ALGEBRAIC NUMBER THEORY Contents Notation 2 Preliminaries I ...

ALGEBRAIC NUMBER THEORY TERUYOSHI YOSHIDA

Contents Notation

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Preliminaries I: Linear Algebra

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a. Sets and Maps

3

b. Algebraic Systems — Structures

4

c. Basis and Dimension

8

d. Linear Maps — Morphisms

11

e. Hom — Classification of Structures

15

f. Matrix Representation of Linear Maps — Representations of Structures

18

g. Determinants and Linear Equations

22

h. Direct Sum and Diagonalization — Decomposition of Structures

26

Part 1.

Introduction to Number Theory

29

1. Prime Decomposition and Principal Ideal Domains

29

2. Quotient Rings

33

Part 2.

Equations and Fields

37

3. Field Extensions

37

4. Galois Theory (1)

39

5. Cyclotomic Extensions

42

6. Galois Theory (2)

46

Date: October 8, 2007. 1

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TERUYOSHI YOSHIDA

7. Kummer Extensions and Solvability of Equations

50

8. Algebraic Closures, Infinite Galois Extensions

53

Part 3.

Algebraic Integers and Prime Ideal Decomposition

59

9. Rings of Integers of Number Fields — I. Definition

59

10. Rings of Integers of Number Fields — II. First Properties

62

11. Rings of Integers of Number Fields — III. Finiteness

64

12. Localization

68

13. Dedekind Domains I

72

14. Dedekind Domains II

74

Part 4.

Arithmetic Galois Theory — Theory of Hilbert

76

15. Modules over PID

76

16. Equations over Rings of Integers — I. Decomposition of Primes

79

17. Equations over Rings of Integers — II. Theory of Hilbert

82

18. Decomposition laws of primes

85

Preliminaries II: Categories and Functors

89

i. Additive Categories

89

j. Functors

93

k. Direct Limits and Inverse Limits

98

l. ⊗ and Hom for Modules

101

Index

106

Notation ∅ — the empty set (a set without any elements). a ∈ A — a is an element of the set A. (Or: an element a of the set A.) a∈ / A — a is not an element of the set A. (Or: an element a not in the set A.) |A| — the cardinality of the set A.

ALGEBRAIC NUMBER THEORY

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∃x ∈ A — There exists an element x of the set A, (satisfying...). ∀x ∈ A — For all elements x of the set A, (the following holds...). B ⊂ A — The set B is a subset of the set A. (Or: a subset B of the set A.) (We decree ∅ ⊂ A for any set A.) A ∪ B — The union of the set A and the set B. A ∩ B — The intersection of the set A and the set B. S n Ai = A1 ∪ · · · ∪ An . Ti=1 n i=1 Ai = A1 ∩ · · · ∩ An . A \ B — The set of elements of A not in B. f : X → Y — A map f from the set X to Y . f : X 3 x 7−→ y ∈ Y — A map f sending a (general) element x of the set X to the element y of the set Y . g ◦ f — The composite of the maps f and g. id — The identity map f : X 3 x 7−→ x ∈ X. N — { 0, 1, 2, 3, 4, . . . }. Z — The set of all integers. Q — The set of all rational numbers. R — The set of all real numbers. C — The set of all complex numbers. deg P — The degree of the polynomial P . (We let deg 0 = −∞.) Preliminaries I: Linear Algebra a. Sets and Maps Definition a.1. (i) A map f : X → Y from the set X to the set Y is a correspondence sending each element x of X to an element y = f (x) of Y . When we want to make the correspondence explicit, we write as f : X 3 x 7−→ y ∈ Y. (ii) The map idX : X 3 x 7−→ x ∈ X is called the identity map of X. (iii) For two maps f : X → Y and g : Y → Z, the map X 3 x 7−→ g(f (x)) ∈ Z is called the composite of f and g, and denoted by g ◦ f . (iv) For a subset X 0 ⊂ X, the subset {f (x) | x ∈ X 0 } of Y is called an image of X 0 , and denoted by f (X 0 ). In particular, the image of X is called the image of f and denoted by Im f . (v) For a subset Y 0 ⊂ Y , the subset {x | f (x) ∈ Y 0 } of X is called the inverse image of Y 0 , and denoted by f −1 (Y 0 ). In particular, for y ∈ Y , the inverse image {x | f (x) = y} of {y} is called the inverse image of y and denoted by f −1 (y). (vi) For a subset X 0 ⊂ X, the map iX 0 : X 0 3 x 7−→ x ∈ X is called the inclusion map. For a map f : X → Y , the map f |X 0 : X 0 3 x 7−→ f (x) ∈ Y is called the restriction of f to X 0 . We have f |X 0 = f ◦ iX 0 . In this case, we say that f is an extension of f |X 0 .

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Definition a.2. (i) A map f is called a surjection if Im f = Y . (ii) A map f is called an injection if it satisfies f (x) = f (y) =⇒ x = y. (iii) A map f is called a bijection if it is a surjection and an injecton. (iv) For a bijection f , we define a map f −1 : Y → X by f (x) = y ⇐⇒ x = f −1 (y) and call it the inverse map of f . Conversely, if there exists a map f −1 : Y → X satisfying f −1 ◦ f = idX and f ◦ f −1 = idY , then f is a bijection. Definition a.3. For two sets X, Y , the set consisting of all pairs (x, y) of an element x of X and an element y of Y is called the direct product and denoted by X × Y . Similarly, for sets X1 , . . . , Xn , the set consisting of n-tuples (x1 , . . . , xn ) of elements xi ∈ Xi is called the direct product of X1 , . . . , Xn , and denoted by n Y Xi = X1 × · · · × Xn . i=1

b. Algebraic Systems — Structures b.1. Operations, groups/rings/fields. Let A be a set, and let x, y, z, . . . denote arbitrary elements of A. Definition b.1. Assume A 6= ∅. A pair (e, ∗) of an element e ∈ A and a map: ∗ : A × A 3 (x, y) 7−→ x ∗ y ∈ A is called an operation on A when it satisfies the following conditions: (i) x ∗ (y ∗ z) = (x ∗ y) ∗ z; (ii) e ∗ x = x ∗ e = x. (We call e the identity for this operation.) An operation is called commutative if in addition it satisfies: (iii) x ∗ y = y ∗ x. Exercise b.2. For a set with an operation (e, ∗), the identity is uniquely determined by ∗, because if e0 satisfies (ii) we get e = e ∗ e0 = e0 . (So we often denote an operation (e, ∗) just by ∗.) Example b.3. N = {0, 1, 2, 3, . . .} has two operations + (addition) and × (or ·, multiplication). The identity of + is 0, and that of × is 1. Example b.4. We will define an operation written as + (addition) or one written as × (or ·, multiplication) on other sets as well. We will always denote the identity for addition by 0, and the identity for multiplication by 1. (The identity 0 of addition is called the zero element.) The name “addition” is only used for commutative operations. Example b.5. Z, Q, R, C all have addition and multiplication.


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Definition b.6. Let A be a set with the operation ∗. For x ∈ A, an element x−1 ∈ A satisfying the following, if it exists, is called an inverse of x: x ∗ x−1 = x−1 ∗ x = e. The element x is called invertible if an inverse exists. Exercise b.7. If y, y 0 are both inverse elements of x, we have y 0 = y 0 ∗ (x ∗ y) = (y 0 ∗ x) ∗ y = y, hence the inverse element of x is unique if exists. Definition b.8. If the inverse of x exists for all x ∈ A, the set A is said to be a group under the operation ∗. When the operation is commutative, A is called a commutative group or an abelian group. When the operation is denoted by +, we call it an additive group, and we denote the inverse of x by −x, and write x − y for x + (−y). Example b.9. (i) We write 0 for the additive group consisting of one element 0. (ii) The set of all vectors on a real plane is an additive group. (iii) Z, Q, R, C are all additive groups. (iv) For any set A with an operation ∗, if x, y ∈ A are invertible then x ∗ y also is, because (y −1 ∗ x−1 ) ∗ (x ∗ y) = e. Hence the subset A× ⊂ A consisting of all invertible elements of A is a group with respect to ∗. Exercise b.10. If we define an addtion by (x, y) 7−→ x + y − 1 on the set Z, it also becomes an additive group. Definition b.11. If a set A with an addition and a multiplication satisfies the following, it is called a ring: (i) A is an additive group; (ii) x(y + z) = xy + xz; (iii) (x + y)z = xz + yz. If the multiplication is commutative, we call A a commutative ring. Example b.12. (i) The ring consisting of one element 0 is denoted by 0, and called a zero ring. (ii) Z is a commutative ring. (iii) The set R[X]of all polynomials in one variable X with coefficients in R is a commutative ring (polynomial ring in one variable over R). In general, for any commutative ring A, we can consider the set A[X] of all polynomials in X with coefficients in A or the set A[X1 , . . . , Xn ] of all polynomials in n variables X1 , . . . , Xn , and they all become commutative rings. (iv) For an integer n > 1, if we define an addition and a multiplication on the set {0, 1, . . . n − 1} by the residue of the sum or the product after dividing by n, we obtain a commutative ring. This ring is called a residue class ring of Z mod n, and denoted by Z/(n). (v) The set of all continuous functions C(R) on R is a commutative ring with respect to the usual (valuewise) addition and multiplication of functions.

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Exercise b.13. (i) For a ring A, 0 = 1 in A ⇐⇒ A = 0. (ii) On the set R>0 of all positive real numbers, we can define a commutative ring structure as follows: Addition: (x, y) 7−→ xy, Multiplication: (x, y) 7−→ xlog y . Definition b.14. (i) An element a ∈ A of a ring A is called a unit when it has an inverse with respect to the multiplication. The set A× of all units of A is a group under the multiplication (Example b.9(iv)), and is called the unit group or group of units of A. (ii) If all the elements except for 0 are units in a commutative ring A 6= 0, the ring A is called a field. Exercise b.15. Z× = {1, −1}, R[X]× = R× . Example b.16. (i) Q, R, C are all fields. (ii) For a prime number p, the residue class ring Z/(p) of Z mod p is a field, and is denoted by Fp . A field consisting of finite number of elements is called a finite field. Definition b.17. (i) If a subset A0 of a group A is again a group under the operation (e, ∗) of A (in particular e ∈ A0 ), A0 is called a subgroup of A. (ii) If a subset A’ of a ring A is again a ring under the operation (0, +) and (1, ×) of A (in particular 0, 1 ∈ A0 ), A0 is called a subring of A. Exercise b.18. A0 is a subgroup of A if and only if x ∗ y −1 ∈ A0 for all x, y ∈ A0 . Example b.19. (i) Z is a subring of Q, which is in turn a subring of R, which is in turn a subring of C. (ii) R is a subring of R[X]. b.2. Vector spaces over K. Definition b.20. Let V be a set, and A a set with a multiplication. A map A × V 3 (a, x) 7−→ ax ∈ V satisfying the following is called an action of A on V : (i) a(bx) = (ab)x, (ii) 1x = x. Definition b.21. Let K be a field. When a set V which has an addition and an action of K satisfies the following, it is called a vector space over K, or a K-vector space: (i) V is an additive group; (ii) a(x + y) = ax + ay for all a ∈ K and x, y ∈ V ; (iii) (a + b)x = ax + bx for all a, b ∈ K and x ∈ V . Example b.22. (i) The additive group 0 consisting of one element 0 is a vector space over any field.


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(ii) A field K is a vector space over K, by regarding its multiplication as an action on itself. (iii) For an integer n ≥ 1, the set K n of n-tuples of elements of K is a vector space if we define the addition and the action of K componentwise: (a1 , . . . , an ) + (b1 , . . . , bn ) = (a1 + b1 , . . . , an + bn ); c(a1 , . . . , an ) = (ca1 , . . . , can ). More generally, for any vector space V over K, the set V n of n-tuples of elements in V is a vector space under the componentwise addition and K-action. (iv) The polynomial √ ring K[X] over √ a field K is a vector space over K. (v) The subset Q(√2) = {a + b √ 2 | a, b ∈ Q} of R is a vector space over Q. (vi) The subset Q( −1) = {a + b −1 | a, b ∈ Q} of C is a vector space over Q. Definition b.23. If a subset V 0 of a vector space V over K is again a vector space over K by the addition and the action of K on V , V 0 is called a subspace of V . Exercise b.24. A subset V 0 ⊂ V is a subspace if and only if it satisfies the following conditions: (i) x, y ∈ V 0 =⇒ x − y ∈ V 0 ; (ii) a ∈ K, x ∈ V 0 =⇒ ax ∈ V 0 . Example b.25. (i) For any vector space V , 0 and V are subspaces of V . (ii) If we consider C as a vector space over R, then R is a subspace of C. (iii) More generally, if a field K is a subring of a ring L, then L is naturally a vector space over K, and K is a subspace of L. (A ring with a vector space structure over K is called a K-algebra.) b.3. A-modules. Definition b.26. Let A be a ring. When a set V which has an addition and an action of A satisfies the following, it is called an A-module: (i) V is an additive group; (ii) a(x + y) = ax + ay for all a ∈ A and x, y ∈ V ; (iii) (a + b)x = ax + bx for all a, b ∈ A and x ∈ V . Example b.27. (i) The additive group 0 consisting of one element 0 is a modular space over any ring. It is the only module over the zero ring. (ii) A ring A is an A-module, by regarding its multiplication as an action on itself. (iii) Any additive group has an action of Z and is a Z-module. (iv) For a field K, a vector space over K means a K-module. (v) For an integer n ≥ 1, the set An of n-tuples of elements of A is an A-module vector space if we define the addition and the action of K componentwise: (a1 , . . . , an ) + (b1 , . . . , bn ) = (a1 + b1 , . . . , an + bn ); c(a1 , . . . , an ) = (ca1 , . . . , can ).

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√ √ √ √ (vi) The subset Z[ 2] = {a + √ b 2 | a, b ∈ Z} of Q( 2), the subset Z[ −1] = √ {a + b −1 | a, b ∈ Z} of Q( −1) are Z-modules. Definition b.28. If a subset V 0 of an A-module V is again an A-module by the addition and the action of A on V , V 0 is called an A-submodule of V . Definition b.29. When we consider a commutative ring A as an A-module, an Asubmodule of A is called an ideal of A. Exercise b.30. (i) For n ∈ Z, the set (n) = {an | a ∈ Z} of all multiples of n is an ideal of Z. (ii) A field K has only two ideals, namely 0 and K. c. Basis and Dimension c.1. Linear relation and bases. Let K be a field, and V a vector space over K. Definition c.1. Let X be a subset of V . P (i) A finite sum of the form ni=1 ai xi (ai ∈ K, xi ∈ X) is called a linear combination of elements of X with coefficients in K. We consider 0 as a linear combination Pof 0 elements of X, and define 0 as the linear combination of ∅. (ii) A relation ni=1 ai xi = 0 expressing 0 as a linear combination of X is called a linear relation among the elements of X. In particular, when all the coefficients ai are 0, it is called a trivial linear relation. Lemma c.2. The following are equivalent: P (i) There exists a linear relation ni=1 ai xi = 0 where the coefficient a1 of x1 is non-zero. (ii) x1 can be expressed as a linear combination of x2 , . . . , xn . ¶ n µ X ai Proof. (i)⇒(ii): x1 = − xi . (ii)⇒(i): Subtract x1 from both sides. a1

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i=2

c.1.1. Linear independence / generating set. Definition c.3. (i) When there is no non-trivial linear relation among the elments of X, the subset X is called linearly independent. If it is not linearly independent, it is called linearly dependent. The empty set is linearly independent. (ii) If all x ∈ V can be written as a linear combination of elements in X, we say that V is generated by X, and X is called a generating set of V . Lemma c.4. Let X be linearly independent. For any y ∈ V , if y is expressed as a linear combination of elements of X, the expresssion is unique (i.e. if we consider the coefficients of the elements of X that do not appear in the expression as 0, then the coefficients are uniquely determined).


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Proof. If there are two different expresssions, their difference gives a non-trivial linear relation among elements of X. ¤ Proposition c.5. Let X ⊂ Y . (i) Y : linearly independent =⇒ X: linearly independent. (ii) X: generates V =⇒ Y : generates V . Proof. A linear combination of elements of X is also that of Y .

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Proposition c.6. Let x ∈ / X, Y = X ∪ {x}. (i) If X is linearly independent and x is not a linear combination of elements of X, then Y is linearly independent. (ii) If Y generates V and x is a linear combination of X, then X generates V . Proof. (i) If X is linearly independent, in any nontrivial linear relation among elements of Y , the coefficient of x must be nonzero, hence x is a linear combination of elements of Y by Lemma c.2. (ii) A linear combination of elements written as linear combinations of elements of X is again a linear combination of elements of X. ¤ Definition c.7. A linear independent generating set of V is called a basis of V . Example c.8. In K n , if we denote by ei the element whose i-th component is 1 and the rest are 0, {e1 , . . . , en } is a basis of K n . This is called a canonical basis of K n . c.2. Existence of a basis and dimension. c.2.1. Existence of a basis. Definition c.9. A vector space V is finitely generated if there is a generating set of V of finite cardinality. Lemma c.10. Assume that V is finitely generated. For any generating set S of V of finite cardinality, there is a subset of S which is a basis of V . Proof. Let T be a linearly independent subset of S whose cardinality is maximal among such subsets. Then by maximality and Lemma c.6(i), all the elements of S are linear combinations of elements of T , hence by Lemma c.6(ii), T is a basis of V . ¤ This lemma gives the following: Theorem c.11. A finitely generated vector space has a basis of finite cardinality.

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c.2.2. Existence of the dimension. Let V be a finitely generated vector space over K, and fix a basis T = {x1 , . . . , xn } of V (whose existence is assured by Theorem c.11). Lemma c.12. If S = {y1 , . . . , yk } ⊂ V (k ≤ n) is linearly independent, we can renumber the indices of xi ∈ T so that U = {y1 , . . . , yk , xk+1 , . . . , xn } is a basis of V . Proof. (i) We argue by induction on k. It is clear when k = 0. For a general k, take a basis U 0 = {y1 , . . . , yk−1 , xk , . . . , xn } by the inductive hypothesis. As U 0 is a basis, we can write yk (uniquely, by Lemma c.4) as a linear combination of elements of U 0 as: yk =

(1)

k−1 X

ai yi +

i=1

n X

bi xi ,

i=k

which gives a linear relation: (2)

k−1 X

ai yi − yk +

i=1

n X

bi xi = 0.

i=k

If the coefficients bi of xk , . . . , xn are all 0, it contradicts the linear independence of S. Hence we renumber the indices so that the coefficient bk of xk is non-zero. We will show that U = U 0 \{xk } ∪ {yk } is a basis of V . As U 0 is linearly independent, U 0 \{xk } is linearly independent. By the uniqueness of expression (1), yk cannot be a linear combination of U 0 \{xk }, hence U 0 \{xk }∪{yk } = U is linearly independent by Lemma c.6(i). As U 0 generates V , U 0 ∪ {yk } generates V . As bk 6= 0 in the relation (2), xk is a linear combination of elements of U = U 0 ∪ {yk }\{xk } by Lemma c.2(i)⇒(ii). Hence U 0 ∪ {yk }\{xk } = U generates V by c.6(ii). ¤ Proposition c.13. Assume that there is a basis T = {x1 , . . . , xn } of V with finite number of elements. (i) If S is a linearly independent subset of V , then |S| ≤ n. If moreover |S| = n, then S is a basis. (ii) If S generates V , then |S| ≥ n. If moreover |S| = n, then S is a basis. Proof. (i) The second part follows from the case k = n of Lemma c.12. If |S| > n, any subset of S with n elements is a basis of V , hence the rest of S are linear combination of them and contradicts the linear independence of S by Lemma c.6(i). (ii) A subset U of S gives a basis of V by Lemma c.10, hence using (i) we see that |T | = n ≤ |U |. Hence |S| ≥ |U | ≥ n, and if |S| = n we have S = U . ¤ By this proposition we obtain: Theorem c.14. All bases of V have the same number of elements.


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Definition c.15. (i) The cardinality of a basis, which is unique by the above theorem, is called the dimension of V , and denoted by dimK V or dim V . (ii) A finitely generated vector space over K is called finite-dimensional. Otherwise it is called infinite-dimensional and we formally write dim V = ∞. Remark c.16. The vector space 0 is finite-dimensional as it has the empty set as a basis, and dim 0 = 0. Conversely dim V = 0 =⇒ V = 0. Exercise c.17. (i) K n is has dimension n as vector space over K, and in general, n V has dimension n dim V . (ii) K[X] is infinite-dimensional vector space over K. (iii) If we denote K[X]deg≤n for the set of all polynomials in X with degree not greater than n, then it is a (n + 1)-dimensional vector space over K, and {1, X, X 2 , . . . , X n } gives its basis. (iv) The set of all sequences {an }n∈N of real numbers satisfying an+2 = an + an+1 is a 2-dimensional vector space over R. (v) C is √ a 2-dimensional √ vector space over √ R. √ (vi) Q( 2) = {a + b 2 | a, b ∈ Q}, Q( −1) = {a + b −1 | a, b ∈ Q} are both 2-dimensional vector spaces over Q. (vii) The R-vector space C(R) of all continuous functions on R is infinite-dimensional. (viii) If we write C ∞ (R) for the set of all functions on R that are differentiable arbitrary many times, then it is an infinite-dimensional vector space over R. (The above two examples are seen to be infinite-dimensional using the Lemma d.9 from R[X] ⊂ C ∞ (R) ⊂ C(R).) d. Linear Maps — Morphisms d.1. Linear maps. Let K be a field. Definition d.1. For two vector spaces V, W over K, a map f : V → W is called a K-linear map if satisfies the following conditions: (i) f (x + y) = f (x) + f (y) (∀x, y ∈ V ); (ii) f (ax) = af (x) (∀a ∈ K, ∀x ∈ V ). In particular, K-linear map f : V → V is called a linear transformation of V . Exercise d.2. (i) The identity map idV : V → V is K-linear. (ii) The composite of two K-linear maps is again K-linear. (iii) The restriction f |V 0 : V 0 → W of a K-linear map f : V → W to a subspace V 0 ⊂ V is again K-linear. Definition d.3. A K-linear map is called an isomorphism if it is bijective. When an isomorphism f : V → W exists, we say that V and W are isomorphic, and write V ∼ = W. Exercise d.4. The identity, the inverse of an isomorphism, the composite of two isomorphisms are all again isomorphisms.

