 :::t (g)) i + i =i  +  t t D 1 t f D1 + 1 :::t + t (g )

t



The same rules hold of course in the case of polynomial algebras R0X1  : : :  Xt ] (4.2) over an arbitrary ring R0 where D1 t is the dierential operator with D1 t (X11



   

1 Xtt ) = 1



t

t

X11 ;t

 Xt ; t

t

n 4.6.Theorem. If R=R0 is smooth, then PR=R is a projective R {module. In case Q 0

we have

R

0

Di 1 R0 (R) = RDerR0 R]

Proof: The rst statement follows from 4.1 by localization since locally there exists a

diential basis of R=R0 . As for the second we may also assume that algebra, hence by (1) M Di 1 ( R ) = R  D1 t R0

R=R0

is a local

1 :::t 2N

Since Q R0 the iteration rule 4.3 implies that as an R {algebra Di 1 R0 (R) is generated by @ the dierential operators D010 = @ai (i = 1 : : :  t) where fa1  : : :  at g is a dierential i

Lt

basis of R=R0 . Since DerR0 R = R @a@ i the second assertion of the theorem follows. i=1 See the exercises for generating sets of Di 1 R0 (R) as an R {algebra in other cases. 4.7.Remarks: a) Sometimes RDerR0 R] is called algebra of dierential operators of R=R0

(Bj]). If Q R0 and R=R0 is a smooth algebra this notion agrees with ours by the above theorem. This need not be true in the non-smooth case, see x 1, exercise 1). b) If in the situation of theorem 4.1 we have t = 1, then the dierential operators D ( 2 N) form an iterative Hasse-Schmidt derivation (see 1.6b). This situation is given, for example, if R=R0 is a separable algebraic function eld of transcendence degree 1, the case to which Hasse-Schmidt derivations were rst applied (Has], HS]). More generally for i 2 f1 : : :  tg and  2 N let  i] = (0 : : :   : : :  0) 2 Nt where  is at the i {th place. Then the sequence of divided powers fDi] g2N is an iterative Hasse-Schmidt derivation of R=R0 . The Di] ( 2 N i = 1 : : :  t) generate Di 1 R0 (R) as an R {algebra. See also exercise 2). 156

Let us consider now a more general situation. Let R=R0 be an algebra which is of nite type, smooth, and has a dierential basis fa1  : : :  at g . Let S=R0 be another algebra where S is noetherian and set T := S R0 R . Then T =S is smooth with dierential basis f1  a1 : : :  1  at g . By 3.8 and 4.1 we have



n n = Pn PT=R S=R0 R0 PR=R0 0 n R

= PS=R

0

0

X n

(n) k (n) (IS=R )  (IR=R )n+1;k 0 0

k=1

R0 U1  : : :  Ut ]

X n

(n) k (IS=R )  (U1  : : :  Ut )n+1;k 0

k=1

Let ui denote the image of Ui modulo (U1  : : :  Ut )n+1 (i = 1 : : :  t). Then we obtain n R R){modules isomorphisms of (PS=R 0 0



M n n n R R)u  u

M (PS=R (IS=R ) t jjn jjn M n ;j  j  

= (PS=R R R)  u  ut

n = PT=R

(5)

For r 2 R , (6)

1

0

jjn

s

1

0

0

t

1

( )

+1;jj u1 1

 ut

t

t

1

2 S the universal dierential operator dnT=R is given by X n;jj dnT=R (s  r) = (dS=R (s)  D  (r))u  ut 0

jjn

1

t

1

0

t

1

with the dierential operators D1 t from (2). Of course these formulas can be applied to polynomial algebras T = S X1 : : :  Xt ] = S R0 R0 X1  : : :  Xt ] where ai = Xi (i = 1 : : :  t). In this case n PT=R 0

(7) and for

s

n;jj X  : : :  X ]  u  u

= M PS=R t t 1

0

jjn

2 S , (  : : :  t) 2 Nt

t

1

1

dnT=R0 (sX11

(8)

1

X  1   t  jjn

1



t



 Xt ) = t



n;jj dS=R0 (s) X11 ;1





1 Xtt ;t u 1

 ut

t

In the situation of formula (5) we have (9)

n M M n;i R R)u1 n n Di R0 (T ) = HomT (PT=R0  T ) = HomT ((PS=R 0 1 0 i=0 jj=i





157

 ut  T ) t

n  T ). Then ` is uniquely Write D 2 Di nR0 (T ) as D = ` dnT=R0 with ` 2 HomT (PT=R 0 determined by the family f`1t gjjn of its restrictions

 ut  T ) n;jj  R = P n;jj  T , hence n to the direct summands of PT=R in (5). We have PS=R S R S=R n;jj  R)u  u  T ) n;jj  T ) ;jj (S T ) HomT ((PS=R = HomS (PS=R = Di nS=R R t For each D 2 Di nR (T ) there is a unique family fD   gjjn of dierential opera;jj(S T ) such that for r 2 R , s 2 S tors D   2 Di nS=R X   (10) D(s  r) = D (s)  D  (r) `1 t

n;jj  2 HomT ((PS=R R 0

0

1 T )u 1

0

0

0

( 1

1

0

0

t

1

0

0

0

( 1

0

t)

t

t)

0

( 1

t)

In particular, if

R = R0 X1  : : :  Xt ]

n;jj  T ) HomS (PS=R = 0

M

 2N+

t

1

jjn

is a polynomial algebra, then n;jj  S ) = HomS (PS=R 0

M

 2N+

j (S ) Di Rn;j 0

1 :::t ) and D(1 t ) uniquely induces a set of dierential operators D(1 t such that for s 2 S

D(1 t ) (s)

=

X

 2N+

In this situation formula (9) becomes (10) D(sX11



xtt ) =

X  1 

 2Nt jjn

1





D(1 1t t ) (s) X11

  t t

 D ( 1 1

t

t)

2 Di nR;jj(S) 0

 Xt

t

(s)  X11;1 +1  Xtt ;dt +t

Similarly as for the module of dierentials there is a recipe for computing the algebra of principal parts if an algebra is given by generators and relations. We illustrate this for algebras of nite type, the general case being similar. Let R=R0 and S=R be algebras where S=R has a presentation S

= RX1  : : :  Xt ]=I = Rx1  : : :  xt ]

and assume that I

= (ff g2 ) with

f

=

X 

()

r1 t X11

158

(xi := Xi + I )

 Xt

t

(r1 t 2 R)

Since dnRX ]=R0 is a ring homomorphism it is clear that PRnX ]=R0  dnRX ]=R0 I = (fdnRX ]=R0 f g2 ) If we write as in (7) PRnX ]=R0

n;jj X  : : :  X ]  u  u

= M PR=R t t 0

jjn

1

1

1

then dnRX ]=R0 f is identied with

X X  1  1

 jjn



  t t





n;jj () dR=R0 (r1 t ) X11 ;1



t



1 Xtt ;t u 1

 ut

t

Let Rn X ]=R0 f be this expression with Xi replaced by xi (i = 1 : : :  t). Then it follows by 3.5 that (12)

n = PS=R 0

M



 

n;jj (PR=R R S ) u1 1 0

jjn



f

g

n u t t ( RX ]=R0 f 2 )

4.8.Examples: a) Let R=R0 and S=R be algebras where S=R is a local algebra which is n ! P n induces essentially nite and etale. We shall show that the functorial map PR=R S=R0 0 a ring-isomorphism n PR=R 0

n R S ;e! PS=R

0

In fact, by I.4.16 we have a presentation S

(f 2 RX ])

= RX ]n =(f )

0 n being a maximal P ideal of RX ] and f (x) a unit of S where x denotes the image of X

in S . Write f =



r X  ,

then by (12)

n M n; R S ) u =(') n PS=R0 = (PR=R 0 =0

(13) n P





P; 

; (r )x ; ( = 0 : : :  n). Here c0 = f (x) = 0 and where ' = cu , c =  dnR=R 0  =0 P n ; 1  ; 1 c1 = dR=R0 (r )  x which has the image f 0 (x) by the canonical ring homomorphism  n;1 R S ! P 0 R S = S . Since its kernel is nilpotent it follows that c1 is a unit in PR=R R=R0 0 n n; n;1 R S , hence ' = "  u with a unit " of L PR=R (PR=R0 R S )u . Now (13) implies that 0 =0 n = Pn PS=R R=R0 R S 0



159

b) Let

R = R0 X1  : : :  Xt ]=(f1  : : :  fm )

n = PR=R 0

L

jjn

Ru1 1

be a complete intersection. Then





n n u t t ( R=R0 f1  : : :  R=R0 fm )

n fi is the image of where R=R 0 dnR0 X ]=R0 fi

X

=

(D fi )(X1  : : :  Xt)  u1 1  ut t

(i = 1 : : :  n)

jjn

We can also write n PR=R 0

= RU1 : : :  Ut ]=(f1 (x + U ) : : :  fn (x + U )) + (U1 : : :  Ut )n+1

Observe that (f1 (x + U ) : : :  fm (x + U )) is a regular sequence in R R0 RY1  : : :  Yn ], since (f1  : : :  fm ) is in R0 X1  : : :  Xn ].

RU1  : : :  Ut ]

=

What are the two-sided ideals in the Weyl-algebra of a smooth algebra? 4.9.Lemma. Let R=R0 be an algebra, I

R an ideal and W := RDerR R]. Then the 0

following assertions are equivalent: a) I  W is a two-sided ideal of W . b) I is a dierential ideal for R=R0 (I.1.20).

!

2 I we have dx 2 IW \ R = I for each d 2 DerR R , since IW is a W {left ideal. b) ! a) For x 2 I , D 2 W and d 2 DerR R we have d (xD) = dx  D + x  (d D) 2 IW from which W  (IW ) IW follows. 4.10.Lemma. Let R=R be an algebra with a dierential basis fa  : : :  at g . Then for Proof: a) b) For x

0

0

i = 1 : : :  t

and all

Proof: For r

0

j>0

2 R we have

@j (ai r) @aji

;

1

@j

j

j ;1

i

i

i

@ @  a ; a  = j  i i @aj @aj @aj ;

@j r ai j @ai

=

j   X j k=1 k

k

j ;k

1

j ;1

j

@ r @ r @ r  @@aakii  @a j ;k ; ai @aj = j @aj ;

i

160

i

i

1

Q R0 , and let W := Di 1 R0 (R).

4.11.Theorem. Let R=R0 be a smooth algebra with

The two-sided ideals of for R=R0 .

W

are the right ideals IW , where the ideal

I

R is dierential

Proof: By 4.9 it suces to show that two-sided ideals have this form. Without loss of

generality we may assume that R=R0 is a local algebra: If for a two-sided ideal J and all p 2 Spec R we have proved the claim of the theorem, i.e. Jp

= (Jp \ Rp )  Wp

then J = (J \ R)  W follows. Therefore we may assume that R=R0 has a dierential basis fa1 : : :  atg . Given a two-sided ideal J of W with J 6= (0) choose D 2 J , D 6= 0. Since @ ] where  @  @ ] = 0 for all i j 2 f1 : : :  tg we may write W = R @a@ 1  : : :  @a @ai @aj t D

s

@ = 0 + 1  @a@ +  + s @a s

t

t

with i 2 R @a@ 1  : : :  @a@t;1 ] (i = 0 : : :  s). Assume s > 0. Then D  at ; at  D 2 J , hence s

s

t

t

@ @ ( 0 at ; at 0) + ( 1 a@ at ; at 1 @a@ ) +  + ( s @a s at ; at s @as ) 2 J

t

t

since at commutes with the i we obtain by 4.10 that + 

@ s;1 + s s s;1 @at

2J Iterating this process we end up with s! s 2 J , hence s 2 J since Q R . iteration we have k 2 J for k = 0 : : :  s , hence @ @ ::: ])  W J = (J \ R @a @at; @

1 + 2 2 @at

0

1

Again by

1

Repeating with at;1  : : :  a1 we nally obtain J

= (J \ R)  W

q.e.d. 4.12.Corollary. The Weyl algebra of a smooth algebra over a eld of characteristic 0

is simple, i.e. has no two-sided ideals other then (0) and (1). For related results, in particular for prime characteristic, see SvdB]. 161

Exercises:

1) Let R = R0X1 : : :  Xt ]] be the algebra of formal power series in the variables X1 : : :  Xt over a ring R0 . Assume that Q R0 . Then Di 1 R0 (R) = RDerR0 R]. 2) For R=R0 as in theorem 4.1 assume that R0 contains a eld of characteristic p > 0. Show that the dierential operator Dpn i] (n 2 N i = 1 : : :  t) dened in 4.7b) generate Di 1 R0 (R) as an R {algebra. 3) For R=R0 as in exercise 1) assume that R0 contains a eld of characteristic p > 0. Then Di 1 R0 (R) is generated as an R {algebra by the dierential operators Dpn i] dened as in 4.7b) with respect to fa1 : : :  at g = fX1  : : :  Xt g . 4) Under the assumptions of 4.1 let Q R0 . Then gr Di 1 R0 (R) (see 1.7) is a polynomial algebra in t variables over R . Conclude that in Di 1 R0 (R) any left (right) ideal is nitely generated. 5) (Universally nite dierential operators). A dierential operator D : R ! M of order  n into an R {module M is called nite, if RDR is a nitely generated R {module. It is called universally nite, if for any nite  2 Di nR=R0 (R N ) there is a unique R {linear map ` : M ! N such that n := M .  = ` D . If D : R ! M is universally nite we write d~nR=R0 := D and P~R=R 0 n n k k a) If (P~R=R0  d~R=R0 ) exists, so does (P~R=R0  d~R=R= ) for any k  n and so does the universally nite dierential module (~ 1R=R0  d~R=R0 ). n n =U is a b) Let fUg be the family of two-sided ideals U PR=R such that PR=R 0 0 T n n =U is a nite R {module and let U := 2 U . Then P~R=R exists if and only if P R=R0 0 n n n ~ ~ nite R {module. In this case PR=R0 = PR=R0 =U . If PR=R0 exists and R is noetherian, n  R). then Di nR0 (R) = HomR(P~R=R 0 Many properties of ~ 1R=R0 (see I, x 8) and about its existence can be generalized to n . For a semilocal noetherian ring R the existence of P~ n P~R=R R=R0 is even equivalent 0 1 ~ with that of R=R0 .

162

Introduction to the bibliography II

Dierential operators are discussed in scheme-theoretic terms in EGA IV x 16. The rst part of x 4 above about dierential operators of smooth algebras essentially comes from there. In the general case the structure of the algebra of dierential operators is di cult to determine. Hasse-Schmidt derivations were rst introduced in HS], see in particular the appendix added in proof by F.K. Schmidt. Among the earliest applications are those of F.K. Schmidt ( Sch1 ], Sch2 ]) to the theory of Weierstra points on curves in positive characteristic, a subject now relevant for coding theory. In this text and in the exercises only the rudiments of the theory of Hasse-Schmidt derivations are exposed. In N11 ], N12 ], M4 ] and in M5 ], x 27 one can learn more about them. Detailed studies were also carried out, applications were given and connections to other notions of higher derivations were investigated by R.W. Berger ( B5 ], B7 ], B10 ], BR]) and his students ( Heu], O1 ], O2 ], Te], Bau2 ]). The relation to integrable derivations was mentioned in 1.6c). The notion of dierentially closed ideal or dierentially simple algebra (I, x 1, ex.7) may be introduced for Hasse-Schmidt derivations and dierential operators, see M4 ] and Mal1 ]- Mal3 ] for the generalization to characteristic p > 0 of results about ordinary dierentially closed ideals in characteristic 0. In general there are more dierentially closed ideals in a ring R than D {submodules. If R is a localization of a reduced a ne algebra over a eld R0 of characteristic 0 Nakai conjectured that the equation Di1 R0 (R) = R DerR0 R] implies the regularity of R . For positive results to this problem, mainly in the one-dimensional and the graded case, see f.i. Is1 ], Is3 ], Is4 ], MV], Re], Schr], Tr1 ] and Vi]. If the Nakai conjecture is true, then the Zariski-Lipman problem has a positive solution too, see Re], Prop.2. In case R0 is a eld of characteristic p > 0 an analog to Nakai's conjecture is the following question: If Di1 R0 (R) is generated over R by the components of the HasseSchmidt derivations of R=R0 , does it follow that R=R0 is smooth (geometrically regular over R0 )? This problem is studied in Is2 ] and Tr2 ]. A method to prove the Nakai conjecture (in special cases) is to nd dierentially closed ideals which are not D {invariant for the algebra D of all dierential operators. In Tr2 ] a description of Di1 K (R) is given when R is a Stanley-Reisner ring over a eld K and the D {invariant ideals of R are determined. For an introduction to D {module theory in characteristic 0 we refer to the book of Bjork Bj]. The homological properties of the algebra R DerR0 R] in characteristic 0 are discussed there. Applications of D {modules to local cohomology theory in arbitrary characteristic were given by Lyubeznik ( Ly1 ], Ly2 ]). 163

Chapter III. Dierential Forms x 1. Dierential Algebras The algebraic analog of the algebra of dierential forms of analysis is the notion of dierential algebra (deRham algebra). This concept will be introduced here and some of its basic properties will be established. A special case is the exterior algebra of the module of Kahler dierentials with the dierentiation extended to all of the exterior algebra. But it is important to consider also more general cases.

