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University of Utah

Algebraic Number Theory Dick Gross Notes by Sean Sather-Wagsta

Introduction These notes are based on a semester-long course in algebraic number theory given at the University of Utah during the Spring of 1999. The author learned the subject from John Tate and has used many of Tate's unpublished ideas here. Throughout these notes, we shall employ the following notations. For a ring A, we let A denote the multiplicative group of units in A. If A is a subgroup (resp. sub eld) of B , we let (B : A) denote the index (resp. degree) of A in B . For a nite extension of elds k  K , we let Tr and N denote the trace and norm maps, respectively.

Lecture 1 One of our rst goals is to discuss the properties of the ring of integers in a number eld. To accomplish this, we shall rst consider general lattices in a rational vector space. De nition. Let V be a nite dimensional vector space of dimension n over the rational numbers Q . A lattice in V is a free abelian subgroup L  V of rank n. Assume that L is a lattice in V with v1 ; : : : ; vn 2 V such that L = Zv1 + Zv2 +    + Zvn. Then the set fv1 ; : : : ; vn g forms a P basis for V . It suces to show that the vi are linearly independent over Q . Suppose that i ai vi = 0 for some a1 ; :P: : ; an 2 QP. Choose a nonzero integer N so that Nai is an integer for each i. Then 0 = N i ai vi = i Nai vi which is a sum of elements in L. Thus, each Nai = 0 which implies that each ai = 0. We recall that the set of bases of V (over Q ) is in bijection with GL(V ) and the set of bases of L (over Z) is in bijection with Aut(L) = GLn (Z). It follows from the previous comments that the set of lattices in V is in bijection with GL(V )=Aut(L)  = GLn(Q )=GLn (Z). If M is a subgroup of L of nite index, then M is also a lattice in V . This follows from the fact that the quotient L=M is a nitely generated, torsion abelian group, and is therefore of the form Z=a1Z     Z=anZ for some integers ai  1. It follows that there is a basis v1 ; : : : ; vn of L such that a1 v1 ; : : : ; an vn is a basis of M . For any 2 GL(V ), the image (L) is another lattice in V . In particular, if 2 Q  = Q r f0g then L is another lattice in V , with basis f v1 ; : : : ; vn g. If M and L are lattices in V , then there exists ; 2 Q  such that L  M  L where each subgroup has nite index. It should be noted that this fact does not hold in a real vector space. De nition. Assume that h ; i : V V ! Q is a nondegenerate bilinear form. Then there 1

is an isomorphism V ! Hom(V; Q ) which is given by v ! hv; i. Assume that L is a lattice in V and de ne the dual lattice L of L to be L = fv 2 V : hv; Li  Zg. The dual lattice is, in fact, a lattice: if fv1 ; : : : ; vn g is a basis of L, then L has the dual basis (over Z) fv1 ; : : : ; vn g where vi is given by hvi ; vj i = i;j where i;j is the Kronecker delta function. De nition. Under these assumptions let E denote the \inner product matrix" E = (hvi ; vj i). We de ne the discriminant of L as disc(L) = det(E ) 2 Q  . This is independent of the basis chosen. To see this, x a change-of-basis matrix A 2 GLn (Z). After changing basis, the new inner product matrix is E 0 = AEAt whose determinant is det(E 0 ) = det(A)2 det(E ) = det(E ). De nition. By multiplying our bilinear form by a nonzero integer, we may assume that hL; Li  Z. We note that this condition is equivalent to the condition L  L . A lattice satisfying these equivalent conditions is called integral. By de nition, if L is an integral lattice, then disc(L) 2 Z. Furthermore, we have the following proposition. Proposition 1. The index (L : L) = disc(L). Proof. Since L is a subgroup of L , and both groups are free abelian of rank n, the additivity of rank on short exact sequences shows that the rank of the quotient L =L is zero. Since the quotient is nitely generated, the classi cation of nitely generated abelian groups shows that the quotient is a nite group. Thus, the index (L : L) is nite. From the observations above, it follows that there is a basis fv1 ; : : : ; vn g of L and nonzero integers a1 ; : : : ; an such that the set fa1Qv1 ; : : : ; an vn g forms a basis of L. Furthermore, L =L  = ni=1 Z=aiZ, n  so that (L : L) = j i=1 ai j. Let fv1 ; : : : ; vn g denote the dual basis of L. Let E and E 0 denote the inner product matrices E = (hvi ; vj i) and E 0 = (hvi ; aj vj i). By de nition, E 0 is the diagonal matrix

0a 1 B 0 E =@

...

0

0

an

1 CA

There is a matrix A 2 GLn(Z) such that E 0 = EA, and it follows that disc(L) = det(E ) =  det(E 0 ) = a1    an :

Q Thus, (L : L) = j ni=1 ai j = disc(L).

De nition. A lattice L is unimodular if L = L. By the proposition, we see that this is equivalent to the condition that disc(L) = 1.

In the general (i.e., not necessarily unimodular) case, the inner product on V induces a pairing h ; i : L =L L=L ! Q =Z which is a duality of nite abelian groups. It follows that the set of unimodular lattices M which lie between L and L is in bijection with the subgroups A of L=L such that A = A? under h ; i. Example. Consider the lattice Zn  Q n with hx; yi = Pi xi yi . Then Zn is a self-dual lattice and ei = ei . De nition. An integral lattice L is even if, for all v 2 L, hv; vi 2 2Z. Note that the previous example is not even. 2

Example. With Zn as in the previous example, let L = fPiPai ei : PPi ai 2 2Zg. Then L  Zn is a sublattice of index 2. Furthermore, L is even since i a2i  i ai (mod 2). Lemma 2. Assume that M and L are lattices in (V; h ; i) such that M is a sublattice of L of index N . Then disc(M ) = N 2 disc(L).

Example. If L  Zn is the even sublattice of the previous example, then the lemma implies

that disc(L) = 4.

Proof. It is straight-forward to check that we have the following containments: M  L  L  M . We claim that the index (M  ; L) = N . Since h ; i : M  =L L=M ! Q =Z is a duality of nite abelian groups and L=M has order N , it follows that M  =L also has order N by the theory of nite abelian groups. By Proposition 1, we see that

disc(M ) = (M  : M ) = (L : M )(L : L)(M  : L) = N 2 (L : L) = N 2 disc(L): Finally, disc(L) and disc(M ) have the same sign. To see this, choose a basis v1 ; : : : ; vn of L such that there are nonzero integers a1 ; : : : ; an such that a1 v1 ; : : : ; an vn is a basis of M . Then Y 2  Y 2 ai disc(L) ai det(hvi ; vj i) = disc(M ) = det(hai vi ; aj vj i) = det(ai aj hvi ; vj i) = i

i

as desired.

Example. Let L  Zn be the even sublattice from the previous example. Then L  Zn  L . Since Zn is unimodular, the proof of Lemma 2 shows that (Zn : L) = 2 = (L : Zn). One can verify directly that the element  = 12 (e1 +    e2 ) is contained in L, but not in Zn. If n is odd, then 2 62 L and 4 2 L, and it follows that L =L  = Z=4Z. If n is even, then every element of L=L is annihilated by 2, so that L =L  = Z=2Z  Z=2Z. The inner product induces a pairing h ; i : L =L  L =L ! ( 12 Z)=Z  Q =Z. We claim that for v 2 L =L, hv; vi = 0 in Q =Z if and only if n  0 (mod 4). If n 6 0 (mod 4) then h; i = 14 n 62 Z. Assume that n  0 (mod 4). It is always the case that for v 2 Zn, hv; vi 2 Z, so we need only check : h; i = 41 n 2 Z since n is divisible by 4. In the case when L=L  = Z=2Z Z=2Z, there are exactly three distinct nontrivial subgroups of L =L. Thus, there are exactly three lattices in k which live between L and L . Zn is one of these lattices, and we call the others M and M 0 . We claim that n  0 (mod 4) if

and only if M and M 0 are integral and unimodular. (Of course, by checking indexes, we see that M and M 0 are always unimodular, so the second condition is equivalent to M and M 0 being integral.) This follows from the fact that there are elements v 2 M and v0 2 M 0 such that M = L + Zv and M 0 = L + Zv0. Then hv; vi and hv0 ; v0 i are integers if and only if for every v00 2 L , hv00 ; v00 i is an integer. Finally, if n  0 (mod 8), then M and M 0 are both even. (This is the best of all worlds.) The point here is that h; i = 14 n 2 2Z since n is divisible by 8. For more discussion and interesting applications, the interested reader is encouraged to consult Sphere Packing, Lattices and Groups by Conway and Sloan. In order to de ne algebraic numbers and algebraic integers, we need the following propositions. 3

Proposition 3. For any complex number the following conditions are equivalent: 1. satis es a monic polynomial xn + an 1 xn 1 +    + a0 with rational coecients. 2. The smallest sub eld Q ( ) of C containing Q and is of nite degree over Q . 3. There is a eld k of nite degree over Q which contains .

De nition. We say that a complex number is algebraic if one of these equivalent condi-

tions holds.

Proof. \2.) 3." clear \1.) 2." Since satis es a monic polynomial with rational coecients, there exists unique such (monic) polynomial g (x) 2 Q [x] with minimal degree. Then g (x) is the generator of the nonzero ideal I  Q [x] of polynomials satis ed by . I is a maximal ideal and the map Q [x]=I ! Q ( ) given by x + I 7! is an isomorphism. This implies that the degree of Q ( ) over Q is nite. \3.) 1." The map : k ! k given by multiplication by is Q -linear. The monic polynomial f (x) = det(xI ) = xn Tr ( )xn 1 +     det( ) is clearly satis ed by . Corollary 4. The set Q of algebraic numbers forms a sub eld of C . Proof. Assume that ; 6= 0 are algebraic. It suces to show that  , and = are algebraic as well. Let k = Q ( ; ) denote the smallest sub eld of C containing Q ; ; . Then the degree (k : Q ( )) = (Q ( ; ) : Q ( ))  (Q ( ) : Q ) is nite since is algebraic. Thus, (k : Q ) = (Q ( ) : Q )(k : Q ( )) is nite. Since the desired elements are all contained in k, they are all algebraic as desired. Proposition 5. For any complex number the following conditions are equivalent: 1. satis es a monic polynomial xn + an 1 xn 1 +    + a0 with integer coecients. 2. Z[ ] is a Z-module of nite rank. 3. There is a subring M  C containing which has nite rank as a Z-module.

De nition. We say that a complex number is an algebraic integer if one of these equivalent conditions holds.

Proof. \1.) 2." Z[ ] = Z + Z +    + Z n 1. \2.) 3." clear \3.) 1." The argument from Proposition 3 applies. The map : M ! M is an endomorphism of a free Z-module. Therefore, the coecients of the polynomial f (x) = det(xI ) = xn Tr ( )xn 1 +     det( ) are integers.

4

Corollary 6. The set Z of algebraic integers forms a subring of Q . Proof. As before, given algebraic integers ; , let M = Z[ ; ]. This ring has nite rank as a Z-module and contains  ; . Lemma 7. With notation as above: 1. Z \ Q = Z. 2. Q =Z is a torsion group (like Q =Z). Proof. 1. For a 2 Z, a satis es the monic polynomial x a 2 Z[x]. This gives the containment \". For the other containment, x = a=b 2 Z \ Q and suppose that 62 Z. Without loss of generality, we assume that a and b are relatively prime. Since 62 Z, it follows that b 6= 1. If satis es the monic polynomial xn + an 1 xn 1 +    + a0 2 Z[x], then

an =bn + an 1 an 1 =bn 1 +    + a0 = 0: Multiplication by bn yields the equation

an + |ban 1 an 1 {z+    + bn a0} = 0 divisible by b

which implies that an is divisible by b. However, since b is not a unit and (a; b) = 1, we see that this is a contradiction. 2. Assume that satis es the polynomial xn + an 1 xn 1 +    + a0 2 Q [x]. For a positive integer N , let = N . Then 0 = N n (( =N )n + an 1 ( =N )n 1 +    + a0 ) = n + Nan 1 n 1 +    + N n a0 which implies that if we choose N such that N i an i 2 Z, then = N 2 Z. Thus, Q =Z is a torsion group.

Lecture 2 Let k  Q be a number eld. It follows from the de nition that every element of k is algebraic. Let A denote the subring of algebraic integers in k, that is, A = k \ Z. It follows that A \ Q = Z. Theorem 8. A is a lattice in the rational vector space k. Proof. First, we claim that A contains a lattice L in k. To see this, x a basis v1 ; : : : ; vn of k over Q . Since Q =Z is a torsion group, there exists nonzero integer N such that Nv1 ; : : : ; Nvn 2 A. Let L = Nv1 Z +    + Nvn Z, which is a lattice in k contained in A.

5

Now, consider the linear and bilinear forms on k de ned by

k!Q 7! Tr ( )

kk !Q ( ; ) 7! h ; i = Tr ( )

respectively. Recall that the trace is given by considering multiplication by as a Q -linear map k ! K and using the formula det(xI ) = xn Tr ( )xn 1 +    + ( 1)n N ( ): The fact that Tr ( + ) = Tr ( ) + Tr ( ) implies that h ; i is a symmetric, bilinear form on k. Furthermore, it is nondegenerate since h ; 1 i = Tr (1) = deg(k) 6= 0. We need the following. Lemma 9. The image of A under the trace map is contained in Z. Before we prove the lemma, we use it to complete the proof of the theorem. Consider the dual lattice of L with respect to this bilinear form: L = f 2 k : Tr ( L)  Zg. For every 2 A, the fact that L  A implies that L  A  A. Therefore, the lemma implies that 2 L . Thus, L  A  L. Since the index (L : L) is nite, we see that the index (L : A) is also nite, which implies that A is a free abelian group of rank equal to the rank of L . That is, A is a lattice. Proof of Lemma 9. It suces to show that all the coecients of the polynomial f (x) = det(xI ) are integers when 2 A  Z. To see this, let m = (Q ( ) : Q ) and let  = (k : Q ( )), so that n = m. Then a basis for Q ( ) over Q is exactly 1; ; 2 ; : : : ; m 1 . Let g (x) = xm + am 1 xm 1 +    + a0 denote the minimal polynomial of over Q . Let 1 ; : : : ;  be a basis for k over Q ( ). Then f i j : 1  j  ; 0  i  m 1g forms a basis for k over Q . We wish to nd the matrix representing left multiplication by with respect to this basis. Let T be the following m  m matrix:

00 BB1 T =B BB0. @ ..

0 0 1 .. . 0 0

  

0 0 0 .. . 0

...



The matrix for xI is then given in block form:

0xI T 0    BB 0 xI T    B@ ... .. ... . 0

a0 a1 a2

.. .

am

1

1 CC CC CA

1 CC C  A 0 0

   xI T

0

It follows that f (x) = (g (x)) . Thus, it suces to show that for 2 A, the coecients of the minimal polynomial of are in Z.

6

Q (x ). We then have a commuting diagram Over C , we may factor g (x) as g (x) = m i i=1 of Q -isomorphisms of elds:  = Q [x]=(g (x)) Q ( )   =

 x

= ? ? ? i = x ? Q ( i ) Q [x]=(g (x))

Since Q ( i )  = Q ( ) and these are nite extensions of Q , Proposition 5 implies that the i are algebraic integers. Corollary 6 implies that the symmetric functions in the i are algebraic integers. Since these are the coecients of g (x), they are also in Q , and it follows that they are in Q \ Z = Z.

De nition. We de ne the discriminant of k to be the discriminant of the lattice A in k with respect to the bilinear form h ; i = Tr ( ). That is, if 1 ; : : : ; n forms a basis of A over Z, then disc(k) = disc(A) = det(Tr ( i j )). Since Tr (A)  Z, this matrix has integer entries and disc(k) 2 Z. Example. If k = Q , then A = Z = A and disc(Q ) = 1. Next lecture, we shall prove the

following theorem, due to Minkowski. The proof involves the \geometry of numbers". Theorem. If disc(A) = 1, then k = Q . Proposition 10. Let k  Q be a number eld with ring of algebraic integers A, and let d = disc(A). Then d  0; 1 (mod 4). Equivalently, d  a2 (mod 4) for some integer a.

Proof. Since the extension Q  k is nite and separable,Pthere are n = (k : Q ) distinct 1 ; : : : ; n of k into C . Then Tr ( ) = n=1  ( ) forPalln 2 k. Let P1n; : : : ; n 2 A form a basis of A over Z. It follows that Tr ( i j ) = =1  ( i j ) = =1  ( i ) ( j ). Let B = ( ( i )), which is an n  n matrix with entries in Z. In particular, det(B ) 2 Z. Then d = det(Tr ( i j )) = det(B t B ) = det(B )2 . Expanding det(B ) using the formula for Tr ( i ) we have Q -embeddings

det(B ) =

X

2Sn

P

sign()

Q

n Y

=1

 ( () ) =

n XY

2An =1

P

 ( () )

Q

n XY

62An =1

 ( () ) = x y

where x = 2An n=1  ( () ) and y = 62An n=1  ( () ). Since each i is an algebraic integer, it follows that each  ( () ) satis es the same monic polynomial as () and is therefore an algebraic integer. Thus, x and y are both algebraic integers, as are the elements x + y and xy. For any Q -automorphism  of C ,  permutes the zeroes of any given polynomial with integer coecients. In particular, for any xed  2 Sn ,  permutes the set f ( () )gn=1 . It follows that  simply rearranges the terms and factors of x + y and xy, so that  (x + y) = x + y and  (xy) = xy. By taking a nite Galois extension K of Q in C which contains x + y and xy, we see that these elements are in the xed eld of the Galois group of K over Q . That is, x + y; xy 2 Q . Since these elements are also in Z, we see that they are integers. Finally, this shows that d = (det(B ))2 = (x y)2 = (x + y)2 4xy 7

which is of the form d = (integer)2 4(integer). Thus, d  a2 (mod 4) for some integer a, as desired.

Example. We apply the proposition to the case of quadratic elds. Let k be an extension of Q of degree 2. The Theorem of the Primitive Element implies that k = Q ( ) for some 2 k. Let x2 bx + c be the minimal polynomial of , and let = 2b . Then 2 k r Q

implies that k = Q ( ). Furthermore, 2 = ( 2b )2 = 2 b + b4 = b4 c. That is, satis es a polynomial of the form x2 D. By replacing by a rational multiple of (after replacing by ) we may assume that the minimal polynomial of is x2 D where D is a square-free integer. It follows that k  = Q [x]=(x2 D). Let A denote the ring of algebraic integers in k and let L denote the lattice L = Z + Z . Since 2 A, it follows that L  A, and since these are both lattices in k, the index N = (A : L) is nite. By Lemma 2, d = disc(A) = disc(L)=N 2 . To compute disc(L), we let 1 = 1 and 2 = , and compute: Tr (1) = (k : Q ) = 2; Tr ( ) = 0 since the trace is the coecient of x in the minimal polynomial x2 D of ; and Tr ( 2 ) is the sum of the two conjugates of 2 = D which is 2D since D 2 Z. Thus, 2 0  disc(L) = det(Tr ( i j )) = det 0 2D = 4D and d = 4D=N 2 , which is an integer. There are two possibilities since D is square-free and 4D is divisible by N 2 . (1) If N = 1 then L = A, which implies that d = 4D  0 (mod 4). (2) If N = 2 then L has index 2 in A and d = 4D=Np2 = 4D=4 = D and the following argument shows that D  1 (mod 4). We write = Dpand note that since A=L has order 2, 2A  L which implies that A  12 L = Zp21 + Z21 D. We wish to know exactly when an element of 12 L is in A r L. Let = 12 (a + b D) be such an element. Then a and b are not both even. Since 2 A, the proof of Lemma 9 shows that N( ) 2 Z. Direct computation shows that N ( ) = 14 (a2 b2 D) 2 Z, which implies that a2 b2 D 2 4Z. That is, a2  b2 D (mod 4). If b were even, then this would imply that a is also even, a contradiction. Thus, b is odd, and b2  1 (mod 4). It follows that a2  D (mod 4), so that if a were even, then D would be divisible by 4, 2 contradicting the fact that D is square-free. Thus, pd a  1 (mod p4), which implies that 1+ D  1 (mod 4). It also follows that A = Z + Z( 2 ) = Z + Z( d+2 d ). Thus, we have proved the following. Theorem 11. The discriminants of quadratic elds are exactly the integers d  0; 1 (mod 4), d 6= 1 which are as square free as possible. That is, the only square factor of d permitted is 4, and this occurs if and only if d4 6 0; 1 (mod 4). 2

2

It follows that the possible values for d are d > 0 d = 5, 8, 12, 13, 17, 20, 21, : : : d < 0 d = -3, -4, -7, -8, -11, -15, -17, : : :

p

The rst list corresponds to real quadratic elds since k = Q ( D) and D = d > 0 or D = d=4 > 0. The second list corresponds to imaginary quadratic elds. Now we are in the position to completely p describe the ring of integers for a quadratic p eld. If d  0 (mod 4), then A = L = Z + Z D. In this case, D = d=4 so that A = Z + Z 2d = 8

pd

p

Z + Z( d+2

). If d  1 (mod 4), then A = Z + Z( d+2 d ), from the above work. Therefore, we have proved the following. Theorem 12. The ring of algebraic integers of a quadratic eld k is always

d + pd 

A = Z+ Z

2

where d = disc(k).