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Exercise d.5. The complex conjugate map C 3 x 7−→ x ∈ C of C is an isomorphism of vector spaces over R, but not a C-linear map of vector spaces over C. Definition d.6. For a K-linear map f : V → W , the subset {x ∈ V | f (x) = 0} of V is called the kernel of f and denoted by Ker f . Proposition d.7. For a K-linear map f : V → W , Ker f and Im f are subspaces of V and W , respectively. Proof. Ker f is a subspace of V because: f (x1 ) = 0, f (x2 ) = 0 =⇒ f (x1 − x2 ) = f (x1 ) − f (x2 ) = 0, f (x) = 0 =⇒ f (ax) = af (x) = 0. Im f is a subspace of Y because: y1 = f (x1 ), y2 = f (x2 ) =⇒ y1 − y2 = f (x1 ) − f (x2 ) = f (x1 − x2 ), y = f (x) =⇒ ay = af (x) = f (ax). ¤ Proposition d.8. A K-linear map f : V → W is injective if and only if Ker f = 0. Proof. Rewrite the defintion f (x) = f (y) =⇒ x = y of the injection as f (x − y) = 0 =⇒ x − y = 0. ¤ d.2. Dimension formula. Lemma d.9. A subspace V 0 of a finite-dimansional vector space V is again finitedimensional and dim V 0 ≤ dim V . If dim V 0 = dim V then V 0 = V . Proof. By Proposition c.13(i), any linearly independent subset of V 0 has cardinality not greater than dim V , hence there is one with maximal cardinality, say T . By maximality and Lemma c.6(i), any other element of V 0 is a linear combination of T , i.e. T generates V 0 , hence a basis of V 0 . Therefore dim V 0 = |T | ≤ dim V , and if |T | = dim V then T is a basis of V by Proposition c.13(i), hence V 0 = V . ¤ In the following, we assume that vector spaces V, W over K are finite-dimensional. Proposition d.10. Let f : V → W be a K-linear map. (i) If f is surjective, X generates V =⇒ f (X) generates W . (ii) If f is injective, X: linearly independent =⇒ f (X): linearly independent. (iii) If f is an isomorphism, X: a basis of V ⇐⇒ f (X): a basis of W . Proof. (i) The image under f of linear combination of elements of X is a linear combination of elements of f (X).


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(ii) A linear relation among the elements of f (X) is an image under f of a linear relation among the elements of X: n n ³X ´ X ai f (xi ) = 0 =⇒ f ai xi = 0. i=1

i=1

Hence, if f is injective and X is linearly independent, it must be a trivial relation. (iii) Combine (i),(ii) and use X = f −1 (f (X)).

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Corollary d.11. For a K-linear map f : V → W : (i) f : surjective =⇒ dim V ≥ dim W . (ii) f : injective =⇒ dim V ≤ dim W . (iii) f : an isomorphism =⇒ dim V = dim W . Theorem d.12. (dimension formula) For a K-linear map f : V → W : dim V = dim(Ker f ) + dim(Im f ). Proof. Let dim V = n, dim(Ker f ) = k and l = n − k and take a basis {y1 , . . . , yk } of Ker f . Then we can take a basis of V of the form T = {y1 , . . . , yk , x1 , . . . , xl } by Lemma c.12. Let V 0 be the subspace of V generated by {x1 , . . . , xl }, and restrict f to f |V 0 : V 0 → W . As T is linearly independent Ker f |V 0 = Ker f ∩ V 0 = 0, hence f |V 0 is injective (Proposition d.8), and any element in the image of V is the image of an element of V 0 , hence f |V 0 is a surjection onto Im f . Therefore V 0 ∼ = Im f and dim(Im f ) = dim V 0 = l. ¤ Corollary d.13. For a linear transformation f : V → V , f : an isomorphism ⇐⇒ f : injective ⇐⇒ f : surjective. Proof. f : injective ⇐⇒ dim(Ker f ) = 0 ⇐⇒ dim(Im f ) = dim V ⇐⇒ f : surjetive. (These equivalences come from respectively Proposition d.8, dimension formula (Theorem d.12) and Lemma d.9.) ¤ Exercise d.14. (i) Fora a vector space over K and a ∈ K, a-multiplication: V 3 x 7−→ ax ∈ V is a K-linear map, and an isomorphism if a 6= 0. (ii) C 3 x 7−→ x + x ∈ R is a surjective R-linear map. (iii) The map of substituting x ∈ K into polynomials with K-coefficients K[X] 3 P (X) 7→ P (x) ∈ K is a surjective K-linear map. (iv) Taking derivatives of polynomials with R-coefficients R[X] 3 P (X) 7−→ P 0 (X) ∈ R[X] is a surjective R-linear map, and its kernel is the subspace R consisting of all constant functions. (The derivative can be defined on polynomial rings over any field K by X n 7→ nX n−1 . See Definition 5.13.) d : C ∞ (R) 3 f 7−→ f 0 ∈ C ∞ (R) is a surjective R-linear (v) The differentiation dx ³d´ map, and Ker is the subspace R of all constant functions. For more dx

14

TERUYOSHI YOSHIDA

d − id : C ∞ (R) 3 f 7−→ f 0 − f ∈ dx C ∞ (R) is an R-linear map, and its kernel is the subspace consisting of solutions of the differential equation (such as f 0 = f ). (vi) If we denote by C([0, 1]) the set of all continuous function on the closed segment R1 [0, 1], the integration C([0, 1]) 3 f 7−→ 0 f (x)dx ∈ R is a surjective R-linear map. general differential operator, such as

d.3. Homomorphisms. Definition d.15. (i) Let X, Y be sets with operations (eX , ∗X ), (eY , ∗Y ) respectively. A map f : X → Y is called a homomorphism if f (x∗X y) = f (x)∗Y f (y) and f (eX ) = f (eY ). (ii) For a set A with a multiplication and sets X, Y with actions of A, a map f : X → Y is called A-equivariantA-equivariant if it satisfies f (ax) = af (x). Definition d.16. (i) A homomorphism f : X → Y between groups X, Y is called a group homomorphism. (ii) A homomorphism f : X → Y between rings X, Y , with respect to both addition and multiplication, is called a ring homomorphism. (iii) An A-equivariant homomorphism f : X → Y between A-modules X, Y is called an A-homomorphism. Exercise d.17. (i) The identity map is a homomorphism. (ii) The composite of two homomorphisms is again a homomorphism. (iii) For any ring X, there is a unique ring homomorphism Z → X. (iv) For any ring X, there is a unique ring homomorphism X → 0. (v) Any ring homomorphism f : X → Y gives a group homomorphism f |X × : X × → Y ×. (vi) For a field K, a K-homomorphism means a K-linear map. Definition d.18. A homomorphism which is a bijection is called an isomorphism. When an isomorphism f : X → Y exists, we say that X and Y are isomorphic, and write X ∼ =Y. Exercise d.19. The identity, the inverse of an isomorphism, the composite of two isomorphisms are all again isomorphisms. Definition d.20. (i) For a group homomorphism f : X → Y , the subset {x ∈ X | f (x) = eY } of X is called the kernel of f and denoted by Ker f . (ii) For a ring homomorphism or an A-homomorphism f : X → Y , the subset {x ∈ X | f (x) = 0} of X is called the kernel and denoted by Ker f . Exercise d.21. (i) For a group homomorphism f : X → Y , Ker f, Im f are subgroups of X, Y respectively. (ii) For a ring homomorphism f : X → Y , Ker f is an ideal of X, and Im f is a subring of Y . (iii) For an A-homomorphism f : X → Y , Ker f, Im f are A-submodules of X, Y respectively.


Exercise d.22. from Z:

15

(i) If we denote the group defined in Exercise b.10 by Z0 , the map

Z 3 x 7−→ x + 1 ∈ Z0 gives an isomorphism of additive groups. (ii) For n ≥ 1, a map which takes x ∈ Z to the residue x after dividing by n: Z 3 x 7−→ x ∈ Z/(n)

gives a surjective ring homomorphism, and its kernel is (n). (iii) √ The map from the polynomial ring R[X] over R to C defined by substituting −1 into X: √ R[X] 3 P (X) 7−→ P ( −1) ∈ C is a surjective ring homomorphism, and its kernel is the set (X 2 + 1) of all polynomials that are divisible by X 2 + 1. e. Hom — Classification of Structures e.1. Hom, End. Let K be a field. Definition e.1. For vector spaces V1 , V2 over K, the set of all K-linear maps from V1 to V2 is denoted by HomK (V1 , V2 ) (or simply Hom(V1 , V2 ). The sum f1 + f2 and scalar multiple af for f, f1 , f2 ∈ HomK (V1 , V2 ) is defined as (f1 + f2 )(x) = f1 (x) + f2 (x), (af )(x) = af (x), which makes HomK (V1 , V2 ) into a vector space over K. In particular, the set of linear transformations Hom(V, V ) of V is denoted by End(V ). As the composite of K-linear maps was again K-linear, there is a map: Hom(V1 , V2 ) × Hom(V2 , V3 ) 3 (g, f ) 7−→ f ◦ g ∈ Hom(V1 , V3 ), which clearly satisfies f ◦ (g ◦ h) = (f ◦ g) ◦ h. Definition e.2. In End(V ), the composite: End(V ) × End(V ) 3 (g, f ) 7−→ f ◦ g ∈ End(V ) is an operation on End(V ) with the identity being the identity map id ∈ End(V ). This gives a (non-commutative) ring structure on End(V ), and this ring is called the endomorphism ring of V . Exercise e.3. Check that End(V ) satisfies the definition of rings. Exercise e.4. End(V ) turns out to have dimension (dim V )2 as a vector space over K, and is an example of a K-algebra. As the inverse with respect to the multiplication of End(V ) is none other than the inverse map, therefore f ∈ End(V ) is a unit if and only if f is an isomorphism. Definition e.5. The unit group End(V )× (the group of all the isomorphisms f : V → V ) of End(V ) is called the automorphism group of V and denoted by Aut(V ).

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Exercise e.6. If we fix an f ∈ Hom(V1 , V2 ), for any W , there are K-linear maps: f ∗ : Hom(V2 , W ) 3 g 7−→ g ◦ f ∈ Hom(V1 , W ), f∗ : Hom(W, V1 ) 3 g 7−→ f ◦ g ∈ Hom(W, V2 ). We have f : an isomorphism ⇐⇒ f ∗ : bijective for all W ⇐⇒ f∗ : bijective for all W . e.2. Hom(K n , V ) — representation by bases. Let V be a vector space over K. For any element x ∈ V , there is a K-linear map K: ϕx : K 3 a 7−→ ax ∈ V. More generally, for any n-tuples X = {x1 , . . . , xn } ∈ V n of elements in V , there is a K-linear map: ϕX : K n 3 (a1 . . . , an ) 7−→ a1 x1 + . . . + an xn ∈ V. This gives a K-linear map V n 3 X 7−→ ϕX ∈ Hom(K n , V ), and its inverse is given by Hom(K n , V ) 3 ϕ 7−→ (ϕ(e1 ), . . . , ϕ(en )) ∈ V n , where {e1 , . . . , en } is the canonical basis of Example c.8. Proposition e.7. By the above bijection, V n ∼ = Hom(K n , V ). Therefore, giving a K-linear map from K n to V is equivalent to choosing n elements (ordered) from V . Using this correspondence, we can translate the definition of linear independence, generating sets and bases: Lemma e.8. For a subset X ⊂ V consisting of n distinct elements, let ϕX : K n → V be the K-linear map defined above. (i) X: linearly independent ⇐⇒ ϕX : injective. (ii) X: generates V ⇐⇒ ϕX : surjective. (iii) X: basis of V ⇐⇒ ϕX : an isomorphism. By the above, we have: Proposition e.9. Let Basis(V ) be the set of all bases of V consisting of n elements (considered as an ordered set, i.e. we distinguish the permuted bases), and Isom(K n , V ) be the set of all isomorphisms from K n to V . Then there is a bijection: Basis(V ) 3 X 7−→ ϕX ∈ Isom(K n , V ). In other words, fixing a basis X of V consisting of n elements is equivalent to fixing an isomorphism f : K n → V . In particular, as a basis exists for any finite-dimensional vector spaces: Theorem e.10. If V is an n-dimensional vector space over K, then K n ∼ = V . In particular, if V, W are finite-dimensional vector spaces, V ∼ W ⇐⇒ dim V = dim W . =


17

Proof. The first part follows from the existence of a basis. As for the second part, ⇒ follows from Corollary d.11(iii), and ⇐ is because if dim V = dim W = n then Kn ∼ ¤ = V, K n ∼ = W. e.3. Change of bases. Let V be a n-dimensional vector space over K. As the composite of two isomorphisms is again an isomorphism, the group Aut(V ) acts on Isom(K n , V ) as follows: Aut(V ) × Isom(K n , V ) 3 (f, ϕ) 7−→ f ◦ ϕ ∈ Isom(K n , V ). Under the bijection Isom(K n , V ) 3 ϕ 7−→ X = {ϕ(e1 ), . . . , ϕ(en )} ∈ Basis(V ) of Proposition e.9, we have f ◦ ϕ 7−→ f (X), so if we consider the above action as an action on Basis(V ), we get: Aut(V ) × Basis(V ) 3 (f, X) 7−→ f (X) ∈ Basis(V ). Proposition e.11. (change of bases) Let X = {x1 , . . . , xn } ∈ Basis(V ). (i) For any f ∈ Aut(V ), f (X) = {f (x1 ), . . . , f (xn )} is again a basis of V . (ii) For any basis X 0 of V , there is a unique f ∈ Aut(V ) satisfying f (X) = X 0 . (iii) Aut(V ) 3 f 7−→ f (X) ∈ Basis(V ) is a bijection. Proof. (i): Clear by the above action. (ii): By the bijection of Proposition e.9, f (X) = X 0 ⇐⇒ f ◦ ϕX = ϕX 0 ⇐⇒ f = ϕX 0 ◦ ϕ−1 X . (iii) follows immediately from (i),(ii). ¤ f = ϕX 0 ◦ ϕ−1 X is represented by the following diagram: n

K C CC ϕ 0 {{ CC X { { CC { { C! { {} f /V V ϕX

(A diagram consisting of sets and arrows representing the maps between the sets is called a commutative diagram if for any two sets the composite of maps along a path between those two sets does not depend on the path. (In this case, f ◦ ϕX = ϕX 0 .) e.4. Categories. Definition e.12. A category C is defined as follows: (i) There is a notion of X being an object of C . We write X ∈ C . (ii) For any X, Y ∈ C , there is a set HomC (X, Y ) of morphisms from X to Y . (A morphism f ∈ HomC (X, Y ) is denoted by f : X → Y .) (iii) HomC (X, Y ) ∩ HomC (X 0 , Y 0 ) = ∅ unless X = X 0 and Y = Y 0 . (iv) For X, Y, Z ∈ C , there is a map HomC (X, Y ) × HomC (Y, Z) 3 (f, g) 7−→ g ◦ f ∈ HomC (X, Z) called the composition of morphisms. (v) (associativity) h ◦ (g ◦ f ) = (h ◦ g) ◦ f .

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(vi) For all X ∈ C , there is an identity morphism idX ∈ HomC (X, X) of X such that for all Y ∈ C and f ∈ HomC (X, Y ), we have f ◦ idX = idY ◦ f = f . Definition e.13. (i) A morphism f ∈ HomC (X, Y ) of C is called an isomorphism of C if there exists a g ∈ HomC (Y, X) satisfying g ◦ f = idX , f ◦ g = idY . (ii) When an isomorphism f ∈ HomC (X, Y ) exists, we say that X and Y are isomorphic in C , and write X ∼ =C Y . Example e.14. The category of sets Sets has sets as objects and maps as morphisms. The category of vector spaces VectK has K-vector spaces as objects and K-linear maps as morphisms. Similarly we can define the category Groups of groups and group homorphisms, Rings of rings and ring homomorphisms, and A-Mod of A-modules and A-homomorphisms for a ring A. For a field K, the category K-Mod means VectK (Example b.27(iv), d.17(vi)). Exercise e.15. For an object X ∈ C of a category C , the set AutC (X) of all isomorphisms f : X → X is a group under the composition. f. Matrix Representation of Linear Maps — Representations of Structures f.1. Matrices. Let K be a field, and n, m ≥ 1 be integers. As giving a K-linear map f ∈ Hom(K n , K m ) is equivalent to giving f (e1 ), . . . , f (en ), if we set, for 1 ≤ j ≤ n, f (ej ) = (a1j , . . . , amj ), the map f is uniquely represented by the mn numbers aij ∈ K (1 ≤ i ≤ m, 1 ≤ j ≤ n). Definition f.1. The arrangement of mn elements in K in the following from is called an m by n matrix over K:   a11 a12 · · · a1n  a21 a22 · · · a2n    (aij ) =  .. .. ..  .  . . .  am1 am2 · · · amn The aij is called the (i, j)-entry of the matrix (aij ). A matrix with all entries equal to 0 is denoted by 0. The set of all m by n matrices over K is denoted by Mm,n (K). Thus we have a bijection: Hom(K n , K m ) 3 f 7−→ (aij ) ∈ Mm,n (K). Moreover, as addition and scalar multiplication on Hom(K n , K m ) corresponds to the entrywise addition and scalar multiplications of matrices, we define the K-vector space structure on Mm,n (K) by entrywise operations, and:


19

Proposition f.2. As K-vector spaces, Hom(K n , K m ) ∼ = Mm,n (K). By Proposition e.7, we have Hom(K n , K m ) ∼ = (K m )n ∼ = K mn as K-vector spaces, hence Mm,n (K) is a mn-dimensional vector space. There is an obvious canonical basis on Mm,n (K): Definition f.3. A matrix with only (i, j)-entry equal to 1 and rest of the entries equal to 0 is called a matrix element and denoted by (1ij ). Clearly {(1ij )}1≤i≤m,

1≤j≤n

gives a basis of Mm,n (K).

Moreover, if we compose g ∈ Hom(K n , K m ) and f ∈ Hom(K m , K l ) we get f ◦ g ∈ Hom(K n , K l ), but if we represent f, g, f ◦ g respectively by (aij ) ∈ Ml,m (K), (bjk ) ∈ Mm,n (K), (cik ) ∈ Ml,n (K), we have: cik =

m X

aij bjk .

j=1

Hence we define: Definition f.4. P For (aij ) ∈ Ml,m (K), (bjk ) ∈ Mm,n (K), the matrix (cik ) ∈ Ml,n (K) defnied by cik = m j=1 aij bjk is called the product of (aij ), (bjk ) and written as (cik ) = (aij )(bjk ). Note that associativity of additions and compositions of matrices follow from those for Hom(K n , K m ). In particular, the set of all n by n matrices, corresponding to the endomorphism ring End(K n ) = Hom(K n , K n ) of K n , has a ring structure. Definition f.5. An n by n matrix is called a square matrix of degree n, and the set Mn,n (K) of all square matrices of degree n over K is denoted simply by Mn (K). This is a ring by the entrywise addition and the product, and the identity of the multiplication is the following matrix:   1 0 ··· 0  0 1 ··· 0    (δij ) =  .. .. . . . .  . . . ..  0 0 ···

1

This (δij ) is called the identity matrix. The symbol δij (Kronecker’s delta) is generally used for the (i, j)-entry of the identity matrix. The identity matrix is the matrix representing the identity map id ∈ End(K n ). Similarly, we can represent the automorphism group Aut(K n ) of K n by matrices. Definition f.6. A square matrix α ∈ Mn (K) is called invertible when there exists an α−1 ∈ Mn (K) which satisfies αα−1 = α−1 α = (δij ), in which case α−1 is called the inverse matrix of α.

