Notes on Algebra MTH 619/620 [version 29 Apr 2011 ed.]

Table of contents :
Monoids and groups......Page 7
Subgroups......Page 10
Homomorphisms of groups......Page 12
The kernel and the image of a homomorphism......Page 15
Normal subgroups, cosets and quotient groups......Page 17
Isomorphism theorems......Page 22
Index of a subgroup and order of an element......Page 26
Free groups and presentations of groups......Page 29
Direct products, direct sums, and free abelian groups......Page 33
Categories and functors......Page 36
Adjoint functors......Page 41
Categorical products and coproducts......Page 44
More on free abelian groups......Page 48
Finitely generated abelian groups......Page 53
Permutation representations and G-sets......Page 60
Some applications of G-sets......Page 65
The Sylow theorems......Page 68
Application: groups of order pq......Page 72
Group extensions and composition series......Page 82
Simple groups......Page 86
Symmetric and alternating groups......Page 89
Simplicity of alternating groups......Page 95
Solvable groups......Page 99
Nilpotent groups......Page 102
Rings......Page 108
Ring homomorphisms and ideals......Page 112
Principal ideal domains and Euclidean rings......Page 116
Prime ideals and maximal ideals......Page 120
Zorn's Lemma and maximal ideals......Page 123
Unique factorization domains......Page 126
Prime elements......Page 130
PIDs and UFDs......Page 133
Application: sums of two squares......Page 136
Application: Fermat's Last Theorem......Page 140
Greatest common divisor......Page 141
Rings of fractions......Page 143
Factorization in rings of polynomials......Page 149
Irreducibility criteria in rings of polynomials......Page 156
Modules......Page 162
Basic operations on modules......Page 165
Free modules and vector spaces......Page 167
Invariant basis number......Page 171
Projective modules......Page 176
Projective modules over PIDs......Page 180
The Grothendieck group......Page 182
Injective modules......Page 190
Exact functors......Page 197
Tensor products......Page 202
Tensor products of homomorphisms......Page 207
Tensor products and adjoint functors......Page 209
Tensor products for non-commutative rings......Page 211
Restriction and extension of scalars......Page 214
Application: rings with IBN......Page 217
Algebras......Page 218
Tensor product of algebras......Page 219
Fields......Page 220
Field extensions......Page 222
Prime subfield and field characteristic......Page 223
Algebraic and transcendental elements......Page 225
Algebraic extensions......Page 231
Separable elements......Page 234
Derivatives and separable elements......Page 236
Separable extensions......Page 241
Simple extensions......Page 245
Simple extensions and intermediate fields......Page 250
Construction of extensions......Page 253
Algebraically closed fields......Page 256
Roots of unity......Page 262
Finite fields......Page 265
Galois theory - motivation......Page 268
Normal extensions......Page 271
Galois extensions......Page 277
Application: rational symmetric functions......Page 282
The fundamental theorem of Galois theory......Page 284
The fundamental theorem of algebra......Page 290
Infinite Galois extensions......Page 292
Abelian and cyclic extensions......Page 295
Radical extensions......Page 301
Solvable extensions......Page 303
Solvability of polynomials by radicals......Page 307
Straightedge and compass constructions......Page 312
Construction of regular polygons......Page 318
Transcendental extensions......Page 321
Algebraic sets......Page 327
Hilbert basis theorem......Page 329
Radical ideals......Page 334
Integral extensions of rings......Page 336
Noether Normalization Lemma......Page 341
Hilbert Nullstellensatz......Page 345
Zariski topology......Page 350
Algebraic varieties......Page 353
Regular functions......Page 358
Suggested further reading......Page 361

Citation preview

Notes on Algebra MTH 619/620

2011.04.28

Contents 1 Monoids and groups

7

2 Subgroups

10

3 Homomorphisms of groups

12

4 The kernel and the image of a homomorphism

15

5 Normal subgroups, cosets and quotient groups

17

6 Isomorphism theorems

22

7 Index of a subgroup and order of an element

26

8 Free groups and presentations of groups

29

9 Direct products, direct sums, and free abelian groups

33

10 Categories and functors

36

11 Adjoint functors

41

12 Categorical products and coproducts

44

13 More on free abelian groups

48

14 Finitely generated abelian groups

53

15 Permutation representations and G-sets

60

16 Some applications of G-sets

65

17 The Sylow theorems

68

18 Application: groups of order pq

72

19 Group extensions and composition series

82

20 Simple groups

86

2

21 Symmetric and alternating groups

89

22 Simplicity of alternating groups

95

23 Solvable groups

99

24 Nilpotent groups

102

25 Rings

108

26 Ring homomorphisms and ideals

112

27 Principal ideal domains and Euclidean rings

116

28 Prime ideals and maximal ideals

120

29 Zorn’s Lemma and maximal ideals

123

30 Unique factorization domains

126

31 Prime elements

130

32 PIDs and UFDs

133

33 Application: sums of two squares

136

34 Application: Fermat’s Last Theorem

140

35 Greatest common divisor

141

36 Rings of fractions

143

37 Factorization in rings of polynomials

149

38 Irreducibility criteria in rings of polynomials

156

39 Modules

162

40 Basic operations on modules

165

41 Free modules and vector spaces

167

3

42 Invariant basis number

171

43 Projective modules

176

44 Projective modules over PIDs

180

45 The Grothendieck group

182

46 Injective modules

190

47 Exact functors

197

48 Tensor products

202

49 Tensor products of homomorphisms

207

50 Tensor products and adjoint functors

209

51 Tensor products for non-commutative rings

211

52 Restriction and extension of scalars

214

53 Application: rings with IBN

217

54 Algebras

218

55 Tensor product of algebras

219

56 Fields

220

57 Field extensions

222

58 Prime subfield and field characteristic

223

59 Algebraic and transcendental elements

225

60 Algebraic extensions

231

61 Separable elements

234

62 Derivatives and separable elements

236

4

63 Separable extensions

241

64 Simple extensions

245

65 Simple extensions and intermediate fields

250

66 Construction of extensions

253

67 Algebraically closed fields

256

68 Roots of unity

262

69 Finite fields

265

70 Galois theory - motivation

268

71 Normal extensions

271

72 Galois extensions

277

73 Application: rational symmetric functions

282

74 The fundamental theorem of Galois theory

284

75 The fundamental theorem of algebra

290

76 Infinite Galois extensions

292

77 Abelian and cyclic extensions

295

78 Radical extensions

301

79 Solvable extensions

303

80 Solvability of polynomials by radicals

307

81 Straightedge and compass constructions

312

82 Construction of regular polygons

318

83 Transcendental extensions

321

5

84 Algebraic sets

327

85 Hilbert basis theorem

329

86 Radical ideals

334

87 Integral extensions of rings

336

88 Noether Normalization Lemma

341

89 Hilbert Nullstellensatz

345

90 Zariski topology

350

91 Algebraic varieties

353

92 Regular functions

358

93 Suggested further reading

361

6

1

Monoids and groups

1.1 Definition. A monoid is a set M together with a map M × M → M,

(x, y) 7→ x · y

such that (i) (x · y) · z = x · (y · z) ∀x, y, z ∈ M (associativity); (ii) ∃e ∈ M such that

x·e=e·x=x

for all x ∈ M (e = the identity element of M ). 1.2 Examples. 1) Z with addition of integers (e = 0) 2) Z with multiplication of integers (e = 1) 3) Mn (R) = {the set of all n × n matrices with coefficients in R} with matrix multiplication (e = I = the identity matrix) 4) U = any set P (U ) := {the set of all subsets of U }

P (U ) is a monoid with A · B := A ∪ B and e = ∅. 5) Let U = any set

F (U ) := {the set of all functions f : U → U } F (U ) is a monoid with multiplication given by composition of functions (e = idU = the identity function). 1.3 Definition. A monoid is commutative if x · y = y · x for all x, y ∈ M . 1.4 Example. Monoids 1), 2), 4) in 1.2 are commutative; 3), 5) are not. 7

1.5 Note. Associativity implies that for x1 , . . . , xk ∈ M the expression x1 · x2 · · · · · xk has the same value regardless how we place parentheses within it; e.g.: (x1 · x2 ) · (x3 · x4 ) = ((x1 · x2 ) · x3 ) · x4 = x1 · ((x2 · x3 ) · x4 ) etc. 1.6 Note. A monoid has only one identity element: if e, e0 ∈ M are identity elements then e = e · e0 = e0 1.7 Definition. A group is a monoid G such that for any x ∈ G there is y ∈ G satistying x · y = e = y · x. The element y is called the inverse of x and it is denoted by x−1 (or by −x in the additive notation). A group G is commutative (or abelian) if x · y = y · x for all x, y ∈ G. 1.8 Examples. 1) Z, Q, R, C with addition 2) Q∗ = Q − {0}, R∗ = R − {0}, C∗ = C − {0} with multiplication 3) GLn (R) = {A ∈ Mn (R) | det(A) 6= 0} with matrix multiplication (the n × n general linear group) 4) SLn (R) = {A ∈ Mn (R) | det(A) = 1} with matrix multiplication (the n × n special linear group) 5) Let U = be any set and let Perm(U ) := {f : U → U | f is a bijection} Perm(U ) with composition of functions is a group (the group of permutations of U ) Note. If U = {1, 2, . . . , n} then Perm(U ) is called the symmetric group on n letters and it is denoted by Sn . 8

7) Let T = an equilateral triangle GT = {I, R1 , R2 , S1 , S2 , S3 }

I R1

S1

S2

R2

S3

GT = the group of symmetries of T . 1.9 Proposition (Cancellation Law). If G is a group, x, y, x ∈ G and xy = xz then y = z. Proof. xy = xz x xy = x−1 xz y=z −1

1.10 Note. The cancellation law does not hold for monoids. E.g. in M2 (R) take       1 0 0 0 0 0 A= , B= , C= 0 0 0 1 0 0 Then AB = AC but A 6= C. 9

2

Subgroups

2.1 Definition. If G is a group then a subgroup of G is a subset H ⊆ G such that (i) e ∈ H;

(ii) if x, y ∈ H then xy ∈ H;

(iii) if x ∈ H then x−1 ∈ H.

2.2 Note. A subgroup of a group is by itself a group. 2.3 Examples. 1) If G is a group then G, {e} are subgroups of G 2) Z is a subgroup of Q, which is a subgroup of R, which is a subgroup of C. 3) SLn (R) is a subgroup of GLn (R) 4) H = {I, R1 , R2 } is a subgroup of GT 2.4 Note. If {Hi }i∈I is a family of subgroups of G then subgroup of G.

T

i∈I

Hi is also a

2.5 Definition. If G is a group and S is a subset of G then denote hSi = the smallest subgroup of G that contains S hSi is the subgroup of G generated by the set S. 2.6 Proposition. If S ⊆ G then hSi consists of all elements of the form ±1 ±1 x±1 1 x2 · · · · · xk

where x1 , . . . , xk ∈ S. 10

Proof. Exercise. 2.7 Definition. A set S ⊆ G generates G if hSi = G. 2.8 Example. S = {S1 , S2 } generates GT . 2.9 Definition. A group G is finitely generated if it is generated by some finite subset S ⊆ G. 2.10 Note. • Every finite group is finitely generated.

• Some infinite groups are finitely generated; e.g. Z = h1i. 2.11 Definition. A group G is cyclic if G = hai for some a ∈ G 2.12 Note. If G is cyclic, G = hai then every element g ∈ G is of the form g = an for some n ∈ Z (where a−n := (a−1 )n , a0 = e). 2.13 Examples. 1) Z = h1i is cyclic. 2) H := {I, R1 , R2 } ⊆ GT is cyclic: H = hR1 i and H = hR2 i

11

3

Homomorphisms of groups

3.1 Definition. Let G, H be groups. A function f : G → H is a group homomorphism if for any a, b ∈ G we have f (ab) = f (a)f (b) 3.2 Proposition. If f : G → H is a homomorphism of groups and eG , eH denote identity elements in, respectively, G and H then (i) f (eG ) = eH (ii) f (a−1 ) = f (a)−1 for any a ∈ G. Proof. (i) We have f (eG ) = f (eG · eG ) = f (eG ) · f (eG ) Multiplying this equation by f (eG )−1 we obtain eH = f (eG ). (ii) Since by (i) we have f (eG ) = eH therefore f (a) · f (a−1 ) = f (a · a−1 ) = f (eG ) = eH It is now enough to multiply this equation from the left by f (a)−1 . 3.3 Definition. A homomorphism f : G → H is an isomorphism if there is a homomorphism g : H → G such that g ◦ f = idG and f ◦ g = idH . 3.4 Proposition. A map f : G → H is an isomorphism of groups iff f is a homomorphism and a bijection. Proof. Exercise. 3.5 Definition. If there exists an isomorphism f : G → H then we say that the groups G and H are isomorphic and we write G ∼ = H. 12

3.6 Definition. A homomorphism f : G → G is called an endomorphism of G. An isomorphism f : G → G is called an automorphism of G. 3.7 Examples. 1) idG : G → G is an automorphism of G. 2) f : G → G, f (g) = e ∀g∈G is an endomorphism of G. 3) If f : G → H, g : H → K are homomorphisms then so is g ◦ f : G → K. 4) For g ∈ G define

cg : G → G,

cg (a) := gag −1

Check: cg is an automorphism of G. Automorphisms of this form are called inner automorphisms of G. Note. If G is an abelian group then cg = idG for all g ∈ G. 5) Recall: GLn (R) = {A ∈ Mn | det(A) 6= 0}, R∗ = R − {0} We have the determinant function:

det : GLn (R) → R∗ Since det(AB) = det(A) · det(B) this function is a homomorphism. 6) Let G ⊆ GL2 (R)

  1 r G := 0 1

 r∈R

G is a subgroup of GL2 (R):       1 r 1 s 1 r+s · = 0 1 0 1 0 1  −1   1 r 1 −r = 0 1 0 1 We have homomorphisms: f : R → G and g : R → G 13

where 

   1 r 1 r f (r) = , g =r 0 1 0 1 Since g ◦ f = idG , f ◦ g = R we get G ∼ = R. 3.8 Definition. If G is a group then |G| := the number of elements of G |G| is called the order of G. 3.9 Example. |GT | = 6, |Z| = ∞. 3.10 Note. If G ∼ = H then |G| = |H|.

14

4

The kernel and the image of a homomorphism

4.1 Proposition. Let f : G → H be a homomorphism. 1) If G0 is a subgroup of G then f (G0 ) is a subgroup of H. 2) If H 0 is a subgroup of H then f −1 (H 0 ) is a subgroup of G. Proof. Exercise. 4.2 Definition. If f : G → H is a homomorphism then • the image of f is the subgroup Im(f ) := f (G) ⊆ H • the kernel of f is the subgroup Ker(f ) := f −1 (eH ) ⊆ G 4.3 Note. f : G → H is an epimorphism (onto) iff Im(f ) = H. 4.4 Proposition. f : G → H is a monomorphism (1-1) iff Ker(f ) = {eG } Proof. (⇒) We have f (eG ) = eH . Thus if f is 1-1 then f (g) = eH only if g = eH . In other words we have then Ker(f ) = {eH }. (⇐) Assume that Ker(f ) = {eG } and let f (a) = f (b) for some a, b ∈ G. We have: f (ab−1 ) = f (a)f (b)−1 = eH so ab−1 ∈ Ker(f ). Therefore ab−1 = eG , and so a = b.

15

4.5 Problem. Let G be a group, and let H be a subgroup of G. Is there a homomorphism f: G→K such that Ker(f ) = H? 4.6 Note. The dual problem is trivial: if H is a subgroup of G then we have the inclusion homomorphism i : H ,→ G and Im(i) = H. It follows that any subgroup of G is an image of some homomorphism. 4.7 Definition. A subgroup H ⊆ G is a normal subgroup if for every h ∈ H we have aha−1 ∈ H ∀a ∈ G 4.8 Notation. If H is a normal subgroup of G then we write H C G 4.9 Proposition. If f : G → H is a homomorphism then Ker(f ) is a normal subgroup of G. Proof. If a ∈ G, h ∈ Ker(f ) then f (aha−1 ) = f (a)f (h)f (a)−1 = f (a) · e · f (a)−1 = f (a)f (a)−1 = e so aha−1 ∈ Ker(f ). 4.10 Examples. 1) Any subgroup of an abelian group is normal. 2) H := {I, R1 , R2 } is a normal subgroup of GT (check!). 3) K := {I, S1 } is not a normal subgroup of GT (check!). As a consequence K cannot be the kernel of any homomorphism GT → G.

16

5

Normal subgroups, cosets and quotient groups

Recall. If f : G → K is a homomorphism then Ker(f ) is a normal subgroup of G. Next goal: If H is a normal subgroup of G then there is a homomorphism f : G → K such that H = Ker(f ). 5.1 Definition. If H is a subgroup of G then a left coset of H in G is a subset of G of the form aH := {ah | h ∈ H} for some a ∈ G. A right coset of H in G is a subset of G of the form Ha := {ha | h ∈ H} for some a ∈ G. 5.2 Example. Recall: GT = {I, R1 , R2 , S1 , S2 , S3 }. Take H := {I, S1 }. We have: IH = {I · I, I · S1 } = {I, S1 } = H S1 H = {S1 · I, S1 · S1 } = {S1 , I} S2 H = {S2 · I, S2 · S1 } = {S2 , R2 } S3 H = {S3 · I, S3 · S1 } = {S3 , R1 } R1 H = {R1 · I, R1 · S1 } = {R1 , S3 } R2 H = {R2 · I, R2 · S1 } = {R2 , S2 } Note: IH = S1 H, S2 H = R2 H, S3 H = R1 H 5.3 Lemma. If G is a group and H is a subgroup of G then aH = bH

iff a−1 b ∈ H 17

Proof. (⇒) Let aH = bH. Since e ∈ H thus b = be ∈ bH = aH so b = ah for some h ∈ H. Therefore a−1 b = h ∈ H. (⇐) Assume that a−1 b ∈ H. For any h ∈ H we have ah = a(a−1 b)(a−1 b)−1 h = b((a−1 b)−1 )h ∈ bH This gives: aH ⊆ bH. Also for any h ∈ H we have: bh = (aa−1 )bh = a(a−1 b)h ∈ aH so bH ⊆ aH. Therefore aH = bH. 5.4 Proposition. If H is a subgroup of G then for any a, b ∈ G either aH = bH

or aH ∩ bH = ∅

Proof. Let aH ∩ bH 6= ∅ and let c ∈ aH ∩ bH. Then ah1 = c = bh2 for some h1 , h2 ∈ H. This gives a−1 b = h1 h−1 2 ∈ H and so aH = bH by (5.3). 5.5 Corollary. If H is a subgroup of G then every element of G belongs to one and only left coset of H. 5.6 Note. In general aH 6= Ha. For example, If H ⊆ GT , H = {I, S1 } then S2 H = {S2 , R2 }, 18

HS2 = {S2 , R1 }

5.7 Proposition. A subgroup H of G is normal iff aH = Ha ∀a ∈ G Proof. Exercise. 5.8 Notation. If H is a subgroup of G then G/H := the set of all left cosets of H in G 5.9. Multiplication of cosets. Let H ⊆ G, aH, bH ∈ G/H. Define aH · bH := (ab)H 5.10 Note. In general this is not well defined, i.e. we may have aH = a0 H, bH = b0 H but (ab)H 6= (a0 b0 )H. For example, take H = {I, S1 } ⊆ GT . Recall: S2 H = R2 H = {S2 , R2 }, S3 H = R1 H = {S3 , R1 } However: (S2 S3 )H = R1 H = {R1 , S2 } (R2 R1 )H = IH = {I, S1 } 5.11 Proposition. If H is a normal subgroup of G then the multiplication of cosets given in (5.9) is well defined. Proof. If H C G then by (5.7) we have aH = Ha ∀a∈G . Let aH = a0 H, bH = b0 H. Then (ab)H = a(bH) = a(b0 H) = a(Hb0 ) = (aH)b0 = (a0 H)b0 = a0 (b0 H) = (a0 b0 )H

19

5.12 Corollary/Definition. If H C G then G/H is a group with multiplication defined by (5.9). The identity elements in G/H is the coset eH = H ∈ G/H. The inverse of a coset aH is the coset a−1 H. The group G/H is called the quotient group (or the factor group) of G by H. 5.13 Example. Take Z, the additive group of integers. Since Z is abelian every its subgroup is normal. For n ∈ Z, n ≥ 2 define nZ = {na | a ∈ Z} e.g. 2Z = {· · · − 4, −2, 0, 2, 4, . . . }, 5Z = {. . . , −10, −5, 0, 5, 10, . . . } Note: nZ is a subgroup of Z. Cosets of nZ in Z:

k + nZ = {k + na | a ∈ Z}

e.g. 1 + 5Z = {· · · − 9, −4, 1, 6, 11, . . . }, 3 + 5Z = {. . . , −7, −2, 3, 8, 13, . . . } Note: k + nZ = l + nZ iff (k − l) ∈ nZ i.e. iff k = l + na for some a ∈ Z. E.g.: 1 + 5Z = 6 + 5Z = 11 + 5Z = −4 + 5Z Recall: if n, k ∈ Z then there is a unique number l ∈ {0, 1, . . . , n − 1} such that k = l + na for some a ∈ Z. Thus every coset of nZ can be uniquely written as l + nZ where l ∈ {0, 1, . . . , n − 1}. Denote ¯l := l + nZ. Then Z/nZ = {¯0, ¯1, . . . , n − 1} The addition table in Z/5Z:

20

+ ¯0 ¯1 ¯2 ¯3 ¯4

¯0 ¯0 ¯1 ¯2 ¯3 ¯4

¯1 ¯1 ¯2 ¯3 ¯4 ¯0

¯2 ¯2 ¯3 ¯4 ¯0 ¯1

¯3 ¯3 ¯4 ¯0 ¯1 ¯2

¯4 ¯4 ¯0 ¯1 ¯2 ¯3

Recall: A group G is cyclic if it is generated by a single element: G = hai for some a ∈ G. Note: For every n the group Z/nZ is cyclic: Z/nZ = h¯1i. 5.14 Note. If H C G then we have a homomorphism π : G → G/H,

π(a) := aH

This is the canonical epimorphism of G onto G/H. We have: Ker(π) = {a ∈ G = {a ∈ G = {a ∈ G = {a ∈ G =H

| | | |

π(a) = eH} aH = eH} e−1 a ∈ H} a ∈ H}

5.15 Corollary. A subgroup H ⊆ G is the kernel of some homomorphism f : G → K iff H is a normal subgroup

21

6

Isomorphism theorems

6.1 Theorem. If f : G → H is a homomorphism then there is a unique homomorphism f¯: G/ Ker(f ) → H such that the following diagram commutes: f

G π

/

;H

f¯



G/ Ker(f ) Moreover, f¯ is a monomorphism and Im(f¯) = Im(f ). Proof. Denote: K := Ker(f ). Define f¯: G/K → H,

f¯(aK) := f (a)

We have: 1) f¯ is well defined: If aK = bK then a−1 b ∈ K, so f (a−1 b) = e. Thus f (b) = f (aa−1 b) = f (a)f (a−1 b) = f (a) 2) f¯ is a homomorphism (check). 3) f¯ is a unique homomorphism satisfying f¯ ◦ π = f Indeed, if g : G/K → H is some other homomorphism and g ◦ π = f then f (a) = g ◦ π(a) = g(aK) and so g(aK) = f¯(aK) for all aK ∈ G/K. 22

4) f¯ is 1-1: We need: Ker(f¯) = {eK}. We have: if f¯(aK) = e then f (a) = e, so a ∈ K and so aK = eK. 5) Im(f ) = Im(f¯) (obvious).

6.2 First Isomorphism Theorem. If f : G → H is an epimorphism then G/ Ker(f ) ∼ =H Proof. Take the map f¯: G/ Ker(f ) → H. Then Im(f¯) = Im(f ) = H, so f¯ is an epimorphism. Also, f¯ is 1-1. Therefore f¯ is a bijective homomorphism and thus it is an isomorphism. 6.3 Example. Recall: GLn (R) = {A ∈ Mn (R) | det(A) 6= 0} SLn (R) = {A ∈ Mn (R) | det(A) = 1} SLn (R) is a normal subgroup of GLn (R). We have the homomorphism det : GLn (R) → R∗ Since this is an epimorphism and Ker(det) = SLn (R) we get GLn (R)/SLn (R) ∼ = R∗ 6.4 Theorem. If G is a cyclic group then G ∼ = {e} or G ∼ = Z or G ∼ = Z/nZ for some n ≥ 2.

23

Proof. Let G = hai for some a ∈ G. Define f : Z → G,

f (n) := an

Notice that 1) f is a homomorphism 2) f is onto. Thus by the First Isomorphism Theorem G ∼ = Z/ Ker(f ). Check: all subgroups H ⊆ Z are of the form H = nZ for some n ≥ 0. It follows that G ∼ = Z/nZ for some n ≥ 0. Also: • if n = 0 then nZ = 0Z = {0} and G ∼ = Z/{0} ∼ =Z • if n = 1 then nZ = 1Z = Z and G ∼ = Z/Z ∼ = {e} • if n ≥ 2 then G ∼ = Z/nZ.

6.5 Notation. If H, K are subgroups of G then HK := {hk ∈ G | h ∈ H, k ∈ K} 6.6 Lemma. If H, K are subgroups of G then HK is a subgroup of G iff HK = KH Proof. Exercise. 6.7 Second Isomorphism Theorem. If H, K are subgroups of G and H C G then KH is a subgroup of G, (H ∩ K) C K and K/(H ∩ K) ∼ = KH/H 24

Proof. Exercise. 6.8 Third Isomorphism Theorem. Let K ⊆ H ⊆ G. If K, H are normal subgroups of G then K C H, H/K C G/K and (G/K)/(H/K) ∼ = G/H Proof. Exercise.

25

7

Index of a subgroup and order of an element

7.1 Definition. Let H be a subgroup of G. Then [G : H] := the number of distinct left cosets of H in G This number is called the index of H in G.

7.2 Note. 1) The number of left cosets of H in G is the same as the number of right cosets (check!), so also [G : H] = the number of distinct right cosets of H in G 2) If H is a normal subgroup of G then [G : H] = |G/H|.

7.3 Lemma. If H is a subgroup of G then every left (and right) coset of H in G has the same number of elements as H. Proof. If a ∈ G then the map of sets f : H → aH,

f (h) = ah

is a bijection (check!). 7.4 Theorem (Lagrange). If H is a subgroup of G then |G| = [G : H]|H| Proof. Recall: 1) G is a disjoint union of left cosets of H (5.5) 2) each left coset has as many elements as H (7.3) 26

This gives: |G| = (number of left cosets of H) · (number of elements of H) = [G : H] · |H|

7.5 Definition. If a ∈ G then the order |a| of a is the order of the subgroup hai ⊆ G generated by a. 7.6 Proposition. |a| = n if n is the smallest positive integer such that an = e, and |a| = ∞ if such n does not exist. Proof. Take the homomorphism f : Z → hai,

f (k) = ak

Note that f is onto. If an 6= e for all n > 0 then Ker(f ) = {0}. Then f is an isomorphism and so |hai| = |Z| = ∞. If an = e for some n > 0 and n is the smallest positive number with this property then Ker(f ) = nZ (check!). Therefore hai ∼ = Z/nZ and so |hai| = |Z/nZ| = n.

7.7 Proposition. If G is a finite group and a ∈ G then |a| divides |G|. Proof. Follows from Lagrange’s theorem (7.4). 7.8 Note. If a number n divides |G| then G need not contain an element of order n. E.g. |GT | = 6 but GT does not have an element of order 6: |S1 | = |S2 | = |S3 | = 2,

|R1 | = |R2 | = 3,

|I| = 1

We will see later that if p is a prime number that divides |G| then G contains an element of order p. 27

7.9 Definition. An element a ∈ G is a torsion element if |a| < ∞. A group G is a torsion group if all its elements are torsion elements. A group is torsion free if it has not torsion elements. 7.10 Examples. 1) Every finite group is a torsion group. 2) Q/Z is also a torsion group. 3) Z, Q, R, C are torsion free. 4) Q∗ is neither torsion nor torsion free: 2 ∈ Q∗ has infinite order, −1 ∈ Q∗ has order 2.

28

8

Free groups and presentations of groups

8.1. Let S be a set. A word in S is a finite sequence of the form w = xλ1 1 xλ2 2 · · · · · xλk k where xi ∈ S and λi = ±1 for i = 1, 2, . . . , k. Also, take e := “the empty word” (i.e. the word corresponding to the sequence of length 0). We identify two words if one can be obtained from the other by a series of “cancellations” and “insertions” of subwords of the form xx−1 and x−1 x, e.g.: −1 −1 −1 x1 x2 x3 x−1 3 ∼ x1 x2 ∼ x1 x4 x4 x2 ∼ x1 x4 x1 x1 x4 x2

x1 x−1 1 ∼ e Let F (S) be the set of equivalence classes of words under this equivalence relation. Note. A word is reduced if it does not contain any subwords of the form xx−1 or x−1 x. Every equivalence class in F (S) is represented by a unique reduced word. Define multiplication in F (S) by concatenation of words, e.g.: −1 −1 −1 (x1 x2 x−1 1 ) · (x2 x3 ) = x1 x2 x1 x2 x3

F(S) with this multiplication becomes a group: • the identity element in F (S): e −1 −1 −1 • inverses in F (S): (x1 x2 x−1 = x−1 3 x2 ) 2 x3 x2 x1

29

8.2 Definition. F (S) is called the free group generated by the set S. In general, a group G is free if G ∼ = F (S) for some set S. 8.3 Note. • If S = ∅ then F (S) = {e} is the trivial group. • If S consists of a single element, S = {x} then F (S) is an infinite cyclic group, so F (S) ∼ = Z. 8.4 Note. We have a map of sets i : S → F (S),

i(x) = x

8.5 Theorem (The universal property of free groups). Let S be a set and G be a group. For any map of sets f : S → G there exists a unique homomorphism f¯: F (S) → G such that the following diagram commutes: f /G S = i

f¯



F (S) Proof. f¯ is defined by f¯(xλ1 1 xλ2 2 · · · · · xλk k ) := f (x1 )λ1 f (x2 )λ2 · · · · · f (xk )λk

8.6 Corollary. Every group is the homeomorphic image of a free group. Proof. Let G be a group. Take the set S := {xg | g ∈ G} 30

We have a map of sets f : S → G,

f (xg ) := g

This gives a homomorphism f¯: F (S) → G. Since f is onto thus also f¯ is onto, i.e. G = f¯(F (S)). 8.7 Note. By Corollary 8.6 and the First Isomorphism Theorem (6.2) we have G∼ = F (S)/ Ker(f¯) One can show that any subgroup of a free group is free. In particular Ker(f¯) is free. This shows that any group is isomorphic to a quotient of two free groups.

8.8 Definition. Let S be a set and let R be a subset of F (S). Then hS | Ri := F (S)/H where H is the smallest normal subgroup of F (S) such that R ⊆ H. We say then that • elements of S are generators of hS | Ri • elements of R are relations (or relators) in hS | Ri 8.9 Definition. If G is a group and G ∼ = hS | Ri for some set S and some subset R ⊆ F (S) then we say that hS | Ri is a presentation of G. We say that a group G is finitely presentable if it has a presentation such that both S and R are finite sets. 8.10 Examples. 1) F (S) ∼ = hS | ∅i

31

2) Z/nZ ∼ = hx | xn i

Note: in particular Z/6Z ∼ = hx | x6 i. Here is a different presentation of Z/6Z: Z/6Z ∼ = hx, y | x2 , y 3 , xyxy 2 i

3) GT ∼ = hx, y | x2 , y 3 , xyxyi (isomorphsm: x 7→ S1 , y 7→ R1 ) 4) Recall: Sn = the symmetric group on n letters (1.8) Sn ∼ = hx1 , . . . xn−1 | x2i , (xi xi+1 )3 , (xi xj )2 for |i − j| > 1i (isomorphism: xi 7→ σi where σi : {1, . . . , n} → {1, . . . , n}, σi (i) = i + 1, σi (i + 1) = i and σi (j) = j for j 6= i, i + 1).

32

9

Direct products, direct sums, and free abelian groups

9.1 Definition. A direct product of Q a family of groups {Gi }i∈I is a group Q Gi is the cartesian product of the i∈I Gi defined as follows. As a set i∈I Q groups Gi . Given elements (ai )i∈I , (bi )i∈I ∈ i∈I Gi we set (ai )i∈I · (bi )i∈I := (ai bi )i∈I

9.2 Definition. Q A weak direct product of a family of groups {Gi }i∈I is the subgroup of i∈I Gi given by Yw Gi := {(ai )i∈I | ai 6= ei ∈ Gi for finitely many i only} i∈I

If all groups Gi are abelian then direct sum of {Gi }i∈I .

Qw

9.3 Note. If I is a finite set then

i∈I Gi

Q

i∈I

is denoted

Gi =

L

i∈I

Gi and it is called the

Qw

i∈I Gi .

9.4 Example. Z/2Z × Z/2Z = Z/2Z ⊕ Z/2Z = {(0, 0), (0, 1), (1, 0), (1, 1)} Note. Z/2Z ⊕ Z/2Z is a the smallest non-cyclic group. It is called the Klein four group. 9.5 Example. Z/2Z ⊕ Z/3Z = {(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)} Note. Z/2Z ⊕ Z/3Z is a cyclic group ((1, 1) is a generator), thus Z/2Z ⊕ Z/3Z ∼ = Z/6Z 33

9.6. Let S be a set. Denote by Fab (S) the set of all expressions of the form X kx x x∈S

where kx ∈ Z and kx 6= 0 for finitely many x ∈ X only. Fab (S) is an abelian group with addition defined by X X X kx x + lx x := (kx + lx )x x∈S

x∈S

x∈S

9.7 Definition. The group Fab (S) is called the free abelian group generated by the set S. In general a group G is free abelian if G ∼ = Fab (S) for some set S.

9.8 Proposition. If S is a set then Fab (S) ∼ = Proof. The isomorphism is given by M Z, f : Fab (S) →

M

Z

x∈S

X kx x) = (kx )x∈S f(

x∈S

x∈S

9.9 Note. We have a map of sets i : S → Fab (S),

34

i(x) = 1 · x

9.10 Theorem (The universal property of free abelian groups). Let S be a set and G be an abelian group. For any map of sets f : S → G there exists a unique homomorphism f¯: F (S) → G such that the following diagram commutes: f /G S = i

f¯



Fab (S) Proof. Define f¯ by ! f¯

X

kx x

:=

x∈S

X

kx f (x)

x∈S

Note: this is well defined since kx = 0 for almost all x ∈ S.

35

10

Categories and functors

10.1 Definition. A category C consists of 1) a collection of objects Ob(C) 2) for any a, b ∈ Ob(C) a set HomC (a, b) of morphisms from a to b 3) for any a, b, c ∈ Ob(C) a function (“composition law”) HomC (a, b) × HomC (b, c) → HomC (a, c) (f

,

g)

7→

g◦f

such that the following conditions are satisfied: • Associativity.

f ◦ (g ◦ h) = (f ◦ g) ◦ h

for any morphisms f, g, h for which these compositions are defined. • Identity. For any c ∈ Ob(C) there is a morphism idc ∈ HomC (c, c) such that f ◦ idc = f, idc ◦ g = g for any f ∈ HomC (c, d), g ∈ HomC (b, c). 10.2 Examples. 1) Set = the category of all sets. • Ob(Set) = the collection of all sets

• HomSet (A, B) = { all maps of sets f : A → B } 2) Gr = the category of all groups • Ob(Gr) = the collection of all groups

• HomGr (G, H) = { all homomorphisms f : G → H } 36

3) Ab = the category of all abelian groups • Ob(Ab) = the collection of all abelian groups

• HomAb (G, H) = { all homomorphisms f : G → H } 4) Top = the category of all topological spaces • Ob(Top) = the collection of all topological spaces

• HomTop (X, Y ) = { all continuous maps f : X → Y }

5) Let G be a group. Define a category CG as follows: • Ob(CG ) = {∗}

• HomCG (∗, ∗) = { elements of G }

• composition of morphisms = multiplication in G

6) A very small category C: c

f

/

d

• Ob(C) = {c, d}

• HomC (c, d) = {f } HomC (d, c) = ∅ HomC (c, c) = idc HomC (d, d) = idd

10.3 Definition. A morphism f : c → d in a category C is an isomorphism if there exists a morphism g : d → c such that gf = idc and f g = idd . If for some c, d ∈ C there exist an isomorphism f : c → d then we say that the objects c and d are isomorphic and we write c ∼ = d.

37

10.4 Note. For an object c ∈ C define Aut(c) := { all isomorphisms f : c → c } Aut(c) with composition of morphisms is a group. 10.5 Definition. Let C, D be categories. A (covariant) functor F : C → D consists of 1) an assignment Ob(C) → Ob(D),

c 7→ F (c)

2) for every c, c0 ∈ C a function HomC (c, c0 ) → HomD (F (c), F (c0 )),

f 7→ F (f )

such that F (gf ) = F (g)F (f ) and F (idc ) = idF (c) .

10.6 Note. If F : C → D is a functor and f : c → c0 is an isomorphism in C then F (f ) : F (c) → F (c0 ) is an isomorphism in D. In particular if c ∼ = c0 in C then F (c) ∼ = F (c0 ) in D.

10.7 Examples. 1) U : Gr → Set

If G ∈ Gr then U (G) = { the set of elements of G }

If f : G → H is a homomorphism then U (f ) : U (G) → U (H) is the map of sets underlying this homomorphism.

2) U : Ab → Set,

defined the same way as in 1).

Note. The functors U in 1), 2) are called forgetful functors. 38

3) Let G be a group. The commutator of a, b ∈ G is the element [a, b] := aba−1 b−1 Note: [a, b] = e iff ab = ba. The commutator subgroup of G is the subgroup [G, G] ⊆ G generated by the set S = {[a, b] | a, b ∈ G}. Note.

(a) [G, G] = {e} iff G is an abelian group.

(b) [G, G] is a normal subgroup of G (check!). (c) G/[G, G] is an abelian group (check!). (d) If f : G → H is a homomorphism then f ([G, G]) ⊆ [H, H].

(e) If f : G → H is a homomorphism then f induces a homomorphism fab : G/[G, G] → H/[H, H] given by fab (a[G, G]) = f (a)[H, H].

The abelianization functor Ab : Gr → Ab is given by Ab(G) := G/[G, G],

Ab(f ) := fab

4) Recall: if S is a set then F (S) is the free group generated by S. A map of sets f : S → T defines a homomorphism f˜: F (S) → F (T ) given by f˜(xλ1 1 xλ2 2 · · · · · xλk k ) = f (x1 )λ1 f (x2 )λ2 · · · · · f (xk )λk . Check: the assignment S 7→ F (S),

(f : S → T ) 7→ (f˜: F (S) → F (T ))

Defines a functor F : Set → Gr. This is the free group functor. 39

5) Similarly we have the free abelian group functor Fab : Set → Ab where • Fab (S) = the free abelian group generated by the set S • if f : S → T then Fab (f ) : Fab (S) → Fab (T ) is given by ! X X Fab (f ) kx x = kx f (x) x∈S

40

x∈S

11

Adjoint functors

11.1 Definition. Given two functors L: C → D

R: D → C

and

we say that L is the left adjoint functor of R and that R is the right adjoint functor of L if for any object c ∈ C we have a morphism ηc : c → RL(c) such that: 1) for any morphism f : c → c0 in C the following diagram commutes: f

c

/ c0 ηc0

ηc



RL(c)

RL(f )

/



RL(c0 )

2) for any c ∈ C and d ∈ D the map of sets HomD (L(c), d) −→ HomC (c, R(d)) f

is a bijection.

ηc

R(f )

(L(c) → d) 7−→ (c → RL(c) → R(d))

In such situation we say that (L, R) is an adjoint pair of functors. 11.2 Note. 1) The collection of morphisms {ηc }c∈C is called the unit of adjunction of (L, R). 2) For any adjoint pair (L, R) we also have morphisms {εd : LR(d) → d}d∈D satisfying analogous conditions as {ηc }c∈C . This collection of morphisms is called the counit of the adjunction.

41

11.3 Note. The morphism ηc is universal in the following sense. For any d ∈ D and any morphism f : c → R(d) in C there is a unique morphism f¯: L(c) → d in D such that the following diagram commutes: f

c ηc

/

R(d)
0 such that di |di+1 for i = 1, . . . , r. 53

Proof. By (14.1) we have

G∼ = F/H

for some free abelian group F of finite rank and H ⊆ F . Let rank F = n. By Theorem 14.2 there is a basis {x1 , . . . , xn } of F such that {d1 x1 , . . . , dr xr } is a basis of H for some d1 , . . . , dr > 0, di |di+1 . We have an isomorphism f: F →Z · · ⊕ Z} | ⊕ ·{z n times

where f (x1 ) = (1, 0, . . . , 0), f (x2 ) = (0, 1, . . . , 0), . . . , f (xn ) = (0, . . . , 0, 1) Notice that f (H) = d1 Z ⊕ . . . ⊕ dr Z ⊕ {0} ⊕ . . . ⊕ {0} . This gives

G∼ = (Z ⊕ · · · ⊕ Z)/(d1 Z ⊕ . . . ⊕ dr Z ⊕ {0} ⊕ . . . ⊕ {0}) Using (13.4) we obtain G∼ = (Z/d1 Z) ⊕ . . . ⊕ (Z/dr Z) ⊕ Zk where k = n − r.