Lecture 3 Example. We continue with the example of quadratic elds. If d = 4, then with the previous notation, D = 1, and the ring of algebraic integers is the ring of Gaussian integers A = Z + Zi  C = R + Ri. Furthermore, A = fing which is cyclic of order 4. p p If d = 3, then A = Z + Z( 1+ 2 3 ) is the ring of Eisenstein integers. If we let  = 1+ 2 3 , then p p Tr ( ) = 1 + 2 3 + 1 2 3 = 1 and  1 + p 3  1 p 3  1 + 3 N ( ) = = 4 = 1: 2 2 It follows that  2  + 1 = 0. Furthermore, 3

p

= 1+3

p

3 9+( 3 8

3) = 1 p

so that  is a primitive sixth root of unity. Since  2 = 1+2 3 , we see that A = Z + Z = Z + Z 2, and A = f n g is cyclic of order 6. One point of interest here is that the point (;  1 ) 2 C 2 is in the intersection of each Fermat curve xp + yp = 1 for primes p > 3. p p If d = 5, then A = Z + Z, where  = 1+2 5 . Let 0 = 1 2 5 . Then 0 = 1 so that  1 = 0 2 A. It follows that A = f1g  f g. De nition. If k is a number eld, then an order in k is a subring B  k of rank n = (k : Q ) as a free Z-module. An order B in k is always contained in the ring of algebraic integers with nite index. Since B is a lattice, it suces to show that B  A. For b 2 B , the fact that B is a ring implies that the Z-linear map given by multiplication by b maps B ! B . Fixing a basis for B over Z, this map is given by an n  n matrix with integer entries which we also denote b. Then b satis es the polynomial fb (x) = det(xI b) which is monic with integer coecients. Thus, b 2 A. If f = (A : B ), then Lemma 2 implies that disc(B ) = f 2 disc(A). Proposition 13. Assume that k is a quadratic eld and f is a positive integer. Then there is a unique order B = Z + fA of index f in A, the ring of integers. Z

9

Proof. It ispclear that B = Z + fA is a ring. Furthermore, if d = disc(A), then B = Z + fA = Z + f Z( d+2 d ) is a lattice. Thus, B is an order, and this formulation shows that B has index f in A. Let B 0 be any order in k with index f in A. Then f annihilates A=B 0 so that fA  B 0  A. Since B 0 is a ring, Z  B 0 so that B  B 0  A. Since B and B 0 both have index f in A, they must be equal.

If d = disc(A) and dB = disc(B ), then dB = f 2 d  0; 1 (mod 4). It follows that

d + pd 

B = Z + fZ

2

 f 2 d + f pd 

= Z+ Z

2

 d B + pd B 

= Z+ Z

2

:

p

Example. If d = p3, then A = Z + Z( 1+ 2 3 ). The unique order given by f = 2 is B = Z + 2A = Z + Z 3, and B  = f1g.

If d = 4, then A = Z + Zi. In this case ,the order of index 2 is B = Z + Z2i. Again B  = f1g. We note that there is a correspondence between orders B in a quadratic eld k and binary quadratic forms. If q(x; y) = ax2 + bxy + cy2 for integers a; b; c then the discriminant of q is disc(q) = b2 4ac. If M is a free Z-module of rank 2, then choosing a basis e1 ; e2 of M induces a map q : M ! Z given by q(xe1 + ye2) = q(x; y). We can use q to induce a symmetric bilinear form [ ; ] : M  M ! Z by de ning [m; n] = q(m + n) q(m) q(n). It

is straightforward to verify that this operation is symmetric and bilinear. What we notice, however, is that this form is even since [m; m] = q(2m) 2q(m) = 4q(m) 2q(m) = 2q(m). If we compare this to the bilinear form h ; i = Tr ( ) on B  k, we see that it is not p always even, even when d p 1 (mod 4). For example, when d = 5, we have  = 1+2 5 and h; i = Tr (2 ) = Tr ( 3+2 5 ) = 3. From this we see that we should devise an alternate bilinear form on k. p From our notes before, we know that there is a square-free integer D such that k  = Q ( D). Quadratic extensions are always Galois, and the nontrivial Q -automorphism of k is given by p p ( D) = D. We de ne a new bilinear form f ; g on k by the formula fx; yg = Tr (x(y)). Since Tr (x) = x + (x), it is straightforward to verify that this form is symmetric and bilinear. If we restrict to B  B , the form takes its values in Z since B  A and (A) = A. Furthermore, for x 2 B , fx; xg = Tr (x(x)) = 2N(x) 2 2Z. B is a free Z-module of rank 2, and we de ne q : B ! Z by x 7! 12 fx;pxg. With the correct basis, this is a binary quadratic form. Let dB = disc(B ) and = dB +2 dB , so that 1; forms a basis of B over Z. Then

q(x; y) = 21 fx + y ; x + y g = N (x + y )   dB + pdb   dB pdb  x+y = x+y

2 2 2 dB d = x2 + dB xy + B 4 y2 The coecients 1 and dB are clearly integers, and since dB  = 0; 1 (mod 4) we see that dB dB is also an integer. Furthermore, disc(q ) = d2 (d2 d ) = d = disc(B ). It is B B B B 4 important to note that frequently there are many quadratic forms of discriminant dB other than the ones coming from this construction. 2

10

Example. Let d = 15 and consider the quadratic forms q1(x; y) = x2 15xy + 60y2

and q2 (x; y) = 2x2 + xy + 2y2, which both have discriminant 15. These forms are not equivalent under any change of basis. To see this we notice that the rst form comes from our construction. If these were equivalent, then A would have an element of norm 2 since p b then N (e1 ) = q2 (1; 0) = 2. This is impossible, though, since if 2 = N ( a+b 2 15 ) = a +15 4 2 2 a + 15b = 8 which is impossible for integers a and b. It turns out that there is an ideal I  A of index 2 which leads to q2 . 2

2

Now, we discuss the \geometry of numbers" which will help us nd estimates of d from the degree n of k. It will also give an elegant proof of the theorem of Minkowski from last lecture. We have a chain of containments A  k  k R  = Rn so we may view A as a discrete, co-compact subspace of Euclidean space. Let k = Q ( ) = Q [x]=f (x) where f is the minimal polynomial of over Q Q . By de nition, f is irreducible over Q , but over R it splits into a product f (x) = i gi (x) of distinct irreducible factors of degree 1 or 2. (The factors are distinct by separability.) If r1 is the numbers of real zeroes of f and r2 is the number of conjugate pairs of complex zeroes, then r1 + 2r2 = n and the Chinese Remainder Theorem implies that Q

k R = R[x]=f (x) = R[x]=

Y i

gi (x) =

Y i

R[

x]=gi (x) = Rr  C r 1

2

As a side note, another method for obtaining r1 and r2 is to consider the nondegenerate bilinear form hx; yi = Tr (xy) on k R and to observe that this form has signature (r1 +r2 ; r2 ). p Example. For the quadratic elds k = Q ( D) where D is square-free, we see that

k R =

(

R

 R if D > 0 if D < 0

C

because if D > 0 then f (x) = x2 D, and if D < 0 then f (x) = x2 + D. This is a speci c case of the general fact that r1 is the number of distinct real embeddings of a general number eld k, and r2 is the number of distinct conjugate pairs of complex embeddings. We also observe that k C  = C n . Let 1 ; : : : ; n be the distinct n embeddings of k into C . Then the isomorphism k C ! C n is given by 1 7! (1 ( ); : : : ; n ( )). Let L  k be any lattice in k. The fundamental domain for L acting on Rn is

D= as in the gure.

n nX i=1

xi i : 0  xi < 1

1 D 2 11

o

Proposition 14. disc(L) = ( 4)r (vol(D))2 . Corollary 15. The sign of disc(L) is ( 1)r . 2

2

Proof. If r2 = 0, then all of the embeddings 1 ; : : : ; n of k are real. Let 1 ; : : : ; n be a basis of L. The coordinates of i in the standard orthonormal basis of Rn are (1 ( i ); : : : ; n ( i )) and vol(D) = j det(i ( j ))j. If B = (i ( j )) then we already observed that disc(L) = det(B t B ) = (det(B ))2 = (vol(D))2 , as desired. If r2 6= 0, then each conjugate pair of complex embeddings contributes a factor of dz ^ dz = 2idx^dy. Thus, each such pair contributes a factor of ( 2i)2 = 4, for a total contribution of ( 4)r . For more details, the interested reader should consult the book Theorie algebrique des nombres by Samuel. 2

Example. Consider the Gaussian integers A = Z[i]  k R = C = R+R i. The fundamental domain of A

i D 1

has volume 1, and disc(A) = 4. De nition. A subset P of Euclidean space is centrally symmetric if for each point x of P , x is also a point of P . Theorem 16. Suppose that L is a lattice in Rn with fundamental domain D, P is a closed, bounded, convex, centrally symmetric subset of Rn and vol(P )  2n vol(D). Then P contains a nonzero point of L. Proof. We will prove the theorem in the case vol(P ) > 2n vol(D). A simple limiting argument gives the general case since P is closed and L is discrete. If we let 12 P = f 12 x : x 2 P g then the condition vol(P ) > 2n vol(D) is equivalent to the condition vol( 12 P ) > vol(D). We claim that there are distinct points x; y 2 21 P such that x y 2 L. This will establish the theorem since it follows that x = 12 p and y = 21 q for some points p; q 2 P (by central symmetry). Then 0 6= x y = 12 p + 21 q 2 P by convexity, and x y is the desired point. We give an intuitive argument to see that such x and y exist. The interested reader can nd the details in Samuel. We can translate pieces of 12 P mod L to ll up D, like cutting up a round carpet to cover the oor of a square room. Since vol(P ) > 2n vol(D), it follows that some point of D is double-covered in this process. That is, there are distinct points x; y of 1 P whose images via some translation by a vector of L are the same., So, x  y (mod L), 2 as desired.

12

Theorem 17. If L is any lattice in k, then there exists a nonzero element of L such that

 r jN ( )j  nnn! 4 jdisc(L)j1=2 : 2

Proof. For t > 0, let Pt be the following region in Rn  = k R:

n

n X

Pt = 2 k R :

i=1

ji ( )j  t

o

As the sketch suggests

t

R C

t=2 t the volume of Pt is

 r n (1) vol(Pt ) = 2r 2 tn! which can be veri ed directly by integration. Choose t so that vol(Pt ) = 2nvol(D). By Proposition 14, 2

1

(r1 +2r2 ) jdisc(L)j1=2

vol(Pt ) = 2n vol(D) = 2

= 2(r +r ) jdisc(L)j1=2 (2) 2r Then Pt satis es the hypotheses of Theorem 16, and we may x 2 (L r f0g) \ Pt . Then

jN ( )j1=n =

n Y

| i=1

1

2

2

ji ( )j

1=n

{z

geometric mean

n X  n1 ji ( )j  nt

} |

i=1{z

}

arithmetic mean

n

which implies that jN ( )j  nt n . From equations (1), (2) and (3) it follows that

 r n jN ( )j  nt n = nnn! 4 jdisc(L)j1=2 2

as desired.

Corollary 18. If A is the ring of algebraic integers in a number eld k, then n  r jdisc(A)j1=2  nn! 4 : 2

13

(3)

Proof. Every 2 A r f0g has jN ( )j  1, so the theorem implies that for some such ,  r 1  jN ( )j  n! 4 jdisc(A)j1=2

nn

2



from which the result is immediate. Corollary 19. (Minkowski) If disc(A) = 1, then A = Z and k = Q . Proof. The function of n:

n2n   2n  n2n   2r  jdisc(A)j = 1 (n!)2 4 (n!)2 4 is monotone increasing, and for n = 2 n2n   2n > 1: (n!)2 4 It follows that n = 1. 2

Lecture 4 Let k be a number eld with ring of algebraic integers A, and let d = disc(A). We now use the geometry of numbers to give a bound on d in terms of n = (Q : k). By Corollary 18 n  r jdj1=2  nn! 4 : 2

By Stirling's formula, n! = (n + 1), so there exists  in the interval (0; 1) such that

p n! = 2nnn e( n+

and it follows that

(n n  r jdj1=2  nn! 4 = e p 2

Raising both sides to the 2=n power, we nd that

jdj1=n r1

 e2

We can write e2 = (e2 )n=n = (e2 )( n +





jdj1=n

r n )

2 2

) 6 2

2n

pn

2

  r

2

4

 n )

6 2

2n

 (e2 )r1 =n

 2  2r =n e( e 4

2

pn

:

 n )

6 2

2n

jdj1=n 'n!1 (7:3)r =n (5:8)2r =n 1

14

2

:



so that

 n The factor e pn 2n ! 1 as n ! 1 so that (

n )

12

  2r =n  e( 4

n )

12



:

Example. If n = 2r2 (the case when k is totally complex) then asymptotically jdj1=n  5:8 so that jdj  (5:8)n . This inequality actually holds for n  108. Stark-Odlyzko have given the following improvement on the bound. Assuming the Riemann Hypothesis for the zeta function of k:

jdj1=n  (215)r =n (44)2r =n : In the totally complex case n = 2r2 , this gives the bound jdj  (44)n. 1

2

Now we consider the arithmetic properties of the ring of integers A of a number eld k. It is a fact that A is a Dedekind domain, i.e., a smooth ring of dimension 1. In particular, every ideal I in A is a product of prime ideals. However, if we take an arbitrary order B in k, then B will generally not be a Dedekind domain. Among other things, we want to know what keeps B from being a Dedekind domain. We note that there are standard methods of constructing new Dedekind domains from old ones. If A0 is a Dedekind domain, with eld of quotients k0  A0 and k is a nite separable extension of k0 , then we let A be the integral closure of A0 in k. It follows that A is a Dedekind domain, as well. Given two lattices L and M in k, we have ways of combining L and M to get other lattices in k. (We will apply these constructions mostly to ideals of an order B , but they are valid in this more general context.) The intersection L \ M is a lattice in k since there is a nonzero rational number such that L  M . This implies that there are nonzero integers a; b such that aL  bM , and it follows that for all x 2 L, ax 2 aL \ bM  L \ M . That is, L=(L \ M ) is torsion and nitely generated, so that L \ M has nite index in L. The sum L + M is also a lattice because the same argument as above shows that a(L + M )  M so that (L + M )=M is also nitely generated and torsion. The product

LM =

nX i

i mi : i 2 L; mi 2 M

o

is a lattice because if m1 ; : : : ; mn is a basis for M then L  M = Lm1 +    + Lmn is a nite sum of lattices. The quotient

L=M = f 2 k : M  Lg = \ni=1 Lmi 1 is a nite intersection of lattices and therefore a lattice. The quotient lattice should not be confused with the quotient of a group by a subgroup since M is not necessarily contained in L. We call this lattice the quotient and use the fraction notation because the de nition implies that (L=M )  M  L, so that in some sense L=M (almost) behaves like an object which we might call L  M 1 . It is important to note that strict containment may occur here, so this justi cation is, at best, imprecise. We also note that L  M = L M and L=M = hom (M; L). Finally, L=L = f 2 k : L  Lg = End(L) is a ring, and therefore an order in k. If we let B = L=L, then L is a left B -module in k. Fix an order B in k and let I be a nonzero ideal of B . Then I is a lattice in k since for a nonzero element 2 I , B  I  B . B is a lattice, so it follows that I has nite index in B . Furthermore, by de nition, B  I=I . Lemma 20. The index (B : B) = jN( )j. Z

Z

15

Proof. (B : B ) = j det( : B ! B )j = jN ( )j by de nition.

De nition. If I is a nonzero ideal in an order B, then we de ne the norm of I to be the index N (I ) = (B : I ). De nition. If L is any lattice in k (not necessarily contained in B) with End(L)  B, then we say that L is a fractional ideal of B . The condition End(L)  B is equivalent to the condition BL  L. Proposition 21. An order B in k is a domain of dimension 1, i.e., every nonzero prime ideal P  B is maximal. Proof. B is a domain since it is a subring of a eld. Since P has nite index in B , the ring kP = B=P is a nite integral domain. For a nonzero element of kP the multiplication map : kP ! kP is therefore injective, and since kP is nite, the map is also surjective. This implies that kP is a eld since this homomorphism maps onto the identity. Thus, P is

maximal.

De nition. An ideal I of B is called principal if I = B for some 2 B. We notice that a principal ideal I = B satis es the condition End(I ) = B :

End(I ) = f 2 k : B  B g = f 2 k : B  B g = B The last equality follows from two facts: (1) every element of B maps B ! B since B is a ring, so \"; and (2) if 2 k amps B ! B , then = (1) 2 B  B , so \". We claim that non-maximal orders are not principal ideal domains (even though they are all domains of dimension 1). To see this, let B  A be an order with index f = (A : B ) > 1. Then the ideal fA is principal in A and is contained in B (and is an ideal in B since fA  B  A). However, fA is not a principal ideal of B since, if it were, then the above notes would show that End(fA) = B . But End(fA) = A 6= B so this is not possible. Example. Here is a case where the ring of integers A  k is not a principal p ideal domain. Let p be a positive prime in Z p such that p  1 (mod 4) and let k = Q ( p). Then by our work in Lecture 2, A = Z + Z( p) since the discriminant is d = 4p and

 4p + p 4p  p p) = Z + Zp p A = Z+ Z = Z + Z( 2p + 2 p

Let I = fa + b p : a  b p(mod 2)g  A. I is pan ideal of A, as follows. Clearly, I is closed under addition. Take a + b p 2 I and + p 2 A. Then

p p)( + p p) = (a b p) + p p(a + b )

(a + b

The relation between a and b says that a and b are either both odd or both even. If a and b are both even, then the product is in 2A  I . If they are both odd, then since p is odd, a b p  + (mod 2) and a + b  + (mod 2), so mod 2 the coecients of the product are the same and the product is in I . It is straightforward to check (by the de nition of I ) that the containments A  I and I  2A each have index 2. Now, suppose that I were principal, say I = A. Then N ( ) = (A : A) = (A : I ) = 2. However, if = a + bp p, then 2 = N( ) = a2 + pb2 , which can never occur since p  5. Thus, I is 16

not principal. Note that this example encompasses the classical example which states that p Z+ Z 5 is not a principal ideal domain. Next, we want to show that if P  A is a nonzero prime ideal, then P  (A=P ) = A. That is, A=P is a fractional ideal which \inverts P ". Proposition 22. Let B be an order in k and P a nonzero prime in B. Then exactly one of the following holds. 1. P  (B=P ) = B , or 2. P=P = End(P ) ) B .

This proposition follows from the following theorem. Theorem 23. Let B be an order in k and P a nonzero prime in B. Then B=P

)

B.

Example. If p is a prime in Z, then Z=pZ = p1 Z ) Z. Proof of Proposition 22. We have the following containments in general:

P = B  P  (B=P )  P  B: If (B=P )  P = B , then 1. holds. Otherwise, (B=P )  P =( B . It is straightforward to show that (B=P )  P is an ideal in B . The maximality of P implies that P = (B=P )  P , and the theorem implies that there exists 2 (B=P ) r B . Then P  (B=P )  P = P which implies that 2 (P=P ) r B , which implies 2. We need a bit more machinery before we can prove the theorem. Let M be a nonzero, simple, nite B -module. Then M is cyclic, say M = Be. Let P be the annihilator of M , P = Ann(e)  B . Then the map B=P ! M given by b 7! be is an isomorphism of one-dimensional vector spaces over the residue eld B=P . Proposition 24. Any nonzero ideal I in an order B of k contains a product of prime ideals P1 ; : : : ; Pt . The primes which occur in this product are exactly the primes which contain I . Proof. Since I has nite index in B , we can nd a chain of ideals I = I0  I1  I2      It = B such that each quotient Im =Im 1 is a simple B -module. By the previous remark, there are prime ideals P1 ; : : : ; Pt such that Im =Im 1  = B=Pm . It follows that Pm Im  Im 1 , and therefore, P1    Pt  I0 = I . To prove the second claim, let P  I  P1    Pt . By maximality, it suces to show that some Pi  P . Suppose that no Pi is contained in P . Then there are elements i 2 Pi r P , and the product of these elements is in P1    Pt  I  P . However, the product is not in

P since P is prime and no factor is in P , yielding a contradiction.

Q We note, in addition, that the norm N (I ) = ti=1 N (Pi ). The chain of ideals I = I0  I1  I2      It = B shows us that I ) = (B : I ) = (It : It 1 )    (I1 : I0 ) = (B : Pt )    (B : P1 ) = N (Pt )    N (P1 )

N(

17

as claimed.

Example. In the notation of the proposition, I may not equal P1    Pt. Let k be a quadratic eld with ring of integers A. By Proposition 13, for a xed prime p there is a unique order B = Z + pA  A with index (A : B ) = p. By our notes above, the ideal of A P = pA  B is also an ideal in B which is prime and not principal in B . Furthermore, I = pB  pA. P is the unique prime of B containing I and P ) I  P 2 = p2 A, but I 6= P 2 since p 2 I r P 2 . Proof of Theorem 23. It is clear from the de nition that B  B=P . To see that equality does not hold, x a nonzero element of P and let I = B  P  B . By Proposition 24, there are primes P1 ; : : : ; Pt such that I  PP1    Pt . If I  P1    Pt , then P = Pi for some i. Cancel o as many \factors" of P as possible so that we may assume that P1    Pu 6 I and PP1    Pu  I . Fix 2 (P1    Pu ) r I . Then in k, = = 62 B and P  B so that

2 B=P . Corollary 25. If P is a prime ideal in the ring of integers A, then P  (A=P ) = A. Proof. By Proposition 22, if P  (A=P ) 6= A, then End(P ) is a ring in k which properly contains A. But every order in k is contained in A, so this is a contradiction. Theorem 26. (Dedekind) Let A be the ring of integers in a number eld k. 1. Every nonzero prime ideal is invertible, in the sense that there is a fractional ideal P 1  k such that PP 1 = A. 2. Every ideal is uniquely the product of prime ideals I = P1    Pt . 3. The set Ideal of fractional ideals forms an abelian group, freely generated by the prime ideals ofQA. The identity element is A, and for any fractional ideal I , I 1 = A=I . If I = P P ordP (I ) , then the isomorphism Ideal ! P Z is given by I 7! (: : : ; ordP (I ); : : : ). Proof. 1. follows from Corollary 25 since A=P is a fractional ideal. We shall prove 3. in the following lecture. 2. As in the proof of Proposition 24, we have ideals Ij and prime ideals Pj such that

P1    Pt  I = I0  I1      It = B If t = 1, then the only prime containing I is P1 and P1  I  P1 which implies that I = P1 . So, we proceed by induction. The chain I1  I2      It = B satis es the induction hypothesis, so

P2    Pt = I1  I  P1    Pt = P1 I1 Multiplying by P2 1    Pt 1 yields

A  IP2 1    Pt 1  P1 If A = IP2 1    Pt 1 , then multiplying by P2    Pt implies that I = P2    Pt . This implies that A=I has a Jordan-Holder series with t 1 links, a contradiction. Thus, 18

A 6= IP2 1    Pt 1 , and the maximality of P1 implies that IP2 1    Pt 1 = P1 so that multiplying by P1    Pt implies that I = P1    Pt . For uniqueness, we write I = P1    Pt = P10    Pu0 . Then P1  P10    Pu0 which implies that P1 = Pi0 for some i as in the proof of Proposition 24. Multiply by P1 1 and apply induction.