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The set of all n by n invertible matrices is the unit group of the ring Mn (K). This group is called the general linear group of degree n over K, and is denoted by GLn (K) = Mn (K)× . Exercise f.7. When n = 1, M1 (K) ∼ = K, GL1 (K) ∼ = K × . For n > 1, Mn (K) is a non-commutative ring, and GLn (K) is a non-commutative group. f.2. Matrix representation of linear maps. Let V1 , V2 be finite-dimensional vector spaces over K, and let their dimension be n, m respectively. If we fix a basis Y = {yj } of V1 and a basis X = {xi } of V2 , they give isomorphisms ϕY : K n → V1 , ϕX : K m → V2 . For any f ∈ Hom(V1 , V2 ), if we let f 0 = ϕ−1 X ◦ f ◦ ϕY , there is a following commutative diagram: f0

Kn ϕY

²

/ Km

∼ =

∼ = ϕX

² / V2

f

V1

By this correspondence, we have an isomorphism: Hom(V1 , V2 ) 3 f 7−→ f 0 ∈ Hom(K n , K m ) ∼ = Mm,n (K) Definition f.8. For a K-linear map f ∈ Hom(V1 , V2 ), the matrix (aij ) ∈ Mm,n (K) representing f 0 is called the representation matrix of f with respect to bases Y = {yj }, X = {xi }. Exercise f.9. The entries of the representation matrix of f with respect to bases Y = {yj }, X = {xi } are the coefficients aij (1 ≤ i ≤ m, 1 ≤ j ≤ n) appearing in: f (yj ) =

m X

aij xi (aij ∈ K).

i=1

Exercise f.10. The representation matrix of an isomorphism is invertible. The identity and inverse maps are represented respectively by identity and inverse matrices. Exercise f.11. Fix the bases Z = {zi }, Y = {yj }, X = {xk } for the K-vector spaces V1 ∼ = K n , V2 ∼ = K m , V3 ∼ = K l and let the representation matrices of f ∈ Hom(V2 , V3 ), g ∈ Hom(V1 , V2 ) with respect to these bases be (aij ) ∈ Ml,m (K), (bjk ) ∈ Mm,n (K) respectively. Then the composite f ◦ g ∈ Hom(V1 , V3 ) is represented by the product (cik ) = (aij )(bjk ): Kn

g0

ϕZ ∼ =

²

V1

/ Km ϕY

g

²

f0

∼ =

/ V2

/ Kl ϕX ∼ =

f

² / V3


21

f.3. Change of bases for matrix representations. Lemma f.12. Let (pij ) ∈ GLn (K) be the representation matrix of f ∈ Aut(V ) with respect to the basis X = {xi }, and let f (X) = X 0 . Then we have a commutative diagram: (pij )

K nC

CC CC ϕX 0 CCC !

V

/ Kn { { {{ {{ ϕX { {}

Proof. By definition of the representation matrix, we have a commutative diagram: (pij )

Kn

/ Kn

ϕX ∼ =

∼ = ϕX

²

² /V

f

V as f = ϕX 0 ◦ ϕ−1 X , we obtain the lemma.

¤

Remark f.13. If we set X 0 = {x0j } = f (X) then x0j =

Pn

i=1 pij xi

by Exercise f.9.

Let V1 , V2 be K-vector spaces with dimension respectively n, m, and fix the bases Y = {yj }, X = {xi } respectively of V1 , V2 . Consider the change of basis g1 (Y ) = Y 0 , g2 (X) = X 0 under g1 ∈ Aut(V1 ), g2 ∈ Aut(V2 ), and let η = (qij ), ξ = (pij ) ∈ GLn (K) respectively be the representation matrices of g1 , g2 with respect to Y, X. Proposition f.14. If we let α = (aij ) (resp. α0 = (a0ij )) be the representation matrix of a K-linear map f ∈ Hom(V1 , V2 ) with respect to Y, X (resp. Y 0 , X 0 ). Then: α0 = ξ −1 αη. Proof. By Lemma f.12, consider the commutative diagram: K nC

CC CC ϕY 0 CCC !

η

V1

/ Kn { { {{ {{ ϕY { }{

α

f

−1

ξ / Km / Km CC { CC { { C {{ϕ 0 { ϕX CCC }{{ X ! / V2

¤ Corollary f.15. Let V be an n-dimensional vector space. Let the image of the basis X = {xi } under g ∈ Aut(V ) be X 0 = g(X), and let ξ = (pjk ) be the representation matrix of g with respect to X. If we denote the representation matrix of an f ∈ End(V ) with respect to X (resp. X 0 ) by α = (aij ) (resp. α0 = (a0ij )), then: α0 = ξ −1 αξ. Exercise f.16. Represent the following linear maps with respect to the given bases.

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(i) For a K-vector space V and a ∈ K, the a-multiplication V 3 x 7−→ ax ∈ V has the same representation matrix for any bases. √ (ii) The R-linear map C 3 x 7−→ x + x ∈ R, the basis {1, −1} of C, the basis {1} of R. √ √ (iii) The R-linear map C 3 x 7−→ (a + b −1)x ∈ C (a, b ∈ R), the basis {1, −1} of C. √ √ √ √ √ (iv) The Q-linear map Q( 2) 3 x 7−→ 2x ∈ Q( √ 2), the √ basis {1, √2} of Q( 2). (v) The same map, with respect to the basis {1 + 2, 1 − 2} of Q( 2), and check that the formula of Corollary f.15 holds. (vi) The substitution map K[X]deg≤n 3 P (X) 7→ P (x) ∈ K for x ∈ K, the basis {1, X, X 2 , . . . , X n } of K[X]deg≤n , the basis {1} of K. (vii) The derivative map R[X]deg≤n 3 P (X) 7−→ P 0 (X) ∈ R[X]deg≤n−1 , the basis {1, X, X 2 , . . . , X n } of R[X]deg≤n , the basis {1, X, X 2 , . . . , X n−1 } of R[X]deg≤n−1 . (For the definition of K[X]deg≤n , see Exercise c.17.) g. Determinants and Linear Equations g.1. Determinants. g.1.1. Volume Forms. Definition g.1. A map Φ : V n → K is called an n-fold multilinear form on V if it satisfies the following conditions: (i) Φ(x1 , . . . , xi + yi , . . . , xn ) = Φ(x1 , . . . , xi , . . . , xn ) + Φ(x1 , . . . , yi , . . . , xn ); (ii) Φ(x1 , . . . , axi , . . . , xn ) = aΦ(x1 , . . . , xi , . . . , xn ) (∀a ∈ K). If moreover it satisfies (iii) Φ(x1 , . . . , xn ) = 0 if xi = xj for i 6= j, then Φ is called a alternating multilinear form. The set of all alternating n-fold multilinear forms on V is a K-vector space by the addition and scalar multiplication on their values. Lemma g.2. For an alternating multilinear form Φ: (i) Φ(x1 , . . . , xi , . . . , xj , . . . , xn ) = −Φ(x1 , . . . , xj , . . . , xi , . . . , xn ). (ii) For a bijection σ : {1, . . . , n} → {1, . . . , n} (permutation): Φ(xσ(1) , . . . , xσ(n) ) = s(σ)Φ(x1 , . . . , xn ). Q (Here s(σ) = 1≤j 1, say n > 1. Then p1 | q1 · · · qm , and as p1 is prime, there exists a j with which we have p1 | qj , but as p1 , qj are primes and therefore irreducibles, p1 , qj must be associates. Therefore, by Exercise 1.4(ii), p2 · · · pn and q1 · · · qj−1 qj+1 · · · qm are associates, but as max{n − 1, m − 1} < max{n, m} the proposition is proven by the inductive hypothesis. ¤ Definition 1.10. If every element of a domain A, except 0 and units, is a product of primes, A is called a unique factorization domain (UFD). The decomposition of an element into a product of primes is unique up to associate (Proposition 1.9). Definition 1.11. Two elements a, b of a UFD A are said to be relatively prime or coprime to each other if no prime is a divisor of both elements. 1.2. Ideals. 1.2.1. Ideals and principal ideals. Let A be a ring. In the following, we always regard A as an A-module by means of the multiplication of A. (Example b.27(ii)). Recall that an A-submodule of A is called an ideal of A (Definition b.29). Exercise 1.12. Ideals of a field A are only 0 and A. Exercise 1.13. If A is a subring of B, (i) B is naturally an A-module, and (ii) every B-module is naturally an A-module. Definition 1.14. For a ∈ A, the set {ax | x ∈ A} of all multiples of a is an ideal of A. We denote this ideal by (a), and call it the principal ideal generated by a. Exercise 1.15. (i) a = 0 ⇐⇒ (a) = 0. × (ii) a ∈ A ⇐⇒ (a) = A. (iii) For an ideal I, (a) ⊂ I ⇐⇒ a ∈ I. Exercise 1.16. (principal ideals and divisibility) (i) (a) ⊃ (b) ⇐⇒ a | b. (ii) (a) = (b) ⇐⇒ a, b : associate. If we assume A is a domain, we have: (iii) A 6= (a) % (b) ⇐⇒ a : a proper divisor of b. (iv) (a) = (b) ⇐⇒ b = ac, c ∈ A× . 1.2.2. Prime ideals and maximal ideals. Definition 1.17. An ideal I $ A is called: (i) prime if the following holds: a, b ∈ / I =⇒ ab ∈ / I,

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(ii) maximal if no ideal other than A contains I as a proper subset. The set of all prime (resp. maximal) ideals of A is denoted by Spec(A) (resp. m-Spec(A)). Exercise 1.18. (i) A = 0 =⇒ Spec(A) = ∅. (ii) A : domain ⇐⇒ 0 ∈ Spec(A). (iii) In a domain, A 3 a : prime ⇐⇒ (a) ∈ Spec(A)\{0}. (iv) If A is a subring of B, then Q ∈ Spec(B) =⇒ Q ∩ A ∈ Spec(A). (v) More generally, for a ring homomorphism f : A → B, Q ∈ Spec(B) =⇒ f −1 (Q) ∈ Spec(A). Definition 1.19. For ideals I1 , I2 of A, the set {x + y | x ∈ I1 , y ∈ I2 } is another ideal of A, and contains both I1 and I2 . This ideal is called the sum of I1 and I2 , and denoted by I1 + I2 . Example 1.20. In A = Z, (6) + (15) = (3). (In general, (a) + (b) is a principal ideal generated by the g.c.d. of a, b.) Proposition 1.21. m-Spec(A) ⊂ Spec(A). Proof. If a maximal ideal I is not prime, as I 6= A, ∃a, b ∈ / I, ab ∈ I. Then I + (b) contains I as a proper subset, therefore equals A. Hence ∃d ∈ I, ∃c ∈ A, I + (b) 3 1 = d + bc, and as ad, ab ∈ I we have a = ad + abc ∈ I, a contradiction. ¤ 1.3. Principal ideal domains. Definition 1.22. A domain is called a principal ideal domain (PID) if all of its ideals are principal. Example 1.23. A field is a PID. Proposition 1.24. Let A be a PID, and a ∈ A. Then: (i) a: irreducible ⇐⇒ (a) ∈ m-Spec(A), (ii) a: irreducible ⇐⇒ a : prime. Proof. (i) Follows from Exercise 1.16(iii). (ii) ⇐: Proposition 1.8. ⇒: Use (i), Proposition 1.21, and Exercise 1.18(i).

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Proposition 1.25. In a PID, evey element, except 0 and units, is decomposed as a product of irreducible elements. Proof. Let S be the set of products of irreducible elements in A. Assume ∃a0 ∈ / S, a0 6= 0, a0 ∈ / A× . As a0 is not an irreducible, it is a product of two proper divisors. If they both belong to S then a0 ∈ S, so at least one of them, say a1 , does not belong to S. By repeating the same procedure on a1 we get a2 ∈ / S, a proper divisor of a1 . Continuing to get a sequence a0 , a1 , a2 , · · · , by Exercise 1.16(iii) we have (ai ) $ (ai+1 ) (∀i ∈ N). S On the other hand, consider the ideal I = ∞ (a ) of A (see Exercise 1.26 below). As i i=0


33

A is a PID, I = (a) for some a ∈ A. As a ∈ I we have a ∈ (ai ) for some i, but then (a) ⊂ (ai ) $ (ai+1 ) ⊂ (a) is a contradiction. ¤ Exercise 1.26. For an increasing sequence I0 ⊂ I1 ⊂ I2 ⊂ · · · of ideals in A, the union S I= ∞ I i=0 i is an ideal of A. Theorem 1.27. PID =⇒ UFD. Proof. Follows from Proposition 1.25 and Proposition 1.24(ii).

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1.4. Euclidean domains. Definition 1.28. A domain A is called a euclidean domain if there exists a map f : A → N ∪ {−∞} satisfying the following condition: a, b ∈ A, a 6= 0 =⇒ ∃q, r ∈ A b = aq + r, f (r) < f (a). Exercise 1.29. (i) Z is euclidean. In fact, f (x) = |x| satisfies the condition. (ii) For a field K, a one-variable polynomial ring K[X] with coefficients in K is euclidean. In fact, f (P ) = deg P suffices. Proposition 1.30. Euclidean domain =⇒ PID. Proof. For an ideal I 6= 0 of a euclidean domain A, take a non-zero element a of I such that f (a) is minimal. Then ∀b ∈ I, ∃q, r ∈ A b = aq + r, f (r) < f (a), but as r = b − aq ∈ I, we have r = 0 by minimality of f (a), therefore b is a multiple of a, hence I = (a). ¤ Theorem 1.31. (Fundamental theorem of arithmetic) Z is a PID, therefore a UFD. Proof. Follows from Exercise 1.29(i), Proposition 1.30 and Theorem 1.27.

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2. Quotient Rings 2.1. Quotient algebraic systems. Definition 2.1. Let X be a set. Suppose we know, for every pair of elements x, y of X, a relation x ∼ y holds or not. The relation ∼ is called an equivalence relation if the following conditions are satisfied: (i) x ∼ x (reflexive law), (ii) x ∼ y =⇒ y ∼ x (symmetric law), (iii) x ∼ y, y ∼ z =⇒ x ∼ z (transitive law). For an element x ∈ X, a subset {y ∈ X | x ∼ y} of X is called an equivalence class of x, and is denoted x, and x is called a representative element of the class x. By (i),(ii),(iii), X is partitioned into mutually disjoint equivalence classes. The set of equivalence classes is called the quotient set of X by the relation ∼.

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Proposition 2.2. (i) For a subgroup Y of an additive group X, if we define a relation ∼ in X as: x ∼ y ⇐⇒ x − y ∈ Y, it is an equivalence relation, and we can define a natural addition on the quotient set of X by ∼ (denoted by X/Y ) by the addition of representative elements, and it is an additive group. (ii) For an ideal Y of a ring X, we can define a natural multiplication on the additive group X/Y by multiplication of representative elements, and it becomes a ring. (iii) For an A-submodule Y of an A-module X, we can define a natural A-action on the additive group X/Y by A-action on representative elements, and it becomes an A-module. Proof. (i) First, the relation ∼ is an equivalence relations because x−y ∈Y x − y, y − z ∈ Y

x − x = 0 ∈ Y, =⇒ y − x = −(x − y) ∈ Y, =⇒ x − z = (x − y) + (y − z) ∈ Y

hold. Secondly, x ∼ x0 , y ∼ y 0 =⇒ x − x0 , y − y 0 ∈ Y =⇒ (x + y) − (x0 + y 0 ) = (x − x0 ) + (y − y 0 ) ∈ Y =⇒ x + y ∼ x0 + y 0 shows that the operation x + y = x + y on X/Y is well-defined regardless of the choice of representative elements (and 0 is the zero element) which gives the addition on X/Y , and it is a group as −x gives the inverse element of x. (ii) Also for the multiplication, as Y is an ideal of X, x ∼ x0 , y ∼ y 0 =⇒ x − x0 , y − y 0 ∈ Y =⇒ xy − x0 y 0 = (x − x0 )y + x0 (y − y 0 ) ∈ Y =⇒ xy ∼ x0 y 0 holds, and hence x · y = xy is well-defined regardless of the choice of representatives (and 1 is the unity), and X/Y becomes a ring. (iii) Also for the A-action, as Y is an A-submodule of X, x ∼ x0 =⇒ x − x0 ∈ Y =⇒ ax − ax0 = a(x − x0 ) ∈ Y =⇒ ax ∼ ax0 holds, and hence the A-action ax = ax is well-defined, and X/Y becomes an A-module. ¤ Definition 2.3. The X/Y in Proposition 2.2(i),(ii),(iii) are called respectively quotient group, quotient ring, quotient A-module of X by Y . The surjection X → X/Y defined by x 7−→ x is called the canonical surjection. It is a homomorphism by definition. We also write x as x mod Y .


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Exercise 2.4. X/0 ∼ = X, X/X ∼ = 0. Example 2.5. (i) For an ideal (n) (n ≥ 1) of Z, the quotient ring Z/(n) = {0, 1, . . . , n − 1} is “the ring of residues of integers divided by n”. We also denote k = k mod (n) as k mod n. (ii) The quotient ring of R[X] by the ideal (X 2 + 1) is isomorphic to C by the following ring isomorphism (in fact, it is the formal definition of the complex numbers): √ R[X]/(X 2 + 1) 3 a + bX 7−→ a + b −1 ∈ C. Exercise 2.6. For general (non-commutative) group G and its subgroup H, in order to define the quotient group G/H, it is necessary for H to satisfy the following condition: x ∈ G, y ∈ H =⇒ x ∗ y ∗ x−1 ∈ H. A subgroup H satisfying this condition is called a normal subgroup of G, and we denote it as G B H. 2.2. Homomorphism theorem. Proposition 2.7. Consider a homomorphism f : X → Y of groups/rings/A-modules. (i) For groups, Ker f, Im f are subgroups of X, Y respectively. (ii) For rings, Ker f is an ideal of X and Im f is a subring of Y . (iii) For A-modules, Ker f, Im f are A-submodules of X, Y respectively. Proof. (i) By the following: f (x1 ) = 0, f (x2 ) = 0 =⇒ f (x1 − x2 ) = 0, y1 = f (x1 ), y2 = f (x2 ) =⇒ y1 − y2 = f (x1 − x2 ). (ii) By (i) and the following: f (x2 ) = 0 =⇒ f (x1 x2 ) = f (x1 )f (x2 ) = 0, y1 = f (x1 ), y2 = f (x2 ) =⇒ y1 y2 = f (x1 x2 ), f (1) = 1. (iii) By (i) and the following: f (x) = 0 =⇒ f (ax) = af (x) = 0, y = f (x) =⇒ ay = af (x) = f (ax). ¤ Theorem 2.8. (Homomorphism theorem) For a (group/ring/A-) homomorphism f : X → Y , there is a canonical isomorphism X/Ker f ∼ = Im f . Proof. For a homomorphism f , if we denote by x the equivalence class of x in X/Ker f : x = y ⇐⇒ x − y ∈ Ker f ⇐⇒ f (x) = f (y),

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therefore the map: f : X/Ker f 3 x 7−→ f (x) ∈ Im f is well-defined and injective. As f is clearly surjective, it is bijective. Also, as f is a homomorphism, the bijection f is a homomorphism by: f (x + y) = f (x + y) = f (x + y) = f (x) + f (y) = f (x) + f (y), f (0) = f (0) = 0, therefore a group isomorphism. Similarly when f is a ring (resp. A-) homomorphism, we see that the bijection f is a ring (resp. A-) homomorphism, therefore it is a ring (resp. A-) isomorphism. ¤ Exercise 2.9. For homomorphism f : X → Y of general (non-commutative) groups, Ker f is a normal subgroup of X and Im f is a subgroup of Y , and X/Ker f ∼ = Im f . 2.3. Ideals and quotient rings. For an A-homomorphism f : X → Y between Amodules, let S be the set of all A-submodules of X that contain Ker f , and T the set of all A-submodules of Im f . As the image of an A-submodule of X is always an A-submodule of Im f , and the inverse image of an A-submodule of Im f is always an A-submodule of X containing Ker f , we have the following two maps: Φ : S 3 I 7−→ f (I) ∈ T, Ψ : T 3 J 7−→ f −1 (J) ∈ S. Proposition 2.10. The two maps Φ, Ψ are inverse to each other, and therefore both isomorphisms. Proof. These maps clearly preserve the inclusion relations, and as f (f −1 (J)) = J, we have Φ◦Ψ = id. Also, observing that Ψ◦Φ(I) = f −1 (f (I)) ⊃ I, as we have f (x) ∈ f (I) for all x ∈ f −1 (f (I)), ∃y ∈ I, f (y) = f (x). Therefore, as x − y ∈ Ker f ⊂ I shows that x = y + (x − y) ∈ I, we have f −1 (f (I)) ⊂ I, hence Ψ ◦ Φ = id. ¤ Proposition 2.11. For a surjective homomorphism f : A → B, above Φ, Ψ give oneto-one correspondence between (i) the prime ideals of A that contain Ker f and the prime ideals of B, and (ii) the maximal ideals of A that contain Ker f and the maximal ideals of B. Proof. That maximals correspond to each other follows from the fact that Φ, Ψ preserve the inclusion relations. Also, for P ∈ Spec(A), let a, b ∈ / f (P ), and choosing a0 ∈ −1 0 −1 f (a), b ∈ f (b), as a0 , b0 ∈ / P =⇒ a0 b0 ∈ / P =⇒ ab = f (a0 b0 ) ∈ / f (P ), we see that f (P ) ∈ Spec(B). That Q ∈ Spec(B) =⇒ f −1 (Q) ∈ Spec(A) is Exercise 1.18(v). ¤ Corollary 2.12. (i) I ∈ Spec(A) ⇐⇒ A/I : domain. (ii) I ∈ m-Spec(A) ⇐⇒ A/I : field.