Proof of Theorem 14.2. Let {y1 , . . . , yn } be any basis of F and let {h1 , . . . , hm } ⊆ H be any set generating H. We have hi = ai1 y1 + ai2 y2 + ain yn for some aij ∈ Z. Consider the matrix   a11 . . . a1n  ..  A =  ... .  am1 . . . amn Note: columns of A correspond to basis elements of F and rows of A correspond to generators of H. Consider the following operations on matrices: 54

1) interchange of two rows 2) multiplication of a row by (−1) 3) addition of a multiple of one row to another row. These operations are called elementary row operations for matrices of integers. Elementary column operations are defined analogously. Notice that: • application of an elementary row operation to the matrix A corresponds to replacing of the set of generators of H by another set of generators of H; • application of an elementary column operation to the matrix A corresponds to passing to a new basis of F and rewriting the generators of H in terms of this new basis. Key step. Starting with any matrix of integers A and applying a sequence of elementary row and column operations we can obtain a matrix of the form ! D 0 B= 0 0 where 0’s denote zero matrices (of appropriate dimensions) and D is a square diagonal matrix   d1 0 . . . 0  0 d2 . . . 0    D =  .. .. . . ..  . . . . 0 0 . . . dr for some d1 , . . . , dr > 0 such that di |di+1 for all i. Note. The matrix B is called the Smith normal form of the matrix A. Let {x1 , . . . , xn } be the basis of F corresponding to columns of the matrix B. In terms of this basis a set of generators of H is given by {d1 x1 , . . . , dr xr , 0, . . . , 0}. It follows that {d1 x1 , . . . dr xr } is a basis of H.

55

Key step. How to compute the Smith normal form of a matrix A. Step 1. Produce a matrix of the form   a11 0 . . . 0  0    A1 =  .   ..  A01 0 This can be done as follows. (1a) By interchanging rows and columns if necessary we can make sure that a11 6= 0. Also, by multiplying the first row by (−1) we can get a11 > 0.

(1b) By adding multiples of the first row to the other rows (and multiples of the first column to the other columns) we can make all other entries in the first row and the first column positive and smaller than a11 . (1c) If all these other entries of the first row and column are 0 we are done. If some entry is non-zero then by replacing rows (or columns) we can move that entry to the (1, 1)-position. Then we go back to (1b). (1d) After a finite number of iterations we get a matrix of the form of A1 . Step 2. Given a matrix A1 as above make sure that the entry a11 divides all entries of A01 . This can be done as follows. (2a) If all entries of A01 are already divisible by a11 we are done. (2b) If some entry aij is not divisible by a11 add the i-th row to the first row. Then go back to (1b). (2c) After a finite number of iterations we get a matrix of the form of A1 with a11 dividing all entries of A0 . Step 3. We are done with the first row and and the first column. Next we apply the same steps recursively to reduce A01 .

56

14.4 Lemma. Let G be a group and let a1 , . . . , ak ∈ G be elements such that |ai | = ni . Assume that for i = 1, . . . , k we have Y gcd(ni , ni ) = 1 j6=i

Then |a1 · . . .·ak | = n1 · . . . · nk . Proof. Exercise.

14.5 Corollary. If m > 0 is an integer and m = pn1 1 · . . . · pnk k where p1 , . . . , pk are distinct primes then n Z/mZ ∼ = (Z/pn1 1 Z) ⊕ . . . ⊕ (Z/pk k Z)

Proof. It is enough to show that the group (Z/pn1 1 Z) ⊕. . .⊕ (Z/pnk k Z) contains an element of order n. This follows however from Lemma 14.4. 14.6 Theorem. If G is a finitely generated abelian group then G is isomorphic to a finite direct sum groups Z and Z/pn Z where p is a prime. Proof. This follows directly from (14.3) and (14.5).

14.7 Theorem. For any finitely generated abelian group G there are unique (up to order) integers pn1 1 , . . . , pns s > 1 where p1 , . . . , ps are primes, and a unique integer k ≥ 0 such that G ∼ = (Z/pn1 1 Z) ⊕ . . . ⊕ (Z/pnk s Z) ⊕ Zk 14.8 Note. The integers pn1 1 , . . . , pns s are called the elementary divisors of the group G.

57

14.9 Lemma. If p is a prime number and 0 < n1 ≤ · · · ≤ ns

0 < m1 ≤ · · · ≤ mr

be integers such that (Z/pn1 Z) ⊕ · · · ⊕ (Z/pns Z) ∼ = (Z/pm1 Z) ⊕ · · · ⊕ (Z/pmr Z) Then s = r and ni = mi for all i. Proof. Exercise.

Proof of Theorem 14.7. Assume that G ∼ = (Z/pn1 1 Z) ⊕ . . . ⊕ (Z/pns s Z) ⊕ Zk and

G ∼ = (Z/q1m1 Z) ⊕ . . . ⊕ (Z/qrmr Z) ⊕ Zl

(1) (2)

where k, l ≥ 0, pi , qj are primes and and n1 , mj > 0. We want to show that k = l, r = s and (after reordering) pni i = qimi for all i. Define Gt := {a ∈ G | |a| < ∞} This is the torsion subgroup of G. From the isomorphism (1) we obtain: Gt ∼ = (Z/pn1 1 Z) ⊕ . . . ⊕ (Z/pns s Z) while from the isomorphism (2) we get Gt ∼ = (Z/q1m1 Z) ⊕ . . . ⊕ (Z/qrmr Z) If follows that

G/Gt ∼ = Zk

and G/Gt ∼ = Zl

Using Proposition 13.3 we obtain from here that k = l. Next, we want to show that the family of integers {pni i | i = 1, . . . , s} is the m same as the family {qj j | j = 1, . . . , r}. 58

For a prime p define Gp := {a ∈ G | |a| = pl for some l > 0} From the isomorphisms (1) and (2) we get M M m (Z/pini Z) ∼ (Z/qj j Z) = Gp ∼ = pi =p

qj =p

Using Lemma 14.9 we obtain that the family {pni i | pi = p} is the same as n {qj j | qj = p}. Since this holds for all primes p we are done.

59

15

Permutation representations and G-sets

Recall. If C is a category and c ∈ C then Aut(c) = the group of automorphisms of c

15.1 Definition. A representation of a group G in a category C is a homomorphism % : G → Aut(c) Special types of representations: • linear representations = representations in the category of vector spaces • permutation representations = representations in the category of sets.

15.2 Note. Let S be a set and let % : G → Aut(S) be a permutation representation of G. The homomorphism % defines a map G × S → S,

(a, x) 7→ %(a)(x)

Denote a · x := %(a)(x). We have: 1) (ab) · x = a · (b · x) for a, b ∈ G, x ∈ S 2) e · x = x for all x ∈ S

15.3 Definition. An action of a group G on a set S is a map G × S → S,

(a, x) 7→ a · x

satisfying conditions 1) - 2) above. A G-set is a set S equipped with an action of the group G. 60

15.4 Note. A permutation representation G → Aut(S) determines a G-set structure on the set S. Conversely, the structure of a G-set on S determines a permutation representation G → Aut(S). 15.5 Examples. 1) Take S = G. The map G × G → G,

(a, b) 7→ ab

defines a G-set structure on G. We say that the group G acts on itself by left translations. 2) Let H be a subgroup of G. Take S = G/H. The map G × G/H → G/H,

(a, bH) 7→ (ab)H

defines a G-set structure on G/H. 3) Take again S = G. The map G × G → G,

(a, b) 7→ aba−1

defines another G-set structure on G. We say that the group G acts on itself by conjugations. 4) R is a Z/2Z-set with the action Z/2Z × R → R given by 0 · x = x, 1 · x = −x. 15.6 Definition. If S, T are G-sets then a G-equivariant map is a function f: S →T such that f (a · x) = a · f (x) for all a ∈ G, x ∈ S. 61

15.7 Note. For a given group G the collection of G-sets and G-equivariant map forms a category SetG .

15.8 Proposition. A G-equivariant map f: S →T is an isomorphism of G-sets iff f is a bijection.

15.9. Proposition/Definition. Let S be a G-set and let x ∈ G. The set Gx := {a ∈ G | a · x = x} is a subgroup of G. It is called the stabilizer of x (or the isotropy group of x).

15.10 Definition. Let S be a G-set. The action of the group G on S is transitive if for any x, y ∈ S there is a ∈ G such that a · x = y.

15.11 Proposition. For any transitive G-set S we have an isomorphism of Gsets: S∼ = G/Gx where x is any element of S. Proof. We have a map f : G/Gx → S,

f (aGx ) = a · x

Check that • f is well defined

• f is G-equivariant • f is a bijection.

62

It follows that f an isomorphism of G-sets.

15.12 Proposition. If S is a transitive G-set and x, y ∈ S then Gx = a−1 Gy a where y = a · x Proof. For b ∈ Gy we have (a−1 ba) · x = (a−1 b) · y = a−1 · y = x It follows that a−1 Gy a ⊆ Gx . On the other hand, if c ∈ Gx then (aca−1 ) · y = (ac) · x = a · x = y Thus aca−1 ∈ Gy and so c = a−1 (aca−1 )a ∈ a−1 Gy a. This shows that Gx ⊆ a−1 Gy a.

15.13 Definition. If S is a G-set then the orbit of an element x ∈ S is the set Orb(x) := {y ∈ S | y = a · x for some a ∈ G}

15.14 Proposition. Let S be a G-set. 1) If x ∈ S then Orb(x) is the unique subset of S such that • x ∈ Orb(x) • G acts transitively on Orb(x). 2) If x, y ∈ S then either Orb(x) ∩ Orb(y) = ∅ or Orb(x) = Orb(y). 3) If x, y ∈ S then Orb(x) = Orb(y) iff x = a · y for some a ∈ G. Proof. Exercise. 63

15.15 Corollary. Every G-set S is a disjoint union of its orbits. For an element x ∈ S we have an isomorphism of G-sets Orb(x) ∼ = G/Gx Proof. This follows directly from Proposition 15.14 and Proposition 15.11.

15.16 Corollary. Let G be a finite group and let S be a finite G-set . 1) For any x ∈ S we have |Orb(x)| · |Gx | = |G| 2) If Orb(x1 ), . . . Orb(xn ) are all distinct orbits of S then |S| =

n X i=1

|Orb(xi )|

Proof. 1) By Corollary 15.15 we have |Orb(x)| = [G : Gx ], so by Lagrange’s Theorem (7.4) |G| = [G : Gx ] · |Gx | = |Orb(x)| · |Gx | .

64

16

Some applications of G-sets

Recall. If G is a finite group and a ∈ G then |a| divides |G|. 16.1 Theorem (Cauchy). Let G be a finite group. If |G| is divisible by a prime number p then there exists a ∈ G such that |a| = p. Proof. Let S = {(a1 , . . . , ap ) | ai ∈ G, a1 · . . . · ap = e}

Notice that (a1 , . . . , ap ) ∈ S iff a1 , . . . , ap−1 are arbitrary elements of G and ap = (a1 · . . . · ap−1 )−1 . If follows that |S| = |G|p−1 In particular p divides S. Define an action of Z/pZ on S by cyclic permutations: k · (a1 , . . . , ap ) := (ak+1 , . . . , ap , a1 , . . . , ak ) for k ∈ Z/pZ. Notice that: 1) The number of elements of each orbit of this action divides |Z/pZ| = p. Therefore for any (a1 , . . . , ap ) ∈ S we have |Orb((a1 , . . . , ap ))| = p

or

|Orb((a1 , . . . , ap ))| = 1

2) |Orb((a1 , . . . , ap ))| = 1 iff a1 = · · · = ap . We have then ap1 = a1 · . . . · ap = e

3) |Orb((e, . . . , e))| = 1 Assume that Orb((e, . . . , e)) is the only orbit consisting of only one element. Then every other orbit of S has p elements. From part 2) of Corollary 15.16 we obtain then |S| = kp + 1 for some k. This is however impossible since p divides |S|. Therefore there is some element (a1 , . . . , ap ) ∈ S such that (a1 , . . . , ap ) 6= (e, . . . , e) and |Orb((a1 , . . . , ap ))| = 1. It follows that a1 6= e and ap1 = e. Thus |a1 | = p. 65

16.2 Definition. Let p be a prime number. A group G is a p-group if order of every element of G is a power of p. 16.3 Proposition. If G is a finite group then G is a p-group iff |G| = pn for some n. Proof. If |G| = pn and a ∈ G then by (7.7) |a| divides pn , so |a| = pr for some r ≥ 0. Conversely, assume that |G| = qm for some prime q 6= p. Then by Cauchy’s Theorem (16.1) there is an element a ∈ G such that |a| = q. Therefore G is not a p-group.

Recall. If G is a group then the center of G is the subgroup Z(G) = {a ∈ G | ab = ba for all b ∈ G} Note. In general it may happen that Z(G) = {e} (take e.g. G = GT ). 16.4 Theorem. If G is a finite p-group then Z(G) 6= {e}. Proof. Consider the action of G on itself by conjugations: a · b := aba−1

G × G → G, Let |G| = pn . Notice that:

1) If b ∈ G then |Orb(b)| divides pn , so |Orb(b)| = pr for some r ≥ 0. In particular either p divides |Orb(b)| or |Orb(b)| = 1. 2) |Orb(b)| = 1 iff b ∈ Z(G).

66

As a consequence if Z(G) = {e} then |Orb(e)| = 1 and p divides the number of elements in every other orbit. From part 2) of Corollary 15.16 we obtain then pn = |G| = kp + 1 which is impossible. Therefore Z(G) 6= {e}.

16.5 Note. Not every p-group is abelian. For example take the group of quaternions Q8 (see Hungerford p. 33) and the group of symmetries of a square D4∗ (see Hungerford p. 25). These group are non-abelian, but |Q8 | = |D4∗ | = 8, so they are 2-groups. 16.6 Definition. Let S be a G-set. A element x ∈ S is a fixed point of the action of G on S if a · x = x for all a ∈ G. Note. If S is a G-set then x is a fixed point of the action of G iff |Orb(x)| = 1.

67

17

The Sylow theorems

Recall. If G is a finite group and H ⊆ G is a subgroup then |H| divides |G|. Note. 1) If G is an abelian group and n divides |G| then there is a subgroup H ⊆ G such that |H| = n (homework). 2) This is not true in general when G is a non-abelian group (homework).

17.1 Definition. Let G be a finite group such that |G| = pr m where p is a prime and p - m. We say that a subgroup P ⊆ G is a Sylow p-subgroup of G if |P | = pr .

17.2 First Sylow Theorem. If G is a finite group and p is a prime number dividing |G| then G contains a Sylow p-subgroup. Proof. Assume that |G| = pr m where p - m. Let S be the set of all subsets A ⊆ G such that |A| = pr . Define an action of G on S as follows. If A ∈ S, A = {a1 , . . . , apr } and b ∈ G then b · A := {ba1 , . . . , bapr } Notice that  |S| =

 Y pr pr m pr (m − 1) + j = j pr j=1

so p does not divide |S|. A a consequence there is an orbit Orb(A0 ) in S such that p - |Orb(A0 )|. Take the stabilizer GA0 . We want to show that |GA0 | = pr . We have pr m = |G| = |Orb(A0 )| · |GA0 | 68

so pr divides |GA0 |. On the other hand, notice that for any a ∈ G there is some element b · A0 ∈ Orb(A0 ) such that a ∈ b · A0 . Ineed, if A0 = {a1 , . . . , apr } then (aa−1 1 ) · A0 = {a, . . . }

Since G has pr m elements and each set b · A0 contains pr elements, thus we must have |Orb(A0 )| ≥ m, and consequently |GA0 | ≤ pr . It follows that |GA0 | = pr .

17.3 Second Sylow Theorem. If P is a Sylow p-subgroup of a finite group G then for any p-subgroup H ⊆ G we have for some a ∈ G.

aHa−1 ⊆ P

In particular if P , P 0 are two Sylow p-subgroups of G then P = aP 0 a−1 for some a ∈ G.

Proof. Assume that |G| = pr m where p - m and let |H| = ps . Take G/P , the set of left cosets of P . The group H acts on G/P by left translations: H × G/P → G/P, a · (bP ) := (ab)P

Since |H| = ps , for every bP ∈ G/P we have |Orb(bP )| = pk for some k ≤ s. So, either p divides |Orb(bP )| or |Orb(bP )| = 1. On the other hand |G/P | = [G : P ] = m, and p - m, so there must be an orbit whose number of elements is not divisible by p. It follows that there is b0 P ∈ G/P such that |Orb(b0 P )| = 1 As a consequence ab0 P = b0 P for all a ∈ H, i.e. Hb0 P ⊆ b0 P . Since Hb0 ⊆ Hb0 P we obtain Hb0 ⊆ b0 P or equivalently: b−1 0 Hb0 ⊆ P .

69

17.4 Example. Take the group GT . We have: |GT | = 6 = 3 · 2. • Sylow 3-subgroup of GT : {I, R1 , R2 }

• Sylow 2-subgroups of GT : P = {I, S1 }, P 0 = {I, S2 }, P 00 = {I, S3 }.

We have: P = R1 P 0 R1−1 and P = R2 P 00 R2−1 .

17.5 Third Sylow Theorem. Let G be a finite group such that |G| = pr m where r > 0 and p - m. If s is the number of Sylow p-subgroups of G then 1) s | m

2) s ≡ 1 (mod p)

17.6 Definition. Let G be a group and H be a subgroup of G. The normalizer of H in G is the subgroup NG (H) ⊆ G given by NG (H) = {a ∈ G | aHa−1 = H}

17.7 Note. 1) H C NG (H). In fact, NG (H) is the biggest subgroup of G that contains H a its normal subgroup. 2) H C G iff NG (H) = G. 3) Let S be the set of all subgroups of G. The group G acts on S by conjugations: G × S → S, a · H := aHa−1 If H is a subgroup of G then

Orb(H) = { all subgroups of G conjugate to H } NG (H) = stabilizer of H By (15.16) this gives: (number of subgroups conjugate to H) = |Orb(H)| = [G : NG (H)] 70

Proof of Theorem 17.5. Let P be a Sylow p-subgroup of G. By (17.3) all other Sylow p-subgroups of G are conjugate to P , so we get s = (number of subgroups conjugate to P ) = [G : NG (P )] Since P ⊆ NG by Lagrange’s Theorem (7.4) we have |G| = [G : NG (P )] · |NG (P )| = [G : NG (P )] ·[NG (P ) : P ] · |P | |{z} {z } |{z} | q pr m

q s

q pr

It follows that s | m. Next, let S = {P1 , . . . , Ps } be the set of all Sylow p-subgroups of G. The group P1 acts on S by conjugations: a · Pi := aPi a−1

P1 × S → S, Notice that:

• |Orb(Pi )| divides |P1 | = pr , so for i = 1, . . . , s either p divides |Orb(Pi )| or |Orb(Pi )| = 1. • |Orb(P1 )| = 1 • If i > 1 then |Orb(Pi )| > 1.

Indeed, if |Orb(Pi )| = 1 for some i > 1 then aPi a−1 = Pi for all a ∈ P1 . Check: in this case P1 Pi is a p-subgroup of G. Since P1 6= Pi we would also have |P1 Pi | > pr which is impossible.

As a consequence we get s = |S| = |Orb(P1 )| + | {z }

X

q 1

Therefore s ≡ 1 (mod p). 71

|( all other orbits )| | {z } q multiples of p

18

Application: groups of order pq

Recall. If G is a group, |G| = p where p is a prime then G ∼ = Z/pZ. Goal. Classify all groups of order pq where p, q are prime numbers. 18.1 Proposition. If G is a group of order p2 for some prime p then either G∼ = Z/p2 Z or G ∼ = Z/pZ ⊕ Z/pZ. Proof. It is enough to show that G is abelian since then the statement follows from the classification of finitely generated abelian groups (14.7). Since G is a p-group by Theorem 16.4 we have Z(G) 6= {e}. If Z(G) = G then G is abelian. If Z(G) 6= G then G/Z(G) is a group of order p and thus it is a non-trivial cyclic group. This is however impossible by Problem 8 of HW 1.

18.2 Proposition. If G is a group of order pq for some primes p, q such that p > q and q - (p − 1) then G∼ = Z/pqZ Proof. If is enough to show that G contains an element of order pq. Let sp denote the number of Sylow p-subgroups of G. By the Third Sylow Theorem (17.5) we have sp | q

and sp = 1 + kp

Since q is a prime the first condition gives sp = 1 or sp = q. Since p > q the second condition implies then that sp = 1. Similarly, let sq be the number of Sylow q-subgroups of G. We have sq | p

and sq = 1 + kq 72

The first condition gives sq = 1 or sq = p. If sq = p then the second condition gives p = 1 + kq, or p − 1 = kq. This is however impossible since q - (p − 1). Therefore we have sq = 1. We obtain that G has exactly one Sylow p-subgroup P (of order p) and exactly one Sylow q-subgroup Q (of order q). By the Second Sylow Theorem (17.3) every element of G of order p belongs to the subgroup P and every element of order q belongs to the subgroup Q. It follows that G contains exactly p − 1 elements of order p, exactly q − 1 elements of order q, and one trivial element (of order 1). Since for all p, q we have pq > (p − 1) + (q − 1) + 1 there are elements of G of order not equal to 1, p, or q. Any such element must have order pq.

Note. If |G| = pq, and q | (p − 1) then G need not be isomorphic to Z/pqZ (take e.g. G = GT ).

18.3. Defintition. Let N , K be groups and let ϕ : K → Aut(N ) be a homomorphism. The semidirect product of N and K with respect to ϕ is the group N oϕ K such that N oϕ K = N × K as sets. Multiplication in N oϕ K is given by (a1 , b1 ) · (a2 , b2 ) := (a1 (ϕ(b1 )(a2 )), b1 b2 )

73

18.4 Note. 1) If ϕ is the trivial homomorphism then N oϕ K = N × K. 2) If (a, b) ∈ N oϕ K then (a, b)−1 = (ϕ(b−1 )(a−1 ), b−1 ) 3) N oϕ K contains subgroups N∼ = {(a, e) | a ∈ N } and K ∼ = {(e, b) | b ∈ K} We have N C N oϕ K, and N ∩ K = {(e, e)}.

18.5 Examples. 1) Notice that Aut(Z/3Z) ∼ = Z/2Z. Let ϕ : Z/2 → Aut(Z/3Z) be the isomorphism. Check: Z/3Z oϕ Z/2Z ∼ = GT 2) In general if p is a prime then we have Aut(Z/pZ) ∼ = Z/(p − 1)Z For any n | (p − 1) there is a unique cyclic subgroup H ⊆ Aut(Z/pZ) of order n. Let ϕ : Z/nZ → Aut(Z/pZ)

be any homomorphism such that Im(ϕ) = H. Then Z/pZ oϕ Z/nZ. is a non-abelian group of order pn.

3) If N is any abelian group then the map inv : N → N, 74

inv(a) = a−1

is an automorphism of N . This gives a homomorphism ϕ : Z/2Z → Aut(N ) such that ϕ(1) = inv. We obtain in this way a group N oϕ Z/2Z of order 2|N |. Special cases:

• If N = Z/nZ then this group is called the dihedral group of order 2n and it is denoted Dn . The group Dn is isomorphic to the group of all isometries of a regular polygon with n sides (exercise). In particular D3 ∼ = GT . • If N = Z then this group is the infinite dihedral group and it is denoted by D∞ . The group D∞ is isomorphic to the free product Z/2Z ∗ Z/2Z (exercise).

Recall. From HW 1: If G, H are abelian groups and f : G → H, g : H → G are homomorphisms such that f g = idH then G ∼ = Ker(f ) ⊕ H. 18.6 Proposition. If G, H are groups and f : G → H, g : H → G are homomorphisms such that f g = idH then G∼ = Ker(f ) oϕ H where ϕ : H → Aut(Ker(f )) is given by ϕ(b)(a) := g(b)ag(b)−1 for a ∈ Ker(f ), b ∈ H. Proof. Exercise.

18.7 Proposition. Let p, q be prime numbers such that p > q and q | (p − 1). Then, up to isomorphism, there are only two groups of order pq: 75

– the abelian group Z/pqZ ∼ = Z/pZ × Z/qZ – the non-abelian group Z/pZ oϕ Z/qZ where ϕ : Z/qZ → Aut(Z/pZ) is any non-trivial homomorphism.

Proof. By the same argument as in the proof of Proposition 18.2 we get that G has only one Sylow p-subgroup. Call this subgroup P . We have P C G. Let Q be any Sylow q-subgroup. Consider the quotient map f : G → G/P Take the restriction f |Q : Q → G/P . Notice that Ker(f |Q ) = Ker(f ) ∩ Q = P ∩ Q = {e}, so f |Q is a monomorphism. In addition |Q| = q = |G/P |, so f |Q is an isomorphism. As a consequence we have a homomorphism g : G/P

(f |Q )−1

/

Q ,→ G

Since f g = idG/P by Proposition 18.6 we obtain G∼ = P oϕ G/P for some homomorphism ϕ : G/P → Aut(P ). Also, since P ∼ = Z/pZ and ∼ G/P = Z/qZ we get G∼ = Z/pZ oϕ Z/qZ for some ϕ : Z/qZ → Aut(Z/pZ).

If ϕ is the trivial homomorphism then G ∼ = Z/pZ × Z/qZ ∼ = Z/pqZ. If ϕ is non-trivial then G is a non-abelian group. Notice that since q | (p − 1) such non-trivial homomorphism exists by (18.5). It remains to show that for any two non-trivial homomorphisms ϕ, ψ : Z/qZ → Aut(Z/pZ) we have an isomorphism Z/pZ oϕ Z/qZ ∼ = Z/pZ oψ Z/qZ (exercise). 76

18.8 Example. For any odd prime p there are two non-isomorphic groups of order 2p: – the cyclic group Z/2pZ – the dihedral group Dp .

77

H. U. Besche, B. Eick, E. A. O’Brien A millennium project: constructing small groups International Journal of Algebra and Computation 12(5) (2002) 623-644.

19

Group extensions and composition series

19.1 Definition. Let fi+1

fi

. . . −→ Gi −→ Gi+1 −→ Gi+2 −→ . . . be a sequence of groups and group homomorphisms. This sequence is exact if Im(fi ) = Ker(fi+1 ) for all i.

19.2 Definition. A short exact sequence is an exact sequence of the form f

g

1 −→ N −→ G −→ K −→ 1 (where 1 is the trivial group).

19.3 Note. f

g

1) A sequence 1 → N −→ G −→ K → 1 is a short exact sequence iff • f is a monomorphism • g is an epimorphism • Im(f ) = Ker(g).

2) If H C G then we have a short exact sequence 1 −→ H −→ G −→ G/H −→ 1 Morever, up to an isomorphism, every short exact sequence is of this form: /N

1

∼ =

1

/



Ker(g)

f

/

G

=

/



G

82

g

/K 

/

1

∼ =

/ G/ Ker(g)

/

1

19.4 Definition. If a group G fits into a short exact sequence f

g

1 −→ N −→ G −→ K −→ 1 then we say that G is an extension of K by N .

19.5 Example. For any n > 1 the dihedral group Dn and the cyclic group Z/2nZ are non-isomorphic extensions of Z/nZ by Z/2Z.

19.6 Definition. A group G is a simple group if G 6= {e} and the only normal subgroups of G are G and {e}. 19.7 Example. Z/pZ is a simple group for every prime p. Note. A group G is simple iff it is not a non-trivial extension of any group.

19.8 Definition. If G is a group then a normal series of G is a sequence of subgroups {e} = G0 ⊆ G1 ⊆ G2 ⊆ . . . ⊆ Gk = G such that Gi−1 C Gi for all i. A composition series of G is a normal series such that all quotient groups Gi /Gi−1 are simple.

19.9 Example. Take the dihedral group D4 = Z/4Z o Z/2Z. We have a composition series {0} ⊆ Z/2Z ⊆ Z/4Z ⊆ D4 Another composition series of D4 : {0} ⊆ Z/2Z ⊆ Z/2Z × Z/2Z ⊆ D4 83

19.10 Theorem (Jordan – H¨older). If G 6= {e} is a finite group then 1) G has a composition series. 2) All composition series of G are equivalent in the following sense. If we have composition series {e} = G0 ⊆ . . . ⊆ Gk = G

and

{e} = H0 ⊆ . . . ⊆ Hl = G

then k = l and there is a bijection σ : {1, . . . , k} → {1, . . . , k} such that for i = 1, . . . , k we have an isomorphism Gi /Gi−1 ∼ = Hσ(i) /Hσ(i)−1 Proof. Exercise (or see Hungerford p. 111).

Upshot. If G 6= {e} is a finite group then G can be obtained by taking successive extensions of simple groups as follows. 1) Take a composition series {e} = G0 ⊆ G1 ⊆ G2 ⊆ . . . ⊆ Gk = G 2) For every i = 1, . . . , k we have a short exact sequence 1 −→ Gi−1 −→ Gi −→ Gi /Gi−1 −→ 1 where Gi /Gi−1 is a simple group. Therefore G1 is an extension of G1 /G0 by G0 G2 is an extension of G2 /G1 by G1 ... ... ... ... ... ... ... ... ... ... G = Gk is an extension of Gk /Gk−1 by Gk−1

84

The grand plan for classifying all finite groups (The H¨ older Program) 1) Classify all finite simple groups. 2) For any two groups N , K describe all possible extensions of K by N . Good news: part 1) is done. See R. Solomon, A brief history of the classification of the finite simple groups, Bulletin AMS 38 (3) (2001), 315-352. M. Aschbacher, The status of the classification of the finite simple groups, Notices AMS 51(7) (2004), 736-740.

85

20

Simple groups

Recall. A group G 6= {e} is a simple group if the only normal subgroups of G are G and {e}. Note. If G is a simple group then any non-trivial homomorphism f : G → H is a monomorphism. 20.1 Proposition. If G is an abelian group then G is simple iff G ∼ = Z/pZ for some prime p. Proof. Exercise.

20.2 Proposition. If G is a simple p-group then G ∼ = Z/pZ. Proof. If G is a p-group then Z(G) 6= {e} by (16.4). We have Z(G) C G, so if G is simple we must have G = Z(G). Therefore G is a simple abelian p-group, and so G ∼ = Z/pZ.

20.3 Lemma. There are no non-abelian simple groups of order pr m where p is a prime, r ≥ 1, p - m and pr m - m!. Proof. Assume that G is a simple, non-abelian group of such order. We must have m > 1 (since if m = 1 then G is a p-group). Let P be a Sylow p-subgroup of G. Consider the action of G on the left cosets G/P : G × G/P → G/P,

a · bP = (ab)P

This action defines a homomorphism % : G → Perm(G/P ) 86

where Perm(G/P ) is the group of all permutations of the set G/P . Since G 6= P this homomorphism is non-trivial, and so, since G is a simple group, % is a monomorphism. Therefore G can be identified with a subgroup of Perm(G/P ). By Lagrange’s Theorem (7.4) we obtain that |G| divides |Perm(G/P )|. Since |Perm(G/P )| = m! this gives pr m | m! which contradicts assumptions of the lemma.

20.4 Theorem. There are no non-abelian simple groups of order < 60.

Proof. Check: If 1 ≤ n < 60, and n 6= 30, 40, 56 then n is of the form pr m for some prime p, and r, m ≥ 1 such that p - m and pr m - m!. By Lemma 20.3 we obtain then that a non-abelian group G of order n < 60 may be simple only if n = 30, 40 and 56. Assume that |G| = 30 = 2 · 3 · 5. We will show that G cannot be a simple group. We argue by contradiction. Assume that G is simple and let s3 be the number of Sylow 3-subgroups of G. We have s3 | 10 and s3 ≡ 1 (mod 3) It follows that either s3 = 1 or s3 = 10. Since G is simple s3 6= 1, so s3 = 10. Notice that if P , P 0 are two distinct Sylow 3-subgroups of G then P ∩ P 0 = {e}. We obtain: – G contains 10 Sylow 3-subgroups. – each Sylow 3-subgroup contains 2 elements of order 3. It follows that G contains 20 elements of order 3. By a similar argument we obtain that – G must contain 6 Sylow 5-subgroups. 87

– each Sylow 5-subgroup contains 4 elements of order 3. so we have 24 elements of order 5 in G. This is however impossible, since 20 + 24 > 30 = |G|. In a similar way one can show that if |G| = 40 or |G| = 56 then G is not a simple group (exercise) Next goal: there are infinitely many non-abelian simple finite groups. In particular there is a simple group of order 60.

88

21

Symmetric and alternating groups

Recall. The symmetric group on n letters is the group Sn = Perm({1, . . . , n})

21.1 Theorem (Cayley). If G is a group of order n then G is isomorphic to a subgroup of Sn .

Proof. Let S be the set of all elements of G. Consider the action of G on S G × S → S,

a · b := ab

This action defines a homomorphism % : G → Perm(S). Check: this homomorphism is 1-1. It follows that G is isomorphic to a subgroup of Perm(S). Finally, since |S| = n we have Perm(S) ∼ = Sn .

21.2 Notation. Denote [n] := {1, . . . , n}

If σ ∈ Sn , σ : [n] → [n] then we write σ=

1

2

3

...

n

!

σ(1) σ(2) σ(3) . . . σ(n)

21.3 Definition. A permutation σ ∈ Sn is a cycle of length r (or r-cycle) if there are distinct integers i1 , . . . , ir ∈ [n] such that σ(i1 ) = i2 , σ(i2 ) = i3 , . . . , σ(ir ) = i1 and σ(j) = j for j 6= i1 , . . . , ir . A cycle of length 2 is called a transposition. 89

Note. The only cycle of length 1 is the identity element in Sn .

21.4 Notation. If σ is a cycle as above then we write σ = (i1 i2 . . . ir )

21.5 Example. In S5 we have 1 2 3 4 5

!

1 4 2 5 3

= (2 4 5 3)

Note: (2 4 5 3) = (4 5 3 2) = (5 3 2 4) = (3 2 4 5).

21.6 Definition. Permutations σ, τ ∈ Sn are disjoint if {i ∈ [n] | σ(i) 6= i} ∩ {j ∈ [n] | τ (j) 6= j} = ∅

21.7 Proposition. If σ, τ are disjoint permutations then στ = τ σ. Proof. Exercise.

21.8 Proposition. Every non-identity permutation σ ∈ Sn is a product of disjoint cycles of length ≥ 2. Moreover, this decomposition into cycles is unique up to the order of factors.

21.9 Example. Let σ ∈ S9 σ=

1 2 3 4 5 6 7 8 9 4 7 1 3 2 6 5 9 8

Then σ = (1 4 3)(2 7 5)(8 9). 90

!

Proof of proposition 21.8. Consider the action of Z on the set [n] given by k · i = σ k (i) for k ∈ Z, i ∈ [n]. Notice that Orb(i) = {σ k (i) | k ∈ Z} Define σi : [n] → [n] ( σi (j) =

σ(j) if j ∈ Orb(i) j otherwise

Notice that σi is a bijection since σ(Orb(i)) = Orb(i). Thus σi ∈ Sn . Check: 1) σi is a cycle of length |Orb(i)|. 2) if Orb(i1 ), . . . , Orb(ir ) are all distinct orbits of [n] containing more than one element then σi1 , . . . , σir are non-trivial, disjoint cycles and σ = σ i1 · . . . · σ ir Uniqueness of decomposition - easy.

21.10 Proposition. Every permutation σ ∈ Sn is a product of (not necessarily disjoint) transpositions.

Proof. By Proposition 21.8 it is enough to show that every cycle is a product of transpositions. We have: (i1 i2 i3 . . . ir ) = (i1 ir )(i1 ir−1 ) · . . . · (i1 i3 )(i1 i2 )

Note. For σ ∈ Sn we have a bijection σ × σ : [n] × [n] → [n] × [n] 91

given by σ × σ(i, j) = (σ(i), σ(j)). Define Sσ := {(i, j) ∈ [n] × [n] | i > j and σ(i) < σ(j)}

21.11 Definition. A permutation σ ∈ Sn is even (resp. odd) if the number of elements of Sσ is even (resp. odd).

21.12 Theorem. 1) The map sgn : Sn → Z/2Z defined by ( 0 if σ is even sgn(σ) = 1 if σ is odd is a homomorphism. 2) If σ is a transposition then sgn(σ) = 1, so this homomorphism is non-trivial.

Proof. 1) Let σ, τ ∈ Sn . Denote sσ = |Sσ |. We want to show sτ σ ≡ sτ + sσ (mod 2) Let [n]+ := {(i, j) ∈ [n] × [n] | i > j}. Define subsets Pσ , Rσ , Pτ , Rτ ⊆ [n]+ as follows: Pσ Rσ Pτ Rτ

:={(i, j) :={(i, j) :={(i, j) :={(i, j)

| | | |

σ −1 (i) > σ −1 (j)} σ −1 (i) < σ −1 (j)} τ (i) > τ (j)} τ (i) < τ (j)}

Notice that sσ = |Rσ | and sτ = |Rτ |. Notice also that (i, j) ∈ Sτ σ iff either (τ (i), τ (j)) ∈ Pσ ∩ Rτ or (τ (j), τ (i)) ∈ Rσ ∩ Pτ . This gives sτ σ = |Pσ ∩ Rτ | + |Rσ ∩ Pτ | 92

On the other hand we have: sσ = |Rσ | = |Rσ ∩ Pτ | + |Rσ ∩ Rτ | sτ = |Rτ | = |Pσ ∩ Rτ | + |Rσ ∩ Rτ | Therefore sσ + sτ = |Rσ ∩ Pτ | + |Pσ ∩ Rτ | + 2|Rσ ∩ Rτ | = sτ σ + 2|Rσ ∩ Rτ | and so sτ + sσ ≡ sτ σ (mod 2). 2) Exercise.

21.13 Definition/Proposition. The set An = {σ ∈ Sn | σ is even} is a normal subgroup of Sn . It is called the alternating group on n letters.

Proof. It is enough to notice that An = Ker(sgn).

Note. We have Since |Sn | = n! thus |An | =

Sn /An ∼ = Z/2Z n! . 2

21.14 Proposition. If σ ∈ Sn then σ is even (resp. odd) iff σ is a product of an even (resp. odd) number of transpositions.

93

Proof. If σ = τ1 . . .τm where τ1 , . . . , τm are transpositions then sgn(σ) = sgn(τ1 . . .τm ) =

m X

sgn(τi ) =

i=1

m X

1

i=1

Thus sgn(σ) = 0 iff m is even and sgn(σ) = 1 iff m is odd.

Note. If follows that if a permutation σ ∈ Sn is a product of an even number of transpositions then it cannot be written as a product of an odd number of transpositions (and vice versa).

21.15 Corollary. A permutation σ ∈ Sn is even iff σ = σ1 σ2 . . .σr P where σi is a cycle of length mi and ri=1 (mi − 1) is even. Proof. It is enough to notice that by the proof of Proposition 21.10 a cycle of length m is a product of m − 1 transpositions. Note. The usual notation for the sign of a permutation is ( 1 if σ is even sgn(σ) = −1 if σ is odd where {−1, 1} ∼ = Z/2Z is the multiplicative group of units in Z.

94

22

Simplicity of alternating groups

22.1 Theorem. The alternating group An is simple for n ≥ 5.

22.2 Lemma. For n ≥ 3 every element of An is a product of 3-cycles. Proof. It is enough to show that if n ≥ 3 and τ , σ are transpositions in Sn then τ σ is a product of 3-cycles. Case 1) τ , σ are disjoint transpositions: τ = (i j), σ = (k l) for distinct elements i, j, k, l ∈ [n]. Then we have τ σ = (i j k)(j k l) Case 2) τ , σ are not disjoint: τ = (i j), σ = (j k). Then τ σ = (i j k)

22.3 Lemma. If n ≥ 5 and σ, σ 0 are 3-cycles in Sn then σ 0 = τ στ −1 for some τ ∈ An Proof. Check: if (i1 i2 . . . ir ) is a cycle in Sn then for any ω ∈ Sn we have ω(i1 i2 . . . ir )ω −1 = (ω(i1 ) ω(i2 ) . . . ω( ir )) If σ = (i1 i2 i3 ), σ 0 = (j1 j2 j3 ) then take ω ∈ Sn such that ω(ik ) = jk for k = 1, 2, 3. We have σ 0 = ωσω −1 95

If ω ∈ An we can then take τ := ω. Assume then that ω 6∈ An . Since n ≥ 5 there are r, s ∈ [n] such that (r s) and σ = (i1 i2 i3 ) are disjoint cycles. Take τ = ω(r s). Then τ ∈ An . Moreover, since (r s) commutes with σ we have τ στ −1 = ω(r s)σ(r s)−1 ω −1 = ωσω −1 = σ 0

22.4 Corollary. If n ≥ 5 and H is a normal subgroup of An such that H contains some 3-cycle then H = An .

Proof. By Lemma 22.3 H contains all 3-cycles, and so by Lemma 22.2 it contains all elements of An .

Proof of Theorem 22.1. Let n ≥ 5, H C An and H 6= {(1)}. We need to show that H = An . By Corollary 22.4 it will suffice to show that H contains some 3-cycle. Let (1) 6= σ be an element of H with the maximal number of fixed points in [n]. We will show that σ is 3-cycle. Take the decompositon of σ into dosjoint cycles: σ = σ1 σ2 · . . . · σm Case 1) σ1 , . . . , σm are transpositions. Since σ ∈ An we must then have m ≥ 2. Say, σ1 = (i j), σ2 = (k l). Take s 6= i, j, k, l and let τ = (k l s) ∈ An . Since H is normal in An we have τ στ −1 σ −1 ∈ H Check: 1) τ στ −1 σ −1 6= (1) since τ στ −1 σ −1 (k) 6= k 96

2) τ στ −1 σ −1 fixes every element of [n] fixed by σ 3) τ στ −1 σ −1 fixes i, j. Thus τ στ −1 σ −1 has more fixed points than σ which is impossible by the definition of σ. Case 2) σr is a cycle of length ≥ 3 for some 1 ≤ r ≤ m. We can assume r = 1: σ1 = (i j k . . . ). If σ = σ1 and σ1 is a 3-cycle we are done. Otherwise σ must move at least two more elements, say p, q. In such case take τ = (k p q). We have τ στ −1 σ −1 ∈ H Check: 1) τ στ −1 σ −1 6= (1) since τ στ −1 σ −1 (k) 6= k

2) τ στ −1 σ −1 fixes every element of [n] fixed by σ 3) τ στ −1 σ −1 fixes j.