Lecture 5 Proof of Theorem 26, Part 3. We already have unique factorization for integral ideals. To factor and invert an arbitrary fractional ideal, we use the following lemma to reduce to the case of integral ideals. Lemma 27. If L is any fractional ideal of A, there exist integral ideals I; J such that I  L = J. Proof. Let J = L \ I  A, which is an integral ideal. Our result follows from the following sublemma. Lemma 28. If J and L are fractional ideals such that J  L, then there exists an integral ideal I such that I  L = J . Proof. By our notes in Lecture 1, since A and L are both lattices, there exists nonzero rational number such that L  A. It follows that J  L  A. Since J and L are therefore integral ideals in A, there are prime ideals P1 ; : : : ; Pm and Q1 ; : : : ; Qn such that J = P1    Pm and L = Q1    Qn. The containment J  L and the proof of Proposition 24, show that (after rearranging the Pi ) Q1 = P1 . Thus, we multiply by Pi 1 and get J 0  L0 = Q2    Qn  A. Continuing this process, we see that

J = P1    Pm = Q1    QnPn+1    Pm = LPn+1    Pm which implies that J = L  I where I = Pn+1    Pm . The rest of the theorem now follows. We notice that I  J if and only if ordP (I )  ordP (J ) for all prime ideals P .

De nition. The set P of principal fractional ideals P = f A : 2 kg is a subgroup of the group I of all fractional ideals, since ( A)( A) = ( )A and ( A) 1 = 1 A. We de ne the ideal class group of A (or k) as the quotient C = I=P. Proposition 29. If k is a number eld with ring of integers A, and Ideal and C are as above, then we have a long exact sequence

 C !1 1 ! A ! k ! I !

where  is the natural inclusion,  is the natural projection and  is given by 7! A.

19

Proof. It is straightforward to check exactness at each point, except maybe at k. An element 2 k is in ker() if and only if A = A. This occurs if and only if and 1 are in A, since 1 2 A, and this occurs if and only if 2 A .

We note, for the people who know some K -theory, that this comes from the long exact sequence in K -theory.

K1w(A)

- K1w(k)

A

- k

ww

ww

which continues to the left as

K2 (A)

= z - P K0w(}|A=P {) - K0w(A) - K0w(k) - 0 ww ww ww Z

-I

- Z+ C - Z

- K2w(k)

- P K1w(A=P ) -

-0

ww

ww

(k k )=S - P (A=P ) where S is the subgroup generated by the Steinberg relations and is the tame symbol. Theorem 30. If k is a number eld with ring of integers A, and I and C are as above, then C is nite and A is nitely generated.

Example. It is a very special property of number elds that C is nite and A is nitely generated. We give an example to show that this is not true in general. Let   C be a lattice, that is, a discrete, co-compact, additive subgroup. Fix nonzero complex numbers w1 ; w2 such that w1 =w2 62 R and  = Zw1 + Zw2. Let E = C = and let A denote the ring of meromorphic functions E ! P1(C ) whose only poles lie in . Then A = H 0 (E r f0g; OE ) = C [p; p0 ]=((p0 )2 = 4p3 g2 p g3 ) For any z 2 E r f0g, let Pz = ff 2 A : f (z ) = 0g  A: Q m(z) = Each Pz is a maximal ideal in A with A=Pz ! C given by f 7! f (z ). When is z Pz principal, given by gA? If g : E ! P1(C ) has a pole at 0 and zeroes of order m(z ) at z 2 E r f0g, then the divisor of g is div(g) =

X

z2E r0

P

X

m(z )z

z2E r0

m(z )(0):

Abel's Theorem says that div(g) = P 2E mP P satis es the relations X X mP = 0 and mz z  0 (mod ): z

P

P Q Thus, z Pzm(z) = gA if and only if z mz z 2 . The exact sequence from Proposition 29 is then M  0 1 ! C ! C (p; p ) !

C

z2E r0

Z

!E!0

= A is not nitely generated and E = C is not nite. 20

Proof of Theorem 30. The fact that A is nitely generated is the Unit Theorem, which we shall prove in Lecture 8. For now we prove that the ideal class group is nite. The norm which we de ned on integral ideals N (I ) = (A : I ) extends to the set of all fractional Q If I is a fractional ideal with prime factorization Q ideals in a well-de ned manner. I = i Piai , then we de ne N (I ) = N (Pi )ai . This agrees with the original de nition for integral ideals, by a composition series argument as in the proof of Proposition 24. Furthermore, we note that since A=Pi is a nite eld, say with qi = pbi i elements, N (Pi ) = (A : Pi ) = qi . As we observed previously, the norm of an element of k is not necessarily the same as the norm of the principal ideal it generates. However, N ( A) = jN ( )j. For an integral ideal I , we compute the discriminant, as a lattice: disc( A) = disc(A)(N ( A))2 . Finally, it is clear that the norm map, as de ned, extends to a homomorphism of abelian groups N : I ! Q + . Our result will follow from the following theorem. Theorem 31. Every ideal class c 2 C is represented by an integral ideal I such that N(I )  e(k) where  r e(k) = n! 4 jdj1=2

nn

and d = disc(A).



2

Before we prove Theorem 31, we see that it implies Theorem 30, as follows. The cardinality of C is at most the cardinality of the set of integral ideals I such that (A : I ) = N (I )  e(k). In turn, this is at most the cardinality of the set of lattices in A  = Zn of index at most e(k). By the theory of nitely generated abelian groups, this cardinality is nite, so C is nite. Proof of Theorem 31. Let J be a fractional ideal in the class c, and consider the lattice J 1 . By Theorem 17, there exists a nonzero element of the lattice J 1 such that

 r  r 1=2 jN ( )j  nnn! 4 jdisc(J 1 )j1=2 = nnn! 4 jNd(j J ) Since 2 J 1 , A  J 1 and I = J  A. Thus, I and J are in the same class of C , since 2

2

they di er by the principal fractional ideal A. And N(

 r 1=2 I ) = jN ( )j  N (J )  nnn! 4 jNd(j J )  N (J ) = e(k) 2

as desired.

De nition. The class number of a number eld k is denoted h = h(k) and is de ned to be

the cardinality of the ideal class group C of k. Example. If n = 1 (so that k = Q ) then r1 + 2r2 = n = 1 ) r2 = 0 and jdj = 1 so that e(k) = 1. Since every element of C is represented by an integral ideal of norm at most 1, it follows that C = (1). If n = 2 then k is a quadratic extension. If d < 0 then D < 0 so that k is not a real eld. That is, r2 = 1. Then e(k) = 2 jdj1=2 which is strictly less than 2 for the allowable 21

discriminants d = 3; 4; 7; 8. In this case, C = (1) again, and pthis implies that every fractional ideal is principal. In particular, each such A = Z + Z( d+2 d ) is a principal ideal domain. If d > 0, then r2 = 0 and e(k) < 2 for d = 5; 8; 12; 13, and the rings of integers in these cases are also principal ideal domains. If e(k) < 3, then every nontrivial class in C is represented by an integral ideal I such that N (I ) = 2. Since this index is prime, there are no lattices strictly between I and A. In particular, I = P is prime and A=P is the eld of two elements. When n = 2 and d < 0 we see that 2  e(k) < 3 when d = 11; 15; 19; 20. In these cases

 d + pd 

= Z[x]=(x2 dx + d 4 d ) 2 We wish to know when, under these assumptions, there is a prime ideal P with N (P ) = 2. This occurs if and only if there is a surjection A ! Z=2Z. To nd such a surjection, it is equivalent to nd a solution of the congruence x2 dx + d 4 d  0 (mod 2), so that we know where to send x, since the image of Z is already determined. Furthermore, the number of distinct solutions (mod 2) is exactly the number of such maps, and therefore the number of such prime ideals. We check this by cases. (a) If d  0 (mod 4), then the congruence is x2 + d4  0 (mod 2), which has exactly one solution. Otherwise, d  1 (mod 4). (b) If d  1 (mod 8), then the congruence is x2 x  0 (mod 2), which has exactly two solutions. (c) If d  5 (mod 8), then the congruence is x2 x +1  0 (mod 2), which has no solutions. We correlate this with our possibilities for d. (a) If d = 20, then there is exactly one prime ideal with norm 2. (b) If d = 15, then there are exactly two prime ideals with norm 2. (c) If d = 11; 19, then there are no prime ideals with norm 2. Thus, for d = 3; 4; 7; 8; 11; 19 the class number is h = 1. For the cases d = 20; 15, we have more work to do. For d = 20, h = 2. The p generator of the class group in this case is a prime P of norm 2. In this case,p A = Z + Z( 5) and it follows that P is not principal. If P were principal, say P = (a + b 5)A, then 2 = N (P ) = a2 + 5b2 which is impossible since a and b are integers. Since C has order 2, the class of P has order 2, so that P 2 is principal. The congruence we solve to nd P in this case is x2 5  x2 + 1  0 (mod 2). It follows that x 7! 1 2 Z=2Z and the kernel is P = h2; x 1i. It is straightforward to verify that P 2 = 2A which has index 4 in A since N (2) = 4. p For d = 15, we know that h  3. In this case A = Z[ 1+ 2 15 ], and the prime ideals of index 2 are P = h2; xi and P 0 = h2; x 1i. Every class in C is represented by A, P or P 0 , and the same check of norms as above shows that neither prime ideal is principal. Thus, h = 2 or 3. Even though we have two distinct prime ideals of index 2 in A, it pturns out that they represent the same nontrivial class in C and h = 2. Let = 2, = 1+ 2 15 and p

= 1 2 15 in A. Each element has norm 4, and the ideals generated by these elements are all distinct, since none of the pairwise quotients are in A. It ends up that

A = Z+ Z

2

2

PP 0 = A

P 2 = A 22

(P 0 )2 = A

so that in the class group [P ][P 0 ] = [P ][P ] which implies that [P 0 ] = [P ]. Another way to see the fact that C has order 2 is to observe that if it had order 3, then each nontrivial element would have order 3, which does not hold in this case. We should observe that for quadratic elds with negative discriminants, it has been proven that the only examples with trivial class group are those with

d = 3; 4; 7; 8; 11; 19; 43; 67; 163: However, for positive discriminant the list seems in nite. Whether or not the list is actually in nite is still an open question. p The case d = 163 is particularly interesting. If = 1+ 2 163 , then A = Z + Z . The minimal polynomial of is f (x) = x2 + x + 41. The interesting fact here is that the values f (0); f (1); : : : ; f (40) are all prime numbers. Of course, f (41) = 41(41 + 1 + 1) is not prime, but people considered functions of this type when trying to nd prime-generating functions. De nition. We are interested in nding the prime ideals of an order B. We de ne the characteristic of a prime ideal P to be the characteristic of its residue eld B=P .

Lecture 6 Let k be a number eld with ring of integers A, and let B be any order in k. The prime ideals P of B with characteristic p are exactly the maximal ideals containing pB , as char(P ) = p if and only if, under the natural projection B ! B=P , the number p 7! 0, which occurs if and only if p 2 P or equivalently that pB  P . Since the elements which occur in the Jordan-Holder series for pB  B are exactly the prime ideals which contain pB , the above remarks show that the prime ideals in B of characteristic p are exactly the primes which occur in the product P1    Pt  pB from Proposition 24. In particular, there are only nitely many primes in B having a prescribed characteristic. Let P1 ;    ; Pu be the prime ideals of characteristic p. Then there are positive integers e1 ; : : : ; eu such that P1e    Pueu  pB . (ei is the multiplicity with which Pi occurs as a Jordan-Holder factor in B=pB .) Since each Pi has characteristic p, the eld B=Pi has order pfi for some positive integer fi . Since B is a lattice of rank n, pB has index pn in B , so 1

u Y Yu P e n i p = (B : pB ) = N (Pi ) = (pfi )ei = p( i ei fi ) i=1 i=1 P from which it follows that n = ui=1 fi ei  u. That is, there are at most n prime ideals of

characteristic p in B . We consider these ideas in the simplest of cases. Assume that is an algebraic integer with minimal polynomial f (x) of degree n. Then B = Z[ ] = Z + Z + Z 2 +    + Z n 1 is an order p in k = Q ( ). For example, if n = 2, then every order B is of this form with = dB +2 dB where dB = disc(B ). Then B  = Z[x]=(f (x)) and

B=pB = Z[x]=(p; f (x)) = (Z=pZ[k])=(f~(x)) = 23

Y i

(Z=pZ[k])=(gi(x)ei )

Q where f~(x) is the reduction of f (x) modulo p and f~(x) = i gi (x)ei is the prime factorization of f~(x). It follows that the set of maximal ideals Pi of characteristic p with multiplicity ei in the Jordan-Holder decomposition are exactly the ideals hp; g^i (x)i where g^i (x) 2 Z[x] is a lift of gi (x), and that under this correspondence (with the previous notation) fi = deg(gi (x)). Thus, we have all the machinery to prove the following. Corollary 32. If k = Q ( ) where is an algebraic integer,Qande p is a prime whose square does not divide disc( ) = disc(Z[ ]) 2 Z r f0g, then pA = Pi i mirrors the factorization of f (x) modulo p. More precisely, if deg(Pi ) = fi where #(A=Pi ) = pfi , then deg(Pi ) = deg(^gi (x)) and the number of times that Pi occurs as a factor in the Jordan-Holder series for A=pA is exactly ei , the multiplicity of the factor gi (x) in f~(x). Moreover, if p does not divide disc( ), then all the ei = 1. De nition. In this notation, we call ei the rami cation index of Pi over p, and we call fi the residue degree. If ei  2, then we say that Pi is rami ed over p. Proof. We claim that the index N of Z[ ] in A is prime to p. The formula N 2 disc(A) = disc(Z[ ]) = disc( ) shows that if p divides N then p2 divides disc( ), a contradiction. Fixing integers r; s such that 1 = pr + Ns, it is straightforward to check that the map Z[ ]=pZ[ ] ! A=pA is an isomorphism. The previous discussion and the results on prime factorization in A imply the result.

Example. Assume that k is a cubic extension of Q with ring of integers A such that 2A is a product of three distinct primes 2A = P1 P2 P3 . For any 2 A r Z, k = Q ( ) since the

degree is prime. However, Z[ ] ( A, because if they were equal, then the factorization of f (x) with coecients reduced modulo 2 would have three distinct linear factors. This is a contradiction since there are only two distinct linear polynomials in Z=2Z[x]. An example where this occurs is when k is the unique cubic extension of Q contained in the cyclotomic eld Q (31 ), where 31 is a primitive 31st root of unity. The essential reason why 2A factors as claimed is in the congruence 2  73 (mod 31). We shall discuss this example in more detail in the future. Next, we want to investigate the relation between d = dA = disc(A) and the ei . We recall that the dual lattice of A is de ned to be A = f 2 k : Tr ( A)  Zg. By Proposition 1, jdj is the index (A : A). De nition. We de ne the di erent of A to be the integral ideal D = (A) 1  A 1 = A. By Lemma 2, N (D) = (A : D) = jdj. We adopt the following notation. Let P have characteristic p and let eP denote the number of times P occurs as a factor in the JordanHolder series for A=pA. Proposition 33. D = QP P mP for nonnegative integers mP , and mP  eP 1 with equality if and only if char(P ) - eP .

De nition. If mP > eP 1, then we say that P is wildly rami ed. Otherwise, we say that P is tamely rami ed.

Before we prove the proposition, we list some consequences. Corollary 34. If k is quadratic, then the only primes p which have a rami ed factor P in A are the primes dividing d. 24

p

Proof. We assume that k = Q ( D) where D is square-free, and recall from Theorem 11 that d  0; 1 (mod 4) is as square-free as possible. Also, if d  1 (mod 4) then d = Dp=  Qodd pi , and if d  0 (mod 4) then d = 4D = 2a Qodd pi where a is 2 or 3. If  = d+2 d , then A = Z + Z. p We claim that D = dA  A and that the index is N (D) = (A : D) = jdj. The rst claim follows from the fact that A = p1d A, which the interested reader is invited to check. The second claim follows from Proposition 1, since N (D) = (D : A) = (A : A) = jdj. P Q As before, pA = i P1ei where i ei fi = n = 2. Then P is a rami ed factor of p if and only if P = P1 and e1 = 2. This implies that mP  eP 1 = 1 so that P occurs in the factorization of D. Thus, d = N (D) is divisible by N (P ), which is a power of p.

By the proof, if k is quadratic, then D is principal so that its class in the class group C is zero. In general, this is not the case. However, we have the following theorem, which we shall not prove (c.f. Andre Weil Basic Number Theory, Theorem XIII.12.13). Theorem. If k is a number eld, then the class of the di erent D in the class group C is contained in 2C . We shall use the following proposition to prove Proposition 33. Proposition 35. Let B be an order in a number eld with dual module B, p a prime number, and Pi the Jordan-Holder factors of pA with multiplicities ei . Then \i Pi  pB  , and \i6=i Pi  pB  if and only if pjei . 0

0

Proof. Let I = \i Pi . We have the following chain of equivalences: I  pB  i 1p I  B  i Tr ( 1p I )  Z i Tr (I )  pZ i for every 2 I , Tr ( )  0 (mod p) i for every 2 I , the multiplication map ~ : B=pB ! B=pB has trace zero. The map ~ is a linear endomorphism of the n-dimensional vector space B=pB over Z=pZ. Using the Jordan-Holder decomposition of B=pB , we can write a matrix for ~ consisting of diagonal blocks each of which gives the action on the corresponding B=Pi . Thus,

Tr (~ : B=pB ! B=pB ) =

P

X i

ei Tr (~ i : B=Pi ! B=Pi )

and I  pB  i the sum i ei Tr (~ i : B=Pi ! B=Pi ) = 0. Thus, if 2 I then each ~i = 0 so that Tr (~ i )  0 and I  pB  . Let I0 = \i6=i Pi and x 2 I0 . If 2 I , then each TrP(~ i )  0, so assume that 62 I . Then every such is in pB  i the i0 term in the sum i ei Tr (~ i ) is zero. It suces to show that there is some such with Tr (~ i ) 6 0 so that the previous condition is equivalent to ei  0 (mod p), as desired. For i 6= i0 , Pi + Pi = B by maximality, so the Chinese Remainder Theorem implies that I0 + Pi = B . It follows that the projection map I0 ! B=Pi is surjective, and it suces to nd an element 2 B=Pi such that Tr ( : B=Pi ! B=Pi ) 6 0. Since B=Pi is a eld, we just choose to be a primitive element over Z=pZ so that the corresponding trace is 1. 0

0

0

0

0

0

0

0

0

Q

Proof of Proposition 33. Q Let B = A in Proposition 35 so that pA = i Piei . Then Q since the Q e 1  i Pi are maximal, \i Pi = i Pi  pA = i Pi D . This implies that D  i Piei 1 with equality if and only if no ei is divisible by p.

25

Example. We consider the factorization of the ideals p pA in a quadratic eld k. In particular,

we prove the converse of Corollary 34. If = d+2 d , then the minimal polynomial of is f (x) = x2 dx + d 4 d and the ring of integers of k is A = Z + Z . We apply Corollary 32 in two cases. Case 1: p = 2. If d  0 (mod 4), then f (x) reduces to x2 d4 which factors to x2 or (x + 1)2 depending on whether d4 is even or odd. Thus, 2A = P 2 . Also, ord2 (D) = 2 or 3 since this is the exponent for the power of 2 dividing d = (A : D). If d  1 (mod 4), then the reduction of f (x) modulo 2 is 2

(

2 if d  1 (mod 8), f~ (x) = x2 x x x + 1 if d  5 (mod 8) If d  1 (mod 8), then f~ (x) has two distinct factors and 2A = P1 P2 is a product of distinct primes. (In this case, we say that p is split.) If d  5 (mod 8), then f~ (x) is irreducible and 2A is prime. (In this case we say that p is inert.) Notice that since 2 - d, ord2 (D) = 0. Case 2: p > 2. If pjd, then modulo p, f~ (x) = x2 so that pA = P 2 and ordp (D) = 1 since d isp as square-free as possible. If p - d, then we notice that in Z=pZ the roots of f~ (x) are d d , which makes sense if d is a square modulo p, since 2 is a unit in Fp . Thus, if d is a 2 square modulo p, then since d 6 0 and p > 2 we see that the zeros of f~ (x) are distinct and so p splits: pA = P1 P2 . If d is not a square modulo p, then f~ (x) is irreducible and p is inert: pA = P . We should note that we can use the Legendre symbol ( dp ) to summarize. Recall that if p is a prime which does not divide an integer a, then  a  (1 if a is a square modulo p, p = 1 if a is a square modulo p Thus, for any prime p 8 2 >

P1 P2 if p = 2 and d  1 (mod 8), or if p > 2 and ( dp ) = 1, :P if p = 2 and d  5 (mod 8), or if p > 2 and ( dp ) = 1,

The basic properties of the Legendre symbol may be found in almost any basic text on number theory. Probably the most fundamental result is the Law of Quadratic Reciprocity. Theorem. (Gauss) If p and q are distinct odd primes, then

 p  q 

from which it follows that

q

p =

1

( p

2

1

)( q 2 1 )

 p  (( pq ) q =

Furthermore,

if p  1 (mod 4) or q  1 (mod 4), ( pq ) if p  q  3 (mod 4)

2

p =

1 26

( p

2 1 8

)

It is unexpected but true that all cases of quadratic reciprocity are roughly equivalent to the fact that the factorization of p in A depends only on the residue class of p modulo d. p For example, if  is an odd prime such that   1 (mod 4) and k = Q ( ), then the discriminant is d =  and we have the following factorizations:

(

2 2A = P1 P2 if   1 (mod 8) i ( 2 ) = 1 by quadratic reciprocity P if   5 (mod 8) i (  ) = 1 by quadratic reciprocity

and if p is an odd prime distinct from 

(

p  pA = P1 P2 if ( p ) = 1 i (  )p= 1 by quadratic reciprocity P if ( p ) = 1 i (  ) = 1 by quadratic reciprocity

which shows that the factorization of pA (whether pA is split or inert) depends only on p modulo  = d. p If  is an odd prime such that   3 (mod 4) and k = Q ( ), then the discriminant is d =   1 (mod 4) and we have the following factorizations:

pA =

(

P1 P2 if ( p ) = 1 = ( p ) P if ( p ) = 1 = ( p )

which is the same behavior as the previous example. Assume that d = 4, so that k = Q (i) and A = Z[i]. Then for an odd prime p, pA is split i p  1 (mod 4) and is inert i p  3 (mod 4). The ideal 2A is rami ed and D = 2A = P 2 . In fact, for any quadratic eld k, there exists a group homomorphism  : (Z=dp Z) ! f1g such that pA splits i (p) = 1, and pA is inert i (p) = 1. In the cases Q ( ) above, we see that (p) = ( p ) : (Z=Z) ! f1g.  is necessarily unique, by the given property. For any number eld k, let T denote the set of nonzero prime ideals of A. Let U denote the prime numbers, and consider the following function of s:  1 Y Y   1 Y 1 N (P1 )s = 1 N (P1 )s P 2T p2U P pA

|

{z

at most n factors

For example, if n = 1 so that k = Q and A = Z, then this function is Y 1  1 1

ps

p2U

27

}

If n = 2, then there are three options for the factor corresponding to the prime number p:   1  1 1 1 N (P1 )s pA = P 2 = 1 ps (4) 1  11  = 1 p2s pA = P 1 N (P1 )s   1 1  1 = 1 1 1+

ps

pA = P1 P2



1

N(

1

P )s

ps

 1 (p)   1 ps = 1 p1s  1  1  1 1  1 = 1 ps   = 1 p1s

1 ps  1 1 p(sp)

1

1

Thus, in each unrami ed case, the factor corresponding to p is  1  1 (p)  1 1 1

ps

ps

Example. We consider the factorization of the ideals pA in a cyclotomic eld. Let  be an odd prime number,  =  a primitive th root of unity, and k = Q ( ). Let (x) = x 1 + x 2 +    + 1 so that  satis es the polynomial x 1 = (x 1)(x). We claim that

(x) is irreducible over Q . To see this, let  =  1, which satis es the polynomial

  1 y  + y  1 +    + y + 1 1 (1 + y )  1 1 = y + 1 y 2 +    +  (1 + y) = (1 + y) 1 = y

This polynomial is monic,  divides all the non-leading coecients and 2 does not divide the constant term. Therefore, by Eisenstein's criterion, (1 + y) is irreducible over Q , and the same is true for (x). Since (1 + y) is the minimal polynomial of  and  is odd, N ( ) =  the constant term of (1 + y ). Similarly, N ( ) = 1. We note, for historical interest, that Kummer studied cyclotomic elds extensively. He discovered that the ring of integers in this case is not a principal ideal domain (although, not in those terms) and developed the theory of ideals as a result.