37

Proof. Using Proposition 2.11 on the canonical surjection A 3 a − 7 → a ∈ A/I, we see that I ⊂ A and 0 ⊂ A/I correspond with each other, therefore the corollary follows from Exercise 1.18(ii). ¤ Example 2.13. Considering the quotient ring Z/(n) of Z by an ideal (n) (n ≥ 1), Z/(n) : field ⇐⇒ Z/(n) : domain ⇐⇒ n : prime. When n is a prime p, the field Z/(p) is denoted by Fp (Example b.16(ii)). Part 2. Equations and Fields 3. Field Extensions 3.1. Polynomial ring over a field. Let A be a ring. Definition 3.1. (i) The set of all polynomials of one variable X with coefficients in A iss a ring with the usual addition and multiplication. This is the polynomial ring over A, and is denoted by A[X]. We naturally consider A as a subring of A[X]. The ring of multivariable polynomials A[X1 , . . . , Xn ] for n ≥ 1 is defined similarly. (ii) A polynomial in A[X] is called monic if its coefficient of the term with highest degree of X is 1. (iii) An element a ∈ A is called a root of P (X) ∈ A[X] if by substituting we have P (a) = 0. Exercise 3.2. (i) If A is a subring of B, A[X] is naturally a subring of B[X]. (ii) A : domain =⇒ A[X] : domain, A[X]× = A× . In the following, K denotes a field. Proposition 3.3. K : field =⇒ K[X]: PID. Proof. Follows from Exercise 1.29(ii) and Theorem 1.30. K[X]×

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K×

Exercise 3.4. As = = K\{0}, every P ∈ K[X]\{0} is associate to some monic polynomial with the same degree. Therefore, every ideal of K[X] is generated by a monic polynomial. Proposition 3.5. The number of roots of P ∈ K[X]\{0} is not greater than deg P . Proof. For a ∈ K, P ∈ K[X], we have P = (X −a)Q+R, deg R < 1 (Exercise 1.29(ii)). As deg R < 1 means R ∈ K, by substituting X by a, we have P (a) = R. Therefore: a : root of P ⇐⇒ R = 0 ⇐⇒ (X − a) | P, and the number of roots of P is equal to the number of monics with degree one that divides P . As K[X] is a UFD, considering the uniqueness of prime decomposition (factorization into irreducibles) of P , which factors into a constant and monic polynomials, this number is at most deg P . ¤

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Definition 3.6. For a root a ∈ K of P ∈ K[X], the maximal integer n satisfying (X − a)n | P is called the multiplicity, and a root with multiplicity greater than one is called a multiple root. 3.2. Field extensions by adjoining roots. Let K be a field and P ∈ K[X] be a monic irreducible polynomial. By Proposition 3.3 and 1.24(i), (P ) is a maximal ideal of K[X], hence the quotient ring KP := K[X]/(P ) is a field by Corollary 2.12(ii). As theu subring K of K[X] (constant polynomials) are mapped injectively into KP by the canonical surjection K[X] → KP (see Exercise 3.8), we can regard K ⊂ KP , and the image X of X gives a root of P in KP . Definition 3.7. The field KP is called the extension field of K obtained by adjoining a root of P . Exercise 3.8. For a group/ring/A- homomorphism f : X → Y and a subgroup/subring/Asubmodule X 0 of X, Ker (f |X 0 ) = X 0 ∩ Ker f . Therefore, by homomorphism theorem, f (X 0 ) ∼ = X 0 /(X 0 ∩ Ker f ). 3.3. Field extensions. The letters F, K, L always denote fields. Definition 3.9. When a subring K of a field F is a field, we call K a subfield of F , and F an extension field of K. The pair of K and its extension field F is called an extension F/K. A field L satsifying F ⊃ L ⊃ K is called an intermediate field of the extension F/K. In the following, let F be an extension field of K. The field F can naturally be regarded as a vector space over K (Exercise 1.13(i)). Definition 3.10. The dimension of F as a vector space over K is called the extension degree of F/K, and is denoted by [F : K]. When [F : K] = n ∈ N, F/K is called a finite extension [of degree n], and when [F : K] = ∞, F/K is called an infinite extension. Example 3.11. [F : K] = 1 ⇐⇒ F = K. Proposition 3.12. F ⊃ L ⊃ K =⇒ [F : K] = [F : L][L : K]. Proof. If one of F/L, L/K is an infinite extension, this is a formal equality ∞ = ∞. If both are finite extensions, letting a basis of F over L and a basis of L over K respectively by {ai } and {bj }, {ai bj } gives a basis of F over K. ¤ Exercise 3.13. Elaborate the above proof using the definition of bases. Example 3.14. The main object of algebraic number theory is the finite extensions of Q. A finite extension field of Q is called an algebraic number field. Proposition 3.15. [KP : K] = deg P .

ALGEBRAIC NUMBER THEORY 2

39

n−1

Proof. Putting deg P = n, the set {1, X, X , . . . , X } generates KP as a K-module, and as the ideal (P ) of K[X] does not contain polynomials with degree less than n, they are linearly independent over K. ¤

4. Galois Theory (1) 4.1. K-homomorphism. Definition 4.1. For two extension fields F, F 0 of a field K, if a ring homomorphism f : F → F 0 satisfies f |K = id, f is called a K-homomorphism. The set of all K-homomorphisms from F to F 0 is denoted by HomK (F, F 0 ). A bijective Khomomorphism is called a K-isomorphism. When there exists a K-isomorphism F → F 0 , F and F 0 are said to be isomorphic, and we write F ∼ = F 0 . In particular, K-isomorphism F → F is called a K-automorphism of F , and the group consisting of all K-automorphisms of F is denoted by AutK (F ). Exercise 4.2. (i) For a ring homomorphism f : F → F 0 between extension fields of K, f is a K-homomorphism if and only if it is K-linear as a map between K-vector spaces. (ii) F ∼ = F 0 =⇒ [F : K] = [F 0 : K]. Lemma 4.3. Every K-homomorphism f : F → F 0 between extension fields is injective, and the image is an intermediate field of F 0 /K. Proof. As Ker f is an ideal of F , it must be equal to F or 0 by Exercise 1.12, but if Ker f = F we have 1 = f (1) = 0 in F 0 , which cannot happen as F 0 is not a zero ring. ¤ Lemma 4.4. If [F : K] = [F 0 : K] < ∞ for two extension fields F, F 0 of K, any K-homomorphism f : F → F 0 is a K-isomorphism. In particular, HomK (F, F ) = AutK (F ) for a finite extension F/K. Proof. By Lemma 4.3 f is an injection and F ∼ = Im f , hence [Im f : K] = [F : K]. Therefore by Proposition 3.12, we have [F 0 : Im f ] = 1, hence F 0 = Im f by Example 3.11 and f is bijective. ¤ Remark 4.5. In general, any injective linear map f : F → F 0 between two finitedimensional vector spaces is bijective if dim F = dim F 0 , as dimension formula (Theorem d.12) gives dim(Im f ) = dim F = dim F 0 , and then use Lemma d.9 to conclude Im f = F 0 . 4.2. Roots and K-homomorphisms. In the rest of subsection, let F/K be a finite extension.

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4.2.1. Roots and minimal polynomials. For every x ∈ F , the “substituting” map fx : K[X] 3 P 7−→ P (x) ∈ F is a ring homomorphism. We consider this homomorphism, using the homomorphism theorem and the theory of PID. First, Im fx is a subring of F , therefore a domain (Exercise 1.4(i)), hence by Corollary 2.12(i), Ker fx ∈ Spec(K[X]). Also as fx is Klinear, Im fx is a K-subspace of F , therefore finite-dimensional over K by Lemma d.9, and as K[X] has infinite dimension over K, Ker fx 6= 0. Therefore, by Proposition 3.3 and Exercise 1.18(i), Ker fx is a principal ideal (Px ) generated by an irreducible polynomial, and the quotient ring K[X]/(Px ) is the extension field KPx of K obtained by adjoining a root of Px . Consequently, by homomorphism theorem, we have a Khomomorphism between extension fields as follows: fx : KPx

∼ =

/ Im fx ⊂ F .

Definition 4.6. (i) We call the monic gererator Px of Ker fx the minimal polynomial of x over K. It is irreducible. (ii) The intermediate field Im fx of F/K is denoted by K(x), and called the field generated by x over K. (iii) A finite extension field F of K is called a simple extension if F = K(x) for some x ∈ F . In this case, KPx ∼ = F by fx . Exercise 4.7. (i) The Px has the minimal degree among the polynomials in K[X] which has x as a root. (ii) When F = K(x), [F : K] = deg Px . Remark 4.8. For a simple√extension F/K, there can √ √ be many choices of x such that F = K(x): for instance, Q( 2) = Q(− 2) = Q(1 + 2). 4.2.2. Roots and K-homomorphisms. Now we fix an irreducible P ∈ K[X], and consider the set RootP (F )of all roots of P in F . The next proposition is proven simply by checking that the two maps are inverse to each other, and recalling Proposition 3.5. Proposition 4.9. The following maps are bijections that are inverse to each other: RootP (F ) 3 x 7−→ fx ∈ HomK (KP , F ), HomK (KP , F ) 3 f 7−→ f (X) ∈ RootP (F ). In particular, |HomK (KP , F )| = |RootP (F )| ≤ deg P = [KP : K]. Now, when F is a simple extension K(x), or equivalently if there exists a Kisomorphism in HomK (KPx , F ), and we will see that all the elements of HomK (KPx , F ) are K-isomorphisms. This is the basis of interpreting the permutations of roots as Kautomorphisms of a simple extension (Proposition 4.10)


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4.2.3. Automorphisms of a field. Now, the group of K-automorphisms AutK (F ) of F acts on the set HomK (KP , F ) as follows: AutK (F ) × HomK (KP , F ) 3 (σ, f ) 7−→ σ ◦ f ∈ HomK (KP , F ), which can be interpreted as an action on RootP (F ) as follows: as the bijection f 7→ f (X) of Proposition 4.9 sends σ ◦ fx to σ(x) (i.e. σ ◦ fx = fσ(x) ), we have: AutK (F ) × RootP (F ) 3 (σ, x) 7−→ σ(x) ∈ RootP (F ). Proposition 4.10. Assume F ∼ = KP . For any x ∈ RootF (P ), the map AutK (F ) 3 σ 7−→ σ(x) ∈ RootP (F ) is bijective. In particular, |AutK (F )| ≤ [F : K]. Proof. As [F : K] = [KP : K], the map fx ∈ HomK (KP , F ) is a K-isomorphism by Lemma 4.4, and induces a bijection: AutK (F ) = HomK (F, F ) 3 σ 7−→ σ ◦ fx ∈ HomK (KP , F ) (use Lemma 4.4 for the first equality), which, composed with the bijection f 7→ f (X) of Proposition 4.9, gives the desired bijection. The latter part follows from Proposition 3.5, as |RootF (P )| ≤ deg P = [F : K]. ¤ Definition 4.11. A simple extension F/K is called a Galois extension if it satisfies |AutK (F )| = [F : K]. In this case we call AutK (F ) a Galois group of F/K, and denote it by Gal(F/K). By definition, |Gal(F/K)| = [F : K]. When we write F = K(x) ∼ = KP , F/K being Galois means that |RootP (F )| = [F : K] = deg P , i.e. P has deg P distinct roots in F = K(x) (no multiple roots, and all the roots are expressible as a polynomial of x with coefficients in K), by Proposition 4.10. Exercise √ 4.12. (i) Quadratic extensions (extensions of degree 2) of Q are Galois. 3 ∼ (ii) Q( 2) = Q[X]/(X 3 − 2) is not a Galois extension of Q. 4.3. Galois theory of simple extensions. Proposition 4.13. Let L be an intermediate field of a Galois extension F/K. Then F/L is also a Galois extension, and Gal(F/L) is a subgroup of Gal(F/K). Proof. Putting F = K(x) ∼ = KP , we have F = L(x), hence F/L is also simple. Let the kernel of fx : L[X] → F be (Q), to write F ∼ = LQ , by definition we have Q | P , therefore Q has deg Q distinct roots in F and F/L is Galois. The latter part follows from the fact that every L-automorphism of F is also a K-automorphism. ¤ Theorem 4.14. (Fundamental theorem of Galois theory) For a Galois extension F/K, let A be the set of all intermediate fields of F/K, and B be the set of all subgroups of G = Gal(F/K). Then the map A 3 L 7−→ Gal(F/L) ∈ B is bijective. More precisely, the following are inverse to each other (note Φ(L) = Gal(F/L)): Φ : A 3 L 7−→ Φ(L) = {σ ∈ G | ∀x ∈ L σ(x) = x} ∈ B, Ψ : B 3 H 7−→ Ψ(H) = {x ∈ F | ∀σ ∈ H σ(x) = x} ∈ A.

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Proof. By definition we have L ⊂ Ψ(Φ(L)), H ⊂ Φ(Ψ(H)), so in order to show L = Ψ(Φ(L)), H = Φ(Ψ(H)), it is enough to compare the degrees and cardinalities: [F : Ψ(Φ(L))] = [F : L],

|Φ(Ψ(H))| = |H|.

(the first equality and Proposition 3.12 gives [Ψ(Φ(L)) : L] = 1, and use Example 3.11.) These two equalities follow from the following lemma. ¤ Lemma 4.15. |Φ(L)| = [F : L], |H| = [F : Ψ(H)]. Proof. The first equality |Φ(L)| = |Gal(F/L)| = [F : L] is trivial. To show the second, by H ⊂ Φ(Ψ(H)) we have |H| ≤ |Φ(Ψ(H))| = [F : Ψ(H)], therefore it is enough to show the inverse inequality. Let F = K(x), and consider a polynomial: Y P (X) = (X − σ(x)) ∈ F [X]. σ∈H

Then all the coefficients of P are symmetric polynomials of the set {σ(x) | σ ∈ H}, therefore invariant under the action of elements of H, i.e. belongs to Ψ(H). Therefore P ∈ Ψ(H)[X], hence the minimal polynomial Qx of x over Ψ(H) divides P , which shows that [F : Ψ(H)] = deg Qx ≤ deg P = |H|. ¤ 5. Cyclotomic Extensions 5.1. Cyclotomic extensions. Let K be a field. For a polynomial P ∈ K[X], we construct an extension field of K which containsQall roots of P (i.e. where we can factorize P completely into linear factors). Let P = i Pi the factorization into irreducibles in K[X], and if there is any Pi that is not linear, then we adjoin its root to construct K1 = KPi . When we factorize P in K1 [X], the total degree of the product of non-linear factor has decreased, so as we repeat this procedure to construct the extension fields K2 , K3 , . . ., P will completely factorize into linear factors in some F = Kj . As F/K is an iteration of extensions obtained by adjoining a root, by Proposition 3.12 it is a finite extension. Definition 5.1. We call this F the extension field of K obtained by adjoining all roots of P . Remark 5.2. This field F turns out to be determined uniquely by P up to Kisomorphisms, and is called a splitting field of P . Definition 5.3. For an integer n ≥ 1, the extension field of K obtained by adjoining all roots of X n −1 (n-th roots of unity) is denoted by K(µn ), and is called a cyclotomic extension of K. We denote the set of all n-th roots of unity in K(µn ) by µn . 5.2. The group µn of n-th roots of unity. By Proposition 3.5, we have |µn | ≤ n, and clearly µn is a group under multiplication, i.e. it is a finite subgroup of K(µn )× (the multiplicative group of K(µn )).


43

5.2.1. Cyclicity. Definition 5.4. Let G be a finite group, i.e. a group of finite cardinality. For a ∈ G, the minimal positive integer n with the property an = 1 is called the order of a. Exercise 5.5. (i) Every element of a finite group has its order. (ii) If the order of a ∈ G is n, then ak = 1 ⇐⇒ n | k. Definition 5.6. For an element a of a finite group G, hai = {ai | i ∈ N} is a subgroup of G, and is called the subgroup of G generated by a. When G = hai for some a ∈ G, G is called a cyclic group, and a is called a generator of G. The order of a generator is equal to |G|. A cyclic group consisting of n elements (cyclic group of order n) is isomorphic to the addtive group of Z/(n) (often denoted by Z/nZ) by sending a generator to 1 mod n. Exercise 5.7. (i) The number of generators of a cyclic group of order n is ϕ(n) = |{1 ≤ k ≤ n − 1 | (k, n) = 1}| (Euler’s function). (ii) A cyclic group of order n has a unique subgroup of order d for each positive divisor d of n, and there are no other subgroups. Proposition 5.8. For a field K, every finite subgroup G of K × is cyclic. Remark 5.9. By Exercise 5.5(i), every element of a finite subgroup of K × is a root of unity. Proof. Take an element x ∈ G which has the maximal order, and call its order n. We show that the order of any y ∈ G is a divisor of n. If the order m of y does not divide n, there is a prime number p and its power pj divides m but not n. So let k 0 m = pj m0 , n = pk n0 , j > k, (p, m0 ) = (p, n0 ) = 1. Then the order of xp y m is, by: k

0

k

0

(xp y m )i = 1 =⇒ xp i = y −im ( j k xp p i = y −im = 1 ⇒ n | pj+k i ⇒ n0 | i =⇒ 0 0 1 = xni = y −im n ⇒ m | im0 n0 ⇒ pj | i

=⇒ pj n0 | i,

equal to pj n0 , which contradicts the maximality of n. Therefore m | n, but now xin/m (1 ≤ i ≤ m) gives m distinct roots of X m − 1 in K, but by Proposition 3.5, these are all the roots of X m − 1 in K. Therefore y = xin/m for some i, and as y was arbitrary, x is a generator of G. ¤ Corollary 5.10. The group µn is cyclic, as it is a finite subgroup of K(µn )× . 5.2.2. Cardinality. Let K be a field. The image of the natural ring homomorphism ϕ : Z → K (defined by 1 7→ 1) is, being a subring of the field, a domain, and hence its kernel is a prime ideal of Z (Corollary 2.12(i)). Definition 5.11. When Ker ϕ = 0, K is said to have characteristic 0, and when Ker ϕ = (p) for a prime p, K is said to have characteristic p. The characteristic of K is denoted by char K.

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Identifying Z or Fp with a subring of K by homomorphism theorem, char K = 0 ⇐⇒ Z ∼ = Im ϕ ⊂ K ⇐⇒ K : an extension field of Q, char K = p ⇐⇒ Fp ∼ = Im ϕ ⊂ K ⇐⇒ K : an extension field of Fp . (Thus arbitrary field can be regarded as an extension field of Q or Fp . In each case, we call Q, Fp the prime field of K.) Proposition 5.12. When (char K, n) = 1, we have |µn | = n, i.e. µn is a cyclic group of order n. Proof. It suffices to show that X n − 1 does not have a multiple root in K(µn ), but this follows readily from Exercise 5.14(ii) below, as it does not have common roots with its derivative nX n−1 , whose only root is 0 by (char K, n) = 1. ¤ Definition 5.13. The K-linear map D : K[X] → K[X] characterized by the following properties is called derivation of K[X]: (i) D(1) = 0, (ii) D(X n ) = nX n−1 (n = 1, 2, · · · ). Exercise 5.14. (i) For P, Q ∈ K[X], D(P Q) = D(P )Q + D(Q)P . (ii) For P ∈ K[X], α ∈ K : a multiple root of P ⇐⇒ (X − α) | P, D(P ). Remark 5.15. When K = Fp , X p − 1 = (X − 1)p (see below) shows that µp = {1}. Proposition 5.16. For a field F of characteristic p > 0, the map Frq : F ∈ x 7−→ xq ∈ F for q = pf (f ≥ 1) is an injective homomorphism, and if F is a finite field then it is a field automorphism. (We call Frq the q-th power Frobenius map.) xp

p

p

n−1 Xµ

¶ p p−i i x y + y p , but i

Proof. Consider Frp : F ∈ x 7−→ ∈ F . Then (x + y) = x + i=1 µ ¶ p p p as p is prime every is divisible by p. Hence (x + y) = x + y p , and as (xy)p = xp y p i we see that Frp is a ring homomorphism. As F is a field it is injective (Lemma 4.3), hence if F is finite it is bijective. The general case follows by Frq = (Frp )f . ¤ 5.3. Galois theory of cyclotomic extensions. Consider a field K and its cyclotomic extension K(µn ). In the rest of this subsection, we assume (char K, n) = 1. Definition 5.17. A generator of the cyclic group µn (an element with order n) is called a primitive n-th root of unity. There are ϕ(n) primitive n-th roots of unity, and when we denote one of them by ζ, they are written as ζ k (1 ≤ k ≤ n−1, (k, n) = 1) (Exercise 5.7). In the below, we choose and fix an arbitrary primitive n-th root of unity ζ in K(µn ), and let Pζ be its minimal polynomial over K. Proposition 5.18. We have K(µn ) = K(ζ), and K(ζ)/K is a Galois extension.