Let R=R0 be an algebra. 1.1.Definition. A dierential

algebra of R=R0 is an associative (not necessarily comL n mutative) graded R {algebra  =  , on which an R0 {linear map d :  !  of n2N degree 1 is given (i.e. dn  n+1) such that the following axioms are satised:

a) 0 = R and R is contained in the center of . b)  = RdR] (i.e. as an R {algebra  is generated by the elements dr (r 2 R)). c) For all r r0 2 R we have d(rr0 ) = rdr0 + r0 dr . d) For all r r1  : : :  rm 2 R we have d(rdr1 : : : drm ) = drdr1 : : : drm . e) drdr = 0 for each r 2 R . The mapping d is called the dierentiation of  and the elements of n are called dierential forms of degree n or n {forms. For every ring R a di erential algebra of R= Z is called an absolute dierential algebra. Clearly any di erential algebra of R=R0 is also one of R=R00 if R00 is the image of R0 in R . If we do not insist on condition 1.1b) we obtain what is usually called a dierential graded algebra (DG-algebra) (ML2 ], Chap. VI). From the axioms we can derive the following properties of a di erential algebra ( d) of R=R0 . 1.2.Rules:

a) The restriction d : R ! 1 of the di erentiation d to elements of degree 0 is a derivation of R=R0 . b) Each ! 2  is a nite sum of elements rdr1 : : : drm (r r1  : : :  rm 2 R). Such elements have degree m . c) For all rP r0 2 R we have drdr0 + dr0 dr P = 0. d) For ! = rdr1 : : : drm we have d! = drdr1 : : : drm . e)  is anticommutative, i.e. for !m 2 m , !n 2 n

!m  !n = (;1)mn !n  !m 164

Each homogeneous left (right) ideal of  is a two-sided ideal. f) d is an antiderivation, i.e. for !m 2 m , ! 2 

d(!m  !) = d!m  ! + (;1)m !m  d! g) d  d = 0, i.e. ( d) is a complex of R0 {modules d

d

0 ;! 1 ;! 2 ! : : : h) Z () := ker d is a graded subring of . j) B() := im d is a two-sided homogeneous ideal of Z (). Proofs: a) follows from 1.1c) and the fact that d is R0 {linear and of degree 1. Rule b)

is a consequence of 1.1a) and 1.1b). c) From 1.1e) we obtain 0 = d(r + r)  d(r + r0 ) = (dr + dr0 )  (dr + dr0 ) = drdr0 + dr0 dr . d) follows from 1.1d), because d is linear. e) Since the multiplication on  is bilinear, we may assume that !m = rdr1 : : : drm and !n = sds1 : : : dsn (r r1  : : :  rm  s s1 : : :  sn 2 R). From 1.2c) we conclude that !m  !n = rsdr1 : : : drm ds1 : : : dsn = rs  (;1)mn ds1 : : : dsndr1 : : : drm = (;1)mn !n  !m . The second statement of e) follows from 1.3.Lemma. Let I be a homogeneous left (right) ideal in a graded anticommutative alge-

bra A . Then I is a two-sided ideal of A . P P If A  I  I , then also I  A  I . To show this, let xm 2 I , rn 2 A with xi  ri 2 Ai be given. Since I is homogeneous we have xi 2 I for all i . Then

P

P

P P

( xm )( rn) = (

 m+n=

P P

xm rn ) = (

(;1)mnrn xm )

 m+n=

and this element belongs to I since rn xm 2 I for all n m by assumption. f) Since d is R0 {linear and the multiplication on  is bilinear, we may assume that !m = rdr1 : : : drm , ! = sds1 : : : dsn (r r1  : : :  rm  s s1  : : :  sn 2 R). By 1.1c) and d) we have

d(!m !) = d(rsdr1 : : : drm ds1 : : : dsn) = (rds + sdr)dr1 : : : drm ds1 : : : dsn = (drdr1 : : : drm )(sds1 : : : dsn) + (;1)m (rdr1 : : : drm )(dsds1 : : : dsn)= d!m  ! + (;1)m !m d! g) For ! = rdr1 : : : drm we get (d  d)(!) = d(1  drdr1 : : : drm ) = d(1)  drdr1 : : : drm = 0 since d(1) = P 0 by 1.9a). P h) For ! = !n (!n 2 n) we have d! = d!n = 0 if and only if d!n = 0 for all n 2 N , since d is of degree 1. Hence Z () is a graded subgroup of ( +). Moreover, for ! !0 2 Z () one can see from 1.2f) that !  !0 2 Z (), hence Z () is a graded subring of . 165

j) From d  d = 0 we conclude that B()  Z (). Obviously B() is a graded subgroup of (Z () +). By lemma 1.3 it is enough to show Z ()  B()  B(). But for !m 2 Z (), d!n 2 B() (!m 2 m !n 2 n) we have by 1.2f)

!m  d!n = (;1)m d(!m  !n) + (;1)m+1 d!m  !n = d((;1)m !m  !n ) 2 B() 1.4.Definition. The elements of Z () are called closed elements of , those of B ()

are called exact. The graded algebra

HDR() := Z ()=B() is called the cohomology of  (deRham-cohomology). L HDR n () where H n () = Z n ()=B n () with Z n () := We have HDR() = DR n2 Z Z () \ n , Bn() := B() \ n . The algebra HDR () is an important invariant of the di erential algebra  (and of the algebra R=R0 ), and it is an essential tool of singularity theory. For its role in the cohomology theory of algebraic varieties we refer to Gr2 ], Ha2 ], Ha3 ] and Mon]. 1.5.Examples of differential algebras:

a) The trivial dierential algebra of a ring R . It is given by  = 0 = R , n = 0 for n > 0. Obviously d has to be the zero map. b) The algebra of C 1 {dierential forms on a C 1 {manifold. Let R be the ring of C 1 {functions f : X ! R on a C 1 {manifold X and  the algebra of C 1 {di erential forms on X . Let d :  !  be the Cartan di erentiation of di erential forms. Then with R0 := R the pair ( d) is a di erential algebra of R=R0 in the sense of denition 1.1, as is shown in analysis. Poincare's lemma states that for a contractible n () = 0 for n 1. manifold HDR c) Let R=R0 be an algebra and d : R ! M an R0 {derivation of R into an R {module M with M = RdR . Let  := R n M be the ring of dual numbers (see I,x 2), and set 0 := R , 1 := M , i := 0 for i > 1. Extend the given derivation d : 0 ! 1 by zero to . Then ( d) is a di erential algebra of R=R0 . We call ( d) the dierential algebra associated with the derivation d . This example, together with 1.2a), shows the close connection between di erential algebras and derivations of R=R0 . d) Let R0 be a ring, R := R0X1  : : :  Xn ] a polynomial algebra over R0 , and let 1 := RdX1   RdXn V the module of formal di erentials of R=RV0 (I.1.4). Consider the exterior algebra  := 1 of the R {module 1 , andPset p := p1 . Then any ! 2  may be written uniquely as ! = !p with

!p =

P

11 0, then H p ( a) If HDR DR RX ]) = 0 for p > 0. p ( b) HDR R0 X1 :::Xn ]=R0 ) = 0 for p > 0. 3) Let R0 be a noetherian ring and R=R0 a smooth algebra of nite type which is equidimensional of dimension d . Then dR=R0 is isomorphic to an invertible ideal I (R) of R and nR=R0 = 0 for n > d . (The class of I (R) modulo principal fractional ideals in the Picard group of R is called the canonical ideal class of R=R0 .) 4) Let ( d) be a dierential algebra of an algebra R=R0 and M an R0 {module. The tensor product of R0 {algebras (in the category of associative algebras)

V

V

V

A :=  R0 M =  R (R R0 M ) =  R (R R0 M ) is called the algebra of M {valued forms with respect to . L Apq with Apq := p R0 Vq M . a) We have A = pq N 2

198

V V b) For ! 2 p , ! 2 p ,  2 q M ,  2 q M 0

0

0

0

(!  )(!   ) = !!   = (;1)pp +qq (!   )(!  ) 0

0

0

0

0

0

0

0

c) The dierentiation d of  extends to a map d : A ! A with d(!  ) = d!   for V ! 2 ,  2 M satisfying the rules

d(r!) = rda + (df  1)a for r 2 R a 2 A d(a a ) = da a + (;1)p ada for a 2 Apq  a 2 A 0

0

0

0

d) If M = Rn there is a welldened map

V  ] :  R n Rn ! 

(!  a1 ^ ^ an 7! det(a1  : : :  an )!)

where ! 2 , a1 : : :  an 2 R . It satises the rules da] = da] (!  1) a] = ! a]

V

for a 2  R nRn V for ! 2   a 2  R nRn

199

Introduction to the bibliography III In algebraic geometry similarly as in complex analytic geometry dierential forms are important objects of the duality theory and residue theory forged by Serre Se1 ], Grothendieck Gr1 ], Hartshorne Ha1 ], and complemented by many others. The dualizing sheaf of a proper scheme map f : X ! Y which is smooth and equidimensional of dimension d can be identi ed with the sheaf dX=Y . But also in the singular case there is in many situations a canonical sheaf of highest order rational dierential forms which is dualizing, the sheaf of regular dierential forms. It was studied under local and global aspects by many authors (f.i. HKR], H u2 ], H uK1 ], H uSa], H uSe], H uY], H uSu], Ke1 ], KeS], Ku9 ], Ku10 ], KW], Li3 ], LiSa]). The notion of trace of differential forms for nite coverings is relevant for the description of regular dierential forms and for the proof of the duality theorem, see H u1 ], KK], Ku5 ], Ku13 ], Li4 ], SS4 ] for this notion, which is a generalization of the classical trace mapping of nite projective algebras. Dierential forms can also be used in the construction and explicit description of fundamental classes of cycles on schemes (El1 ], Li3 ]). Residues are local invariants associated with a dierential form at a point of a scheme, generalizing the residues of dierentials on curves and Riemann surfaces. Numerous algebraic constructions of residues are known in various contexts, and their relations are investigated (Bei], EZ1 ], GH], Ho], HoLi], H u4 ], H uK2 ], KU1 ], KU2 ], Ku11 ], Ku14 ], Li4 ], LiSa1 ], Lo], Pa], SS6 ], T], Ye1 ]). The residue theorem is a global relation among residues generalizing the classical theorem on curves and having many geometric applications. The relevance of deRham cohomology for local and global questions was already mentioned in the text. The torsion of A=K for ane and analytic algebras A over a eld K of characteristic 0 has been studied f.i. by Ara], H u3 ], Ve1 ]. If A is a reduced hypersurface algebra which has only isolated singularities, then the torsion of pA=K vanishes except possibly for p = n ; 1 or p = n where n ; 1 = dim A (L]). For the discussion of the ;1 in this case, see Mi1 ]-Mi3 ]. minimal number of generators of the torsion of nA=K Dierential forms will show up again in the next chapter where algebraic versions of concepts and results from dierential geometry will be given.

200

Chapter IV. Connections x 1. Connections on Modules In dierential geometry connections are an important tool for studying dierential manifolds and in particular Riemannian manifolds. We introduce here the analogous algebraic notions. The discussion is quite similar as for dierential manifolds though neither of the two theories is an immediate specialisation of the other. For example localization in algebra is dierent from localization on dierential manifolds.

Let S=R be an algebra, M an S {module. For example we may have R = R and we can take for S the R {algebra of C 1 {functions on a dierential manifold. 1.1.Definition. A connection on M is an R {bilinear map

r : DerR S  M ! M which as a function on DerR S is S {linear and as a function on M satises the (product)rule

r(V fm) = f  r(V m) + V (f )  m for V 2 DerRS f 2 S m 2 M We sometimes write rV m for r(V m) and call rV m the covariant derivative of m with respect to V . Equivalently r is an S {linear map DerR S ! EndRM satisfying the product rule. An important special case is M = DerRS .

If r and r0 are two connections on M , then r;r0 is S {bilinear. Conversely, for any connection r on M and any S {bilinear map  : DerRS  M ! M the sum r +  is a connection on M . Thus the set of connections on M is either empty or can be identied with HomS (DerR S S M M ). For the relation of connections to D {modules see exercise 3) below. 1.2.Examples:

a) This example indicates a relation between systems of linear dierential equations and connections. Let S = K

t]] be the algebra of formal power series in a variable t over a eld K . Then2 Der3K S = S  dtd . Let A 2 M (r  r S2) be1 an 3 r  r {matrix with entries df 1 f dt . 7 6 6 df r . in S , let f = 4 . 5 2 S be a column, and set dt = 4 ... 75 . Then the system of linear df r fr dt dierential equations dfdt = A  f gives rise to a connection: Dene

r : DerK S  S r ! S r 201

by

r( dtd  f ) =  df dt ; A  f

(for each  2 S f 2 S r )

Then for  2 S and V =  dtd r(V f ) =  d(dtf ) ; A  (f ) = V ( )f + V (f ) ; A  f = V ( )f +  r(V f ) The solutions f 2 S r of the dierential equation dfdt = A  f are exactly the elements with r(V f ) = 0 for each V 2 DerK S . b) Suppose M = S R N with an R {module N . Then the map r : DerRS  M ! M with r(V f  n) = V (f )  n for V 2 DerR S f 2 S n 2 N is a connection on M , called the constant connection with respect to N . We have L this situation, if M has a basis B = fbg2 , because then M = S R N with N = Rb . 2 In this case we will speak of the trivial connection with respect to the basis B . For S = R X1  : : :  Xn ] and M = DerRS let r be the trivial connection with respect to n P ( @X@ 1  : : :  @X@ n ). Then the covariant derivative of W = f i @X@ i (f i 2 S ) with resprect i=1 to V 2 DerRS is given by

@ (coordinatewise application of V ) rV W = P V (f i ) @X n

i

i=1

c) (Divergence of vector elds). Let r : DerR S  DerR S ! DerR S be a connection on DerRS . Assume 1S=R is nitely presented and locally free (hence projective). Then so is DerRS . For V 2 DerR S the function r(; V ) is an endomorphism of DerRS . Its trace divr V is called the divergence of V with respect to r . For S = R X1  : : :  Xn] and the trivial connection with respect to ( @X@ 1  : : :  @X@ n ) the n P divergence of the vector eld V = f k @X@ k (f k 2 S ) is given by the usual formula k=1

divr V = P

n @f k P

k=1 @Xk

@fk  @ (i = 1 : : :  n). since r( @X@ i  V ) = @X k i @Xk r L If r is a connection on a free S {module M = Sbi we can write for each V 2 DerR S i=1

(1)

r

rV bi = P !ki(V )  bk k=1

202

(!ki (V ) 2 S )

Then !ki as a function of V is S {linear, therefore it can be regarded as an element of the module (DerR S ) = (1S=R ) of Zariski dierentials. If 1S=R is reexive the !ki may be looked at as Kahler dierentials. The !ki 2 (DerRS ) are called the connections forms and ! = !ki ]ik=1:::n is called the connection matrix of r with respect to the basis B = (b1  : : :  br ). Clearly the trivial connection with respect to Br has connection matrix 0. P In the situation of (1) let m = f i bi (f i 2 S ). Then i=1

r r rV m = P f i  !ki (V )  bk + P V (f k )  bk

(2)

ik=1

k=1

Hence r is uniquely determined by its connection matrix ! . Conversely for an arbitrary r  r {matrix ! = !ki] with !ki 2 (DerRS ) formula (2) denes a connection on M with connection matrix ! . Thus 1.3.Proposition. On a free module of rank r the connections are in one-to-one corre-

spondence with the r  r {matrices whose entries are elements of (DerR S ) .