Lecture 7 We continue with the notation from the previous lecture. Let 1 ; : : : ;  1 be the distinct of kQinto C . If we assume that  2 C , then we may assume that i ( ) =  i . Since  = N () = i=11 i () = ( 2 1)    (  1 1), we see that  divides  in A. In particular, A  A  A. Proposition 36. The ring of integers in k is A = Z[ ] = Z[]. Q -embeddings

28

Proof. The second equality is clear. Let B = Z[ ] = Z + Z +    + Z  2. Then dB = disc(B ) = det(i ( j ))2 as we noted in the proof of Proposition 10. This is exactly

01  BB1  2 dB = det B BB1. .3 @ .. .. 1 

  

2 4 6 1

.. .

 2(

1)



1 2(   1)  3( 1)

...

    (

.. .

1)( 1)

12 CC Y CC = ( i  j )2 CA i 2, a contradiction. 29

It follows from the proof that D = (  )A. We now return to the task of factoring the ideals pA. First, the ideal A is completely rami ed, in the sense that A = P  1 where P has residue degree f = 1 and rami cation index e =  1 = n, andQN (P ) = . Let P = A, which is prime since A=P = Z[]=() = F . Suppose that A = P e i Piei with ei  1. Since P  A, we know that e  1. Furthermore, since P and each Pi has characteristic

 1 = N ( ) = N ( P e

Y i

Piei ) = N (P )e

Y i

N(

P Pi )ei  (e+ i ei )

so that e; ei < . In particular, e and the ei are not divisible by . By Proposition 33, we see that the factorization of D contains P e 1 and Piei 1 so that

Pe

1 Y P ei

i

i =

 e A = D  P

1 Y P ei 1

i

i

so that each ei  ei 1, a contradiction. Thus, A = P e and  1 = N () = N (P )e = e so that e =  1, as claimed. If p 6= , then by our computation in Proposition 37, jdj =  2 so that p - d. By Corollary 32, Q the ideal pA completely splits as pA = i Pi and the factorization of pA mirrors the factorization of (x) modulo p. When does (x) have a linear factor modulo p? This occurs i there is an th root of unity in Fp . Since Fp is cyclic of order p 1, such an element exists i Fp contains a subgroup of order  i jp 1 i p  1 (mod ). In this case, though, Fp has  1 distinct elements of order  and so (x) splits into distinct linear factors modulo p and pA = P1    P 1 . In general, let f be the smallest integer such that j(pf 1), that is, f is the order of p in F . Then pA = P1    Pg for distinct Pi such that N (Pi ) = pf and g =  f 1 . The same reasoning as above shows that (x) splits completely over Fpf . Each zero of (x) in Fpf has f distinct conjugates, and the coecients of the factors of (x) modulo p will be the symmetric polynomials in these conjugates. It follows that, modulo p, (x) factors into g distinct factors, each of degree f . p Example. If  = 3, then  = 1+2 3 so that k = Q ( ) = Q (p 3). By our work above, we know that 3A = P 2 . If p  1 (mod 3), then pA = P1 P2 . If p  1 (mod 3), then f = 2 and g = 1 so that pA = P . Example. If  = 5, then F5 = Z=4Z. As before, 5A = P 4. If p  1 (mod 5), then pA splits completely as pA = P1 P2 P3 P4 . If p  1 (mod 5), then f = 2 and g = 2 so that pA = P1 P2 . If p  2; 3 (mod 5), then f = 4 and g = 1 so that pA = P . We observe that a cyclotomic eld k is always Galois over Q . The extension is separable because char(Q ) = 0, and it is normal since the conjugates of  are simply  i 2 k. Furthermore, the Galois group G = Gal(K=Q ) = (F ) . The homomorphism : G ! (F ) is given by  7! a (mod ) where ( ) =  a . The map is always an injection, and the irreducibility of (x) implies that is also a surjection. Back in the case  = 5, G = Z=4Z has a unique subgroup of index 2. The Galois correspondence implies that there is a unique quadratic sub eld L of k. If =  +  1 , then Q ( ) ( k since is real and Q ( Q ( ) since is irrational. Thus, Q ( ) must be the p unique quadratic sub eld of k. It will follow from the following theorem that Q ( ) = Q ( 5). 30

Theorem 38. (Gauss) Assume that  is a prime number,  > 2, and  =  2 C is a

primitive thqroot of unity. Let k = Q ( ). Then there is a unique quadratic sub eld L of k, and L = Q ( ( 1)(  ) ). 2

1

Compare this to the theorem which states that every quadratic eld is contained in a cyclotomic eld (cf. Lang, Algebra, Theorem VI.3.3). Proof. As we noted before, k is Galois over Q with Galois group G = F = Z=( 1)Z. Since  is odd, G has a unique subgroup of index 2, which implies that k has a unique quadratic sub eld. Let

D = ( 1)(  )  2

1

and

g=

1 a  X a a=1

 

It is not dicult to show that g2 = D (cf. Lang, Algebra, Theorem VI.3.3). Thus, L = p Q (g ) = Q ( D ) is the desired sub eld. Of course, the elds we have studied in depth thus far have all been Galois and cyclic over Q . This has made our computations relatively easy. The following example involves an extension with Galois group S3 , and we omit proofs of many of the facts. Example. Let f (x) = x3 x 1 and 2 C a zero of f (x). Any rational zero of f (x) would necessarily be 1, so f (x) is irreducible. Let k = Q ( ) and let K be a splitting eld of f (x) over Q . The discriminant of f (x) is 23 and therefore not a square, so that k is not Galois over Q and the Galois group of f (x) is G = Gal(K=Q ) = S3 . The ring of algebraic integers in k is A = Z[ ] and d = disc(A) = 23. Modulo 23, f (x) factors as (x 3)(x 10)2 . Thus, by Corollary 32, 23A = P1 P22 where the residue degrees are f1 = f2 = 1. In fact, P2 = D has index 23 in A. If p 6= 23 then there is no rami cation. When ( 23p ) = 1, i.e. p is not a square modulo 23, pA = P1 P2 where f1 = 1 and f2 = 2. When ( 23p ) = 1, either pA = P1 P2 P3 with the fi = 1 or pA = P with f = 3. These di erent possibilities mirror the di erent quadratic forms x2 + xy + 6y2 and 2x2 + xy + 3y2 with discriminant 23 as in Lecture 3: pA splits completely when p = x2 + xy + 6y2 for integers x; y, and pA is inert when p = 2x2 + xy + 3y2 for integers x; y. We note that prime factorization is known explicitly in relatively few cases. For example, if k has degree 5 over Q and has nonsolvable Galois closure, then a simple recipe for prime factorization is not known. Now, we are ready to build the machinery to nish the proof of Theorem 30. We state the result in more generality, as a separate theorem. Theorem 39. (Unit Theorem) Assume that k is a number eld and that B is an order in k. Then the group of units B  is a nitely generated abelian group. Furthermore, the torsion subgroup of B  is cyclic of nite order, and the rank of B  is r = r1 + r2 1. Finally, if  = f1g. r1 6= 0, then Btor It should be noted that the torsion subgroup of B  is exactly the set of roots of unity in B , since an element 2 B is torsion i s = 1 for some integer s. Example. If k = Q , then r1 = 1 and r2 = 0 so that A = f1g, not that this is a surprise. 31

p

If k = Q ( D) for a positive, square-free integer D, then r1 =p0 and r2 = 1. Thus, A is torsion. If D = 1, then A is cyclicp of order 4, generated by 1. If D = 3, then A is cyclic of order 3, generated by 1+2 3 . In all other cases, A = f1g. p If k = Q ( D) for a positive, square-free integer D, then r1 = 2 and p r2 = 0. Thus, Ap has rank 1, and since k  R, the torsion subgroup is f1g. If B = Z[ D] and  = a + b D is a unit in B , then a2 Db2 = N () = 1. The equation a2 Db2 = 1 is Pell's equation, whichp was studied in depth by Fermat, Wallis and Brounker, using continued fractions to nd D. Lemma 40. Assume that k is a number eld and that B is an order in k. Then  2 B is a unit in B i N () = 1. Proof. If  is a unit, then 1 = N (1) = N ()N ( 1 ), and since N () and N ( 1 ) are integers, N () = 1. Conversely, assume that N () = 1. The minimal polynomial of  over Q is xm Tr ()xm 1 +    + N (). Thus, (m 1 Tr ()m 2 +    ) = N() = 1. Since the element m 1 Tr ()m 2 +    is in B , we see that  is a unit in B .

Given a number eld k, there are r1 distinct embeddings of k in R, and there are 2r2 distinct embeddings of k in C whose images are not contained in R. We construct r1 + r2 valuations v : k ! R corresponding to the embeddings. Recall that a valuation on k is a homomorphism of multiplicative groups j j : k ! R such that j + j  j j + j j. We shall use the notation j jv in place of v( ) to remind us that v is a valuation. If  : k ! R, then de ne j jv = j( )j where j j represents the usual absolute value on R. If  : k ! C is a strictly complex embedding, then let j jv = j( )j2 where j j is the usual absolute value on C . If  : k ! C is the conjugate embedding, then it follows that j jv = ( ) ( ), so this does not depend on the choice of representative of conjugacy class. We refer to the valuations constructed in this manner as the in nite valuations, and we say that the valuation divides 1. Also, an in nite place is one of the following: a real embedding of k or a conjugate pair of complex embeddings. Q Claim: For any nonzero 2 A, r +r j jv = jN ( )j 2 Q + , where the product is taken over the distinct in nite valuations of k. To see this, we compute 1

jN ( )j =

n Y

i=1

j( )j =

2

Y

R

plc.

j( )j 

Y

C

plc.

( ) ( ) =

Y

r1 +r2

j jv

as claimed.

Corollary 41. A nonzero element 2 A is a unit i Qr +r j jv = 1. p Example.p If k = Q ( D) for a positive, square-free integer D, then r1 = 2 and rp2 = 0. If  = a + b pD = 6 1 is a unit in A, then jj1 > 1 and jj2 < 1 so that either ja + b Dj > 1 and ja b Dj < 1 or vice versa. We shall generalize this example in the next lecture. We can also nd valuations corresponding to nonzero prime ideals P of A. For 2 k , let ordP ( ) = ordP ( A) as in Lecture 4, and de ne j jP = N (P ) ord P ( ) = qordP ( ) where q = N (P ) = pf . The image of j jP in R+  = R is exactly q . In this case, we have j + jP  maxfj j; j jg. A valuation which satis es this property is called non-archimedean. 1

2

Z

The valuations constructed in this manner are the nite valuations on k. 32

Theorem 42.Q(Product Formula) If 2 k, then j jv = 1 for almost all valuations de ned thus far, and all v j jv = 1. Proof. For any such there exist an integer N and nonzero Q 2 A such that = N , so we m may assume without loss of generality that 2 A. Then A = i Pi , so the only possible valuations v with j jv 6= 1 are the in nite valuations (a nite number) and those from the product with mi  1. Furthermore, 1

Y

all v

j jv =

Y

vj1

j jv 

Y P

j jP = jN ( )j 

Y i

N(

Pi ) m

= jN ( )j  (A : A) = jN ( )j  jN ( )j 1

1

1

=1

as claimed. It follows from the proof that 2 A i j jP  1 for all prime ideals P , and 2 A i j jP = 1 for all prime ideals P . Corollary 43. For all 2 A, Qvj1 j jv = 1.

Lecture 8 We continue with the notation from last lecture. Proposition 44. Assume that r1 + r2  2, and x an in nite valuation v. Let B be an order in k. Then there exists a unit  2 B  such that jjv < 1 and for all in nite valuations w 6= v, jjw < 1. Proof. For any t = (t1 ; : : : ; tr +r ) 2 Rr+ +r let Pt  k R = Rr  C r be the region 1

1

2

2

2

1

Pt = fx 2 Rr  C r : jxjv < tv for all vg The numbers t1 ; : : : ; tr are radii ofpintervals. The tr +1 ; : : : ; tr +r are square roots of radii of circles, as zz < tv ) jz j < tv , which implies that the area of the circle is tv . Then P t is a closed, symmetric subset of Euclidean space. Also, Q Q bounded, convex, centrally Vol(Pt ) = 2r r r +r tv . Let M = r +r tv and choose t and t0 so that P t0  Pt and M 0 = ( 2 )r jdB j1=2 = M + 1. Then Vol(Pt0 ) = 2r +r jdB j1=2 = 2n Vol(D) where D is the fundamental domain of B acting on k R. It follows from Theorem 16 that P t0 contains a nonzero element of B , and by our choices 2 Pt \ B . Since is an algebraic integer, jN ( )j  1. By construction jN ( )j < M +1. So, is an element of small norm. There is no reason a priori why we should expect to satisfy the properties we desire from . However, we apply the following strategy to generate . (We say \generate" instead of \construct" because this method is not an ecient way to calculate  in practice.) We shall construct elements i of B with small norm, and it will follow that i B = j B for some i 6= j . Then i = j  for some unit . For  > 0, let tw =  for w 6= v and tv = M=(r +r 1) , and let P () denote the corresponding Pt . (Here, we are taking \small" values of .) These choices for t give us a nonzero 2

1

1

1

1

2

1

2

1

2

1

2

1

33

2

2

1

2

2 P () \ B . By construction, if w 6= v, then 0 < j jw < . We claim that j jw  M , so that these values are bounded away from zero. Fix some w0 6= v, and suppose that j jw < M . Then   Y j jw < M (r +Mr 1) (r +r 2) = 1 1  jN ( )j = j jw  j jv  w6=v;w 0

1

0

1

2

2

0

a contradiction. Let c be the number of ideals I of B such that N (I )  M + 1. (As noted before, c is nite.) For i = 0; 1; 2; : : :; c x nonzero i 2 P ( M i ) \ B . Since the set f i B g consists of c + 1 ideals of norm N ( i B ) = jN ( )j < M + 1, it follows that some i B = j B for 0  i < j  c. Let  = i = j , which is an element of B  . Then, for w 6= v, j i jw  M i and j j jw < M j , which implies that +1

i+1 ) Mj  1 jjw = jj i jjw > (=M = (=M j M i+1 jw Q Furthermore, by Corollary 43, 1 = j jv  w6=v j jw < j jv , as desired.

If B is an order in a number eld k, de ne a group homomorphism

 : B  ! vj1 Rv = R(r +r ) 1

as

2

() =

X v

log jjv  v:

Since R(r +r ) is torsion-free, it is clear that the torsion subgroup of B  is in the kernel of . Q Also, the fact that that the image of  is contained in the hyperplane vj1 jjv = 1 implies P P P H = f v av v : v av = 0g, since v log jjv = 0. We can now formulate the Unit Theorem in terms of . Theorem 45. (Unit Theorem) Assume that k is a number eld and that B is an order in k. Then the group of units B  is a nitely generated abelian group. In fact, 1

2

1. The torsion subgroup of B  is cyclic of nite order and equal to ker(). 2. The image of  is a lattice in H , that is, a discrete, co-compact subgroup. Therefore, the rank of B  is r = r1 + r2 1.  = f1g. 3. If r1 6= 0, then Btor  is nite, then the fact that it consists Proof. First, we note that, if we can prove that Btor exactly of the roots of unity in B implies that it is cyclic. Second, if r1 6= 0, then there is a real embedding of k. Since roots of unity map to roots of unity, this implies that the only roots of unity in k are 1. Let S be a compact subset of H . Then  1 (S ) is contained in the intersection of B with a bounded subset of Rr  C r = k R. Since B is discrete in Rr  C r , it follows that  is nite. The fact that ker() is  1 (S ) is nite. Let S = f0g to see that ker()  Btor nite implies that it is torsion, so that it is contained in (and therefore equal to) the set of roots of unity in B . This implies 1. 1

1

2

34

2

The argument of the preceding paragraph implies that Im() is discrete in H . To show that Im() is co-compact in HP , we must show that the R-span of Im() is H . It suces to show that if a linear form  =P v av xv vanishes on Im(), then it vanishes on H , i.e., all av are equal since H = f(xv ) : v xv = 0g. If r1 + r2 < 2 then there are two cases. Case 1: r1 = 1 and r2 = 0. In this case k = Q and the result is trivial. Case 2: r1 = 0 and r2 = 1. In this case, k is a purely imaginary quadratic eld, which we checked explicitly in the previous lecture. Thus, we assume without loss of generality that k has at least two places. P If the form  vanishes on Im(), then v av log jjv P = 0 for all  2 B  . Let v0 be a place 0 such that av  av for all v, and let  =  av v xv . Then all the coecients of 0 are nonnegative, and the coecient of xv is 0. Furthermore, 0 vanishes on Im() because P  vanishes on Im() and av v xv vanishes on H . It suces to show that 0 = 0. By Proposition 44, there is an element  2 B  such that jjv < 1 and for w 6= v0 , jjw > 1. If some a0w > 0, then X X 0 aw log jjw > 0 0 = 0 (()) = a0v log jjv = 0

0

0

0

0

w6=v0

v

We sketch a generalization of the ideas behind the Unit Theorem. Recall that A = f 2 k : j jP  1 for all P g and A = f 2 k : j jP = 1 for all P g = f 2 k : j jv = 1 for all nite valuations vg: Let S be a nite set of valuations of k which contains the in nite valuations. That is, S consists of all r1 + r2 in nite valuations and a nite number of the nite valuations. Let AS denote the set AS = f 2 k : j jv  1 for all v 62 S g. In other words, AS consists of the elements of k such that the the only primes with negative rami cation index in the prime decomposition of A must be in S . AS is a ring containing A, and we sometimes denote AS by A[1=P ]P 2S . The Generalized Unit Theorem states that the group of units of AS is exactly A = f 2 k : j jv = 1 for all v 62 S g. The proof is similar to that of the Unit Theorem. The main tool is the homomorphism of groups X S : AS ! v2S Rv given by S () = log jjv v:

v2S  The kernel of S is (AS )tor , which is nite and consists P of roots of unity. The image of S is a lattice in the hyperplane HS = f(xv )v2S : v2S xv = 0g and therefore the rank of A is dim(H ) = #(S ) 1. The details are left to the interested reader. We will, however, S

consider a few examples. Example. Let k = Q so that A = Z, and let S = f1; pg for some prime number p. (By saying that 1 2 S , we mean that the unique (real) embedding of Q is in S .) Then AS = Z[1=p] = f par : r 2 Zg. A contains the units 1 as well as the new unit p. It follows that AS = f1g  p . Example. Assume that A is the ring of algebraic integers in an arbitrary number eld k, and S consists of the in nite valuations and a single nite valuation, say at the prime ideal P . Fix an element  2 AS r A . Then for Q 6= P , ordQ () = 0 and ordP () < 0. It follows from the niteness of the class group of A that P n = A for some n  1. Z

35

Lecture 9 We continue our discussion of the units in the ring of integers A of a number eld k. Corollary 46. (Kronecker) If  is an algebraic integer such that jjv = 1 for all in nite valuations v, then  is a root of unity.

Q

Proof. Since jN ()j = vj1 jjv = 1,PCorollary 41 implies that  is a unit in A. Letting  : A ! vj1 Rv be given by 7! v log j jv v as in the previous lecture, it follows that  is in the kernel of , which is exactly the set  of roots of unity in A.