Thus τ στ −1 σ −1 has more fixed points than σ which is again impossible by the definition of σ. As a consequence σ must be a 3-cycle.

22.5. Classification of simple finite groups. 1) cyclic groups Z/pZ, p – prime 2) alternating groups An , n ≥ 5

3) finite simple groups of Lie type, e.g. projective special linear groups P SLn (F) := SLn (F)/Z(SLn (F)) F -finite field, n ≥ 2 (and n > 2 if F = F2 or F = F3 ). 97

4) 26 sporadic groups (the smallest: Mathieu group M11 , |M11 | = 7920, the biggest: the Monster M , |M | ≈ 8 · 1053 ).

98

23

Solvable groups

Recall. Every finite group G has a composition series: {e} = G0 ⊆ . . . ⊆ Gk = G where Gi−1 C Gi and Gi /Gi−1 is a simple group. 23.1 Definition. A group G is solvable if it has a composition series {e} = G0 ⊆ . . . ⊆ Gk = G such that for every i the group Gi /Gi−1 is a simple abelian group (i.e. Gi /Gi−1 ∼ = Z/pi Z for some prime pi ).

23.2 Example. 1) Every finite abelian group is solvable. 2) For n ≥ 5 the symmetric group Sn has a composition series {(1)} ⊆ An ⊆ Sn and so Sn is not solvable.

23.3 Proposition. A finite group G is solvable iff it has a normal series {e} = H0 ⊆ . . . ⊆ Hl = G such that Hj /Hj−1 is an abelian group for all j. Proof. Exercise.

99

Recall. 1) If G is a group the [G, G] is the commutator subgroup of G [G, G] = h{aba−1 b−1 | a, b ∈ G}i 2) [G, G] is the smallest normal subgroup of G such that G/[G, G] is abelian: if G/H for some H C G then [G, G] ⊆ H.

23.4 Definition. For a group G the derived series of G is the normal series · · · ⊆ G(2) ⊆ G(1) ⊆ G(0) = G where Gi+1 = [G(i) , G(i) ] for i ≥ 1. The group G(i) is called the i-th derived group of G.

23.5 Theorem. A group G is solvable iff G(n) = {e} for some n ≥ 0. Proof. Exercise.

23.6 Theorem. 1) Every subgroup of a solvable group is solvable. 2) Ever quotient group of a solvable group is solvable. 3) If H CG, and both H and G/H are solvable groups then G is also solvable. Proof. 1) If H ⊆ G then H (i) ⊆ G(i) . Thus if G(n) = {e} then H (n) = {e}. 2) For H C G take the canonical epimorphism f : G → G/H. We have f (G(i) ) = (G/H)(i) 100

Therefore if G(n) = {e} then (G/H)(n) = {e}. 3) Assume that H C G, and that H (m) , (G/H)(n) are trivial groups. Consider the canonical epimorphism f : G → G/H. We have f (G(n) ) = (G/H)(n) = {e} Therefore G(n) ⊆ Ker(f ) = H. As a consequence we obtain G(n+m) = G(n)


(m)

⊆ H (m) = {e}

23.7 Theorem (Feit-Thompson). Every finite group of odd order is solvable. Proof. See: W. Feit, J.G. Thompson, Solvability of groups of odd order, Pacific Journal of Mathematics 13(3) (1963), 775-1029.

23.8 Corollary. There are no non-abelian finite simple groups of odd order.

Proof. Let G 6= {e} be a simple group of odd order. By Theorem 23.7 G is solvable so [G, G] 6= G. Since [G, G] C G, by simplicity of G we must have [G, G] = {e}, and so G is an abelian group.

101

24

Nilpotent groups

24.1. Recall that if G is a group then Z(G) = {a ∈ G | ab = ba for all b ∈ G} Note that Z(G) C G. Take the canonical epimorphism π : G → G/Z(G). Since Z (G/Z(G)) C G/Z(G) we have: π −1 (Z (G/Z(G))) C G Define: Z1 (G) :=Z(G) Zi (G) :=πi−1 (Z (G/Zi−1 (G)))

for i > 1

where πi : G → G/Zi−1 (G). We have Zi (G) C G for all i.

24.2 Definition. The upper central series of a group G is a sequence of normal subgroups of G: {e} = Z0 (G) ⊆ Z1 (G) ⊆ Z2 (G) ⊆ . . .

24.3 Definition. A group G is nilpotent if Zi (G) = G for some i. If G is a nilpotent group then the nilpotency class of G is the smallest n ≥ 0 such that Zn (G) = G.

24.4 Proposition. Every nilpotent group is solvable. Proof. If G is nilpotent group then the upper central series of G {e} = Z0 (G) ⊆ Z1 (G) ⊆ . . . ⊆ Zn (G) = G is a normal series. 102

Moreover, for every i we have Zi (G)/Zi−1 (G) = Z(G/Zi−1 (G)) so all quotients of the upper central series are abelian.

24.5 Note. Not every solvable group is nilpotent. Take e.g. GT . We have Z(GT ) = {I}, and so Zi (GT ) = {I}

for all i. Thus GT is not nilpotent. On the other hand GT is solvable with a composition series {I} ⊆ {I, R1 , R2 } ⊆ GT

24.6 Proposition. 1) Every abelian group is nilpotent. 2) Every finite p-group is nilpotent.

Proof. 1) If G is abelian then Z1 (G) = G. 2) If G is a p-group then so is G/Zi (G) for every i. By Theorem 16.4 if G/Zi (G) is non-trivial then its center Z(G/Zi (G)) a non-trivial group. This means that if Zi (G) 6= G then Zi (G) ⊆ Zi+1 (G) and Zi (G) 6= Zi+1 (G). Since G is finite we must have Zn (G) = G for some G.

24.7 Definition. A central series of a group G is a normal series {e} = G0 ⊆ . . . ⊆ Gk = G such Gi C G and Gi+1 /Gi ⊆ Z(G/Gi ) for all i.

103

24.8 Proposition. If {e} = G0 ⊆ . . . ⊆ Gk = G is a central series of G then Gi ⊆ Zi (G) Proof. Exercise.

24.9 Corollary. A group G is nilpotent iff it has a central series.

Proof. If G is nilpotent then {e} = Z0 (G) ⊆ Z1 (G) ⊆ . . . ⊆ Zn (G) = G is a central series of G. Conversely, if {e} = G0 ⊆ . . . ⊆ Gk = G is a central series of G then by (24.9) we have G = Gk ⊆ Zk (G), so G = Zk (G), and so G is nilpotent.

24.10 Note. Given a group G define Γ0 (G) :=G Γi (G) :=[G, Γi−1 (G)]

for i > 0.

We have . . . ⊆ Γ1 (G) ⊆ Γ0 (G) = G

24.11 Proposition. If G is a group then 1) Γi (G) C G for all i 104

2) Γi (G)/Γi+1 (G) ⊆ Z(G/Γi+1 (G)) for all i Proof. Exercise.

24.12 Definition. If Γn (G) = {e} then {e} = Γn (G) ⊆ . . . ⊆ Γ0 (G) = G is a central series of G. It is called the lower central series of G.

24.13 Proposition. A group G is nilpotent iff Γn (G) = {e} Proof. Exercise.

24.14 Theorem. 1) Every subgroup of a nilpotent group is nilpotent. 2) Ever quotient group of a nilpotent group is nilpotent. 3) If H C G, and both H and G/H are nilpotent groups then G is also nilpotent.

Proof. Similar to the proof of Theorem 23.6.

24.15 Corollary. If G1 , . . . , Gk are nilpotent groups then the direct product G1 × · · · × Gk is also nilpotent. Proof. Follows from part 3) of Theorem 24.14.

105

24.16 Corollary. If p1 , . . . , pk are primes and Pi is a pi -group then P1 × . . . × Pk is a nilpotent group.

Proof. Follows from (24.6) and (24.15).

24.17 Theorem. Let G be a finite group. The following conditions are equivalent. 1) G is nilpotent. 2) Every Sylow subgroup of G is a normal subgroup. 3) G isomorphic to the direct product of its Sylow subgroups.

24.18 Lemma. If G is a finite group and P is a Sylow p-subgroup of G then NG (NG (P )) = NG (P )

Proof. Since P ⊆ NG (P ) ⊆ G and P is a Sylow p-subgroup of G therefore P is a Sylow p-subgroup of NG (P ). Moreover, P C NG (P ), so P is the only Sylow p-subgroup of G. Take a ∈ NG (NG (P )). We will show that a ∈ NG (P ). We have aP a−1 ⊆ aNG (P )a−1 = NG (P ) As a consequence aP a−1 is a Sylow p-subgroup of NG (P ), and thus aP a−1 = P . By the definitions of normalizer this gives a ∈ NG (P ).

24.19 Lemma. If H is a proper subgroup of a nilpotent group G (i.e. H ⊆ G, and H 6= G), then H is a proper subgroup of NG (H).

106

Proof. Let k ≥ 0 be the biggest integer such that Zk (G) ⊆ H. Take a ∈ Zk+1 (G) such that a 6∈ H. We will show that a ∈ NG (H). We have H/Zk (G) ⊆ G/Zk (G) and Zk+1 (G)/Zk (G) = Z(G/Zk (G)) If follows that for every h ∈ H we have ahZk (G) = (aZk (G))(hZk (G)) = (hZk (G))(aZk (G)) = haZk (G) Therefore ha = ahh0 for some h0 ∈ Zk (G) ⊆ H, and so a−1 ha = hh0 ∈ H. As a consequence a−1 Ha = H, so a−1 ∈ NG (H), and so also a ∈ NG (H).

Proof of Theorem 24.17. 1) ⇒ 2) Let P be a Sylow p-subgroup of G. It suffices to show that NG (P ) = G. Assume that this is not true. Then NG (P ) is a proper subgroup G, and so by Lemma 24.19 it is also a proper subgroup of NG (NG (P )). On the other hand by Lemma 24.18 we have NG (NG (P )) = NG (P ), so we obtain a contradiction. 2) ⇒ 3) Exercise. 3) ⇒ 1) Follows from Corollary 24.16.

107

25

Rings

25.1 Definition. A ring is a set R together with two binary operations: addition (+) and multiplication (·) satisfying the following conditions: 1) R with addtion is an abelian group. 2) multiplication is associative: (ab)c = a(bc) 3) addition is distributive with respect to multiplication: a(b + c) = ab + ac

(a + b)c = ac + bc

The ring R is commutative if ab = ba for all a, b ∈ R. The ring R is a ring with identity if there is and element 1 ∈ R such that a1 = 1a = a for all a ∈ R. (Note: if such identity element exists then it is unique)

25.2 Examples. 1) Z, Q, R, C are commutative rings with identity. 2) Z/nZ is a ring with multiplication given by k(nZ) · l(nZ) := kl(nZ) 3) If R is a ring then R[x] = {a0 + a1 x + . . . + an xn | ai ∈ R, n ≥ 0} is the ring of polynomials with coefficients in R and R[[x]] = {a0 + a1 x + . . . | ai ∈ R} is the ring of formal power series with coefficients in R. If R is a commutative ring then so are R[x], R[[x]]. If R has identity then R[x], R[[x]] also have identity. 108

4) If R is a ring then Mn (R) is the ring of n × n matrices with coefficients in R. 5) The set 2Z of even integers with the usual addition and multiplication is a commutative ring without identity. 6) If G is an abelian group then the set Hom(G, G) of all homomorphisms f : G → G is a ring with multiplication given by composition of homomorphisms and addition defined by (f + g)(a) := f (a) + g(a) 7) If R is a ring and G is a group then define X R[G] := { ag g | ag ∈ R, ag 6= 0 for finitely many g only } g∈G

addition in R[G]: X

ag g +

g∈G

X

bg g =

g∈G

X

(ag + bg )g

g∈G

multiplication in R[G]: ! X g∈G

ag g

! X

bg g

g∈G

! =

X

X

g∈G

hh0 =g

ah ah0

g

The ring R[G] is called the group ring of G with coefficients in R.

25.3 Definition. Let R be a ring. An element 0 6= a ∈ R is a left (resp. right) zero divisor in R if there exists 0 6= b ∈ R such that ab = 0 (resp. ba = 0). An element 0 6= a ∈ R is a zero divisor if it is both left and right zero divisor. 25.4 Example. In Z/6Z we have 2 · 3 = 0, so 2 and 3 are zero divisors. 109

25.5 Definition. An integral domain is a commutative ring with identity 1 6= 0 that has no zero divisors.

25.6 Proposition. Let R be an integral domain. If a, b, c ∈ R are non-zero elements such that ac = bc then a = b.

Proof. We have (a − b)c = 0. Since c 6= 0 and R has no zero divisors this gives a − b = 0, and so a = b.

25.7 Definition. Let R be a ring with identity. An element a has a left (resp. right) inverse if there exists b ∈ R such that ba = 1 (resp. there exists c ∈ R such that cb = 1). An element a ∈ R is a unit if it has both a left and a right inverse.

25.8 Proposition. If a is a unit of R then the left inverse and the right inverse of a coincide.

Proof. If ba = 1 = ac then b = b · 1 = b(ac) = (ba)c = 1 · c = c 25.9 Note. The set of all units of a ring R forms a group R∗ (with multiplication). E.g.: Z∗ = {−1, 1} ∼ = Z/2Z R∗ = R − {0} (Z/14Z)∗ = {1, 3, 5, 9, 11, 13} ∼ = Z/6Z 110

25.10 Definition. A division ring is a ring R with identity 1 6= 0 such that every non-zero element of R is a unit. A field is a commutative division ring.

25.11 Examples. 1) R, Q, C are fields. 2) Z is an integral domain but it is not a field. 3) The ring of real quaternions is defined by H := {a + bi + cj + dk | a, b, c, d ∈ R}

Addition in H is coordinatewise. Multiplication is defined by the identities: i2 = j 2 = k 2 = −1, ij = −ji = k, jk = −kj = i, ki = −ik = j

The ring H is a (non-commutative) division ring with the identity 1 = 1 + 0i + 0j + 0k The inverse of an element z = a + bi + cj + dk is given by z −1 = (a/kzk) − (b/kzk)i − (c/kzk)j − (d/kzk)k √ where kzk = a2 + b2 + c2 + d2 25.12 Proposition. The following conditions are equivalent. 1) Z/nZ is a field. 2) Z/nZ is an integral domain. 3) n is a prime number. Proof. Exercise.

111

26

Ring homomorphisms and ideals

26.1 Definition. Let R, S be rings. A ring homomorphism is a map f: R →S such that 1) f (a + b) = f (a) + f (b) 2) f (ab) = f (a)f (b)

26.2 Note. If R, S are rings with identity then these conditions do not guarantee that f (1R ) = 1S . Take e.g. rings with identity R1 , R2 and define R1 ⊕ R2 = {(r1 , r2 ) | r1 ∈ R1 , R2 } with addition and multiplication defined coordinatewise. Then R1 ⊕ R2 is a ring with identity (1R1 , 1R2 ). The map f : R1 → R1 ⊕ R2 ,

f (r1 ) = (r1 , 0)

is a ring homomorphism, but f (1R1 ) 6= (1R1 , 1R2 ). 26.3 Note. Rings and ring homomorphisms form a category Ring.

26.4 Proposition. A ring homomorphism f : R → S is an isomorphism of rings iff f is a bijection.

Proof. Exercise.

112

26.5 Definition. If f : R → S is a ring homomorphism then Ker(f ) = {a ∈ R | f (a) = 0}

26.6 Proposition. A ring homomorphism is 1-1 iff Ker(f ) = {0} Proof. The same as for groups (4.4).

26.7 Definition. A subring of a ring R is a subset S ⊆ R such that S is an additive subgroup of R and it is closed under the multiplication. A left ideal of R is a subring I ⊆ R such that for every a ∈ I and b ∈ R we have ab ∈ I. A right ideal of R is defined analogously. A ideal of R is a subring I ⊆ R such that I is both left and right ideal.

26.8 Notation. If I is an ideal of R then we write I C R.

26.9 Proposition. If f : R → S is a ring homomorphism then Ker(f ) is an ideal of R.

Proof. Exercise.

26.10 Definition. If I is an ideal of a ring R then the quotient ring R/I is defined as follows. R/I := the set of left cosets of I in R Addition: (a + I) + (b + I) = (a + b) + I, multiplication: (a + I)(b + I) = ab + I. 113

26.11 Note. If I C R then the map π : R → R/I,

π(a) = a + I

is a ring homomorphism. It is called the canonical epimorphism of R onto R/I. 26.12 Theorem. If f : R → S is a homomorphism of rings then there is a unique homomorphism f¯: R/ Ker(f ) → S such that the following diagram commutes: f

R π

/S ;

f¯



R/ Ker(f ) Moreover, f¯ is a monomorphism and Im(f¯) = Im(f ).

Proof. Similar to the proof of Theorem 6.1 for groups.

26.13 First Isomorphism Theorem. If f : R → S is a homomorphism of rings that is an epimorphism then R/ Ker(f ) ∼ =S Proof. Take the map f¯: R/ Ker(f ) → S. Then Im(f¯) = Im(f ) = S, so f¯ is an epimorphism. Also, f¯ is 1-1. Therefore f¯ is a bijective homomorphism and thus it is an isomorphism.

26.14 Note. Let I, J C R. Check: 114

1) I ∩ J C R 2) I + J C R where I + J = {a + b | a ∈ I, b ∈ J}

26.15 Second Isomorphism Theorem. If I, J are ideals of R then I/(I ∩ J) ∼ = (I + J)/J Proof. Exercise.

26.16 Third Isomorphism Theorem. If I, J are ideals of R and J ⊆ I then I/J is a ideal of R/J and (R/J)/(I/J) ∼ = R/I Proof. Exercise.

115

Note. From now until Section 39 all rings are commutative rings with identity 1 6= 0 unless stated otherwise. Also, all ring homomorphisms preserve the identity.

27

Principal ideal domains and Euclidean rings

27.1 Definition. If R is a ring and S is a subset of R then denote hSi = the smallest ideal of R that contains S We say that hSi is the ideal of R generated by the set S. 27.2 Note. We have hSi = {b1 a1 + . . . bk ak | ai ∈ I, bi ∈ R, k ≥ 0} 27.3 Definition. An ideal I C R is finitely generated if I = ha1 , . . . , an i for some a1 , . . . , an ∈ R. An ideal I C R is a principal ideal if I = hai for some a ∈ R. 27.4 Definition. A ring R is a principal ideal domain (PID) if it is an integral domain (25.5) such that every ideal of R is a principal ideal.

27.5 Proposition. The ring of integers Z is a PID. Proof. Let I C Z. If I = {0} then I = h0i, so I is a principal ideal. If I 6= {0} then let a be the smallest integer such that a > 0 and a ∈ I. We will show that I = hai. 116

Since a ∈ I we have hai ⊆ I. Conversely, if b ∈ I then we have b = qa + r for some q, r ∈ Z, 0 ≤ r ≤ a − 1. This gives r = b − qa, so r ∈ I. Since a is the smallest positive element of I, this implies that r = 0. Therefore b = qa, and so b ∈ hai. 27.6 Proposition. If F is a field then F is a PID. Proof. If I C F and 0 6= a ∈ I then for every b ∈ F we have b = (ba−1 )a ∈ I And so I = F. As a consequence the only ideals of F are {0} = h0i and F = h1i. 27.7 Proposition. If F is a field then the ring of polynomials F[x] is a PID.

Proof. Let I ∈ R. If I = {0} then I = h0i. Otherwise let 0 6= p(x) be a polynomial such that p(x) ∈ I and deg p(x) ≤ deg q(x) for all q(x) ∈ I − {0}. Check that I = hp(x)i. 27.8 Note. Z[x] is not a PID. E.g. the ideal h2, xi is not a principal ideal of Z[x] (check!).

27.9 Definition. A Euclidean domain is an integral domain R equipped with a function N : R − {0} −→ N = {0, 1, . . . } such that 117

1) N (ab) ≥ N (a) for all a, b ∈ R − {0}

2) for any a, b ∈ R, a 6= 0 there exist q, r ∈ R such that b = qa + r and either r = 0 or N (r) < N (a). The function N is called the norm function on R.

27.10 Examples. 1) Z is a Euclidean domain with the norm function given by the absolute value. 2) Any field F is a Euclidean domain with N (a) = 0 for all a ∈ F − {0}. 3) If F is a field then F[x] is a Euclidean domain with N (p(x)) = deg p(x) for p(x) ∈ F[x], p(x) 6= 0. Note: Z[x] is not a Euclidean domain with N (p(x)) = deg p(x). E. g. there are no q(x), r(x) ∈ Z[x] such that either r(x) = 0 or deg r(x) < 0 and that x = 2q(x) + r(x) 4) The ring of Gaussian integers is the subring Z[i] ⊆ C given by Z[i] := {a + bi ∈ C | a, b ∈ Z} Z[i] is a Euclidean domain with N (a + bi) = a2 + b2 = (a + bi)(a + bi) (exercise). 5) Define

√ √ Z[ −5] := {a + b 5i | a, b ∈ Z}

√ Exercise: Z[ −5] is not a Euclidean domain. 118

27.11 Theorem. If R is a Euclidean domain then R is a PID.

Proof. Let I CR, I 6= {0}. Choose a ∈ I such that a 6= 0 and that N (a) ≤ N (b) for all b ∈ I − {0}. Check that hai = I.

27.12 Note. It is not true that every PID is a Euclidean domain. Take e.g. √ α = 12 + 219 i and let Z[α] = {a + bα | a, b ∈ Z}

Then Z[α] is a PID, but it is not a Euclidean domain. See J. C. Wilson, A principal ideal ring that is not a euclidean ring, Mathematics Magazine, 46 (1) (1973), 34-38. (Note: this link requires a JSTOR access)

119

28

Prime ideals and maximal ideals

28.1 Definition. Let R be a ring. 1) An ideal I C R is a prime ideal if I 6= R and for any a, b ∈ R we have ab ∈ I

iff either a ∈ I or b ∈ I

2) An ideal I C R is a maximal ideal if I 6= R and for any J C R such that I ⊆ J ⊆ R we have either J = I or J = R.

28.2 Examples. 1) The zero ideal {0} ∈ R is a prime ideal iff R is an integral domain, and it is a maximal ideal iff R is a field. 2) Recall that if I C Z then I = nZ for some n ≥ 0. Check: (nZ is a prime ideal) iff (nZ is a maximal ideal) iff (n is a prime number) 3) hxi C Z[x] is prime ideal (check!) but it is not a maximal ideal: hxi ⊆ h2, xi C Z[x]

28.3 Proposition. Let I C R 1) The ideal I is a prime ideal iff R/I is an integral domain. 2) The ideal I is a maximal ideal iff R/I is a field.

28.4 Corollary. Every maximal ideal is a prime ideal.

120

Proof. If I C R is a maximal ideal then R/I is a field. In particular R/I is an integral domain, and so I is a prime ideal.

Proof of Proposition 28.3. 1) Assume that I C R is a prime ideal. For a + I, b + I ∈ R/I we have (a + I)(b + I) = ab + I Thus if (a + I)(b + I) = 0 + I then ab + I = 0 + I i.e. ab ∈ I. Since I is a prime ideal we get that either a ∈ I (and so a + I = 0 + I) or b ∈ I (and so b + I = 0 + I). Therefore R/I is an integral domain. The other implication follows from a similar argument.

2) Assume that I C R is a maximal ideal. Let a + I ∈ R/I, a + I 6= 0 + I. We want to show that there exists b + I ∈ R/I such that (a + I)(b + I) = 1 + I Take the ideal J = hai + I. We have I⊆J ⊆R and I 6= J (since a 6∈ I). Since I is a maximal ideal we must have J = R. In particular 1 ∈ J, so 1 = ab + c for some b ∈ R, c ∈ I. This gives 1 + I = (ab + c) + I = ab + I = (a + I)(b + I) Conversely, assume that R/I is a field, and let J C R be an ideal such that I⊆J ⊆R 121

We will show that either J = I or J = R. Take the canonical epimorphism π : R → R/I. Since π(J) is an ideal of R/I (check!) and R/I is a field we have either π(J) = {0 + I} or π(J) = R/I. Also, since I ⊆ J, we have π −1 (π(J)) = J. If follows that either J = π −1 ({0 + I}) = I

or J = π −1 (R/I) = R

28.5 Examples. 1) Since an ideal nZ of Z is prime (and maximal) iff n is a prime number therefore Z/nZ is an integral domain (and in fact a field) iff n is a prime number. 2) Take x2 + 1 ∈ R[x]. We have an epimorphism of rings f : R[x] −→ C

f (p(x)) = p(i)

Check: Ker(f ) = hx2 + 1i. By the First Isomorphism Theorem (26.13) we get R[x]/hx2 + 1i ∼ = C. Since C is a field this shows that hx2 + 1i is a maximal ideal of R[x].

28.6 Note. For I, J C R define IJ := {a1 b1 + . . . + ak bk | ai ∈ I, bi ∈ J, k ≥ 0} Check: IJ is an ideal of R.

28.7 Proposition. Let I C R, I 6= R. The ideal I is a prime ideal iff for any ideals J1 , J2 such that J1 J2 ⊆ I we have either J1 ⊆ I or J2 ⊆ I.

Proof. Exercise. 122

29

Zorn’s Lemma and maximal ideals

Goal: 29.1 Theorem. If R is a ring, I C R, and I 6= R then there exists a maximal ideal J C R such that I ⊆ J.

29.2 Definition. A partially ordered set (or poset) is a set S equipped with a binary relation ≤ satisfying 1) x ≤ x for all x ∈ S (reflexivity)

2) if x ≤ y and y ≤ z then x ≤ z (transitivity)

3) if x ≤ y and y ≤ x then y = x (antisymmetry).

29.3 Example. If A is a set and S is the set of all subsets of A then S is a poset with ordering given by the inclusion of subsets.

29.4 Definition. A linearly ordered set is a poset (S, ≤) such that for any x, y ∈ S we have either x ≤ y or y ≤ x.

29.5 Definition. If (S, ≤) is a poset then an element x ∈ S is a maximal element of S if we have x ≤ y only for y = x.

29.6 Example. Let S be the set of all proper subsets of a set A ordered with respect to the inclusion. For every a ∈ A the set A − {a} is a maximal element of S.

123

29.7 Note. If (S, ≤) is a poset and T ⊆ S then T is also a poset with ordering inherited from S. 29.8 Definition. Let (S, ≤) is a poset and let T ⊆ S. An upper bound of T is an element x ∈ S such that y ≤ x for all y ∈ T . 29.9 Definition. If (S, ≤) is a poset. A chain in S is a subset T ⊆ S such that T is linearly ordered.

29.10 Zorn’s Lemma. If (S, ≤) is a non-empty poset such that every chain in S has an upper bound in S then S contains a maximal element.

Proof of Theorem 29.1. Let I 6= R be an ideal of R, and let S be the set of all ideals J C R such that I ⊆ J and J 6= R. Notice that S 6= ∅ since I ∈ S. The set S is a poset ordered with respect to inclusion of ideals. We will show that every chain in S has an upper bound in S. Let T = {Ji }i∈A S be a chain in S. Check: JT := i∈A Ji is an ideal of R. Moreover, I ⊆ JT . Finally, we have JT 6= R. Indeed, otherwise 1 ∈ JT , and so 1 ∈ Ji for some i ∈ A. This would give Ji = R, which contradicts our assumptions. It follows that JT ∈ S. Since Ji ⊆ JT for all i ∈ A, we obtain that JT is an upper bound of the chain T . By Zorn’s Lemma (29.10) there is a maximal element J ∈ S. This means, in particular, that J is an ideal of R such I ⊆ J. Moreover, let K C R be any ideal such that K 6= R and J ⊆ K. We have I⊆J ⊆K which means that K ∈ S. Maximality of J is S implies then that J = K. This shows that J is a maximal ideal of R. 124

29.11 Corollary. Every ring contains a maximal ideal.

Proof. If R is a ring then by Theorem 29.1 there exists a maximal ideal I C R such that I contains the zero ideal {0} C R.

125

30

Unique factorization domains

Motivation: 30.1 Fundamental Theorem of Arithmetic. If n ∈ Z, n > 1 then

n = p1 p2 · . . . · pk

where p1 , . . . , pk are primes. Moreover, this decomposition is unique up to reordering of factors.

Goal. Extend this to other rings.

30.2 Definition. Let R be an integral domain. An element a ∈ R is irreducible if a 6= 0, a is not a unit and if a = bc for some b, c ∈ R then either b or c is a unit.

30.3 Examples. 1) n ∈ Z is irreducible iff n = ±p where p is a prime number. 2) A field has no irreducible elements. 3) Take p(x) ∈ R[x], p(x) = x2 + 1. Then p(x) is irreducible in R[x]. 4) Take p(x) ∈ C[x], p(x) = x2 + 1. Then p(x) is not irreducible in C[x]: p(x) = (x − i)(x + i)

30.4 Note. If a ∈ R is an irreducible element and u ∈ R is a unit then ua is irreducible. 126

30.5 Definition. Let R be an integral domain. Elements a, b ∈ R are associates if a = ub for some unit u ∈ R. We write: a ∼ b. 30.6 Examples. 1) If m, n ∈ Z then m ∼ n iff m = ±n. 2) Check: units in R[x] are non-zero polynomials of degree 0. It follows that if p(x), q(x) ∈ R[x] then p(x) ∼ q(x) iff p(x) = aq(x) for some a ∈ R−{0}. 30.7 Definition. A unique factorization domain (UFD) is an integral domain R that satisfies the following conditions: 1) if a ∈ R is a non-zero, non-unit element then a = b 1 · . . . · bk

for some irreducible elements b1 , . . . , bk ∈ R

2) if b1 , . . . , bk , c1 , . . . , cl are irreducible elements such that b1 · . . . · bk = c 1 · . . . · c l

then k = l and for some permutation σ : {1, . . . , k} → {1, . . . , k} we have b1 ∼ cσ(1) , . . . , bk ∼ cσ(k) . 30.8 Examples. 1) Z is a UFD by the Fundamental Theorem of Arithmetic (30.1). 2) If F is a field then F is a UFD since all non-zero elements of F are units.

√ Recall. Z[ −5] is the subring of C given by √ √ Z[ −5] = {a + b 5i | a, b ∈ Z}

127

√ 30.9 Proposition. Z[ −5] is not a UFD. √ √ Proof. For a + b 5i ∈ Z[ −5] define

√ √ √ N (a + b 5i) = (a + b 5i)(a + b 5i) = a2 + 5b2 ∈ N

Notice that 1) N (α) = 1 iff α = ±1 2) N (α) = 0 iff α = 0

√ 3) N (α) 6= 3 for all α ∈ Z[ −5]

4) N (αβ) = N (α)N (β)

√ Observation 1. The only units in Z[ −5] are 1 and −1. As a consequence for any α, β we have α ∼ β iff α = ±β. √ Indeed, if α ∈ Z[ −5] is a unit then N (α)N (α−1 ) = N (αα−1 ) = N (1) = 1 Therefore N (α) = 1, and so α = ±1. √ Observation 2. If α ∈ Z[ −5] is an element such that N (α) = 9 then α is irreducible. Indeed, if α = ββ 0 then N (β)N (β 0 ) = N (α) = 9 Therefore N (β) must be either 1 (and so β is a unit), 3 (impossible), or 9 (and then N (β 0 ) = 1, i.e. β 0 is a unit) .

128

√ Take 9 ∈ Z[ −5]. We have

√ √ 3 · 3 = 9 = (2 + 5i)(2 − 5i) √ √ √ By Observation 2 the elements 3, 2 + 5i , 2 − 5i are irreducible in Z[ −5]. On the other hand by Observation 1 we obtain √ √ 3 6∼ (2 + 5i), 3 6∼ (2 − 5i) √ As a consequence 9 does not have a unique factorization in Z[ −5]

129

31

Prime elements

31.1 Definition. Let R be an integral domain, and let a, b ∈ R. We say that a divides b if b = ac for some c ∈ R. We then write: a | b.

31.2 Proposition. If R is an integral domain and a, b ∈ R then a ∼ b iff a | b and b | a. Proof. Exercise.

31.3 Definition. Let R is an integral domain. An element a ∈ R is a prime element if a 6= 0, a is a non-unit and if a | bc then either a | b or a | c.

31.4 Example. 1) In Z we have {prime elements} = {± prime numbers} = {irreducible elements} √ √ 2) By the proof of Proposition 30.9 in Z[ −5] the element α = 2 + 5i is irreducible. On the other hand α is not a prime element: α | (3 · 3)

but

α-3

31.5 Proposition. If R is an integral domain and a ∈ R is a prime element then a is irreducible.

Proof. Let a ∈ R be a prime element and let a = bc. We want to show that either b or c must be a unit in R. 130

We have a | (bc), and since a is a prime element it implies that a | b or a | c. We can assume that a | b. Since also b | a, thus by (31.2) we obtain that a ∼ b, i.e. a = bu for some unit u ∈ R. Therefore we have bc = a = bu By (25.6) this gives u = c, and so c is s unit.

31.6 Proposition. If R is a UFD and a ∈ R then a is an irreducible element iff a is a prime element.

Proof. (⇐) Follows from Proposition 31.5. (⇒) Assume that a ∈ R is irreducible and that a | (bc). We want to show that either a | b or a | c. If b = 0 then b = a · 0 so a | 0. If b is a unit then c = b−1 bc so a | c. As a consequence we can assume that b, c are non-zero, non-units. Since a | (bc) there is d ∈ R such that bc = ad. Assume that d is not a unit. Since R is a UFD we have decompositions: b = b1 · . . . · bm ,

c = c1 · . . . · cn , d = d 1 · . . . · d p

where bi , cj , dk are irreducible. This gives b1 · . . . ·bm · c1 · . . . ·cn = a · d1 · . . . ·dp By the uniqueness of decomposition in UFDs this implies that either a ∼ bi for some i or a ∼ cj for some j. In the first case we get a | b, and in the second case a | c. If d is a unit the argument is similar

131

31.7 Theorem. An integral domain R is a UFD iff 1) every non-zero, non-unit element of R is a product of irreducible elements 2) every irreducible element in R is a prime element. Proof. (⇒) Follows from the definition of UFD (30.7) and Propositon 31.6. (⇐) Assume that R satisfies conditions 1)-2) of the theorem. We only need to show that if b1 , . . . , bk , c1 , . . . , cl are irreducible elements in R such that b1 · . . . · bk = c 1 · . . . · c l

then k = l, and after reordering factors we have b1 ∼ c1 , . . . , bk ∼ ck . We argue by induction with respect to k. If k = 1 then we have b1 = c1 · . . . · cl . Since b1 is irreducible this implies that l = 1, and so b1 = c1 . Next, assume that the uniqueness property holds for some k and that we have b1 · . . . · bk · bk+1 = c1 · . . . · cl

where bi , cj are irreducible elements. This implies that bk+1 | (c1 · . . . · cl ). By condition 2) of the theorem bk+1 is a prime element. It follows that bk | cj for some 1 ≤ j ≤ l. We can assume that bk+1 | cl . Then cl = abk+1 for some a ∈ R. Also, since cl , bk+1 are irreducible a must be a unit. This shows that bk+1 ∼ cl . Furthermore, we obtain from here that b1 · . . . · bk · bk+1 = c1 · . . . · cl−1 · abk+1

Since R is an integral domain we get

b1 · . . . · bk = c1 · . . . · cl−1 a

Since bk is irreducible and a is a unit the product cl−1 a is an irreducible element. Therefore by the inductive assumption k = l − 1, and after reordering of factors we have b1 ∼ c1 , . . . , bk−1 ∼ ck−1 , bk ∼ cl−1 a ∼ cl−1

132

32

PIDs and UFDs

32.1 Theorem. If R is a PID then it is a UFD.

32.2 Lemma. If R is a PID and I1 , I2 , . . . are ideals of R such that I1 ⊆ I2 ⊆ . . . then there exists n ≥ 1 such that In = In+1 = . . . .

S Proof. Take I = ∞ i=1 Ii . Check: I is an ideal of R. Since R is a PID we have I = hai for some a ∈ I. Take n ≥ 1 such that a ∈ In . Then we get I ⊆ In ⊆ In+1 ⊆ . . . ⊆ I It follows that In = In+1 = . . . = I.

32.3 Lemma. Let R be a PID. An element a ∈ R is irreducible iff hai is a maximal ideal of R. Proof. Exercise.

32.4 Lemma. If R is an integral domain then a ∈ R is a prime element iff hai is a non-zero prime ideal of R. Proof. Exercise.

Proof of Theorem 32.1. Let R be a PID. By Theorem 31.7 it suffices to show that 133

1) every non-zero, non-unit element of R is a product of irreducible elements 2) every irreducible element in R is a prime element. 1) We argue by contradiction. Assume that a0 ∈ R is a non-zero, non-unit element that is not a product of irreducibles. This implies that a0 = a1 b 1 for some non-zero, non-unit elements a1 , b1 ∈ R. Next, if both a1 and b1 were products of irreducibles, then a0 would be also a product of irreducibles, contradicting our assumption. We can then assume that a1 is not a product of irreducibles, and so in particular we have a1 = a2 b 2 for some non-zero, non-unit elements a2 , b2 ∈ R. By induction we obtain that for i = 1, 2, . . . there exists non-zero, non-unit elements ai , bi ∈ R such that ai = ai+1 bi+1 for all i ≥ 0. Consider the chain of ideals ha0 i ⊆ ha1 i ⊆ . . . By Lemma 32.2 we obtain that han i = han+1 i for some n ≥ 0. This means that an = an+1 u for some unit u ∈ R (check!). As a consequence we obtain an+1 bn+1 = an = an+1 u and so bn+1 = u. This is a contradiction, since bn+1 is not a unit.

2) Let a ∈ R be an irreducible element and let a | (bc). We need to show that either a | b or a | c. Assume that a - b. This implies that b 6∈ hai and so hai = 6 hai + hbi 134

Since by Lemma 32.3 the ideal hai is a maximal ideal we obtain then that hai + hbi = R, and so in particular 1 ∈ hai + hbi. Therefore 1 = ar + bs for some r, s ∈ R, and so

c = a(rc) + (bc)s

Since a | a(rc) and a | (bc)s we obtain from here that a | c.

135

33

Application: sums of two squares

Recall. The ring of Gaussian integers: Z[i] = {a + bi | a, b ∈ Z} Z[i] is a Euclidean domain with the norm function N (a + bi) = a2 + b2 . In particular it follows that Z[i] is a UFD. 33.1 Note. For α, β ∈ Z[i] we have N (αβ) = N (α)N (β)

33.2 Proposition. An element α ∈ Z[i] is a unit iff N (α) = 1. Proof. Exercise.

33.3 Corollary. The only units in Z[i] are ±1 and ±i.

33.4 Proposition. If p ∈ Z is a prime number and for some α ∈ Z[i] we have N (α) = p then α is an irreducible element in Z[i].

Proof. Assume that α = βγ. Then we have p = N (α) = N (β)N (γ) Since p is a prime number we get that either N (β) = 1 or N (γ) = 1, and so either β or γ is a unit in Z[i].

136

33.5 Theorem. Let p be an odd prime. The following conditions are equivalent. 1) p = a2 + b2 for some a, b ∈ Z 2) p ≡ 1 (mod 4)

33.6 Lemma. If p ∈ Z is a prime number and p ≡ 1 (mod 4) then there is m ∈ Z such that m2 ≡ −1 (mod p)

33.7 Proposition. If p ∈ Z is a prime number then the groups of units (Z/pZ)∗ of the ring Z/pZ is a cyclic group of order (p − 1). Proof. See Propositon 38.8.

Proof of Lemma 33.6. By Proposition 33.7 we have (Z/pZ)∗ ∼ = Z/(p − 1)Z Since p ≡ 1 (mod 4), thus 4 | (p − 1), and so the group (Z/pZ)∗ has a cyclic subgroup of order 4. Let a ∈ Z/pZ be a generator of this subgroup. Then a2 is an element of order 2 in (Z/pZ)∗ , so a2 = −1. Take the canonical epimorphism π : Z → Z/pZ and let m ∈ π −1 (a). Then π(m2 ) = a2 = π(−1) which gives m2 ≡ −1 (mod p)

137

Proof of Theorem 33.5. 1) ⇒ 2) Assume that p = a2 + b2 . Since p is odd we can assume that a is even, a = 2m, and b is odd, b = 2n + 1. Then we have p = (2m)2 + (2n + 1)2 = 4m2 + 4n2 + 4n + 1 and so p ≡ 1 (mod 4).

2) ⇒ 1) Since p ≡ 1 (mod 4) by Lemma 33.6 there exists m ∈ Z such that p | (m2 + 1). In Z[i] we have m2 + 1 = (m + i)(m − i)

Therefore in Z[i] we have p | (m + i)(m − i). On the other hand p - (m ± i) since otherwise for some a, b ∈ Z we would have m ± i = p(a + bi) = pa + pbi and so pb = ±1 which is impossible. As a consequence we obtain that p is not a prime element in Z[i]. Since Z[i] is a UFD, prime elements in Z[i] coincide with irreducible elements and so p is not irreducible. Therefore p = αβ for some non-units α, β ∈ Z[i]. We have

p2 = N (p) = N (α)N (β)

By Lemma 33.2 we have N (α) 6= 1 6= N (β) so N (α) = N (β) = p. Therefore, if α = a + bi then p = a2 + b2 .