45

Proof. As µn = {1, ζ, ζ 2 , . . . , ζ n−1 }, the first part follows. Also, as Pζ divides X n − 1 = Qn−1 i ¤ i=0 (X − ζ ), it has deg Pζ distinct roots in K(ζ). Lemma 5.19. All the roots of Pζ in K(ζ) are primitive n-th roots of unity. Proof. By Pζ | X n − 1, all roots of Pζ belong to µn . A non-primitive α ∈ µn with its order d < n is a root of X d − 1. As ζ is not a root of X d − 1, Pζ does not divide X d − 1, i.e. Pζ and X d − 1 are relatively prime, and hence α is not a root of Pζ . ¤ Remark 5.20. The quotient Φn (X) of X n − 1 by (l.c.m. of) all the X d − 1 with d < n, d | n has coefficients in Z (or its image in K) by definition, and has ϕ(n) primitive n-th roots of unity as its roots. It is called the cyclotomic polynomial. By the previous lemma, Pζ | Φn (X). The first few are: Φ2 (X) = X + 1, Φ3 (X) = X 2 +X +1, Φ4 (X) = X 2 +1, Φ5 (X) = X 4 +X 3 +X 2 +X +1, Φ6 (X) = X 2 −X +1, . . .. Definition 5.21. A Galois extension with an abelian Galois group is called an abelian extension. Theorem 5.22. (Galois group of cyclotomic extension) There is an injective homomorphism as follows, and in particular K(µn )/K is an abelian extension: Gal(K(µn )/K) 3 (ζ 7→ ζ k ) 7−→ k mod n ∈ (Z/(n))× . Remark 5.23. The (Z/(n))× = {k mod n | (k, n) = 1} is the group of units of the ring Z/(n). |(Z/(n))× | = ϕ(n). Proof. By Proposition 4.10, Gal(K(ζ)/K) 3 σ 7−→ σ(ζ) ∈ RootPζ (K(ζ)) is a bijection, and as we know by Lemma 5.19 that all the elements of RootPζ (K(ζ)) are primitive n-th roots of unity, it is contained in {ζ k | k ∈ (Z/(n))× }. This gives an injection which does not depend on the choice of ζ, and as the composition of ζ 7→ ζ k and ζ 7→ ζ l is ζ 7→ ζ kl , it is a group homomorphism. ¤ Remark 5.24. The above homomorphism is not a surjection in general, but it is when K = Q. It is equivalent to say that Φn (X) is irreducible in Q[X], and is a deep theorem called irreducibility of cyclotomic polynomials (Theorem 18.4(ii)). Exercise 5.25. Let K = Q, n = 5, and describe the Galois group of Q(ζ5 )/Q. 5.4. Example: Finite Fields. Proposition 5.26. Let F be a finite field with |F | = q. (i) F is an extension of Fp of degree f for some p and f , and q = pf . (ii) (existence of a primitive root) F × = µq−1 , i.e. if we take a generator ζ (primitive root of F ) then F = {0, 1, ζ, ζ 2 , . . . , ζ q−2 }. Proof. (i): By |F | < ∞ the prime field of F is Fp for some p, and F must be a finite extension of Fp . If [F : Fp ] = f we have |F | = pf . (ii): Follows from F × = F \{0} and Proposition 5.8. ¤

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Theorem 5.27. Let p a prime. For each f ≥ 1 there is a unique finite field Fq = µq−1 ∪ {0} with q = pf elements in Fp , and these are all the finite fields of charactecteristic p. Proof. Let Fq = µq−1 ∪ {0} be the set of all roots of X q − X in Fp . As (p, q − 1) = 1 we have |Fq | = q by Proposition 5.12. As Fq = {x ∈ Fp | Frq (x) = x} we see that Fq is a subfield of Fp by Proposition 5.16, i.e. Frq (x ± y) = Frq (x) ± Frq (y) = x ± y, Frq (xy) = Frq (x) Frq (y) = xy, Frq (x−1 ) = Frq (x)−1 = x−1 for x, y ∈ Fq . Conversely, by Proposition 5.26 the elements of a degree f extension of Fp inside Fp must coincide with the q roots of X q − X, namely Fq . ¤ By Proposition 5.26, any finite extension of finite fields is cyclotomic, hence Galois, hence separable. (Therefore finite fields are perfect.) Theorem 5.28. Any finite extension Fqn /Fq of finite fields of degree n is a Galois extension. Its Galois group Gal(Fqn /Fq ) is a cyclic group of order n with a generator Frq : Fqn → Fqn : ∼ =

ϕn : Z/nZ 3 1 mod n 7−→ Frq ∈ Gal(Fqn /Fq ) Proof. By Proposition 5.16, Frq is a field automorphism of Fqn fixing the elements of Fq = µq−1 ∪ {0}, hence a Fq -automorphism of Fqn . Taking the primitive root ζ of Fqn 2 n−1 by Proposition 5.26, then ζ, ζ q , ζ q , . . . , ζ q are all distinct, hence Frq is a generator of Gal(Fqn /Fq ). ¤ Exercise 5.29. Draw a diagram of all the intermediate fields of the extension F4096 /F2 of degree 12 and their inclusions, together with corresponding subgroups of the Galois group Z/12Z. We have an immediate corollary of the above theorem: Corollary 5.30. For any positive integer m, n with m | n we have a commutative diagram: Z/nZ ²

Z/mZ

ϕn ∼ = ϕm ∼ =

/ Gal(Fqn /Fq ) ² / Gal(Fqm /Fq )

where the right vertical map is the natural restriction σ 7−→ σ|Fqm and the left vertical map is the natural surjection 1 mod n 7−→ 1 mod m. 6. Galois Theory (2) 6.1. Simple extension theorem. A natural thing to ask for an irreducible polynomial P is that it will not have multiple roots when you extend the field to see its roots.


47

In view of Proposition 4.9, it can be interpreted as a property of KP , which has natural generalization to arbitrary finite extensions. It turns out that this condition is enough to ensure the simplicity of extensions, to see that the theory developed in Section 4 had enough generality. Definition 6.1. (i) A polynomial P ∈ K[X]\K is called separable if |RootP (E)| = deg P for some extension E/K. (ii) A finite extension F/K is called separable if |HomK (F, E)| = [F : K] for some extension E/K. Definition 6.2. For a finite extension F/K, we inductively define the intermediate field K(x1 , . . . , xn ) of F/K generated by x1 , . . . , xn ∈ F as follows: K0 = K, Ki+1 = Ki (xi+1 ) (0 ≤ i ≤ n − 1), Kn = K(x1 , . . . , xn ). The field K(x1 , . . . , xn ) is the minimal intermediate field of F/K that contains x1 , . . . , xn . By iterating simple extensions in this way, we can always recover F after finite iteration. For example, take a basis {ei } of F over K and letting xi = ei , we see: Proposition 6.3. ∃x1 , . . . , xn ∈ F, F = K(x1 , . . . , xn ). Theorem 6.4. A separable finite extension F/K is simple. Proof. If K is a finite field, F is also finite, therefore F × is a cyclic group by Proposition 5.8, hence its generator generates F/K. Assume K is infinite. By Proposition 6.3, if we show the claim for F = K(x, y), the general case follows by induction. Let F = K(x, y), [F : K] = n, and for an E/K, we have HomK (F, E) = {σ1 , . . . , σn }. As any element of HomK (F, E) is determined by images of x, y, for i 6= j we have σi (x) 6= σj (x) or σi (y) 6= σj (y). Consider a polynomial: Y {(σi (x) − σj (x))X + (σi (y) − σj (y)} ∈ E[X]. Q(X) = X i6=j

As K is infinite, it contains an element z which is not a root of Q. Putting w = xz + y, we see that i 6= j =⇒ σi (w) 6= σj (w), therefore σ1 , . . . , σn restricts to n distinct K-homomorphisms of K(w) into E. But Proposition 4.9 shows that n ≤ |HomK (K(w), E)| ≤ [K(w) : K], and as K(w) ⊂ F , [K(w) : K] ≤ [F : K] = n, so these are all equalities, hence F = K(w). ¤ Corollary 6.5. A finite extension F/K is called a Galois extension if it satisfies |AutK (F )| = [F : K]. In particular it is separable, hence simple, therefore it agrees with the previous Definition 4.11. 6.2. Separable extensions. We will see that separability is not a special condition, but rather non-separability is an anomaly, by showing that any finite extension in characteristic 0 is separable. For polynomials this is easily seen:

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Proposition 6.6. For an irreducible P ∈ K[X], the following are equivalent: (i) for the extension field E obtained by adjoining all roots of P , |RootP (E)| = deg P , (ii) P is separable, (iii) D(P ) 6= 0. Proof. (i)⇒(ii) is clear. (ii)⇒(iii): If D(P ) = 0, all roots of P are multiple roots in any field by Exercise 5.14(ii). (iii)⇒(i): If D(P ) 6= 0, as deg D(P ) < deg P and D(P ) ∈ / (P ) in K[X], (P ) + (D(P )) = K[X] 3 1 as (P ) is a maximal ideal (Proposition 3.3, Proposition 1.24(i)). As 1 ∈ (P ) + (D(P )) remains true in E[X], P and D(P ) cannot have a common divisor (X − α), i.e. there is no multiple root of P in E. ¤ Corollary 6.7. If char K = 0, all P ∈ K[X]\K are separable. Exercise 6.8. When K = Fp (T ), X p − T ∈ K[X] is a non-separable irreducible polynomial. To prove the fundamental property of separable extensions (Proposition 6.11), we need the following lemma, which generalizes Proposition 4.9 (which is a special case when L = K and τ = id), and will be a key lemma in perfecting the Galois theory in the next subsection: Lemma 6.9. Let L/K, E/K be any extensions, F = L(x) a simple extension of L ∼ = with the minimal polynomial P , so that fx : LP − → F . Consider the restriction map ResFL : HomK (F, E) 3 f 7−→ f |L ∈ HomK (L, E). Then for any τ ∈ HomK (L, E), we have a bijection : (ResFL )−1 (τ ) 3 f 7−→ f (x) ∈ Rootτ (P ) (E). Here τ (P ) ∈ E[X] is defined by extending τ to a ring homomorphism τ : L[X] → E[X] by sending each coefficients by τ . Proof. Note that if f |L = τ , then τ (P )(f (x)) = f (P (x)) = 0, hence f (x) ∈ Rootτ (P ) (E). We construct the inverse as follows. First observe that τ : L[X] → τ (L)[X] is an ring isomorphism with τ |K[X] = id, hence τ (P ) is irreducible in τ (L)[X], and τ induces a K-isomorphism τ : LP → τ (L)τ (P ) with τ |L = τ , and any y ∈ Rootτ (P ) (E) defines a τ (L)-homomorphism fy : τ (L)τ (P ) → E by 4.2.1. Then set: fx−1

τ

fy

gy = fy ◦ τ ◦ fx−1 : F −−→ LP − → τ (L)τ (P ) −→ E. This gy is a K-homomorphism with gy |L = τ , and the map y 7→ gy is the desired inverse because (i) gy (x) = y (as fy ◦ τ ◦ fx−1 sends x 7→ X 7→ X 7→ y), and (ii) gf (x) = f (both sides restrict to τ on L, and agree on x, as ff (x) ◦τ ◦fx−1 sends x 7→ X 7→ X 7→ f (x)). ¤ Corollary 6.10. In this situation, |HomK (F, E)| ≤ deg P · |HomK (L, E)|, and if both sides are finite, the equality holds if and only if |Rootτ P (E)| = deg P for all τ ∈ HomK (L, E). (Note deg P = [F : L].) Strategy to compute |HomK (F, E)| for a general finite extension F/K is to write it as a successive simple extensions and see how they vary at each extensions.


49

Proposition 6.11. For a finite extension F = K(x1 , . . . , xn )/K, the following are equivalent: (i) The minimal polynomial Px over K for all x ∈ F are separable, (ii) for the extension field E obtained by adjoining all roots of P1 , . . . , Pn , the minimal polynomials of x1 , . . . , xn , |HomK (F, E)| = [F : K], (iii) F/K is separable. In general, for arbitrary extension E/K, |HomK (F, E)| ≤ [F : K]. Proof. By expressing F/K by successive simple extensions as K = K0 ⊂ K1 ⊂ · · · ⊂ Kn = F, Ki = Ki−1 (xi ) and applying Corollary 6.10 to F = Ki , L = Ki−1 , x = xi for i = 1, . . . , n shows |HomK (F, E)| ≤ [F : K] in general, by Proposition 3.12. For (i)⇒(ii), we check the equality condition in the corollary for each simple extension. As minimal polynomial P of xi over Ki−1 divides Pi by definition, sending by τ gives τ P | τ Pi = Pi , which shows |Rootτ P (E)| = deg P as Pi is separable and E has all roots of Pi . (ii)⇒(iii) is clear. (iii)⇒(i): For any x ∈ F , writing F/K(x) by successive simple extensions and applying Corollary 6.10 repeatedly to compute |HomK (F, E)|, and if it equals [F : K] for an E, the equality condition forces |RootPx (E)| = deg Px . ¤ Corollary 6.12. If char K = 0, any finite extension of K is separable. Remark 6.13. When K is a finite field, any finite extension is cyclotomic, hence Galois, hence separable (Section 9). If any finite extension of K is separable, K is called a perfect field. Proposition 6.14. Any finite separable extension F/K is an intermediate field of some Galois extension E/K. Proof. Take the E of Proposition 6.11(ii). It is generated by all roots of P1 , . . . , Pn over K, therefore repeating the procedure to E instead of F will give back E again. Therefore (i)⇒(ii) of Proposition 6.11 gives |HomK (E, E)| = [E : K], which shows that E is Galois by Lemma 4.4 and Corollary 6.5. ¤ Remark 6.15. This E/K is called the Galois closure of F/K. 6.3. Properties of Galois extensions. Now we are ready to generalize the argument of 4.2.2 to a more general setting. Assume L/K is finite and F/K is Galois, and we return to the action of 4.2.3, generalizing KP to L: Gal(F/K) × HomK (L, F ) 3 (σ, f ) 7−→ σ ◦ f ∈ HomK (L, F ). The generalization of Proposition 4.10 is as follows: Proposition 6.16. Assume HomK (L, F ) 6= ∅. Then for any f ∈ HomK (L, F ), the map Gal(F/K) 3 σ 7−→ σ◦f ∈ HomK (L, F ) is surjective, and |HomK (L, F )| = [L : K]. Proof. As f (L) is an intermediate field of F/K, the map is decomposed as: ResF f (L)

∼ =

→ HomK (L, F ) Gal(F/K) = HomK (F, F ) −−−−−→ HomK (f (L), F ) −

50

TERUYOSHI YOSHIDA

where the second map g 7→ g ◦ f is bijective as f : L → f (L) is isomorphism, it suffices to show the surjectivity of ResFL when L ⊂ F (i.e. the case f = id). In this case, as F/L is Galois (Proposition 4.13), hence simple F ∼ = LP (Corollary 6.5), we can apply Lemma 6.9 by setting E = F . For any τ ∈ HomK (L, F ), as F/τ (L) is Galois (Proposition 4.13) and F ∼ = LP ∼ = τ (L)τ (P ) , |(ResFL )−1 (τ )| = |Rootτ (P ) (F )| = deg P = [F : L]. Thus ResFL is surjective and |HomK (L, F )| = [L : K] by Proposition 3.12. ¤ Proposition 6.17. For a finite extension F/K, the following are equivalent: (i) For any x ∈ F and its minimal polynomial Px over K, |RootPx (F )| = deg Px , (ii) F/K is Galois. Proof. (i)⇒(ii): As F/K is separable by Proposition 6.11(i)⇒(iii), so write L = F (x) by Theorem 6.4, and apply (i) to this x. (ii)⇒(i): For any x, apply Proposition 6.16 to L = K(x) and use Proposition 4.9. ¤ Proposition 6.18. Let F/K be Galois, G = Gal(F/K), L be an intermediate field of F/K, and H = Gal(F/L): (i) (ii) (iii) (iv)

G 3 σ 7−→ σ|L ∈ HomK (L, F ) is surjective. Gal(F/σ(L)) = σHσ −1 = {στ σ −1 | τ ∈ H} (∀σ ∈ G). G B H ⇐⇒ σ(L) = L (∀σ ∈ G) ⇐⇒ L/K : Galois. When L/K is Galois, G/H 3 σ 7−→ σ|L ∈ Gal(L/K) is an isomorphism.

Proof. (i): Proposition 6.16. (ii): Putting H 0 = Gal(F/σ(L)), σHσ −1 ⊂ H 0 is clear. Similarly, L = σ −1 (σ(L)) gives σ −1 H 0 σ ⊂ H, hence H 0 ⊂ σHσ −1 . (ii) By (ii), G B H ⇐⇒ σHσ −1 = H (∀σ ∈ G) ⇐⇒ σ(L) = L (∀σ ∈ G). This last condition is equivalent to say that the map (i) factors through HomK (L, L) ⊂ HomK (L, F ), but by its surjectivity, it only happens when HomK (L, L) = HomK (L, F ), which in turn means that L/K is Galois as |HomK (L, F )| = [L : K]. (iii) By Gal(L/K) = HomK (L, F ), the surjection (i) is a homomorphism with the kernel H. ¤

7. Kummer Extensions and Solvability of Equations The contents of this section are not used in Parts 3 and 4. 7.1. Preliminaries: minimal polynomial of linear transformations. Let K be a field, V an n-dimensional vector space over K. The ring of endomorphisms End(V ) of V is a K-algebra which is non-commutative ring in general and has dimension n2 as a vector space over K (Exercise e.4). For any ϕ ∈ End(V ), consider the ring homomorphism of “substituting ϕ into polynomials with coefficients in K”: fϕ : K[X] 3 P 7−→ P (ϕ) ∈ End(V )


51

(set fϕ (1) = id to make it into a ring homomorphism). This fϕ is K-linear and Im fϕ is finite-dimensional, being a subspace of End(V ) (Lemma d.9), and as K[X] is infinitedimensional, Ker fϕ 6= 0. Therefore, by Proposition 3.3, Ker fϕ is a principal ideal (Qϕ ) generated by Qϕ 6= 0, and Im fϕ is a subring of End(V ) isomorphic to K[X]/(Qϕ ). Definition 7.1. The monic generator Qϕ (Exercise 3.4) of Ker fϕ is called the minimal polynomial of f over K. Remark 7.2. This Qϕ has minimal degree among the polynomials with coefficients in K which have ϕ as a “root”, but it is not necessarily irreducible. Proposition 7.3. For c ∈ K, c : an eigenvalue of ϕ ⇐⇒ Qϕ (c) = 0. Proof. Recall that, by Proposition h.14 and Corollary d.13, c : an eigenvalue of ϕ ⇐⇒ Ker (ϕ − c · id) 6= 0 ⇐⇒ ϕ − c · id ∈ / Aut(V ). ⇒: If Qϕ (c) 6= 0, Qϕ ∈ / (X −c) in K[X], and as (X −c) ∈ m-SpecK[X], (Qϕ )+(X −c) = K[X]. Hence there exist R1 , R2 such that Qϕ R1 + (X − c)R2 = 1. Then fϕ (R2 ) gives the inverse of fϕ (X − c) = ϕ − c · id in End(V ), hence ϕ − c · id ∈ Aut(V ). ⇐: If Qϕ (c) = 0, we can write Qϕ = (X − c) · R, so by applying fϕ , 0 = (ϕ − c · id) ◦ fϕ (R). Now if ϕ−c·id ∈ Aut(V ), we can multiply its inverse ψ = (ϕ−c·id)−1 to get 0 = fϕ (R), hence Qϕ | R, which contradicts deg R < deg Qϕ . ¤ Remark 7.4. The same proposition holds for the characteristic polynomial Pϕ of ϕ (Proposition h.14), but in general Qϕ 6= Pϕ . (Qϕ | Pϕ by definition, and they have same sets of roots, but their multiplicities can be different.) 7.2. Kummer extensions. Definition 7.5. A Galois extension is called a cyclic extension if its Galois group is cyclic. In the rest of this subsection, let n be a positive integer and K a field with (char K, n) = 1, and which contains the group of n-th roots of unity µn . Over such a field K, for a ∈ K × such that X n − a is irreducible in K[X], we consider the extension of degree √ n − a) obtained by adjoining a n-th root of a. Denoting n n L = K( a) ∼ = K[X]/(X √ n one root in L by x = a, and fixing a primitive n-th root of unity ζ = ζn ∈ K, ζ i x ∈ L (0 ≤ i ≤ n − 1) are all roots of X n − a, and as they are distinct by Proposition 5.12, they exhaust the set of roots, hence L/K is a Galois extension. We have the following isomorphism, which in particular shows that it is a cyclic extension: √ √ Gal(L/K) 3 ( n a 7→ ζ i n a) 7−→ i mod n ∈ Z/nZ. Definition 7.6. For a field K = 1, µn ⊂ K, a cycylic extension of √ with (char K, n) degree n of the form L = K( n a) ∼ = K[X]/(X n − a) is called a Kummer extension. Exercise 7.7. For any field K with char K 6= 2, any quadractic extension is a Kummer extension, because µ2 = {±1} ⊂ K.