De nition. A unit  of A is called a Lehmer unit if all but two of its conjugates lie on the unit circle in C . More precisely, if 1 ; : : : ; n are the distinct embeddings of k in C and j j is the standard absolute value on C , then for some i 6= j , ji ()j; jj ()j 6= 1 and for l 6= i; j , jl ()j = 1. Q If  is a Lehmer unit, then the fact that 1 = vj1 jjv = i ()j () implies that j () = i () 1 = i ( 1 ). For example, if k is a real quadratic eld, and is a unit in the interval (0; 1), then for the nontrivial embedding , ( ) = 1= is in the interval (1; 1) and is a Lehmer unit. In fact, if  is a Lehmer unit and  and  are the embeddings such that j()j; j ()j 6= 1, then () and  () must be real. If not, say () is not real, then () is another conjugate of . Since j()j = j()j 6= 1, it follows that () =  (). However, the fact that j ()j = j(1)j = 1 6= 1 j()j implies that j ()j 6= j()j. It is natural to ask whether or not there are Lehmer units which are not roots of unity and whose real conjugates get arbitrarily close to 1. Lehmer conjectured that this could not occur. Example. Let  be a prime number   3,  a primitive th root of unity, and k = Q ( ). By Proposition 36, the ring of integers of k is A = Z[ ]. k is Galois over Q with Galois group G which is cyclic of order n =  1 = 2r2 since k has no real embeddings. Then r1 = 0 and r2 =  2 1 , and by the Unit Theorem, A has rank r1 + r2 1 =  2 3 . Since G is cyclic of even order, it has a unique subgroup of order 2 whose nontrivial element is given by complex conjugation. Under the Galois Correspondence, the xed eld k+ of this subgroup is the unique intermediate eld of degree  2 1 over Q . Let A+ denote the ring of algebraic integers in k+ . Clearly, Z[ +  1 ]  A+ , and a discriminant computation shows equality. (One must verify directly that disc(A+ ) = (  ) .) Since k+ is the xed eld of complex conjugation, k+ is a real eld. It follows from a little Galois theory that, r1+ = n+ =  2 1 and r2+ = 0. We verify this in more generality below. Again by the Unit Theorem, the rank of (A+ ) is  2 3 . Since (A+ )  A and these nitely generated groups have the same rank, the index (A : (A+ ) ) is nite. More generally, let k be a number eld which is totally complex (i.e., r1 = 0) which therefore has degree n = 2r2 over Q . Furthermore, assume that k is a quadratic extension of a totally real eld k+ . (k is called a CM eld in this event.) Assuming that k  C , we let  : k ! C denote the embedding of k corresponding to complex conjugation. That is, if  is a xed 2

36

3

embedding of k, then  =  . We let k+ denote the xed eld of  , or equivalently, k+ = k \R. Note that if  is any embedding of k in C , then  =    is the embedding of k conjugate to  and f; g are the same real embedding of k+ . That is, there is a 1{1 correspondence between real embeddings of k+ and conjugate pairs of embeddings of k. Furthermore, (A+ ) has nite index in A , as both groups have rank r2 1. Let u be a unit in A, and let  = u= (u). Then for any in nite place v of k, (u) jjv = = 1 (u) so that  is a root of unity by Corollary 46. It follows that  (u) = u for a root of unity . If is another root of unity in k, then  ( u) = 1 u =  ( u). Of course,  is another root of unity. It follows that if we let  = 2 and u0 = u, then u0 is in the xed eld of  , that is, u0 2 k+ . Since u0 is an algebraic integer, u0 2 A+ . Also, u0 is a unit in A+ , as is shown by Lemma 40 and the correspondence between real embeddings of k+ and conjugate pairs of embeddings of k. Let  denote the set of roots of unity in k, and consider the map  : A ! =2 given by u 7! u= (u). The previous paragraph shows  is well-de ned and has kernel exactly (A+ )  . Since  is nite cyclic, it follows that =2  = Z=2Z. Thus, the index of (A+ )    in A is 1 or 2, depending on whether  is trivial or not. For example, if k = Q ( ) is the cyclotomic eld from above, then (A+ )   = A . Since (A+ ) \  = f1g, this is not a direct product, but if 0 denotes the set of th roots of unity, then (A+ )  0 = A , as  is odd by assumption. Recall the notation from Lectures 6 and 7. If  is a primitive th root of unity, let  =  1. For integers a such that 2a 6 0 (mod ), let (a) =  a  a = ( 2a 1)( a ). We claim that (a) has norm . From our earlier computations, N ( ) = 1 and N () = . Since  2a 1 is a conjugate of  and  a is a conjugate of  we see that N ( (a)) = N ()N ( ) = . Also, it is straightforward to check that  ( (a)) = (a). Let a a a) ua =   1 = ((1) (We call ua a circular unit because if  = e2i= then 2ai= 2ai= a=) ua = ee2i= ee 2i= = sin(2 sin(2=) which is an expression in terms of the circular function sin.) We claim that ua is a unit in A+ . To see this, it suces to show that ua is a unit in A which is xed by  , since then ua is in A+ and  (ua 1 ) =  (ua ) 1 = ua 1 so that ua 1 2 A+ . Since N ( (a)) = , we see that N (ua ) = = = 1. Furthermore, ua can be expressed as a polynomial in the algebraic integers ;  1 so that ua is an algebraic integer. Thus, ua is a unit in A. Finally, since  ( (a)) = (a), it follows that  (ua ) = ua . To get distinct (a) up to sign, we can allow a to run from 1 to  2 1 . The choice a = 1 gives ua = 1, so we have  2 3 reasonable choices for a. In fact, we have the following theorem. Theorem 47. (Kummer) Let U3 be the subgroup of (A+ ) generated by the ua. Then U is a free abelian group of rank 2 and the index of U in (A+ ) is nite and divisible by 2. Furthermore, if h+ is the class number of A+ , then the index is exactly 2h+. 2

37

2

The proof of this theorem involves the theory of L-functions, so we omit it here (cf. Borevich and Shafarevich Number Theory, Chapter 5). However, we check an example. Example. Let  = 5. Then u2 p=   =  +  1 . Since k has degree 4 over Q , the eld k+ is real quadratic, so k+ = Q ( D) for some square free positive integer D. The nontrivial element of Gal(k+ =Q ) = Gal(k=Q )=Gal(k=k+) is given by sending  7!  2 . Thus, 2

2

1

Tr (u2 ) = ( +  1 ) + ( 2 +  2 ) =  +  2 +  3 +  4 = 1 by examining the minimal polynomial of  , and

u2 ) = ( +  1 )( 2 +  2 ) =  3 +  1 +  +  3 =  +  2 +  3 +  4 = 1 2 so that the formula tells us that p5 minimal polynomial of u2 is x + x 1. The quadratic 1+ p 1  u = 2p and by xing an embedding, we may assume that u = 2 5 . It follows that k+ = Q ( 5), and a computation from Lecture 5 shows that h+ = 1. N(

Next, we will discuss the tools developed by Chevalley, Artin, and Weil: completions of elds, the Adeles and the Ideles. These tools will give us a succinct reinterpretation of the niteness of the ideal class group and the Unit Theorem. In much of the following discussion, only sketches of proofs will be given. The interested reader is invited to supply the missing details. Recall that we construct in nite valuations on a number eld from its embeddings in C . These valuations satisfy the Archimedean property: for all x; y 2 k there exists an integer n such that jnxjv > jyjv . This follows from the fact that the absolute values on C and R are Archimedean. The nite valuations are those induced by nonzero prime ideals P in A by the formula jxjP = N(P ) ord P (x) . It follows that j + jP  maxfj jP ; j jP g  j jP + j jP . Another way to express the rst inequality is to say that all triangles are isosceles, since equality holds in the rst inequality exactly when j jP 6= j jP . However, the nite valuations do not satisfy the Archimedean property: since Z  A, any integer n has jnjP  1 so that jnxjP = jnjP jxjP  jxjP . We recall the construction of the real numbers from the rationals. With the standard absolute value on Q

(

jxj = x x ifif xx 00 we say that a sequence fan g is Cauchy if, for every  > 0 there exists N such that n; m  N ) jam an j < . The problem with Q is that there exist (lots of) Cauchy sequences in

which do not converge to elements of Q . Basically, R is constructed by adjoining all the limits of Cauchy sequences in Q to Q . We will be more speci c below. De nition. A valued eld is a eld k with valuation j j : k ! R+ . A valued eld k is complete if every Cauchy sequence in k converges to an element of k. A completion of a valued eld k is an extension eld K satisfying the following properties: (1) K is a valued eld which is complete and whose valuation extends the valuation on k; and (2) K contains k as a dense sub eld. We require property (2) for uniqueness purposes: without it, C would be a completion of Q . Q

38

Theorem 48. For a valued eld k with valuation j jv , a completion kv exists and is unique up to unique continuous isomorphism. More precisely, if kv and kv0 are completions of k, then there exists a unique continuous isomorphism kv ! kv0 making the following diagram

commute.

kv

@I@

k

- kv0 

Sketch of proof. Let R be the ring of all Cauchy sequences fang with an 2 k. Let m denote the (maximal) ideal of R consisting of all null sequences, that is, sequences fang such that limn!1 jan jv = 0. We de ne kv to be the quotient kv = R=m, which is the eld of limits of Cauchy sequences in k. We claim that this is a completion of k. We rst extend the valuation from k to kv in the natural way

j jv : kv ! R+

jfangjv = nlim !1 jan jv 2 R

as

We embed k in kv as the constant sequences

k ! kv

a 7! fa; a; a; : : : g

as

since the constant sequences are trivially Cauchy. Under this embedding, k is dense in kv . Fix any = fan g 2 kv , and de ne a Cauchy sequence 1 ; 2 ; : : : of elements in the image of k in kv as i = fai ; ai ; : : : g. Then = limn!1 i . Finally, kv is complete. Fix a Cauchy sequence in kv , i = f in g = f i1 ; i2 ; : : : g. We want to show that this sequence converges to an element of kv . Let denote the \diagonal" sequence = f 11 ; 22 ; : : : g. It is straightforward to show that = limn!1 n . The uniqueness result is left as an easy exercise.

Example. Let k = Q and P = pZ for a prime number p. The completion of Q with respect to the valuation j jP is denoted Q p and called the p-adic numbers. It will follow from more

general work that the ring of integers in Q p is exactly Zp = lim (Z=pnZ). More generally, if k is a number eld and P is a nonzero prime ideal in A of characteristic p, then the completion of k with respect to the valuation j jP is denoted kP . We shall see that kP is a nite extension of Q p of degree eP fP , where eP is the rami cation index of P and fP is the residue degree.

Lecture 10 We continue with the notation from the previous lecture. Given a nonzero prime ideal P in A, let q = N (P ). Then the valuation j jP on k takes its values in q . Since the valuation is non-Archimedean, if the sequence fang is Cauchy with respect to j jP , then the sequence fjan jP g in q  Q must stabilize. It follows that the valuation on kP also takes its values in q . It also follows that the extended valuation is non-Archimedean. De ne Z

Z

Z

AP = f 2 kP : j jP  1g 39

and

PP = f 2 kP : j jP < 1g = f 2 kP : j jP  q1 g The fact that the valuation is non-Archimedean implies that AP is a subring of kP and that PP is an ideal in AP . Furthermore, restricting the embedding k ,! kP to A and P induces

a commuting diagram

PP  AP  kP

6

6

[

6

[

[

P  A  k

Proposition 49. AP is a discrete valuation ring with maximal ideal PP . Proof. Since P 6= 0, we know that P 2 ( P . Fix an element  2 P r P 2 , that is, such that jjP = q1 . The fact that PP = AP will follow immediately from the following theorem. Theorem 50. Let I be any nonzero ideal of AP and suppose that

maxfj jP : 0 6= 2 I g = q m for some m  0. Then I = AP for some 2 I such that j jP = q m . Proof. First, we note that maxfj jP : 0 6= 2 I g is always of the form q m . This follows from the fact that fordP ( ) : 0 6= 2 I g is a nonempty set of natural numbers. Furthermore, there is an element of I with j jP = q m and we x such an element. Clearly, I  AP . Conversely, if 2 I , then j = jP  1 so that = = 2 AP which implies that = 2 AP .

From the proof of the theorem, we see that PP = AP . Furthermore, if PP were not maximal, then there would be some proper ideal I of AP such that PP ( I . By the theorem, I = AP for some 2 I . If were in PP , then I = PP , so that 62 PP . It follows that 1  j jP  1 so that j jP = 1. This implies that j 1 jP = j jP 1 = 1 so that 1 2 AP which implies that I = AP . It follows exactly from this logic that the units of AP are exactly those elements of kP such that j jP = 1. To nish the proof of the proposition, it suces to show that every nonzero ideal I of AP is of the form m AP . Let m and be as in the theorem. Then j =m jP = 1 so that =m is a unit in AP and I = AP = m AP . Restricting the embedding k ,! kP to the ideals P n induces another commuting diagram

  n AP      2 AP  AP  AP  kP

6

[

 Pn   

6

[ P2

Proposition 51. With notation as above. 1. The ring A is dense in AP .

40

6

[

6

[

6

[

 P  A  k

2. The inclusion A ,! AP induces an isomorphism A=P n  = AP =n AP for n = 1; 2; : : : . 3. AP = limn (A=P n ). Proof. 1. For each x 2 AP , the fact that k is dense in kP implies that for every  > 0 there are ; 2 A such that j = xjP < . Assuming that   1, the fact that jxjP  1 implies that j = jP < 1 since otherwise

 > j = xjP = maxfj = jP ; jxjP g = j = jP  1 Thus, ordP ( )  ordP ( ) so that 2 P ordP ( ) . It suces to nd an element 2 A such that j = jP < , since then j xjP  j = jP + j = xjP < 2 It is equivalent to nd 2 A such that j jP < j jP . Furthermore, it suces to show that 2 A + P n for n  0, as this implies that for each such n there exists n 2 A such that 2 P n (so that ordP ( )  n) and if we choose n such that q n < j jP then

j n jP = q ordP ( n )  q n < j jP as desired. Claim: for n  ordP ( ), A + P n = P ordP ( ) . Once the claim is proved, we are done, since we have already shown that 2 P ordP ( ). Since P is maximal, the only ideal

containing A+P n is P . Thus, the prime decomposition of A+P n is of the form P m. Since 2 A + P n = P m , we see that m  ordP ( ). And the fact that P m = A + P n  P ordP ( ) implies that m  ordP ( ). 2. The ideal n AP is both open and closed in AP , as

n AP = f 2 AP : j jP  q1n g = f 2 AP : j jP < qn1 g Since A is dense in AP and the coset + n AP is open for each 2 AP , it follows that A \ ( + n AP ) 6= ;. This implies that the map A ! AP =n AP is surjective for every n. Furthermore, the kernel is exactly n AP \ A. If we can show that this is exactly P n , then we are done. This is clear, though, since n AP \ A is exactly the collection of elements of A with j jP  q n , and this is exactly P n . 3. By the completeness of AP we see that AP = limn (AP =n AP ) = limn (A=P n ). 1

It follows from 3. that AP is a compact topological ring, as it is the projective limit of nite (and therefore compact) topological rings. (We say that AP is pro nite.) Furthermore, kP is locally compact as for every 2 kP , the translate + AP is a compact neighborhood. The advantage of working in kP instead of R or C is that, in kP , A is contained in the compact subring AP . The images of A in R and C have no hope of being compact, as they are unbounded. RecallQthat the Tychono Theorem implies that, with the product topology, the topological ring P 6=0 AP is compact, where theQproduct is taken over all nonzero prime ideals P of A. The diagonal embedding A ! P AP given by a 7! (a; a; : : : ) is a topological ring homomorphism. 41

Proposition 52. The image of A under the diagonal embedding A ! QP AP is a dense

subring.

Q

Proof. By the de nition of the product topology, to approach an element x 2 P AP , we must nd 2 A which is arbitrarily close to x in any number of coordinates, indexed by a nite set of prime ideals. That is, if P1 ; : : : ; Pm are prime ideals and mi > 0 for i = 1; : : : ; m, then we need to nd an element 2 A such that j i jPi < qi mi for i = 1; : : : ; m. This mi i is equivalent to nding an element such that  i in APi =(Pi )m Pi = A=PQi for meach i. The Chinese Remainder Theorem tells us that the homomorphism A ! i A=Pi i is surjective, since i 6= j ) Pi + Pj = A. The result now follows.

We want a ring R which has the same properties with respect to k. That is, R should be locally compact and k Qshould embed in R naturally. With the proposition in mind, our rst candidate for R is v kv where the product is taken over all ( nite and in nite) places. However, since the kv are only locally compact (and not compact) and the product  is in nite, this ring is no longer locally compact. The next  reasonable candidate is v kv , which is locally compact. However, the diagonal map k ! v kv is not well-de ned. Therefore, the ring we want should be somewhere in between the coproductQand the product. This leads us to the Adeles A which is the locally compact subring of v kv consisting Q of those (av ) such that aP 2 AP  kP for almost all P . The diagonal embedding k ! v kv lands in A since, for 2 k, ordP ( ) = 0 for all but nitely many P . We describe the construction in a slightly more general context. Assume that V is some set indexing a collection of locally Q compact (abelian) groups fGv g. How do we construct a locally compact subgroup ofQ v Gv ? Under an additional hypothesis we will do this via a \restricted" direct product v Gv . Assume that, for all but nitely many v 2 V , there is a distinguished compact, open subgroup Hv of Gv . Let V1 denote the nite collection of indices which there is no Hv . The restricted direct product is then the subgroup Q for Q G = v Gv of v Gv consisting of those (gv ) such that gv 2 Hv for almost all v 2 V r V1 . If S is a nite subset of V containing V1 , then let

GS =

Y  Y  v2S

Gv 

v62S

Hv

Q

Since each Gv is locally compact Q and S is nite, v2S Gv is locally compact as well. And since each Hv is compact, v62S Hv is also compact. It follow immediately that G = [S GS where the union is taken over all such nite sets S . We also notice that if S  T , then GS  GT is an open subgroup. Thus, the collection of GS give an open system of neighborhoods for a topology on G which isQlocally compact. (Note that this is not the topology induced by the product topology on v Gv .) If V now denotes the collection of ( nite and in nite) valuations on k, we may take Gv = kv and for nonzero prime ideals P of A, HP = AP . In the previous notation, V1 is the collection of in nite valuations on k. For any nite subset S of V containing V1 , we then have AS

=

Y   Y v 2S

kv 

42

P 62S

AP



The bene t of the generality of the previous paragraph is that we can apply the same construction to the units of A , taking Gv = kv and HP = AP . The resulting restricted direct product J is called the ideles. As abelian groups, J = A  . To convey somewhat the structure of kP , consider the short exact sequence of abelian groups 1 ! AP ! kP ord !P Z ! 1

which necessarily splits. We see that a ltration on AP will then induce a ltration on kP . Recall that a ltration of the ring AP was given by powers of the maximal ideal PP

    PP3  PP2  PP  AP and since each PPi = i , it follows that each quotient in the ltration is a one dimensional vector space over the nite eld AP =PP  = A=P . The ltration of AP is given by translates of powers of the maximal ideal

    1 + PP3  1 + PP2  1 + PP  AP The rst quotient is and for i  1,

AP =(1 + PP )  = (A=P ) (1 + PPi )=(1 + PPi+1 )  = A=P

For more details, the interested reader should consult Serre Local Fields, Chapter IV, SS2.. Q As we noted above, the diagonal embedding k ! v kv is a continuous homomorphism of rings which lands in A . The restriction to k is a continuous homomorphism of abelian groups which lands in A  . In addition, we have the following. Proposition 53. With notation as above, 1. k is a discrete subring of A and the quotient A =k is compact. 2. k is a discrete subgroup of J , and the quotient J=k is almost compact.

De nition. A space X is almost compact if it is of the form C  R where C is compact. In the situation of the proposition, we can be moreQspeci c. There is a canonical surjective homomorphism k k : J ! R+ given by (av ) 7! v jav jv . This is well-de ned since the product is always nite. It is a multiplicative homomorphism, as each j jv is multiplicative. It is surjective, as k has at least one in nite place v. The completion kv is isomorphic to R or C as a valued eld, and therefore j jv : kv ! R+ is surjective. To see that k k is surjective, we apply it to an element of the form (1; : : : ; av ; 1; : : : ) where the non-identity entry is in the place indexed by the distinguished v. Furthermore, the kernel, which we shall denote by J1 , contains the image of k , by the product formula (Theorem 42). Since the quotient J=J1  = R+ is isomorphic to the additive group of real numbers, part 2 of Proposition 53 is equivalent to the fact that J1 =k is compact. 43

Sketch of proof. If we write

then A

Q

Y all v

all v kv

kv =

Y vj1

kv 

Y v-1

kv

has two distinguished subsets, namely

Y

Q

vj1

kv  1

1

and

Y

v -1

Av

As we noted above, vj1 kv = Rr  C r  = Rn , which contains A as Qa discrete, cocompact subset. Let D1 denote the fundamental domain for A acting on vj1 kv , which has compact closure, by co-compactness. Let U1  Rn be an open subset containing 0 such that A \ U1 = (0), which exists by discreteness. Consider the subset 2

1

D = D1 

Y

v -1

Av  A

which has compact closure. 1. D is a fundamental domain for k acting on A , i.e., A is the disjoint union of translates A = [ 2k ( + D), so the quotient A =k is compact. Let

U = U1 

Y

v-1

Av  A

which is an open subset. Then U \ k = (0), as



k\ 1 Thus, k is discrete in A . 2. If we now let

U=

Y vj1

Y 

v-1

Av = A

kv 

Y v -1

Av

Q

then U is an open subgroup of J and U \ k = A . Using a bounded region in vj1 kv , we can nd a smaller open subset U0 such that U0 \ k = (0). Thus, k is discrete in J . Since U \ k = A , it follows that k  U=U = k =A . If we let I denote the group of all fractional ideals of A, let P denote the subgroup of principal fractional ideals and let C denote the ideal class group, then we have the following commuting diagram

44

1 1

1

1

J=(k  U )

C

- Uw - k ? U - k=A? === P - 1 ww w ? -U - J? - a Z ==== I - 1 P ? ?  ?

?

1 1 in which the top two rows and the right-hand two columns are exact. It follows from the snake lemma that J=(k  U )  = C , which is nite. The fact that J=(k  U ) is nite and the following short exact sequence 1 ! U=A ! J=k ! J=(k  U ) ! 1 show that the index of U=A in J=k is nite. Thus, to prove that J=k is almost compact, it suces to prove that U=A is almost compact. We have an inclusion Y Y A ,! U = kv  Av vj1

v -1

and a short exact sequence  Y  Y kv =Im(A ) ! 1 1 ! 1  Av ! U=A !

vj1 | v{z1 } compact Q so that we need only show that ( vj1 kv )=Im(A ) is almost compact. The short exact sequence Y Y 1 -Y   -M -1 Rv 1 kv f1g  S v cpx. v real v j1 vj1 -X -

( v )

v

log j v jv v

induces a short exact sequence on quotients Y M   Y 1   Y  Rv =(A ) ! 1 1! kv =Im(A ) ! f1g  S =Ator ! v real

Q Q Since f1g  S 1

v cpx.

vj1

vj1

Q Q S 1)=A . The Unit is compact, so is the quotient ( f1g  Q tor k =Im(A ) is Theorem implies that ( Rv )=(A ) is almost compact. Thus, vj1

vj1 v

almost compact and the proof is complete.