33.8 Note. 1) One can use factorization in Z[i] to show that, in general, a positive integer n is a sum of two squares iff n is of the form mr 2n1 2ns i n = 2k pm 1 . . .pr q1 . . .qs

where k, mi , nj ≥ 0, and pi , qj are prime numbers such that pi ≡ 1 (mod 4) and qj ≡ 3 (mod 4). 138

2) The ring Z[i] can be also used to describe all Pythagorean triples, e.i. triples of positive integers (x, y, z) that satisfy the equation x2 + y 2 = z 2 .

139

34

Application: Fermat’s Last Theorem

34.1 Fermat’s Last Theorem. For n ≥ 3 there are no integers x, y, z > 0 such that xn + y n = z n . Kummer’s idea of the proof. 1) It is enough to show that xn + y n = z n has no integral solutions for n = 4 and for n = p where p is an odd prime. Indeed, if n ≥ 3 is any integer then n = mk where either k = 4 or k is an odd prime, and if x = a, y = b, z = c would be a solution of xn + y n = z n then x = am , y = bm , z = cm would be a solution of xk + y k = z k

2) The case m = 4 was proved by Fermat. 3) Take an odd prime p. Let ζ = e2πi/p be a pth primitive root of 1, and let Z[ζ] be the smallest subring of C that contains ζ. For any x, y ∈ Z we have a factorization in Z[ζ]:

p−1 Y x +y = (x + ζ i y) p

p

i=0

Therefore, if x + y = z then in Z[ζ] we have p

p

p

p−1 Y z = (x + ζ i y) p

i=0

Assuming that Z[ζ] is a UFD we can compare these factorizations and try to get a contradiction. Problem. For some values of p (e.g. p = 23) the ring Z[ζ] is not a UFD. 140

35

Greatest common divisor

35.1 Definition. Let R be a ring and let a1 , . . . an ∈ R. We say that b ∈ R is a greatest common divisor of a1 , . . . , an if 1) b | ai for i = 1, . . . , n

2) if c | ai for i = 1, . . . , n then c | b. In such case we write b ∼ gcd(a1 , . . . , an ).

35.2 Note. 1) If b is a greatest common divisor of a1 , . . . , an and b0 ∼ b then b0 is also a greatest common divisor of a1 , . . . , an . 2) In general gcd(a √ 1 , . . . , an ) need not exists. E.g. gcd(9, 3(2 + not exist in Z[ −5].

√

5i)) does

35.3 Theorem. If R is a UFD then gcd(a1 , . . . , an ) exists for any a1 , . . . , an .

35.4 Lemma. If R is a ring such that gcd(a1 , a2 ) exists for any a1 , a2 ∈ R then gcd(a1 , . . . , an ) exists for any a1 , . . . , an ∈ R. Proof. Check: gcd(a1 , . . . , an ) ∼ gcd(gcd(a1 , . . . , an−1 ), an )

Proof of Theorem 35.3. Let a, b ∈ R. By Lemma 35.4 it is enough to show that gcd(a, b) exists. 141

If a = 0 then gcd(a, b) ∼ b (check!). If a is a unit then gcd(a, b) ∼ a (check!). Therefore we can assume that a, b are non-zero, non-unit elements of R. Since R is a UFD in such case we have a = uck11 ck22 . . .ckmm ,

b = vcl11 cl22 . . .clmm

where u, v are units, c1 , . . . , cm are distinct irreducible elements, and ki , li ≥ 0. Check: we have gcd(a, b) ∼ cn1 1 cn2 2 . . .cnmm where ni = min{ki , li } for i = 1, . . . m.

35.5 Definition. Let R be a ring. Elements a1 , . . . , an ∈ R are relatively prime if gcd(a1 , . . . , an ) ∼ 1.

35.6 Note. 1) The elements a1 , . . . , an are relatively prime iff every common divisor of a1 , . . . , an is a unit (check!). 2) If R is a UFD, a, b, c ∈ R, gcd(a, b) ∼ 1 and a | bc then a | c (check!).

142

36

Rings of fractions

Recall. If R is a PID then R is a UFD. In particular • Z is a UFD • if F is a field then F[x] is a UFD. Goal. If R is a UFD then so is R[x]. Idea of proof. 1) Find an embedding R ,→ F where F is a field. 2) If p(x) ∈ R[x] then p(x) ∈ F[x] and since F[x] is a UFD thus p(x) has a unique factorization into irreducibles in F[x]. 3) Use the factorization in F[x] and the fact that R is a UFD to obtain a factorization of p(x) in R[x].

36.1 Definition. If R is a ring then a subset S ⊆ R is a multiplicative subset if 1) 1 ∈ S

2) if a, b ∈ S then ab ∈ S

36.2 Example. Multiplicative subsets of Z: 1) S = Z

143

2) S0 = Z − {0} 3) Sp = {n ∈ Z | p - n} where p is a prime number. Note: Sp = Z − hpi.

36.3 Proposition. If R is a ring, and I is a prime ideal of R then S = R − I is a multiplicative subset of R.

Proof. Exercise.

36.4. Construction of a ring of fractions. Goal. For a ring R and a multiplicative subset S ⊆ R construct a ring S −1 R such that every element of S becomes a unit in S −1 R. Consider a relation on the set R × S: (a, s) ∼ (a0 , s0 )

if s0 (as0 − a0 s) = 0 for some s0 ∈ S

Check: ∼ is an equivalence relation. Elements of S −1 R = (equivalence classes of R × S under the relation ∼). Notation. a/s := the equivalence class represented by (a, s) Note. 1) If 0 ∈ S then (a, s) ∼ (a0 , s0 ) for all (a, s), (a0 , s0 ) ∈ R × S, and so S −1 R consists of only one element. 2) If R is an integral domain and 0 6∈ S then (a, s) ∼ (a0 , s0 ) iff as0 = a0 s. 144

Multiplication in S −1 R: (a/s)(a0 /s0 ) = (aa0 )/ss0 Addition in S −1 R: a/s + a0 /s0 = (as0 + a0 s)/ss0

Check: 1) The operations of addition and multiplication in S −1 R are well defined. 2) These operations define a commutative ring structure on S −1 R. 3) The map ϕS : R → S −1 R,

ϕS (r) = r/1

is a ring homomorphism.

Note. If s ∈ S then ϕS (s) = s/1 is a unit in S −1 R:

(s/1)(1/s) = s/s = 1/1 = 1S −1 R

36.5 Definition. If S is a multiplicative subset of a ring R then S −1 R is called the ring of fractions of R with respect to S.

36.6 Theorem. If S is a multiplicative subset of a ring R and f : R → R0 is a ring homomorphism such for ever s ∈ S the element f (s) ∈ R0 is a unit, then there exists a unique homomorphism f¯: S −1 R → R0 such that the following diagram commutes: R ϕS

f

/

f¯



S −1 R 145

=R

0

Proof. Define f¯(r/s) = f (r)f (s)−1 . Check that 1) f¯ is a well defined ring homomorphism 2) f¯ is the only homomorphism such that f = f¯ϕS .

36.7 Examples. 1) If S ⊆ Z, S = Z then S −1 Z ∼ = {0}. 2) If S0 ⊆ Z, S0 = Z − {0} then S0−1 Z ∼ = Q. 3) If Sp ⊆ Z, Sp = Z − hpi then Sp−1 Z is isomorphic to the subring of Q consisting of all fractions m such that p - n. n 4) If R = Z/6Z, S = {1, 3} ⊆ R then S −1 R ∼ = Z/2Z (check!).

36.8 Proposition. If S is a multiplicative subset of a ring R then Ker(ϕS ) = {a ∈ R | sa = 0 for some s ∈ S} Proof. We have ϕS (r) = 0/1 iff s(r · 1 − 0 · 1) = 0 for some s ∈ S.

36.9 Corollary. If R is an integral domain and S ⊆ R is a multiplicative subset such that 0 6∈ S then the homomorphism ϕS : R → S −1 R is 1-1. As a consequence in this case we can identify R with a subring of S −1 R. 146

36.10 Note. 1) If R is an integral domain and S = R − {0} then the ring S −1 R is a field. In this case S −1 R is called the field of fractions of R. 2) If I is a prime ideal of a ring R then the set S = R − I is a multiplicative subset of R. In this case the ring S −1 R is called the localization of R at I and it is denoted RI .

36.11 Definition. A ring R is a local ring if R has exactly one maximal ideal.

36.12 Examples. 1) If F is a field then it is a local ring with the maximal ideal I = {0}. 2) Check: if F is a field then F[[x]] is a local ring with the maximal ideal hxi. 3) Check: if F is a field then F[x] is not a local ring since for every a ∈ F the ideal hx − ai is a maximal ideal in F[x]. 4) Z is not a local ring.

36.13 Proposition. If R is a local ring then the maximal ideal J C R consists of all non-units of R.

Proof. Since J 6= R thus J does not contain any units. Conversely, if a ∈ R is a non-unit then 1 6∈ hai so hai 6= R. Therefore by Theorem 29.1 hai is contained in some maximal ideal of R. Since J is the only maximal ideal we obtain hai ⊆ J, and so a ∈ J.

147

36.14 Proposition. If R is a ring and I C R is a prime ideal then the ring RI is a local ring, and the maximal ideal J C RI is given by J := {a/s | a ∈ I, s 6∈ I} Proof. Exercise.

148

37

Factorization in rings of polynomials

Goal: 37.1 Theorem. If R is a UFD then so is R[x].

37.2 Lemma. If R is an integral domain then p(x) ∈ R[x] is a unit in R[x] iff deg p(x) = 0 and p(x) = a0 where a0 is a unit in R.

Proof. Exercise.

37.3 Lemma. If R is an integral domain and p(x) = a0 is a polynomial of degree 0 in R[x] then p(x) is irreducible in R[x] iff a0 is irreducible in R.

Proof. Exercise.

37.4 Definition. If R is a UFD and p(x) = a0 + a1 x + . . . an xn ∈ R[x], then p(x) is a primitive polynomial if gcd(a0 , . . . , an ) ∼ 1.

37.5 Lemma. If R is a UFD and p(x) ∈ R[x] is an irreducible polynomial such that deg p(x) > 0 then p(x) is a primitive polynomial.

Proof. If p(x) = a0 + a1 x + . . . an xn is not primitive then p(x) = gcd(a0 , . . . an )q(x) for some q(x) ∈ R[x], deg q(x) > 0. Since both gcd(a0 , . . . , an ) and q(x) are non-units in R[x] the polynomial p(x) is not irreducible. 149

37.6 Lemma (Gauss). If R is a UFD and p(x), q(x) ∈ R[x] are primitive polynomials then p(x)q(x) is also primitive. Proof. Assume that r(x) = p(x)q(x) is not primitive. Then we have r(x) = c · r˜(x) where c ∈ R is an irreducible element, and r˜(x) ∈ R[x]. Take the canonical epimorphism π : R −→ R/hci

This defines a homomorphism of rings of polynomials π ¯ : R[x] −→ (R/hci)[x] given by

π ¯ (a0 + a1 x + . . . + an xm ) := π(a0 ) + π1 (a)x + . . . + π(an )xn Note: 1) Since c is irreducible element thus hci is a prime ideal and so R/hci is a domain. As a consequence (R/hci)[x] is a domain. 2) We have π ¯ (p(x))¯ π (q(x)) = π ¯ (c · r˜(x)) = 0 · π ¯ (˜ r(x)) = 0 so either π ¯ (p(x)) = 0 or π ¯ (q(x)) = 0. We can assume that π ¯ (p(x)) = 0. Then p(x) = c · p˜(x) for some p˜(x) ∈ R[x]. Since p(x) is primitive we get a contradiction.

37.7 Lemma. Let R be a UFD and K be the field of fractions of R. 1) For any non-zero polynomial p(x) ∈ K[x] there is an element c(p) ∈ K and a primitive polynomial p∗ (x) ∈ R[x] such that p(x) = c(p)p∗ (x) 150

2) If p(x) = c(p)p∗ (x) and p(x) = c˜(p)˜ p∗ (x) for some c(p), c˜(p) ∈ K, and ∗ ∗ some primitive polynomials p (x), p˜ (x) ∈ R[x] then there exists a unit u ∈ R such that p˜∗ (x) = u−1 p∗ (x)

c˜(p) = uc(p),

3) If p(x), q(x) ∈ K[x] are non-zero polynomials then for some unit u ∈ R we have c(pq) = uc(p)c(q)

(p(x)q(x))∗ = u−1 p∗ (x)q ∗ (x)

37.8 Definition. If p(x) ∈ K[x] then the element c(p) ∈ K is called the content of p(x).

Proof of Lemma 37.7. 1) If p(x) ∈ K[x] then there is there exists 0 6= c ∈ R such that cp(x) ∈ R[x]. Let b ∈ R be a greatest common divisor of coefficients of cp(x). Take p∗ (x) = c/b · p(x),

c(p) = b/c

Check that p∗ (x) ∈ R[x] and p∗ (x) is a primitive polynomial. 2) We have c(p)p∗ (x) = p(x) = c˜(p)˜ p∗ (x) where c(p), c˜(p) ∈ K are non-zero elements and p∗ (x), p˜∗ (x) ∈ R[x] are primitive polynomials. Let p∗ (x) = a0 + a1 x + . . . + an xn . We have p˜∗ (x) = c(p)˜ c(p)−1 p∗ (x) Since c(p)˜ c(p)−1 ∈ K there b, d ∈ R such that c(p)˜ c(p)−1 = b/d. We can assume that gcd(b, d) ∼ 1. We have d˜ p∗ (x) = bp∗ (x) = ba0 + ba1 x + . . . + ban xn 151

This gives: d | bai for i = 1, . . . , n. By (35.6) this gives that d | ai for all i. Since p∗ (x) is a primitive polynomial it implies that d is a unit in R. By an analogous argument we obtain that b is also a unit in R. As a consequence u = db−1 is a unit in R and we have c˜(p) = uc(p),

p˜∗ (x) = u−1 p∗ (x)

3) For p(x), q(x) ∈ K[x] we have c(p)c(q)p∗ (x)q ∗ (x) = p(x)q(x) = c(pq)(p(x)q(x))∗ We have c(p)c(q), c(pq) ∈ K, (p(x)q(x))∗ ∈ R[x] is primitive by construction and p∗ (x)q ∗ (x) ∈ R[x] is primitive by Lemma 37.6. Applying part 2) we obtain that there is a unit u ∈ R such that c(pq) = uc(p)c(q)

(p(x)q(x))∗ = u−1 p∗ (x)q ∗ (x)

37.9 Note. Check: 1) Let p(x) ∈ K[x]. Then p(x) ∈ R[x] iff c(p) ∈ R. 2) p(x) ∈ R[x] is primitive iff p(x) = up∗ (x) for some unit u ∈ R.

37.10 Proposition. Let R be a UFD and K be the ring of fractions of R. 1) A polynomial p(x) ∈ K[x] of non-zero degree is irreducible in K[x] iff p∗ (x) is irreducible in R[x]. 2) A polynomial p(x) ∈ R[x] of non-zero degree is irreducible in R[x] iff p(x) is primitive and it is irreducible in K[x].

152

Proof. 1) (⇐) If p(x) is not irreducible in K[x] then p(x) = q(x)r(x) for some q(x), r(x) ∈ K[x], deg q(x), deg r(x) > 0. By Lemma 37.7 we have p∗ (x) = uq ∗ (x)r∗ (x) for some unit u ∈ R. Therefore p∗ (x) is not irreducible in R[x]. (⇒) If p∗ (x) is not irreducible in R[x] then p∗ (x) = q(x)r(x) where q(x), r(x) are non-units in R[x]. Since p∗ (x) is primitive we must have deg q(x), deg r(x) > 0. Then p(x) = c(p)q(x)r(x) and so p(x) is not irreducible in K[x]. 2) (⇒) If p(x) ∈ R[x] is irreducible then it must be a primitive polynomial. Therefore p(x) = up∗ (x) for some unit u ∈ R. Since p(x) is irreducible in R[x], thus p∗ (x) is also irreducible in R[x], and so by part 1) p(x) is irreducible in K[x]. (⇐) Since p(x) is irreducible in K[x] by part 1) we get that p∗ (x) is irreducible in R[x]. Also, since p(x) is primitive we have p(x) = up∗ (x) for some unit u ∈ R. It follows that p(x) is irreducible in R[x].

Proof of Theorem 37.1. By Theorem 31.7 we need to show that: 1) Every non-zero, non-unit p(x) ∈ R[x] is a product of irreducible polynomials. 2) Every irreducible polynomial p(x) ∈ R[x] is a prime element of R[x]. 153

1) Recall that irreducible polynomials of degree 0 in R[x] correspond to irreducible elements of R. Since R is a UFD it follows that if p(x) ∈ R[x] has degree 0 then p(x) is a product of irreducibles in R[x]. If deg p(x) > 0 then p(x) is a non-zero non-unit element of K[x]. Since K[x] is a UFD we have p(x) = q1 (x)q2 (x). . .qk (x) where q1 (x), . . . , qk (x) are irreducible polynomials in K[x]. We have p(x) = (c(q1 )c(q2 ). . .c(qk ))q1∗ (x)q2∗ (x). . .qk∗ (x) By Proposition 37.10 qi∗ (x) are irreducible polynomials in R[x]. Moreover, by Lemma 37.7 we have c(q1 )c(q2 ). . .c(qk ) = uc(p) for some unit u ∈ R, and also since p(x) ∈ R[x] we have c(p) ∈ R. Therefore c(q1 )c(q2 ). . .c(qk ) ∈ R, and since R is a UFD we have c(q1 )c(q2 ). . .c(qk ) = a1 a2 . . .al where a1 , . . . , al ∈ R are irreducible elements in R[x]. As a consequence we obtain a factorization of p(x): p(x) = a1 a2 . . .al q1∗ (x)q2∗ (x). . .qk∗ (x) where a1 , . . . , al , q1∗ (x), . . . , qk∗ (x) are irreducible elements in R[x]. 2) Let p(x) ∈ R[x] be an irreducible polynomial. We need to show that if for some q(x), r(x) ∈ R[x] we have p(x) | q(x)r(x) then either p(x) | q(x) or p(x) | r(x). Exercise: this holds if deg p(x) = 0. If deg p(x) > 0, then since p(x) is irreducible in R[x] by Proposition 37.10 we obtain that p(x) is primitive and it is irreducible in K[x]. Since K[x] is a UFD we obtain that p(x) is a prime element of K[x], and so p(x) | q(x) or p(x) | r(x) in K[x]. We can assume that p(x) | q(x) in K[x]. Then there exists h(x) ∈ K[x] such that p(x)h(x) = q(x) 154

We have c(p)c(h) = uc(q) for some unit u ∈ R. Also, since q(x) ∈ R[x] we have that c(q) ∈ R and therefore c(p)c(h) ∈ R. Finally, since p(x) is primitive thus c(p) is a unit in R. We obtain: c(h) = c(p)−1 uc(q) ∈ R Therefore h(x) ∈ R[x], and so p(x) | q(x) in R[x]. 37.11 Corollary. Z[x] is a UFD

37.12 Corollary. If R is a UFD then the ring of polynomials of n variables R[x1 , . . . , xn ] is also a UFD.

Proof. It is enough to notice that R[x1 , . . . , xn ] ∼ = R[x1 , . . . , xn−1 ][xn ] (check!).

155

38

Irreducibility criteria in rings of polynomials

38.1 Theorem. Let p(x), q(x) ∈ R[x] be polynomials such that p(x) = a0 + a1 x + . . . + an xn ,

q(x) = b0 + b1 x + . . . + bm xm

and an , bm 6= 0. If bm is a unit in R then there exist unique polynomials r(x), s(x) ∈ R[x] such that p(x) = s(x)q(x) + r(x) and either deg r(x) < deg q(x) or r(x) = 0.

Proof. Exercise (or see Hungerford p.158).

38.2 Definition. If R is a ring and p(x) ∈ R[x] then p(x) defines a function p : R −→ R,

a 7→ p(a)

A function of this form is called a polynomial function.

38.3 Note. Different polynomials may define the same polynomial function. E.g. if p(x) = x + 1, q(x) = x2 + 1 are polynomials in Z/2Z[x] then p(x), q(x) define the same function Z/2Z → Z/2Z: p(0) = q(0) = 1,

p(1) = q(1) = 0

38.4 Definition. Let p(x) ∈ R[x]. An element a ∈ R is a root of p(x) if f (a) = 0.

156

38.5 Proposition. An element a ∈ R is a root of p(x) ∈ R[x] iff (x − a) | p(x). Proof. (⇐) If (x − a) | p(x) then p(x) = q(x)(x − a) for some q(x) ∈ R[x] so p(a) = q(a)(a − a) = 0. (⇒) Assume that p(a) = 0. By Theorem 38.1 we have p(x) = s(x)(x − a) + r(x) where deg r(x) = 0, so r(x) = b for some b ∈ R. This gives 0 = p(a) = s(a)(a − a) + b Thus b = 0, and so p(x) = s(x)(x − a).

38.6 Corollary. If R is an integral domain and 0 6= p(x) ∈ R[x] is a polynomial of degree n then R[x] has at most n distinct roots in R.

Proof. Let a1 , . . . , ak ∈ R be all distinct roots of p(x). By (38.5) we have p(x) = (x − a1 )q1 (x) for some q1 (x) ∈ R[x]. Also we have 0 = p(a2 ) = (a2 − a1 )q1 (a2 ) Since R is an integral domain and a2 − a1 6= 0 we obtain q1 (a2 ), and so q2 (x) = (x − a2 )q3 (x) for some q3 (x) ∈ R[x]. This gives p(x) = (x − a1 )(x − a2 )q3 (x) 157

By induction we obtain p(x) = (x − a1 ) · . . .·(x − ak )qk (x) for some 0 6= qk (x) ∈ R[x]. This gives deg p(x) = deg((x − a1 ) · . . .·(x − ak )qk (x)) ≥ k

38.7 Note. 1) Corollary 38.6 in not true if R is not an integral domain. E.g. if R = Z/6Z and p(x) = x2 + x then 0, 2, 3 ∈ Z/6Z are roots of p(x). 2) Corollary 38.6 is not true is R is a non-commutative ring (even if R has no zero divisors). For example, if R = H and p(x) = x2 +1 then ±i, ±j, ±k ∈ H are roots of p(x).

38.8 Proposition. Let R is an integral domain. If G is a finite subgroup of the multiplicative group of units of R then G is a cyclic group.

Proof. Let a ∈ G be an elements of a maximal order in G. We will show that hai = G. We argue by contradiction. Assume that there exists an element b ∈ G − hai, and let |b| = m. By assumption m ≤ |a|. If m - |a| then |ab| > |a| (check!) which is impossible by the choice of a. Therefore |a| = km for some k > 0. This shows that there are k + 1 distinct elements of order m in G: b, ak , a2k , . . . , a(m−1)k . This is however impossible since each of these elements would be a root of the polynomial f (x) = xm − 1 ∈ R[x] and by Corollary 38.6 this polynomial has at most m roots in R.

38.9 Proposition. If F is a field and p(x) ∈ F[x] is a polynomial such that deg p(x) > 1 and p(x) has a root in F then p(x) is not irreducible in F[x]. 158

Proof. By (38.6) we have p(x) = q(x)(x − a) for some q(x) ∈ F[x]. Since deg p(x) > 1 we have deg q(x) > 0, so q(x) and (x − a) are not units in R[x].

38.10 Corollary. Let R be a UFD and let K be the field of fractions of R. If p(x) ∈ R[x] is a polynomial such that deg p(x) > 1 and p(x) has a root in K then p(x) is not irreducible in R[x].

Proof. By (38.6) p(x) is not irreducible in K[x], so by (37.10) it is also not irreducible in R[x].

38.11 Proposition (Integral root test). Let R be a UFD, let K be the field of fractions of R and let p(x) ∈ R[x] be a polynomial p(x) = a0 + a1 x + . . . + an xn where an 6= 0. If a ∈ K is a root of p(x) then a is of the form a = b/s where b, s ∈ R, gcd(b, s) ∼ 1, b | a0 and s | an . In particular, in an = 1 then a ∈ R and a | a0 . Proof. Exercise.

38.12 Theorem (Eisenstein Irreducibility Criterion). Let R be a UFD. If p(x) = a0 + a1 x + . . . + an xn is a primitive polynomial in R[x] such that deg p(x) > 0, and b ∈ R is an irreducible element b ∈ R such that 1) b - an 2) b | ai for all i < n 159

3) b2 - a0 then p(x) is irreducible in R[x]. Proof. Assume that p(x) is not irreducible in R[x]. Then we have p(x) = q(x)r(x) for some non-units q(x), r(x) ∈ R[x]. Since p(x) is primitive we must have deg q(x), deg r(x) > 0. Let q(x) = c0 + c1 x + . . . + ck xk ,

r(x) = d0 + d1 x + . . . + dl xl

Notice that since b is irreducible it is a prime element of R and so by (32.4) the ideal hbi is a prime ideal of R. As a consequence R/hbi is an integral domain. Consider the canonical epimorphism π : R → R/hbi and the induced homomorphism of rings of polynomials π ˜ : R[x] −→ R/hbi[x] By assumption on p(x) we have π ˜ (p(x)) = π(an )xn On the other hand we have π ˜ (p(x)) = π ˜ (q(x))˜ π (r(x)) Check: since R/hbi is an integral domain we must have π ˜ (q(x)) = π(ck )xk ,

π ˜ (r(x)) = π(dl )xl

In particular π(c0 ) = π(d0 ) = 0, so b | c0 and b | d0 . On the other hand a0 = c0 d0 , so b2 | a0 which contradicts the assumption on a0 . 38.13 Example. If p ∈ Z is a prime number then q(x) = xn − p is an irreducible polynomial in Z[x]. Note: by (37.10) q(x) is also irreducible in Q[x]. This shows in particular that √ q(x) has no roots in Q, and so that n p is an irrational number for all primes p and all n > 1. 160

38.14 Proposition. Let R be an integral domain and let c ∈ R. Pn PnA polynomial i p(x) = i=0 ai x is irreducible iff the polynomial p(x − c) = i=0 (x − c)i is irreducible.

Proof. It is enough to notice that the map f : R[x] → R[x],

f (p(x)) = p(x − c)

is an isomorphism of rings.

38.15 Example. Let p ∈ Z be a prime number, and let q(x) = xp−1 + xp−2 + . . . + x + 1 We will show that q(x) is irreducible in Z[x]. We have q(x) =

xp − 1 x−1

This gives (x + 1)p − 1 (x + 1) − 1 (x + 1)p − 1 = x      p p−2 p p−3 p p−1 =x + x + x + ... + 1 2 p−1

q(x + 1) =



p Since p | kp for k = 1, . . . , p − 1 and p2 - p−1 the polynomial q(x + 1) is irreducible in Z[x], and so q(x) is also irreducible.

161

39

Modules

39.1 Definition. Let R be a (possibly non-commutative) ring. A left R-module is an abelian group M together with a map R × M → M,

(r, m) 7→ rm

satisfying the following conditions: 1) r(m + n) = rm + rn 2) (r + s)m = rm + sm 3) (rs)m = r(sm) 4) If R is a ring with identity 1 ∈ R then 1m = m for all m ∈ M . A right R-module is defined analogously.

39.2 Definition. If M, N are left R-modules then a map f : M −→ N is a left R-modules homomorphism if f is a homomorphism of abelian groups and f (rm) = rf (m) for all r ∈ R, m ∈ M .

39.3 Note. Left R-modules and their homomorphisms form a category R-Mod. Analogously, right R-modules form a category Mod-R.

39.4 Examples. 1) If I is a left ideal of R then I is a left module of R. In particular R is a left R-module. 2) If F is a field then left (or right ) F-modules are vector spaces over F, and homomorphisms of F-modules are F-linear maps. 162

3) The category of left (or right) Z-modules is isomorphic to the category of abelian groups: if G is an abelian group then G has a natural Z-module structure such that for n ∈ Z and g ∈ G we have  n times   z }| {   g + ··· + g for n > 0    0 for n = 0 ng =    (−g) + · · · + (−g) for n < 0   {z }   | |n| times

4) Let G be an abelian group, and let R = Hom(G, G) be the ring of homomorphisms of G (with the usual addition of homomorphims and multiplication given by composition of homomorphisms). We have a map R × G → G,

ϕ · g = ϕ(g)

Check: this defines a left R-module structure on G. Note: the multiplication G × R → G,

g · ϕ = ϕ(g)

does not define a right module structure on G. Indeed, for ϕ, ψ ∈ R we have: (g · ψ) · ϕ = (ψ(g)) · ϕ = ϕ(ψ(g)) One the other hand g · (ψ · ϕ) = ψ(ϕ(g)). Since in general ψ(ϕ(g)) 6= ϕ(ψ(g)) we get that (g · ψ) · ϕ 6= g · (ψ · ϕ)

39.5 Note. For a ring R define a ring Rop as follows: • Rop = R as abelian group • r ·Rop s := sr

163

We have: (left R-modules) = (right Rop -modules) (check!). In particular, if R is a commutative ring then R = Rop , and so (left R-modules) = (right R-modules)

Note. From now on by an R-module we will understand a left R-module.

164

40

Basic operations on modules

40.1 Definition. If M in an R-module then a submodule of M is an additive subgroup N ⊆ M such that if r ∈ R and n ∈ N then rn ∈ N .

40.2 Note. If f : M → N is a homomorphism of R-modules then Ker(f ) := f −1 (0) is a submodule of M and Im(f ) := f (M ) is a submodule of N .

40.3 Definition. If M is an R-module and S is a subset of M then the module generated by S is the submodule hSi ⊆ M that is the smallest submodule of M containing S. If hSi = M then we say that the set S generates M . A module M is finitely generated if M = hSi for some finite set S.

40.4 Note. If M is an R-module and S ⊆ M then hSi = {r1 m1 + . . . + rk mk | ri ∈ R, mi ∈ S}

40.5 Definition. If M is an R-module and N ⊆ M is a submodule then the quotient module M/N is the quotient abelian group with multiplication defined by r(m + N ) := rm + N for r ∈ R, m + N ∈ M/N .

40.6 First Isomorphism Theorem. If f : M → N is a homomorphism of R-modules that is onto then M/ Ker(f ) ∼ =N 165

Proof. Similar to the proof of Theorem 6.1 for groups.

40.7 Definition. If {Mi }i∈I is a family of R-modules then the direct product of {Mi }i∈I is the module Y Mi = {(mi )i∈I | mi ∈ Mi } i∈I

with addition and multiplication by R defined coordinatewise. L Q The direct sum of {Mi }i∈I is the submodule i∈I Mi of i∈I Mi given by M i∈I

Mi := {(mi )i∈I | mi 6= 0 for finitely many i only }

40.8 Note. Recall the notions of categorical products and copruducts (Section 12). Check: Q i∈I Mi is the categorical product of the family {Mi }i∈I in the category R-Mod. L i∈I Mi is the categorical coproduct of {Mi }i∈I in the category R-Mod.

166

41

Free modules and vector spaces

41.1 Definition. Let M be an R-module. A set S ⊆ M is linearly independent if for any distinct m1 , . . . , mk ∈ S we have r1 m1 + . . . + rk mk = 0 only if r1 = . . . = rk = 0.

41.2 Definition. Let M be an R-module. A set B ⊆ M is a basis of M if B is linearly independent and B generates M .

41.3 Definition. An R-module M is a free module if M has a basis.

41.4 Theorem. Let R be a ring with identity 1 6= 0 and let F be an R-module. The following conditions are equivalent. 1) F is a free module. L 2) F ∼ = i∈I R for some set I. 3) There is a non-empty subset B ⊆ F satisfying the following universal property. For any R-module M and any map of sets f : B → M there is a unique R-module homomorphism f¯: F → N such that the following diagram commutes: f

B i

/M >

f¯



F Here i : B ,→ F is the inclusion map. Proof. Exercise. 167

41.5 Note. Let R be a ring with a an identity and let U : R-M od → Set be the forgetful functor. Check: U has a left adjoint functor FRM od : Set → R-M od One can show that an R-module M is free iff M ∼ = FRM od (S) for some set S (exercise).

41.6 Note. We have: (free Z-modules) = (free abelian groups)

41.7 Theorem. If R is a division ring then every R-module is free.

Proof. Let M be an R-module. It is enough to show that M has a basis. Let S be the set of all linearly independent subsets of M ordered with respect to inclusion of subsets. Claim 1. S has a maximal element. Indeed, by Zorn’s Lemma (29.10) it is enough to show that every chain in S has an upper bound. Let then T = {Bi }i∈I S be a chain in S. Take B := i∈I Bi . We have Bi ⊆ B for all i ∈ I, so it suffices check that B is a linearly independent. Let then b1 , . . . , bk ∈ B, and assume that r1 b1 + . . . + rk bk = 0 We need to show that r1 = . . . = rk = 0. We have b1 ∈ Bi1 , . . . , bk ∈ Bik for some i1 , . . . , ik ∈ I. Since {Bi }i∈I is a chain we can assume that Bi1 ⊆ Bi2 ⊆ . . . ⊆ Bik 168

As a consequence b1 , . . ., bk ∈ Bik , and since Bik is a linearly independent set we get that r1 = . . . = rk = 0. Claim 2. If B is a maximal element in S then B is a basis of M . Indeed, by the definition of S the set B is linearly independent so it is enough to show that hBi = M . Assume that this is not true, and let m ∈ M − hBi. Take the set B 0 = B ∪ {m}

Notice that the set B 0 is linearly independent. To see this, assume that for some b1 , . . . , bk ∈ B and r1 , . . . rk , s ∈ R we have r1 b1 + . . . + rk bk + sm = 0 If s 6= 0 then s is a unit (since R is a division ring) and so m = (−s−1 r1 )b1 + . . . + (−s−1 rk )bk This is however impossible since m 6∈ hBi. Therefore s = 0, and so r1 b1 + . . . + rk bk = 0 Linear independence of B gives then s = r1 = . . . = rk = 0 As a consequence we get that B 0 ∈ S and B ( B 0 . This is however a contradiction since B is a maximal element of S.

41.8 Note. Let R be a division ring and let M be an R-module. By a similar argument as in the proof of Theorem 41.7 we can show that: 1) if V ⊆ M is a linearly independent set then there is a basis B of M such that V ⊆ B; 2) if V ⊆ M is a set generating M then then there is a basis B of M such that B ⊆ V . 169

41.9 Note. 1) For a general ring R it is not true that a linearly independent subset V of a free R-module F can be always extended to a basis. Take e.g. R = Z,

F = Z,

V = {2}

2) It is also not true in general that if V is a set generating a free R-module then V contains a basis of F . Take e.g. R = Z,

F = Z,

170

V = {2, 3}

42

Invariant basis number

42.1 Definition. A ring R has the invariant basis number (IBN) property if for any free R-module F and for any bases two B, B 0 of F we have |B| = |B 0 |.

42.2 Definition. If a ring R has IBN then for a free R-module F the rank of F is the cardinality of a basis of F .

42.3 Example. Since free Z-modules correspond to free abelian groups by Proposition 13.3 the ring of integers Z has IBN.

42.4 Notation. For a ring R and n > 0 denote Rn :=

Ln

i=1

R.

42.5 Example. Let F be a field and let V be an F-vector space with an infinite, countable basis. Let R be the ring of all linear maps V → V : R = HomF (V, V ) with pointwise addition and with multiplication given by composition. We have Rn ∼ = Rm for every m, n ≥ 0 (exercise). Thus R does not have IBN.

42.6 Theorem. Let R be a ring with identity and let F be a free R-module. If F has an infinite basis B then for any other basis B 0 of F we have |B| = |B 0 |.

42.7 Corollary. Let R be a ring with identity. The following conditions are equivalent. 171

1) R has IBN 2) If F is a free R module with two finite bases B and B 0 then |B| = |B 0 |. 3) For any m, n > 0 if Rm ∼ = Rn then m = n.

Proof. Follows directly from Theorem 42.6.

Proof of Theorem 42.6. Let F be a free R module with an infinite basis B. Let B 0 be any other basis of F . Claim 1. The basis B 0 is infinite. Indeed, assume that B 0 is finite. Since F = hBi thus every element of B 0 is a linear combination of a finite number of elements of B and so B 0 ⊆ {b1 , . . . , bn } where {b1 , . . . , bn } is some finite subset of B. This gives hb1 , . . . , bn i ⊇ hB 0 i = F so hb1 , . . . , bn i = F . Since B is an infinite set there is b ∈ B such that b 6∈ {b1 , . . . , bn }. On the other hand b ∈ F = hb1 , . . . , bn i. This is a contradiction since B is a linearly independent set.

Next, assume that B, B 0 are two infinite bases of F . We can also assume that |B 0 | ≤ |B|. Claim 2. Let T = {b01 , . . . , b0k } be a finite subset of B 0 and let BT := {b ∈ B | b ∈ hT i} Then BT is a finite subset of B. Indeed, each b0i is a linear combination of a finite number of elements of B and so T ⊆ hb1 , . . . , bn i where {b1 , . . . , bn } is some finite subset of B. This gives BT ⊆ hT i ⊆ hb1 , . . . , bn i 172

By linear independence of B we must then have BT ⊆ {b1 , . . . , bn }

Claim 3. |B| ≤ |B 0 |. Indeed, let SB 0 be the set of all finite subsets of B 0 . Note that since B 0 is an infinite set we have |SB 0 | = |B 0 |. We have a map of sets f : B → SB 0

such that f (b) = {b01 , . . . , b0k } if we have

b = r1 b01 + · · · + rk b0k for some non-zero elements r1 , . . . , rk ∈ R. Since B 0 is a basis of F this map is well defined. Notice that for T ∈ SB 0 we have b ∈ f −1 (T ) iff b ∈ hT i and so by Claim 2 the set f −1 (T ) is finite for all T ∈ SB 0 . As a consequence we obtain [ [ |B| = | f −1 (T )| ≤ | N| = |SB 0 | · ℵ0 = |B 0 | · ℵ0 T ∈SB 0

T ∈SB 0

Since B 0 is an infinite set we have |B 0 | · ℵ0 = |B 0 |, and so |B| ≤ |B 0 |.

Since by assumption we had |B| ≤ |B 0 | and by Claim 3 we have |B| ≤ |B 0 | we obtain that |B| = |B 0 |.

42.8 Theorem. If R is a division ring then R has IBN.

173

Proof. By Corollary 42.7 it is enough to show that if F is a free R-module with two finite bases B = {b1 , . . . , bn } and B 0 = {b01 , . . . , b0m } then n = m. We will argue by induction with respect to n Assume that n = 1, and so B = {b1 }. If B 0 = {b01 , . . . , b0m } for some m > 1 then we have b01 = r1 b1 and b02 = r2 b1 for some r1 , r2 ∈ R, r1 , r2 6= 0. Therefore r1−1 b01 − r2−1 b02 = 0 which contradicts the assumption that B 0 is a inearly independent set. As a consequence we must have m = 1. Next, assume that n ≥ 1 is a number such that if a free R-module has a basis consisting of n elements then every other basis of that module also has n elements. Let F be a free R-module with a basis B = {b1 , . . . , bn+1 } consisting of n + 1 elements and let B 0 = {b01 , . . . , b0m } be another basis of F . Since hB 0 i = F we have bn+1 = r1 b01 + . . . + rm b0m for some r1 , . . . , rm ∈ R. Also, since bn+1 6= 0 we have ri 6= 0 for some i. We can assume that rm 6= 0. Let B 00 := {b01 , . . . , b0m−1 , bn+1 }. Check: B 00 is a basis of F . Take the canonical epimorphism π : F → F/hbn+1 i Check: since F is a free module with basis B := {b1 , . . . , bn , bn+1 }, thus F/hbn+1 i is a free module with basis {π(b1 ), . . . , π(bn )}. On the other hand, since F has a basis B 00 := {b01 , . . . , b0m−1 , bn+1 } therefore {π(b01 ), . . . , π(b0m−1 )} is a basis of F/hbn+1 i. By the inductive assumption we obtain than n = m − 1, and so n + 1 = m

174

42.9 Note. Let I be an ideal of R and let M be an R-module. Define: IM := {rm | r ∈ I, m ∈ M } Check: 1) IM a submodule of M . 2) M/IM has a structure of a R/I-module with the multiplication given by (r + I)(m + IM ) = rm + IM

42.10 Theorem. Let R be a ring with identity and let I 6= R be an ideal of R. If R/I has IBN then R also has IBN. Proof. Let F be a free R-module and let B = {b1 , . . . , bn } be a basis of F . Check: F/IF is a free R/I-module with basis {b1 + IF, . . . , bn + IF }. Since R/I has IBN any basis of F/IF has n elements. As a consequence any basis of F also has n elements.

42.11 Corollary. If R is a commutative ring with identity 1 6= 0 then R has IBN. Proof. Let I be a maximal ideal in R. Then R/I is a field and we have the canonical homomorphism π : R → R/I By Theorem 42.8 R/I has IBN, so by Theorem 42.10 R also has IBN.

42.12 Note. Corollary 42.11 can be generalized as follows. If f: R →S is an epimorphism of rings of identity such that S is a division ring then R has IBN. 175

43

Projective modules

43.1 Note. If F is a free R-module and P ⊆ F is a submodule then P need not be free even if P is a direct summand of F . Take e.g. R = Z/6Z. Notice that Z/2Z and Z/3Z are Z/6Z-modules and we have an isomorphism of Z/6Z-modules: Z/6Z ∼ = Z/2Z ⊕ Z/3Z Thus Z/2Z and Z/3Z are non-free modules isomorphic to direct summands of the free module Z/6Z.

43.2 Definition. An R-module P is a projective module if there exists an Rmodule Q such that P ⊕ Q is a free R-module.

43.3 Examples. 1) If R is a ring with identity then every free R-module is projective. 2) Z/2Z and Z/3Z are non-free projective Z/6Z-modules.