Suppose in addition that DerR S has a basis (V1  : : :  Vn), too, and let (1  : : :  n ) be the dual basis of (DerRS ) , i.e. i (Vk ) = ik (i k = 1 : : :  n). Then the connection forms can be written (3)

!ki =

n k j P ; ij 

j =1

(;kij 2 S i k = 1 : : :  r j = 1 : : :  n)

The ;kij are called Christoel symbols of r with respect to B and (V1  : : :  Vn ). They determine r according to the formula (4)

P j P i PP i j k P i k  r( g Vj  f bi ) = f g ; ij + g Vi(f )  bk j

i

k ij

i

For a connection r on an S {module M let M r be the set of all m 2 M such that rV m = 0 for all V 2 DerR S . Such elements are called horizontal sections, and M r is an R {submodule of M . Suppose M has a basis B = (b1  : : :  br ), and let !ki ]ik=1:::r be r P the connection matrix of r with respect to B . Then m = f i bi (f i 2 S i = 1 : : :  r) i=1 is a horizontal section if and only if f 1  : : :  f r satisfy the \dierential equations" 2 13 2 13 f Vf 6 64 .. 7 5 5 = ; ! k (V )] 4 .. 7

. V fr

i

203

. fr

for all V 2 DerR S . If moreover DerR S has a basis (V1 : : :  Vn) with dual basis (1  : : :  n ) and ;kij 2 S (i k = 1 : : :  r j = 1 : : :  n) are the Christoel symbols with respect to these bases, then m is a horizontal section of r if and only if 2 13 2 13 Vj f f .7 64 .. 75 = ; ;k ]  6 ij 4 .. 5 .

Vj f r

(j = 1 : : :  n)

fr

A connection is said to have suciently many solutions, if M r generates M as an S {module. In the situation of example 1.2a) the horizontal section of r are the solutions of the dierential equation dfdt = A  f . If Char K = 0, then each initial value problem for this dierential equation 2 1 3 2 13 f (0) a 64 .. 75 = 4 ... 5

(a 1  : : :  a r 2 K ) . ar f r (0) has a unique solution, hence the solution space of the dierential equation is a K {vector space of dimension r . A matrix whose columns form a basis of the solution space has a determinant which is a unit of K

t]], hence the columns form even a basis of S r . Thus r has suciently many solutions in this case. To the contrary, if Char K = p > 0 the dierential equation dfdt = f has only the trivial solution in K

t]].

Let again M be an arbitrary S {module and r a connection on M . Then for V 2 DerR S , m 2 M and a unit " of S we have a quotient rule (5)

r(V ";1  m) = ";1  r(V m) + V (";1 )  m = ";2 ("  r(V m) ; V (")  m)

Let N  S be a multiplicatively closed subset and let

: DerR S ! DerRSN   : M ! MN be the canonical maps (see Chap.I.1.17). 1.4.Proposition. Suppose induces an isomorphism (DerR S )N ;e! DerR SN . Then

there is exactly one connection rN on MN such that the following diagram commutes DerRS  M ;



r

M 

DerR SN  MN rN MN 204

Proof: If V

2 DerRS then VN := (V ) is given by the formula

VN ( s ) = 12 (V (s) ; sV ( )) for s 2 SN a) If rN exists, then for V 2 DerR S and m 2 M we have by (5) rN (VN  m ) = 12 ( rN (VN  m1 ) ; VN ( 1  m1 ) = 12 (  r(V m) ; V ( )  m)

and hence rN is unique, since it is SN {linear on DerR SN . b) In order to prove existence consider V N 2 DerRSN with V 2 DerR S and m 2 MN . Dene rN ( V N m ) by the formula

rN ( VN  m ) = 1 2 (  r(V m) ; V ( )  m) This expression clearly does not depend on the special respresentation of the fraction V N . Assume m = m , hence that there is a  00 2 N with  00  0m =  00 m0 . It suces to show 0

0

(6) We have and

1 (  r(V m) ; V ( )  m) = 1 ( 0  r(V m) ; V ( 0 )  m0) 2 2 0

 00 r(V  00  0m) =  002  0 r(V m) +  002 V ( 0 )m +  00  0 V ( 00 )m  00 r(V  00 m0) =  002  r(V m) +  002 V ( )m0 +  00 V ( 00 )m0

hence

 0  002 ( 0 r(V m) + V ( 0 )m) =  0  002 ( r(V m0 ) + V ( )m0 ) from which (6) easily follows. It is clear that rN is SN {linear on DerR SN and makes the diagram of the proposition commute. Further for f 2 SN , m 2 MN rN (VN  f0 m ) = (10 )2 ( 0  r(V fm) ; V ( 0 )  fm) = f0  12  (  r(V m) ; V ( )  m) + 1 0  V (f )  m ; (f0 )2  V ( 0 )  m = f0  12 (  r(V m) ; V ( )  m) + V ( f0 )  m     m f m f =  0  rN (VN   ) ; V (  0 )   0

205

hence rN is a connection on DerRSN Under the assumptions of the proposition we call rN the localization of r with respect to N and we write rf , if N = f1 f f 2  : : : g for f 2 S resp. rp , if N = S n p for p 2 Spec S . Sometimes we set X := Spec S and write rx instead of rp if x 2 X is the point corresponding to p . Then rx is called the localization of the connection at x . 1.5.Proposition. (Local-global principle for connections). Let 1S=R and M be nitely

presented and let f1  : : :  fn 2 S be given such that S = (f1  : : :  fn ). Then for two connections r1 r2 on M , the following assertions are equivalent: a) (r1)fi = (r2)fi for i = 1 : : :  n . b) (r1)p = (r2)p for all p 2 Spec S . c) r1 = r2 . Proof: Since

r1 ; r2 can be regarded as an element of HomS (DerRS S M M ) the

claim is a consequence of the local-global-principle for linear maps.

It follows that connections of nitely presented locally free S {modules can be described locally by connection matrices and, if DerR S is locally free, also by Christoel symbols. Moreover connections which \t together" can be pasted: 1.6.Proposition. (Pasting of connections). Under the assumptions of 1.5 let connections

ri on Mfi be given (i = 1 : : :  n) such that (ri)fj = (rj )fi

for i j = 1 : : :  n

Then there is exactly one connection r on M such that rfi = ri for i = 1 : : :  n . Proof: For V

2 DerR S , m 2 M let Vi 2 DerR Sfi and mi 2 Mfi be their images by the

canonical maps. The canonical diagram

ri

DerRSfi  Mfi 

r)

Mfi 'ij

i fj DerRSfi fj  Mfi fj Mfi fj commutes for i j = 1 : : :  n . Since (ri)fj = (rj )fi we obtain (

'ij (ri(Vi  mi)) = 'ji (rj (Vj  mj ))

(i j = 1 : : :  n)

Therefore the elements ri(Vi  mi) (i = 1 : : :  n) uniquely determine an element r(V m) in M with image ri(Vi  mi) in Mfi for i = 1 : : :  n . We thus have a canonical map 206

r : DerR S  M ! M for which DerRS  M



r

M

DerR Sfi  Mfi ri Mfi commutes (i = 1 : : :  n). One easily checks that r is a connection on M with rfi = ri (i = 1 : : :  n). Uniqueness follows from the local-global-principle 1.5. Consider the localization rp of a connection r : DerR S  M ! M at a point p of Spec S =: X . Let Y := Spec R so that X is a scheme over Y . Assume 1S=R is nitely presented and 1Sp =R a free Sp {module. Then rp induces an R {bilinear map

r( p) : DerR Sp = pDerRSp  Mp ! M ( p ) where M ( p ) := Mp = pMp . Further we have canonical isomorphism DerR Sp = pDerR Sp = HomSp (1Sp =R  Sp )= p HomSp (1Sp =R  Sp ) = TX=Y ( p) Hence r( p) can be regarded as an R {bilinear map

r( p ) : TX=Y ( p )  Mp ! M ( p ) Since it is induced by rp it satises the rule (7)

r( p )( fz) = f ( p )  r( p)( z) + (f )  z( p )

for any tangent vector  2 TX=Y ( p) and \germs" f 2 Sp , z 2 Mp . We sometimes write r( p) (z) instead of r( p )( z) and call this element the directional derivative of z with respect to  . Clearly (8)

r( p)V (p)m = (rV m)( p ) for V 2 DerRS and m 2 M

Intuitively the connection r measures the rate of change of m at p in the direction of V ( p ). If S is reduced and M locally free, then it can be easily shown that r(V m) is uniquely determined by the set of its directional derivatives fr( p)V (p) mgp2X . In case S is the R {algebra of the C 1 {functions on a dierential manifold X and P 2 X , then each connection r : DerRS  DerRS ! DerRS induces a connection rP : DerREP  DerREP ! DerREP , where EP is the algebra of germs of C 1 {functions 207

at P . Here rP is compatible with the canonical map DerRS ! DerREP and induces an R {bilinear map r(P ) : TP X  DerRS ! TP X satisfying a rule analogous to (7) and associating with a vector eld V 2 DerRS and a tangent vector  2 TP X the \directional derivative" r(P )( V ) 2 TP X of V in direction of  . In the algebraic case let S := S=I be a residue class algebra of S and X := Spec S . We have the canonical maps

: DerRS ! DerR (S S) and  : DerR S ! DerR (S S ) where associates to each vector eld on X the tangential vector eld along X and  to each vector eld on X its restriction to a vector eld along X (see III,x 4). In the following we assume that 1S=R is nitely generated and projective, hence  is surjective, i.e. each vector eld along X is the restriction of some vector eld on X . 1.7.Proposition. For V

2 DerRS and W 2 DerR(S S ) choose V W 2 DerR S with

images V and W by  . Then

r : DerRS  DerR (S S ) ! DerR (S S ) with r(V  W ) =  (r(V W )) is a well-dened map and a connection on the S {module DerR(S S ). For each p 2 X with preimage p 2 X we have

r( p )V (p)W = r( p )V (p)W Proof:  (r(V W )) depends only on V since

r(V W ) is S {linear in V . Suppose 0 W 2 DerRS is another representative of W . Write W ; W0 =

m i P f Vi with f i 2 I Vi 2 DerR S (i = 1 : : :  m)

i=1

P

P

P

Then  (r(V W ; W 0 )) =  ( f i r(V Vi ) + V (f i )  Vi ) = V (f i )  Vi where the bar i i i denotes canonical images in S resp. DerR (S S ). Since V 2 DerR S = HomS (1S=R  S ) is mapped onto V 2 DerR S = HomS (1S=R  S )  HomS (1S=R  S) the diagram

S

V

S

S

V

S



208

is commutative, hence V (f i ) = 0 (i = 1 : : :  n) and  (r(V W )) only depends on W , too. It is easy to show that r is a connection on DerR (S S ). By (8) we have

r( p )V (p)W = (rV W )( p ) = (rV W )( p ) = r( p)V (p) W

q.e.d. Sometimes a dierent notion of connection is useful. Let S=R and M be as before and let ( d) be a dierential algebra of S=R . We consider  S M as a left {module. In case S is the R {algebra of C 1 {functions on a dierential manifold ( d) can be taken to be the algebra of C 1 {dierential forms on the manifolds (III. 1.5b). 1.8.Definition. A connection

on M with respect to  is an R {linear map re :  S M !  S M

with the following properties: e = Lr e p with R {linear maps a) re is homogeneous of degree 1, e.i. we have r p2N

re p : p S M ! p+1 S M .

b) For ! 2 p , m 2 M the product rule

re p(!  m) = d!  m + (;1)p!  re 0m holds. In particular for f 2 S , m 2 M (9)

re 0(f  m) = df  m + f  re 0m

and for ! 2 p , !0 2 q , m 2 M (p q 2 N) (10)

re p+q (!0 !  m) = d!0  (!  m) + (;1)q !0 re p(!  m)

e with respect to  in Similarly as in 1.2a) we have the notion of constant connection r case M = S R N with an R {module N . It is given by

(11)

re (!  n) = d!  n

(! 2  n 2 N )

Let us call for the moment a connection in the sense of denition 1.1 a geometric connection. 209

Assume that the canonical map

: HomS (1  S ) ! DerRS

(` 7! `  d)

is an isomorphism. The canonical S {bilinear map h  i : 1  DerRS ! S with h V i = `V () for  2 1 , V 2 DerR S where `V  d = V extends to an S {bilinear map

h  i : (1 S M )  DerR S ! M with h  m V i = h V i  m for m 2 M . In particular for  = df with f 2 S we have (12) hdf  m V i = V (f )  m e on M with respect to  induces then a geometric connection A connection r r : DerR S  M ! M as follows: For V 2 DerR S and m 2 M set (13) rV m := hre 0m V i Clearly r is R {bilinear and as a function of V even S {linear. Moreover for f 2 S rV (fm) = hre 0(fm) V i = hdf  m + f re 0 m V i = V (f )m + f  rV m hence r is a geometric connection on M . We say that it is associated to re . Clearly it e0. depends only on r Assume that in addition the scalar product h  i : 1  DerR S ! S induces an isomorphism

 : 1 S M ! HomS (DerR S M ) with  (  m) = h  m ;i for  2 1 , m 2 M . This is the case, for example, if 1 is a e 0 m 2 1 S M to be the unique element with nite projective S {module. Dene r (14)

hre 0m V i = rV m

(m 2 M V 2 DerRS )

Then re 0 : M ! 1 S M is R {linear and for f 2 S

hre 0 (fm) V i = rV (fm) = V (f )  m + f  rV m = hdf  m V i + f  hre 0 m V i = hdf  m + f  re 0m V i hence (15)

re 0(fm) = df  m + f  re 0m 210

e 0 to all of  S M . For p 2 N the map b) Now we extend r ' : p  M ! p+1 S M e 0m (! 2 p  m 2 M ) is R {bilinear, hence it induces with '(!  m) = d!  m + (;1)p !r an R {linear map  : p R M ! p+1 S M An easy calculation shows that (f!  m ; !  fm) = 0 for f 2 S , ! 2 p , m 2 M , therefore an R {linear map re p : p S M ! p+1 S M with (16) re p(!  m) = d!  m + (;1)p!re 0m e := L r e p is the desired extension of r e 0 and it is a connection on M is induced. Then r p2N with respect to  with which r is associated. Thus we have shown 1.9.Proposition. Assume that DerR S = HomS (1 S ) and the natural bilinear map

h  i : 1  DerR S ! S induces an isomorphism  : 1 S M ! HomS (DerR S M ). Then the geometric connections r : DerR S  M ! M are in one-to-one correspondence with e :  S M !  S M on M with respect to . the connections r In the following we use the canonical S {bilinear map (discussed in Chap. III,x 4) h  i : p  VpDerR S ! S with h1 ^    ^ p V1 ^    ^ Vpi := det hi Vj i]ij=1:::p for i 2 1 , Vi 2 DerR S (i = 1 : : :  p). It induces an S {bilinear map h  i : (p S M )  VpDerR S ! M V with h!  m i = h! i  m for ! 2 p ,  2 pDerRS and m 2 M .

e be a connection on M with re1.10.Proposition. Under the assumptions of 1.8 let r

spect to  where DerRS = HomS (1  S ) and let r be its associated geometric connection. Then for  2 p S M and V0  : : :  Vp 2 DerR S the following formula holds hre p V0 ^    ^ Vpi = p X

X (;1)i rVi (h V0 ^  ^V^i^  ^Vpi)+ (;1)i+jh Vi  Vj ]^V0^  ^V^i ^  ^V^j ^  ^Vpi

i=0

0

i . Since DerRS is Symmetry: r(g V W )> ; r(g W V )> = (r(g V W ) ; r(g W V ))> = V gRS commutes with the brackets we a Lie-algebra with respect to   ] and DerRS ! Der have VgW ]> = Vg W ] = V  W ] Compatibility: Let Z 2 DerR S be represented by Z 2 DerR S . Then with obvious notation ~ >Z W ) + g(W  r~ Z>V ) = g(V  (rg g (V  r Z W )> ) + g (W  (rg Z (W )> ) = g~(V  rg Z W ) + g~(W  rg Z V ) = g (V rZ W ) + g (W rZ V ) = Z (g(V W )) = Z (g(V  W ))

where we have used that g(V  (rg Z W )> ) = g~(V  rg Z W ) since V ?(rg Z W )? . 242

4.8.Examples: a) Let S = RX1  : : :  Xn ], and assume that S = S=I is smooth over R .