Notice that we used the niteness of the class group and the Unit Theorem at key steps of the proof. It can be shown that the conclusions of the proposition are actually equivalent to these two results. (For example, see A. Weil Basic Number Theory, Champter V.) 45

Lecture 11 Now we introduce the zeta function of a number eld and study its properties. In particular, we discuss the classical Riemann zeta function. For a number eld k and integer n  1, let an denote the number of integral ideals I of A with N (I ) = n. For example, p if k = Q , then A = Z and an = 1 for all n since the unique such ideal is nZ. If k = Q ( D), then to calculate an , we must know the factorization of primes p in A. For example, if pA = P1 P2 for distinct primes P1 and P2 , then the ideals of index p are exactly the primes of characteristic p, namely, the Pi . The other cases are similar, and we summarize.

8 > 1 :0

if pA = P1 P2 is split if pA = P 2 is rami ed if pA = P is inert

We can apply the same methods to ap .

8 > 1 :1 2

2

if pA = P1 P2 is split: the ideals are P12 , P22 , and P1 P2 = pA if pA = P 2 is rami ed: the ideal is P 2 = pA if pA = P is inert: the ideal is P = pA

De nition. Using the above notation, we de ne the zeta function of k as the formal series k (s) =

1 a X n s

n=1 n

where s is a complex variable. Example. If k = Q , then we have the Riemann zeta function

 (s) =  (s) = Q

1 X

1

s

n=1 n

Most properties of general zeta functions will be proved for the Riemann zeta function rst, and the general case will follow from the speci c case. P an with all a 2 C . (A series of First, we study general series of the form f (s) = 1 i n=1 ns this form is called a Dirichlet series.) Assume that there are xed real numbers 0  0 all n. (We say that an = O(n ).) Claim: Under and K > 0 such that jan j  Kn for P 1 these hypotheses, the Dirichlet series n=1 anns converges absolutely and de nes an analytic function of s in the open half-plane Re(s) > 0 + 1. In particular, the derivative of f may be taken term by term, and the series may be treated as a nite sum in terms of products, on the set of convergence. We prove this by the comparison test. We have 0

0

jn s j = je

s log n j = eRe( s log n) = e  log n = n 

where  = Re(s). By assumption then, we have the inequality an K 1 =K 1

ns

n(  ) 0

46

n

P

P

1 1 an where  =  0 . Thus, we compare the series 1 =1 ns to the series n=1 n = P 1 1 . The second series converges if  > 1 (by thenintegral test) and diverges if  = 1 n=1 n (as this is the harmonic series). It follows that the original series converges absolutely for  0 =  > 1, i.e., for  > 1 + 0 . Furthermore, for  > 0 and c0  1 + , the second series converges uniformly for 1 +     c0 . Thus, f (s) converges uniformly on compact subsets. Since each function in the sum is analytic, this proves the claim (cf. W. Rudin Real and Complex Analysis, Theorem 10.28). rephrase thePstatement regarding P aWe 1 bn n products more precisely. Assume that f (s) = 1 n=1 ns and g (s) = n=1 ns are Dirichlet series which converge absolutely and de ne an analytic function of s in the open half-plane Re(s) > 0 + 1. Then the product f (s)g(s) is another Dirichlet series 1 c 1 b  X 1 a X X n n = n s

s

n=1 n

s

n=1 n

n=1 n

which converges absolutely and de nes an analytic function of s in the same open half-plane. The coecients cn are given by the formula

cn =

X

de=n1 d;e

as expected. We apply the claim to the Riemann zeta function. Since an = 1 for all n, the constants 0 = 0 and K = 1 tell us that  (s) is analytic on the open half-plane Re(s) > 1. We shall prove later that the same convergence result holds for k (s) where k is any number eld. As noted,  (s) does not converge at s = 1. We would like to know \how bad" the singularity is at s = 1. Lemma 54. The function f (s) =  (s) s 1 1 has an analytic continuation to the open halfplane Re(s) > 0. That is,  (s) is meromorphic on Re(s) > 0 with unique pole at s = 1 with residue Ress=1  (s)ds = 1. Proof. For Re(s) > 1 we have

Thus, for Re(s) > 1

Z1 1

1 s 1 t s dt = 1t s 1 = s 1 1

11 X  (s) s 1 1 = s n=1 n

= =

R

1 Z n+1 X

n=1 n

1 X

n=1

Z n+1 n

 t s dt

(n s t s ) dt

n (s)

where n (s) = nn+1 (n s t s ) dt. Note that each n (s) is de ned for all s, and analytic on C . As above, then , we need only show uniform convergence on compact subsets of Re(s) > 0. 47

It is immediate that jn (s)j  maxntn+1 jn s t s j. For xed n, let fs (t) = n s t s . The Taylor series for fs (t) at t = n is given by fs (t) = n s t s = fs (n) + fs0 (n)(t n) + R(t) = nss+1 (t n) + R(t)

where R(t) = 21 fs00 (c)(t n)2 for some c in the interval [n; n + 1]. The fact that jt nj  1 on this interval implies that jf (t)j  1 (jsj + 1 js(s + 1)j) s

n+1

2

where  = Re(s) > 0. As before, we have absolute convergence for  > 0 and uniform convergence on compact subsets of the half-plane, establishing the result. Lemma 55. For Re(s) > 1 we have an in nite product expansion Y  (s) = (1 p s ) 1 p

Essentially, this result is a restatement of the unicity of prime factorization. This product is nice because each factor is analytic and nonzero for Re(s) > 0. This implies that  (s) has no zeroes in the open half-plane Re(s) > 1. Proof. Let S be a nite set of prime numbers, and let N (S ) denote the in nite set of integers n  1 whose prime factorization involves only the primes in S . The factor (1 p s ) 1 has an expansion as a geometric series 1 s 2s 3s 1 p s = 1+p +p +p + so that formal multiplication yields 1 Y X 1 X (1 p s ) 1 =  1 =  (s) s

n2N (S ) n

p2S

s

n=1 n

for Re(s) > 1. Notice for all such nite sets S , and as S becomes P that this inequality P 1 holds suciently large, n2N (S) n1s ! 1 n=1 ns . Corollary 56. As s ! 1 inside Re(s) > 1  1  X1  log s s 1 p p where  means that the quotient tends to 1. Proof.

log( (s)) = log = =

Y

X p

p

(1 p s )

1



( log(1 p s ))

X

 2s 3s p s + p 2 + p3 + p 48

using the Taylor expansion log(1 x) = x + x2 + x3 +    3

2

=

PP

X 1 XX p ps +

p

p k2

ks

k

ks

The series p k2 p k is nite as s ! 1 and is bounded by

XX p p k2

k

k 

X p

1

X

1 p(p 1)  n2 n(n 1) < 1

By Lemma 54, log( 1s )  log( (s)) 

X1 p

ps

as desired.

Corollary 57. The sum Pp p1 is in nite. In particular, there are an in nite number of prime numbers.

Proof. Immediate.

It has been shown that as x ! 1, the function X1  log(log(x)) px p

P

Let (x) denote the number of prime numbers p  x, or equivalently, (x) = px 1. Theorem 58. (Prime Number Theorem) As x ! 1,

(x)  log(x x)

Riemann proved that the Prime Number Theorem is equivalent to the condition that  (s) has no zeros with R Re(s) = 1, s 6= 1. Hadamard then proved this statement. Furthermore, if we de ne lix = 2x logdt t , then it has been shown that

(x) lix = O(xe c(log x) = ) for a constant c, provided that  (s) has no zeroes for Re(s)  . Along these lines, we have 1 2

the Riemann Hypothesis, which is still unsolved. Conjecture 59. (Riemann) The only zeroes of  (s) occur on the line Re(s) = 21 .

There is much computational evidence for this conjecture. Also, it has been shown that #f  t :  ( 12 + i) = 0g = 2t log 2t 49

t

2 + O(log t)

(c.f., Davenport, Multiplicative Number Theory, p. 59). Q Since we have been studying the properties of  (s) = p (1 p s ) 1 , it seems reasonable Q for us to ask \What is p (1 p s )?" We de ne the Mobius function  : Z+ ! f0; 1g as

(

if n is not square-free ( 1)t if n is square-free with t distinct prime factors

(n) = 0

P

Q

(n) Then we have the formula p (1 p s ) = 1 n 1 ns . The proof of this is an exercise in bookkeeping, and is left to the interested reader. From this, we can prove the following. Theorem 60. (Mobius Inversion) Assume that g : Z+ ! C is a function such that for a xed real number 0  0, g(n) = O(n ). De ne f : Z+ ! C by the formula 0

f (n) =

X djn d1

g(d)

Then we can recover g from f by the formula

g(n) =

X

nd=de 1

f (d)(e)

P

Proof. First, we observe that, given a sequence an , the formal series n anns satis es the following property X X f (n) X g(n) X an = i f (n) = g(d)a n

ns

n

ns n ns

n=de

e

This equivalence follows from the following expansion X g(n) X an X g(d) X ae X g(d)ae X Pn=de g(d)ae s s= s s= s = ns n n n n n d d e e d;e (de)

P

If we write f (n) = n=de g(d)1e , then it follows that X f (n) X g(n) X 1 X g(n) = =  (s) n

ns

n

ns n ns

ns

n

P (n) yields Q Multiplication by  (s) 1 = p (1 p s ) = 1 n 1 ns 1 (n) X g(n) X f (n) X n

ns = n ns n

1

so that our rst observation implies the desired formula.

50

ns

Lecture 12 We now return to the general zeta function k (s) for a number eld k. As above, for n = 1; 2; : : : , let an denote the number of integral ideals I of A with norm N (I ) = n. If P is a prime ideal of A of characteristic p, then P occurs in the factorization of pA, so we write P jp. Formally, we have  1 YY  1 X X Y k (s) = N (1I )s = anns = 1 N (P1 )s = 1 N (P1 )s p P jp I A n1 P

by the unicity of prime factorization of ideals. We want to show that k (s) converges absolutely on the open half-plane Re(s) > 1 and describes an analytic function there. To accomplish this, we notice that the nal expression may be written as Y Y 1  1 1 lim x!1 px N (P )s P jp We shall show that the nite product in the limit satis es Y Y  1 1 N(P1 )s   (s)n px P jp

And therefore, the desired properties of k (s) follow from the corresponding properties of  (s). It should be noted that the equalities for the in nite sums and products are purely

formal. However, taking limits of nite sums and products shows that the series also converge, and they converge to the same function as the in nite products. The equality of the sum with the product will be crucial in proving certain properties. In addition, we note that the individual factors (1 (P1 )s ) 1 are all holomorphic on the region Re(s) > 0. Assume that p is an unrami ed prime, that is, p - d = disc(k). (We shall only consider this case, as the rami ed primes only contribute a nite number of factors to the product, each of which is holomorphic on the region Re(s) >Q0. Thus, with respect to questions of N P , then pn = QN N (P ). Thus, convergence, these factors may be ignored.) If pA = i i i=1 =1 QN (1 p fi s) is a polynomial in p s of degree n. If iwe s , then let x = p i=1 N

N Y

i=1

Y (1 p fi s ) = (1 xfi ) N

i=1

Q

i (1  x) for f th roots is a polynomial in x of degree n. We may factor each 1 xfi as fj=1 j i of unity j . By renumbering, we see that there are roots of unity 1 ; : : : ; n such that

N Y

It follows that

(1 xfi ) =

i=1 N Y i=1

n Y

(1 i x)

i=1

Y (1 p fi s ) 1 = (1 i p s ) n

1

i=1 s ) 1 . If  = Re(s) > 1, then the distance from 1 to i p s

We estimate the size of (1 i p is minimized when i = 1, as the sketch shows. 51

i p  p  1

It follows that so that

1 1 i p s  1 1p Y 1 P jp

and therefore



 1  ((1 p N (P )s 1

Y Y 1 px P jp

 ) 1 )n

 1   (s)n N (P )s 1

as desired. Next, we wish to understand the nature of the singularity of k (s) at s = 1. First, recall some notions from the Unit Theorem. Let w denote the number of roots of unity in k, so that the group P of roots ofPunity in K is Ator  = Z=wZ. Let H denote the hyperplane in vj1 Rv, H = f v av v : v av = 0g. We de ne the map  : A ! H as  7! Pv log jjv v. Let E denote the image of  in H , which is free of rank r = r1 + r2 1. This gives a short exact sequence of abelian groups 1 ! Z=wZ ! A ! E ! 1 which necessarily splits. Let 1 ; : : : ; r 2 A such that the images of the i in E form a basis. Then the matrix of  is 1 0 log j j log j2 jv    log jr jv 1v CA .. .. M =B @ ... . . log j1 jvr log j2 jvr    log jr jvr 1

1

1

+1

+1

+1

which is of order (r + 1)  r. Let Mj denote the matrix obtained by deleting the j th row of M . We claim that j det(Mj )j is independent of j . To see this, observe that since E  H , the sum of elements of any column of M is zero. It follows that Mj is obtained from any Ml by elementary row operations, so that the determinants di er only up to sign. De nition. With the notation of the previous paragraph, we de ne the regulator of k as R = j det(Mj )j. 52

There is an equivalent de nition of R which is more geometric. Let  denote the Haar measure on H induced by the short exact sequence 0 ! H ! Rr+1 ! R ! 0 and the normalized Haar measures on R and Rr+1 . Let G be a free abelian subgroup of A which has nite index g in A . It can be shown that R = (H=(G))

w

g

We shall not prove that these de nitions are equivalent. Theorem 61. The function

k (s) s h1

has an analytic continuation to the open half-plane Re(s) > 1 number of k and r r  = 2 p(2) R

1 n where

h is the class

2

1

jdj w

In particular, k (s) has a meromorphic extension to this region with unique pole at s = 1 which is simple with residue Ress=1 k (s)ds = h.

Below, we o er a sketch of the proof. For a more complete treatment, the interested reader should consult Borevich and Shafarevich Number Theory, Chapter 5. Sketch of proof. Let C denote the ideal class group of k and x and ideal class c 2 C . For m 2 Z+ we de ne

Sc (m) = #fI  A : cl(I ) = c and N (I )  mg and

S (m) =

X c2C

Sc (m) = #fI  A : N (I )  mg

The key proposition one proves for this theorem is the following. Proposition 62. With notation as above

Sc (m) = m + O(m(1 n ) ) In particular, the leading term in the expansion of Sc(m) is independent of c. 1

Sketch of proof in a simple case. Assume that R = 1, r2 = 1 and r1 = 0, so that k is an imaginary quadratic eld. Furthermore, assume that c = 1 in C . Then S (m) = #f 2 A : jN ( )j  mg c

w

53

In this computation, we divide by w because every I = A has w distinct generators, as A is nite of order w. If  : k ! C is an embedding, then our assumptions imply that N ( ) =  ( ) ( ). It follows that

p

The following sketch

Sc(m) = #f 2 A : jw( )j  mg

pm

p volume of disc m #f 2 A : j( )j  mg  volume of fundamental domain for A = 1 1=2 2 jdj

and this estimate is accurate with error O(m1=2 ). With this estimate m + O(m1=2 ) S (m) = 2p c

w jdj

with our assumptions on R, r1 and r2 this is exactly the statement

Sc (m) = m + O(m(1

n1 ) )

which is the desired conclusion. The following is an immediate consequence of Proposition 62. Corollary 63. With notation as above

S (m) = hm + O(m(1

n1 ) )

The nal piece of the puzzle is supplied by a lemma. Lemma 64. Assume that f (s) = Pn nbns is a Dirichlet series. 1. If jbn j = O(n ), then the series converges absolutely and de nes an analytic function f (s) on the open half-plane Re(s) > 1 + 0 . P b and suppose that jT (m)j = O(m ) for some xed real number 2. Let T (m) = m n=1 n   0. Then the Dirichlet series converges and de nes a holomorphic function f (s) in the region Re(s) > . (It will not necessarily converge absolutely in this region.) 0

54

Part 2 of the lemma is more subtle than our previous convergence results, as it is not saying that the series isPabsolutely convergent. Even though the condition jbn j = O(m )  1 ), when we consider conditional convergence, implies that jT (m)j  m n=1 jbn j = O(m cancellation can occur. Example. Let  : (Z=f Z) ! C  be a nontrivial character (i.e., homomorphism of abelian groups). ExtendP to a map  : Z ! C by de ning (n) = 0 if (n; f ) 6= 1, and consider the (n) Dirichlet series 1 n, so 0 = 0 and n j  1 for all P n=1 ns . In the notation of the lemma, jbP f we have absolute convergence for Re(s) > 1. However, 0 = n=1 (n) = (n;f )=1 (n), so jT (m)j = O(1) and the original series converges in Re(s) > 0, although not absolutely. 0

0

Proof. We proved part 1 in Lecture 11. For part 2, let Ap;m denote the partial sum

Ap;m =

m X

n=p

bn = T (m) T (p 1)

for all m > p. Then jAp;m j = O(m ) because m  (p 1) . Let cn = n s . Then m X

n=p

X X bn n s = bn cn = (Ap;n Ap;n 1 )cn = =

The following bounds

imply that

m

m

n=p mX1

n=p

n=p mX1

Ap;n (cn cn+1 ) + Ap;m cm

 Ap;n n1s n=p

1 ns

1 +A 1 p;m ms (n + 1)s

1  jsj (n + 1)s n(Re(s)+1)

jAp;n j  Kn Ap;m Km ms  mRe(s)

m hmX1 X i bnn s  K n(Re(sjs) j +1) + m(Re(1s) ) n=p

n=p

Since jsj is bounded in any compact region, we see that as p; m ! 1 the terms mX1 n=p

j sj

n(Re(s) +1)

and

1

m(Re(s) )

both tend to zero, by the absolute convergence of  (s) in the region Re(s) > 1. This implies uniform convergence in compact regions, so f (s) is analytic in Re(s) > , as desired. 55

To see how this completes the proof of the theorem, consider the function

f (s) = k (s) h (s) =

X bn n

ns

where bn = an h. By Corollary 63 we see that

T (m) =

m X

n=1

bm = (hm hm) + O(m(1

n1 ) )

so that Lemma 64 implies that f (s) is holomorphic on the region Re(s) > 1 n1 . The fact that i h k (s) s h1 = [k (s) h (s)] + h  (s) s 1 1 with Lemma 54 give the desired region of holomorphicity and the correct residue. At the point s = 1 we may consider the formal product  1 YY N(P ) 1  YY 1 N (1P ) = p P jp p P jp N (P )

1

=

YY

P)

N(

p P jp N ( P )

1



obtained by \evaluating" k (1). IfQwe again disregard the nite number of rami ed prime numbers, and consider the factor char(P )=p (P(P) ) 1 where p - d, then we have N

Y

N

pn pn P ) = Q pn Q = =  #(A=pA) P jp (N (P ) 1) P jp #(A=P ) P jp N (P ) 1 N(

Q

since the Chinese Remainder Theorem implies that A=pA = P jp A=P .

Lecture 13 Given a nontrivial character  : (Z=f Z) ! C  , we can extend  to a function  : Z ! C by de ning (n) = 0 if (n; f ) 6= 1. (We shall applyP this convention frequently below without s formally restating it each time.) Let L(; s) = 1 n=1 (n)n . As we noted in Lecture 12, Lemma 24 implies that L(; s) converges and de nes an analytic function in the region Re(s) > 0. This allows us to restate an improvement of the converse to Corollary 34 which we proved in Lecture 6. Theorem 65. (Dirichlet's Reinterpretation of Quadratic Reciprocity) Let k be a quadratic eld with discriminant d. Then there is a character  : (Z=dZ) ! f1g (i.e., a character  : (Z=dZ) ! C  such that  6= 1 and 2 = 1) such that for every prime number p 1. p splits in k i (p) = 1 2. p is inert in k i (p) = 1

56

3. p is rami ed in k i (p) = 0 Moreover, we have a factorization k (s) =  (s)L(; s) in the region Re(s) > 1, and k (s) has a meromorphic continuation to the region Re(s) > 0 whose unique pole is at the point s = 1 and is simple.

Example. If d =   1 (mod 4) is a prime, then, as noted in Lecture 6, the character is exactly (n) = ( n ), as 1. p splits in k i ( p ) = ( p ) = 1

2. p is inert in k i ( p ) = ( p ) = 1 3. p is rami ed in k i p = . Proof. The algebraic properties were proved in Lecture 6. Furthermore, we demonstrated that for any prime number p (rami ed or unrami ed) the factor of k (s) corresponding to p is  1  1 (p)  1 1 1

ps

so that

k (s) =

ps

 1 p1s p

Y

Since

 (s) = it suces to show that

1

 1 p(sp)

Y

 1 p1s p

1

1

1 1 1 = s ps n n p Q Q Factor any positive integer n = i pai i so that (n) = i (pi )ai . Since

L(; s) =

X (n) Y

X (1 (p)p s ) 1 = 1 + (p)p s + (p2 )p 2s +    = (pn )p ns 1

n=0

a bit of bookkeeping implies the desired result. Corollary 66. (Class Number Formula for Quadratic k) Let k be a quadratic eld with discriminant d. If d > 0, then the regulator is given by R = log jj for some  > 1 generating A =f1g. If d < 0, then R = 1. Moreover,

8 2h log jj < p L(; 1) = h = : 2phd w jdj 57

if d > 0 if d < 0

Proof. If d > 0 then r1 = 2 and r2 = 0, so A has rank 1. With the notation of Lecture 12, let H  R2 denote the hyperplane which contains the image E of A under the map  : A ! R2 given by  7! (log jjv ; log jjv ). Fix an element  2 A which maps to a basis element of E . Without loss of generality, assume that jjv > 1, as  6= 1. Then R = log jjv , by de nition. If d < 0 then A has rank 0, and the de nition of R shows that R is the \empty determinant" which is 1. The nal claim follows from considering the Taylor expansions of  (s) and k (s), and the facts that Ress=1  (s) = 1 and Ress=1 k (s) = h by Lemma 54 and Theorem 61. Corollary 67. L(; 1) > 0 2

1

1

1

Proof. Immediate.