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Theorem 7.8. (Kummer theory) For n ≥ 1, every cyclic extension of degree n of a field K with (char K, n) = 1, µn ⊂ K is a Kummer extension. Proof. Fix a generator σ of the cyclic group Gal(L/K), and consider it as an endomorphism σ ∈ End(L) of L as a vector space over K. Lemma 7.9. The set of all eigenvalues of σ is µn ⊂ K. Proof. First, the set Xσ of all eigenvalues of σ is a subgroup of µn because of the following. If eigenvalues c, d ∈ K have x, y ∈ L× as eigenvectors (σ(x) = cx, σ(y) = dy), by σ n = id we have x = σ n (x) = cn x, hence c ∈ µn , and as σ is a K-isomorphism, σ(xy) = σ(x)σ(y) = (cd)(xy), σ(x−1 ) = σ(x)−1 = c−1 x−1 , hence cd, c−1 ∈ K are also eigenvalues of σ with eigenvectors xy, x−1 ∈ L, respectively (of course 1 = cc−1 is an eigenvalue). Therefore Xσ is equal to µd ⊂ µn for some d | n (Exercise 5.7(ii)). Now, by Proposition 7.3, Xσ = µd is equal to the set of roots of the minimal polynomial Qσ of σ. But as σ n = id we have Qσ (X) | X n − 1, therefore Qσ (X) = X d − 1 which implies σ d = id, but as σ is a generator, d = n. ¤ Now we return to the proof of Theorem 7.8. By the above lemma, σ has the primitive n-th root of unity ζ as an eigenvalue, therefore letting x ∈ L be its eigenvector, σ(xn ) = σ(x)n = (ζx)n = xn . So if we put a = xn , a ∈ K by the fundamental theorem of Galois theory. Therefore the minimal polynomial P of x over K divides X n − a. But σ i (x) = ζ i x for 0 ≤ i ≤ n − 1 are all distinct, and as σ i are K-isomorphisms, they are all roots of P . Therefore [K(x) : K] = deg P = n, hence P = X n − a, L = K(x). ¤ Remark 7.10. Kummer theory can be generalized to any abelian extension where all the elements of its Galois group has order dividing n, and by introducing the absolute Galois group (See section 8.3), it acquires a stronger formulation. 7.3. Solvable extensions and radical extensions. Definition 7.11. (i) A finite group G is called a solvable group if there exists a decreasing sequence of subgroups G = G0 ⊃ G1 ⊃ · · · ⊃ Gk−1 ⊃ Gk = {e} such that Gi−1 B Gi and Gi−1 /Gi is cyclic for all 1 ≤ i ≤ k. (ii) A finite separable extension F/K is called a solvable extension if its Galois closure E/K has a solvable Galois group. Exercise 7.12. When G B H, G : solvable ⇐⇒ H, G/H : solvable. Lemma 7.13. (i) A finite abelian group is solvable. Hence abelian extensions are solvable. (ii) For any intermediate field L of a solvable extension F/K, L/K is solvable. (iii) If L/K is solvable and F/L is abelian, then F/K is solvable. Proof. (i) Let G be a finite abelian group. If we take any element a which is different from identity and consider H = hai, G B H B {e}. Do the same procedure for G/H, and by repeating this, finite number of steps lead to the desired sequence because


53

|G/H| < |G|. (ii) Proposition 6.11(i)⇒(iii) shows that L/K is separable. If we let EL /K, EF /K be the Galois closures of L/K, F/K we have EL ⊂ EF , which proves the claim by Proposition 6.18(iv) and Exercise 7.12. (iii) Let EL /K, EF /K be as above, write F = K(x) and let x = x1 , . . . , xk be the roots of the minimal polynomial of x over K. Then EF = K(x1 , . . . , xk ). Let EL = E0 , Ei = Ei−1 (xi ) for 1 ≤ i ≤ k (here EF = Ek ), and if we show that each Ei /Ei−1 is abelian, (i) and Exercise 7.12 show the solvability of Gal(EF /Ei ) inductively from i = k to i = 0, and as Gal(EL /K) is solvable, Gal(EF /K) also is. Now, define σi ∈ HomK (F, EF ) by σi (x) = xi , and letting Li = σi (L), Fi = σi (F ), by an isomorphism Gal(F/L) 3 τ 7−→ σi τ σi−1 ∈ Gal(Fi /Li ), we see that Fi /Li is abelian. By F = L(x) we have Fi = Li (xi ), and noting that Li ⊂ EL ⊂ Ei−1 , the next Lemma 7.14 shows that Ei /Ei−1 is abelian with Galois group isomorphic to a subgroup of Gal(Fi /Li ). ¤ Lemma 7.14. When a Galois extension F = K(x) and a finite extension L/K are contained in a same field, L(x)/L is Galois and the restriction map Gal(L(x)/L) 3 σ 7−→ σ|F ∈ Gal(F/K) is injective. (For general field extension L/K and a finite separable F = K(x)/K contained in a same field E, we denote the field L(x) ⊂ E by F L, called the composite field of F and L in E.) Proof. As the minimal polynomial of x over L divides that of x over K, its roots all belong to F , therefore L(x), which shows the first part. The injectivity is seen by σ|F = id =⇒ σ(x) = x =⇒ σ = id. ¤ Definition 7.15. A finite extension obtained by succession of cyclotomic extensions and Kummer extensions is called a radical extension. Theorem 7.16. If char K = 0, for a finite extension L/K: L/K : solvable ⇐⇒ L : an intermediate field of some radical extension F/K. Proof. ⇐: As radical extensions are solvable by Lemma 7.13(iii), the Lemma 7.13(ii) proves the claim. ⇒: It suffices to show that a Galois extension F/K with a solvable Galois group is contained in some radical extension. Take the sequence of normal subgroups Gal(F/K) = G0 B · · · B Gk = {e}, and let K = K0 ⊂ · · · ⊂ Kk = F be the sequence of corresponding intermediate fields. Let ni be the order of each cyclic group Gi−1 /Gi (1 ≤ i ≤ k), and put n = n1 · · · nk , ζ = ζn . By Lemma 7.14, Ki (ζ)/Ki−1 (ζ) is a cyclic extension with its Galois group isomorphic to a subgroup of Gal(Ki /Ki−1 ), and its degree divides ni (Exercise 5.7(ii)). Therefore Theorem 7.8 shows that Ki (ζ)/Ki−1 (ζ) is a Kummer extension, hence F (ζ) = Kn (ζ) is a radical extension of K. ¤

8. Algebraic Closures, Infinite Galois Extensions The contents of this section, except for Zorn’s lemma, are not used in Parts 3 and 4.

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8.1. Zorn’s lemma. Definition 8.1. Let X be a set. Suppose we know, for every pair of elements x, y of X, a relation x ≤ y holds or not. We call X an ordered set by the order ≤ if the following conditions are satisfied: (i) x ≤ x (reflexive law), (ii) x ≤ y, y ≤ x =⇒ x = y (antisymmetric law), (iii) x ≤ y, y ≤ z =⇒ x ≤ z (transitive law). If moreover either x ≤ y or y ≤ x hold for all pairs x, y ∈ X, X is called a totally ordered set. Exercise 8.2. (i) We usually consider a subset of an ordered set with the naturally inherited order. (ii) For a set X and a set Y whose elements are subsets of X, we can define a natural order by inclusions on Y , by A ≤ B ⇐⇒ A ⊂ B for A, B ∈ Y . Definition 8.3. Let X be an ordered set. (i) An element x ∈ X with the property x ≤ y =⇒ x = y is called a maximal element of X. (ii) For a subset Y ⊂ X and x ∈ X, if y ≤ x for all y ∈ Y , x is called an upper bound of Y . (iii) If X 6= ∅ and all non-empty totally ordered subset of X has an upper bound in X, X is called inductive. Theorem 8.4. (Zorn’s lemma) Any inductive ordered set has a maximal element. Proposition 8.5. For a ring A and an ideal I 6= A, there exists a maximal ideal which contains I. In particular (taking I = 0), m-Spec(A) 6= ∅ if A 6= 0. Proof. The set of all ideals containing I and not equal to A is inductive with the order by inclusions (shown as in Exercise 1.26), therefore has a maximal element. ¤ Remark 8.6. We used the axiom of choice in the proof of Proposition 1.25 when we took an infinite sequence (if we wanted to prove it for euclidean domains and not generals PID’s, we did not need it). Rewrite this proof by using the Zorn’s lemma. 8.2. Infinite algebraic extensions, Algebraic closure. Definition 8.7. An extension field L of a field K is called an algebraic extension field when every x ∈ L is a root of a polynomial in K[X]. An extension field which is not algebraic is called a transcendental extension field. Exercise 8.8. (i) A finite extension is algebraic (Section 4, 4.2.1). (ii) L/K : algebraic ⇐⇒ L : a union of finite extensions of K. Definition 8.9. When every irreducible element of K[X] is linear, K is called an algebraically closed field. An algebraically closed field has no algebraic extensions other than itself.


55

Example 8.10. The field of complex numbers C is algebraically closed (fundamental theorem of algebra). Theorem 8.11. (Steinitz’ theorem) (i) For any field K, there exists an algebraic extension K of K which is algebraically closed, and every algebraic extension of K is isomorphic to some intermediate field of K/K. (ii) Any field isomorphism σ : K → K 0 has an extension to a field isomorphism σ : K → K 0 . In particular, K is uniquely determined up to K-isomorphisms. This K is called an algebraic closure of K. We will make use of the polynomial ring A[S] for a set of variables with infinite cardinality, but note that each of its elements (a polynomial) can contain only finitely many variables in S. Proof. (i) Consider a variable XP for each element P of K[X]0 = K[X]\K, and denote the set of all those variables by S. In the polynomial ring K[S], consider the ideal I generated by the subset {P (XP ) | P ∈ K[X]0 }. We first show I 6= K[S]. Assume I = K[S], or 1 ∈ I. Then: ∃a1 , . . . , an ∈ K[S],

n X

ai Pi (XPi ) = 1 (as a polynomial).

i=1

Now let K0 = K, Ki+1 = Ki [X]/(Pi ) for each i ∈ N. Then Kn includes the roots x1 , . . . , xn respectively of P1 , . . . , Pn . In this ring Kn , substitute xi to the XPi of the left hand side of the above identity, and 0 to all the other variables appearing in ai . Then we get 0 = 1, a contradiction. We conclude I 6= K[S]. Hence take a maximal ideal Q of K[S] containing I by Proposition 8.5 and consider the field K = K[S]/Q, which is naturally an extension field of K. For each P ∈ K[X]0 , if we denote the image of XP in K by xP = XP , by definition it is a root of P . As any x ∈ K is expressed in terms of a polynomial of finite numbers of xPi , hence it is contained in the finite extension K(xP1 , . . . , xPn ) of K. Therefore K is an algebraic extension of K, and each P ∈ K[X]0 has a root xP in K. Now we show that K is an algebraically closed field. Let R be an irreducible element of K[X]. The field K 0 obtained by adjoining all the coefficients of R to K is a finite 0 = K 0 [X]/(R) is a finite extension of extension of K, and R ∈ K 0 [X]. The field KR 0 of R 0 0 K , hence KR /K is also finite by Proposition 3.12, therefore a root X ∈ KR has a minimal polynomial P over K. Then by definition R | P , so it suffices to show that any irreducible element P of K[X] decomposes into a product of linear factors. If P is separable, then the field F/K obtained by adjoining all roots of P is a Galois extension, and if we write F ∼ = KP 0 by Theorem 6.4, then P 0 has a root xP 0 in K, hence K/K has an intermediate field which is K-isomorphic to F . Therefore K contains all the roots of P . If P is not separable, by Proposition 6.6 we have D(P ) = 0, hence

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we can write P (X) = Q1 (X p ) (p = char K). We repeat this to get P (X) = Q(X q ) for q = pf , f ≥ 1 with a separable polynomial Q, hence there is an intermediate field F ∼ = KP 0 which contains all the roots of Q as above. Now take a root z ∈ K 0 of P (X q ) and let z q = y, then it is a root of P 0 hence F = K(y), and all the roots y = y1 , . . . , yn ∈ F of Q can be written as yi = Pi (y) with Pi ∈ K[X]. Now take a polynomial Pi0 ∈ K[X] whose coefficients are the q-th roots of the coefficients of Pi q (these are contained in K) and letting zi = Pi0 (z) ∈ K, we have Proposition Q ziq = yi byQ q q q q 5.16, and as (X −zi ) = X −zi = X −yi we have P (X) = i (X −yi ) = i (X −zi )q . (ii) Let X be the set of all pairs (L, τ ) where L is an intermediate field of K/K and τ : L → K 0 is a ring homomorphism with τ |K = σ. It is an ordered set if we define (L1 , τ1 ) ≤ (L2 , τ2 ) ⇐⇒ L1 ⊂ L2 , τ2 |S L1 = τ1 . For any totally ordered subset Y of X, the element (LY , τY ) defined as LY = (L,τ )∈Y L and (L, τ ) ∈ Y =⇒ τY (x) = τ (x) (x ∈ L) is an upper bound of Y , hence X is inductive. Hence we can take a maximal element (L0 , τ0 ) of X by the Zorn’s lemma (Theorem 8.4), and let τ0 (L0 ) = L00 . For all x ∈ K, there exists a τ0 (x) ∈ HomK (L0 (x), K 0 ) with τ0 (x)|L0 = τ0 by Lemma 6.9, as K 0 is algebraically closed. By the maximality of (L0 , τ0 ) we have L0 (x) = L0 , i.e. L0 = K. As L00 is algebraically closed because it is isomorphic to L0 = K, and as K 0 is an algebraic extension of L00 we conclude L00 = K 0 . ¤ 8.3. Infinite Galois extensions. In the following we will fix an algebraic closure of K of a field K, and regard any algebraic extension of K as an intermediate field of K/K. Definition 8.12. An algebraic extension F/K is called a Galois extension if it is a union of finite Galois extensions of K. In this case the group AutK (F ) of all Kautomorphisms of F is called the Galois group of F/K, and denoted by Gal(F/K). For example, the union Ksep of all finite separable extensions of K inside K is a Galois extension of K by the next lemma and Proposition 6.14: Lemma 8.13. The composite LL0 /K of two finite Galois extensions L/K, L0 /K is a Galois extension. If L/K, L0 /K are both abelian, so is LL0 /K. Proof. As the Galois closure of LL0 /K (see Remark 6.15 – the extension field obtained by the procedure of Proposition 6.11(ii)) coincides with LL0 , it is a Galois extension. The latter part follows from the injectivity of the group homomorphism Gal(LL0 /K) 3 σ 7−→ (σ|L , σ|L0 ) ∈ Gal(L/K) ⊕ Gal(L0 /K). ¤ Definition 8.14. We call Ksep the separable closure of K, the Galois group GK = Gal(Ksep /K) is called the absolute Galois group of K. If K is perfect then Ksep = K. Exercise 8.15. (i) By Lemma 8.13, the union Kab of all abelian extensions of K contained in K is a Galois extension of K. This is called the maximal abelian extension of K.


57

S (ii) The union K(µ∞ ) = n≥1 K(µn ) of all cyclotomic extensions of K contained in K is a Galois extension of K. This is called the maximal cyclotomic extension of K. We have K(µ∞ ) ⊂ Kab by Theorem 5.22. Proposition 8.16. Let F/K be a Galois extension, and a finite extension L/K be its intermediate field. Then: (i) F/L is Galois and Gal(F/K) 3 σ 7−→ σ|L ∈ HomK (L, F ) is surjective. (ii) If L/K is also Galois, then H = Gal(F/L) is a normal subgroup of G = Gal(F/K), and we have a group isomorphism: G/H 3 σ 7−→ σ|L ∈ Gal(L/K). S 0 Proof. (i) Write F = L as a union ofS finite Galois extensions L0 /K. Then by 0 Lemma 7.14 LL /L is Galois and F = LL0 , hence F/L is Galois. For each L0 , by the latter part of Proposition 6.11 we have Gal(L0 /K) = HomK (L0 , K), hence Gal(F/K) = HomK (F, K). Therefore if we extend an arbitrary element of HomK (L, F ) to an element of HomK (K, K) by Theorem 8.11(ii) and restrict it to F we get an element of Gal(F/K), hence the surjectivity. (ii) By the latter part of Proposition 6.11 we have Gal(L/K) = HomK (L, F ) hence the surjection in (i) is a group homomorphism, and as H is its kernel it is normal. The second part follows from homomorphism theorem. ¤ Definition 8.17. For a family {Xi }i∈Λ of groups (resp. rings), indexedQby the elements of a set Λ, if we define componentwise operations on the product set i∈Λ Xi as: (x1 , . . . , xn )(y1 , . . . , yn ) = (x1 y1 , . . . , xn yn ), ( resp. and (x1 , . . . , xn ) + (y1 , . . . , yn ) = (x1 + y1 , . . . , xn + yn ) ), Q then it becomes a group (resp. ring). This i∈Λ Xi is called the direct product of groups (resp. rings). Exercise 8.18. An integral domain cannot be isomorphic to a direct product of more than one rings. Proposition 8.19. For an infinite Galois extension F/K, if we denote the set of all intermediate finite Galois extensions of F/K by Λ, then we have the following group isomorphism: ¯ © ª Y Gal(F/K) 3 σ 7−→ (σ|L ) ∈ (σL )L∈Λ ¯ L ⊂ L0 =⇒ σL0 |L = σL ⊂ Gal(L/K). L∈Λ

The set in the right hand side is a subgroup of the direct product, and is called an inverse limit lim Gal(L/K) of {Gal(L/K)}L∈Λ (see section k). ←−

Proof. The group homomorphism is defined as HomK (L, F ) = Gal(L/K) for each S L ∈ Λ. As F = L, an element σ is determined by (σ|L )L∈Λ , hence the map is injective. Conversely any element (σL )L∈Λ of the right hand side defines an element σ ∈ Gal(F/K) by x ∈ L =⇒ σ(x) = σL (x), hence it is also surjective. ¤

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Remark 8.20. For an infinite Galois extension F/K, by introducing a topology called a Krull topology in Gal(F/K), a bijective correspondence between the closed subgroups and the intermediate fields (in particular, open subgroups corresponds to finite intermediate fields) can be proven. 8.4. Example: Absolute Galois group of finite fields. We would like to represent the fact (Corollary 5.30) that there are isomorphisms between Galois groups and cyclic groups for all finite extensions simultaneously and compatibly, using the absolute Galois group. We will introduce the inverse limit of Z/nZ (n ∈ N\{0}). This is defined from the cyclic groups by the same procedure as we found the Galois group of infinite Galois extension in Proposition 8.19. b of Z as follows: Definition 8.21. We define the profinite completion Z ¯ n o Y b = (an )n≥1 ¯¯ m | n =⇒ an ≡ am mod m ⊂ Z Z/nZ. n≥1

Q

Exercise 8.22. (i) n≥1 Z/nZ is a ring by componentwise addition and multiplib cation, and Z is its subring. b is an injective ring homomorphism (ii) The natural map Z 3 1 7−→ (1)n≥1 ∈ Z b which identifies Z with a subring of Z. b 3 (an )n≥1 7−→ an ∈ Z/nZ with (iii) For each n ≥ 1 there is a natural surjection Z b (Use (ii).) kernel (n) ⊂ Z. b as an additive group, and also Z ⊂ Z, b in particular In what follows, we regard Z b 1 = (1)n≥1 ∈ Z. Now we consider the absolute Galois group of Fq . By the definition of Frobenius map, if we restrict Frq ∈ Gal(Fqn /Fq ) to Fqm for m | n we get Frq ∈ Gal(Fqm /Fq ). This defines, by Proposition 8.19, the Frobenius map as an element of the absolute Galois group as follows: Y Frq = (Frq )n≥1 ∈ Gal(Fq /Fq ) ⊂ Gal(Fqn /Fq ). n≥1

Hence defining ϕ((an )n≥1 ) = (ϕn (an ))n≥1 by means of Corollary 5.30, we have: Theorem 8.23. For any finite field Fq , there is an isomorphism: b 3 1 7−→ Frq ∈ Gal(Fq /Fq ). ϕ:Z Proposition 8.24. For a finite field Fq and a positive integer n, the diagram below is commutative: n /b / Z/nZ b Z Z ∼ = ϕ

∼ = ϕ

Gal(Fq /Fqn )

/ Gal(Fq /Fq )

²

²

∼ = ϕn

² / Gal(Fqn /Fq )


59

where the lower left horizontal map is the natural inclusion, the upper left horizontal map is the multiplication by n. The horizontal maps on the right is the canonical surjection to the quotient group by the image of the horizontal maps on the left. Proof. The first part follows from Frqn = (Frq )n and the definitions. The second part ∼ b follows from Z/(n) ¤ = Z/nZ (Exercise 8.22(iii)). Part 3. Algebraic Integers and Prime Ideal Decomposition 9. Rings of Integers of Number Fields — I. Definition 9.1. Generating sets of modules. Definition 9.1. Let A be a ring, M an A-module, X a subset of M . P (i) A finite sum of the form ni=1 ai xi (ai ∈ A, xi ∈ X) is called a linear combination of elements of X with coefficients in A. We consider 0 as a linear combination Pof 0 elements of X, and define 0 as the linear combination of ∅. (ii) A relation ni=1 ai xi = 0 expressing 0 as a linear combination of X is called a linear relation among the elements of X. In particular, when all the coefficients ai are 0, it is called a trivial linear relation. (iii) When there is no non-trivial linear relation among the elments of X, the subset X is called linearly independent. If it is not linearly independent, it is called linearly dependent. The empty set is linearly independent. (iv) If all x ∈ M can be written as linear combinations of elements in X, we say that M is generated by X, and X is called a generating set of M . (v) The subset of M consisting of all the elements which are linear combinations of elements of X is clearly an A-submodule of M , and is called the A-submodule generated by X. (vi) If M has a linearly independent generating set, M is called a free A-module, and a linearly independent generating set is called a basis of M . Exercise 9.2. The A-submodule N generated by X is the minimal A-submodule of M containing X, as any A-submodule of M containing X also contains N . Exercise 9.3. (i) For a ring A, its ideal is free if and only if it is a principal ideal generated by an element which is not a zero divisor. (ii) For a ring A and its ideal I (6= 0, 6= A), the quotient ring A/I is not a free A-module (as there it has no linearly independent element). Definition 9.4. An A-module M is finitely generated if there is a generating set of M of finite cardinality. The module 0 is finitely generated as it is generated by ∅. 9.2. Algebras over rings. Definition 9.5. Let A be a ring.