Let g be the standard metric on X := AnR and g the metric induced by g on X := Spec S . Then the Levi-Civita connection r of (S=R g) acts on vector elds V  W 2 DerRS as follows: Choose vector elds V W 2 DerRS which restrict to V  W . Take the directional derivative rV W of W with respect to V (i.e. apply the constant connection) and restrict it to X . Then take the tangential component of this restriction. The result is r(V  W ). b) For (S=R g) as in example 3.16b) set i := dxi and Vi := @x@ (i = 1 : : :  n ; 1). Write nP ;1 d'i = 'ik k (i = 1 : : :  n ; 1) with i

k=1

'ik

=

@'i @xk

=

@ @xk



f ; xi fxn



=;

fxi xk fx2n

; fx fx fx x ; fx fx fx x + fx fx fx x i

n

k n

f3

k

n

i n

i

k

n n

xn

Since gij = ij + 'i 'k Christoel's equation (2) yields ;ijk = 21

 @ (' ' ) @ (' ' ) @ (' ' )  j k ; i k + i j ='' i ik (i j k = 1 : : :  n ; 1) @x @x @x i

j

k

and for the connection forms of the Levi-Civita connection we obtain !ij

= 'id'j

(i j = 1 : : :  n ; 1)

c) For the (n ; 1){sphere S := Spec RX1  : : :  Xn ]=(f ) with f = 'j

= ; xj

xn

and !ij

 'jk

n 2 P X ; 1 we have

i=1

i

j = ; 1 (x x +  x2 ) = ; 1 g = @' j k ik n jk @x x3 x k

= 'id'j = xx2i

n

n

nX ;1

n k=1

gjk k

= xi G 

nX ;1 k=1

gjk k

gR S ! Der gR S be as above. Set Under the assumptions 3.13 let r~ : DerR S  Der X := Spec S , X := Spec S , and assume that g is non-degenerate. 4.9.Definition. The map

: DerR S  (DerR S)? ! DerR S (V  N ) 7;! ;(r~ V N )> is called the second fundamental form of X X . For a xed N 2 (DerR S)? the map II

~ V N )>) SN : DerR S ! DerR S (V 7! ;(r 243

is called shape operator (or Weingarten map) with respect to

N.

Clearly SN is S {linear since r~ V N is S {linear in V and so is the projection to (DerR S)> . The form II is S {bilinear: For N 1  N 2 2 (DerR S)> , V 2 DerRS and '1  ' 2

2S

~ V ('1 N 1)> ; r~ V ('2 N 2)> II (V  '1 N 1 + '2 N 2 ) = ;r

=

~ V N 1)> ; V ('2)  ; '2  (rV N 2 )> = ; '1  (r ; V ('1 )  '1  II (V  N 1 ) + '2 II (V  N 2 ) since N i = 0 (i = 1 2) : > N1

> N2 >

The shape operator with respect to `N

N

denes an S {bilinear map

: DerR S  DerR S ! S ((V  W ) 7! g(SN V  W ))

which uniquely determines SN since g is non-degenerate. 4.10.Proposition.

a) `N is symmetric: `N (V  W ) = `N (W  V ). Moreover `N (V  W ) = g~(r~ V W  N ). b) Let V W 2 DerR S be representatives of V  W 2 DerR S and let ' 2 I . Then for N := Grad ' (which is in (DerR S )? by 3.14) `N (V  W ) = ;(hess ')(V W )

Proof: Let N 2 DerR S be a representative of N .

a) `N (V  W ) = g(;(r~ V N )> W ) = ;g~(r~ V N  W ) = 4:1b) ~ V W N) ;g (rV N W ) = g (N rV W ) ; V (g (N W )) = g (r~ V W  N ) ; V (~g (N  W )) = g~(r since N ?W . Further by 4.1a) ~ V W  N ) = g(N V W ]) + g(N rW V ) = g~(r~ W V  N ) g (r

which proves the symmetry of `N . b) `Grad ' (V  W ) = g(;(r~ V Grad ')> W ) = ;g(rV Grad ' W ) = ;(hess')(V W ). The symmetric tensor IIN 2 1 S 1 corresponding to `N is called the 2nd fundamental tensor of X 1X with respect to N . Suppose I=I 2 and  have a basis as S {modules. Then 1 has a basis ( 1  : : :  n ) such that for the dual basis (V1  : : :  Vn) of DerR S the residues (V~1  : : :  V~m ) form a basis of DerR S and (V~m+1  : : :  V~n ) a basis of (DerR S)? . Write N

=

n X k=m+1

k V~k

244

( k 2 S )

let !ji ]i=1:::m j=1:::n be the connection matrix of r~ with respect to (V~1  : : :  V~m ) and g

=

m X k`=1

gk`  k

 ` (g k` 2 S )

the fundamental tensor of S=R where k is the image of k in 1 . The second fundamental tensor and the shape operator can be expressed in terms of these bases as follows: 4.11.Proposition. m

a)

IIN

=

P

ij =1

bij (N )  i  j

with bij (N ) := ;

n P

s=m+1

s !js (V~i ).

b) The matrix of SN with respect to (V~1  : : :  V~m ) is gij ]  bij (N )] Proof: a) We have

~ V~i N )> V~j ) IIN (V~i  V~j ) = `N (V~i  V~j ) = ;g~((r

= ;g~((r~ V~ (

n X

i

s V~s ))>  V~j ) =

s=m+1 n n X X s s t s ~ ~ ~ ~ ~ ; g~( Vi ( )  Vs + ! s (Vi )  Vt  Vj ) = ; gtj ! ts (V~i ) = s=m+1 t=1 s=m+1 t=1 n X s ;  !js (V~i ) = bij (N ) s=m+1 n X

m X

b) Similary ~ V~ N )> = ; SN (V~i ) = ;(r i

n X s=m+1

s

n X t=1

! ts (V~i )  V~t

=

m X

n X

t=1

s=m+1

(;

s  ! ts (V~i ))  V~t

from which b) follows. Exercises:

Let (S=R g) be a Riemannian algebra. Assume S=R has a dierential basis fx1  : : :  xng . In the following exercises the fundamental matrix, the Christoel symbols and connection forms are taken with respect to the basis B = fdx1  : : :  dxn g . 245

1) Formula for the connection forms: n @g X 1 jk ; @gik )dxk ) !ij = (dgij + ( i 2 @xj k=1 @x a1  : : :  an

2) Formula for the divergence of a vector eld: For div(

n X j =1

aj

@ @ai ) = @xj @xi

+

n X j =1

2S

aj ;iji

3) Formula for the Laplace operator: For f 2 S f =

n  X

it=1

 n @g it X jt i @f @ 2f it g + ( @xi + g ; ji ) @xt @xi @xt j =1

4) Formula for the Hesse form: For f 2 S 2f (hessf )( @x@ i  @x@ k ) = @x@i @x k

n jt @f X t + (; ik + gjk @g i ) @xt @x t=1 j =1

246

n X

x 5. Curvature of Riemannian Algebras The curvature of the Levi-Civita connection on a Riemannian algebra or a (pseudo-) Riemannian manifold leads to the curvature tensor and various other notions of curvature which will be introduced and studied in this section.

Let a Riemannian algebra (S=R g) be given such that 2 is a unit in S and let r : Der S  Der S ! Der S be its Levi-Civita connection. For example we can take R = R and S = E (X ), the algebra of C 1 {functions of a pseudo-Riemannian manifold X . R

R

R

5.1.Definiton. The curvature

curvature of (S=R g).

R := R(r) of the Levi-Civita connection is called the

By de nition 2.1 it is the element

R 2 Hom (Der S Hom (V2Der S Der S )) S

with

R

S

R

R(V )(W ^ Z ) = r  r ]V ; r W

Z

WZ

]

R

V (V W Z 2 Der S ) R

The following proposition holds for any connection with torsion 0. 5.2.Proposition. For V W Z 2 Der S R

R(V )(W ^ Z ) + R(W )(Z ^ V ) + R(Z )(V ^ W ) = 0 Proof: r  r ]V

; r ] V + r  r ]W ; r ] W + r  r ]Z ; r r (r V ) ; r (r V ) ; r ] V + r (r W ) ; r (r W ) ; r ] W + r (r Z ) ; r (r Z ) ; r ] Z = r (r V ) ; r (r W ) ; r W V ] ; r ] V + r (r W ) ; r (r Z ) ; r Z W ] ; r ] W + r (r Z ) ; r (r V ) ; r V Z ] ; r ] Z = r V W ] ; r ] V + r W Z ] ; r ]W + r Z V ] ; r ] Z = W

W

Z

Z

Z

V

W

Z

Z

W

WZ

W

Z

Z

WZ

W

V

V

V

Z

V

Z

VW

]

Z=

ZV

WZ

Z

V

V

W

V

V

W

W

Z

W

WZ

V

W

VW

Z

V

V

ZV

ZV

VW

W

ZV

VW

Z V W ]] + V W Z ]] + W Z V ]] = 0 by the Jacobi identity. The map K : (Der S )4 ! S (U V W Z ) 7! g(U R(V )(W ^ Z )) is S {linear in its four variables. Assume that 1 is a nitely generated projective S { module. Then K can be regarded as a tensor K 2 T4(1 ). R

247

5.3.Definition.

of (S=R g).

K is called the curvature tensor (or Riemann-Christoel tensor)

5.4.Theorem.(Symmetry properties of the curvature tensor). For U V W Z 2 Der S we

have a) K(U V W Z ) + K(U W Z V ) + K(U Z V W ) = 0. b) K(U V W Z ) = ;K(V U W Z ) = ;K(U V Z W ). c) K(W Z U V ) = K(U V W Z ). If 3 is not a zerodivisor of S , then any tensor K0 2 T4 (1) satisfying a)-c) and

K0 (U V U V ) = K(U V U V )

R

for U V 2 Der S R

coincides with K . Proof: a) follows from 5.2 and K(U V W Z ) = ;K(U V Z W ) is clear by the de nition

of K . Moreover K(V V W Z ) = g(V R(V )(W ^ Z )) = g(V r (r V )) ; g(V r (r V )) ; g(V r( ]V ) 4.1b) = W (g(V r V )) ; g(r V r V ) ; Z (g(V r V )) + g(r V r V ) ; 12 W Z ](g(V V )) = 1 2 (W  Z )(g (V V )) ; (Z  W )(g (V V )) ; W Z ](g (V V ))] = 0 Hence K is alternating with respect to U and V which proves the rst statement of b). Formula c) and the last assertion of the theorem follow from W

Z

Z

Z

Z

W

W

WZ

W

W

Z

5.5.Lemma. Let S be a ring in which 2 is a unit. Let M be an S {module and f : M 4 ! S

a four times linear map satisfying a) f (x y z t) + f (x z t y) + f (x t y z) = 0. b) f (x y z t) = ;f (y x z t) = ;f (x y t z) for x y z t 2 M . Then f (x y z t) = f (z t x y), for x y z t 2 M . If 3 is not a zerodivisor of S and f (x y x y) = 0 for all x y 2 M , then f = 0. Proof: f (x y z t) = ;f (x z t y ) ; f (x t y z ) = a)

b)

a) f (z x t y) + f (t x y z) = ;f (z t y x) ; f (z y x t) ; f (t y z x) ; f (t z x y) a),b) = a) ;2f (t z x y) + f (y z x t) + f (y t z x) = ;2f (t z x y) ; f (y x t z) ; f (y t z x) + f (y t z x) = 2f (z t x y) ; f (y x t z) = 2f (z t x y) ; f (x y z t) hence f (x y z t) = f (z t x y). Suppose now that 3 is not a zerodivisor of S and f (x y x y) = 0 for all x y 2 M . Then for x y z 2 M

0 = f (x + z y x + z y) = f (z y x y) + f (x y z y) = 2f (x y z y) 248

and

0 = f (x y + t z y + t) = f (x y z t) + f (x t z y)

hence

f (x y z t) = f (x t y z) = f (x z t y) Now a) implies that 3f (x y z t) = 0, hence f = 0. Assume 1 has a basis B = (1  : : :   ) and let B = (V1 : : :  V ) be the dual basis of Der S . Regard R(r) as an element of Hom (Der S 2 Der S ) as in 2.6 and write n

n

R

S

R(V ) =

n X

j

 V k

=1

S

R

(j = 1 : : :  n)

k

j

k

R

with the curvature forms  2 2 (j k = 1 : : :  n). Then k

j

K(V  V  V  S ) = g(V  R(V )(V ^ V )) = i

j

r

s

i

j

r

n X

s

n P

ij

k

=1

k

ik

k

With  :=

g  (V ^ V )

=1

j

r

s

g  (i j = 1 : : :  n) we obtain ik

k

j

K(V  V  V  V ) =  (V ^ V )

(1)

i

j

r

s

ij

r

(i j r s = 1 : : :  n)

s

and 5.4b) implies is a skew-symmetric matrix:  = ; (i j = 1 : : :  n) In terms of  ] the relation between curvature forms and connection forms is as follows:

5.6.Corollary.  ]

ij ij

=1:::n

ij

ji

ij

5.7.Proposition. For i j = 1 : : :  n let ! and ! be the connection forms of the rst i j

ij

and second kind of r with respect to B . Then P a)  = d! ; ! ^ ! . n

ij

b) d =

ij

n P

ij

=1

` j

( ^ ! ; ! ^  ).

=1

`

`i

`

i`

` j

` i

`j

Proof: We use the formulas

(2)

 = d! + k

j

k j

n X

`

=1

! ^! k `

` j

and (3)

dg = ! + ! ik

ik

ki

249

(4.5a)

(2.12)

a)  =

n P

ij

k

=1

g  (2) =

n P

k j

ik

=

k

=1

P n

k

=1

g d! + P g n P

ij

= d! ;

k

P

=;

(2)

=

5:6

=1

`

P n

`

=1

`

n P

=1

n P

ij

ij

=1

=1

`

k

=1

! ^! ; k j

ik

! ^! `i

` j

 ^! + ` j

`i

=1

`

P n

=1

`t

n

n P k

=1

`

=1

i`

! ^! + k j

ki

n P k

=1

` j

! ^! ik

k j

` j

n

` j

`i

k `

k j

ik

d! ^ ! + P ! ^ d! `i

=1

`

! ^!

n

k j

ik

= d! ;

a) b) d = ;

k

ik

d(g ! ) ; P dg ^ ! + P ! ^ !