We desire an independent evaluation of L(; 1) for any nontrivial character  : (Z=f Z) ! C  . Our rst tool shall be the Gauss sum of  which we de ne to be the complex number

g() =

Lemma 68. If (n; f ) = 1, then

f X a=1

(n)g() =

(a)e2ia=f

f X b=1

(b)e2ibn=f

where  denotes the conjugate character, i.e.,  =  1 as the image of  is contained in the set of roots of unity in C . Proof. By making the substitution b  an 1 (mod f ) so that nb  a (mod f ) we have

(n)g() =

f X a=1

(n)(a)e2ia=f =

f X b=1

(b)e2ibn=f

as desired. By de nition,

g()L(; s) =

1 X n=1

g()(n)n s

If (n; f ) 6= 1 then (n) = 0 and the corresponding terms in the series vanish. If (n; f ) = 1, then the lemma applies, and it follows that

g()L(; s) =

f X b=1 (f;b)=1

(b)

1 e2ibn=f  X n=1

ns

58

=

f X b=1 (f;b)=1

(b)

1 (e2ib=f )n  X n=1

ns

and we have absolute convergence in the region Re(s) > 1. If  = e2i=f , then the formula becomes f 1 ( b )n  X X g()L(; s) = (b) (5) n=1

b=1 (f;b)=1

ns

We recall that the branchPof log(1 x) de ned on the region jxj < 1 has power series expansion log(1 x) = n1 xnn and that the series actually converges to log(1 x) for jxj = 1 such that x 6= 1. A lemma of Abel states that we may substitute the value s = 1 into equation (5) which yields

g()L(; 1) =

f X

b=1 (f;b)=1

(b)( log(1  b ))

(6)

Lecture 14 De nition. A character  : (Z=f Z) ! C  is primitive if there is no proper divisor f 0 of f such that  is induced by a character 0 : (Z=f 0Z) ! C  . It follows immediately that, if

f is prime, then  is primitive i  is nontrivial. A character  (not necessarily primitive) is said to be even if ( 1) = 1, and is said to be odd if ( 1) = 1. Lemma 69. (Gauss' Formula) If  is primitive, then g()g() = ( 1)f . In particular, p g() 6= 0 and jg()j = f . We shall prove this lemma in the case where f is prime. Proof. Assume that f =  is prime. For any function F : Z=Z ! C , the Fourier transform of F is the function F^ : Z=Z ! C given by the formula

F^ (b) =

 X a=1

F (a)e2iab=

In particular, a character  : (Z=Z) ! C  can be extended to such a function, and we have

^(b) =

X a=1

(a)e2iab=

For b = 0 the fact that  is a nontrivial character implies that

^(0) =

1 X a=1

(a) = 0

If b 6= 0, then we make the substitution c  ab (mod ) so that a  cb

^(b) =

X c6=0

(cb 1 )e2ic= = (b) 59

1X

c6=0

1

(mod ) and

(c)e2ic= = (b) 1 g()

(7)

Thus, ^ = g() 1 . If g() = 0, then ^ = 0, but this can not hold because of the Fourier inversion formula (d F^ )( a) = F (a) Applying formula (7) twice yields

b^ = g()d1 = g()g( 1 )

and Fourier inversion implies that g()g( 1 )( a) = (a) Letting a = 1 we get ( 1) = g()g( 1 )(1) = g()g() as desired. If we conjugate the de nition of g() and make the substitution a 7! a, this yields

g() = so that

X a=1

X X (a)e 2ia= = ( a)e2ia= = ( 1) (a)e 2ia= = ( 1)g()



a=1

a=1

p jg()j = [g()g()]1=2 = [g()( 1)g()]1=2 = [( 1)( 1)]1=2 = 

as desired. Assume that  is a prime number and that  is nontrivial. Then

(n)g() =

 X b=1

(b)e2ibn=

for all n  1: if (n; ) = 1 then this is Lemma 68, otherwise each side is zero. Thus

g()L(; s) =

XX

 X X e2ibn=  (b)e2ibn= n s = (b) ns 

n1 b=1  1 X ( b )n  X = (b) s n1 n b=1

b=1

n1

(8)

where  = e2i= . This series converges (conditionally) for Re(s) > 0 by Lemma 64. The power series log(1 x) = gives the branch of log given by log(z ) = log jz j + i arg(z )

for 60

X xn

n1

n

=2 < arg(z ) < =2

Thus, substituting s = 1 into equation (8) yields

g()L(; 1) =

1 X b=1

(b)

 1 X ( b )n  X

=

n

n1

(b)( log(1  b ))

b=1

To simplify the formula for L(; 1), we consider two cases. If  is even, then ( b) = (b) for all b so that

g()L(; 1) =

1 X b=1

(b) log(1  b ) =

 1 X b=1

(b) log j1  b j

by our choice of the branch of log x, since the argument terms cancel. By Lemma 69, this implies that  1 X L(; 1) = g() (b) log j1  b j b=1

(9)

If  is odd, then ( b) = (b) for all b so that

g()L(; 1) = i as in the even case. Thus,

L(; 1) = ig()

 1 X b=1

1 X b=1

(b) arg(1  b )

(b) arg(1  b )

This is not quite as nice as the nal formula in the even case since arg(1  b ) is more complicated than log j1  b j. For b = 1; : : : ;  1 the fact that arg is additive (up to multiple of 2) implies that arg(1  b ) = arg( b=2 ( b=2  b=2 )) = arg( b=2 ) + arg( b=2  b=2 ) The complex number  b=2  b=2 is purely imaginary, so its argument is =2. By de nition,  b=2 = e2ib=2 = eib= and therefore has argument b= which is in the interval (0; ). It follows that, in order for arg(1  b ) to fall in the correct interval,  b=2  b=2 must have argument =2. Thus, arg(1  b ) = ( b 21 ) which implies that

L(; 1) = ig() We note that, if B1 (x) = x

1 is 2  1 X

b=1

1 X b=1

(b)( b

1 2)

the rst Bernoulli polynomial, then

(b)( b

1 2) =

61

 1 X b=1

(b)B1 (b=)

(10)

which we shall denote by B1; , and we may write

L(; 1) = ig() B1;

p Example. Let k = Q ( ) where  is a positive prime such that   1 (mod 4). Then

Corollary 66 implies that

log jj = L(; 1) Ress=1 k (s) = 2h p



where (b) = ( b ), as we noted in Lecture 6. Since  is real-valued, we see that  =  and has order 2. By equation (9) it follows that

Q b (1  b ) g (  ) L(; 1) =  log Q (b )=1 (1  b ) (  )= 1

p

We know that g()2 = , so the coecient is  p1 . In fact, g() = + , but this is not easy to verify. (For a proof, see Borevich and Shafarevich, Number Theory Theorem 5.4.7.) If we let Q b (1  b ) u = Q (b )=1 (1  b ) (  )= 1 then this implies that 2h p log jj = g() log juj = p1 log juj 



Cancellation implies that 2h log jj = log juj and exponentiation yields

u =  2h

Example. Let k = Q (p ) where  is a positive prime such that   3 (mod 4). Then  X p = L(; 1) = ig() (b)( b Ress=1 k (s) = 2h

as

1)

2 w  b=1     X X X (b)(b=) (b)(1=2) = ig() (b)(b=) = ig() b=1 b=1 b=1

X b=1

(b)(1=2) = 12 62

 X b=1

(b) = 0

p

p

Here (b) = ( b ) as above, and g()2 =  so that g() = i . In fact, g() = +i , so

ig() = p1   and therefore

X

2h =

w

(b)(b=)

b=1

It should be noted that w = 2 unless  = 3. Example. Let k = Q (i). We can recover the fact that the class number of k is h = 1 from the ideas above. Let  : (Z=4Z) ! f1g be the character described in Theorem 65, so that k (s) =  (s)L(; s). By Corollary 66, L(; 1) = 2ph = h w jdj 4 By de nition

L(; 1) = 1 13 + 15 71 + 19    which lies between 1 and 2=3. It follows that h = 1. Proposition 70. Let  be an odd prime number,  =  a primitive th root of unity, and k = Q ( ). Then k (s) =  (s)

Y

6=1

L(; s)

where the product is taken over all nontrivial characters  : (Z=Z) ! C  . Proof. The Euler products of each function show that the desired equation is

YY p P jp

(1

Y YY P ) s ) 1 = (1 p s ) (1 (p)p s )

N(

1

6=1 p6=

p

and we check equality factor-by-factor at each prime. The prime  is completely rami ed, so that A = P  1 and N (P ) = . In the in nite product on the left, this contributes a factor of (1  s ) 1 . In the factorization of k (s) we have a factor of (1  s ) 1 contributed by , and no factor contributed to any of the L(; s). Thus, \over " the left and right-hand sides agree. Let p 6= , so that p is unrami ed. Then pA = P1    Pg , and N (Pi ) = pf where f is the order of p in (Z=Z). In particular, g =  f 1 . In k (s), p contributes factors

Yg

i=1

(1 p fs ) 1 = ((1 p fs )( 1)=f ) 63

1

On the right-hand side of the desired equation, p contributes (1 p s )

1

Y

6=1

(1 (p)p s )

1

and we want to know that these expressions are equal. Let X = p s . Then the desired equality is (1 X f )( 1)=f =

Y

all

(1 (p)X )

which we shall show is an identity on polynomials. We note that the zeroes of each side are f th roots of unity, so this is reasonable. The subgroup hpi  (Z=Z) has order f . Let G denote the quotient (Z=Z)=hpi, which has order g =  f 1 . The short exact sequence

1 ! hpi ! (Z=Z) ! G ! 1 implies that the right-hand side of the desired equality is

Y

(1 (p)X ) =

all chars.

of (Z=Z)

Y

(1

all chars. of hpi

(p)X )( 1)=f

and the factorization of cyclotomic polynomials implies that this is exactly the left-hand side of the desired equality. Next, we have some consequences of the proposition. Corollary 71. For nontrivial characters  : (Z=Z) ! C  , L(; 1) 6= 0. Proof. Immediate from the proposition, as L(; s) is regular at s = 0, and the fact that k (s) and  (s) both have simple pole at s = 1. Corollary 72. (Dirichlet) If  - a, then there are in nitely many prime numbers p such that p  a (mod ).

This is the statement that there are in nitely many primes in an arithmetic progression modulo . Before we prove the corollary, we need a pair of lemmas. Lemma 73. For a nontrivial character , the function X f (s) = (p)

is bounded as s ! 1.

p6=

ps

Proof. For Re(s) > 1, the power series expansion of log(1 x) implies that Y (p)  1 X X (p)n = log(L(; s)) = log 1

ps

p6=

The rest of the proof is similar to that of Corollary 56. 64

p6= n1

npns

De nition. Let G be a nite abelian group and x g 2 G. De ne the characteristic function of g, char(g) : G ! C , as g 7! 1 and h 7! 0 for h = 6 g. Lemma 74. The characteristic function of g can be described by the formula char(g) =

1 X (g) #G all

Proof of Lemma 74. For h 2 G

X X (h=g) char(g)(h) = #1G (g)(h) = #1G all

If h = g, then this is

all

X char(g)(g) = #1G (1) = 1 all

since the number of characters of G is exactly #G. If h 6= g, then h=g 6= 1 and there exists a xed character 0 of G such that 0 (h=g) 6= 1 (c.f. Borevich and Shafarevich Number Theory, p. 417), and of course 0 (h=g) 6= 0. As  runs through all characters of G, so does 0 , so that

X

all

(h=g) =

X

all

0 (h=g)(h=g) = 0 (h=g)

X

all

(h=g)

which implies that the sum is zero. Proof of Corollary 72. In the notation of the lemmas, let G = (Z=Z). Then 1 X (a)f (s) = 1 X (a) X (p) = X 1 1 X (a)(p)  s s  1 all  #G all  p6= p p6= p #G all  X1 X 1 = char(a)(p) = s p6= p

pa(mod ) p

s

For each  6= 1, f (s) is bounded as s ! 1, and f1 (s)  log( s 1 1 ). Thus, up to a bounded function X 1 1  1  s   1 log s 1 p pa(mod )

P

as s ! 1, which implies that pa p1 is in nite. In particular, the number of terms in the sum must be in nite, as desired. Notice that the proof of the corollary gives us a stronger result. Up to a bounded function X 1 1  1  s   1 log s 1 pa(mod ) p as s ! 1. This says that the density of such primes is 1=( 1). 65

Corollary 75. (Kummer's Class Number Formula) For the cyclotomic eld k = Q ( )   Y ig()   Y g() X  ( 1)=2 hR b j = (2 )  ( b ) log j 1   B 1 ; (     2)=2  2  odd

even 6=1

b=1

Proof. The right-hand side of the desired equality is



Ress=1 k (s) = Ress=1  (s)

Y

6=1



L(; s) = Ress=1

Y all

L(; s)



which is the left-hand side by Theorem 61.

Lecture 15 We de ne the gamma function (s) =

Z1 0

e t ts dtt

for Re(s) > 0. By integrating by parts, we see that (s) satis es the functional equation (s + 1) = s (s). Thus, for Re(s) > 1, we may de ne (s) = (s + 1)

s

to obtain a meromorphic continuation to the region Re(s) > 1, with simple pole at s = 0. We may continue this process inductively to obtain a meromorphic continuation to all of C with only simple poles located at s = 0; 1; 2; : : : . Note that (1) = 1, so that (m) = (m 1)! for m = 1; 2; 3; : : : . Theorem 76. (Hecke's Thesis) Let k be a number eld. 1. k (s) has a meromorphic continuation all of C , with a unique pole at s = 1 which is simple. 2. If we de ne

k (s) = k (s)( s=2 (s=2))r ((2) s (s))r 1

2

then k (s) has a meromorphic continuation to all of C , with unique poles at s = 0; 1 which are simple. Furthermore, k (s) satis es the functional equation

k (s) = jdj s k (1 s) where d is the discriminant of k. 1 2

For a complete proof, see Weil Basic Number Theory, Chapter VII. We shall continue with these ideas below. First, we give some motivation for the Gamma factors in this product by considering curves over nite elds. 66

Let Fq denote the nite eld with q = pm elements, and let X be a complete, nonsingular curve over Fq . Let k = Fq (X ) be the eld of rational functions on X , and let A  k be the integral domain of functions regular outside a single place 1. We de ne the zeta function of A as Y X 1 1  1 = A (s) = 1 s N (P )s P A I A N (I ) P 6=0

for Re(s) > 1, where the product is taken over all nonzero prime ideals of A. For example, if X = P1 = P1q , then k = Fq (X ) = Fq (t)  Fq [t] = A. The nonzero prime ideals P of A are in bijection with the collection of irreducible, monic polynomials f (t) over F q . Under this correspondence, if f (t) has degree d, then N (P ) = q d . Since F q sits inside A=P for every such prime, we see that every P has characteristic p. (This is true in general, of course.) Furthermore, A = Fq . If q is odd and X is an elliptic curve, say given by the equation y2 = x3 + Ax + B , then F

A = Fq [x; y]=(y2 x3 Ax B ) and k = Fq (X ) is the quotient eld of A. Let X , k and A be as above. Since Fq ,! A  K , A has characteristic p. For every prime ideal P , N (P ) = qd for some positive integer d, which we call the degree of P and denote by d = deg(P ). We de ne the expression ZA (T ) by ZA (T ) =

Y

(1 T deg(P ) )

1

P 6=0 and note that, by de nition, A (s) = ZA(q s ). As in the number eld situation, we can use

P to de ne a valuation on k by

j jP = N (P )

ord P ( )

For a valuation v on k, let Av denote the ring of elements of k such that v( )  1. Then Av is a local ring with maximal ideal equal to the set of elements of k such that v( ) < 1. In addition, Av is a discrete valuation ring so that the maximal ideal is principal, generated by a uniformizing parameter which we denote by v . The degree of v is de ned to be the dimension of the quotient Av =v Av over Fq , and this is exactly the degree of the prime ideal P which induced v. We de ne the complete zeta function of X

ZX (T ) =

Y v

(1 T deg(v) ) 1 = ZA (T )(1 T deg(1) )

1

by using all the valuations of k, including the valuation atP1, which n does not correspond to a prime ideal of A. Using the power series log(1 x) = n1 xn , which holds for jxj < 1, we see that in Q [ T ] log(ZX (T )) =

X X T n deg(v) X X deg(v)T n deg(v) X T n n = n deg(v) = n An v n1

v n1

67

n1

where

An =

X deg(v)jn

deg(v) 2 Z

A prime ideal P is determined by an orbit of the action of G = Gal(F q =Fq ) on sets of points of X = X (F q ) (the points of X de ned over F q ) and the degree of P is the number of elements in the orbit. Fix a point x of X . G acts on X and therefore determines an orbit of x.PThe orbit is nite and appears as a set of points in X (Fqm ) for some m. The sum d = 2G x is a prime divisor on X and therefore determines a prime ideal of A. It follows that An = #X (Fqn ) and so log(ZX (T )) = Exponentiation yields the equation

ZX (T ) = exp

X Tn

n1

n n (#X (Fq ))

X T n

(#X (Fqn )) n1 n



For example, if X = P1, then #X (Fqn ) = qn + 1 and n X n X n X n log(Z (T )) = (q + 1)T = T + (qT ) X

so that

n1

n

n

n

  1 = log(1 T ) 1 + log(1 qT ) 1 = log (1 T )(1 qT ) 1 Z (T ) = (1 T )(1 qT ) P1

and the substitution T = q s implies that

1 X (s) = (1 q s )(1 q1 s ) Notice that this is a rational function in q s , which is meromorphic with poles only where q s = 1 and q1 s = 1. This is similar to Theorem 76. (T ) , where P (T ) is a polynomial in T with integer coecients In general, ZX (T ) = (1 TP)(1 qT ) having degree 2g where g is the genus of X . More speci cally,

P (T ) = 1 +    + qg T 2g = for some algebraic integers i , and we may write exp

2g Y

i=1

(1 i T )

 X #X (Fqn )T n  (# X ( Fq n )) = ZX (T ) = exp n n1 n1 n

X T n

  2 = exp #X (Fq )T + #X (Fq ) T2 +    2

68

If X is an elliptic curve, then P (T ) = 1 aT + qT 2 since gX = 1. The only unknown coecient is a. In fact, a = 1 + q #X (Fq ), as can easily be veri ed by checking the coecient of T in ZX (T ). Now, we return to the case where k is a number eld. The de nition of k (s) may be restated, as follows. Every valuation v on k corresponds to a factor v (s) in the Euler product of k (s). For the nite valuations (those corresponding to nonzero prime ideals of A) v (s) = (1 N (P ) s ) 1 , which is exactly the factor of k (s) corresponding to P . For a valuation arising from a real place, v (s) =  s=2 (s=2), which is meromorphic in C with (simple) poles at s = 0; 2; 4; : : : . For a valuation arising from a complex place, v (s) = (2) s (s), which is meromorphic in C with (simple) poles at s = 0; 1; 2; : : : . We shall prove part of Theorem 76 for the case k = Q . In this case, d = 1, r1 = 1 and r2 = 0. As with the zeta function of Q , we denote  (s) as  (s). Theorem 77. The function  (s) =  (s) s=2 (s=2) has a meromorphic continuation to all of C with simple poles at s = 0; 1. Furthermore,  (s) satis es the functional equation  (1 s) =  (s). Q

Proof. For a xed positive integer, we make the substitution t = n2 x into the de nition of (s=2). Here, dtt = dxx so that

Z1

ts=2 e t dtt Z01 = (n2 x)s=2 e n x dx x Z01 = ns s=2 xs=2 e n x dx x Z0 1 n s  s=2 (s=2) = xs=2 e n x dx x (s=2) =

2

2

2

0

and summing over all n  1 yields

 (s) =  (s) s=2 (s=2) X =  s=2 (s=2) n1s = =

X

n1



n s=2 (s=2)n s

XZ 1

n1 0

xs=2 e n x dx x 2

which converges absolutely for Re(s) > 1, so we may interchange the sum and integral

 (s) =

Z1 0

69

xs=2

X

n1

 e n x dx x 2

P

Let !(x) = n1 e n x. Here, we introduce Jacobi's theta function. For x > 0, let X (x) = e n x = 1 + 2!(x) 2

2

If we write f (t) = e summation formula

n2Z t2 x and let f^(t) denote the Fourier transform of f , then by the Poisson

(x) =

Furthermore, the identity

X

f (n) =

n2Z

X^

f (n)

n2Z

t2 =x

f^(t) = e x1=2

implies that

(x) =

X^

1

f (n) = (xx1=2 )

n2Z

(11)

which is Jacobi's famous theta identity. Thus,

xs=2 !(x) dx x Z1 Z0 1 dx s= 2 xs=2 !(x) dx = x !(x) x + x 1 0

 (s) = we substitute x 7! x

1

into the rst integral =

The fact that !(x) = so that

Z1

1 ((x) 2

Z1 1

x

s=2 ! (x 1 ) dx +

x

Z1 1

xs=2 !(x) dx x

1) implies that 1 =2 !(x 1 ) = 21 ((x 1 ) 1) = 21 (x1=2 (x) 1) = x1=2 !(x) 12 + x 2

Z1

Z1

+ xs=2 !(x) dx x1=2 !(x)x s=2 dx x x 1 1 Z 1 1 xs=2 dx + 1 Z 1 x1=2 s=2 dx x 2 1 x Z 12 1 dx 1 = (x(1 s)=2 + xs=2 )!(x) x s + s 1 1 1 Z1 1 = s(s 1) + (x(1 s)=2 + xs=2 )!(x) dx x 1

 (s) =

To see that  (s) has the proper meromorphic continuation, it suces to show that the nal integral has an analytic continuation to all of C . The essential fact is that, as x ! 1, !(x) decays exponentially so the integral converges. To see that  (s) satis es the desired functional equation, we simply observe that if we replace s by 1 s in the nal expression for  (s), it remains the same. 70

Lecture 16 Next, we will discuss the analytic continuation of L(; s) and a functional equation satis ed by L(; s). Let  = x + yi be a complex number with y > 0, and let q = ei . We de ne a function ( ) similar to Jacobi's theta function, as ( ) =