43.4 Definition. Let fi+1

fi

. . . −→ Mi −→ Mi+1 −→ Mi+2 −→ . . . be a sequence of R-modules and R-module homomorphisms. This sequence is exact if Im(fi ) = Ker(fi+1 ) for all i.

43.5 Definition. A short exact sequence is an exact sequence of R-modules the form f g 0 −→ N −→ M −→ K −→ 0

(where 0 is the trivial R-module).

176

43.6 Note. f

g

1) A sequence 0 → N −→ M −→ K → 0 is short exact iff • f is a monomorphism • g is an epimorphism • Im(f ) = Ker(g).

2) If M 0 is a submodule of M then we have a short exact sequence 0 −→ M 0 −→ M −→ M/M 0 −→ 0 Morever, up to an isomorphism, every short exact sequence is of this form: /

0

∼ =

/

0

f

N

/

=



/

Ker(g)

g

M 

/

M

/

/

K 

0

∼ =

/

M/ Ker(g)

0

43.7 Definition. A short exact sequence f

g

0 −→ N −→ M −→ K −→ 0 ∼ =

is split exact if there is an isomorphism ϕ : M −→ N ⊕K such that the following diagram commutes: 0

/

N

=

0

/



N

f

/M

g

/

/

ϕ ∼ =



N ⊕K

/

0 /

0

K /



=

K

f

g

43.8 Proposition. Let R be a ring and let 0 → N −→ M −→ K → 0 be a short exact sequence of R-modules. The following conditions are equivalent. 1) The sequence is split exact. 177

2) There exists a homomorphism h : K → M such that gh = idK .

3) There exists a homomorphism k : M → N such that kf = idN Proof. Exercise.

43.9 Theorem. Let R be a ring with identity and let P be an R-module. The following conditions are equivalent. 1) P is a projective module. 2) For any homomorphism f : P → N and an epimorphism g : M → N there is a homomorphism h : P → M such that the following diagram commutes: P h

M

~

f g

/

f



N g

3) Every short exact sequence 0 → N −→ M −→ P → 0 splits. Proof. (1) ⇒ (2) Let Q be a module such that P ⊕ Q is a free module, and let B = {bi }i∈I be a basis of P ⊕ Q. Since g is an epimorphism for every i ∈ I we can find mi ∈ M such that g(mi ) = f (bi ). Define ˜ : P ⊕ Q −→ M h by ! ˜ h

X

r i bi

:=

i

X i

178

ri mi

˜ is a well defined R-module Check: since B is a basis of P ⊕ Q the map h ˜ = f . Then we can take h = h| ˜ P. homomorphism and g h (2) ⇒ (3) We have a diagram

P idP g

M

/



P

Since g is an epimorphism there is h : P → M such that gh = idP . Therefore f g by (43.8) the sequence 0 → N −→ M −→ P → 0 splits. (3) ⇒ (1) We have the canonical epimorphism of R-modules: M f: R→P p∈P

This gives a short exact sequence 0 −→ Ker(f )−→

M p∈P

f

R −→ P −→ 0

By assumption on P this sequence splits. so we obtain M R P ⊕ Ker(f ) ∼ = p∈P

and thus P is a projective module.

43.10 Corollary. If R is a ring with identity, P is a projective R-module and f : M → P is an epimorphism of R-modules then M ∼ = P ⊕ Ker(f ). Proof. We have a short exact sequence f

0 −→ Ker(f )−→M −→ P −→ 0 which splits by Theorem 43.9. 179

44

Projective modules over PIDs

44.1 Theorem. If R is a PID, F is a free R-module of a finite rank, and M ⊆ F is a submodule then M is a free module and rank M ≤ rank F . 44.2 Corollary. If R is a PID then every finitely generated projective R-module is free. Proof. If P is a finitely generated projective R-module then we have an epimorphism f : Rn → P for some n > 0. By Corollary 43.10 we have an isomorphism P ⊕ Ker(f ) ∼ = Rn

Therefore we can identify P with a submodule of Rn , and thus by Theorem 44.1 P is a free module.

44.3 Note. Theorem 44.1 is true also for infinitely generated free modules over PIDs. As a consequence Corollary 44.2 is true for all (non necessarily finitelly generated) projective modules over PIDs.

Proof of Theorem 44.1 (compare with the proof of Theorem 13.6). We can assume that F = Rn . We want to show: if M ⊆ Rn then M is a free R-module and rank M ≤ n. Induction with respect to n: If n = 1 then M ⊆ R, so M is an ideal of R. Since R is a PID we have M = hai for some a ∈ R. If a = 0 then M = {0} is a free module of rank 0. Otherwise we have an isomorphism of R-modules ∼ =

f : R −→ M, 180

f (r) = ra

and so M is a free module of rank 1. Next, assume that for some n every submodule of Rn is a free R-module of rank ≤ n, and let M ⊆ Rn+1 . Take the homomorphism of R-modules f : Rn+1 → R, We have:

f (r1 , . . . , rn+1 ) = rn+1

Ker(f ) = {(r1 , . . . , rn , 0) | ri ∈ R} ∼ = Rn

We have an epimorphism: f |M : M → Im(f |M ) Since Im(f |M ) ⊆ R, thus Im(f |M ) is a free R-module, and so by Corollary 43.10 we have M∼ = Im(f |M ) ⊕ Ker(f |M ) We also have: Ker(f |M ) = Ker(f ) ∩ M It follows that that Ker(f |M ) is a submodule of Ker(f ), and since Ker(f ) is a free R-module of rank n by the inductive assumption we get that Ker(f |M ) is a free R-module of rank ≤ n. Therefore M∼ = Im(f |M ) ⊕ Ker(f |M ) | {z } | {z } free rank ≤ 1

free rank ≤ n

and so M is a free R-module of rank ≤ n + 1.

181

45

The Grothendieck group

Recall. If R is a ring with IBN and F is a free, finitely generated R-module then rank F = number of elements of a basis of F Goal. Extend the notion of rank to finitely generated projective modules. Idea. 1) Rank should be additive: rank(P ⊕ Q) = rank P + rank Q. 2) Rank of a module need not be an integer. Each ring determines a group K0 (R) such that for each finitely generated projective module rank of P is an element [P ] ∈ K0 (R).

Recall. A commutative monoid is a set M together with addition M × M → M,

(x, y) 7→ x + y

and with a trivial element 0 ∈ M such that the addition is associative, commutative and 0 + x = x for all x ∈ M . 45.1 Example. Let ProjfRg be the set of isomorphism classes of finitely generated projective R-modules. For a projective finitely generated R-module P denote [P ] = the isomorphism class of P The set ProjfRg is a commutative monoid with addition given by [P ] + [Q] := [P ⊕ Q] The identity element in ProjfRg is [0], the isomorphism class of the zero module.

182

45.2 Theorem. Let M be a commutative monoid. There exists an abelian group Gr(M ) and a homomorphism of monoids αM : M → Gr(M ) that satisfies the following universal property. If G is any abelian group and f : M → G is a homomorphism of monoids then there exists a unique homomorphism of groups f¯: Gr(M ) → G such that the following diagram commutes: f

M αM

/

=G

f¯



Gr(M ) Moreover, such group Gr(M ) is unique up to isomorphism.

45.3 Note. Let CMono denote the category of commutative monoids. We have the forgetful functor U : Ab −→ CMono Theorem 45.2 is equivalent to the statement that this functor has a left adjoint Gr : CMono → Ab,

M 7→ Gr(M )

45.4 Definition. Let M be a commutative monoid. The group Gr(M ) is called the group completion or the Grothendieck group of the monoid M .

Proof of Theorem 45.2. Construction of Gr(M ). Let M be a commutative monoid. Define Gr(M ) := M × M/ ∼ 183

where (x, y) ∼ (x0 , y 0 ) iff x + y 0 + t = x0 + y + t for some t ∈ M Check: ∼ is an equivalence relation on M × M . Notation: [x, y] := the equivalence class of (x, y) (Intuitively: [x, y] = x − y) Note: for any x ∈ M we have [x, x] = [0, 0] since x + 0 = 0 + x. Addition in Gr(M ): [x, y] + [x0 , y 0 ] = [x + x0 , y + y 0 ] Check: this operation is well defined, it is associative, and it has [0, 0] as the identity element. Additive inverses in Gr(M ): −[x, y] = [y, x] Indeed: [x, y] + [y, x] = [x + y, y + x] = [0, 0]

Construction of the homomorphism αM : M → Gr(M ). Define αM : M → Gr(M ),

x 7→ [x, 0]

The universal property of Gr(M ). Let G be an abelian group and let f : M → G be a homomorphism of commutative monoids. Define f¯: Gr(M ) → G, f¯([x, y]) := f (x) − f (y) Check: 184

1) f¯ is a well defined group homomorphism. 2) f¯αM = f 3) f¯ is the only homomorphism Gr(M ) → G satisfying condition 2). Uniqueness of Gr(M ) follows from the universal property.

45.5 Examples. 1) Gr(N) ∼ =Z 2) Let M = N ∪ {∞} with n + ∞ = ∞ for all n ∈ M . Then Gr(M ) is the trivial group. Indeed, for any m, n ∈ M we have [m, n] = [∞, ∞] since m + ∞ = ∞ + n. 3) If G is an abelian group then Gr(G) ∼ = G.

45.6 Definition. If R is a ring then K0 (R) := Gr(ProjfRg )

45.7 Notation. For [P ], [Q] ∈ ProjfRg denote [P ] − [Q] :=[P, Q] ∈ K0 (R) [P ] :=[P, 0] −[Q] :=[0, Q] 45.8 Proposition. Let R be a ring with identity. If P , Q are finitely generated projective R-modules then [P ] = [Q] in K0 (R) iff there exists n ≥ 0 such that P ⊕ Rn ∼ = Q ⊕ Rn . 185

Proof. (⇒) If P ⊕ Rn ∼ = Q ⊕ Rn then in K0 (R) we have [P ] + [Rn ] = [P ⊕ Rn ] = [Q ⊕ Rn ] = [Q] + [Rn ] and so [P ] = [Q] (⇒) If [P ] = [Q] in K0 (R) then P ⊕S ∼ =Q⊕S for some finitely generated projective R-module S. Exercise: If S is a finitely generated projective R-module then there is a finitely generated projective R-module T such that S ⊕ T ∼ = Rn for some n ≥ 0. We obtain

P ⊕ Rn ∼ =P ⊕S⊕T ∼ =Q⊕S⊕T ∼ = Q ⊕ Rn

45.9 Definition. Let R be a ring with identity. We say that R-modules M , N are stably isomorphic if M ⊕ Rn ∼ = N ⊕ Rn for some n ≥ 0.

45.10 Definition. Let R be a ring with identity. We say an R-module M is stably free if M ⊕ Rn ∼ = Rm for some m, n ≥ 0. 45.11 Note. Let R be a ring with identity. We have a homomorphism ϕ : Z → K0 (R) given by ( [Rn ] for n ≥ 0 ϕ(n) := −n −[R ] for n < 0

186

45.12 Proposition. Let R be a ring with identity, and let ϕ : Z → K0 (R) be the homomorphism as in (45.11). 1) ϕ is 1-1 iff R has IBN. 2) ϕ is an epimorphism iff every finitely generated projective R-module is stably free. Proof. 1) By Proposition 45.8 for n ≥ 0 we have n ∈ Ker(ϕ) iff Rn ⊕ Rm ∼ = 0 ⊕ Rm for some m ≥ 0. If R has IBN this is possible only if n = 0, and so Ker(ϕ) = {0}. Conversely, assume that R does not have IBN. Then Rn ∼ = Rm for some n > m. This gives Rn−m ⊕ Rm ∼ = 0 ⊕ Rm , and so ϕ(n − m) ∈ Ker(ϕ). 2) (⇒) Assume that ϕ is an epimorphism. Then for every finitely generated projective R-module P we have [P ] = [Rn ] for some n ≥ 0 or [P ] = −[Rn ] for some n ≥ 0. If [P ] = [Rn ] then by Proposition 45.8 we have P ⊕ Rm ∼ = Rn ⊕ Rm and so P is a stably free module. If [P ] = −[Rn ] then

[0] = [P ] + [Rn ] = [P ⊕ Rn ] Again by Proposition 45.8 this gives 0 ⊕ Rm ∼ = P ⊕ Rn ⊕ Rm , and again we obtain that P is stably free. (⇐) The group K0 (R) is generated by elements [P ] where P is a finitely generated projective R-module, so it is enough to show that for any such P we have [P ] = ϕ(k) for some k ∈ Z. Since P is stably free we have P ⊕ Rn ∼ = Rm for some n, m ≥ 0. This gives [P ] + [Rn ] = [P ⊕ Rn ] = [Rm ]

Therefore [P ] = [Rm ] − [Rn ] = ϕ(m) − ϕ(n) = ϕ(m − n).

187

45.13 Example. Here is an example of a stable free module that is not free. For details see: R. G. Swan, Vector bundles and projective modules, Transactions AMS 105 (2) (1962), 264-277. Let B be a compact, normal, topological space and let p : E → B be a real vector bundle over B. Define: C(B) = {f : B → R | f - continuous } C(B) is a ring (with pointwise addition and multiplication). Let Γ(p) be the set of all continuous sections of p: Γ(p) = {s : B → E | ps = idB } Note: Γ(p) is an C(B)-module with poinwise addition and pointwise multiplication by elements of C(B). Fact 1. The module Γ(p) is free iff p is a trivial vector bundle. Fact 2. If p : E → B and q : E 0 → B are real vector bundles over B then we have an isomorphism of C(B)-modules: Γ(p ⊕ q) ∼ = Γ(p) ⊕ Γ(q)

Upshot. If p : E → B, q : E 0 → B are bundles such that p is non-trivial, but both q and p ⊕ q are trivial bundles then Γ(p) is stably free C(B)-module that is not free. Indeed, in such case we have: Γ(p) ⊕ Γ(q) ∼ = Γ(p ⊕ q) |{z} | {z } free

188

free

Fact 3. It is possible to find vector bundles as above. Take e.g. p : T S 2 → S 2 to be the tangent bundle of the 2-dimensional sphere, and q : S 2 × R1 → S 2 to be the 1-dimensional trivial bundle over S 2 .

Note: one can also show that a C(B)-module M is finitely generated projective module iff M ∼ = Γ(p) for some vector bundle p : E → B.

189

46

Injective modules

Recall. If R is a ring with identity then an R-module P is projective iff one of the following equivalent conditions holds: 1) For any homomorphism f : P → N and an epimorphism g : M → N there is a homomorphism h : P → M such that the following diagram commutes: P h

M

~

f g



/

f

N g

2) Every short exact sequence 0 → N −→ M −→ P → 0 splits.

46.1 Proposition. Let R be a ring and let J be an R-module. The following conditions are equivalent. 1) For any homomorphism f : N → J and an monomorphism g : M → N there is a homomorphism h : M → J such that the following diagram commutes: JO ` h

f

M

g

f

/

N g

2) Every short exact sequence 0 → J −→ M −→ N → 0 splits. Proof. Exercise.

190

46.2 Definition. An R-module J is an injective module if J satisfies one of the equivalent conditions of Proposition 46.1.

46.3 Theorem (Baer’s Criterion). Let R be a ring with identity and let J be an R-module. The following conditions are equivalent. 1) J is an injective module. 2) For every left ideal I C R and for every homomorphisms of R-modules f : I → J there is a homomorphism f¯: R → J such f¯|I = f . Proof. 1) ⇒ 2) Given a homomorphism f : I → J we have a diagram JO f

 I 

i

/R

where i : I ,→ R is the inclusion homomorphism. By the definition of an injective module there is a homomorphism f¯: R → J such that f = f¯i = f¯|I 2) ⇐ 1) Assume that J is an R-module satisfying 2). It is enough to show that if M is an R-module, N is a submodule of M , and f : N → J is an Rmodule homomorphism then there exists a homomorphism f¯: M → J such that f¯|N = f Let S be a set of all pairs (K, fK ) such that (i) K is a submodule of M such that N ⊆ K ⊆ M

(ii) fK : K → J is a homomorphism such that fK |N = f

191

Define partial ordering on S as follows: (K, fK ) ≤ (K 0 , fK 0 )

if K ⊆ K 0 and fK 0 |K = fK

Check: assumptions of Zorn’s Lemma 29.10 are satisfied in S, and so S contains a maximal element (K0 , fK0 ). It will suffice to show that K0 = M . Assume, by contradiction, that K0 6= M , and let m0 ∈ M − K0 . Define I := {r ∈ R | rm0 ∈ K0 } Check: I is an ideal of R and the map g : I → J,

g(r) = fK0 (rm0 )

is a homomorphism of R-modules. By the assumptions on J we have a homomorphism g¯ : R → J such that g¯|I = g. Define K0 + Rm0 := {k + rm0 | k ∈ K, r ∈ R} Check: K0 + Rm0 is a submodule of M and the map f 0 : K0 + Rm0 → J,

f 0 (k + rm0 ) = fK0 (k) + g¯(r)

is a well defined homomorphism of R-modules such that f 0 |N = f . This shows that (K0 + Rm0 , f 0 ) ∈ S. We also have (K0 , fK0 ) < (K0 + Rm0 , f 0 ) This is impossible since by assumption (K0 , fK0 ) is a maximal element in S.

46.4 Corollary. Let R be an integral domain and let K the field of fractions of R. Then K is an injective R-module.

Proof. Let I be an ideal of R and let f : I → K be a homomorphism of Rmodules. For 0 6= r, s ∈ I we have rf (s) = f (rs) = sf (r) 192

As consequence in K we have f (r)/r = f (s)/s for any 0 6= r, s ∈ I. Denote this element by a. Define f¯: R → K,

f¯(r) := ra

Check: f¯ is a homomorphism of R-modules and f¯|I = f . By Baer’s Criterion (46.3) it follows that K is an injective R-module.

46.5 Example. Q is an injective Z-module.

46.6 Definition. Let R be an integral domain. An R-module M is divisible if for every r ∈ R − {0} and for every m ∈ M there is n ∈ M such that rn = m.

46.7 Theorem. If R is a PID then an R-module J is injective iff J is divisible.

Proof. Exercise.

46.8 Example. Since Z is a PID injective Z-modules are divisible Z-modules (i.e. divisible abelian groups). Exercise: an abelian group G is divisible iff G is isomorphic to a direct sum of copies of Q and Z(p∞ ) for various primes p.

46.9 Corollary. If R is a PID, J is an injective R-module and K is a submodule of J then J/K is injective.

193

Proof. Since J is divisible thus so is J/K (check!).

46.10 Note. If R is not a PID then a quotient of an injective R-module need not be injective.

Note. If R is a ring with identity then for any R-module M there exists an epimorphism of R-modules: f : P −→ M L where P is a projective module (take e.g. P = m∈M R). 46.11 Theorem. If R is a ring with identity then for any R-module M there exist a monomorphism j : M −→ J where J is an injective R-module.

46.12 Lemma. For any abelian group G there exists a monomorphism i : G −→ D where H is a divisible abelian group.

Proof. We have an epimorphism f : ∼ =

L

ϕ : G −→

g∈G

M

Z → G which gives an isomorphism

Z/ Ker(f )

g∈G

L L Moreover, the monomorphism g∈G Z → g∈G Q induces a monomorphism M M ψ: Z/ Ker(f ) −→ Q/ Ker(f ) g∈G

We can take D :=

L

g∈G

g∈G

Q/ Ker(f ) and i := ψϕ. 194

46.13 Note. Let G is an abelian group, let R let be a ring, and let HomZ (R, G) be the set of all homomorphisms of abelian groups ϕ : R → G. Check: HomZ (R, G) is an R-module with pointwise addition and with multiplication by elements of R given by (r · ϕ)(s) := ϕ(sr) for r, s ∈ R.

46.14 Lemma. If D is a divisible abelian group and R is a ring with identity then HomZ (R, D) in an injective R-module. Proof. Exercise.

Proof of Theorem 46.11. Let M be an R-module. Consider M as an abelian group. By Lemma 46.12 we have a monomorphism of abelian groups i : M −→ D where D is a divisible abelian group. Consider the induced map i∗ : HomZ (R, M ) → HomZ (R, D),

i∗ (ϕ) = i ◦ ϕ

Check: 1) i∗ is a monomorphism. 2) i∗ is a homomorphism of R-modules. Since M is an R-module we also have a map f : M → HomZ (R, M ), Check: 1) f is a monomorphism. 195

f (m)(r) = rm

2) f is a homomorphism of R-modules. As a consequence we obtain a monomorphism of R-modules i∗ f : M −→ HomZ (R, D) Moreover, by Lemma 46.12 HomZ (R, D) is an injective R-module.

196

47

Exact functors

47.1 Definition. A chain complex of R-modules is a sequence of R-modules and R-homomorphisms di+1

di−1

d

i . . . −→ Mi+1 −→ Mi −→ Mi−1 −→ Mi−2 −→ . . .

such that di di+1 = 0 for all i.

47.2 Note. If M∗ = (Mi , di ) is a chain complex then Im(di+1 ) ⊆ Ker(di ).

47.3 Definition. If M∗ = (Mi , di ) is a chain complex of R-modules then the i-th homology module of M∗ is the module Hi (M∗ ) = Ker(di )/ Im(di+1 )

Recall. A sequence di+1

di−1

d

i Mi−1 −→ Mi−2 −→ . . .) M∗ = (. . . −→ Mi+1 −→ Mi −→

is exact if Ker(di ) = Im(di+1 ) for all i. Therefore M∗ is exact iff Hi (M∗ ) = 0 for all i.

47.4 Definition. Let R, S be rings. A functor F : R-Mod → S-Mod is exact if for every short exact sequence of R-modules f

g

0 −→ N −→ M −→ K −→ 0 the sequence F (f )

F (g)

0 −→ F (N ) −→ F (M ) −→ F (K) −→ 0 is short exact. 197

47.5 Note. If F : R-Mod → S-Mod is a functor such that F (0) = 0 and di+1

d

i M∗ = (. . . −→ Mi+1 −→ Mi −→ Mi−1 −→ . . .)

is a chain complex of R-modules then F (di+1 )

F (di )

F (M∗ ) = (. . . −→ F (Mi+1 ) −→ F (Mi ) −→ F (Mi−1 ) −→ . . .) is a chain complex of S-modules. Moreover, the functor F is exact iff for every chain complex M∗ we have isomorphisms F (Hi (M∗ )) ∼ = Hi (F (M∗ )) for all i.

47.6 Note. For a ring R and R-modules L, M let HomR (L, M ) be the set of all R-module homomorphisms ϕ : L → M . Notice that HomR (L, M ) is an abelian group (with respect to the pointwise addition of homomorphisms). Moreover, for any homomorphism of R-modules f : M → N the map f∗ : HomR (L, M ) → HomR (L, N ),

f∗ (ϕ) = f ◦ ϕ

is a homomorphism of abelian groups. This defines a functor HomR (L, −) : R-Mod −→ Ab This functor is in general not exact. Take e.g. R = Z, L = Z/2Z. We have a short exact sequence of abelian groups: ·2

0 → Z −→ Z −→ Z/2Z → 0 On the other hand the sequence 0 → HomZ (Z/2Z, Z) −→ HomZ (Z/2Z, Z) −→ HomZ (Z/2Z, Z/2Z) → 0 is not exact since HomZ (Z/2Z, Z) ∼ = 0 and HomZ (Z/2Z, Z/2Z) ∼ = Z/2Z.

198

47.7 Proposition. Let R be a ring and let L be an R-module. If f

g

0 −→ N −→ M −→ K −→ 0 is a short exact sequence of R-modules then f∗

g∗

0 −→ HomR (L, N ) −→ HomR (L, M ) −→ HomR (L, K) is an exact sequence of abelian groups.

Proof. Exercise.

47.8 Theorem. Let R be a ring with identity and let P be an R-module. The functor HomR (P, −) is exact iff P is a projective module. Proof. By Proposition 47.7 is suffices to show that P is a projective module iff for every epimorphism of R-modules g : M → K the map g∗ : HomR (P, M ) −→ HomR (P, K) is an epimorphism. This follows directly from Theorem 43.9.

47.9 Definition. Let C, D be categories. A contravariant functor F : C → D consists of 1) an assignment Ob(C) → Ob(D),

c 7→ F (c)

2) for every c, c0 ∈ C a function HomC (c, c0 ) → HomD (F (c0 ), F (c)),

f 7→ F (f )

such that F (idc ) = idF (c) and F (gf ) = F (f )F (g).

199

47.10 Example. Let R be a ring and let L be an R-module. For any homomorphism of R-modules f : M → N we have a map f ∗ : HomR (N, L) −→ HomR (M, L),

f ∗ (ϕ) = ϕ ◦ f

Moreover, f ∗ is a homomorphism of abelian groups. This defines a contravariant functor HomR (−, L) : R-Mod −→ Ab

47.11 Note. The functor HomR (−, L) is in general not exact. Take e.g. R = Z, L = Z. We have a short exact sequence of abelian groups: ·2

0 → Z −→ Z −→ Z/2Z → 0 On the other hand the sequence 0 → HomZ (Z/2Z, Z) −→ HomZ (Z, Z) −→ HomZ (Z, Z) → 0 is not exact.

47.12 Proposition. Let R be a ring and let L be an R-module. If f

g

0 −→ N −→ M −→ K −→ 0 is a short exact sequence of R-modules then g∗

f∗

0 −→ HomR (K, L) −→ Hom(M, L) −→ Hom(M, L) is an exact sequence of abelian groups.

Proof. Exercise.

200

47.13 Theorem. Let R be a ring with identity and let J be an R-module. The functor HomR (−, J) is exact iff J is an injective module.

Proof. By Proposition 47.12 is suffices to show that J is an injective module iff for every monomorphism of R-modules f : N → M the map f ∗ : HomR (M, J) −→ HomR (N, J) is an epimorphism. This follows directly from Proposition 46.1.

47.14 Note. Let d

i M∗ = (. . . −→ Mi −→ Mi−1 −→ . . .)

be a chain complex of R-modules. For any R-module L we have the induced chain complex of abelian groups d∗

i HomR (Mi , L) −→ . . .) HomR (M∗ , L) = (. . . −→ HomR (Mi−1 , L) −→

Homology groups of the complex HomR (M∗ , L) are called cohomology groups of M∗ with coefficients in L. We denote: H i (M∗ , L) := Hi (HomR (M∗ , L)) If L is an injective module then by Theorem 47.13 the functor HomR (−, L) is exact. By (47.5) in such case we have H i (M∗ , L) ∼ = HomR (Hi (M∗ ), L)

201

48

Tensor products

Note. For now all rings are commutative rings with identity.

48.1 Definition. Let M, N, K be R-modules. A function f : M × N −→ K is bilinear if 1) f (m1 + m2 , n) = f (m1 , n) + f (m2 , n) f (m, n1 + n2 ) = f (m, n1 ) + f (m, n2 ) 2) f (rm, n) = rf (m, n) = f (m, rn)

48.2 Theorem. Let M , N be R-modules. There exists an R-module M ⊗R N and a bilinear map η : M × N → M ⊗R N

that satisfies the following universal property. For any R-module K and a bilinear map f : M × N → K there exists a unique homomorphism of R-modules f¯: M ⊗R N → K such that the following diagram commutes: M ×N

f

η

/

0, gcd(m, n) = 1 Z/mZ ⊗Z Z/nZ = ? Note: if f : Z/mZ × Z/nZ → K is a bilinear map then f (k, l) = f (k · 1, l · 1) = (kl) · f (1, 1) Therefore the map f is determined by the value of f (1, 1). 204

Moreover, since gcd(m, n) = 1 we have am + bn = 1 for some m, n ∈ Z. This gives: f (1, 1) = f (1, (am + bn) · 1) = f (1, (am) · 1) + f (1, (bn) · 1) = f ((am) · 1, 1) + f (1, (bn) · 1) = f (0, 1) + f (1, 0) = f (0 · 1, 1) + f (1, 0 · 1) = 0 · f (1, 1) + 0 · f (1, 1) =0 Therefore for any bilinear map f we have f (1, 1) = 0, and so f = 0. It follows that Z/mZ ⊗Z Z/nZ = 0 since the zero module satisfies the universal property: Z/mZ × Z/nZ

f =0

/

:K

f¯=0



0

48.5 Proposition. If M is an R-module then R ⊗R M ∼ =M Proof. We have a bilinear map η : R × M → M,

η(r, m) = rm

(note: this works only if R is commutative!). Moreover, if f : R × M → K is any bilinear map then define f¯: M → K,

f¯(m) = f (1, m) 205

Check: f¯ is the unique R-module homomorphism such that f¯η = f . By the universal property of tensor products this gives M ∼ = R ⊗R M .

48.6 Proposition. 1) If M , N are R-modules then we have an isomorphism ϕ : M ⊗R N → N ⊗R M,

ϕ(m ⊗ n) = n ⊗ m

2) If M , N , K are R-modules then M ⊗R (N ⊗R K) ∼ = (M ⊗R N ) ⊗R K 3) If Mα , N are R-modules then M M Mα ) ⊗R N ∼ (Mα ⊗R N ) ( = α

α

Proof. Exercise.

206

49

Tensor products of homomorphisms

Let f : M → K, g : N → L be homomorphisms of R-modules. Notice that we have a bilinear map f × g : N : M × N → K ⊗R L,

f × g(m, n) = f (m) ⊗ g(n)

By the unniversal property of tensor products f × g induces a unique homomorphism of R-modules f ⊗ g : M ⊗R N → K ⊗R L,

f ⊗ g(m ⊗ n) = f (m) ⊗ g(n)

49.1 Proposition. 1) Tensor product preserves composition of homomorphisms: if we have homomorphisms of R-modules f

f0

g

g0

M −→ K −→ K 0 N −→ L −→ L0 then (f 0 ◦ f ) ⊗ (g 0 ◦ g) = (f 0 ⊗ g 0 ) ◦ (f ⊗ g) 2) Tensor product preserves identity maps: idM ⊗ idN = idM ⊗R N Proof. Exercise.

49.2 Corollary. If f : M → K, g : N → L are isomorphisms of R-modules then f ⊗ g : M ⊗R N → K ⊗R L is also an isomorphism. 207

Proof. Exercise.

49.3 Proposition. If F , F 0 are free R-modules then F ⊗R F 0 is also a free module. Proof. We have

F ∼ =

M

R,

i∈I

F0 ∼ =

M

R

j∈J

Therefore M M R) R) ⊗R ( F ⊗R F 0 ∼ =( j∈J

i∈I

∼ =

MM

∼ =

M

i∈I j∈J

(R ⊕R R) R

(i,j)∈I×J

49.4 Note. If F , F 0 are free modules, {bi }i∈I is a basis of F and {b0j }j∈J is a basis of F 0 then {bi ⊗ bj }(i,j)∈I×J is a basis of F ⊗R F 0 (exercise). 49.5 Corollary. If P , Q are projective R-modules then P ⊗R Q is also projective. Proof. Since P , Q are projective R-modules there exists modules P 0 , Q0 such that P ⊕ P 0 , Q ⊕ Q0 are free modules. We have (P ⊕ P 0 ) ⊗R (Q ⊕ Q0 ) ∼ = (P ⊗R Q) ⊕ (P ⊗R Q0 ) ⊕ (P 0 ⊗R Q) ⊕ (P 0 ⊗R Q0 ) | {z } free

Therefore P ⊗R Q is a direct summand of a free module, and so it is projective.

208

50

Tensor products and adjoint functors

Recall. Let R be a commutative ring. If M , N are R-modules then HomR (M, N ) = (the set of all R-module homomorphisms M → N ) Note. 1) HomR (M, N ) is an abelian group with addition given by (f + g)(m) := f (m) + g(m) 2) HomR (M, N ) is an R-module with multiplication by elements of R given by (rf )(m) := r · f (m)

(Note: rf need not be an R-module homomorphism if R is a non-commutative ring.)

3) If g : N → N 0 is a homomorphism of R-modules then the map g∗ : HomR (M, N ) → HomR (M, N 0 ),

g∗ (f ) = g ◦ f

is a homomorphism of R-modules.

Upshot. For any R-module M we have a functor Hom(M, −) : R-Mod −→ R-Mod

50.1 Proposition. For any R-module M the functor − ⊗R M : R-Mod −→ R-Mod is left adjoint to the functor HomR (M, −).

209

50.2 Note. Proposition 50.1 says that for any R-modules N , K we have a natural bijection of sets ∼ =

HomR (N ⊗R M, K) −→ HomR (N, HomR (M, K)) Compare this with the exponential law for in set theory: if A, B, C are sets then we have a bijection ∼ =

Map(A × B, C) −→ Map(A, Map(B, C))

Sketch of proof of Proposition 50.1. Denote: Hom2R (N, M ; K) = (set of all bilinear maps N × M → K) By the universal property of tensor ptoducts we have a bijection HomR (N ⊗R M, K) ∼ = Hom2R (N, M ; K) Therefore we only need a bijection Φ : Hom2R (N, M ; K) −→ HomR (N, HomR (M, K)) For f ∈ Hom2R (N, M ; K) define: Φ(f ) : N → HomR (M, K),

210

Φ(f )(n) = f (n, −)

51

Tensor products for non-commutative rings

Recall. If R is a commutative ring and M , N , K are R-modules then we have a bijection 

bilinear maps M ×N →K



∼ =



R-linear maps M ⊗R N → K



Note. If R is a non-commutative ring then there are few bilinear map since for any such map f : M × N → K we have (sr) · f (m, n) = s · f (rm, n) = f (rm, sn) = r · f (m, sn) = (rs) · f (m, n) A more appropriate notion in this setting is a middle linear map.

51.1 Definition. Let R be a ring with identity. Let M be a right R-module, N be a left R-module, G be an abelian group. A map f: M ×N →G is middle linear if 1) f (m1 + m2 , n) = f (m1 , n) + f (m2 , n) f (m, n1 + n2 ) = f (m, n1 ) + f (m, n2 ) 2) f (mr, n) = f (m, rn)

51.2 Examples. 1) If R is a ring then the map µ : R × R → R, is middle linear 211

µ(r, s) = rs

2) In general, if N is a left R-module then the map µ : R × N → N,

µ(r, n) = rn

is middle linear.

51.3 Definition. If R is a ring with identity, M is a right R-module and N is a left R-module then M ⊗R N is the abelian group given by M ⊗R N := Fab (M × N )/S Here Fab (M × N ) is the free abelian group generated by the set M × N and S is the subgroup of Fab (M × N ) generated by all elements of the following forms: (i) (m1 + m2 , n) − (m1 , n) − (m2 , n) (m, n1 + n2 ) − (m, n1 ) − (m, n2 ) (ii) (mr, n) − (m, rn) 51.4 Note. Let m ⊗ n denote the element of M ⊗R N represented by (m, n). We have a middle linear map η : M × N → M ⊗R N,

η(m, n) = m ⊗ n

51.5 Theorem. Let R be a ring with identity, let M be a right R-module, N be a left R-module. If G be an abelian group and and f : M × N → G is a middle linear map then there exists a unique homomorphism of abelian groups f¯: M ⊗R N → G such that the following diagram commutes: M ×N

f

η

/

|Q|

Exercise: show that the subset √ { p | p – prime } ⊆ R

is linearly independent over Q. 2) [C : R] = 2 since {1, i} is a basis of C over R.

57.5 Note. [L : K] = 1 iff L = K.

222

58

Prime subfield and field characteristic

58.1 Proposition. If L is a field and {Ki }i∈I is a family of subfields of L then T i∈I Ki is a subfield of L.

Proof. Exercise.

58.2 Definition. The prime subfield of a field L is the intersection of all subfields of L.

58.3 Definition. If L is a field then the characteristic of L is the smallest positive integer χ(L) such that 1L + · · · + 1L = 0 {z } | χ(L) times

If 1L + · · · + 1L 6= 0 for all n > 0 then χ(L) = 0. | {z } n times

58.4 Example. χ(Q) = χ(R) = χ(C) = 0. χ(Fp ) = p.

58.5 Proposition. If χ(L) 6= 0 then it is a prime number. Proof. Exercise.

223

58.6 Theorem. Let L be a field and let K be the prime subfield of L. 1) If χ(L) = 0 then K ∼ = Q. 2) if χ(L) = p > 0 then K ∼ = Fp . Proof. Exercise.

224

59

Algebraic and transcendental elements

59.1 Notation. If L/K is a field extension, and S is a subset of L then K(S) is the smallest subfield L such that K ⊆ K(S) and S ⊆ K(S).

Note. 1) If a ∈ L then  r0 + r1 a + · · · + rn an K(a) = s 0 + s 1 a + · · · + s m am

ri , sj ∈ K,

 s 0 + · · · + sm a = 6 0 m

2) If a1 , . . . , an ∈ L then K(a1 , . . . , an ) = K(a1 , . . . , an−1 )(an )

59.2 Definition. Let L/K be a field extension. An element a ∈ L is algebraic over K is there exists r0 , . . . , rn ∈ K, rn 6= 0 such that r0 + r1 a + · · · + rn an = 0 Equivalently, a ∈ L is algebraic over K if there exists a non-zero polynomial p(x) = r0 + r1 x + · · · + rn xn ∈ K[x] such that p(a) = 0. An element a ∈ L is transcendental over K if it is not algebraic over K.

59.3 Example. √ √ 1) 2 ∈ R is a root of p(x) = x2 − 2 ∈ Q[x], so 2 is algebraic over Q. 2) e, π ∈ R are transcendental over R (harder to show).

225

59.4 Proposition. If L/K is a field extension and a ∈ L is a transcendental element over K then K(a) ∼ = K(x) where K(x) is the field of rational functions of variable x with coefficients in K.

Proof. We have a homomorphism of rings ϕa : K[x] → L,

ϕa (f ) = f (a)

Since a is transcendental over K this map is a monomorphism, so ϕa (f ) is an invertible element for all f 6= 0. Therefore, by the universal property of localizations of rings (36.6) there exists a homomorphism ϕ¯a : K(x) → L,

ϕ¯a (f /g) = f (a)/g(a)

We have Im(ϕ¯a ) = K(a), so K(a) ∼ = K(x).

59.5 Example.

Q(π) ∼ = Q(x) ∼ = Q(e)

59.6 Proposition. Let L/K be a field extension and let a ∈ L be an algebraic element over K. 1) There exists an irreducible polynomial p(x) ∈ K[x] such that p(a) = 0 and if q(a) = 0 for some q(x) ∈ K[x] then p(x) | q(x). 2) We have an isomorphism K(a) ∼ = K[x]/hp(x)i.

Proof. 1) Take the homomorphism of rings ϕa : K[x] → L, 226

ϕa (f ) = f (a)

Since a is algebraic over K we have Ker(ϕa ) 6= {0}. Also, since K[x] is a PID thus Ker(ϕa ) = hp(x)i for some p(x) ∈ K[x]. If q(a) = 0 for some q(x) ∈ K[x] then q(x) ∈ hp(x)i, so p(x) | q(x). In order to see that p(x) is irreducible assume that p(x) = g(x)h(x) for some g(x), h(x) ∈ K[x]. Then 0 = p(a) = g(a)h(a) so either g(a) = 0 or h(a) = 0. We can assume that g(a) = 0. Then p(x) | g(x). Since also g(x) | p(x) we obtain that h(x) must be a unit in K[x]. 2) We have

K[x]/hp(x)i ∼ = Im(ϕa )

It is then enough to show that Im(ϕa ) = K(a). It is clear that Im(ϕa ) ⊆ K(a). On the other hand, since p(x) is an irreducible element of K[x] then by (32.3) hp(x)i is a maximal ideal of K[x] and so K[x]/hp(x)i is a field. As a consequence Im(ϕa ) is a subfield of L, and K ∪ {a} ⊆ Im(ϕa ). Since K(a) is the smallest subfield containing K ∪ {a} we get that K(a) = Im(ϕa ).

59.7 Notation. Let L/K be a field extension and let a ∈ L be an algebraic element over K. By Proposition 59.6 there is a unique irreducible polynomial p(x) ∈ K[x] of the form p(x) = xn + rn−1 xn−1 + . . . + r1 x + r0

such that p(a) = 0. Denote this polynomial by irrK a (x).

Note. Polynomials of the form f (x) = xn + rn−1 xn−1 + . . . + r1 x + r0 (with the highest degree coefficient equal to 1) are called monic polynomials.

227

59.8 Proposition. Let L/K, M/K be extensions of a field K and let a ∈ L, b ∈ M be algebraic elements over K. The following conditions are equivalent. K 1) irrK a (x) = irrb (x)

2) there exists an isomorphism ϕ : K(a) → K(b) such that ϕ|K = idK and ϕ(a) = ϕ(b).