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(i) If a ring B is also an A-module and the A-homomorphism A 3 a 7−→ a · 1 ∈ B is a ring homormorphism, B is called an A-algebra. (ii) A ring homomorphism B → B 0 between A-algebras is called an A-algebra homomorphism or morphism of A-algebras if it is also an A-homomorphism of A-modules. We denote the category of A-algebras and A-algebra homomorphisms by A-Alg and its set of morphisms by HomA-Alg (B, B 0 ), so that we can distinguish it from the set of A-homomorphisms HomA (B, B 0 ) = HomA-Mod (B, B 0 ). A morphism of A-algebras is an isomorphism if it is an isomorphism either as sets, rings, or A-modules. Remark 9.6. This definition makes sense for non-commutative rings B as well (but it is better to assume that A is commutative): e.g. EndA (M ) := HomA (M, M ) for an A-module M is a non-commutative A-algebra in general (we have seen the case over a field in K in Definition e.2). Example 9.7. (i) Any ring has a unique structure of Z-algebra, and every ring homomorphism is a morphism of Z-algebras. Therefore Z-Alg = Rings. (ii) A field extension L of K is a K-algebra, and K-homomorphisms are K-algebra homomorphisms between the extension fields. (iii) Polynomial rings A[X], A[X1 , . . . , Xn ] are A-algebras. (iv) A quotient ring A/I for an ideal I ⊂ A is an A-algebra. More generally, for any A-algebra B, its quotient rings are A-algebras. (v) If A is a subring of a ring B, then B is an A-algebra (Exercise 1.13(i)). (vi) Defining an A-algebra structure on a ring B is equivalent to specify a ring homomorphism A 3 a 7−→ a · 1 ∈ B. (vii) If a subring of an A-algebra B is an A-submodule, it is called an A-subalgebra. Th image of an A-algebra homomorphism B → B 0 is an A-subalgebra of B 0 . (viii) If B is an A-algebra, every B-module is naturally an A-module (compare Exercise 1.13(ii)). In particular, every B-algebra is naturally an A-algebra. Definition 9.8. If an A-algebra is finitely generated as an A-module, it is called a finite A-algebra. (This is the same as finite extensions (Definition 3.10) in the case of extension fields.) 9.3. Integral closure. Let B be an A-algebra. For any x ∈ B, the image of the A-algebra homomorphism “substituting x into polynomials with coefficients in A”: A[X] 3 X 7−→ x ∈ B is denoted by A[x]. More generally, for x1 , . . . , xn ∈ B, the image of the A-algebra homomorphism A[X1 , . . . , Xn ] → B substituting x1 , . . . , xn into X1 , . . . , Xn is denoted by A[x1 , . . . , xn ]. By Example 9.7(vii), A[x], A[x1 , . . . , xn ] are A-subalgebras of B. Definition 9.9. An x ∈ B is called integral over A if it is a root of a monic of A[X] (see Definition 3.1). If all x ∈ B are integral over A, then B is called integral over A. Remark 9.10. When B is integral over A and A is a subring of B, then B is called an integral extension of A — this is the same as algebraic extensions (Definition 8.7) in the case of extension fields.


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Proposition 9.11. Let B an A-algebra. For an x ∈ B, the following are equivalent: (i) (ii) (iii) (iv)

x is integral over A; A[x] is a finite A-algebra; x belongs to a finite A-subalgebra M of B; There exists an A[x]-module M such that (1) it is finitely generated as an Amodule, and (2) if a ∈ A[x] \ {0}, there exists m ∈ M such that am 6= 0. (This means that the A-algebra homomorphism A[x] → HomA (M, M ) is injective).

Proof. (i)⇒(ii): If xn + a1 xn−1 + · · · + an = 0 (ai ∈ A), the ring A[x] is generated by {1, x, x2 , . . . , xn−1 } as an A-module. (ii)⇒(iii): Take M = A[x]. (iii)⇒(iv): Take m = 1. (iv)⇒(i): If M is generated Pn by m1 , . . . , mn and we represent x ∈ HomA (M, M ) as a matrix (aij ), i.e. xmi = j=1 aij mj , aij ∈ A for 1 ≤ i ≤ n. Then (xδij − aij )(mj ) = (0) and by multiplying the cofactor matrix of (xδij − aij ) from the left gives det(xδij − aij )mj = 0 (Proposition g.17 – it is an equality of polynomials with Zcoefficients, hence holds for matrices with entries in any rings). Then det(xδij −aij ) = 0 follows from condition (2). ¤ Remark 9.12. (ii)⇒(i) is easily seen as follows: if {P1 (x), . . . , Pm (x)} generates A[x] as an A-module and we take n greater than all deg Pi (1 ≤ i ≤ m), then xn is a linear combination of Pi , hence x is a root of a monic. Exercise 9.13. (Compare Exercise 8.8.) (i) A finite A-algebra is integral over A. (ii) B : integral over A ⇐⇒ B : a union of finite A-algebras. Lemma 9.14. If B is a finite A-algebra, then a finitely generated B-module is also finitely generated as an A-module. (Compare Proposition 3.12.) Proof. Let x1 , . . . , xn be a generating set of M as a B-module, and b1 , · · · , bm be a generating set of B as an A-module. Then bi xj (1 ≤ i ≤ m, 1 ≤ j ≤ n) gives a generating set of M as an A-module. ¤ Proposition 9.15. a1 , . . . , an : integral over A ⇐⇒ A[a1 , . . . , an ]: finite A-algebra. Proof. ⇐: A[a1 , . . . , an ] satisfies Proposition 9.11(iii) for every ai . ⇒: We prove by induction on n. As an is integral over A, it is also integral over A[a1 , . . . , an−1 ], hence by Proposition 9.11 A[a1 , . . . , an ] = (A[a1 , . . . , an−1 ])[an ] is a finite A[a1 , . . . , an−1 ]algebra. Therefore it is finite A-algebra by the inductive hypothesis and Lemma 9.14. ¤ Proposition 9.16. A0 = {x ∈ B | x: integral over A} is an A-subalgebra of B. This A0 is called the integral closure of A in B. Proof. For x, y ∈ A0 , by Proposition 9.15, A[x, y] is a finite A-algebra, hence integral over A by Proposition 9.11. Therefore x ± y, a · x, xy ∈ A[x, y] ⊂ A0 , 1 ∈ A ⊂ A0 . ¤

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Definition 9.17. Let K be an algebraic number field (Example 3.14). The integral closure of Z in K is called the ring of integers of K, and denoted by OK . The elements of OK are called the integers of K. In general, the integers of algebraic number fields are called algebraic integers. √ Example 9.18. OQ = Z (see Theorem 10.8). For K = Q( d) (d is a square-free integer):  √ (d ≡ 2, 3 (mod 4)) Z[ d] √ · ¸ OK = 1+ d Z (d ≡ 1 (mod 4)) 2

10. Rings of Integers of Number Fields — II. First Properties 10.1. Integrally closed domains. Let A be a domain. Definition 10.1. In the product set A × (A \ {0}), the relation ∼ defined as (a, b) ∼ (c, d) ⇐⇒ ad = bc is an equivalence relation. If we denote the quotient set by K and the equivalence class of (a, b) by ab , we can define ad + bc a c ac a c + = , = b d bd bd bd independently on the choice of representatives and K becomes a field. This field is called the fraction field of A, and denoted by Frac(A). The map A 3 x 7−→ x1 ∈ K is an injective ring homomorphism (with which we always regard A as a subring of K). Example 10.2. Q is (defined as!) the fraction field of Z. Exercise 10.3. For a subring A of a field K, the fraction field of A is contained in K. In particular, the fraction field of a field K is K itself. More precisely, any ring homomorphism f : A → X satisfying f (A \ {0}) ⊂ X × will extend uniquely to an A-algebra morphism K → X. Definition 10.4. Let K = Frac(A). If the integral closure of A in K coincides with A, the domain A is called integrally closed (or normal). Proposition 10.5. A : UFD =⇒ A : integrally closed domain. Proof. If A is a UFD, every element of K = Frac(A) is represented as x = ab with a, b coprime. If xn + a1 xn−1 + · · · + an = 0 (a1 , . . . , an ∈ A), then by multiplying by bn we have an + a1 an−1 b + · · · + an bn = 0, which shows b | an , but as a, b are coprime, b ∈ A× and x ∈ A. ¤


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10.2. Rings of integers are integrally closed. Lemma 10.6. Let B an integral A-algebra. Then for an element x of any B-algebra, x : integral over B =⇒ x : integral over A. In particular, an integral B-algebra is also integral as an A-algebra. Proof. If xn + a1 xn−1 + · · · + an = 0 (a1 , . . . , an ∈ B), then M = A[a1 , . . . , an ] is a finite A-algebra by Proposition 9.15 (as B is integral over A). As x is integral over M , the ring M [x] = A[a1 , . . . , an , x] is a finite M -algebra, hence a finite A-algebra by Lemma 9.14. Therefore x is integral over A by Proposition 9.15. ¤ Proposition 10.7. Let K = Frac(A) and L an algebraic extension of K, and B be the integral closure of A in L. Then: (i) B is integrally closed. (ii) For ∀x ∈ L, there is a ∈ A \ {0} such that ax ∈ B. In particular, L = Frac(B). Proof. (i) As Frac(B) is contained in L by Exercise 10.3, and if an element of L is integral over B, it is integral over A by Lemma 10.6, hence contained in B. (ii) If x ∈ L satisfies xn + pq11 xn−1 + · · · + pqnn = 0 (pi , qi ∈ A), multiplying by n-th power of q = q1 · · · qn ∈ A \ {0}, we have (qx)n + pq11 q(qx)n−1 + · · · + pqnn q n = 0, pqii q i ∈ A, hence qx ∈ B. ¤ Theorem 10.8. The ring of integers OK of an algebraic number field K is integrally closed, and K = Frac(OK ). Also OK ∩ Q = Z. Proof. The first part follows from Proposition 10.7. Latter part is because Z is integrally closed by Theorem 1.31 and Proposition 10.5. ¤ 10.3. Krull dimension. Let A be a ring. Definition 10.9. For P ∈ Spec(A), if a maximum length n of chains P0 $ P1 $ · · · $ Pn = P of prime ideals of A exists, then it is called the height of P , and denoted by htP . If there is no maximum, we set htP = ∞.) For a ring A, if the maximum of htP for P ∈ Spec(A) exists, it is called the (Krull) dimension of A, and denoted by dim A. We set dim 0 = −∞, and if there is no maximum dim A = ∞. Example 10.10. (i) A : a field ⇐⇒ 0 ∈ m-Spec(A) ⇐⇒ Spec(A) = {0} ⇐⇒ A : domain of dimension 0. (ii) A : a domain of dimension 1 ⇐⇒ 0 ∈ Spec(A) and m-Spec(A) = Spec(A)\{0} (cf. Proposition 8.5). In particular, a PID is of dimension 1 if it is not a field, by Proposition 1.24. (iii) The polynomial ring K[X1 , . . . , Xn ] of n variables over a field K is n-dimensional. (Not so easy to prove!)

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Definition 10.11. Let A be a subring of B. Recall that Q ∈ Spec(B) =⇒ P = Q∩A ∈ Spec(A) (Exercise 1.18(iv)). When this holds, we say that Q lies above P and P lies below Q. Proposition 10.12. If B is integral over A, there is no inclusion between two distinct Q, Q0 ∈ Spec(B) lying above the same P ∈ Spec(A). In particular, dim B ≤ dim A. Proof. Assume Q ⊂ Q0 , and we will show Q = Q0 . For all x ∈ Q0 , take a monic f (X) ∈ A[X] of the minimal degree with the property f (x) ∈ Q (it exists as x ∈ B is integral over A, and has positive degree as 1 ∈ / Q). If we let f (X) = Xg(X) + a, as 0 0 a = f (x) − xg(x) ∈ Q we see that a ∈ Q ∩ A = P ⊂ Q, hence xg(x) = f (x) − a ∈ Q. But g(x) ∈ / Q by our assumption, and as Q is prime, x ∈ Q. For the latter part, for any chain Q0 $ Q1 $ · · · $ Qn of prime ideals of B, letting Pi = Qi ∩ A gives P0 $ P1 $ · · · $ Pn , hence n ≤ dim A. ¤ Proposition 10.13. Let K = Frac(A) and L an algebraic extension of K, and B be the integral closure of A in L. Then: A : integrally closed of dimension 1 =⇒ B : integrally closed of dimension 1. Proof. B is integrally closed by Proposition 10.7(i), and by Proposition 10.12 we have dim B ≤ dim A = 1. As A is integrally closed, B ∩ K = A, therefore B is not a field as K 6= A. Therefore dim B = 1. ¤ Corollary 10.14. The ring of integers OK of an algebraic number field K is an integrally closed domain of dimension 1. 11. Rings of Integers of Number Fields — III. Finiteness 11.1. Noetherian rings. 11.1.1. Definition. Definition 11.1. A ring A is called noetherian if all of its ideals are finitely generated A-modules. Example 11.2. (i) PID is noetherian. (ii) The polynomial ring of n-variables over a field (or even over any noetherian ring) is noetherian (Hilbert basis theorem). Proposition 11.3. For a ring A, the following are equivalent: (i) A is noetherian; (ii) For any chain I0 ⊂ I1 ⊂ I2 ⊂ · · · of ideals of A, there exists k ∈ N and Ik = Ik+1 = · · · . (iii) Any nonempty set X consisting of ideals of A has a maximal element with respect to the inclusions.


65

S Proof. (i)⇒(ii): For a chain I0 ⊂ I1 ⊂ I2 ⊂ · · · , the union I = ∞ i=0 Ii is an ideal of A, hence finitely generated. Take an Ik which contains the finite generating set of I. (ii)⇒(iii): If X does not have a maximal element, for all I0 ∈ X, there is an I1 ∈ X with I0 ⊂ I1 and I0 6= I1 . Repeating this, we obtain a chain I0 ⊂ I1 ⊂ I2 ⊂ · · · with Ii 6= Ii+1 for all i. (iii)⇒(i): Take any ideal I of A. The set X of all finitely generated ideals contained in I is nonempty as 0 ∈ X, hence has a maximal element J by (iii). If J 6= I, there is an a ∈ I \ J and J ⊂ J + (a) ∈ X which contradicts the maximality by J 6= J + (a). Hence J = I and I is finitely generated. ¤ Exercise 11.4. In a noetherian domain, any element, which is neither 0 nor a unit, is a product of irreducible elements. 11.1.2. Modules over noetherian rings. Proposition 11.5. Let A be a noetherian ring. If N is an A-submodule of an A-module M , then: M : finitely generated =⇒ N : finitely generated. Proof. We prove by induction on the cardinality n of a generating set of M . If n = 1, there is a surjective A-homomorphism f : A 3 a 7−→ ax ∈ M for the generator x of M , hence N = f (J) for some ideal J of A by Proposition 2.10. As A is noetherian J is finitely generated, and the images under f of the generators of J generates N . For n > 1, let x1 , . . . , xn be a generating set of M . Consider M 0 = {ax1 | a ∈ A} and the surjective A-homomorphism f : M → M/M 0 . As f (x2 ), . . . , f (xn ) generates M/M 0 , the A-submodule f (N ) of M/M 0 is finitely generated by the inductive hypothesis. Also the kernel of the surjective A-homomorphism f |N : N → f (N ) is N ∩ M 0 , which is, being an A-submodule of M 0 , finitely generated by the case n = 1. Take a generating set y1 , . . . , yk of N ∩M 0 andPa generating set f (yk+1 ), . . . , fP (ym ) of f (N ), and for any x ∈ N , m 0 write f (x) ∈ f (N ) as m a f (y ) and write x − i i i=k+1 i=k+1 ai yi ∈ Ker f |N = N ∩ M Pk P ¤ as i=1 ai yi . Then x = m i=1 ai yi , which shows that y1 , . . . , ym generates N . Proposition 11.6. If B is a finite A-algebra, then A : noetherian =⇒ B : noetherian. Proof. Regarding any ideal of B as an A-module by Exercise 1.13(i), it is finitely generated A-module by Proposition 11.5. A generating set as an A-module also generates the ideal as a B-module. ¤ 11.1.3. Dedekind domains. Definition 11.7. An integrally closed noetherian domain of dimension 1 is called a Dedekind domain. Example 11.8. PID which is not a field =⇒ Dedekind domain (Proposition 10.10(ii), Theorem 1.27, Proposition 10.5, Exercise 11.2(i)).

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The goal of this section is to show the following: Theorem 11.9. The ring of integers OK of an algebraic number field K is a finite Z-algebra, hence a Dedekind domain. 11.2. Proof of finiteness. 11.2.1. Preliminaries: Trace and dual basis. Proposition 11.10. (Artin’s lemma) Let K be a field, and L, E be two extension fields of K. If σ1 , . . . , σm are mutually distinct K-homomorphisms of L into E, then they are linearly independent over E as K-linear maps, i.e.: m X ai σi (x) = 0 (∀x ∈ L) for ai ∈ E =⇒ ai = 0 (∀i). i=1

Proof. Assume the contrary, and take a linearly dependent subset of {σ1 , . . . , σm } with the minimal cardinality k. As σ1 6= 0, we have k > 1, and by suitably changing the P indices, let this set be {σ1 , . . . , σk }. If ki=1 ai σi = 0 is a nontrivial linear relation, then ai 6= 0 for all i by the minimality of k. Denoting a(i) := σi (a), there is an a ∈ L Pk (i) (i) = with a(k−1) 6= a(k) , because σk−1 6= σk . Then for all x ∈ L, by i=1 ai a x Pk Pk (i) i=1 ai σi (ax) = 0, we have i=1 ai a σi = 0, hence k−1 X i=1

ai (a(i) − a(k) )σi =

k X i=1

ai a(i) σi − a(k)

k X

ai σi = 0,

i=1

which is a nontrivial linear relation because ak−1 (a(k−1) − a(k) ) 6= 0, contradicting the minimality of k. ¤ Now let L/K be a separable extension of degree n, and denote the n distinct Khomomorphisms of L into some extension field E/K by: σi : L 3 x 7−→ σi (x) = x(i) ∈ E (1 ≤ i ≤ n). We let indices i, j, k, l run through 1, . . . , n, and adopt the notation σj (ei ) = eij for a basis {ei } of L/K. Proposition 11.11. For a basis {ei } of L/K, we have det(eij ) 6= 0. Proof. If det(e P ij ) = 0, by Proposition P g.14(i), there exists a1 , . . . , an ∈ E, not all equal to 0, and j eij aj = 0, which gives j aj σj = 0 and contradicts Proposition 11.10. ¤ P Definition 11.12. Define the trace of x ∈ L by TL/K (x) = ni=1 x(i) . TL/K : L → E is a K-linear map of vector spaces over K. Proposition 11.13. For x ∈ L and a basis {ei } of L/K, if xei = P then TL/K (x) = ni=1 xii ∈ K.

P j

xij ej (xij ∈ K),


67

P P Proof. Letting σk act on xei = j xij ej (xij ∈ K), we get x(k) eik = j xij ejk , hence (eik )(δkj x(j) ) = (xij )(ejk ) as matrices. By Proposition 11.11 and Exercise g.12 we have (eik ) ∈ GLn (E), hence if we let (eik )−1 = (e0ki ) then (δkj x(j) ) = (e0ki )(xij )(ejk ) P (k) P P P P P and TL/K (x) = x = e0ki xij ejk = xij ( k ejk e0ki ) = k k i,j i,j i,j xij δji = Pn ¤ i=1 xii . Proposition 11.14. For a basis {ei } of L/K, there is a basis {e0i } of L/K which satisfies TL/K (ei e0j ) = δij . (This {e0i } is called the dual basis of {ei }. ) Proof. As TL/K (ei ek ) =

P

l eil ekl ,

by Exercise g.12(i),(iii) and Proposition 11.11,

det(TL/K (ei ek )) = det(eil ) det t (elk ) = det(eij )2 6= 0, hence the system of linear equation (for each j): X TL/K (ei ek )xk = δij k

P has a solution xk = ckj by Proposition g.13. Then letting e0j = k ek ckj , it satisfies P P the condition by TL/K (ei e0j ) = TL/K (ei k ek ckj ) = k TL/K (ei ek )ckj = δij . P P ¤ As j aj e0j = 0 =⇒ 0 = TL/K ( j aj ei e0j ) = aj , {e0j } is a basis of L/K. 11.2.2. Extension of integrally closed domains. In the rest of this section, let A be an integrally closed domain, K = Frac(A), L/K a finite separable extension, and B the integral closure of A in L. Proposition 11.15. x ∈ B =⇒ TL/K (x) ∈ A. Proof. For the K-homomorphisms σi (1 ≤ i ≤ n) of L into some field, σi (x) are all roots of the same monic with coefficients in A as x, hence integral over A, hence their sum TL/K (x) is also integral over A (Proposition 9.16). As TL/K (x) ∈ K by Proposition 11.13, it is in A as A is integrally closed. ¤ For a A-submodule M of L, define M ∗ = {x ∈ L | ∀y ∈ M TL/K (xy) ∈ A}. As TL/K is a K-linear map, M ∗ is an A-submodule of L. M ∗ is called the dual module of M . We immediately see that M ⊂ N =⇒ M ∗ ⊃ N ∗ . Lemma 11.16. If M is a free A-module generated by a basis {ei } of L/K, the dual module M ∗ is a free A-module generated by its dual basis {e0i }. P P Proof. For x, y ∈ L, x = i ai e0i , y = i bi ei (ai , bi ∈ K), as TL/K is K-linear and by definition of the dual basis, ³X ´ X X X TL/K (xy) = TL/K ai bj e0i ej = ai bj TL/K (e0i ej ) = ai bj δij = ai bi . i,j

i,j

i,j

i

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Therefore: x ∈ M ∗ =⇒ TL/K (xei ) = ai ∈ A (∀i), X X ¡ ¢ ai ∈ A (∀i) =⇒ TL/K (xy) = ai bi ∈ A ∀y = bi ei ∈ M =⇒ x ∈ M ∗ , which shows

M∗

={

P

0 i ai ei

i

i

| ai ∈ A (∀i)}.