X

ein  = 2

n2Z

X

qn

2

n2Z

It readily follows from the de nitions that (iy) = (y). Example. Let k = Q (i). Then A = Z[i] and X 1 1 X 1 k (s) = =4 2 + b2 )s N (a)s ( a aA (a;b)6=(0;0) a=(a+bi)

as N (a) = a2 + b2 . If we de ne f ( ) as

f ( ) =

X (a;b)

qa +b = (( ))2 2

2

then f satis es an equation similar to the equation satis ed by : f (iy) = y1 f (iy 1 ). Assume that  : (Z=f Z) ! C  is a primitive character. In order to uncover the functional equations for L(; s) we introduce functions  (; s) as we did for k (s). There are two cases. If  is even, then we de ne  (; s) =  s=2 (s=2)L(; s)f s=2 p In this case, let w = g()= f , where g() is the Gauss sum from Lecture 13. Then jw j = 1 and  (; s) has a holomorphic continuation to all of C which satis es the functional equation  (; s) = w  (; 1 s) If  is odd, then we de ne  (; s) =  (s+1)=2 ((s + 1)=2)L(; s)f s=2 p Let w = g()=i f . Again jw j = 1 and  (; s) satis es the functional equation  (; s) = w  (; 1 s) If 2 = 1, then we say that  is a quadratic character. In this case  =  and it can be shown that (p g() = pf if  is even i f if  is odd (c.f., Borevich and Shafarevich Number Theory, Theorem 5.4.7) so that w = 1. Thus, the simpler functional equation  (; s) =  (; 1 s) 71

is satis ed. We shall only prove that  satis es the desired functional equation when  is an even character. First, however, we note that there are ways of slightly changing the functions k (s) to get the same meromorphic continuation properties and simpler functional equations. For example, assume that k is a quadratic eld, with character  as is Theorem 65. It can be shown that k is real if and only if  is even (c.f., Borevich and Shafarevich Number Theory, Chapter 5,SS4). Furthermore, it can be shown that every quadratic character (Z=f Z) ! f1g corresponds to a quadratic eld in this manner. If k is a real eld, then we de ne

k (s) = ( s=2 (s=2))2 f s=2 k (s) which is very similar to the de nition of k (s). It is straightforward to check that k (s) satis es the functional equation

k (s) = jdj s k (1 s) 1 2

if and only if k (s) satis es the functional equation

k (s) = k (1 s)

which is simpler. To see that this second equation is satis ed, notice that the de nitions of

k (s),  (s) and  (; s) imply that

k (s) =  (s) (; s)

In the quadratic case,  (; s) =  (; 1 s) and by Theorem 77,  (1 s) =  (s). Thus,

k (s) =  (s) (; s) =  (1 s) (; 1 s) = k (1 s) as desired. A similar change works for the case when k is imaginary. Before we prove the real case, we need some notation and a lemma. Let g be an element of the real Schwarz space, so that g is C 1 , and g and all its derivatives decrease rapidly at 1. The Fourier transform of g, denoted g^, is given by

g^(y) =

Z

R

g(x)e

2ixy dx =

Z

R

g(x)hx; yi; dx

where hx; yi = e2ixy . (The assumption that g is in the real Schwarz space implies that we may take iterated Fourier transforms of g.) De ne g (t; x; y) as g (t; x; y) =

X

g((x + n)t)hy; ni

n2Z

P

It is immediate from this de nition that g (t; 0; 0) = n g(nt). Lemma 78. (Generalized Jacobi Identity) g (t; x; y) = hx;t yi g^(t 1 ; y; x) 72

Proof. To simplify notation, let G(w) = g((x + w)t)hy; wi. Then by the Poisson summation formula X X g = G(n) = G^ (n) n2Z

n2Z

so we calculate G^ . By de nition

G^ (u) = =

Z

Z

R

R

g((x + w)t)hy; wihu; wi dw g((x + w)t)hu y; wi dw

and we make the substitution z = (x + w)t so that w = zt x, dw = dzt and

Z

E D g(z ) u y; zt x dzt Z D E = 1t hx; yihu; xi g(z ) z; u t v x dz

G^ (u) =

R

  = hx; yihu; xi g^ u y R

t

t

Substituting u = n and summing over all n yields g (t; x; y) =

X^

G(n) =

X hx; yihn; xi  n y  g^

t   X = hx;t yi hn; xig^ n t y = hx;t yi g^(t 1 ; y; x) n

as desired.

n2Z

t

n

Corollary 79. For real numbers y and t

X

X (ny) =t e( n t+2iny) = p1 e t n2 n2 2

2

Z

Z

Proof. Let g( ) = e  . Then it is a standard fact that g^ = g, and Lemma 78 implies that 2

X

p e( n t+2iny) = g ( t; 0; y)

n2Z

2

= h0p; yi g^(t 1=2 ; y; 0)

t

= h0p; yi  (t

1=2 ; y; 0) t g X (ny) =t = p1 e t n2 2

Z

as desired. 73

Theorem 80. Assume that  : (Z=f Z) ! f1g is a primitive, even character. Let  (; s) =  s=2 (s=2)L(; s)f s=2 p and let w = g()= f , where g() is the Gauss sum. Then  (; s) has a holomorphic continuation to all of C , and satis es the functional equation

 (; s) = w  (; 1 s) Proof. Start with the identity

Z1 f s=2  s=2 (s=2)n s = e n x=f xs=2 dx x 2

0

which is proved as in the proof of Theorem 77. Let

 (x) =

X

(n)e n x=f 2

n2Z

so that, when we sum over all n  1 we have

 (; s) =  s=2 (s=2)L(; s)f s=2 X = (n)n s  s=2 (s=2)f s=2 n1

= =

Z1 0

Z1 0

as ( n) = (n) and (0) = 0 = 21

xs=2

X

n1

Z 1 0

2

 dx  ( x )  2 x

 s=2 1

x

 (n)e n x=f dx x

Z1  s=2  (x) dx + x xs=2  (x) dx x x 1

(12)

The rst integral converges for Re(s) > 0, while the second converges for all s. To obtain the desired properties, we need to nd an equation relating  (x) to  (x 1 ). By Lemma 68,

(n)g() = so that

g() (x) = =

f XX

f X b=1

 (b)e2ibn=f e n x=f 2

n2Z b=1 f X X b=1

(b)

(b)e2ibn=f

 e( n x=f )+(2ibn=f )

n2Z

74

2

and the substitutions t = x=f and y = b=f yield = =

f X

(b)



b=1 r f fX

x b=1

1 px=f

(b)

X

X

 e (n+b=f ) f =(xf ) 2 2

n2Z

e (nf +b) =(xf ) 2

n2Z

and the replacement m = nf + b gives =

r X f

(m)e m =(fx) 2

x n2 = fx  (x 1 )

r

Z

If we substitute this nal expression into (12), we nd

Z1  s=2  (x) dx x + x(1 s)=2 g(f)  (x) dx x 1 x 1 which gives our analytic continuation. Furthermore, the substitution s 7! 1 s gives the  (; s) = 21

Z 1

p

desired functional equation.

We have shown that k (s) has order of vanishing 1 at s = 1 and residue r r Ress=1 k (s)ds = 2 p(2) hR w 2

1

jdj

We shall use Hecke's functional equation for k (s) to calculate the order of vanishing and residue at s = 0. By Theorem 76, we know that

k (s)( s=2 (s=2))r ((2) s (s))r = k (s) = jdj s k (1 s) and that k (s) has a simple pole (i.e., vanishing of order 1) at s = 0. Also, (s) has a simple pole at s = 0. Thus, 2

1

1 2

1 = ords=0 k (s) = ords=0 k (s) + r1 ords=0 (s) + r2 ords=0 (s) = ords=0 k (s) r1 r2 so that ords=0 k (s) = r1 + r2 1 = rank(A ) and

k (s) = csr +r 1

2

A further computation shows that c = hR w. 75

1 + O(sr1 +r2 )

Lecture 17 In the nal lectures, we shall give a basic treatment of Artin L functions to introduce the reader to the ideas behind Stark's Conjectures. For a very thorough treatment of these ideas, see Tate Les conjectures de Stark sur les fonctions L d'Artin en s=O: notes d'un cours Orsay [de] John Tate, Birkhauser, Progress in Mathematics. Let Q  k be a Galois extension of degree n with Galois group G which is necessarily nite of order n. Let A denote the ring of algebraic integers in k. G acts on A, so for every prime number p, G acts on A=pA. Therefore, G permutes the prime ideals P of A of characteristic p. Proposition 81. G permutes the primes P of characteristic p transitively.

Q

Proof. Fix a prime ideal P of A of characteristic p. Let I = Pi 6=P Pi where the product is taken over all prime ideals Pi of characteristic p such that Pi 6= P . Since each prime ideal of characteristic p is maximal, we see that P + I = A. Fix 2 P and 2 I such that + = 1. It follows that 62 Pi for any i. For any  2 G, ( ) 2 (PQ) and if Pi 6= (P ) then ( ) 62 Pi . Now, pZ = P \ Z and this ideal contains a = N ( ) = 2G ( ). Thus, a is contained in exactly the primes Pi which are conjugate to P . Since a 2 pZ, we see that a lies in all the Pi , as desired. Corollary 82. In the prime factorization pA = Qgi=1 Piei , let fi = deg(Pi ) = (A=Pi : Z=pZ). Then the ei and fi are independent of i and n = efg where e and f denote the common value of ei and fi , respectively. Proof. The fact that the ei and fi are independent of i follows immediately from the fact that the Pi are all conjugate. The fact that n = efg follows from the fact that the extension Q  k is Galois. Corollary 83. Let P be a xed prime ideal of A dividing p. Let GP = f 2 G : (P ) = P g which is a subgroup of G of order ef . For an element  of the coset space G=GP , the ideal P 0 =  (P ) is a well-de ned conjugate of P , and GP 0 = GP  1 in G. Proof. This follows from the standard results regarding the transitive action of a group on a set.

With the notation of the corollary, we see that GP acts on the nite eld A=P since it stabilizes P . Let IP denote the inertia subgroup, i.e., the kernel of this action. Then there is an exact sequence Aut(A=P ) 1 ! IP ! GP ! by de nition of IP . Since A=P is nite of order pf , the Frobenius automorphism F : A=P ! A=P given by 7! p is a generator of Aut(A=P ). Thus, Aut(A=P ) is cyclic of order f . Theorem 84. The map is surjective. Consequently, IP has order e. In particular, if p does not divide the discriminant dk of k, then IP has order 1.

76

Proof. We shall only prove the result under the additional hypotheses of Corollary 32, i.e., when k = Q ( ) with 2 A and p does not divide the index (A : Z[ ]). By Corollary 32 A=pA = Z[ ]=pZ[ ] = (Z[x]=h(x))=p(Z[x]=h(x)) = (Z=pZ)[x]=h~(x)

where h~ (x) denotes the reduction of h(x) modulo p. The irreducible factorization of h~ (x) over Fp = Z=pZ has the form

h~ (x) =

Yg

i=1

hi (x)e

(13)

where each hi (x) has degree f . Since every ( ) satis es h(x), G permutes the zeroes of h(x). G acts transitively on the zeroes: if 0 is another root, then k = Q ( 0 ) and the map 7! 0 determines an element of G. Furthermore, G acts simply-transitively on the zeroes, that is, for each root i there exists a unique  2 G such that ( ) = i . This follows from the fact that h(x) has exactly n distinct zeroes where n = (k : Q ) = #(G), and G acts

transitively on the zeroes. There are n = efg distinct zeroes of h(x) over Q . The fact that Fp is perfect and the factorization 13 implies that there are exactly fg distinct zeroes of h~ (x) over F p , each of multiplicity e. By Corollary 32, the prime ideal P corresponds to one of the hi (x). Assume without loss of generality that P corresponds to h1 (x). Let ~1 denote the reduction modulo p of to a zero of h1 (x). For  2 G,  is an element of GP if and only if ( ) reduces to a zero of h1 (x). Also,  2 IP if and only if ( ) is one of the e zeroes in characteristic 0 which reduces to ~1 . Since G acts simply transitively, this implies that #(IP )  e. #(IP )  e because #(GP ) = ef and #(Aut(A=P )) = f . Thus, #(IP ) = e and must be surjective. If p does not divide dk , then Corollary 32 implies that e = 1, proving the nal claim. Corollary 85. There exists a coset IP F of IP in GP such that the action of any  2 IP F on A=P is given by 7! p . Proof. This follows from the fact that the Frobenius automorphism generates Aut(A=P ) and the sequence Aut(A=P ) ! 1 1 ! IP ! GP !

is exact.

Corollary 86. If p does not divide dk , so that e = 1 and IP = 1, then the decomposition group GP is cyclic, with a canonical generator FP which satis es the condition

FP ( )  p (mod P ) Proof. Immediate.

If P 0 =  (P ) for some  2 G, then GP 0 = GP  1 and FP 0 = FP  1 2 G. Thus, for all p - dk we get a conjugacy class FP = fFP g in G. Example. Let  = m = e2i=m be a xed primitive mth root of unity. The minimal polynomial of  is denoted m (x) and divides xm 1. If k = Q ( ) then the extension 77

 k is Galois with Galois group G ,! (Z=mZ). (Note that if F is any eld such that char(F ) - m then we may consider the extension F  F (m ) then the Galois group G embeds in (Z=mZ) by mapping  ! a if (m ) = ma .) Gauss showed that over Q the embedding G ,! (Z=mZ) is a surjection, i.e., that m (x) is irreducible over Q . (This is not true for Q

an arbitrary eld F .) We know that G is abelian, so for unrami ed p we get an element Fp which is FP for any P dividing p. For 2 A, the fact that FP ( )  p (mod P ) for every P j p implies that Fp ( )  p (mod pA). If p - m, then p is unrami ed in k, in fact, p - disc(Z[m]). Via the map G ,! (Z=mZ), Fp is some class in (Z=mZ). It is the class of a where Fp ( ) =  a . Since Fp ( )   p (mod pA) it follows that a = p so that Fp is the class of p in (Z=mZ). The Dirichlet Density Theorem states that, as s ! 1, X 1 1 log 1   s m (s) s 1 pa (mod m) p and it follows from the above discussion that we can state this theorem in terms of the Galois group G, as X 1 X 1 = pa (mod m) p

s

s p with p Fp =a2G

We return to the general case where Q  k is any ( nite) Galois extension. Fix a general conjugacy class C  G. Can we nd an unrami ed prime p such that FP 2 C ? Theorem 87. (Cebotarev) There are in nitely many primes p - dk with FP 2 C and as s!1 X 1 #(C )  1  s  #(G) log s 1 p with p FP 2C

Corollary 88. There are in nitely many primes p - dk with FP = 1 in G for some (equivalently, for every) P dividing p, i.e., p = P1    Pn splits completely. These have density

1=#(G).

Next, we develop the higher analogs of the L(; s). De nition. Let k be a nite Galois extension of Q . We de ne the Artin L-function of k, as follows. Let G be the Galois group of k over Q and let (V; ) be a nite-dimensional, complex representation of G. That is, V is a nite-dimensional, complex vector space and  : G ! GL(V ) is a homomorphism of groups. For a prime number p - dk , we have the conjugacy class Fp in G. For any element FP 2 Fp , the element (FP ) in GL(V ) has nite order. This implies that (FP ) is semisimple and determined up to conjugacy by its characteristic polynomial. Thus, for each such p the assignment det(1 Fp tjV ) = 1 Tr (Fp jV )t +    + ( 1)n det(Fp jV )tn is well-de ned, where n = dim(V ) = deg(det(1 Fp tjV )). 78

We de ne L(V; s) as a product over all primes p,

L(V; s) =

Y p

det(1 FP ps jV IP )

1

where V IP denotes the subspace of V xed by (IP ). The product converges absolutely and is nonzero in the region Re(s) > 1. Example. Let V = C and () = 1 for all  2 G. (This is the trivial representation of G.) Then for every p det(1 FP p s jV IP ) 1 = (1 p s )

1

as every V IP = C . Thus, L(C ; s) =  (s). Example. Let k = Q ( ) where  =  is a primitive th root of unity. Then G = (Z=`Z). If  : G ! C  is a character, then the fact that GL1 (C ) = C  implies that we may think of  as a 1-dimensional representation. It follows that the L function of the representation is exactly the L function of the character, as de ned before. In particular, we see that the L function of the representation has an analytic continuation to all of C and satis es a functional equation. Artin Conjecture. With notation as above, L(V; s) has a meromorphic continuation to all of C with a pole of order dim(V G ) at s = 1 and no other poles. Brauer proved that L(V; s) has a meromorphic continuation to all of C . There are a few facts which allow us to reduce to certain basic cases. Fact 1. If V = V1 + V2 as a linear representation of G, then

L(V; s) = L(V1 ; s)L(V2 ; s) To see this, note that V IP = V1IP + V2IP since the representation on V takes the form

A

1

0

 7! 0 A 2



This form also shows that the characteristic polynomial of FP jV IP is the product of the characteristic polynomials of the FP jViIP . Fact 2. L(C ; s) =  (s). Fact 3. Let V = g2GC eg denote the regular representation of G. Then

L(V; s) = k (s) To see this, note that for p - dk , the factor of k (s) corresponding to p can be written ((1 p fs )n=f ) 1 where f = #(GP ) and n = #(G). As noted previously, FP has order f in G: as a permutation of G, left multiplication by FP has a cycle decomposition with n=f cycles, each of length f . Fact 4. (Frobenius) The regular representation of G can be expressed in the form

V = irred. V dim(V )V 79

where the sum is taken over all irreducible representations of G. It follows from Facts 1 and 3 that

k (s) =

Y

irred. V

L(V; s)dim(V ) =  (s)

Y

V 6=C

L(V; s)dim(V )

This is known as the Artin factorization of k (s). Example. The symmetric group S3 has three irreducible representations: the trivial representation, the one-dimensional representation given by  = sgn, and a two-dimensional representation which we denote V . Frobenius' formula tells us that (in the representation ring) the regular representation is of the form 1 +  + 2V . If k is a number eld with Galois group G = S3 , then Fact 4 implies that k (s) =  (s)L(; s)L(V; s)2 . Theorem 89. (Brauer) k (s)= (s) is an entire function of s.

Lecture 18 Last lecture, we restricted our attention to the case of Galois extensions. Now, we wish to see what we can say in the general case. Let k be a number eld. Let K denote the Galois closure of k in Q , that is, K is the smallest sub eld of Q which contains all the conjugates of k. Then K is nite and Galois over Q , say with Galois group G. Let H denote the subgroup of G which xes k. The coset space G=H is a nite set which is in bijection with the set Hom (k; C ) = Hom (k; Q ) = Hom (k; K ) Q

Q

Q

Note that the bijection depends on a xed inclusion of k in K . This set has order n = (k : Q ). Also, there is an action of  2 G = Aut (K ) on 2 Hom (k; K ) given by ( ) =   . If k = Q ( ), then the coset space G=H is in bijection with the n zeroes of f (x) in Q by the assignment  7! ( ). If we assume that p - disc(f ), then distinct zeroes in characteristic zero pass to distinct zeroes modulo p, so this set is in bijection with the n zeroes of f (x) (mod p) in F p . In this set of zeroes, FP acts by i 7! pi . Assume that p is a prime number such that p - dK so that p is unrami ed in K and pAK = P1    Pg . As in the discussion of Lecture 17, G permutes the Pi transitively, and if P = P1 , then the stabilizer of P is Gp = hFP i  = Z=f Z. We claim (without proof) that the factors of p in k are in bijection with the elements in the ( nite) double-coset space GP nG=H . We may think of GP nG=H either as the set of GP -orbits on G=H or as the H orbits on GP nG. Under this correspondence, if a factor p of pAk corresponds to the double coset GP gH , then the degree of p is exactly the number of single G=H cosets in GP gH . We rewrite the Euler factor of p in k (s). Assume that P1 ; : : : ; Pg are the factors of pAk with degrees f1 ; : : : ; fg , respectively. Then the Euler factor corresponding to p is Q

Q

Yg

(1 p fi s ) 1 = det(1 FP p s jC [G=H ])

i=1

80

1

as the Frobenius element Fp has a block form

0A 0    BB 01 A2    B@ .. .. . . . . .

0 where Ai is the fi  fi square matrix

1

0 0C C .. C .A

0    Ag

00 BB1 BB0 B@ .. .

1

0  0  1  .. . . . . 0 0  We de ne the induced representation of G as

1 0C C 0C C .. C .A 0

IndGH (C ) = C [G=H ] Then IndGH (C ) = irred. reps. dim(V H )V by Frobenius reciprocity, as

V of G

HomG(V; IndGH (C )) = HomH (ResV; C ) It follows that

k (s) = L(IndGH (C ); s) =

Y irred. reps. V of G

L(V; s)dim(V H )

Problem. Can we nd two non-isomorphic (i.e., not conjugate in Q ) number elds k and k0 with the same zeta function? This occurs if and only if every p decomposes identically in Ak and Ak0 . Suppose that two such elds exist with the same Galois closure K . We must have (k; Q ) = (k0 : Q ) and they must have the same number of real places, complex places and roots of unity. By the class number formula, this implies that hR = h0 R0 . It is not necessarily true, however, that h = h0 . We know that k and k0 are isomorphic sub elds of K if and only if the corresponding subgroups H and H 0 are conjugate in G. (Under these conditions, we say that H and H 0 are globally conjugate.) By our previous work, k (s) = k0 (s) if and only if IndGH (C )  = IndGH 0 (C ) as representations of G. This occurs if and only if, for every  2 G, Tr ( : IndGH 0 (C )) = Tr ( : IndGH (C )). If we write IndGH (C ) = G=H C egH , then (egH ) = egH so that Tr ( : IndGH (C )) = (the number of cosets gH such that gH = gH ) = (the number of cosets gH such that g 1 gH = H ) = #(fg : g 1 g 2 H g=fg  g0 , gH = g0 H g) = #(C \ H )=(#(H )=#(Z (H ))) 81

Thus, k (s) = k0 (s) if and only if, for every conjugacy class C in G, #(C \ H ) = #(C \ H 0 ). (Under these conditions, we say that H and H 0 are locally conjugate.) We now give two examples of subgroups H; H 0  G which are locally conjugate but not globally conjugate. Example. Let G = S6 and consider the subgroups

H = fe; (1 2)(3 4); (1 3)(2 4); (1 4)(2 3)g H 0 = fe; (1 2)(3 4); (1 2)(5 6); (3 4)(5 6)g The subgroups each have order 4, and they have the same cycle structure (2,2,1,1), and are therefore locally conjugate. (It is straightforward to verify this fact directly, as a conjugacy class is uniquely determined by the cycle structure of a representative.) However, H xes the numbers 5 and 6, while H 0 xes nothing, so H and H 0 are not globally conjugate. Example. Let G = GL3(F2 ) = GL(V ) where V is a three-dimensional vector space over F 2 . Then G has order 168. If we let P denote the stabilizer of a line and P 0 the stabilizer of a plane, then P and P 0 each have index 7 in G. In fact, P and P 0 are locally conjugate, but they are not globally conjugate.

82