Proof. K (1) ⇒ (2) If irrK a (x) = irrb (x) = p(x) then we have isomorphisms ϕ

ϕa

Take ϕ = ϕb ◦ ϕ−1 a .

b K(b) K(a) ←− K[x]/hp(x)i −→

K n n−1 (2) ⇒ (1) It is enough to check that irrK + a (b) = 0. If irra (x) = x + rn−1 x . . . + r1 x + r0 Then we have n n−1 irrK + . . . + r1 b + r0 a (b) = b + rn−1 b n = ϕ(a) + rn−1 ϕ(a)n−1 + . . . + r1 ϕ(a) + r0

Since ri ∈ K we have ri = ϕ(ri ), so

n n−1 irrK + . . . + ϕ(r1 )ϕ(a) + ϕ(r0 ) a (b) = ϕ(a) + ϕ(rn−1 )ϕ(a) = ϕ(an + rn−1 an−1 + . . . + r1 a + r0 ) = ϕ(0) =0

59.9 Note. If L/K is a field extension, a, b ∈ L and irrK (x) = irrK we a√ b (x) then √ √ 3 1 3 3 may have K(a) 6= K(b). Take e.g. K/L = C/Q, a = 2, b = 2( 2 + 2 i). Then Q 3 irrQ a (x) = irrb (x) = x − 2 but Q(a) ⊆ R, Q(b) 6⊆ R, so Q(a) 6= Q(b). 228

59.10 Proposition. Let L/K be a field extension and let a ∈ L be an algebraic element over K, and let p(x) = irrK a (x). If deg p(x) = n then 1) K(a) = {s0 + s1 a + . . . + sn−1 an−1 | si ∈ K}

2) the set {1, a, . . . , an−1 } is a basis of K(a) over K 3) [K(a) : K] = n

Proof. 1) We have the isomorphism ϕa : K[x]/hp(x)i −→ K(a),

ϕa (f ) = f (a)

Also, for f (x) ∈ K[x] we have f (x) = q(x)p(x) + r(x) for some q(x), r(x) ∈ K[x], deg r(x) < n. This gives f (a) = q(a)p(a) + r(a) = 0 + r(a) = r(a) Ii follows that K(a) = {r(a) | r(x) ∈ K[x], deg r(x) < n}. 2) By part 1) we have K(a) = SpanK (1, a, . . . an−1 ) so it suffices to show that the set {1, a, . . . , an−1 } is linearly independent over K. Assume that s0 · 1 + s1 a + . . . + sn−1 an−1 = 0

for some si ∈ K. Then a is a root of the polynomial g(x) = s0 + s1 x + . . . + sn−1 xn−1 . Since deg g(x) < deg p(x) we must have g(x) = 0, so s0 = s1 = . . . = sn−1 = 0. 3) Obvious by part 2).

229

59.11 Example. √ √ (x) = x2 − 2, thus Take 2 ∈ R. Since irrQ 2

√ √ Q( 2) = {a + b 2 | a, b ∈ Q}

Notice that

√ (a + b 2)−1 =

√ 1 2) (a − b a2 − 2b2

230

60

Algebraic extensions

60.1 Notation. If L/K is a field extension and a ∈ L then degK (a) := [K(a) : K] Note. 1) If a is a transcendental element over K then K(a) ∼ = K(x), so degK (a) = ∞. 2) If a is an algebraic element over K then degK (a) = deg irrK a (x) < ∞.

Upshot. If L/K is a field extension then an element a ∈ L is algebraic over K iff degK (a) < ∞. 60.2 Definition. A field extension L/K is an algebraic extension if all elements of L are algebraic over K.

60.3 Proposition. If [L : K] < ∞ then L/K is an algebraic extension.

√ √ √ √ Note. L = Q( 2, 3, 5, 7, . . . ) is an algebraic extension of Q even though [L : Q] = ∞. 60.4 Lemma. Let K ⊆ L ⊆ M be field extensions. If {ai }i∈I is a basis of M over L and {bj }j∈J is a basis of L over K then {ai bj }(i,j)∈I×J is a basis of M over K. As a consequence [M : K] = [M : L] · [L : K]

231

Proof. Claim 1. The set {ai bj }(i,j)∈I×J spans M over K. Indeed, if m ∈ M then

m=

X

li ai

i

for some li ∈ L. Also, for each li we have X li = kij bj j

for some kij ∈ K. Therefore we obtain m=

X X i

!

kij bj

ai =

j

X

kij (ai bj )

ij

Claim 2. The set {ai bj }(i,j)∈I×J is linearly independent over K. P Indeed, if ij kij (ai bj ) = 0 for some kij ∈ K then ! X X 0= kij ai bj i

j

P

Since j kij ai ∈ L and the set {bj }j∈J is linearly independent over L we have P k a j ij i = 0 for each j. Then, since the set {ai }i∈I is linearly independent over K we have kij = 0 for all i, j.

Proof of Proposition 60.3. Let a ∈ L. We have K ⊆ K(a) ⊆ L, so by Lemma 60.4 we get [L : K(a)] · [K(a) : K] = [L : K] < ∞ Therefore [K(a) : K] < ∞, and so a is an algebraic element over K.

232

60.5 Corollary. If L/K is a field extension and a1 , . . . , an ∈ L then [K(a1 , . . . , an ) : K] ≤ degK (a1 ) · . . . · degK (a1 )

In particular, if a1 , . . . , an are algebraic elements over K then [K(a1 , . . . , an ) : K] < ∞ and so K(a1 , . . . , an ) is an algebraic extension of K. Proof. Exercise.

60.6 Corollary. If M/L, L/K are algebraic extensions then M/K is also an algebraic extension. Proof. Let a ∈ M . We need to show that a is algebraic over K. Since a is algebraic over L there is a polynomial f (x) = l0 + l1 x + . . . + ln xn in L[x] such that f (a) = 0. Let N = K(l0 , . . . , ln ). Notice that f (x) ∈ N [x], so a is an algebraic element over N . In particular we have [N (a) : N ] < ∞. Moreover, since l0 , . . . , ln are algebraic elements over K by Corollary 60.5 we have [N : K] < ∞. This gives [N (a) : K] = [N (a) : N ] · [N : K] < ∞

Finally, since K(a) ⊆ N (a) we have

[K(a) : K] ≤ [N (a) : K] < ∞

so a is an algebraic element over K.

60.7 Proposition. Let L/K be a field extension and let Kalg (L) = {a ∈ L | a is an algebraic element over K}

Then Kalg (L) is a subfield of L.

Proof. We need to show that if a, b ∈ L are algebraic elements over K, then a ± b, ab, a/b are also algebraic over K. This holds since by (60.5) K(a, b)/K is an algebraic extension and a ± b, ab, a/b ∈ K(a, b). 233

61

Separable elements

61.1 Example. √ √ √ √ Note: since 2, 3 ∈ R are algebraic elements over Q, thus Q( 2, 3)/Q is an algebraic extension. We have: √ √ √ √ Q ⊆ Q( 2 + 2) ⊆ Q( 2, 3) √ √ √ √ Claim: Q( 2 + 2) = Q( 2, 3) √ √ √ √ Indeed, it is enough to show that [Q( 2, 3) : Q( 2 + 3)] = 1 √ (x) = x2 − 3, so by (60.5) √ (x) = x2 − 2 and irrQ We have irrQ 2 2

√ √ √ · degQ √ = 4 [Q( 2, 3) : Q] ≤ degQ 2 3

√ √ On the other hand 2 + 3 is a root of p(x) = x4 − 10x2 + 1, and since p(x) is irreducible in Q[x] (check!) we have √ √ [Q( 2 + 3) : Q] = deg p(x) = 4 This gives √ √ √ √ √ √ √ √ [Q( 2, 3) : Q( 2 + 3)] · [Q( 2 + 3) : Q] = [Q( 2, 3) : Q] | {z } | {z } = 4

≤ 4

√ √ √ √ so [Q( 2, 3) : Q( 2 + 3)] = 1.

Goal. If L/K is a field extension and a1 , . . . , an ∈ L are algebraic elements over K then there exists an element c ∈ L such that K(a1 , . . . , an ) = K(c) provided that a1 , . . . , an are separable over K. 234

Recall. Let L/K be a field extension. An element a ∈ L is a root of f (x) ∈ K[x] iff f (x) = (x − a)g(x) for some g(x) ∈ L[x]. 61.2 Definition. If L/K is a field extension and a ∈ L then a is a multiple root of f (x) ∈ K[x] if f (x) = (x − a)k g(x) for some k > 1 and g(x) ∈ L[x].

61.3 Definition. A polynomial f (x) ∈ K[x] is separable if it has no multiple roots in any extension of K.

61.4 Definition. Let L/K be a field extension and let a ∈ L be an algebraic element over K. The element a is separable over K if irrK a (x) ∈ K[x] is a separable polynomial.

235

62

Derivatives and separable elements

62.1 Definition. If f (x) ∈ K[x], f (x) = a0 + a1 x + . . . + an xn then the derivative of f (x) is the polynomial f 0 (x) = a1 + 2a2 x + . . . + nan xn−1

62.2. Properties of derivatives. 1) (f + g)0 = f 0 + g 0 2) (cf )0 = c · f 0 for c ∈ K 3) (f g)0 = f 0 g + f g 0 4) (f n )0 = nf n−1 · f 0 5) (f ◦ g)0 = (f 0 ◦ g) · g 0

62.3 Proposition. A polynomial f (x) ∈ K[x] is separable iff f and f 0 have no common roots in any extension of K.

Proof. (⇒) Assume that there exists an extension L/K such that f (a) = f 0 (a) = 0 for some a ∈ L. We have f (x) = (x − a)g(x) for some g(x) ∈ L[x]. It follows that f 0 (x) = 1 · g(x) + (x − a)g 0 (x) and so g(x) = f 0 (x) − (x − a)g 0 (x) 236

This gives g(a) = f 0 (a) − (a − a)g 0 (a) = 0 Therefore g(x) = (x − a)h(x) for some h(x) ∈ L[x], and f (x) = (x − a)2 h(x) Thus a is a multiple root of f (x). (⇐) Assume that f (x) is not a separable polynomial. Then there exists an extension L/K and a ∈ L such that f (x) = (x − a)k g(x) for some k > 1 and g(x) ∈ L[x]. This gives f 0 (x) = k(x − a)k−1 g(x) + (x − a)k g 0 (x) so f 0 (a) = f (a) = 0.

62.4 Proposition. Let L/K be a field extension, let a ∈ L be an algebraic element over K and let p(x) = irrK a (x). The element a is separable over K iff p0 (x) 6= 0. Proof. (⇐) Assume that p0 (x) = 0. Then p(a) = p0 (a) = 0 and by Proposition 62.3 p(x) is not a separable polynomial. (⇒) Assume that a is not separable over K. Then p(x) is not a separable polynomial, so by Proposition 62.3 there exists an extension M/K and b ∈ M such that p(b) = p0 (b) = 0 237

Since p(x) is an irreducible polynomial in K[x] we have p(x) = irrK b (x). It 0 0 follows that p(x) | p (x). However, deg p(x) > deg p (x), so we must have p0 (x) = 0.

62.5 Corollary. Let K be a field such that χ(K) = 0. If L/K is a field extension then every algebraic element a ∈ L is separable over K. 0 Proof. If p(x) = irrK a (x) then deg p(x) ≥ 1, so p (x) 6= 0.

62.6 Example. Let p be a prime number, and let K = Fp (t) be the field of rational functions of variable t with coefficients in Fp . Check: p(x) = xp − t is an irreducible polynomial in K[x]. Let L/K be an extension such that p(a) = 0 0 for some a ∈ L. Then p(x) = irrK a (x), and since p (x) = 0 then element a is not separable over K.

62.7 Proposition. If K is a field, χ(K) = p 6= 0 and f (x) ∈ K[x] then f 0 (x) = 0 iff f (x) ∈ K[xp ]. Proof. (⇒) If f (x) ∈ K[xp ] then f (x) = a0 + ap xp + a2p x2p + . . . + anp xnp 238

for some akp ∈ K. Then f 0 (x) = pap xp−1 + 2pa2p x2p−1 + . . . + npanp xnp−1 = 0

(⇐) If f (x) = a0 +a1 x+. . .+an xn and f 0 (x) = 0 then kak = 0 for k = 0, . . . , n. Therefore either ak = 0 or p | k. It follows that f (x) ∈ K[xp ].

62.8 Corollary. Let L/K be a field extension, let χ(K) = p 6= 0 and let a ∈ L p be an algebraic element over K. Then a is separable over K iff irrK a (x) 6∈ K[x ]. Proof. Follows from (62.4) and (62.7).

62.9 Proposition. Let L/K be a field extension such that χ(K) = p 6= 0, and let a ∈ L be an algebraic element over K. The element a is separable over K iff K(a) = K(ap ).

Proof. (⇐) If a is separable over K then it is separable over K(ap ) (check!). p) Also, a is a root of g(x) = xp − ap ∈ K(ap )[x]. Let f (x) = irrK(a (x). Then a f (x) | g(x). On the other hand we have g(x) = (x − a)p , so f (x) = (x − a)k for some 1 ≤ k ≤ p. Since f (x) is a separable polynomial we must have f (x) = x−a, and since f (x) ∈ K(ap )[x] this gives that a ∈ K(ap ). It follows that K(a) = K(ap ). (⇒) Assume that a is not separable over K, and let f (x) = irrK a (x). By Corollary 62.8 we have f (x) ∈ K[xp ], so f (x) = r0 + rp xp + . . . + rnp xnp

239

for some r0 , . . . , rnp ∈ K. Let g(x) = r0 + rp x + . . . + rnp xn . We have g(ap ) = f (a) = 0, so degK (ap ) ≤ deg g(x) < deg f (x) = degK (a) This gives [K(ap ) : K] = degK (ap ) < degK (a) = [K(a) : K] so K(ap ) 6= K(a).

62.10 Corollary. Let K be a field such that χ(K) = p 6= 0. If L/K is a field extension and a1 , . . . , an ∈ L are elements separable over K then K(a1 , . . . , an ) = K(ap1 , . . . , apn )

62.11 Notation. If K, L ⊆ M are field extensions then   the smallest subfield of L KL := containing K ∪ L

Note: if L/K is a field extension and a1 , . . . , an ∈ L then K(a1 , . . . , an ) = K(a1 )K(a2 ) · . . . · K(an )

Proof of Corollary 62.10. Using Proposition 62.9 we obtain: K(a1 , . . . , an ) = K(a1 ) · . . . · K(an ) = K(ap1 ) · . . . · K(apn ) = K(ap1 , . . . , apn )

240

63

Separable extensions

63.1 Definition. A field extension L/K is separable if every element a ∈ L is separable over K.

63.2 Note. By (62.5) if χ(K) = 0 then every algebraic extension of K is separable.

63.3 Notation. If L is a field and χ(L) = p 6= 0 then Lp := {ap | a ∈ L}

Note. 1) Lp is a subfield of L since if ap , bp ∈ Lp then ap bp = (ab)p ∈ Lp and ap + bp = (a + b)p ∈ Lp . 2) The map ϕ : L → Lp ,

ϕ(a) = ap

is a field isomorphism.

63.4 Proposition. Let K be a field such that χ(K) = p 6= 0. If L/K is a field extension and [L : K] < ∞ then L/K is separable iff KLp = L.

63.5 Lemma. If K ⊆ M ⊆ L and N ⊆ L are field extensions then [N M : N K] ≤ [M : K] Proof. Exercise. 241

Proof of Proposition 63.4. (⇐) If L/K is separable and a ∈ L then by Proposition 62.9 we have a ∈ K(a) = K(ap ) ⊆ KLp This gives L ⊆ KLp , and since also KLp ⊆ L we obtain KLp = L. (⇒) Assume that L = KLp , and let b ∈ L. We need to show that b is a separable element over K. By Proposition 62.9 it suffices to show that K(b) = K(bp ). Take the isomorphism ϕ : L → L,

ϕ(a) = ap

We have [ϕ(L) : ϕ(K(b))] = [L : K(b)] ≤ [L : K] < ∞ Consider the extensions ϕ(K(b)) ⊆ ϕ(L) ⊆ L and K ⊆ L. By Lemma 63.5 we have [Kϕ(L) : Kϕ(K(b))] ≤ [ϕ(L) : ϕ(K(b))] Notice that Kϕ(L) = KLp = L and Kϕ(K(b)) = KK p (bp ) = K(bp ), so this gives [L : K(bp )] ≤ [ϕ(L) : ϕ(K(b))] = [L : K(b)] One the other hand we have [L : K(bp )] = [L : K(b)] · [K(b) : K(bp )] Therefore we must have [L : K(bp )] = [L : K(b)] and [K(b) : K(bp )] = 1. This gives K(bp ) = K(b)

63.6 Theorem. If L/K is a field extension, χ(K) = p 6= 0 and a1 , . . . , an ∈ L are elements separable over K then K(a1 , . . . , an ) is a separable extension of K.

242

Proof. Denote M := K(a1 , . . . , an ). Since [M : K] < ∞ by Proposition 63.4 it suffices to show that KM p = M . We have KM p = K(K(a1 , . . . , an ))p = K(K p (ap1 , . . . , apn )) = K(ap1 , . . . , apn ) By (62.10) we also have K(ap1 , . . . , apn ) = K(a1 , . . . , an ) = M Therefore KM p = M .

63.7 Theorem. If M/L and L/K are separable extensions then M/K is also a separable extension.

Proof. Let a ∈ M we need to show that a is separable over K. Let f (x) = irrLa (x), f (x) = xn + rn−1 xn−1 + . . . + r1 x + r0 . Define N := K(r0 , . . . , rn−1 ) ⊆ L Notice that f (x) = irrN a (x), and since f (x) is a separable polynomial the element a is separable over K. Moreover, we have 1) N/K is a separable extension (since L/K is separable and N ⊆ L) 2) [N : K] < ∞ Therefore by Proposition 63.4 we have KN p = N . Similarly, since 1) N (a)/N is a separable extension (by (63.6), since a is separable over N ) 2) [N (a) : N ] < ∞ thus N (N (a))p = N (a). Finally we have: 1) [N (a) : K] = [N (a) : N ] · [N : K] < ∞ 243

2) K(N (a))p = KN p (N (a))p = N (N (a))p = N so by Proposition 63.4 N (a)/K is a separable extension. Since a ∈ N (a) it is a separable element over K.

63.8 Proposition. Let L/K be a field extension and let Ksep (L) = {a ∈ L | a is separable over K} Then Ksep (L) is a subfield of L.

Proof. We need to show that if a, b ∈ L are separable elements over K, then a ± b, ab, a/b are also separable over K. This holds since by (63.6) K(a, b)/K is an separable extension and a ± b, ab, a/b ∈ K(a, b).

63.9 Definition. The separability degree of an extension L/K is the number [L : K]sep := [Ksep (L) : K]

244

64

Simple extensions

64.1 Definition. A field extension L/K is simple if L = K(a) for some a ∈ L.

64.2 Theorem. If K is an infinite field and L/K is a separable extension such that [L : K] < ∞ then L/K is a simple extension.

64.3 Note. Theorem 64.2 is true also if K is a finite field (later, see (69.5)).

Recall. If K is a field then K[x] is a Euclidean domain, so 1) for every f (x), g(x) ∈ K[x] there exists gcd(f (x), g(x)) ∈ K[x] 2) there exist λ(x), µ(x) ∈ K[x] such that gcd(f (x), g(x)) = λ(x)f (x) + µ(x)g(x)

64.4 Lemma. Let L be a field, let f (x), g(x), h(x) ∈ L[x] be polynomials such that h(x) ∼ gcd(f (x), g(x)) If K ⊆ L is a subfield such that f (x), g(x) ∈ K[x] then h(x) = ah1 (x) for some 0 6= a ∈ L and some h1 (x). In particular is h(x) is a monic polynomial then h(x) ∈ K[x]. Proof. Exercise.

245

64.5 Lemma. Let K be an infinite field, let L/K be a field extension and let f (x), g(x) ∈ L[x]. If f (0) 6= 0 and g(x) 6= 0 then there exists 0 6= d ∈ K such that gcd(f (x), g(dx)) ∼ 1 Proof. If deg f (x) = 0 or deg g(x) = 0 the statement is obvious. Otherwise we have f (x) = a · f1 (x) · . . . · fn (x) g(x) = b · g1 (x) · . . . · gn (x) where 0 6= a, b ∈ L and fi (x), gj (x) ∈ L[x] are irreducible monic polynomials: fi (x) = ai,0 + ai,1 x + . . . + xri gj (x) = bj,0 + bj,1 x + . . . + xsj Note: 1) Since f (0) 6= 0 thus ai,0 6= 0 for all i. 2) For any 0 6= d ∈ K we have g(dx) = b · g1 (dx) · . . . · gn (dx) Also, g1 (dx), . . . , gn (dx) are irreducible polynomials in L[x] since gj (dx) = ϕ(gj (x)) where ϕ is the ring isomorphism ϕ : L[x] → L[x],

ϕ(q(x)) := q(dx)

Let d ∈ K be an element such that gcd(f (x), g(dx)) 6∼ 1. Then for some i, j we have fi (x) ∼ gj (dx), i.e. fi (x) = cgj (dx) foe some 0 6= c ∈ L. This gives: ai,0 + ai,1 x + · · · + xri = cbj,0 + cbj,1 dx + · · · + cdsj xsj As a consequence we obtain: 246

1) ri = sj 2) ai,0 = cbj,0 3) 1 = cds

j

Therefore dsj = c−1 = bj,0 a−1 i,0 Since K is an infinite field we can find d ∈ K such that dsj 6= bj,0 a−1 i,0 for all i, j. Then fi (x) 6∼ gj (dx) for all i, j, and so gcd(f (x), g(dx)) ∼ 1.

64.6 Lemma. Let K be an infinite field and let L/K be a separable extension such that L = K(a, b) let a, b ∈ L. There exists c ∈ L such that K(a, b) = K(c)

K Proof. Let f (x) = irrK a (x), g(x) = irrb (x). Since f (a) = 0 we have

f (x) = (x − a)f1 (x) for some f1 (x) ∈ L[x]. Moreover, since a is separable over K we have f1 (a) 6= 0. Denote: F (x) := f1 (x + a),

G(x) := g(x + b)

We have F (0) = f1 (a) 6= 0 and G(x) 6= 0, so by Lemma 64.5 there exists 0 6= d ∈ K such that gcd(F (x), G(dx)) ∼ 1 Take c := b − da. We will show that K(a, b) = K(c).

It is enough to show that a ∈ K(c). Indeed, we have K(c) ⊆ K(a, b), and if a ∈ K(c) then also b = c + da ∈ K(c), so in such case K(a, b) ⊆ K(c). To see that a ∈ K(x) consider the ring isomorphism ϕ : L[x] → L[x],

ϕ(q(x)) = q(x − a) 247

Since gcd(F (x), G(dx)) ∼ 1, thus also gcd(ϕ(F (x)), ϕ(G(dx))) ∼ 1. Moreover we have ϕ(F (x)) = ϕ(f1 (x + a)) = f1 ((x − a) + a) = f1 (x) ϕ(G(dx)) = ϕ(g(dx + b)) = g(d(x − a) + b) = g(dx − da + b) = g(dx + c) Denote h(x) := g(dx + c) ∈ K(c)[x]. We have: h(a) = g(da − da + b) = 0 so h(x) = (x − a)h1 (x) for some h1 (x) ∈ L[x]. This gives: gcd(f (x), h(x)) = gcd((x − a)f1 (x), (x − a)h1 (x)) = (x − a) · gcd(f1 (x), h1 (x)) Since gcd(f1 (x), h(x)) ∼ 1 and h1 (x) | h(x) we have gcd(f1 (x), h1 (x)) ∼ 1. Therefore we get gcd(f (x), h(x)) ∼ (x − a) Finally, notice that f (x), h(x) ∈ K(c)[x]. Thus, by Lemma 64.4, (x − a) ∈ K(c)[x], and so a ∈ K(c).

Proof of Theorem 64.2. Since [L : K] < ∞ we have L = K(a1 , . . . , an ) for some a1 , . . . , an ∈ L. We will show that L/K is a simple extension by induction with respect to n. If n = 1 then L = K(a1 ), so L/K is simple. 248

Next, assume that for some n any extension K(a1 , . . . , an )/K is simple, and let L = K(a1 , . . . , an+1 ). Then we have L = K(a1 , . . . , an )(an+1 ) By the inductive assumption K(a1 , . . . , an ) = K(b) for some b ∈ L, so we obtain L = K(b)(a) = K(b, a) Finally, by Lemma 64.6 we have K(a, b) = K(c) for some c ∈ L, and so L/K is simple.

249

65

Simple extensions and intermediate fields

65.1 Definition. If L/K is a field extension then an intermediate field of L/K is a field M such that K⊆M ⊆L 65.2 Theorem. If L/K is a field extension, K is an infinite field and [L : K] < ∞ then L/K is a simple extension iff L/K has finitely many intermediate fields. 65.3 Note. Theorem 65.2 is true also when K is a finite field. If K is finite and [L : K] < ∞ then L is also a finite field, and so it has L finitely many subfields. Also, every finite extension of a finite field is simple (see Theorem 69.5).

65.4 Corollary. If L/K is a separable extension, K is an infinite field and [L : K] < ∞ then L/K has finitely many intermediate fields. Proof. By (64.2) L/K is a simple extension, so this follows from Theorem 65.2.

Proof of Theorem 65.2. (⇒) Let L/K be a simple extension: L = K(a) for some a ∈ L, and let f (x) = irrK a (x). Let M be an intermediate field of L/K and let g(x) = irrM (x). Since K ⊆ M we have g(x) | f (x) in L[x]. a Claim. If g(x) = r0 + r1 x + · · · + rn−1 xn−1 + xn then M = K(r0 , . . . , rn−1 ) 250

Indeed, denote N := K(r0 , . . . , rn−1 ). Notice that g(x) = irrN a (x). Since M (a) = L = N (a) this gives [L : M ] = degM (a) = deg g(x) = degN (a) = [L : N ] On the other hand we have [L : N ] = [L : M ] · [M : N ] Therefore [M : N ] = 1, and so M = N . We obtained that if K ⊆ M ⊆ L then M = K(r0 , . . . , rn−1 ) where g(x) = r0 + r1 x + · · · + rn−1 xn−1 + xn is some monic polynomial in L[x] such that g(x) | f (x). Since there are only finitely many such polynomials there are only finitely many such fields M . (⇐) Assume that L/K has finitely many intermediate fields. We want to show: L = K(c) for some c ∈ L. Since [L : K] < ∞ we have L = K(a1 , . . . , an ) for some a1 , . . . , an ∈ L. We can assume that n = 2, since then we can argue by induction as in the proof of (64.2). Thus, assume that we have L = K(a1 , a2 ). For k ∈ K define ck := a1 + ka2 ∈ L We have K ⊆ K(ck ) ⊆ L

Notice that since K is an infinite field there are infinitely many elements of the form ck . On the other hand, by assumption L/K has finitely many intermediate fields, so there exist k, k 0 ∈ K, k 6= k 0 such that K(ck ) = K(ck0 ) 251

We will show K(ck ) = K(a1 , a2 ) = L. We have K(ck ) ⊆ K(a1 , a2 ). Also, since ck , ck0 ∈ K(ck ) we have K(ck ) 3 ck − ck0 = (a1 + ka2 ) − (a1 + k 0 a2 ) = (k − k 0 )a2 Since 0 6= k − k 0 ∈ K this gives a2 ∈ K(ck ), and then also a1 ∈ ck − ka2 ∈ K(ck ) Therefore K(a1 , a2 ) ⊆ K(ck ).

252

66

Construction of extensions

66.1 Proposition. Let K be a field and let p(x) ∈ K[x] be an irreducible polynomial. There exists a filed extension L/K such that p(x) has a root in L.

Proof. Let K[t] be the ring of polynomials of variable t. Since p(t) ∈ K[t] is an irreducible polynomial the ideal I = hp(x)i is a maximal ideal of K[t], so K[t]/I is a field. Denote L := K[t]/I. We have an embedding i : K → L,

i(a) = a + I

so we can identify K with a subfield of L. Assume that p(x) = a0 + a1 x + . . . + an xn Under the above identification we have p(x) = (a0 + I) + (a1 + I)x + . . . + (an + I)xn Let β := t + I ∈ L. We have p(β) = (a0 + I) + (a1 + I)(t + I) + . . . + (an + I)(t + I)n = (a0 + a1 t + . . . + an T n ) + I = p(t) + I =0+I Therefore p(β) = 0 in L.

66.2 Definition. Let L/K be a field extension and let p(x) ∈ K[x]. We say that p(x) splits over L if p(x) decomposes into a product of polynomials of degree 1 in L[x] 253

66.3 Proposition. Let K be a field and let p(x) ∈ K[x] be a polynomial such that deg p(x) > 0. There exists a field extension M/K such that p(x) splits over M . Proof. We argue by induction with respect to n = deg p(x). If n = 1 then p(x) = ax + b, and we can take M = K. Next, assume that the statement holds for some n and let deg p(x) = n + 1. We have p(x) = p1 (x) · . . . · pk (x) where p1 (x), . . . , pk (x) ∈ K[x] are irreducible polynomials. By Proposition 66.1 there exists an extension L/K such that p1 (x) has a root a ∈ L. Then we have p1 (x) = (x − a)q1 (x) for some q1 (x) ∈ L[x]. Take p¯(x) := q1 (x)p2 (x) · . . . · pk (x) ∈ L[x] We have p(x) = (x − a)¯ p(x). Moreover, since deg p¯(x) = n by inductive assumption there exists a field extension M/L such that p¯(x) splits over M . If follows that p(x) also splits over M .

66.4 Definition. If K is a field and p(x) ∈ K[x] then we say that a field L is a splitting field of p(x) if 1) K ⊆ L

2) p(x) splits over L 3) p(x) does not split over any proper subfield of L.

66.5 Proposition. Let K be a field. For any polynomial p(x) ∈ K[x], such that deg p(x) > 0 there exists a splitting field of p(x).

254

Proof. By Proposition 66.3 there exists a field extension L/K such that p(x) splits over L: p(x) = b(x − a1 )(x − a2 ) · . . . · (x − an ) for some b ∈ K, a1 , . . . , an ∈ L. Then K(a1 , . . . , an ) is a splitting field of p(x).

66.6 Example. Take p(x) = x4 − 10x2 + 1 ∈ Q[x]. The polynomial p(x) splits over R[x]: √ √ √ √ √ √ √ √ p(x) = (x−( 2+ 3))·(x−(− 2− 3))·(x−(− 2+ 3))·(x−( 2− 3)) √ √ √ √ Therefore L = Q(±( 2 + 3), ±( 2 − 3)) is a splitting field of p(x). √ √ Note: L = Q( 2, 3).

66.7 Proposition. Let K be a field, let p(x) ∈ K[x], deg p(x) > 0 and let L, L0 be splitting fields of p(x). There exists an isomorphism ϕ : L → L0 such ϕ|K = idK .

Proof. Exercise.

255

67

Algebraically closed fields

67.1 Definition. A field is algebraically closed if the only irreducible polynomials in K[x] are polynomials of degree 1.

67.2 Proposition. Let K be a field. The following conditions are equivalent. (i) K is algebraically closed. (ii) If p(x) ∈ K[x] and deg p(x) > 0 then p(x) splits over K.

(iii) If p(x) ∈ K[x] and deg p(x) > 0 then p(x) has a root in K. (iv) K is the only algebraic extension of K.

Proof. Exercise.

67.3 Proposition. Every algebraically closed field is infinite.

Proof. Let K be a finite field, K = {a1 , . . . , an }. Take the polynomial p(x) ∈ K[x] given by p(x) = (x − a1 )(x − a2 ) · . . . · (x − an ) + 1 We have p(ai ) = 1 for all ai ∈ L, so p(x) has no roots in K. Therefore K is not algebraically closed.

67.4 Definition. Let L/K be a field extension. The field L is an algebraic closure of K if L is an algebraically closed field and L/K is an algebraic extension.

256

67.5 Proposition. Let L/K be an algebraic field extension. The following conditions are equivalent (i) L is an algebraic closure of K. (ii) If p(x) ∈ K[x] is a polynomial such that deg p(x) > 0 then p(x) splits over L.

Proof. (i) ⇒ (ii) If L is an algebraic closure of K, then L is an algebraically closed field and so by Proposition 67.2 every polynomial p(x) ∈ L[x] such that deg p(x) > 0 splits over L. Since K[x] ⊆ L[x] the statement of (ii) follows. (iI) ⇒ (i) Since L/K is an algebraic extension we only need to show that L is an algebraically closed field. By Proposition 67.2 it will suffice to show that for any algebraic extension M/L we have M = L. Let then M/L be an algebraic extension. In such case the extension M/K is also algebraic so if a ∈ M then a is a root of some polynomial p(x) ∈ K[x]. Since by assumption p(x) splits over L we get that a ∈ L. Therefore M = L.

¯ ¯ is 67.6 Theorem. For any field K there exists an extension K/K such that K an algebraic closure of K.

67.7 Lemma. Let M/K be an algebraic extension. 1) If K is an infinite field then |K| = |M |. 2) If K is a finite field then |M | ≤ ℵ0 .

Proof. Exercise. 257

Proof of Theorem 67.6. Let U be a set such that K ⊆ U and |U | > |K|. If K is a finite field assume also that U is an uncountable set. Let S be the set of all fields L such that L ⊆ U and L is an algebraic extension of K. Define a partial ordering on S where L ≤ L0 if L0 is a field extension of L. Check: assumptions of Zorn’s Lemma 29.10 are satisfied in S, and so S contains ¯ a maximal element K. ¯ is an algebraic closure of K. Since the extension K/K ¯ We will show that K is ¯ algebraic we only need to show that K is an algebraically closed field. ¯ is not algebraically closed. By Proposition Assume, by contradiction, that K ¯ where M 6= K. ¯ Then M/K is 67.2 there is then an algebraic extension M/K also an algebraic extension and so by Lemma 67.7 we have |M | < |U |. As a consequence there exists a monomorphism of sets ϕ: M → U such that ϕ|K¯ = idK¯ . We can introduce a field structure on the set ϕ(M ) such ¯ < ϕ(M ) which is that ϕ becomes a field isomorphism. Then ϕ(M ) ∈ S and K ¯ impossible by maximality of K.

¯ 67.8 Proposition. Let K/K be an algebraic closure and let L/K be any algebraic extension of K. There exists an embedding ¯ ϕ: L → K such that ϕK = idK .

67.9 Lemma. Let L/M be a field extension, let a ∈ L be an algebraic element over M , and let ¯ ψ: M → K ¯ is algebraically closed. Then there exists be a field homomorphism such that K ¯ M = ψ. ¯ such that ψ| a homomorphism ψ¯ : M (a) → K 258

Proof. Exercise (compare with (59.8)).

Proof of Proposition 67.8. Let S be a set of all pairs (M, ψ) such that (i) M is a field such that K ⊆ M ⊆ L ¯ is a homomorphism such that ψ|K = idK (ii) ψ : M → K Define partial ordering on S as follows: (M, ψ) ≤ (M 0 , ψ 0 )

if M ⊆ M 0 and ψ 0 |M = ψ

Check: assumptions of Zorn’s Lemma 29.10 are satisfied in S, and so S contains a maximal element (M0 , ψ0 ). Assume that M0 6= L and let a ∈ L/M0 . By Lemma 67.9 there exists a homomorphism ¯ ψ¯0 : M0 (a) → K such that ψ¯0 |M0 = ψ0 . It follows that (M0 , ψ0 ) < (M0 (a), ψ¯0 ) which is impossible since (M0 , ψ0 ) is a maximal element. Therefore M0 = L and we can take ϕ = ψ0 .

67.10 Note. Proposition 67.8 can be generalized as follows. Let K ⊆ L ⊆ M ¯ be an algebraic closure of K, and let be algebraic extensions, let K ¯ ϕ˜ : L → K be an embedding such that ϕ|K = idK . Then there exists an embedding ¯ ϕ: M → K such that ϕL = ϕ˜ (exercise).

259

67.11 Corollary. If L/K, L0 /K are algebraic closures of K then there exists an isomorphism ϕ : L → L0 such that ϕ|K = idK .

Proof. By Proposition 67.8 there exists a homomorphism ϕ : L → L0 such that ϕ|K = idK . It suffices to show that ϕ is an epimorphism. We have K = ϕ(K) ⊆ ϕ(L) ⊆ L0 Since L is algebraically closed and ϕ(L) ∼ = L we get that ϕ(L) is also algebraically closed and so (by Proposition 67.2) the only algebraic extension of ϕ(L) is ϕ(L) itself. On the other hand, since L0 /K is an algebraic extension and K ⊆ ϕ(L) the extension L0 /ϕ(L) is algebraic. This gives that L0 = ϕ(L).

67.12 Fundamental Theorem of Algebra. The field C of complex numbers is algebraically closed.

Proof. Later.

Recall. If L/K is a field extension then Kalg (L) = {a ∈ L | a is an algebraic element over K} is a subfield of L.

260

67.13 Proposition. If L/K is a field extension and L is an algebraically closed field then the field Kalg (L) is also algebraically closed. As a consequence Kalg (L) is an algebraic closure of K.

Proof. Let p(x) ∈ Kalg (L)[x], deg p(x) > 0. It suffices to show that p(x) has a root in Kalg (L). The field L is algebraically closed, so p(x) has a root a ∈ L. Consider the extensions Kalg (L)(a)/Kalg (L) and Kalg (L)/K These extensions are algebraic, so by Proposition 60.6 Kalg (L)(a)/K is also an algebraic extension. This means in particular that a ∈ L is an algebraic element over K, so a ∈ Kalg (L). ¯ := Qalg (C) is an 67.14 Note. Since C is an algebraically closed field thus Q algebraic closure of Q.

261

68

Roots of unity

68.1 Definition. Let K be a field. We say that a ∈ K is an n-th root of unity if an = 1, i.e. if a is a root of the polynomial fn (x) = xn − 1 ∈ K[x].

68.2 Proposition. Let K be a field and let µn (K) := {a ∈ K | an = 1} Then µn (K) is a subgroup of the multiplicative group K ∗ = K − {0}. Proof. Exercise.

68.3 Note. 1) If m|n then µm (K) is a subgroup of µn (K). 2) If χ(K) = p 6= 0 and n = pk m then µn (K) = µm (K). Indeed, we have fn (x) = xp

km

k

k

− 1p = (xm − 1)p = (fm (x))p

k

so roots of fn are the same as roots of fm .

68.4 Theorem. For any field K the group µn (K) is cyclic and |µn (K)| | n. Proof. Since µn (K) is a finite subgroup of K ∗ it is cyclic by 38.8. so µn (K) = hai for some a ∈ µn (K). This gives |µn (K)| = |a|. Also, since an = 1, we have |a| | n.

262

68.5 Corollary. Let K be an algebraically closed field. If χ(K) = p 6= 0 assume also that p - n. Then we have µn (K) ∼ = Z/nZ Proof. By Theorem 68.4 it is enough to show that |µn (K)| = n. Let fn (x) = xn − 1 ∈ K[x]. We have fn0 (x) = nxn−1 Since fn (x) and fn0 (x) have no common roots thus by Proposition 62.3 fn (x) is a separable polynomial. As a consequence fn (x) has n distinct roots in K, and so |µn (K)| = n. 68.6 Definition. An element a ∈ K is a primitive n-th root of unity if an = 1 and am 6= 1 for all 0 < m < n.

68.7 Note. 1) A primitive n-th root of unity exists in K iff |µn (K)| = n. In such case a ∈ K is primitive n-th roots of unity if a generates µn (K). 2) If |µn (K)| = m then µn (K) = µm (K) and µn (K) is generated by a primitive m-th root of unity in K.

68.8 Examples. 1) In Q for any n ≥ 1 we have µ2n−1 (Q) = {1} = µ1 (Q),

µ2n (Q) = {1, −1} = µ2 (Q)

Thus 1 is a primitive 1-st root of unity and −1 is a primitive 2-nd root of unity. For n > 2 there are no primitive n-roots of unity in Q. 263

2) Since C is an algebraically closed field we have |µn (C)| = n for all n > 0, and so C contains a primitive n-th root of unity for every n. We have µn (C) = {εk := cos(2πk/n) + i sin(2πk/n) | k = 0, . . . , n − 1} and εk is a primitive n-th root of unity iff gcd(k, n) = 1.

264

69

Finite fields

69.1 Proposition. If K is a finite field and χ(K) = p then |K| = pn for some n ≥ 1. Proof. Since χ(K) thus by Theorem 58.6 Fp can be identified with the primitive subfield of K. Since K is a finite field we have [K : Fp ] = n for some n ≥ 1. This means that K is an n dimensional vector space over Fp , and so K has pn elements.

69.2 Proposition. If K is a finite field, such that χ(K) = p then |K| = pn iff K is a splitting field of the polynomial n

f (x) = xp − x ∈ Fp [x]

69.3 Lemma. If K be a field and ϕ : K → K be a field homomorphism then L := {a ∈ K | ϕ(a) = a} is a subfield of K.

Proof. Exercise.

Proof of Proposition 69.2. (⇒) We will show that if |K| = pn then every element of K is a root of f (x). Since f (x) can have at most pn roots it will follow that K is a splitting field of f (x).

265

Let a ∈ K. If a = 0 then a is a root of f (x). Assume then that a 6= 0. Then a is an element of the multiplicative group K ∗ = K − {0}. Since |K ∗ | = pn − 1 n thus we have we have ap −1 = 1. This gives n

f (a) = ap − a = a(ap

n −1

− 1) = 0

(⇐) Assume that K is a splitting field of f (x). Consider the field homomorphism ϕ(a) = ap

ϕ : K → K,

n

By Lemma 69.3 the set L = {a ∈ K | ϕ(a) = a} is a subfield of K. On the other hand L consists of all roots of f (x) in K, and so since f (x) splits over K it also splits over L. Since K is a splitting field of f (x) this implies that K = L, i.e. every element of K is a root of f (x). It suffices to show then that f (x) has pn distinct roots, i.e. that it is a separable polynomial. This follows from Proposition 62.3, since f 0 (x) = −1, and so f (x) and f 0 (x) have no roots in common.

69.4 Corollary. For every prime p and n ≥ 1 there exists a field K such that χ(K) = p and |K| = pn . Moreover such field is unique up to isomorphism. n

Proof. Take the polynomial f (x) = xp − x ∈ Fp [x]. By Proposition 66.5 there exists a splitting field K of f (x) and by Proposition 69.2 we have |K| = pn . Assume now that K 0 is another field such that |K 0 | = pn . By Proposition 69.2 we have that K 0 is also a splitting field of f (x). Uniqueness of splitting fields (66.7) gives then that K 0 ∼ = K.