(3)

n

n P

n

k j

ik

` j

! ^! ^! + P ! ^ ; t `

ti

n

` j

=1

` j

`i

`

n P

=1

`t

! ^! ^! ` t

`i

t j

( ^ ! ; ! ^  )

=1

` j

i`

` i

j`

5.8.Proposition. Suppose 1 is a nitely generated projective S {module. Let N

S

be multiplicatively closed and assume that (Der S ) ! Der S is bijective. The curvature R of (S =R g ) is the image of the curvature R of (S=R g) by the canonical map V V Hom (Der S Hom ( 2 Der S Der S )) ! Hom N (Der S  Hom N ( 2 Der S  Der S )) and the curvature tensor K of (S =R g ) is the image of K by the canonical map T4(1 ) ! T4(1 ). R

N

S

N

R

N

R

N

N

S

R

R

N

S

N

R

N

S

R

N

R

N

N

N

V

Proof: Since Der S = (Der S )

V

and 2Der S = ( 2 Der S ) in order to prove the statement about R it suces to show that for V W Z 2 Der S with images V 0  W 0  Z 0 in Der S the derivation R (V 0 )(W 0 ^ Z 0 ) is the image of R(V )(W ^ Z ) in Der S . But this follows from the fact that R

N

R

N

R

N

R

N

N

R

R

R

N

N

N

r

Der S  Der S R

R

;

Der S R

Der S  Der S rN Der S commutes and that V 0  W 0 ] is the image of V W ] in Der S . The statement about the curvature tensor is then clear. Assume that 1 is nitely generated and projective. For V W 2 Der S de ne Ric(V W ) := Trace(U 7! R(W )(U ^ V )) (U 2 Der S ) Since Ric is linear in V and W it de nes a tensor Ric 2 1 1 which is called the Ricci tensor of (S=R g). For (S =R g ) as in 5.6 the Ricci tensor Ric is the canonical image of Ric in 1 N 1 . R

N

R

N

R

R

N

N

R

R

S

N

N

S

N

N

N

250

5.9.Proposition. Ric is a symmetric tensor. If ( 1  : : :   ) is a basis of 1 and n

(V1  : : :  V ) its dual basis, then n

Ric =

n P

=1

(

rs

n P

ij

g K(V  V  V  V ))  ij

=1

i

r

j

r

s

s

R(V )(V ^ V ) = P  V with  2 S . Then K(V  V  V  V ) = g(V  R(V )(V ^ V )) = P  g and

Proof: Write i

r

j

r

j

k

s

k

s

i

j

r

k

k

rjs

rjs

k

s

k



k rjs

P

=

i

ik

rjs

g K (V  V  V  V ) ik

i

r

j

s

On the other hand, by the de niton of Ric Ric(V  V ) = r

P

s

j



j rjs

=

P ij

g K(V  V  V  V ) ij

i

r

j

s

By 5.4c) and the local global principle Ric is a symmetric tensor. The composed map

 g;1 1 Trace :   ;;;;;!  Der S ! S !1 !2 ;7 ! !1  (!2) 7!  (!2)(!1 ) is called the g {trace of 2{tensors. For example in the situation of 5.9 g

1

1

S

id



S

R

g

(4)

P

g

Trace (  ) = ( g V )( ) = g g

i

j

jk

i

k

ij

k

and Trace (g) =

(5)

P

g

g g =n ij

ij

ij

5.10.Definition. := Trace (Ric) is called the scalar g

curvature of (S=R g).

With respect to a basis (V1  : : :  V ) as above it is given by the formula n

(6)

=

n P

ij

=1

ijrs

5.11.Definition. (S=R g ) is called

g g K (V  V  V  V ) rs

i

r

j

s

Einstein algebra or g an Einstein metric, if

Ric =  g for some  S . This notion is analogous to that of an Einstein manifold. 251

5.12.Proposition. Let 1 have constant rank n . If (S=R g ) is an Einstein algebra as

in 5.11 and is its scalar curvature then

=n

Proof: = Trace (Ric) = Trace (  g ) =  Trace (g ) = n  . g

g

The two-dimensional case is of special interest:

g

5.13.Proposition. Let 1 have constant rank 2. Then (S=R g ) is always an Einstein

algebra. If 1 has a basis (1  2 ) with dual basis (V1  V2), and g ] mental matrix of (S=R g) with respect to (1  2 ) then (V1 V2  V1 V2 ) Ric =  g with = Kdet g ] =1 2

ij ij

ij ij

=12

is the funda-



= 12 is independent of the choice of the basis (1  2 ) of 1 . Proof: It suces to prove the second assertion of the proposition. Observe that 

g11 g12  = 1  g22 ;g12  g21 g22 detg ] ;g21 g11 ij

By 5.9 and the symmetry properties of K we have Ric =

2 X 2 X rs

=1 ij =1

g K(V  V  V  V )    ij

i

r

j

s

r

s

= K(V1 V2  V1 V2 )  (g22  1 1 ; g21  1 2 ; g12  2 1 + g11  2 2 )  V2  V1 V2 )  g = K(V1det g ] In the situation of the proposition is called the Gaussian curvature of (S=R g) and is denoted by K . ij

5.14.Corollary. Suppose (V1  V2 ) is an orthonormal basis of Der S and write 12 = R

K   ^  (K 2 S ). Then K is the Gaussian curvature of (S=R g). 1

2

Proof: By (1)

K(V1 V2  V1 V2 ) = 12(V1 ^ V2 ) = K and = 2K follows from 5.12 and

5.14. Let (V1 V2 ) be an orthonormal basis of Der S . Then ! = ! (i k = 1 2) and ! ] is skew-symmetric by 4.5a). Further 5.6c) implies R

i k

d1 = 1 ^ !11 + 2 ^ !12 = 2 ^ !12 d2 = 1 ^ !21 + 2 ^ !22 = ;1 ^ !12 252

ik

ik

Hence, if !12 = 1 + 2 (  2 S ), we have

d1 = ;  1 ^ 2  d2 = ;  1 ^ 2

(7)

If 1  2 are explicitly known  and  can be computed by these formulas, hence !12 can be computed. Then using 5.7 we obtain 12 = d!12 ; (!11 ^ !12 + !21 ^ !22) = d!12

(8) 5.15.Examples:

a) If 1 is projective of rank 1 then, since 2 = 0, the curvature R(r) vanishes and so do the curvature and Ricci tensors and the scalar curvature. b) Let R be an integral domain and S the quotient eld of RX1  X2]. For 2 S set

g := 2  (dX1 dX1 + dX2 dX2) = 1 1 + 2 2 with  := dX (i = 1 2). Then with i = i

i

X

@ @X

i

(i = 1 2)

1 ^ 2 = 2  dX1 ^ dX2 d1 = ; 2  dX1 ^ dX2 = ; 22  1 ^ 2 d2 = 1  dX1 ^ dX2 = 21  1 ^ 2 X

X

X

X

and in (7)

 = 22   = ; 21 X

X

By (2) and (1)



 1 2 2 1 12 = d 2  ; 2      2 2 1 1 1 = d 2 ^  + 2 d ; d 2 ^ 2 ; 21 d2 = 14 ( 2 1 + 2 2 ; ( 1 1 + 2 2 ))  1 ^ 2 X

X

X

X

X

and

X

X X

X



X X

2 1 + 2 2 =2 4 ; X

X1 X1

X

X

+

3

X2 X2



In case = X2;1 (the hyperbolic plane) we have for the scalar curvature 

2 = 2 42 ; 23 X

X X2



and K = ;1 for the Gaussian curvature.

253

= 2(1 ; 2) = ;2

c) For (S=R g) as in 3.16 and 4.8 we compute the connection forms by 5.7a) using ! = ' d' and g =  ; 1 ' ' . ij

i

i`

j

i`

i

G

i

`

;

= d' ^ d' ;

=1 n 1 X

=1

;

k

' d' ^ `

j

g ' d' `k

i

=1 n 1 X `

;1 X

n

' d' ^

j

i

;1 X

;1 X

n

 = d' ^ d' ; ij

`

k



j

`k

i

=1



 ; G1 ' ' `

k

 ' d' k



j

;1  X ' = 1 ; ' ' ; G '2  d' ^ d' =1 =1   ; 1 ; 1 X 2 1 X 2  = 1; ' + G '2  d' ^ d' 

`

k



n

n

`

`

`

i

k

`

j

k

n

n

`

i

`

=1

`

`

j

=1

= (1 + (1 ; G) + G1 (1 ; G)2 )  d' ^ d' = G1 d' ^ d'   X '  1 '   ^ = i

G 1  ;1 r))> = (r (g r Z ))> ; r~ ((rgZ )?)> r (r Z ) = (r g (r Z ))> ; (r~ ((rgZ )?))> r ] Z = (r g ] Z )> V

W

V

W

V

W

V W

W

V

V

W

VW

255

W

V

V

W

from which

W )(Z ))> + (r~ ((rgZ )?)> ; (r~ ((rgZ )?) op R(v ^ W )(Z ) = (R(V ^g W

V

t

W

V

follows. Then

K(U V  W Z ) = g(U R(V ^ W )(Z )) = g~(U R((V ^g W )(Z )) + g~(U r~ ((rgZ )?)) ; g~(U r~ ((rgZ )?)) W

V

W

V

and nally 4.10a) implies

K(U V  W Z ) = K(U V W Z ) + g~((rgU )? (rgZ )?) ; g~((rgU )? (rgZ )?) W

V

V

W

q.e.d. Suppose we are in a situation in which the curvature tensor K of a Riemannian algebra (S=R g) is de ned, i.e. 1 is a nitely generated projective S {module. For U V W Z 2 Der S set R

  W) ;(U V W Z ) := det  gg((U V W )



g(U Z )  g(V Z ) 

Then ; is S {linear in each of its four variables and it de nes a tensor ; 2 T4(1 ). It has the same symmetry properties as K , as is easily checked, namely a) ;(U V W Z ) = ;;(V U W Z ) = ;;(U V Z W ) b) ;(U V W Z ) = ;(W Z U V ) c) ;(U V W Z ) + ;(U W Z V ) + ;(U Z V W ) = 0 Further ;(U V U V ) is a unit of S if and only if U V are linearly independent over S and the restriction of g to the submodule hU V i Der S is non-degenerate. If (U 0  V 0 ) = (U V )  A where A is a 2  2{matrix with entries in S , then R

K(U 0  V 0  V 0 ) = (det A)2  K(U V U V ) and ;(U 0  V 0  U 0  V 0 ) = (det A)2  ;(U V U V ) Hence ;(U 0  V 0  U 0  V 0 ) is a unit in S if and only if det A and ;(U V U V ) are. If

;(U V U V ) is so, then (9)

K(U V U V ) ;(U V U V )

depends only on the module E := hU V i and not on the choice of a special basis of E . The quotient (9) will then be denoted by K(E ). 256

5.17.Definition.

K(E ) is called the sectional curvature (or Riemannian curvature)

of (S=R g) with respect to E .

5.18.Example: Assume 1 is free of rank 2. Then E = Der S for each submodule R

E = hU V i of Der S for which K(E ) is de ned. In this situation K(E ) = K is the Gaussian curvature of (S=R g) (see 5.13). R

Exercises:

Under the assumptions of the exercises to x 4 show the following formulas: 1) Formula for the curvature tensor: Its coecients with respect to B are given by

K

ijk`

=

n  X

t

t

it

=1

t



@; @; g ( @x ; @x ) + ; ; ; ; ; j`

kj

k

`

t j`

t k`

itk

it`

2) Formula for the scalar curvature:

=

n X

jrs

=1

g

 rs

 X @; @; @x ; @x + =1(; ; ; ; ; ) j

j

n

js

rj

r

s

j

t

js

t rs

tr

j

ts

t

3) How does the curvature tensor of a Riemannian algebra (S=R g) change if its metric g is replaced by a conformal metric "g with a unit " of S ? 4) Let R be a ring. In RX1 X2  X3 ] consider the polynomials f = X3 ; X1 X2 (the \saddle") and s = 1 + X12 + X22 . Then S := RX1  X2  X3 ] =(f ) is a Riemannian algebra with the metric induced by the standard metric of A3 . Compute its Gaussian curvature. 5) For a Riemannian algebra (S=R g) there are wellde ned bilinear maps V2 V2   : Der S  Der S ! S where (U ^ V W ^ Z ) = K (U V W Z ) and W ) g(U Z ) for U V W Z 2 Der S . There is a unique  (U ^ V W ^ Z ) = det gg((U V W )V g(V Z ) V endomorphism : 2Der S ! 2Der S such that S

R

R

R

R

R

R

K (U V W Z ) =  ( (U ^ V ) W ^ Z ) is called curvature operator of (S=R g).

257

x 3, ex.1):

Solutions to some of the problems

X

X

n

n

@ ) = gik dxk (i = 1 : : :  n) and g;1 (dxk ) = g ik (k = 1 : : :  n). i @x @xi i=1 k=1 Hence for f 2 S by de nition 3.10

We have g (

@

X n

Grad f = g;1(

@f @xk

k=1

x 4, ex.1):

 dxk ) =

X n

k=1

g ik

X = ; n

By the de nition of the Christo el symbols !ij

X

i=1

n 1 @g jk becomes 2 dgij + ( @g ; ikj )dxk ] i @x @x k=1

@f @xk

X X =1 =1 X X = ( ( ) + =1 =1 X X X + = ( ; )

The divergence (1.2c) is the trace of the linear map sending n n @aj @ @ j @ r@x@ i ( a j ) = (aj r@x@ j ( i ) + i  j ) @x @x @x @x j

n

j

n

aj

n

n

k

k=1 j =1

j

@

! kj

@

@xi

@xk n

@ aj kji @xk

k=1

@aj @xi

@ak @xi

X( X ; =1 =1 X , hence div(

)= @xj

@



n

@



@xk

j

k n

X X ) )=  = div(grad ) = div( ( X (X ) + X=1 =1 ; = X (  x 4, ex.3):

n

f

f

n

i=1

n

@

@xi

n

t=1

j n

@f g it t @x

k

jt=1

x 4, ex.4):

g jk

@f @xk

@f g jt t iji @x

n

aj

j =1

k

ji

X =1 X )+

X X=1( n

jt n

jt=1

n

jt

@f @ g jt r@x@ i j t @x @x

(

g jt

@ 2f @xi @xt

+

g jt

n

@f @xt



@

(

@xj n @2f it g @xi @xt it=1

@xj

)=

@f @ g jt t i @x @x @xj

jt=1 @f @g jt @ @xt @xi @xj

)

@

X + jt

@

j

@ + @a  ) @xi @xj

)=

@ai @xi

X +

@

(hessf )( @x@ i  @x@ k ) = g(r@x@ i (Grad f @x@ k )). By x 3, ex.1) r@x@ i (Grad f ) = r@x@ i (

@xi

to

@ @xi

@xk

@ aj @xj

@

and by x 4, formula (2) this

k ijk dx

x 4, ex.2):



)

@f g jt t @x

=

X; n

`=1

258

@ ` ij @x`

+

@g it @xi



@f @xt

n

j =1

aj ;iji

X +( n

j =1

@f g jt ;ij i ) t ) @x

X )= ( X ; +

@

@

hence (hessf )( @xi  @xk

X +

@ 2f @xi @xk

n

jt=1

gjk

jt n

@f g jt kij @xt

jt=1

g jt

@2f @xi @xt

@f @g jt gjk @xt @xi

X + =1 X X + (; +

@f @g jt + gjk @x t @xi

=

@ 2f @xi @xk

@f g jt t @x

n

t=1

t

ik

n

` n

j =1

g`k ;`ij ) gjk

=

@g jt @f ) @xi @xt

x 5, ex.1):

= K ( @x@ i  @x@ j  @x@ k  @x@ ` ) = g( @x@ i  R( @x@ j )( @x@ k ^ @x@ ` )) We have n @ R( j )( @ k ^ @ ` ) = r @k  r @ ` ]( @ j ) = r @ k ( ;tj` @ t ) ; r @ ` ( @x @x @x @x @x @x @x @x @x Kijk`

X( n

t=1

@xk

hence Kijk`

X n

t=1

@

@

@

;tj` ; @x` ;tkj ) @xt

X = n

@ ;tj` git ( @xk

t=1 @ ;tj` git ( @xk

X

;

;

X

X + (; n

t=1

t

u=1

j`

@

;utk ; ;tkj ;utj ) @xu ]

X X

n n @ ;tkj t ;u ) + giu (; j` tk ` @x u=1 t=1

@ ;tkj t ; ) + ; j` itk ` @x

X;

; ;tkj ;ut` ) =

; ;tkj ;it` ]

x 5, ex.2): 

=

X X X

n

g ij g rs Kijrs

=

ijrs=1 n @ tjs @ trj t g ij g rs git ; js itr ; @xr @xs ijrst=1 n n t @ trj j rs @ js ; g ij g rs g t r s @x @x ijrst=1 jrst=1 j j n n @ js @ j g rs ; srj g rs tjs tr r @x @x jrs=1 jrst=1

 (

(

(

;

;

;

;

;

;

)+

)+; ;

)+

X

;trs ;its] =

X

(;tjs ;itr ; ;trs;its ) =

(; ; ; ;trs ;j ts)

259

n

t=1

t

kj

@

@xt

)=

Appendices Our general references for results of commutative algebra are the books of Bourbaki Bour], Bruns-Herzog BH], Eisenbud E], or Matsumura M3 ], M5 ]. Many results of this text can be found, sometimes in more general form and usually formulated in the language of schemes, in EGA]. The present book uses the language of commutative rings and their spectra no global results are proved. The rst appendix describes the terminology and the notations used without further explanation in the whole text.