266

Recall: 64.2 Theorem. If K is an infinite field and L/K is a separable extension such that [L : K] < ∞ then L/K is a simple extension. 69.5 Theorem. If K is an finite field and L/K is a field extension such that [L : K] < ∞ then L/K is a simple extension.

69.6 Note. If K is a finite field then any algebraic extension L/K is a separable extension (exercise).

Proof of Theorem 69.5. Assume that χ(K) = p, and that Fp ⊆ K. We will show that a L = Fp (c) for some c ∈ L. Since Fp (c) ⊆ K(c) this will imply that K(c) = L. Since K is a finite field and [L : K] < ∞ the field L is finite, and so |L| = pn for some n ≥ 1. Let L∗ = L − {0} be the multiplicative group of L. By the proof of Proposition 69.2 we have L∗ = {a | ap

n −1

= 1}

i.e. L∗ is the group of (pn − 1)-st roots of unity in L: L∗ = µpn −1 (L) By Theorem 68.4 µpn −1 (L) is a cyclic group. Let c be a generator of µpn −1 (L). Then we have L = Fp (c).

267

70

Galois theory - motivation

70.1 Proposition. Let L/K be a field extension and let ϕ: L → L be an automorphism such that ϕ|K = idK . If a ∈ L is a root of a polynomial f (x) ∈ K[x] then ϕ(a) is also a root of f (x). Proof. If f (x) = r0 + r1 x + . . . + rn xn then we have f (ϕ(a)) = r0 + r1 ϕ(a) + . . . + rn ϕ(a)n Also, since ri ∈ K we have ri = ϕ(ri ), so we obtain f (ϕ(a)) = ϕ(r0 ) + ϕ(r1 )ϕ(a) + . . . + ϕ(rn )ϕ(a)n = ϕ(f (a)) = ϕ(0) =0

70.2 Corollary. Let L/K be an algebraic extension, let f (x) ∈ K[x], and let S = {a1 , . . . , an } ⊆ L be the set of all roots of f (x) in L. If ϕ: L → L is an automorphism such that ϕ|K = idK then ϕ|S is a permutation of S. Moreover, if L is a splitting field of f (x) then the automorphism ϕ is uniquely determined by this permutation.

Proof. By Proposition 70.1 we have ϕ(S) ⊆ S. Also, since ϕ is a bijection and S is a finite set thus ϕ|S : S → S is also a bijection i.e. it is a permutation of the set S. 268

Recall that if L is a splitting field of f (x) we have L = K(a1 , . . . , an ) Thus, since ϕ|K = id, the automorphism ϕ is determined by the restriction ϕ|{a1 ,...,an } .

Note. Let L be a splitting field of a polynomial f (x) ∈ K[x], and let S ⊆ L be the set of all roots of f (x). It is not true that every permutation of S can be extended to an automorphism ϕ : L → L such that ϕ|K = idK . 70.3 Example. Let f (x) = x4 − 10x2 + 1 ∈ Q[x]. Roots of f (x) in C are: √ √ √ √ √ √ √ √ 2 + 3, 2 − 3, − 2 + 3, − 2 − 3 √ √ It follows that the field L := Q( 2, 3) is a splitting field of f (x). Let ϕ: L → L

√ be√an automorphism such that ϕ|Q = idQ . Notice that we must have ϕ( 2) = √ 2 ± √2 since ± √ 2 are the only roots of g(x) = x − 2. Similarly we must have ϕ( 3) = ± 3. √ √ √ √ Assume that ϕ( 2 + 3) = 2 − 3. Then we must have √ √ √ √ ϕ( 2) = 2, and ϕ( 3) = − 3 This gives that √ √ √ √ ϕ( 2 − 3) = 2 + 3 √ √ √ √ ϕ(− 2 + 3) = − 2 − 3 √ √ √ √ ϕ(− 2 − 3) = − 2 + 3

269

By this argument one gets that there are only 4 permutations of the set of roots f (x) that can extend to an automorphism L, and that each of these √ of √ permutations is determined by the value of ϕ( 2 + 3). Note. Let L be a splitting field of f (x) ∈ K[x], and let S ⊆ L be the set of all roots of f (x). Permutations of S that extend to automorphisms of L form a subgroup of the group of permutations of S (check!). We will see that properties of this group provide information about the polynomial f (x) and its roots.

270

71

Normal extensions

71.1 Definition. Let L/K be an algebraic extension and let Gal(L/K) := {ϕ : L → L | ϕ is an automorphism and ϕ|K = idK } Then Gal(L/K) is a group (with composition of automorphism), and it is called the Galois group of the extension L/K.

71.2 Proposition. Let L be a field and let Aut(L) be the group of all automorphisms of L. If H is a subgroup of Aut(L) then the set LH = {a ∈ L | ϕ(a) = a for all ϕ ∈ H} is a subfield of L.

Proof. Exercise.

71.3 Definition. If L is a field and H ⊆ Aut(L) is a subgroup then the subfield LH ⊆ L is called the fixed field of H.

71.4 Note. Let L be a field. We have maps     subfields subgroups Φ:  :Ψ of L of Aut(L) where Φ(K) = Gal(L/K) and Ψ(H) =√LH . In general these maps are not inverses of each other. Take e.g. L = Q(3 2) and let K = Q ⊆ L. We have Φ(K) = Gal(L/K) = {idL } (check!). This gives Ψ(Φ(K)) = LGal(L/K) = L 6= K 271

71.5 Definition. An algebraic extension L/K is called a Galois extension if K = LGal(L/K) .

Goal. An extension L/K is a Galois extension iff it is a normal and separable extension.

71.6 Definition. An algebraic extension L/K is a normal extension if for every a ∈ L the polynomial irrK a (x) splits over L.

71.7 Example. ¯ is the algebraic closure of K then the extension K/K ¯ 1) If K is normal. √ √ √ 2) The extension Q(3 2)/Q is not normal. Indeed, 3 2 ∈ Q(3 2), but the polynomial Q 3 irr√ 3 2 (x) = x − 2 √ does not split over Q(3 2).

¯ be an algebraic closure of a field K. An algebraic 71.8 Proposition. Let K extension L/K is normal iff for any two homomorphisms ¯ ϕ, ψ : L → K such that ϕ|K = ψ|K = idK we have ϕ(L) = ψ(L).

Proof. ¯ be homomorphisms satisfying ϕ|K = ψ|K = idK . It is (⇒) Let ϕ, ψ : L → K enough to show that for any a ∈ L we have ϕ(a) ∈ ψ(L).

272

Let a ∈ L and let f (x) = irrK a (x). Since L/K is a normal extension f (x) splits over L: f (x) = (x − a1 )(x − a2 ) · . . . · (x − an ) where ai ∈ L and a1 = a. We have ϕ(f (x)) = (x − ϕ(a1 ))(x − ϕ(a2 )) · . . . · (x − ϕ(an )) ψ(f (x)) = (x − ψ(a1 ))(x − ψ(a2 )) · . . . · (x − ψ(an )) On the other hand, since f (x) ∈ K[x] and ϕ|K = ψ|K = idK we have ϕ(f (x)) = f (x) = ψ(f (x)) ¯ Since K[x] is a UFD and x − ϕ(ai ), x − ψ(ai ) are irreducible polynomials in ¯ K[x] we must have x − ϕ(a1 ) = x − ψ(ai ) for some i. Since a1 = a this gives ϕ(a) = ψ(ai ) ∈ ψ(L)

(⇐) Assume that L/K is an algebraic extension which is not normal. Then there exists a ∈ L such that the polynomial f (x) = irrK a (x) does not split over ¯ L. Let ϕ : L → K be a homomorphism such that ϕ|K = idK . Since L ∼ = ϕ(L) we have that ϕ(a) ∈ ϕ(L) is a root of f (x), but f (x) does not split over ϕ(L). ¯ so there exists b ∈ K ¯ − ϕ(L) such that On the other hand f (x) splits over K, f (b) = 0. By Proposition 59.8 there exists an isomorphism ψ˜ : K(a) → K(b) ˜ K = idK . By (67.10) ψ˜ can be extended to a homomorphism such that ψ| ¯ ψ: L → K Since b ∈ ψ(L) and b 6∈ ϕ(L) we get that ϕ(L) 6= ψ(L).

71.9 Proposition. Let L/K be a field extension such that [L : K] < ∞. The extension L/K is normal iff L is a splitting field of some polynomial f (x) ∈ K[x]. 273

Proof. (⇒) Since [L : K] < ∞ we have L = K(a1 , . . . , an ) for some a1 , . . . , an ∈ L. Let fi (x) = irrK ai (x). Since L/K is a normal extension fi (x) splits over L for each i, so the polynomial f (x) = f1 (x) · . . . · fn (x) also splits over L. Moreover, since a1 , . . . , an are roots of f (x) and they generate L, thus L is a splitting field of f (x). (⇐) Let L be a splitting field of some polynomial f (x) ∈ K[x]. Then we have f (x) = (x − a1 )(x − a2 ) · . . . · (x − an ) for some a1 , . . . , an ∈ L, and L = K(a1 , . . . , an ). ¯ be an algebraic closure of K and let Let K ¯ ϕ, ψ : L → K be field homomorphisms such that ϕ|K = ψ|K = idK . We have ϕ(f (x)) = (x − ϕ(a1 ))(x − ϕ(a2 )) · . . . · (x − ϕ(an )) ψ(f (x)) = (x − ψ(a1 ))(x − ψ(a2 )) · . . . · (x − ψ(an )) On the other hand, since f (x) ∈ K[x] e have ϕ(f (x)) = f (x) = ψ(f (x)) It follows that for every 1 ≤ i ≤ n there exists 1 ≤ j ≤ n such that ϕ(ai ) = ψ(aj ). This gives ϕ(L) = K(ϕ(a1 ), . . . , ϕ(an )) ⊆ K(ψ(a1 ), . . . , ψ(an )) = ψ(L) Similarly we get ψ(L) ⊂ ϕ(L). Therefore ϕ(L) = ψ(L), and so by Proposition 71.8 L/K is a normal extension.

274

71.10 Proposition. Let K, L, M, N be fields such that K, L, M ⊆ N , N/K is an algebraic extension and L/K is a normal extension. Then LM/KM is also a normal extension.

71.11 Corollary. If K ⊆ L ⊆ M are field extensions and M/K is normal then M/L is also normal.

Proof. We have M = M L, L = KL, and by Proposition 71.10 the extension M L/KL is normal.

Proof of Proposition 71.10. Let KM be an algebraic closure of KM , and let ϕ, ψ : LM → KM be two embeddings of LM such that ϕ|KM = ψ|KM = id|KM . By Proposition 71.8 it is enough to show that ϕ(LM ) = ψ(LM ). Notice that since KM/K and N/KM are algebraic extensions we have ¯ =K ¯ KM = N Since ϕ|K = ψ|K = id|K and L/K is a normal extension we have ϕ(L) = ψ(L). Moreover, ϕ|M = ψM = idM , so ϕ(M ) = ψ(M ) = M . This gives ϕ(LM ) = ϕ(L)ϕ(M ) = ψ(L)ψ(M ) = ψ(LM )

71.12 Note. It is not true that if K ⊆ L ⊆ M and M/L are L/K are normal extensions then M/K is also normal. Take e.g. √ √ Q ⊆ Q( 2) ⊆ Q(4 2)

275

71.13 Definition. Let L/K be an algebraic extension. An normal closure of L/K is an extension M/K such that 1) L ⊆ M

2) M/K is a normal extension 3) M does not have any proper subfields satisfying 1) and 2).

71.14 Proposition. For any L/K is any algebraic extension there exists a normal closure M/K of L/K. Moreover, if M , M 0 are normal closures of L/K then there exists an isomorphism ϕ: M → M0 such that ϕ|L = idL . ¯ be an algebraic closure of L, and let S = {Mα } be the set of all Proof. Let L ¯ such that L ⊆ Mα and that Mα /K is a normal extension. The subfields of L ¯ ∈ S. Take M = T Mα . Check: M/K is a normal set S is non-empty since L α extension, and so it is a normal closure of L/K. Let M 0 /K be another normal closure of L/K. Since M 0 /L is an algebraic extensions by (67.8) there exists an embedding ¯ ϕ: M0 → L ¯ and ϕ(M )/K is a normal such that ϕ|L = id|L . Since L ⊆ ϕ(M 0 ) ⊆ L extension, thus by the construction of M we have M ⊆ ϕ(M 0 ). On the other hand ϕ(M 0 ) is a normal closure of L/K. This implies that ϕ(M ) = M 0 .

276

72

Galois extensions

72.1 Theorem. An algebraic extension L/K is a Galois extension iff L/K is separable and normal.

Proof. (⇒) Let L/K be a Galois extension let a ∈ L, and let f (x) = irrK a (x). It is enough to show that f (x) splits over L and that it is a separable polynomial. Let S = {a1 , . . . , an } be the set of all distinct roots of f (x) in L where, say, a = a1 . Let g(x) = (x − a1 )(x − a2 ) · . . . · (x − an ) ∈ L[x] Notice that g(x) is a separable polynomial that splits over L. Therefore it suffices to show that f (x) = g(x). Furthermore, it is enough to show that f (x)|g(x). Indeed since g(x)|f (x) and since f (x) and g(x) are monic polynomials this will imply that f (x) = g(x). Let ϕ ∈ Gal(L/K). By Corollary 70.2 ϕ permutes the set {a1 , . . . , an }. This gives ϕ(g(x)) = (x − ϕ(a1 ))(x − ϕ(a2 )) · . . . · (x − ϕ(an )) = (x − a1 )(x − a2 ) · . . . · (x − an ) = g(x) Therefore, if g(x) = r0 + r1 x + · · · + rn xn then we have ϕ(ri ) = ri , i.e. ri ∈ LGal(L/K) for i = 0, . . . , n. Since L/K is a Galois extension we have LGal(L/K) = K, and so we obtain that g(x) ∈ K[x]. Since g(a) = g(a1 ) = 0 and f (x) = irrK a (x) this gives f (x)|g(x). (⇐) Let L/K be a separable and normal extension and let a ∈ L − K. We need to show that there exists ϕ ∈ Gal(L/K) such that ϕ(a) 6= a. 277

¯ be an algebraic closure of K such that L ⊆ K. ¯ If f (x) = irrK (x) ∈ K[x] Let K a then deg f (x) > 1 (since a 6∈ K), and f (x) has no multiple roots (since a is ¯ such that b 6= a and separable over K). It follows that there exists b ∈ K f (b) = 0. By Proposition 59.8 there exists an isomorphism ¯ ϕ˜ : K(a) → K(b) ⊆ K ¯ such that ϕ| ˜ K = idK . By (67.10) we can extend ϕ˜ to an embedding ϕ : L → K. ¯ be the inclusion homomorphism. Since L/K is a normal exLet i : L ,→ K tension thus by (71.8) we have ϕ(L) = i(L) = L. It follows that ϕ gives an automorphism ϕ : L → L. Since ϕ|K = idK we have ϕ ∈ Gal(L/K) and ϕ(a) = b 6= a.

72.2 Corollary. If K ⊆ L ⊆ M are fields and M/K is a Galois extension then M/L is also a Galois extension.

Proof. The extension M/K is normal and separable, so M/L is also normal (by (71.11)) and separable.

72.3 Theorem. Let L be a field, let Aut(L) be the group of automorphisms of L, and let H be a finite subgroup of Aut(L). Then: 1) L/LH is a Galois extension 2) [L : LH ] = |H| 3) Gal(L/LH ) = H

278

72.4 Note. Recall that for a field L we have maps of sets:     subgroups subfields :Ψ  Φ: of Aut(L) of L where Φ(K) = Gal(L/K) and Ψ(H) = LH . By part 3) of Theorem 72.3 we obtain that if H ⊆ Aut(L) is a finite subgroup then Φ(Ψ(H)) = H

Proof of Theorem 72.3. 1) We need to show that L/LH is an algebraic extension and that LH = LGal(L/L

H)

Notice that if G, H are subgroups of Aut(L) and G ⊆ H then LH ⊆ LG . Since H H ⊆ Gal(L/LH ) this gives LGal(L/L ) ⊆ LH . On the other hand, by definition H of Gal(L/LH ), if ϕ ∈ Gal(L/LH ) then ϕ|LH = idLH , so LH ⊆ LGal(L/L ) . H Thus we obtain LH = LGal(L/L ) . It remains to show that L/LH is an algebraic extension. Let a ∈ L and let Ha := {ϕ(a) | ϕ ∈ H} Since H is a finite group the set Ha is finite, say Ha = {a1 , . . . , an } where a1 = a. Let g(x) = (x − a1 )(x − a2 ) · . . . · (x − an ) ∈ L[x] Notice that if ϕ ∈ H then ϕ|Ha is a bijection: ϕ|Ha : Ha → Ha As a consequence we have ϕ(g(x)) = g(x) for all ϕ ∈ H. This means that g(x) ∈ LH [x]. Since g(a) = g(a1 ) = 0 we get that a is an algebraic element over LH .

279

2) Let a ∈ L. The argument in part 1) shows that a is a root of a polynomial g(x) ∈ LH [x] such that deg g(x) ≤ |H|. It follows that for any a ∈ L we have [LH (a) : LH ] ≤ |H|

(∗)

Claim. [L : LH ] ≤ |H|. Indeed, if [L : LH ] > |H| then we can find an intermediate field LH ⊆ K ⊆ L such that |H| < [K : LH ] < ∞. By part 1) L/LH is a Galois extension, so it is separable. This implies that K/LH is also a a separable extension. By (64.2) and (69.5) we obtain then that K/LH is a simple extension i.e. K = LH (a) for some a ∈ K. This gives [LH (a) : LH ] = [K : LH ] > |H| which contradicts the inequality (∗). We obtain that L/LH is a separable extension, and [L : LH ] ≤ |H| < ∞. It follows that L/LH is a simple extention, L = LH (a) for some a ∈ L. Let f (x) = irrK a (x). We have deg f (x) = [L : LH ] ≤ |H| On the other hand each automorphism ϕ ∈ Gal(L/LH ) is uniquely determined by the value of ϕ(a). Since ϕ(a) is a root of f (x) we obtain that | Gal(L/LH )| ≤ deg(f (x)) Since H ⊆ Gal(L/LH ) this gives |H| ≤ | Gal(L/LH )| ≤ deg f (x) ≤ |H| Therefore |H| = | Gal(L/LH )| = [L : LH ]. 280

3) We have H ⊆ Gal(L/LH ), and by the proof of part 2), |H| = | Gal(L/LH )|. Since H is a finite group this implies that H = Gal(L/LH ).

72.5 Corollary. Let L/K be a field extension such that [L : K] < ∞. The extension L/K is a Galois extension iff [L : K] = | Gal(L/K)|. Proof. (⇐) If L/K is a Galois extension then K = LGal(L/K) . Since [L : K] < ∞ by part 2) of Theorem 72.3 we obtain [L : K] = | Gal(L/K)|. (⇒) Assume that [L : K] = | Gal(L/K)|. We want to show that K = LGal(L/K) . We have K ⊆ LGal(L/K) ⊆ L, so | Gal(L/K)| = [L : K] = [L : LGal(L/K) ] · [LGal(L/K) : K] By part 2) of Theorem 72.3 we also have [L : LGal(L/K) ] = | Gal(L/K)|. It follows that [LGal(L/K) : K] = 1, and so K = LGal(L/K) .

281

73

Application: rational symmetric functions

Let L = K(t1 , . . . , tn ) be the field of rational functions of variables t1 , . . . , tn with coefficients in a field K, and let Sn be the symmetric group on n letters. The group Sn can be identified with a subgroup of the group Aut(L) of automorphims of L as follows. If σ ∈ Sn then σ induces an automorphism σ : L −→ L

such that σ|K = id|K and σ(ti ) = tσ(i) . 73.1 Definition. If function.

f (t1 ,...,tn ) g(t1 ,...,tn )

∈ LSn then we say that

f (t1 ,...,tn ) g(t1 ,...,tn )

is symmetric

73.2 Example. Let f (x) = (x − t1 )(x − t2 ) · · · · · (x − tn ) ∈ L[x]

For σ ∈ Sn we have

σ(f (x)) = (x − tσ(1) )(x − tσ(2) ) · · · · · (x − tσ(n) ) = f (x)

Therefore f (x) ∈ LSn [x]. Notice that we have

f (x) = xn − σ1 xn−1 + σ2 xn−2 − · · · + (−1)n−1 σn−1 x + (−1)n σn where σ1 =

X

ti

i

σ2 =

X

ti tj

i 0 f (0) = −1 < 0 so f (x) has a real root in each of the intervals (−q, −1) and (−1, 0). It follows that Gal(L/K) cannot be a solvable group.

311

81

Straightedge and compass constructions

81.1. Motivation. Three great problems of antiquity. 1) Squaring of a circle. Using a straightedge and a compass construct a square whose area is equal to the area of a given circle. 2) Doubling of a cube. Using a straightedge and a compass construct a cube whose volume is double the volume of a given cube. 3) Trisection of an angle Using a straightedge and a compass construct an angle whose measure is one third of the measure of a given angle.

Cartesian reformulation. 1) Squaring of a circle. We can assume that the given circle has its center at the point (0, 0) and that it passes through the point (1,√ 0). Squaring of this circle amounts to constructing the point with coordinates ( π, 0)

(0, 0)

(1, 0)

√ ( π, 0)

2) Doubling of a cube. We can assume that two vertices of the front face of the cube have coordinates (0, 0) and (1,√ 0). Doubling of this cube amounts to 3 constructing the point with coordinates ( 2, 0) 312

(0, 0)

√ (3 2, 0)

(1, 0)

3) Trisection of an angle. We can assume that one of the arms of the given angle θ coincides with the x-axis. Being given such an angle is equivalent to being given the point with coordinates (cos θ, 0). Trisection of this angle is equivalent to constructing the point with coordinates (cos 13 θ, 0)

θ 1 θ 3

(0, 0)

(cos θ, 0) (cos

313

1 θ, 0) 3

(1, 0)

81.2. General constructibility problem. Assume that on the coordinate plane R2 we are given points a set of points S0 = {P1 , . . . , Pn } where Pi = (ai , bi ). Assume also that (0, 0), (1, 0) ∈ S0 . We can perform two operations: • draw a line through any two points in S0 • draw a circle with the center in one point of S0 and passing through another point of S0 . Let S1 be the set of all intersection points of these lines and circles. Recursively, for each k ≥ 1 let Sk be the set of all points obtained in this way from the set Sk−1 . Denote ∞ [ S= Sk k=0

81.3 Definition. A point Q ∈ R2 is constructible from the set S0 if Q ∈ S. A point Q ∈ R2 is constructible if it is constructible from the set {(0, 0), (1, 0)}. Problem. Given a set S0 ∈ R2 determine which points of R2 are constructible from S0 .

81.4 Note. √ 1) Squaring of the circle can be performed if the point ( π, 0) is constructible. √ 2) Doubling of a cube can be performed if the point (3 2, 0) is constructible. 3) Trisection of the angle θ can be performed if the point (cos 13 θ, 0) is constructible from the set {(0, 0), (1, 0), (cos θ, 0)}.

314

Constructible points and radical extensions. 81.5 Definition. A field extension L/K is a degree 2 radical extension if there exist elements a1 , . . . , an ∈ L such that L = K(a1 , . . . , an ) and for every i = 1, . . . , n we have a2i ∈ K(a1 , . . . , ai−1 )

81.6 Theorem. Let S0 = {P1 , . . . , Pn } be set of points in R2 where Pi = (ai , bi ). Let K = Q(a1 , b1 , . . . , an , bn ) A point Q = (c, d) is constructible from the set S0 iff there exists a degree 2 radical extension L/K, L ⊆ R such that c, d ∈ L. In particular a point Q = (c, d) is constructible iff there exists a degree 2 radical extension L/Q, L ⊆ C such that c, d ∈ L. Proof. Exercise.

81.7 Corollary. Let S0 = {P1 , . . . , Pn } be a set of points in R2 where Pi = (ai , bi ). and let K = Q(a1 , b1 , . . . , an , bn ) If a point Q = (c, d) is constructible then [K(c, d) : K] = 2k for some k ≥ 1. Proof. By Theorem 81.6 we have K(c, d) ⊆ L where L is some degree 2 radical extension of K. We have [L : K] = 2m for some m ≥ 1, so [K(c, d) : K] = 2k for some 1 ≥ k ≥ m.

315

81.8 Example. √ √ Since π is a transcendental number thus√ π 6∈ L for any degree 2 radical extension of Q. It follows that the point ( π, 0) is not constructible and it is impossible to square a circle using straightedge and compass.

81.9 Example. √ √ We have [Q(3 2) : Q] = 3. By Corollary 81.7 we obtain that the point (3 2, 0) is not constructible. As a consequence it is impossible to double a cube using straightedge and compass.

81.10 Example. Recall that we can perform a trisection of an angle θ iff the point (0, cos 13 θ) can be constructed from the set S0 = {(0, 0), (1, 0), (cos θ, 0)}. By Theorem 81.6 this is possible iff cos 13 θ ∈ L for some degree 2 radical extension L of the field K := Q(cos θ). Claim. cos 31 θ ∈ L where L is some degree 2 radical extension of K iff the polynomial 3 1 f (x) = x3 − x − cos θ 4 4 is not irreducible in K[x]. Indeed, by de Moivre formula for any angle α we have: cos 3α = 4 cos3 α − 3 cos α It follows from here that cos 13 θ is a root of f (x). If f (x) is irreducible over K then [K(cos 13 θ) : K] = 3, and so by Corollary 81.7 the point (cos 31 θ, 0) is not constructible from S0 . Conversely, if f (x) is not irreducible then we have f (x) = (x − a)g(x) 316

where a ∈ K, g(x) ∈ K[x], deg g(x) = 2. Since cos 13 θ is a root f (x) we obtain that either cos 31 θ = a or cos 31 is a root of g(x), and so it can be expressed by elements of K and square roots of such elements. In both cases cos 13 θ belongs to some degree 2 radical extension of K. Some special cases: •θ=

π 4 √

We have cos θ =

2 , 2

√ and so Q(cos θ) = Q( 2). Moreover

√ 3 2 f (x) = x3 − x − 4 8 √ √ Check: f (x) has a root − 22 ∈ Q( 2) so it is not irreducible. Thus the angle θ = π4 can be trisected using straightedge and compass. •θ=

π 3

We have cos θ = 12 , so Q(cos θ) = Q and 3 1 f (x) = x3 − x − 4 8 Check: f (x) has no roots in Q, and so it is irreducible in Q[x]. As a consequence the angle θ = π3 cannot be trisected using straightedge and compass.

317

82

Construction of regular polygons

Problem. For which n it is possible to construct a regular polygon with n sides using a straightedge and a compass? Note. Construction of a regular polygon with n sides is equivalent to the con, sin 2π ). struction of the point with coordinates (cos 2π n n

(cos 2π , sin 2π ) n n

2π n

(0, 0)

(1, 0)

82.1 Lemma. The numbers cos 2π , sin 2π belong to some degree 2 radical n n 2π extension L of Q iff ζn = cos n + i sin 2π ome degree 2 radical extension of Q. n Proof. , sin 2π ∈ L where L is a degree 2 radical extension of Q then (⇒) If cos 2π n n ζn ∈ L(i), and L(i) is a degree 2 radical extension of Q. (⇐) Assume that ζn ∈ L for some degree 2 radical extension L of Q. We can assume that i ∈ L. We have ζ −1 = cos 2π − i sin 2π , which gives n n     2π 1 1 2π 1 1 cos = ζn + , sin = ζn − n 2 ζn n 2i ζn 318

, sin 2π ∈ L. Therefore cos 2π n n

82.2 Note. The element ζn is a primitive root of unity of degree n, so Q(ζn )/Q is the cyclotomic Galois extension of degree n (77.3). By (77.6) we have Gal(Q(ζn )/Q) ∼ = (Z/nZ)∗

82.3 Lemma. If L/Q is a Galois extension and [L : Q] < ∞ then L ⊆ M for some degree 2 radical extension M iff Gal(L/Q) = 2m for some m ≥ 0 Proof. Exercise.

82.4 Corollary. A regular polygon with n sides can be constructed using a straightedge and a compass iff |(Z/nZ)∗ | = 2m for some m.

82.5 Note. 1) We have |(Z/nZ)∗ | = ϕ(n) where ϕ(n) is the Euler function: ϕ(n) = |{m | 1 ≥ m ≥ n, gcd(m, n) = 1}| 2) If n = pk11 · . . . · pkr r where p1 , . . . , pr are distinct primes then ϕ(n) = ϕ(pk11 ) · . . . · ϕ(pkr r ) and ϕ(pki i ) = (pi − 1) · pki i −1 .

319

Upshot. The number ϕ(n) is a power of two iff n = 2k p1 · . . . · pr where p1 , . . . , pr are distinct primes such that pi = 2mi + 1 for some mi ≥ 1.

Note. If 2m + 1 is a prime then m must be of the form 2s for some s ≥ 0. Prime numbers of the form s Fs = 22 + 1 are called Fermat primes.

We obtain: 82.6 Theorem. A regular polygon with n sides can be constructed using a straightedge and a compass iff n = 2k p1 · . . . · pr where p1 , . . . , pr are distinct Fermat primes.

82.7 Note. The only known Fermat primes are F0 = 3, F1 = 5, F2 = 17, F3 = 257, F4 = 65537 It is not known whether there are any other prime numbers of this form.

320

83

Transcendental extensions

83.1 Definition. Let L/K be a field extension. A set S ⊆ L is algebraically independent over K if for every finite sequence (s1 , . . . , sn ) of distinct elements of S and for every non-zero polynomial f (x1 , . . . , xn ) ∈ K[x1 , . . . , xn ] we have f (s1 , . . . , sn ) 6= 0 Otherwise we say that the set S is algebraically dependent over K.

83.2 Note. If L/K is a field extension then an element s ∈ L is transcendental over K iff the set S = {s} is algebraically independent over K.

83.3 Proposition. Let L/K be a field extension. A set S ⊆ L is algebraically independent over K iff every element s0 ∈ S is transcendental over the field K(S − {s0 }). Proof. (⇒) Assume that there exists an element s0 ∈ S which is algebraic over the field K(S − {s0 }). Then there exists a finite subset {s1 , . . . , sn } such that s0 is algebraic over K(s1 , . . . , sn ). Let 0 6= f (x) ∈ K(s1 , . . . , sn )[x] be a polynomial such that f (s0 ) = 0. Notice that K(s1 , . . . , sn ) is the field of fractions of the ring K[s1 , . . . , sn ]. As a consequence we can assume that f (x) ∈ K[s1 , . . . , sn ][x]: f (x) = fr (s1 , . . . , sn )xr + . . . + f1 (s1 , . . . , sn )x + f0 (s1 , . . . , sn ) for some fr , . . . , f0 ∈ K[s1 , . . . , sn ]. Take F ∈ K[x1 , . . . , xn , xn+1 ] given by F (x1 , . . . , xn+1 ) = fr (x1 , . . . , xn )xrn+1 + . . . + f1 (x1 , . . . , xn )xn+1 + f0 (x1 , . . . , xn ) Then F 6= 0 and F (s1 , . . . , sn , s0 ) = 0. This shows that the set S is not algebraically independent over K. 321

(⇐) Assume that the set is not algebraically independent over K, and let n be the minimal number such that there exists a polynomial 0 6= F ∈ K[x1 , . . . , xn ] satisfying F (s1 , . . . , sn ) = 0 for some distinct elements s1 , . . . , sn ∈ S. We have F (x1 , . . . , xn ) = fr (x1 , . . . , xn−1 )xrn + . . . + f1 (x1 , . . . , xn−1 )xn + f0 (x1 , . . . , xn−1 ) for some fr , . . . , f0 ∈ K[x1 , . . . , xn−1 ]. Since F 6= 0 we must have fi 6= 0 for some 0 ≤ i ≤ r. Then by the minimality of n we have fi (s1 , . . . , sn−1 ) 6= 0. Take G(x) ∈ K(s1 , . . . , sn−1 )[x] defined by G(x) = fr (s1 , . . . , sn−1 )xr + . . . + f1 (s1 , . . . , sn−1 )x + f0 (s1 , . . . , sn−1 )

We have G(x) 6= 0 and G(sn ) = F (s1 , . . . , sn ) = 0. Therefore sn is an algebraic element over K(s1 , . . . , sn−1 ), and so it is also algebraic over K(S − {sn }).

83.4 Note. If L = K(x1 , . . . , xn ) is the field of rational functions in variables x1 , . . . , xn with coefficients in K then the set S = {x1 , . . . , xn } is algebraically independent over K.

83.5 Proposition. If L/K is a field extension and S = {s1 , . . . , sn } is an algebraically independent set over K then we have an isomorphism K(x1 , . . . , xn ) ∼ = K(s1 , . . . , sn ) Proof. Define ϕ : K(x1 , . . . , xn ) → K(s1 , . . . , sn ) by   f (s1 , . . . , sn ) f (x1 , . . . , xn ) = ϕ g(x1 , . . . , xn ) g(s1 , . . . , sn ) Check that this an isomorphism of fields. 322

83.6 Definition. Let L/K be a field extension. A transcendence basis of the extension L/K is a set S ⊆ L such that 1) S is algebraically independent over K 2) if S 0 ⊆ L is an algebraically independent set such that S ⊆ S 0 then S = S 0 . 83.7 Proposition. For any field extension L/K there exists a transcendence basis of L/K.

Proof. Exercise.

83.8 Theorem. Let L/K be a field extension. A subset S ⊆ L is a transcendence basis of L/K iff S a set algebraically independent over K and L/K(S) is an algebraic extension.

83.9 Lemma. Let L/K be a field extension. If S, T ⊆ L are sets such that S is algebraically independent over K and T is algebraically independent over K(S) then S ∪ T is algebraically independent over K. Proof. Let s1 , . . . , sm ∈ S, t1 , . . . , tn ∈ T be distinct elements, and let 0 6= F ∈ K[x1 , . . . , xm , y1 , . . . , yn ] We need to show that F (s1 , . . . , sm , t1 , . . . , tn ) 6= 0 We have X F = fi1 ,...,in y1i1 · . . . · ynin i1 ,...,in

for some fi1 ,...,in ∈ K[x1 , . . . , xm ]. Since F 6= 0 we have fi1 ,...,in 6= 0 for some i1 , . . . in , and so by the algebraic independence of the set S over K we get that fi1 ,...,in (s1 , . . . , sm ) 6= 0. As a consequence the polynomial X G(y1 , . . . , ym ) = fi1 ,...,in (s1 , . . . , sm )y1i1 · . . . · ynin i1 ,...,in

323

is a non-zero polynomial in K(S)[y1 , . . . , yn ]. By the algebraic independence of the set T over K(S) we obtain F (s1 , . . . , sm , t1 , . . . , tn ) = G(t1 , . . . , tn ) 6= 0

Proof of Theorem 83.8. (⇒) Let S be a transcendence basis of L/K. It is enough to show that any element a ∈ L − K(S) is algebraic over K(S). Assume by contradiction that a is transcendental over K(S). Then the set {a} is algebraically independent over K(S), and so, by Lemma 83.9 the set S ∪ {a} is algebraically independent over K. This is impossible since S is a proper subset of S ∪ {a}. (⇐) Assume that S is a set algebraically independent over K and that L/K(S) is an algebraic extension. It is enough to show that for any a ∈ L − S the set S ∪ {a} is not algebraically independent over K. Assume, by contradiction, that for some a ∈ L − S the set S ∪ {a} is algebraically independent over K. By Proposition 83.3 we would have then that a is a transcendental element over K(S). This contradicts the assumption that all elements of L are algebraic over K(S).

83.10 Definition. A field extension L/K is purely transcendental if L = K(S) where S ⊆ L is a set algebraically independent over K.

83.11 Corollary. For any field extension L/K there exists an intermediate field K ⊆ M ⊆ L such that M/K is a purely transcendental extension and L/M is an algebraic extension.

324

Proof. By Theorem 83.8 is it enough to take M = K(S) where S ⊆ L is a transcendence basis of L/K.

83.12 Proposition. If L/K is a field extension and L = K(A) for some set A ⊆ L then there exists S ⊆ A such that S is a transcendence basis of L/K. Proof. By Zorn’s Lemma 29.10 we can find a maximal subset S ⊆ A that is algebraically independent over K. We will show S is a transcendence basis of L/K. By Theorem 83.8 it suffices to show that L/K(S) is an algebraic extension. Since L = K(A) = K(S)(A − S) this amounts to showing that every element a ∈ A − S is algebraic over K(S). Assume, by contradiction, that there exists an element a ∈ A − S that is transcendental over K(S). Then the set {a} is algebraically independent over K(S), and so by Lemma 83.9 the set S ∪ {a} ⊆ A is algebraically independent over K. This however is impossible by the maximality of S.

83.13 Proposition. If If K ⊆ L ⊆ M are field extensions, S ⊆ L be a transcendence basis of L/K and T ⊆ M be a transcendence basis of M/L then S ∪ T is a transcendence basis of M/K. Proof. The set T is algebraically independent over L, so it is also algebraically independent over K(S) ⊆ L. By Lemma 83.9 it follows that the set S ∪ T is algebraically independent over K. By Theorem 83.8 the field L is an algebraic extension of K(S). It follows that L(T ) is an algebraic extension of K(S)(T ) = K(S ∪ T ) (check!). Moreover, by Theorem 83.8 again, M is an algebraic extension of L(T ). As a consequence M is an algebraic extension of K(S ∪ T ). Applying Theorem 83.8 one more time we get from here that S ∪ T is a transcendence basis of M over K.

325

83.14 Corollary. If L/K is a field extension and S ⊆ L is a set algebraically independent over K then there exists a transcendence basis S 0 of L/K such that S ⊆ S 0. Proof. The set S is a transcendence basis of the extension K(S)/K. Let T be a transcendence basis of the extension L/K(S). By Proposition 83.13 S ∪ T is a transcendence basis of L/K. Therefore we can take S 0 = S ∪ T .

83.15 Theorem. If L/K is a field extension then any two transcendence bases of L over K have the same cardinality.

Proof. See Hungerford Theorem 1.8 and Theorem 1.9.

83.16 Definition. The transcedence degree of a field extension L/K is the cardinality of a transcendence basis of L over K. It is denoted by tr(L/K).

83.17 Proposition. If K ⊆ L ⊆ M are field extensions then tr(M/K) = tr(L/K) + tr(M/L)

Proof. Let S ⊆ L be a transcendence basis of L/K and let T ⊆ M be a transcendence basis of M/L over L. By Proposition 83.13 the set S ∪ T is a transcendence basis of M/K. Since S ∩ T = ∅ we obtain tr(M/K) = |S ∪ T | = |S| + |T | = tr(L/K) + tr(M/L)

326

84

Algebraic sets

84.1 Definition. Let K be an algebraically closed field, and let Kn = K . . × K} | × .{z n times

A subset A ⊆ K n is an algebraic set if A is the set of solution of some system of polynomial equations    f1 (x1 , . . . , xn ) = 0 .. .   fr (x1 , . . . , xn ) = 0 for some f1 , . . . , fr ∈ K[x1 , . . . , xn ]. In such case we write: A = V (f1 , . . . , fr )

84.2 Note. Every finite subset A = {a1 , . . . , am } ⊆ K is an algebraic set since this is the set of solutions of the equation (x − a1 ) · . . . · (x − am ) = 0

84.3 Note. Different sets of equations may give the same algebraic set. Take e.g. f1 , f2 , g1 , g2 ∈ K[x1 , x2 ] given by: f1 (x1 , x2 ) = x1 , f2 (x1 , x2 ) = x2 , g1 (x1 , x2 ) = x1 +x2 , g2 (x1 , x2 ) = x1 −x2 Then V (f1 , f2 ) = V (g1 , g2 ) = {(0, 0)} ⊆ K 2

84.4 Definition. Let K be a field and let I be an ideal of K[x1 , . . . , xn ]. Denote: V (I) := {(a1 , . . . , an ) | f (a1 , . . . , an ) = 0 for all f ∈ I} 327

84.5 Proposition. Let K be a field, let f1 , . . . , fr ∈ K[x1 , . . . , xn ], and let I = hf1 , . . . , fr i be the ideal generated by f1 , . . . fr . Then V (I) = V (f1 , . . . , fr )

Proof. Since {f1 , . . . , fr } ⊆ I we have V (I) ⊆ V (f1 , . . . , fr ). Conversely, assume that (a1 , . . . , an ) ∈ V (f1 , . . . , fr ). If g ∈ I then we have g=

r X

hi fi

i=1

for some gi ∈ K[x1 , . . . , xn ]. Therefore g(a1 , . . . , an ) =

r X

gi (a1 , . . . , an )fi (a1 , . . . , an ) = 0

i=1

Therefore (a1 , . . . , an ) ∈ V (I), and as a consequence V (f1 , . . . , fr ) ⊆ V (I).

84.6 Corollary. If K is a field, f1 , . . . , fr , g1 , . . . , gs ∈ K[x1 , . . . , xn ] and hf1 , . . . , fr i = hg1 , . . . , gs i then V (f1 , . . . , fr ) = V (g1 , . . . , gs ).

328

85

Hilbert basis theorem

Goal: 85.1 Theorem. If K is a field and I C K[x1 , . . . , xn ] then the ideal I is finitely generated: I = hf1 , . . . , fr i for some f1 , . . . , fr ∈ K[x1 , . . . , xn ]

85.2 Note. Let K be an algebraically closed set. If I C K[x1 , . . . , xn ] and I = hf1 , . . . , fr i then by Proposition 84.5 we have V (I) = V (f1 , . . . , fr ) i.e. V (I) is an algebraic set. Thus by Theorem 85.1 we obtain an epimorphism:     ideals algebraic sets V: −→ I C K[x1 , . . . , xn ] A ⊆ Kn which sends an ideal I to the algebraic set V (I).

85.3 Definition. Let R be a commutative ring with identity. The ring R is a Noetherian ring if every ideal I C R is finitely generated: I = hr1 , . . . , rn i for some r1 , . . . , rn ∈ R.

85.4 Example. If R is a PID then R is a Noetherian ring. In particular 1) Z is Noetherian 2) If K is a field then K[x] is Noetherian.

329

85.5 Theorem. Let R be a commutative ring with identity. The following conditions are equivalent. 1) R is a Noetherian ring. 2) If I1 ⊆ I2 ⊆ I3 ⊆ . . .

is any increasing sequence of ideals of R then there exists n ≥ 1 such that In = Im for all m ≥ n. 3) Every family of ideals of R contains a maximal element.

85.6 Note. The condition 2) is called the ascending chain condition on ideals of R.

Proof of Theorem 85.5. 1) ⇒ 2) Let I1 ⊆ I2 ⊆ I3 ⊆ . . . be a chain of ideals in R. Take I :=

∞ [

Ik

k=1

Then I is an ideal of R, and since R is a Noetherian ring this I = hr1 , . . . , rn i for some r1 , . . . , rn ∈ R. For j = 1, . . . , n there exists kj such that rj ∈ Ikj . Take n = max{k1 , . . . , kn } For any m ≥ n we have r1 , . . . , rn ∈ In and so I ⊆ Im ⊆ I As a consequence for any m ≥ n we get In = I = Im . 2) ⇒ 3) Let J be a family of ideals in R. Assume that J does not contain a maximal element. Take any I1 ∈ J. Since I1 is not a maximal element of J 330

therefore there exists I2 ∈ J such that I1 ⊂ I2 and I1 6= I2 . By induction we can find in J an infinite sequence of ideals I1 ⊂ I2 ⊂ I3 ⊂ . . . such that Ik 6= Ik+1 for all k ≥ 1. This however contradicts the assumption that ideals of R satisfy the ascending chain condition. 3) ⇒ 2) Let I be an ideal of R. We need to show that I is finitely generated. Let J be the family of all finitely generated ideals J such that J ⊆ I. By assumption J contains a maximal element J0 = hr1 , . . . , rn i. Assume the J0 6= I and let s ∈ I − J0 . Then J1 := hr1 , . . . , rn , si is an ideal such that J1 ∈ J, J0 ⊆ J1 , and J0 6= J1 . This is impossible since J0 is a maximal element of J. As a consequence we get that I = J0 , and so I is fintely generated.

85.7 Hilbert Basis Theorem. Let R be a commutative ring with identity. If R is a Noetherian ring then R[x] is also Noetherian.

Proof. Let I C R[x]. We will show that I is generated by a finite set. Let In ⊆ R be the set such that a ∈ In if either a = 0 or if there exists f (x) = an xn + · · · + a1 x + a0 such that f (x) ∈ I and a = an . Check: In C R. Moreover we have I0 ⊆ I1 ⊆ . . . Since R is a Noetherian ring we obtain: 1) for any n ≥ 0 there exists an,1 , . . . , an,mn ∈ R such that In = han,1 , . . . , an,mn i 2) there exists N ≥ 0 such that IN = IN +1 = . . . .

331

By the definition of In for any 0 ≤ n ≤ N and 1 ≤ i ≤ mn there exists a polynomial fn,i (x) ∈ I such that fn,i (x) = an,i xn + . . . . Define S := {fn,i (x) | 0 ≤ n ≤ N, 1 ≤ i ≤ mn } Notice that S is a finite set and S ⊆ I. We will show that I = hSi. It suffices to show that if g(x) ∈ I then g(x) ∈ hSi. We will argue by induction with respect to the degree of g(x). If deg g(x) = 0 then g(x) = b0 for some b0 ∈ I0 . Then b0 =

m0 X

ri a0,i

i=1

for some ri ∈ R. Since f0,i (x) = a0,i this gives g(x) =

m0 X

ri f0,i (x)

i=1

and so g(x) ∈ hSi. Next, assume that for some n > 0 if h(x) ∈ I and deg h(x) ≤ n − 1 then h(x) ∈ hSi. Let g(x) ∈ I be a polynomial of degree n: g(x) = bn xn + . . . b1 x + b0 Assume first that n ≤ N By the definition of In we have bn ∈ In , and so bn =

mn X

ri an,i

i=1

for some ri ∈ R. Define f (x) :=

mn X

ri fn,i (x)

i=1

Notice that f (x) = bn xn + . . . , and so deg(g(x) − f (x)) < n. Also, since g(x), f (x) ∈ I we have g(x) − f (x) ∈ I and thus by the inductive assumption g(x) − f (x) ∈ hSi. Therefore g(x) = f (x) + (g(x) − f (x)) {z } |{z} | ∈hSi

∈hSi

332

is an element of hSi. If n > N then In = IN so we have bn =

mN X

ri aN,i

i=1

Then define f (x) := xn−N

mN X

! ri fN,i (x)

= bn x n + . . .

i=1

Then f (x) ∈ hSi and deg(g(x) − f (x)) < n. Similarly as before we obtain from here that g(x) ∈ hSi. 85.8 Corollary. If R is a Noetherian ring then the ring R[x1 , . . . , xn ] is also Noetherian for any n ≥ 0. Proof. This follows from Theorem 85.7 by induction with respect to n.

Proof of Theorem 85.1. Since K is a field it is a Noetherian ring. Thus we can apply Corollary 85.8

333

86

Radical ideals

86.1 Note. Let K be an algebraically closed field. For any subset X ⊆ K n define J(X) := {f ∈ K[x1 , . . . , xn ] | f (a1 , . . . , an ) = 0 for all (a1 , . . . , an ) ∈ X} Check: J(X) is an ideal of K[x1 , . . . , xn ]. We get maps of sets:     ideals algebraic sets V:  :J I C K[x1 , . . . , xn ] A ⊆ Kn Notice that 1) if I, I 0 C K[x1 , . . . , xn ] and I ⊆ I 0 then V (I) ⊇ V (I 0 ) 2) if A, A0 ⊆ K n are algebraic sets and A ⊆ A0 then J(A) ⊇ J(A0 ). 86.2 Proposition. If A ⊆ K n is an algebraic set then V (J(A)) = A. Proof. If (a1 , . . . , an ) ∈ A then f (a1 , . . . , an ) = 0 for all f ∈ J(A). Therefore (a1 , . . . , an ) ∈ V (J(A)). This show that A ⊆ V (J(A)). On the other hand, A is an algebraic set, so A = V (I) for some ICK[x1 , . . . , xn ]. We have I ⊆ J(A) which gives A = V (I) ⊇ V (J(A))

86.3 Note. It is not true that for any ideal I ∈ K[x1 , . . . , xn ] we have J(V (I)) = I. Take e.g. I = hx2 i C K[x]. We have V (I) = {a ∈ K | a2 = 0} = {0} 334

Then J(V (I)) = {g(x) ∈ K[x] | g(0) = 0} = hxi Therefore J(V (I)) 6= I.

86.4 Definition. Let R be a commutative ring with identity. An ideal I C R is a radical ideal if for every a ∈ R such that an ∈ I for some n ≥ 1 we have a ∈ I.

86.5 Proposition. If K is an algebraically closed field and A ⊆ K n is an algebraic set then J(A) is a radical ideal of K[x1 , . . . , xn ].

Proof. Exercise.

Goal. The following maps are inverse bijections of sets:  V:

radical ideals I C K[x1 , . . . , xn ]



 

335

algebraic sets A ⊆ Kn

 :J

Note. In Sections 87-88 all rings are commutative rings with identity 1 6= 0.

87

Integral extensions of rings

87.1 Definition. Let R, S be rings. We say that S is a ring extension of R if R is a subring of S.

87.2 Notation. If R ⊆ S is a ring extension and A ⊆ S is a subset of S then R[A] is the smallest subring of S such that R ∪ A ⊆ R[A]. If A is a finite set, A = {a1 , . . . , an } then we write R[A] = R[a1 , . . . , an ] 87.3 Note. We have R[a] = {b ∈ S | b = r0 + r1 a + . . .rk ak for some r0 , . . . , rk ∈ R} and R[a1 , . . . , an ] = R[a1 , . . . , an−1 ][an ].

87.4 Definition. Let R ⊆ S be a ring extension. An element a ∈ S is an integral element over R if a is a root of a monic polynomial f (x) ∈ R[x].

87.5 Example. 1) If K, L are fields and K ⊆ L then an element a ∈ L is an integral element over L iff a is an algebraic element over K. 2)

√

2 is integral over Z since it is a root of the monic polynomial f (x) = x − 2 ∈ Z[x]. 2

336

3)

1 2

is not integral over Z. Indeed, if f (x) ∈ Z[x] is a monic polynomial f (x) = xn + an−1 xn−1 + . . . + a0

then 2n f ( 12 ) = 1 + 2an−1 + · · · + 2n a0 6= 0 | {z } odd

so

f ( 12 )

= 0.

87.6 Definition. Let R be a ring. An R-module M is faithful if for every 0 6= r ∈ R there exists m ∈ M such that rm 6= 0.

87.7 Theorem. Let R ⊆ S be a ring extension and let a ∈ S. The following conditions are equivalent. 1) The element a is integral over R. 2) R[a] is a finitely generated R-module. 3) There exists M ⊆ S such that M is a a faithful R[a]-module and it is finitely generated as an R-module.

87.8 Lemma. Let S be a ring and let A = (aij ) be an n × n matrix with coefficients in S. If s1 , . . . , sn ∈ S are elements such that   s1  ..  A .  = 0 sn then (det A) · si = 0 for i = 1, . . . , n. Proof. See Hungerford p.354 Exercise 8.

337

Proof of Theorem 87.7. 1) ⇒ 2) Since the element a is integral over R there exists a monic polynomial f (x) ∈ R[x] such that f (0). Assume that deg f (x) = n. If b ∈ R[a] then b = g(a) for some g(x) ∈ R[x]. By (38.1) there exists q(x), r(x) ∈ R[x] such that deg r(x) ≤ n − 1 and g(x) = q(x)f (x) + r(x) This gives g(a) = r(a). As a consequence we obtain R[a] = {r(a) | r(x) ∈ R[x], deg r(x) ≤ n − 1} = {b0 + b1 a + . . . + bn−1 an−1 | b1 , . . . , bn−1 ∈ R} Therefore the elements 1, a, . . . , an−1 generate R[a] as an R-module. 2) ⇒ 3) R[a] is a faithful R[a]-module and by assumption it is finitely generated as an R-module. Thus we can take M = R[a]. 3) ⇒ 1) Assume that the set {s1 , . . . sn } generates M as an R-module. For every i = 1, . . . n there exists ai,j ∈ R such that asi = ai1 s1 + . . . + ain sn This gives a system of equations: (a11 − a)s1 + a12 s2 + . . . a1n sn = 0 a21 s1 + (a22 − a)s2 + . . . a2n sn = 0 .. .. .. an1 s1 + an2 s2 + . . . (ann − a)sn = 0 In the matrix notation this gives:   s1  ..  ((aij ) − aI)  .  = 0 sn

338

By Lemma 87.8 we get det((aij ) − aI)si = 0 for i = 1, . . . , n. Since the elements s1 , . . . , sn generate M this shows that det((aij ) − aI)m = 0 for all m ∈ M . Furthermore, since det((aij ) − aI) ∈ R[a] and M is a faithful R[a]-module we obtain that det((aij ) − aI) = 0. Take f (x) = det((aij ) − xI) ∈ R[x]. This is a monic polynomial and f (a) = 0. This shows that a is an integral element over R.

87.9 Definition. A ring extension R ⊆ S is an integral extension if every element of S is integral over R.

87.10 Lemma. Let R ⊆ S ⊆ T be ring extensions. If T is finitely generated as an S-module and S is finitely generated as an R-module then T is finitely generated as an R-module.

Proof. Exercise.

87.11 Proposition. Let R ⊆ S be a ring extension and let a1 , . . . , an ∈ S be elements integral over R. Then R[a1 , . . . , an ] is finitely generated R-module and it is an integral extension of R.

Proof. We argue by induction with respect to n. If n = 1 then R[a1 ] is finitely generated R-module by Theorem 87.7. Also, if b ∈ R[a1 ] then R[a1 ] is a faithful R[b]-module. Using Theorem 87.7 again we obtain that b is integral over R. Assume the statement of the proposition holds for some n, and let a1 , . . . , an+1 ∈ S be elements integral over R. Then an+1 is integral over R[a1 , . . . , an ], so

339

R[a1 , . . . , an+1 ] is a finitely generated R[a1 , . . . , an ]-module. By Lemma 87.10 we obtain that R[a1 , . . . , an+1 ] is then a finitely generated R-module. If b ∈ R[a1 , . . . , an+1 ] then R[a1 , . . . , an+1 ] is a faithful R[b] module, so by Theorem 87.7 b is integral over R.

87.12 Corollary. Let R ⊆ S be a ring extension and let Rint := {a ∈ S | a is integral over R} then Rint is a subring of S. Proof. Let a, b ∈ Rint . It is enough to show that a ± b, ab are integral elements over R. This holds since a ± b, ab ∈ R[a, b], and by Proposition 87.11 R[a, b] is an integral extension of R.

87.13 Proposition. Let R ⊆ S ⊆ T be ring extensions. If T is an integral extension of S and S is an integral extension of R then T is an integral extension of R. Proof. It is enough to show that if b ∈ T then b is integral over R. Since b is integral over S there exists a monic polynomial f (x) ∈ S[x] f (x) = xn + an−1 xn−1 + . . . + a0 such that f (b) = 0. Since f (x) ∈ R[an−1 , . . . , a0 ] it follows that b is integral over R[an−1 , . . . , a0 ]. By Theorem 87.7 R[an−1 , . . . , a0 , b] is then a finitely generated R[an−1 , . . . , a0 ]-module. Also, since an−1 , . . . , a0 are integral over R by Proposition 87.11 R[a1 , . . . , an−1 ] is a finitely generated R-module. Therefore, by Lemma 87.10 R[an−1 , . . . , a0 , b] is a finitely generated R-module. Since R[an−1 , . . . , a0 , b] is a faithful R[b] module applying Theorem 87.7 we get that b is integral over R.

340

88

Noether Normalization Lemma

88.1 Noether Normalization Lemma. Let K ⊆ S be a ring extension such that K is a field and S is finitely generated over K: S = K[a1 , . . . , an ] for some a1 , . . . , an ∈ S. There exists a ring K ⊆ R ⊆ S such that R∼ = K[x1 , . . . , xm ] for some m ≤ n and that R ⊆ S is an integral extension

88.2 Note. Compare this with the decomposition of field extensions into a purely transcendental extension and an algebraic extension (Corollary 83.11).

Proof of Theorem 88.1. We argue by induction with respect to n. For n = 1 we have S = K[a1 ]. Consider the homomorphism ϕ : K[x] → K[a1 ] given by ϕ(f (x)) = f (a). This is an epimorphism, so if Ker(ϕ) = {0} we get that K[x] ∼ = K[a1 ], and then we can take R = S. If Ker(ϕ) 6= {0} then f (a) = 0 for some 0 6= f (x) ∈ K[x]. Since K is a field we can assume that f (x) is a monic polynomial. It follows that a is an integral element over K, and so by (87.11) S is an integral extension of K. In this case take R = K. Next, assume that the statement of the theorem holds for some n − 1 and let S = K[a1 , . . . , an ]. We have a homomorphism ϕ : K[x1 , . . . , xn ] → K[a1 , . . . , an ] 341

given by ϕ(f (x1 , . . . , xn )) = f (a1 , . . . , an ). If Ker(ϕ) = {0} then as before we obtain K[x1 , . . . , xn ] ∼ = K[a1 , . . . , an ], so we can take R = S. Assume then that there exists 0 6= f ∈ Ker(ϕ). We have X f (x1 , . . . , xn ) = bi1 ,...,in xi11 . . . xinn (i1 ,...,in )∈I

for some 0 6= bi1 ,...,in ∈ K. Let (j1 , . . . , jn ) be the element of I that is the biggest with respect to the lexicographic order. Since K is a field we can assume that aj1 ,...,jn = 1. Take d = max { ik | 1 ≤ k ≤ n, (i1 , . . . , in ) ∈ I }

and let h ∈ K[x1 , . . . , xn ] be a polynomial given by n

n−1

h(x1 , . . . , xn ) = f (x1 + xdn , x2 + xdn

, . . . , xn−1 + xdn , xn )

Claim. Consider h as a polynomial of the variable xn with coefficients in the ring K[x1 , . . . , xn−1 ]. Then h is a monic polynomial. Indeed, notice that X h=

n

n−1

bi1 ,...,in (x1 + xdn )i1 (x2 + xdn

)i2 . . . xinn

(i1 ,...,in )∈I

=

X

n +i

bi1 ,...,in (xin1 d

2d

n−1 +...i

n

+ lower powers of xn )

(i1 ,...,in )∈I

By the choice of (j1 , . . . , jn ) the highest degree monomial of h is then n +j

bj1 ,...,jn (xjn1 d

n−1 +...j n 2d

)

and by assumption bj1 ,...,jn = 1

Next, let g(x) be the polynomial given by n

n−1

g(x) = h(a1 − adn , a2 − and Notice that 342

, . . . , an−1 − adn , x)

(i) g(x) ∈ K[a1 − adnn , . . . , an−1 − adn ][x]

(ii) g(x) is a monic polynomial (since h is monic in xn ) (iii) g(an ) = f (a1 , . . . , an ) = 0 (since f ∈ Ker(ϕ)) Denote S 0 := K[a1 − adnn , . . . , an−1 − adn ]. By (i)-(iii) above we obtain that an is an integral element over S 0 , and so S 0 [an ] = K[a1 , . . . , an ] is an integral extension of S 0 . On the other hand S 0 is generated over K by n − 1 elements, so by the inductive assumption there is a ring K ⊆ R ⊆ S0 such that R ∼ = K[x1 , . . . xm ] for some m ≤ n − 1 and that S 0 is an integral extension of R. Consider the extensions K⊆R⊆S To finish the proof it is enough to notice that since R ⊆ S 0 and S 0 ⊆ S are integral extensions thus, by Proposition 87.13, the extension R ⊆ S is also integral.

88.3 Corollary. Let K be a field and let L be a finitely generated ring extension of K: L = K[a1 , . . . , an ] If L is a field then it is an algebraic extension of K.

Proof. By the Noether Normalization Lemma 88.1 there exists a ring R such that K ⊆ R ⊆ L, R ∼ = K[x1 , . . . , xm ] for some 0 ≤ m ≤ n, and that R ⊆ L is an integral extension. It is enough to show that R is a field. Indeed, this will imply that m = 0, (i.e. K = R), and that as a consequence K ⊆ L is an integral (and thus algebraic) extension. Take b ∈ R. Since L is a field we have b−1 ∈ L. We need to show that b−1 ∈ R. 343

Since L is an integral extension of R there exists a monic polynomial f (x) = xk + rk−1 xk−1 + . . . + r1 x + r0 ∈ R[x] such that f (b−1 ) = 0. This gives b−k = −rk−1 b−(k−1) − rk−2 b−(k−2) . . . − r1 b−1 − r0 Multiplying both sides by bk−1 we obtain b−1 = −rk−1 − rk−2 b − . . . − r1 bk−2 − r0 bk−1 Since ri , b ∈ R therefore b−1 ∈ R.

344

89

Hilbert Nullstellensatz

Recall. 1) Let K be an algebraically closed field. We have maps  V:

   algebraic sets ideals :J  A ⊆ Kn I C K[x1 , . . . , xn ]

where V (I) = {(a1 , . . . , an ) ∈ K n | f (a1 , . . . , an ) = 0 ∀f ∈ I} J(A) = {f ∈ K[x1 , . . . , xn ] | f (a1 , . . . , an ) = 0 ∀(a1 , . . . an ) ∈ A } 2) For any algebraic set A ⊆ K n we have V (J(A)) = A (86.2). 3) If A ⊆ K n is an algebraic set then J(A) is a radical ideal (86.5).

Goal. If I CK[x1 , . . . , xn ] is a radical ideal then J(V (I)) = I. As a consequence we have inverse bijections of sets     radical ideals algebraic sets V:  :J I C K[x1 , . . . , xn ] A ⊆ Kn

89.1 Theorem (Weak Nullstellensatz). Let K be an algebraically closed field. If I ∈ K[x1 , . . . , xn ] then V (I) = ∅ iff I = K[x1 , . . . , xn ]

89.2 Note. If I C K[x] then I = hf (x)i for some f (x) ∈ K[x] since K[x] is a PID. This gives V (I) = {a ∈ K | f (a) = 0}

345

Therefore in the case of a single variable Theorem 89.1 says that a polynomial f (x) ∈ K[x] has no roots iff hf (x)i = K[x], i.e. iff f (x) is a non-zero constant polynomial. This property defines algebraically closed fields, so in this case Theorem 89.1 is tautologically true.

Proof of Theorem 89.1. If I = K[x1 , . . . , xn ] then 1 ∈ I, and so V (I) = ∅. Assume then that I 6= K[x1 , . . . , xn ]. By (29.1) there exists a maximal ideal P such that I ⊆ P . We have V (P ) ⊆ V (I), so it will suffice to show that V (P ) 6= ∅. Let ϕ : K[x1 , . . . , xn ] → K be a ring homomorphism and let ai = ϕ(xi ) for i = 1, . . . , n. Notice that if ϕ|K = idK then ϕ is given by ϕ(f (x1 , . . . , xn )) = f (a1 , . . . , an ) Moreover, (a1 , . . . , an ) ∈ V (P ) iff P ⊆ Ker ϕ. This shows that we have a bijection 

ring homomorphisms





points



    Φ :  ϕ : K[x1 , . . . , xn ] → K  −→  (a1 , . . . , an ) ∈ V (P ) ϕ|K = idK , P ⊆ Ker(ϕ)

where Φ(ϕ) = (ϕ(x1 ), . . . , ϕ(xn )). In order to show that V (P ) 6= ∅ is suffices then to prove that there exists a homomorphism ϕ : K[x1 , . . . , xn ] → K satisfying the above conditions. Take the quotient ring K[x1 , . . . , xn ]/P and let π : K[x1 , . . . , xn ] → K[x1 , . . . , xn ]/P 346

be the canonical epimorphism. Identifying a ∈ K with π(a) = a + P we get that π|K = idK and we can consider K[x1 , . . . , xn ]/P as a ring extension of K. This extension is finitely generated: K[x1 , . . . , xn ]/P = K[π(x1 ), . . . , π(xn )] Since P is a maximal ideal K[x1 , . . . , xn ]/P is a field. By Corollary 88.3 we obtain that K[x1 , . . . , xn ]/P is an algebraic extension of K. Since K is an algebraically closed field this implies that K[x1 , . . . , xn ]/P = K. Therefore we can take ϕ := π.

89.3 Corollary. If K is an algebraically closed field and P C K[x1 , . . . , xn ] is a maximal ideal then P = hx1 − a1 , . . . , xn − an i for some a1 , . . . , an ∈ K. Proof. Since P 6= K[x1 , . . . , xn ] by Theorem 89.1 we get V (P ) 6= ∅. Let (a1 , . . . , an ) ∈ V (P ). Denote I := hx1 − a1 , . . . , xn − an i Notice that the set V (I) = {(a1 , . . . , an )}. Since V (I) 6= ∅ the ideal I is a proper ideal of K[x1 , . . . , xn ]. Also, since V (I) ⊆ V (P ) we have P ⊆ I. By maximality of P we obtain P = I.

89.4 Definition. Let R be a commutative ring and let I C R. The radical of I is the ideal √ I = {a ∈ R | an ∈ I for some n ≥ 1}

89.5 Note. If I C R then

√

I = I iff I is a radical ideal. 347

89.6 Theorem (Hilbert Nullstellensatz). Let K be an algebraically closed field. If I C K[x1 , . . . , xn ] then √ J(V (I)) = I In particular if I is a radical ideal then J(V (I)) = I. Proof. Since J(V (I)) is a radical ideal and I ⊆√ J(V (I)) we have J(V (I)). It remains to show then that J(V (I)) ⊆ I.

√

I ⊆

Let 0 6= F ∈ J(V (I)). It will be enough to show that F N ∈ I for some N ≥ 1. Consider the inclusion homomorphism K[x1 , . . . , xn ] ,→ K[x1 , . . . , xn , y] Let g := (1 − yF ) ∈ K[x1 , . . . , xn , y] and let J C K[x1 , . . . , xn , y] be the ideal given by the set I ∪ {g}. Claim. V (J) = ∅. Indeed, assume that (a1 , . . . , an , b) ∈ V (J). Since F ∈ J(V (I)) we have F (a1 , . . . , an ) = 0. Therefore g(a1 , . . . , an , b) = 1 − bF (a1 , . . . , an ) = 1 which is a contradiction since g ∈ J. By the Weak Nullstellensatz (89.1) we obtain that J = K[x1 , . . . , xn , y]. This means that there exists f1 , . . . , fk ∈ I and h0 , h1 , . . . hk ∈ K[x1 , . . . , xn , y] such that k X gh0 + fi hi = 1 (∗) i=1

Consider the ring homomorphism ϕ : K[x1 , . . . , xn , y] → K(x1 , . . . , xn ) such that ϕ|K = idK , ϕ(xi ) = xi , ϕ(y) = F1 . Notice that ϕ(g) = 1 − F1 F = 0. Therefore from the equation (∗) we obtain 1 = ϕ(1) =

k X i=1

348

fi ϕ(hi )

P i For i = 1, . . . k we have hi = nj=0 hij y j for some hij ∈ K[x1 , . . . , xn ]. This gives: ! ni k k X X X 1 1= fi ϕ(hi ) = fi hij j F i=1 i=1 j=1 Take N = max{n1 , . . . , nk }. We obtain FN = FN

k X i=1

fi

ni X

1 hij j F j=1

!! =

k X i=1

fi

ni X

! hij F N −j

j=1

where N − j ≥ 0. Since fi ∈ I and hij F N −j ∈ K[x1 , . . . , xn ] for all i, j this shows that F N ∈ I.

89.7 Corollary. Let K be an algebraically closed field. We inverse bijections of sets:     radical ideals algebraic sets V:  :J I C K[x1 , . . . , xn ] A ⊆ Kn Proof. This follows directly from Proposition 86.2 and Theorem 89.6.

89.8 Corollary. If I1 , I2 C K[x1 , . . . , xn ] then V (I1 ) = V (I2 ) iff Proof. If I C K[x1 , . . . xn ] then by (86.2) and (89.6) we have √ V (I) = V (J(V (I))) = V ( I) √ √ Therefore if I1 = I2 then V (I1 ) = V (I2 ). Conversely, if V (I1 ) = V (I2 ) then p p I1 = J(V (I1 )) = J(V (I2 )) = I2

349

√ √ I1 = I2 .

90

Zariski topology

Recall. If R is a commutative ring and I, I C R then I + J := {a + b | r ∈ I, s ∈ J} In general, if {Ii }i∈S then ( X X Ii := ai i∈S

P

i∈S

i∈S

) ai ∈ Ii , ai 6= 0 for finitely many i only

Ii is the smallest ideal of R containing Ii for all i ∈ S.

90.1 Proposition. Let K be an algebraically closed set. 1) If {I1 , . . . , Ik } is a finite family of ideals of K[x1 , . . . , xn ] then V (I1 ) ∪ · · · ∪ V (Ik ) = V (I1 ∩ · · · ∩ Ik ) 2) If {Ii }i∈S is an arbitrary family of ideals of K[x1 , . . . , xn ] then [ P V (Ii ) = V ( i∈S Ii ) i∈S

Proof. 1) Denote J := I1 ∩ · · · ∩ Ik . Since J ⊆ Ii for i = 1, . . . , k thus V (Ii ) ⊆ V (J) and so n [ V (Ii ) ⊆ V (J) i=1

Sn

Conversely, assume that a 6∈ i=1 V (Ii ). Then for i = 1, . . . , k there exists fi ∈ Ii such that fi (a) = 0. Take f = f1 · . . . · fn . We have f ∈ J, and f (a) 6= 0. Thus a 6∈ V (J). Therefore we obtain V (J) ⊆

n [ i=1

350

V (Ii )

2) Denote J :=

P

i∈S

Ii . Since Ii ⊆ J for all i ∈ S thus V (J) ⊆ V (Ii ) and so V (J) ⊆

\

V (Ii )

i∈S

T

Conversely, if a ∈ i∈S V (Ii ) then f (a) = 0 for all f ∈ Ii Since the set generates J we get that f (a) = 0 for all f ∈ J, and so a ∈ V (J).

S

i∈S

Ii

90.2 Corollary. Let K be an algebraically closed field. 1) ∅ and K n are algebraic sets in K n . 2) If {A1 , . . . An } is a finite family of algebraic sets in K n then also an algebraic set.

Sn

i=1

Ai is

3) If {Ai }i∈S is an arbitrary family of algebraic sets in K n then also an algebraic set.

T

i∈S

Ai is

Proof. 1) We have ∅ = V (h1i) and K n = V ({0}). 2) – 3) This follows directly from Proposition 90.1.

90.3 Definition/Proposition. Let K be an algebraically closed field. There exists a topology on K n such that closed sets in this topology are algebraic sets A ⊆ K n . This topology is called the Zariski topology on K n . Proof. This follows directly from Corollary 90.2.

351

90.4 Note. A set A ⊆ K 1 is algebraic iff A is the set of zeros of some polynomial f (x) ∈ K[x]. If f (x) = 0 then A = K 1 , if f (x) = 1, then A = ∅, and if deg f (x) > 0 then A is a finite set. Thus the only closed sets in the Zariski topology on K 1 are K 1 , ∅ and all finite subsets A ⊆ K 1 .

352

91

Algebraic varieties

91.1 Definition. Let K be an algebraically closed field. An algebraic set A ⊆ K n is irreducible if for any algebraic sets A1 , A2 ⊆ K n satisfying A = A1 ∪ A2 we have either A = A1 or A = A2 . An irreducible algebraic set is called an algebraic variety.

91.2 Theorem. Let K be an algebraically closed field. Every algebraic set in K n is a union of a finite number of algebraic varieties.

Proof. We argue by contradiction. Let J be the family of all radical ideals I C K[x1 , . . . , xn ] such that V (I) is not a union of a finite number of algebraic varieties. Assume that J 6= ∅. Since the ring K[x1 , . . . , xn ] is Noetherian, by Theorem 85.5 there exists an ideal J that is a maximal element in J. The set V (J) is not an algebraic variety (since by the definition of J it is not a union of a finite number of algebraic varieties), so V (J) = V (I1 ) ∪ V (I1 ) for some radical ideals I1 , I2 C K[x1 , . . . , xn ]. such that V (I1 ) 6= V (J) 6= V (I2 ). Since V (Ii ) ⊆ V (J) for i = 1, 2 we have J ⊆ Ii . Also, since V (J) 6= V (Ii ) and J, Ii are radical ideals, thus by Corollary 89.7 we have J 6= Ii for i = 1, 2. Since J is a maximal element in J this implies that I1 , I2 6∈ J. As a consequence the sets V (I1 ) and V (I2 ) are finite unions of algebraic varieties. Since V (J) = V (I1 )∪V (I2 ) this implies that V (J) is also a union of a finite number of algebraic varieties which is a contradiction.

91.3 Definition. Let K be an algebraically closed field, let A ⊆ K n be an algebraic set and let A = A1 ∪ · · · ∪ Ar (∗) 353

where A1 , . . . , Ar are algebraic varieties. The decomposition (∗) is irredundant if A 6= A1 ∪ · · · ∪ Ai−1 ∪ Ai+1 ∪ · · · ∪ Ar for all i = 1, . . . , r.

91.4 Theorem. Let K be an algebraically closed field and let A ⊆ K n be an algebraic set. Assume we have two irredundant decompositions of A into algebraic varieties: B1 ∪ · · · ∪ Br = A = C1 ∪ · · · ∪ Cs Then s = r and there exists a permutation σ : {1, . . . , r} → {1, . . . , r} such that Bi = Cσ(i) for i = 1, . . . , r.

Proof. Let 1 ≤ i ≤ r. We have Bi = (Bi ∩ C1 ) ∪ · · · ∪ (Bi ∩ Cs ) and since Bi is an algebraic variety thus Bi = Bi ∩ Cσ(i) for some 1 ≤ σ(i) ≤ s. Therefore Bi ⊆ Cσ(i) . By the same argument we obtain that Cσ(i) ⊆ Bk for some 1 ≤ k ≤ r. This gives Bi ⊆ Cσ(i) ⊆ Bk Since the decomposition A = B1 ∪ · · · ∪ Br is irredundant we have i = k, and so Bi = Cσ(i) . Since the decomposition A = C1 ∪ · · · ∪ Cs is irredundant the map σ : {1, . . . , r} → {1, . . . , s} is onto, and in particular s ≤ r. On the other hand, since the decomposition A = B1 ∪ · · · ∪ Br is irredundant σ must be 1-1, and so r ≤ s. As a consequence r = s and σ is a bijection.

91.5 Theorem. Let K be an algebraically closed field. An algebraic set V ⊆ K n is an algebraic variety iff J(V ) C K[x1 , . . . , xn ] is a prime ideal. 354

91.6 Lemma. Let K be an algebraically closed field. If I1 , I2 C K[x1 , . . . , xn ] then p p I1 I2 = I1 ∩ I2 As a consequence we have V (I1 I2 ) = V (I1 ∩ I2 ) = V (I1 ) ∪ V (I2 ) Proof. Exercise.

Proof of Theorem 91.5. (⇒) Assume that V is an algebraic variety, and let I1 , I2 C K[x1 , . . . , xn ] be ideals such that I1 I2 ⊆ J(V ). By Proposition 28.7 it suffices to show that either I1 ⊆ J(V ) or I2 ⊆ J(V ). By Lemma 91.6 we have V ⊆ V (I1 I2 ) = V (I1 ∩ I2 ) = V (I1 ) ∪ V (I2 ) As a consequence V = (V ∩ V (I1 )) ∪ (V ∩ V (I1 ))

Since V is a variety we obtain that either V = V ∩ V (I1 ) or V = V ∩ V (I2 ). We can assume that V = V ∩ V (I1 ). This gives V ⊆ V (I1 ), and so I1 ⊆ J(V ). (⇐) Assume that J(V ) is a prime ideal and let V = V1 ∪ V2 where V1 , V2 are algebraic sets. For i = 1, 2 let Ji = J(Vi ). Since J1 , J2 are radical ideals the ideal J1 ∩ J2 is also radical (check). By Proposition 90.1 we have V (J1 ∩ J2 ) = V (J1 ) ∪ V (J2 ) = V1 ∪ V2 = V so the bijective correspondence (89.7) implies that J(V ) = J1 ∩ J2 . This in turn gives J1 J2 ⊆ J(V ) 355

Since J(V ) is a prime ideal by (28.7) we have that either J1 ⊆ J(V ) or J2 ⊆ J(V ). This gives that either V ⊆ V (J1 ) = V1 or V ⊆ V (J2 ) = V2 . Therefore either V = V1 or V = V2 .

91.7 Corollary. Let K be an algebraically closed set. We have inverse bijections of sets:     prime ideals algebraic varieties V:  :J I C K[x1 , . . . , xn ] A ⊆ Kn Proof. This follows from Corollary 89.7 and Theorem 91.5.

91.8 Corollary. Let K be an algebraically closed field and let I C K[x1 , . . . , xn ] be a radical ideal. Then there exist prime ideals P1 , . . . , Pm C K[x1 , . . . , xn ] such that I = P 1 ∩ · · · ∩ Pn Proof. By Theorem 91.2 we have V (I) = V1 ∪ · · · ∪ Vk where V1 , . . . , Vk are algebraic varieties. Let Pi = J(Vi ). By Theorem 91.5 P1 , . . . , Pn are prime ideals. Since Vi = V (Pi ) we obtain V (I) = V (P1 ) ∪ · · · ∪ V (Pk ) = V (P1 ∩ · · · ∩ Pk ) Both I and P1 ∩ · · · ∩ Pk are radicals ideals so by (89.7) we have I = P1 ∩ · · · ∩ Pk

356

91.9 Corollary. Let K be an algebraically closed field and let I C K[x1 , . . . , xn ]. Then \ √ I= Pi i∈S

where {Pi }i∈S is the family of all prime ideals of K[x1 , . . . , xn ] such that I ⊆ Pi . Proof. Since prime √ ideals are radical ideals, for any prime ideal P such that I ⊆ P we have I ⊆ P . Therefore \ √ I⊆ Pi i∈S

Conversely, by Corollary 91.8 there exist ideals Pi1 , . . . , Pik ∈ {Pi }i∈S such that √ I = Pi1 ∩ · · · ∩ Pik . This gives \ √ Pi ⊆ Pi 1 ∩ · · · ∩ Pi k = I i∈S

As a consequence

√ T I = i∈S Pi .

91.10 Note. 1) The statement of Corollary 91.9 holds if we replace K[x1 , . . . , xn ] by an arbitrary commutative ring with identity (exercise). 2) As a consequence for the zero ideal {0} C R we obtain that intersection of all prime ideals of R. Notice that we have p {0} = {a ∈ R | ak = 0 for some k ≥ 0} p The ideal {0} is called the nilradical of R. We denote: p nil(R) := {0}

357

p {0} is the

92

Regular functions

92.1 Definition. Let K be an algebraically closed field and let V ⊆ K n be an algebraic set. A regular function is a function V → K given by (a1 , . . . , an ) 7→ f (a1 , . . . , an ) where f is some polynomial in K[x1 , . . . , xn ].

92.2 Note. 1) Different polynomials may define the same regular function. E.g. if V = {0} ⊆ K 1 then f (x) = x and g(x) = x2 + x define the same regular function since f (0) = g(0) = 0. 2) Polynomials f, g ∈ K[x1 , . . . , xn ] define the same regular function on V iff for every a ∈ V we have f (a) − g(a) = 0 Equivalently, f, g define the same regular function on V iff f − g ∈ J(V ). This gives a bijection of sets:     regular functions elements of the rings  V →K K[x1 , . . . , xn ]/J(V )

92.3 Definition. Let K be an algebraically closed field and let V ⊆ be an algebraic set. The coordinate ring of V is the ring K[V ] := K[x1 , . . . , xn ]/J(V )

358

92.4 Note. 1) The ring K[V ] is a finitely generated K-algebra. 2) Check: since J(V ) is a radical ideal we have nil(K[V ]) = {0} (see 91.10). 3) If V is an algebraic variety then J(V ) is a prime ideal and so K[V ] is an integral domain.

92.5 Proposition. Let K be an algebraically closed field. If A is a finitely generated K-algebra such that nil(A) = {0} then there exists an algebraic set V such that A = K[V ]. Moreover, if A is in integral domain then V is an algebraic variety. Proof. Since A is finitely generated we have A = K[a1 , . . . , an ] for some elements a1 , . . . , an ∈ A. We have a ring epimorphism ϕ : K[x1 , . . . , xn ] → A given by ϕ(xi ) = ai for i = 1, . . . , n and ϕ|K = idK . Take V := V (Ker ϕ). Since nil(A) = {0} thus Ker(ϕ) is a radical ideal (check!) and so by Theorem 89.6 we have J(V ) = Ker(ϕ). Therefore we have K[V ] = Ker[x1 , . . . , xn ]/ Ker ϕ ∼ =A If A is an integral domain then Ker(ϕ) is a prime ideal and so V is an algebraic variety.

92.6 Definition. Let K be an algebraically closed field and let V ⊆ K n , W ⊆ K m be algebraic sets. Let pi : W → K be the map given by pi (a1 , . . . , am ) = ai A map ϕ: V → W

is a morphism of algebraic sets if pi ϕ : V → K is a regular function for i = 1, . . . , m. 359

92.7 Note. If V is an algebraic set over K then morphisms V → K coincide with regular functions on V (check).

92.8 Proposition. If K is an algebraically closed field and ϕ : V → W , ψ : W → Z are morphism of algebraic sets over K then ψϕ : V → Z is also a morphism of algebraic sets. Proof. Exercise.

92.9 Definition. For an algebraically closed field K denote by AlgSet(K) the category of algebraic sets over K and morphisms of algebraic sets. Let Var(K) denote the subcategory of algebraic varieties over K.

92.10 Note. Let K be an algebraically closed field and let ϕ : V → W be a morphism of algebraic sets over K. For any regular function f : W → K the map f ϕ : V → K is a regular function on V . Using the bijective correspondence between regular functions on W and elements of the coefficient ring K[W ] we obtain a map ϕ] : K[W ] → K[V ] Check: ϕ] is a homomorphism of K-algebras.

92.11 Proposition. Let K be an algebraically closed field and let FDK be the category of finitely generated K-algebras that are integral domains. The functor F : Var(K) → FDK given by F (V ) = K[V ] and F (ϕ) = ϕ] is an equivalence of categories.

92.12 Corollary. If K is an algebraically closed field and V, W are algebraic varieties over K then V ∼ = W in Var(K) iff K[V ] ∼ = K[W ] as K-algebras. 360

93

Suggested further reading

1) M. Aschbacher, Finite group theory, 2000. 2) J.P. Serre, Linear representations of finite groups, GTM 42, 1977 3) J.P. Serre, Trees, 2003 4) K. S. Brown, Cohomology of groups, GTM 87, 1982. 5) Ch. Weibel, Homological algebra, 1995. 6) M.I. Atiyah, I.G. MacDonald, Introduction to commutative algebra, 1969. 7) D. Eisenbud, Commutative algebra with a view toward algebraic geometry, GMT 150, 1995. 8) I.N. Herstein, Noncommutative rings, 1968. 9) J.W. Milnor, Introduction to algebraic K-theory, 1972. 10) S. Mac Lane, Categories for the working mathematician, GTM 5, 1998.

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