A. Commutative Algebras If not stated otherwise a ring R is always an associative commutative ring with 1, and a ring homomorphism  : R ! S maps 1R to 1S . Occasionally also the notion of a Lie algebra occurs. Modules M are left modules with 1  m = m for all m 2 M . For a ring R its spectrum is denoted by Spec R , its maximal spectrum by Max R , and

its set of minimal prime ideals by Min R . If the kernel of the natural ring homomorphims  : Z ! R is generated by n 2 N , then n is called characteristic of R . The full ring of quotients of R is always denoted by Q(R). An algebra is a triple (S R ) where S and R are (commutative) rings and  : R ! S is a ring homomorphism, called the structure homomorphism of the algebra. The algebra is usually denoted by S=R , or simply by S if it is clear which base ring R is considered. If R and S are local rings and  : R ! S is a local homomorphism we call S=R a local algebra. If S=R and S =R are two algebras with structure homomorphims  and  , then an R {homomorphism  : S ! S (or an R {algebra homomorphism) is a ring homomorphism with '   =  . A subalgebra S =R of S=R is a subring S such that (R)  S its structure homomorphism is given by  . It is clear how the residue class algebra S=I for an ideal I  S and the algebra of fractions (quotient algebra) SN for a multiplicatively closed set N  S are dened. We always assume that 1 2 N for each multiplicative set N . We assume that the reader is familiar with basic concepts of multilinear algebra such as the tensor product of algebras or modules, the tensor algebra, exterior and symmetric algebra of a module, and with basic notions of category theory such as diagrams, functors, limits and colimits. They are applied mainly in the category of rings, algebras or modules. Let S=R be an algebra and fxg a family of elements x 2 S . We write R fx g ] or simply R fxg] for the subalgebra of S=R generated by fxg , i.e. the intersection of the subalgebras containing all x . In case S = R fx g] we call fx g a system of generators of S=R . If in this case I is the kernel of the R {homomorphism 0

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2

2

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2

' : R fXg ] ! S 2

260

(X 7! x )

of the polynomial algebra over R in the indeterminates X ( 2 ) onto S , the induced isomorphism

S = R fXg]=I

(A.1)

is called the presentation of S=R given by the generators x . If ff g M is a system of generators of the ideal I we say that S=R is given by the generators x and the relations f . It is frequently desirable to have a description of an algebra in terms of generators and relations. General results of this form are frequently applied in the text. In concrete situations computer algebra supplies algorithms for their explicit computation. An algebra S=R is said to be nitely generated or of nite type if it has a nite system of generators. Algebras of nite type over a eld K are called ane K {algebras. For example an ane algebra of the form A = K X1  : : :  Xn ]=(F ) with a non-constant polynomial F 2 K X1 : : :  Xn] is called a hypersurface algebra. Observe that in eld theory the notions \generating system" and \nitely generated" have a slightly dierent meaning to that of algebras. We call a eld extension L=K with L = K (x1  : : :  xn) an algebraic function eld L over K . As an algebra it is essentially of nite type: by this we mean an algebra S=R such that S = R x1  : : :  xn ]N where x1  : : :  xn 2 S and N is a multiplicatively closed set in R x1 : : :  xn ]. It has a presentation 2

S = R X1  : : :  Xn ]N =I

(A.2)

with the preimage N of N in the polynomial algebra R X1  : : :  Xn ]. The algebra S=R is called nite (free, projective, at, faithfully at) if S as an R { module has the corresponding property. Here nite R {modules are modules with a nite system of generators (modules of nity type). An algebra S=R is said to be essentially nite if S = TN with a nite algebra T=R and a multiplicatively closed set N  T . The notion of quasinite algebra, equidimensional algebra and various notions of complete intersection algebras are discussed to some extend in the appendices B and C. We assume that the reader is familiar with the basic concepts of eld theory such as transcendence degree, which is denoted by Trdeg, and with the dimension theory of rings. We write dim S for the Krull dimension of a ring S . For algebras S=R and R =R their tensor product S := R R S can be regarded as an R {algebra via R ! R R S (a 7! a  1). We say that S =R is obtained from S=R by the base change R ! R . If T=S is another algebra, then T is an R {algebra via the composition of the structure homomorphisms R ! S and S ! T . It is usually an important question which properties of algebras are preserved by the fundamental operations base change, localization, product decomposition, and which are transitive. In the text such properties are sometimes proved with \dierential methods". If S=R is an algebra and P 2 Spec S , then P \ R will always denote the preimage of P by the structure homomorphism  : R ! S , even if  is not injective. With p := P \ R 

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261

0

there is an induced local homomorphism P : Rp ! SP . We call the local algebra SP =Rp the localization of S=R at P , and we write k( P ) for the residue eld of SP . It is an extension eld of k( p), and an algebraic function eld over k( p ) if S=R is essentially of nite type. For any p 2 Spec R we call

Sp = pSp = k( p ) R S or Spec Sp = pSp

the ber of S=R at p . Here Sp = pSp is an algebra over k( p ), and an ane k( p ){ algebra in case S=R is of nite type. It is wellknown that Spec Sp = pSp can be identied set-theoretically with the preimage of p by Spec  : Spec S ! Spec R . It is a basic problem to decide how the properties of Sp = pSp change as p varies. Many notions are dened \locally" at the P 2 Spec S . For example S=R is at at P if SP is a at Rp {module for p := P \ R . Further P is a Cohen-Macaulay point (Gorenstein point) if S=R is at at P and SP = pSP a Cohen-Macaulay ring (Gorenstein ring). S=R is unramied at P if P SP = pSP ( p := P \ R) and k( P)=k( p ) is a separable algebraic extension. S=R is called etale at P if it is unramied and at at P . Algebras that are smooth at P are studied in Chap. I,x 6 and various other sections of the book, and those which are complete intersections at P in appendix C and in Chap. I,x 7. For an S {module M the minimal number of generators of M is denoted by (M ). For P 2 Spec S we write P (M ) instead of (MP ). The set of associated prime ideals of M is denoted by AssV(M ). Further S(M ) or more precisely SS (M ) is the symmetric V algebra of M and M or S M the exterior algebra of M . For an ideal I of S and an S {module M we can speak of the I {topology of M ( I {adic topology), and we denote by M^ the completion of M with respect to I . For a local ring S the completion of M is to be understood as the one with respect to the topology induced by the maximal ideal of S . A.3.Definiton. Let R  S be an extension of rings. A noetherian normalization of S=R is a subalgebra R X1  : : :  Xn]  S such that a) X1 : : :  Xn are algebraically independent over R . b) S is nite over R X1 : : :  Xn]. A related notion is that of a quasi-normalization with will be introduced in appendix B. If S=R has a noetherian normalization, then S=R is of nite type. The converse is not true in general if R is not a eld. However the following version of the classical normalization theorem is sometimes useful. It is easily obtained from the normalization theorem for K R S=K where K := Q(R). A.4.Proposition. Let S=R be of nite type where R is a domain and R  S . Then there are elements X1  : : :  Xn 2 S which are algebraically independent over R and there is an f 2 R n f0g such that Sf =Rf X1 : : :  Xn] is nite. 262

Let S=R be an algebra with structure homomorphism  , and set K := Q(R), L := Q(S ). If for each non-zero-divisor x 2 R the image (x) is a non-zero-divisor of S , then  induces a ring homomorphism K ! L , and we may regard L as a K {algebra. A.5.Lemma. Suppose R and S are noetherian rings. a) If for each P 2 Ass (S ) we have P \ R 2 Ass (R), then R ! S induces ring homomorphisms K ! L and K R R ! L . b) Assume that Ass (S ) = Min (S ), Ass (R) = Min(R) and further that for P 2 Spec S we have P \ R 2 Min (R) if and only if P 2 Min(S ). Then K R S ! L is an isomorphism. c) Under the assumptions of b) the algebra L=K is nite if and only if k( P )=k( p ) is nite for each P 2 Min(S ), p := P \ R . Proof: a) Under the assumptions of a) the image of a non-zero-divisor of R is one of S .

b) Under the assumptions of b) the condition of a) holds, hence K R S ! L is dened. The only prime ideals of S that survive in K R S are the elements of Min (S ). Therefore K R S ! L is locally an isomorphism, hence an isomorphism. Q SP and K = Q Rp by the Chinese Remainder Theorem. c) We have L = p Min (R) P Min (S ) For P 2 Min (S ), p := P \ R , the local rings SP and Rp are artinian, hence complete. Then SP =Rp is nite if and only if k( P )=k( p) is. From this statement c) follows. As pointed out in the introduction algebraic dierential calculus is to some extend similar to calculus on dierential and complex manifolds or spaces. We frequently compare the algebraic concepts with those of geometry. Though logically independent of these it is useful for a better understanding to be familiar with notions like dierential manifold, germs of functions, tangent space and tangent bundle, vector elds on manifolds, their integral manifolds, Riemannian manifold , curvature etc. 2

2

263

B. Dimension Formulas in Algebras of Finite Type. Quasinite and Equidimensional Algebras For a ring R and P 2 Spec R the dimension of R at P (dimP R) is dened as the supremum of the lengths of all chains of prime ideals of R that contain P . It follows from the denition that dimP R = dim RP + dim R= P

(1)

If R is an a ne algebra over a eld K , then dimP R = dim RP + Trdeg(k( P )=K )

(2)

since dim R= P = Trdeg(k( P )=K ). A ring R is called equidimensional, if dim R= p for p 2 Min (R) is independent of the p chosen. It is known that for an equidimensional a ne algebra R over a eld K , and a eld extension K =K , the ring R := K K R is equidimensional too. An a ne K {algebra R is equidimensional if and only if dimP R = dim R for all P 2 Spec R . If S=R is a local algebra where R and S are noetherian, then 0

0

0

dim S  dim R + dim S= mS

(3)

where m denotes the maximal ideal of R . Here equality holds e.g. if S=R is at (M3 ], Thm.16). Let R=K be an a ne algebra, and let R := K K R be as above. For each P 2 Spec R , P := P \ R 0

0

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dimP R = dimP R

(4)

0

0

In fact, dim RP = dim RP + dim RP = P RP , since RP =RP is at. Here we have dim RP = P RP = dim(K K R= P )P , where P is the image of P in K K R= P . This ring is equidimensional, hence dim(K K R= P )P = dim R= P ; dim(K K R= P)= P = dim R= P ; dim R = P = Trdeg(k( P )=K ) ; Trdeg(k( P )=K ), and we obtain 0

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dimP R = dim RP + Trdeg(k( P )=K ) = dim RP + Trdeg(k( P )=K ) = dimP R 0

0

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0

.

Consider now an algebra S=R of nite type where R and S are noetherian domains and R  S . Let K := Q(R) and L := Q(S ). The ring K R S or its spectrum is called the generic ber of S=R . It is, of course, an a ne K {algebra. The dimension of the generic ber will be denoted by d . 264

B.1.Theorem. For P 2 Spec S , p := P \ R , we have

dim SP + Trdeg(k( P )=k( p ))  dim Rp + Trdeg(L=K ) If Rp is universally catenary, or S a polynomial algebra over R , the formula holds with the equality sign. For a proof, see Matsumura M3 ], 14.C. Write dimP Sp = pSp for dimP Sp = pSp , if P is the prime ideal of the ber Sp = p Sp over p corresponding to P . Since d = Trdeg(L=K ) and dimP Sp = pSp = dim SP = p SP + dim Sp = P Sp = dim SP = pSP + Trdeg(k( P )=k( p )) the formula of B.1 can also be written (5)

dim SP  dim Rp + dim SP = pSP ; (dimP Sp = pSp ; d)

The equality sign holds under the same conditions as in the theorem. B.2.Corollary. If R p is universally catenary, or S=R a polynomial algebra, then dimP Sp = pSp  d .

Since (5) holds with the equality sign this formula is a consequence of (3). If p 2 Spec R is given and p Sp 6= Sp , then B.2 implies that the dimension of all irreducible components of the ber Spec (Sp = pSp ) is at least d . (This can also be shown, if R is not universally catenary, see EGA] IV, 13.1.1.) B.3.Corollary. a) dim S  dim R + Trdeg(L=K ).

b) If R is universally catenary and dim R = Supfdim Rp j pSp 6= Sp g , then dim S = dim R + Trdeg(L=K ) Proof: We may assume that dim R < 1 . Then dim S < 1 as well. a) Choose P 2 Spec S with dim SP = dim S . Now apply the formula of B.1. b) Consider p 2 Spec R with pSp 6= Sp and dim Rp = dim R . Choose P 2 Spec S lying over p such that the prime ideal corresponding to P in Sp = p Sp is maximal. Then Trdeg(k( P )=k( p )) = 0. By B.1 and a) we have

dim SP = dim R + Trdeg(L=K )  dim S Since always dim SP  dim S we obtain the formula of b). 265

B.4.Corollary. Assume there exists P 2 Spec S with S = R + P , P \ R = (0). Then

dim S = dim R + Trdeg(L=K ) Proof: We have dim S  dim S= P + dim S P = dim R + dim S P . Since S P is a localization of the a ne K {algebra K R S , which is a domain with K R S= P (K R S )  = K,

we obtain

dim SP = dim K R S ; dim(K R S= P (K R S )) = Trdeg(L=K ) hence

dim S  dim R + Trdeg(L=K )

and we have equality by B.3a). B.4 is applicable, if S is a positively graded domain with degree zero component R and P the ideal generated by the elements of positive degree, hence B.5.Corollary. Let S be a positively graded ring with degree zero component R , which

is supposed to be noetherian with nite dimension, and let S be of nite type over R . For a homogeneous prime ideal P of S and p := P \ R we have dim S= P = dim R= p + Trdeg(k( P )=k( p )) B.6.Definition: An algebra S=R has dim Sp = pSp = d for all p 2 Spec R .

constant ber dimension d 2 N if we have

B.7.Corollary. Let R be a noetherian ring which is universally catenary, and let S=R

be an algebra of nite type. If S is a domain and S=R has constant ber dimension d 2 N , then dim S = dim R + d Proof: Since p S p 6= S p for all p 2 Spec R the mapping Spec S ! Spec R is surjective,

and therefore the kernel of R ! S is nilpotent. Hence we may assume that R  S . Now B.3b) can be applied.

B.8.Proposition. Under the assumptions of B.1 let R be universally catenary. Then there exists an f 2 R n f0g such that for all p 2 Spec R with f 2= p and all P 2 Spec S with P \ R = p we have dimP Sp = pSp = d . (This is also true when R is not universally

catenary.)

266

Proof: By B.2 we have dimP S p = p S p  d for all P 2 Spec S , p := P \ R . By

the noetherian normalization theorem (A.4) there are elements Y1 : : :  Yd 2 S that are algebraically independent over R , and there is an f 2 R n f0g such that Sf is nite over Rf Y1 : : :  Yd ]. For each p 2 Spec R with f 2= p the ring Sp = pSp is nite over k( p)Y1  : : :  Yd ], hence dim Sp = pSp  d . Now the claim follows.

B.9.Theorem. (Semicontinuity theorem of Chevalley). Let R be noetherian and univer-

sally catenary, and let S=R be an algebra of nite type. Then Spec S ! N

( P 7! dimP Sp = pSp )

with p := P \ R is an upper semicontinuous function. In other words: For each n 2 N the set Fn(S=R) := f P 2 Spec S j dimP Sp = p Sp  ng is closed in Spec S . (The theorem holds for arbitrary rings R , see EGA] IV, 13.1.3.) Proof: We may assume that R and S are reduced rings. Using noetherian recursion,

it is su cient to prove the theorem under the assumptions that it already holds for all R=I {algebras S=IS where I is an ideal = 6 (0) for R . Let P 1  : : :  P s be the minimal prime ideals of S , pi := P i \ R , Si := S= P i , and Ri := R= p i (i = 1 : : :  s). The formula (6) shows that

dimP Sp = pSp = PMax fdim P (Si ) p = p (Si ) p g P 

Fn (S=R) =

i

Ss F (S =R ) n i i

i=1

We may therefore assume that R and S are domains and that R  S . Let d be the dimension of the generic ber of S=R . For n  d we have Fn(S=R) = Spec S by B.2. Assume now that n > d . By B.8 there exists an f 2 R n f0g such that, for all p 2 Spec R with f 2= p and all P 2 Spec S with P \ R = p , we have dimP S p = p S p = d . Then Fn (S=R)  Spec (S=fS )  Spec S . By hypothesis Fn (S=R) is closed in Spec (S=fS ) and consequently also in Spec S . B.10.Corollary. If ' : Spec S ! Spec R is a closed mapping, then

Spec R ! Z

( p 7! dim Sp = pSp )

is an upper semicontinuous function. In fact, for each n 2 N we have f pj dim Sp = pSp  ng = '(Fn(S=R)). 267

Following Huneke-Rossi HR] we can apply the above formulas to determine the dimension of symmetric algebras. For a noetherian ring R with nite Krull dimension and a nitely generated R {module M let S := S(M ) be the symmetric algebra of M . For p 2 Spec R let t( p ) be the kernel of the canonical map S ! Sp = pSp . Since Sp = p Sp  = Sk(p)(Mp = pMp ) is a domain t( p ) is a homogeneous prime ideal of S with t( p ) \ R = p . Its canonical image in S= pS is the R= p {torsion of that ring. Hence by B.5 dim S=t( p) = dim R= p + (Mp )

(7)

due to the fact that Sk(p)(Mp = pMp ) is a polynomial algebra over k( p) in p (M ) variables. For P 2 Spec S let p := P \ R . Then t( p)  P since there is a canonical commutative diagram

S ;

S= pS

S= P 

Sp = p Sp

Sp = P Sp = (S= P )p

where is injective. B.11.Theorem. a) All minimal prime ideals of S(M ) have the form t( p ) for some p 2 Spec R and moreover for any p 2 Min R the prime ideal t( p ) is minimal in S(M ). b) dim S(M ) = Max fdim R= p + (Mp )g .

p

Spec R

2

Proof: a) Suppose P 2 Spec S is contained in t( p ) for some p 2 Spec R . Then P \ R = p and t( p )  P , hence P = t( p ). Now let p 2 Min R and Q  t( p ) a prime ideal of S . Then Q \ R = p , since p is minimal, hence t( p )  Q .

b) follows from a) and formula (7). The number fM := p Max fdim R= p + (M p )g Spec R 2

is called Forster-number of M . By a theorem of Forster For] M can be generated by fM elements. B.12.Remarks. a) The "dimension components" of S(M ) are given by the t( p ) for which dim R= p + (Mp ) is maximal.

b) S(M ) is a domain if and only if t(0) = (0), i.e. if and only if R is a domain and Si(M ) is a torsion free R {module for all i 2 N+ . 268

B.13.Definition. An algebra S=R of nite type is called quasinite

dimP Sp = pSp = 0

at

P 2 Spec S if

( p := P \ R)

It is called quasinite, if dimP Sp = pSp = 0 for all P 2 Spec S . Of course, if S=R is nite or, more generally, essentially nite, then it is also quasinite. B.14.Remark: The set of all P 2 Spec S at which S=R is quasinite is open in Spec S ,

since it is the complement of F1(S=R) in Spec S .

B.15.Definition: A module M over a local ring (R m ) is called quasinite if M= m M is a nite dimensional vector space over R= m .

It is known and easy to show that, if R is complete and M is quasinite and separated in the m {topology, then M is even nite. B.16.Proposition. Let S=R be an algebra of nite type, P 2 Spec S , p := P \ R .

Then the following statements are equivalent: a) S=R is quasinite at P . b) P is an isolated point of Spec (Sp = p Sp ) (that is, the prime ideal of Sp = pSp corresponding to P is both maximal and minimal). c) SP is a quasinite Rp {module. d) p SP is a P SP {primary ideal and k( P )=k( p) is nite. Proof: For an algebra S=R of nite type dimP S p = p S p is the supremum of the dimensions of the irreducible components of Spec (Sp = pSp ) that contain P . From this a) $ b) immediately follows. The formula dimP Sp = pSp = dim SP = p SP + Trdeg(k( P )=k( p))

shows the equivalence of a) with d) and with c).

B.17.Corollary. a) Let S=R be quasinite at P and R p a noetherian ring. Let ScP

and Rcp denote the completions of SP and Rp . Then ScP =Rcp is nite. b) S=R is quasinite if and only if Sp = pSp is a nite dimensional k( p ){vector space for all p 2 Spec R . Proof: a) Clearly ScP is a quasinite Rcp {module. Since S P is noetherian, ScP is separated in the p ScP {topology, and the niteness of ScP =Rcp follows. b) Sp = pSp is a nite k( p ){algebra if and only if all prime ideals of Sp = pSp are both

maximal and minimal. Now use the equivalence of the statements a) and b) of B.16. 269

B.18.Example: Let S=R be a local algebra, and let n and m be the maximal ideals of S and R respectively. Suppose S= mS is noetherian and S= n is nite over R= m . Let t1 : : :  td 2 n be elements whose images in S= m S form a system of parameters of this ring. Then S is a quasinite module over Rt1 : : :  td]p where p := n \ Rt1  : : :  td]. The

Rb {homomorphism RbT1 : : :  Td]] ! Sb sending Ti to ti (i = 1 : : :  d) is nite.

Let us collect some properties of quasinite algebras that easily follow from the denition. B.19.Rules: Let S=R be a quasinite algebra of nite type. a) For each f 2 S the algebra Sf =R is quasinite. b) If T=S is another quasinite algebra of nite type, then T=R is quasinite. c) For an arbitrary algebra R =R the algebra R R S=R is quasinite. d) Suppose R is noetherian, Ass (S ) = Min (S ), Ass (R) = Min (R), and each P 2 Min (S ) contracts to an element of Min (R). Let L := Q(S ), K := Q(R). Then R ! S induces a ring homomorphism K ! L . The algebra L=K is nite, and L  = K R S . 0

0

0

Proof: a) is clear, since quasiniteness is dened locally. b) For Q 2 Spec T let P := Q \ S , p := Q \ R = P \ R . Since p SP is primary for P SP and P TQ is primary for Q TQ , we have that pTQ is Q TQ {primary. Clearly k( Q )=k( p ) is nite, since k( Q )=k( P ) and k( P)=k( p ) are. Now we can apply B.16. c) Write S := R R S . For p 2 Spec R and p := p \ R , we then have Sp = p Sp = k( p ) k(p) Sp = p Sp . The claim follows from B.17b). d) For each P 2 Spec S , p := P \ R , the Rp {module SP is quasinite. If p 2 Min(R), 0

0

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0

0

0

0

0

0

0

then dim Rp = 0, and Rp is an artinian ring, so it is complete. But then SP is nite over Rp and P 2 Min (S ). The claim follows now from A.5b). One of the most important results about quasinite algebras is B.20.Zariski's Main Theorem. Let S=R be a quasinite algebra of nite type. Then

there exists a subalgebra T=R of S=R which is nite, such that Spec S ! Spec T is an open immersion, i.e. a homeomorphism onto an open subset of Spec T such that for P 2 Spec S and P := P \ T the canonical homomorphism T P ! S P is an isomorphism. For proofs in the language of algebras we refer to Peskine Pe], Raynaud Ray], Evans Ev], or Iversen Iv]. Another basic notion of algebras is that of equidimensionality. B.21.Definition. An algebra S=R is said to be equidimensional at P 2 Spec S of dimension d if there is an f 2 S n P such that the following holds: a) Each minimal prime of Sf contracts to a minimal prime of R . b) All irreducible components of the non-empty bers of Sf =R have dimension d . 270

S=R is said to be equidimensional of dimension d if it is so at all P 2 Spec S . In case S=R is of nite type condition b) of this denition is equivalent with dimQ (Sf )q = q (Sf )q = d for all Q 2 Spec Sf , q := Q \ R . B.22.Examples: a) If S=R is of nite type and quasinite, then S=R is equidimensional (of dimension 0) if and only if each P 2 Min S contracts to a minimal prime of R .

b) If S = RX1  : : :  Xd ] is a polynomial algebra, then S=R is equidimensional of dimension d. The following denition leads to another class of equidimensional algebras and in fact to a characterization of such algebras, if they are of nite type. B.23.Definition. Let S=R be of nite type and P 2 Spec S . A quasinormalization of S=R in the neighbourhood of P is an R {homomorphism RX1  : : :  Xd ] ! Sf where f 2 S n P and the following conditions are satised:

a) Each minimal prime of Sf contracts to a minimal prime of RX1  : : :  Xd]. b) Sf =RX1  : : :  Xd ] is quasinite. For example, if A is an equidimensional a ne algebra over a eld K , then a noetherian normalization K X1  : : :  Xd ]  A denes everywhere in Spec A a quasinormalization of A=K . B.24.Proposition. Let S=R be of nite type and assume that a quasinormalization RX1  : : :  Xd ] ! Sf in the neighbourhood of P exists. Then S=R is equidimensional at P of dimension d . Proof: For Q 0 2 Min Sf , Q 0 := Q 0 \ RX1  : : :  Xd ] and q 0 := Q 0 \ R we have Q 0 = q 0 RX1  : : :  Xd ], since Q 0 is a minimal prime of RX1  : : :  Xd ]. Therefore q 0 2 Min R . Moreover R= q 0X1  : : :  Xd ]  Sf = Q 0 , hence Trdeg(k( Q 0)=k( q 0))  d , and by 0

0

0

B.2

dimQ (Sf )q = q (Sf )q  d for all Q 2 Spec Sf  q := Q \ R On the other hand, let S := RX1  : : :  Xd ] and q := Q \ S . Then by formula (3) dim(Sf )Q = q(Sf )Q  dim Sq = q Sq + dim(Sf )Q = q (Sf )Q Here dim(Sf )Q = q (Sf )Q = 0 and Trdeg(k( Q )=k( q )) = 0 since Sf =S is quasinite (B.16d). We conclude that dimQ (Sf )q = q (Sf )q = dim(Sf )Q = q (Sf )Q + Trdeg(k( Q )=k( q ))  dim S q = q S q + Trdeg(k ( q )=k ( q )) = d 0

0

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0

271

0

0

0

0

Therefore dimQ (Sf )q = q (Sf )q = d for all Q 2 Spec Sf , q := Q \ R , and S=R is equidimensional of dimension d at P . In the next proposition we show that if S=R is equidimensional at P 2 Spec S , then a noetherian normalization of the ber of P can be \lifted". B.25.Proposition. Let S=R be of nite type and equidimensional of dimension d at P 2 Spec S . Let p := P \ R . Choose f 2 S n P so that conditions a) and b) of denition B.21 are satised, and let k( p ) 1 : : :  d ]  (Sf )p = p (Sf )p be a noetherian

normalization such that 1 : : :  d are the images of T1  : : :  Td 2 S by the canonical map S ! (Sf )p = p (Sf )p . Then there exists a g 2 S n P such that RT1  : : :  Td ] ! Sg is a quasinormalization of S=R in the neighbourhood of P . Proof: For simplicity let us write S for Sf and P for P Sf . There is certainly a noetherian normalization k( p) 1 : : :  d]  Sp = p Sp such that 1 : : :  d are images of elements T1  : : :  Td 2 S . The R {homomorphism RT1  : : :  Td ] ! S is quasinite at P since Sp = pSp is nite over k( p) 1  : : :  d]. By B.14 there is a g 2 S n P such that

Sg =RT1 : : :  Td] is quasinite. In order to prove that RT1 : : :  Td] ! Sg is a quasinormalization in the neighbourhood of P it remains to be shown that each minimal prime Q of Sg contracts to a minimal prime of RT1  : : :  Td]. By assumption we have that q := Q \ R 2 Min R . Assume that Q := Q \ RT1  : : :  Td ] is not minimal in RT1  : : :  Td ]. Then the prime ideal in k( q )T1 : : :  Td] which corresponds to Q is not zero. Since k( Q )=k( Q ) is nite (B.16d) 0

0

0

this would imply that

Trdeg(k( Q )=k( q )) = Trdeg(k( Q )=k( q )) < d 0

and hence

dimQ (Sg )q = q (Sg )q = Trdeg(k( Q )=k( q )) < d a contradiction. This proves the proposition. Combining B.24 and B.25 we obtain

B.26.Corollary. An algebra S=R of nite type is equidimensional at P 2 Spec S if and only if there exists a quasinormalization of S=R in the neighbourhood of P . B.27. Proposition. Let S=R and T=S be algebras of nite type where S=R is equidi-

mensionel of dimension d and T=S is equidimensionel of dimension d ( R noetherian and universally catanary). Then T=R is equidimensionel of dimension d + d . 0

0

Proof: We shall use the criterion B.26. Consider P 2 Spec (T ) and p := P \ S . Let

S Y1 : : :  Yd ] ! Tf and RX1  : : :  Xd ] ! Sg be quasinormalizations in the neighbourhood 0

272

of P resp. p with f 2 T n P and g 2 S n p . It will turn out that the composed map

RX1 : : :  Xd  Y1 : : :  Yd ] ! Sg Y1 : : :  Yd ] ! Tfg 0

0

is a quasinormalization of T=R in the neighbourhood of P . Since quasiniteness is compatible with base change(B.15c) and is a transitive notion (B.15b) we see that Tfg is quasinite over RX Y ]. For Q 2 Min(Tfg ) we have Q \ Sg Y ] 2 Min(Sg Y ]), since Tf =S Y ] has the corresponding property. Then Q \ Sg Y ]) = q Sg Y ] with some q 2 Min (Sg ) and q \ RX Y ] 2 Min (RX Y ]). We conclude that Q \ RX Y ] = q Sg Y ] \ RX Y ] 2 Min (RX Y ])

and we have shown that Sfg =RX Y ] is indeed a quasinormalization. B.28. Proposition. Let S=R be an algebra of nite type and R =R an arbitrary algebra. Set S := R R S , and assume that each minimal prime of S contracts to a minimal prime ideal of S . If S=R is equidimensional of dimension d at P 2 Spec (S ), then S =R is equidimensional of dimension d at all P 2 Spec (S ) lying over P . 0

0

0

0

0

0

0

0

Proof: Replacing S by Sf with a suitable f 2 S n P we may assume that the conditions a) and b) of denition B.21 are already satised for S=R . For Q 2 Spec (S ) let Q := Q \S , q := Q \R and q := Q \R = Q \R . Then S q = q S q = k ( q )k( q ) S q = q S q and all irreducible components of Spec (Sq = q Sq ) have dimension d , since this is the case for Spec (Sq = q Sq ), see formula (4). If Q 2 Min (S ), then Q 2 Min (S ) by assumption, and q 2 Min (R) since S=R is equidimensional. As Sq = qSq is a localization of Rq = q Rq k(q) SQ = q SQ it is at over Rq = q Rq . Since dim SQ = qSQ = 0 we also have dim Rq = dim Rq = qRq = 0 and q 2 Min (R ), q.e.d. 0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

An example (EGA IV, 13.3.9) shows that the condition about the minimal primes of S' is necessary for the proposition to be valid. It is satises e.g. if R'/R is at, in other words: The notion of equidimensionality is compatible with at base change. Let us study the presentations of an equidimensional algebra as a homomorphic image of a polynomial algebra. Consider such a presentation (7)

S = RX1  : : :  Xn ]=I

of an algebra S=R of nite type, where R is a noetherian ring. B.29.Proposition. Let P be a minimal prime ideal of S such that p := P \R 2 Min(R), and let Q be a preimage of P in RX1 : : :  Xn ]. Then

h(IQ ) = n ; dimP Sp = pSp 273

If S=R is equidimensional of dimension d , then all minimal prime divisors of I have height n ; d. Proof: The ideal Q is a minimal prime divisor of I with Q \ R = p . Therefore we have

h(IQ ) = dim RX ]Q = dim k( p )X ]Q where Q is the prime ideal corresponding to Q in k( p)X ]. To obtain this formula we have used that dim Rp = 0 and that RX ]Q R is at over Rp . It implies the desired relation

h(IQ ) = n ; dim k( p )X ]= Q = n ; dim Sp = PSp = n ; dimP Sp = pSp The second statement of the proposition now follows at once from the denition of an equidimensional algebra. B.30.Proposition. Given a presentation (7) assume that each minimal prime ideal of S

contracts to a minimal prime ideal of R . Then the following statements are equivalent: a) S=R is equidimensionel of dimension d . b) For each p 2 Spec (R) with p Sp 6= Sp all minimal prime divisors of the image I p of Ip in k( p)X1  : : :  Xn ] have height n ; d . Proof: The presentation induces a presentation

Sp = p Sp = k( p)X1  : : :  Xn ]=I p Since the minimal prime divisors of I p are in one-to-one correspondence with the irreducible components of Spec Sp = p Sp the claim follows immediately. B.31. Corollary. Let S=R be equidimensional of dimension d and set h := n ; d . Then for all prime ideals p of R with p Sp 6= Sp the ideal I p contains elements t1  : : :  th having the following property: For each P 2 Spec (RX ]) lying over p and containing I

the sequence ft1 : : :  th g is RX ]P -regular and RX ]P =(t1 : : :  th ) is at over Rp .

Proof: Since I p is by B.30b) an ideal of height h in the Cohen-Macaulay-ring k( p)X1  : : :  Xn] it contains a regular sequence ft1 : : :  th g . Let ft1 : : :  th g be a se-

quence of preimages of these elements in Ip . They have the desired property by the following lemma which is a consequence of the local criterion of atness (M3 ],20.F): 274

B.32. Lemma. Let R and P be noetherian local rings and R ! P a at local homomorphism. For an ideal a of R set R := R= a and P := P= aP . Given a sequence

ft1  : : :  th g of elements of P let ti be the image of ti in P ( i = 1 : : :  h ). Then the

following conditions are equivalent: a) ft1 : : :  th g is a P -regular sequence and P=(t1 : : :  th ) a at R -module. b) ft1  th g is a P -regular sequence and P=(t1 : : :  th ) a at R -module.

In connection with quasinormalizations the following fact is sometimes of interest. B.33. Proposition. Let S=R and P=R be algebras where R S and P are noetherian rings, and let : P ! S an R -homomorphism. For P 2 Spec (S ) set Q := P \ P and p := P \ R . Assume that S P is a quasinite P Q -module, P=R is regular at Q and dim SP = p SP = dim PQ = pPQ . If P is a Cohen-Macaulay-point of S=R , then it is also a Cohen-Macaulay-point of S=P , in particular SP =PQ is at. Proof: In order to show the atness of S P =P Q it su ces to prove (see: Bour], Chap.III,

x 5, Thm.1) that a) SP = pSP is at over PQ = pPQ .

b) the canonical homomorphism grpPQ (PQ ) PQ SP ! grp SP (SP ) is an isomorphism. In our situation the Cohen-Macaulay-ring  := SP = pSP is a quasinite module over the regular local ring ; := PQ = pPQ . Then the completion of  is a Cohen-Macaulay-ring which is nite over the completion of ;, because quasiniteness implies niteness for the completions (B.17a). Since the completion of ; is regular the completion of , being a Cohen-Macaulay-ring, is free over it, and a) follows. The mapping of b) is the composition of natural homomorphisms grpPQ (PQ ) PQ SP ! (grpRp (Rp ) Rp PQ ) PQ SP ! grpRp (Rp ) Rp SP ! grpSP (SP ) . These are all isomorphisms since PQ =Rp and SP =Rp are at algebras. Hence b) is satised, and the atness of SP =PQ follows. A regular system of parameters of PQ = pPQ is by a) a regular sequence of SP = pSP . Therefore with SP = pSP the ring SP = Q SP is Cohen-Macaulay as well, q.e.d. Exercises:

1) Let R be a noetherian ring and S=R an algebra of nite type. Then dim S  dim R + Supfdim Sp = pSp j p 2 Spec (R)g . 275

2) Let S=R be an algebra of nite type where R  S are noetherian domains. For P 2 Spec (S ) let p := P \ R . a) If S=R is equidimensional at P , then dim SP = dim Rp + dim SP = pSP b) If this formula holds and Rp is universally catenary, then S=R is equidimensional at P . 3) Let S=R be an algebra of nite type, P 2 Spec (S ) p := P \R . Let P 1 : : :  P h be the minimal prime ideals of S that are contained in P , further let p i := P i \R Si := S= Pi and Ri := R= pi ( i = 1 : : :  h ). Write di for the dimension of the generic ber of Si=Ri ( i = 1 : : :  h ). Then the following statements are equivalent. a) S=R is equidimensional at P of dimension d . b) p i is a minimal prime ideal of R , di = d and Si=Ri is equidimensional of dimension d at P for i = 1 : : :  h .

276

C. Complete Intersections In this appendix various notions of "complete intersections" will be discussed. We rst consider a local ring ( m ) with m {adic completion ^ . C.1.Definition: is called a complete intersection if ist noetherian and if ^ = where is a regular local ring and the ideal  can be generated by an {regular sequence. It is wellknown that this last condition is equivalent with ( ) = ( ) = dim ; dim or, in other words, with being a complete intersection ideal. Clearly a noetherian local ring is a complete intersection if and only if ^ is. If is a complete intersection and = with a regular local ring , then it is known that is a complete intersection ideal of . In this case, with the ring p is likewise a complete intersection for any p 2 Spec ( ). By Avramov Av] this is true in general, i.e. if is not necessarily a homomorphic image of a regular local ring. Ferrand Fe] and Vasconcelos Va1 ] have shown R

R

R

R

S

I

R

S

S=I

S

 I

h I

S

R

I

R

R

R

R

T =J

J

T

T

R

R

R

R

C.2.Theorem. Let

be an ideal of a noetherian local ring and := . Let pdS ( ) be the projective dimension of the {module . Then the following statements are equivalent: a) can be generated by an {regular sequence. a') Each minimal system of generators of is an {regular sequence. b) pdS ( ) 1 , and 2 is a free {module. c) pdS ( ) 1 , and 2 = t  for some {module and some 2 N such that does not contain an {regular sequence of length + 1. I

S

R

S

I

R

S=I

R

S

I

R


< Fi (M ) = > (e1  : : :  ej ) for i = r + s ; j (j = 1 : : :  s) :R for i  r + s = (M ) Thus the Fitting ideals allow us to determinate the rank r and the elementary divisors (ei ) (invariant factors) of M in this case. Moreover, if M has nite length `(M ), then (3) `(M ) = `(R=F0 (M )) In fact, with the notation above, we have r = 0 and `(M ) =

s X i=1

`(R=(ei )) = `(R=(e1  : : :  es )) = `(R=F0 (M ))

By the local-global principle and D.4.d) formula (3) holds true if R is a Dedekind domain and M an R {module of nite length. For a more general case, see exercise 2). 289

has a rank, if Q(R) R M is a free Q(R){module. Its rank is then also called the rank of M .

D.6.Definition. We say that M

;

6

D.7.Proposition. If M has rank r , then Fi (M ) = (0) for i = 0 : : :  r 1 and Fi (M ) = (0)

for i  r . By D.4d) we have Q(R)  Fi (M ) = Fi(Q(R) R M ) for i 2 N . Since Q(R) R M is free of rank r , we obtain Fi (Q(R) R M ) = (0) for i = 0 1 : : :  r ; 1 and Fi(Q(R) R M ) = Q(R) for i  r . Since R ! Q(R) is injective, the claim follows. D.8.Proposition. Let (R m ) be a local ring. Then

(M ) = Min fnjFn(M ) = Rg Moreover, the following statements are equivalent: a) M is free of rank r . b) Fi(M ) = (0) for i = 0 : : :  r ; 1, and Fi(M ) = R for i  r .

f

g

Proof: Let m1  : : :  mn be a minimal system of generators of M , and let (1) be the

presentation dened by this system. The coecients of the relation matrix (2) are elements of m because the generating system was minimal. Therefore Fn;1(M )  m and Fn(M ) = R . This proves the formula for (M ). It suces now to show that b) ! a). But b) implies that n = r and Fn;1(M ) = (0). Then the relation matrix (2) must be the zero matrix, and M = Rr . D.9.Corollary. For an arbitrary ring R and p

2 Spec (R) the following conditions are

equivalent: a) p (M ) = n . b) Fn;1(M )  p and Fn(M ) 6 p . D.10.Corollary. (Semicontinuity of  ) For n 2 N the set of all p 2 Spec (R) with p (M )  n is open. In fact, this is the set of all p with Fn(M ) 6 p . For, if p (M ) = m  n , then Fm (M ) 6 p , and hence Fn(M ) 6 p as well.

f 2 Spec (R)jF (M )  p g . D.12.Corollary. If M is nitely presentable, then the set of all p 2 Spec (R) for which V

D.11.Corollary. Supp(M ) = (F0 (M )) = p

0

Mp is a free Rp {module (of rank r ) is open in Spec (R). By D.8 the Rp {module Mp is free of rank r if and only if p 2= Supp (Fi (M )) for i = 0 : : :  r ; 1 and Fr (M ) 6 p . Since Fi (M ) is nitely generated (D.4a), Supp(Fi (M )) is closed in Spec (R), and so is V (Fr (M )). This proves the corollary, since a union of open sets is open. 290

D.13.Corollary. For a ring R and a nite R {module M the following statements are

equivalent: a) M is locally free of rank r . b) Fi(M ) = (0) for i = 0 : : :  r ; 1, and Fi(M ) = R for i  r . This is a consequence of D.8 and the local-global principle. D.11 results also from the following proposition. D.14.Proposition. If (M ) = n , then

f

g

(Ann M )n  F0 (M )  Ann M

Proof: Let m1  : : :  mn be a minimal system of generators of M and a1  : : :  an

2 Ann M . From then relations aimi = 0 (i = 1 : : :  n) we obtain a  : : :  an 2 F (M ), and hence (Ann M )  F (M ). i i 1

0

0

Let (xj )ij=1:::n be an n {rowed submatrix of a relation matrix (2) and  := det(xj ). Let ji be the subdeterminant of (xij ) adjoint to xij . By Cramer's rule

0 0 1 (ji )  (xij ) = @ . . . A 0

Pn



From xij mj = 0 (i = 1 : : :  n) it follows that mj = 0 (j = 1 : : :  n), hence j =1 F0(M )  Ann M . Let us now look at the multiplicative properties of the Fitting ideals. D.15.Proposition. Let M1 and M2 be nite R {modules. Then

Fi(M1 M2 ) = and in particular

P F(M1 )  F (M2 ) + =i

(i 2 N)

F0 (M1 M2) = F0(M1 )  F0(M2 )

Proof: There is a relation matrix of M1

M

"A

2

0 0 B

of the form

#

where A is a relation matrix of M1 and B a relation matrix of M2 . The claim follows immediately. 291

D.16.Corollary. Let M be a nitely generated R {module and m

 (0)

i = 0 : : :  m ; 1 Fi;m (M ) for i  m D.17.Proposition. For an exact sequence Fi(Rm

M) =

2 N . Then

0 ! M1 ! M2 ! M3 ! 0 of nite R {modules we have

F0(M1 )  F0(M3 )  F0(M2 ) If M3 has a system of n generators such that the kernel of the corresponding presentation is also generated by n elements, then F0(M3 ) is a principal ideal and

F0(M1 )  F0(M3 ) = F0(M2 ) Proof: Start with presentations

0 ! Ki ! Rni ! Mi ! 0 for Mi ( i = 1 or 3). As usual, one constructs a commutative diagram with exact rows and columns 0 0 0 ;  

0

K1

K2

K3

0

0

Rn1

Rn1 +n3

Rn3

0

0

M1

M2

M3

0

0 0 0 We obtain a system of generators of K2 by adding to a system of generators of K1 a system of representatives in K2 for a generating system of K3 . The relation matrix corresponding to this system is of the form 292

n n z }| { z }| { # " 1

3

A

0  B where A is a relation matrix of M1 and B a relation matrix of M3 . We now see that F0(M1 )  F0(M3 )  F0(M2 ), and it is clear that equality holds, if B is a square matrix.

Let T (M ) be the torsion of M , i.e. the set of all m 2 M which are annihilated by a non-zero divisor of R (the kernel of the canonical map M ! Q(R) R M ). D.18.Theorem. (Lipman Lip2 ]). For a local ring R and r

2 N the following conditions

are equivalent: a) Fi (M ) = 0 for i = 0 : : :  r ; 1, and Fr (M ) is a principal ideal generated by a non-zero divisor of R . b) M=T (M ) = Rr , and there exists an exact sequence 0 ! P1 ! P0 ! M ! 0 with free R {modules P0 P1 of nite rank.

!

Proof: b) a). Since M = Rr

T (M ), we have Fi (M ) = 0 for i = 0 : : :  r ; 1 and

Fr (M ) = F0(T (M )) by D.16. Moreover, there is an exact sequence 0 ! P10 ! P0 ! T (M ) ! 0

with a free R {module P10 of nite rank (snake lemma). In fact, rank P0 = rank P10 , since T (M ) is a torsion module (tensorize with Q(R)). Necessarily F0 (T (M )) is a principal ideal, and since Q(R)  F0 (T (M )) = F0(Q(R) R T (M )) = Q(R), any generator of F0(T (M )) is a non-zero divisor of R . a) ! b). Let fm1 : : :  mng be a system of generators and A a relation matrix of M with respect to fm1 : : :  mng . By Nakayama's lemma the principal ideal Fr (M ) is already generated by an (n ; r){rowed minor  of A which is not a zero divisor of R . Then there are relations n X xij mj = 0 (i = 1 : : :  n ; r xij 2 R) j =1

and we can assume  = det(xij )ij=1:::n;r . By Cramer's rule there are equations mh =

n X j =n;r+1

jh mj 293

(h = 1 : : :  n ; r)

where jh are certain determinants contained in Fr (M ). We have jh = rhj   with certain rhj 2 R and, since   (mh ;

n X

rhj mj ) = 0

(h = 1 : : :  n ; r)

j =n;r+1 P n j the elements mh j =n;r+1 rh mj belong to T (M ). Hence M=T (M ) is generated by the images mn;r+1 : : :  mn of mn;r+1 : : :  mn in M=T (M ). From Q(R) R M = Q(R) R (M=T (M )), Fi (Q(R) R M ) = Q(R) Fi (M ) = 0 (i = 0 : : :  r 1), and Fr (Q(R) R M ) = Q(R) Fr (M ) = Q(R) it follows, as in D.8, that Q(R) R M is free of rank r . Since M=T (M ) is generated by r elements, we necessarily have M=T (M ) = Rr .

;



;





 







Consider now the presentation 0 ! K ! Rn ! M ! 0 belonging to fm1 : : :  mng . The elements vj := (x1j  : : :  xnj ) 2 Rn (j = 1 : : :  n ; r) are in K , and they are linearly independent over R , since  := det(xij )ij=1:::n;r is not a zero divisor of R . We shall show that they span K , which will give us the last statement of b). For any (b1  : : :  bn ) 2 K and h = 1 : : :  n we have 2 x11  : : :  x1n;r x1h 3 .. .. .. 7 6 . . . 7=0 6 det 4 n;r n n ; r x1  : : :  xn;r xh;r 5 b1  : : :  bn;r bh because Fr;1 (M ) = 0. The expansion of the determinant with respect to the last column yields an equation nX ;r bh = 0i  aih i=1

with certain i 2 Fr (M ) which do not depend on h . Since  divides all 0i , we see that (b1  : : :  bn) is a linear combination of v1 : : :  vn;r , q.e.d. 0

D.19.Corollary. For a noetherian ring R with Ass (R) = Min (R) and a nite R {

module M the following statements are equivalent: a) Fi(M ) = 0 for i = 0 : : :  r ; 1, and Fr (M ) is an invertible ideal. b) M=T (M ) is a projective R {module of rank r , and pdR(M )  1. Since Ass M = Min M we have T (M )p = T (Mp ) for each p 2 Spec (R). Since an ideal in a noetherian ring is invertible if and only if it is locally generated by a non-zero divisor, D.19 follows from D.18 by the local-global principle. In particular, F0 (M ) is an invertible ideal if and only if M is a torsion module with pdR(M )  1. 294

Exercises:

1) Let R be a principal ideal domain. Two nitely generated R {modules M and N are isomorphic if and only if Fi (M ) = Fi (N ) for all i 2 N . 2) Let R be an integral domain such that R=(a) has nite length for all a 2 R n f0g . Let M be an R {module with a presentation M = Rn=K where K  Rn is a free submodule of rank n . Then `(M ) = `(R=F0(M )). 3) Let R be a ring. Assume there is an exact sequence of R {modules 0 ! F1 ! F0 ! M ! 0

R . Let T (M ) be the torsion of M and R : Fn;1(M ) := with F0 = Rn , F1 = fx 2 Q(R)jx  Fn;1(M )  Rg . Show that T (M ) = R : Fn;1(M )=R

295

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