¤

Proposition 11.17. A : noetherian =⇒ B : finite A-algebra. Proof. By Proposition 10.7(ii), we can take a basis {ei } of L/K contained in B after multiplying by some element in A. As the basis is linearly independent over A, the Asubmodule M generated by {ei } in L is a free A-module. As M ⊂ B we have B ∗ ⊂ M ∗ , and moreover B ⊂ B ∗ by Proposition 11.15, hence B ⊂ M ∗ . Now Proposition 11.16 shows that M ∗ is a finitely generated A-module, and as A is noetherian B is also finitely generated by Proposition 11.5. ¤ Theorem 11.18. Let A a domain, K = Frac(A), L/K a finite separable extension and B the integral closure of A in L. Then: (i) A : integrally closed noetherian =⇒ B : finite A-algebra. (ii) A : integrally closed noetherian =⇒ B : integrally closed noetherian. (iii) A : Dedekind domain =⇒ B : Dedekind domain. Proof. (i): Proposition 11.17. (ii): Proposition 10.7(i), Proposition 11.17 and Lemma 11.6. (iii): (i) and Proposition 10.13. ¤ Proof. (of Theorem 11.9) As Z is a Dedekind domain by Theorem 1.31 and Exercise 11.8, and K/Q is separable by Proposition 6.12, it follows from Theorem 11.18(iii). ¤

12. Localization 12.1. Ring of fractions. Let A be a domain. 12.1.1. Definition. Definition 12.1. (i) A subset S ⊂ A is called a multiplicative subset of A if it is closed under the multiplication (in particular 1 ∈ S) and 0 ∈ / S. (ii) For a multiplicative subset S ⊂ A, the subset of K = Frac(A) consisting of a elements that can be written as (s ∈ S, a ∈ A) is a subring (domain) of K. s We denote this ring by S −1 A and call it the ring of fractions of A by S. Exercise 12.2. A ⊂ S −1 A and K = Frac(S −1 A).


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For an ideal I ⊂ A, we denote by S −1 I the ideal of S −1 A generated by the elements a of I. It is the subset of S −1 A consisting of the elements that can be writeen as (s ∈ s S, a ∈ I). If write I (A) for the set of all ideals of A, we obtain the following inclusionpreserving maps: f : I (S −1 A) 3 I 7−→ I ∩ A ∈ I (A) g : I (A) 3 J 7−→ S −1 J ∈ I (S −1 A) Proposition 12.3. (i) g ◦ f = id. In particular, f is injective and g is surjective. (ii) I $ S −1 A =⇒ f (I) = I ∩ A ⊂ A \ S. Proof. (i): For all I ∈ I (S −1 A), It is clear that S −1 (I ∩ A) ⊂ I. Conversely for any x ∈ I is written as x = pq (p ∈ A, q ∈ S), but as p = qx ∈ I we see that p ∈ I ∩ A, hence x ∈ S −1 (I ∩ A). (ii): s ∈ S ∩ I ⇒ 1 = s−1 s ∈ I. ¤ Now we restrict this correspondence to prime ideals. By Exercise 1.18(iv) Q ∈ Spec(S −1 A) =⇒ Q ∩ A ∈ Spec(A), and (Q ∩ A) ∩ S = ∅ by Proposition 12.3(ii), hence we have the map: f 0 : Spec(S −1 A) 3 Q 7−→ Q ∩ A ∈ {P ∈ Spec(A) | P ∩ S = ∅}. Proposition 12.4. Let P ∈ Spec(A) such that P ∩ S = ∅. Then: (i) For s ∈ S and p ∈ A, we have (ii) S −1 P ∈ Spec(S −1 A). Proof. (i): p ∈ P =⇒

p s

p s

∈ S −1 P ⇐⇒ p ∈ P .

∈ S −1 P is clear. Conversely, for

p s

∈ S −1 A:

p p0 p ∈ S −1 P =⇒ = 0 (p0 ∈ P, s0 ∈ S) =⇒ ps0 = sp0 ∈ P (s0 ∈ / P ) =⇒ p ∈ P. s s s (ii): By (i),

p p0 p p0 pp0 , 0 ∈ / S −1 P =⇒ p, p0 ∈ / P =⇒ pp0 ∈ / P =⇒ · 0 = 0 ∈ / S −1 P . s s s s ss

¤

From what is said above, we obtain the map: g 0 : {P ∈ Spec(A) | P ∩ S = ∅} 3 P 7−→ S −1 P ∈ Spec(S −1 A). Proposition 12.5. f 0 , g 0 are the inverse of each other, and in particular bijections. Proof. For P ∈ Spec(A), P ∩ S = ∅, we have a = a1 ∈ (S −1 P ) ∩ A ⇐⇒ a ∈ P by Propositioin 12.4(i), hence (S −1 P ) ∩ A = P , i.e. f 0 ◦ g 0 = id. By Proposition 12.3(i), g 0 ◦ f 0 = id. ¤

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12.1.2. Properties. Proposition 12.6. (i) A : noetherian =⇒ S −1 A : noetherian. −1 (ii) dim S A ≤ dim A. Proof. (i): Proposition 12.3(i) and Proposition 11.3. (ii): Proposition12.5 and Definition 10.9. ¤ If A is a subring of B, then a multiplicative subset of A is also that of B. Then we have: Proposition 12.7. For a multiplicative subset S ⊂ A: A0 : integral closure of A in B =⇒ S −1 A0 : integral closure of S −1 A in S −1 B. Proof. It suffices to show that x ∈ S −1 B is integral over S −1 A if and only if x ∈ S −1 A0 . p1 pn x : integral over S −1 A =⇒ xn + xn−1 + · · · + = 0 (pi ∈ A, qi ∈ S) q1 qn =⇒ Letting q = q1 · · · qn ∈ S, ³p ´ p1 pn i i (qx)n + q(qx)n−1 + · · · + q n = 0 q ∈A q1 qn qi 0 −1 0 =⇒ qx ∈ A =⇒ x ∈ S A . −1 0 x ∈ S A =⇒ (qx)n + a1 (qx)n−1 + · · · + an = 0 (q ∈ S, ai ∈ A) ³a ´ a1 an i −1 =⇒ xn + xn−1 + · · · + n = 0 ∈ S A q q qi =⇒ x : integral over S −1 A. ¤ 12.2. Localization. Let A be a domain. 12.2.1. Localization. Definition 12.8. For a P ∈ Spec(A), the complement S = A \ P is a multiplicative subset of A. The ring of fractions S −1 A is called the localization of A at P , and denoted by AP . For an ideal I ⊂ A, we write IP for S −1 I. Example 12.9. K = Frac(A) is the localization of A at 0 ∈ Spec(A). Definition 12.10. A ring (resp. domain) is called a local ring (resp. local domain) if it has only one maximal ideal P . The field A/P is called the residue field of A. Exercise 12.11. If A is local with the maximal ideal P , we have A = A× ∪ P . (If a∈ / A× , then (a) 6= A, so use Proposition 8.5.) Proposition 12.12. Let AP be the localization of A at P ∈ Spec(A). (i) AP is a local domain with the maximal ideal PP .


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(ii) dim AP = htP . ¯ © ª Proof. (i): By Proposition 12.4(i), we have PP = ps ∈ AP ¯ p ∈ P , hence a s ³ a ´−1 a ∈ AP \ PP ⇐⇒ a ∈ / P ⇐⇒ = ∈ AP ⇐⇒ ∈ A× P, s a s s therefore AP \ PP = A× P , which shows that PP ∈ m-Spec(A) and that PP contains every ideal I of AP except for AP . (ii): Proposition 12.5. ¤ 12.2.2. Local properties of rings. Proposition 12.13. For a domain A, the following are equivalent: (i) dim A = 1. (ii) dim AP = 1 for all P ∈ m-Spec(A). Proof. Proposition 12.12(ii), Exercise 10.10.

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Proposition 12.14. For a x ∈ K = Frac(A), the following are equivalent: (i) x ∈ A. (ii) x ∈ AP for all P ∈ m-Spec(A). Proof. (i)⇒(ii): Clear. (ii)⇒(i): For an x satisfying (ii), consider the ideal I = {a ∈ A | ax ∈ A} of A. Then for all P ∈ m-Spec(A), we have x ∈ AP =⇒ x = as (s ∈ A \ P ) =⇒ s ∈ I \ P , hence I is not contained in P . Therefore I = A by Proposition 8.5, and 1 ∈ I shows that x = 1x ∈ A. ¤ Proposition 12.15. For a domain A, the following are equivalent: (i) A : integrally closed. (ii) AP : integrally closed for all P ∈ m-Spec(A). Proof. (i)⇒(ii): Let K = Frac(A). By Proposition 12.7, the integral closure of AP in KP = K is AP . (ii)⇒(i): For any P ∈ m-Spec(A), if x ∈ K is integral over A, then it is integral over AP , hence x ∈ AP by (ii). Therefore x ∈ A by Proposition 12.14 (ii)⇒(i). ¤ Proposition 12.16. For a noetherian domain A, the following are equivalent: (i) A : Dedekind domain. (ii) AP : Dedekind domain for all P ∈ m-Spec(A). Proof. Proposition 12.13 (dimension), Proposition 12.15 (integrally closed), Proposition 12.6(i) (noetherian). ¤

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13. Dedekind Domains I 13.1. Fractional ideals. Let A be a noetherian domain of positive dimension (i.e. not a field), and let K = Frac(A). 13.1.1. Definition. Definition 13.1. A finitely generated A-submodule of K is called a fractional ideal of A. We denote the set of all nonzero fractional ideals by FA . Ideals of A are the fractional ideals that are contained in A. Lemma 13.2. An A-submodule I ⊂ K is a fractional ideal if and only if there exists a ∈ A \ {0} such that aI = {ax ∈ K | x ∈ I} ⊂ A. Proof. For a fracional ideal I of A, the product of denominators of the generators of I works as an a. Conversely, if aI ⊂ A, it is an ideal of A, hence finitely generated, and as dividing its generators by a gives the generators of I, the module I is finitely generated as well. ¤ 13.1.2. Product and invertibility. Definition 13.3. For I, J ∈ FA , denote the A-submodule of K generated by the set {ab | a ∈ I, b ∈ J} by IJ. Then IJ ∈ FA as aI ⊂ A, a0 J ⊂ A =⇒ aa0 IJ ⊂ A. We call it the product of I, J. Exercise 13.4. (i) A product of ideals is again an ideal. (ii) This defines a commutative multiplication on FA , i.e. (1) IA = I for I ∈ FA , (2) IJ = JI, and (3) I1 (I2 I3 ) = (I1 I2 )I3 . Definition 13.5. An I ∈ FA is called invertible if there exists an I −1 ∈ FA such that II −1 = A. We denote by FA× the set of all invertible fractional ideals, which is an abelian group by Example b.9(iv). Proposition 13.6. For I ∈ FA , let I 0 = {x ∈ K | xI ⊂ A}. Then: (i) I 0 ∈ FA . (ii) II 0 ⊂ A, I ⊂ A ⇒ A ⊂ I 0 . (iii) I ∈ FA× ⇐⇒ II 0 = A. Proof. (i): I 0 is an A-submodule of K by definition, and aI 0 ⊂ A for any for an a ∈ I, hence is a fractional ideal by Lemma 13.2. (ii): Clear. (iii): ⇒: If II −1 = A, then I −1 ⊂ I 0 by definition, hence A = II −1 ⊂ II 0 ⊂ A which shows II 0 = A. ⇐ is clear. ¤ 13.1.3. Principal fractional ideals. Definition 13.7. For x ∈ K, the fractional ideal (x) = {ax | a ∈ A} of A is called the principal fractional ideal generated by x. We denote by HA the set of all nonzero principal fractional ideal of A. (x)(y) = (xy), (x)(x−1 ) = A shows that HA is a subgroup of FA× .


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Proposition 13.8. A : PID ⇐⇒ FA = HA . Proof. ⇒: For any¡I ¢∈ FA , for some a ∈ K × , aI ⊂ A is an ideal, hence principal. If ¤ aI = (x) then I = xa . ⇐: Clear. 13.2. Invertibility is local. Let S ⊂ A be a multiplicative subset. Note that K = Frac(S −1 A). For a fractional ideal I of A, let S −1 I be the S −1 A-submodule of K generated by the elements of I. In particular if P ∈ Spec(A) and S = A \ P write IP = S −1 I. The S −1 I is generated by the generators of I as an A-module, hence finitely generated, i.e. a fractional ideal of S −1 A. Proposition 13.9. For I, J ∈ FA : (i) S −1 (IJ) = S −1 I · S −1 J. (ii) S −1 (I 0 ) = (S −1 I)0 . Proof. (i): S −1 I · S −1 J is generated by the elements of IJ as an S −1 A-module. (ii): For an x ∈ S −1 (I 0 ), we have xS −1 I ⊂ S −1 (II 0 ) ⊂ S −1 A, hence S −1 (I 0 ) ⊂ (S −1 I)0 . To show (S −1 I)0 ⊂ S −1 (I 0 ), let a1 , . . . , an be the generators of I. For an x ∈ (S −1 I)0 , as xai ∈ S −1 A can be written as xai = pqii (pi ∈ A, qi ∈ S), letting q = q1 · · · qn ∈ S we have qxai ∈ A, i.e. qx ∈ I 0 , therefore x ∈ S −1 (I 0 ). ¤ Proposition 13.10. For I ∈ FA , the following are equivalent: (i) I ∈ FA× . (ii) IP ∈ FA×P for all P ∈ m-Spec(A). Proof. (i)⇒(ii): If II 0 = A then IP (I 0 )P = (II 0 )P = AP by Proposition 13.9(i). (ii)⇒(i): If II 0 6= A, take P ∈ m-Spec(A) such that II 0 ⊂ P by Proposition 8.5. Then (II 0 )P ⊂ PP , which contradicts (ii) as (II 0 )P = IP (I 0 )P = IP (IP )0 by Proposition 13.9. ¤ Corollary 13.11. For a noetherian domain A, the following are equivalent: (i) FA = FA× . (ii) FAP = FA×P for all P ∈ m-Spec(A). Proof. (i)⇒(ii): For any nonzero ideal I ⊂ AP , we have I = (I ∩ A)P by Proposition 12.3(i)), hence invertible by Proposition 13.10. Hence any I ∈ FAP is invertible by aI ⊂ A ⇒ I −1 = a(aI)−1 . (ii)⇒(i): Proposition 13.10. ¤ Proposition 13.12. A : noetherian local domain =⇒ FA× = HA .

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Proof. It suffices to show FA× ⊂ HA . Let P be the maximal ideal of A. Take any I ∈ FA× . By II 0 = A, ∃a1 , . . . , an ∈ I, ∃a01 , . . . , a0n ∈ I 0 ,

n X

ai a0i = 1 ∈ / P,

i=1

hence ak a0k ∈ / P for some k, i.e. ak a0k ∈ A× by Exercise 12.11. Therefore for all a ∈ I writing a = (a(ak a0k )−1 a0k )ak and a(ak a0k )−1 a0k ∈ II 0 = A shows that a ∈ (ak ), hence I ⊂ (ak ), and as ak ∈ I, in fact I = (ak ). ¤ Corollary 13.13. For a noetherian domain A, the following are equivalent: (i) FA = FA× . (ii) AP : PID for all P ∈ m-Spec(A). Proof. Corollary 13.11, Proposition 13.12, Proposition 13.8.

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14. Dedekind Domains II 14.1. DVR / Dedekind domains and fractional ideals. 14.1.1. DVR. Definition 14.1. If a local domain A, not a field, is a PID, then A is called a discrete valuation ring (DVR). Proposition 14.2. For a domain A, which is not a field: A : DVR ⇐⇒ ∃π ∈ A, ∀x ∈ A \ {0} x = u · π n (u ∈ A× , n ∈ N). Proof. ⇒: By Proposition 1.24, the only nonzero prime ideal of A is the maximal ideal, and let P = (π). As A is a UFD by Theorem 1.27, and π is the unique prime element of A up to associate by Proposition 1.24. Hence the prime decomposition of any x ∈ A \ {0} reads x = u · π n , u ∈ A× . ⇐: Any nonzero ideal I ⊂ A is written as I = (π n ) by the minimal value of v(x) for nonzero x ∈ I, hence A is a DVR. ¤ Definition 14.3. For a DVR A, a generator of its maximal ideal P , i.e., any element of P \ P 2 , is called a uniformizer of A. The map v : A \ {0} 3 uπ n 7→ n ∈ N (we set v(0) := ∞) is independent of the choice of π, and is called the valuation of A. Proposition 14.4. For a noetherian local domain A, not a field, with maximal ideal P: A : DVR ⇐⇒ P ∈ HA . T T Proof. ⇒: Clear. ⇐: Let P = (π). We show n∈N (π n ) = 0. If x ∈ n∈N (π n ), let x = π n xn for all n. Then xn = πxn+1 and we obtain the chain of ideals (x0 ) ⊂ (x1 ) ⊂ (x2 ) ⊂ (x3 ) ⊂ · · · , and by Proposition 11.3(ii) there exists k with xk+1 ∈ (xk ), i.e. there is a y ∈ A such that xk+1 = yxk = yπxk+1 , (1 − yπ)xk+1 = 0, but 1 − yπ 6= 0


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gives xk+1 = 0, i.e. x = 0. Hence for all x ∈ A \ {0} there is a maximal n ∈ N such that x ∈ (π n ), and x = π n u, u ∈ A \ (π) = A× . Now appeal to Proposition 14.2. ¤ 14.1.2. Fractional ideals of DVR / Dedekind domains. Proposition 14.5. For a noetherian local domain A, the following are equivalent: (i) A : DVR. (ii) A : Dedekind domain. Proof. (i)⇒(ii): Exercise 11.8. (ii)⇒(i): Let P be the unique nonzero prime ideal. We first show P 0 6= A. The set {I ∈ FA | I ⊂ A, I 0 % A} is nonempty as it contains HA , hence has a maximal element J by Proposition 11.3. If J is not prime, there are a, b ∈ /J 0 with ab ∈ J. But then for all x ∈ J , we have xb(J +(a)) = x(bJ +(ab)) ⊂ xJ ⊂ A which shows that xb ∈ (J +(a))0 = A, hence x(J +(b)) = xJ +(xb) ⊂ A and x ∈ (J +(b))0 = A. Therefore J 0 = A, a contradiction. We conclude J = P , hence P 0 6= A. By A ⊂ P 0 we have P ⊂ P P 0 , and as P P 0 is an ideal of A and P is maximal, either P P 0 = A or P P 0 = P . If P P 0 = P , by Proposition 9.11(iv)⇒(i) all elements of P 0 are integral over A, hence in A as A is integrally closed, a contradiction. Therefore P P 0 = A, and by Proposition 13.12 P ∈ HA , and we conclude the proof by Proposition 14.4. ¤ Theorem 14.6. For a noetherian domain A, not a field, the following are equivalent: (i) FA = FA× . (ii) AP : DVR for all P ∈ m-Spec(A). (iii) A : Dedekind domain. Proof. (i)⇔(ii): Corollary 13.13. (ii)⇔(iii): Proposition 12.16, Proposition 14.5.

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14.2. Prime ideal decomposition. Lemma 14.7. For two ideals I, J and a P ∈ Spec(A): IJ ⊂ P =⇒ I ⊂ P or J ⊂ P. Proof. ∃a ∈ I \ P, ∃b ∈ J \ P =⇒ ab ∈ IJ \ P .

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Theorem 14.8. For a Dedekind domain A: (i) Any I ∈ FA is uniquely decomposed in the following form: e

I = P1e1 · · · Pg g (ei ∈ Z). The exponent of P in this decomposition is denoted by vP (I) (if I does not appear, vP (I) = 0.) (ii) This vP (I) is equal to the valuation in AP of a generator of IP . (iii) For I, J ∈ FA , we have I ⊂ J ⇐⇒ vP (I) ≥ vP (J) for all P ∈ Spec(A).

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Proof. (i): We first show the existence and uniqueness of the decomposition for the ideals I ⊂ A. Take a maximal ideal P containing I by Proposition 12.11. Then I = P (IP −1 ), P −1 6= A (as P −1 = A ⇒ A = P P −1 = P A = P ), hence I ⊂ IP −1 ⊂ A, I 6= IP −1 (as I = IP −1 ⇒ A = I −1 I = I −1 IP −1 = P −1 ). Repeating this, we obtain the decomposition by Proposition 11.3. If P1 · · · Pn = Q1 · · · Qm , then as Q1 · · · Qm ⊂ Pn and Pn is prime, Qi ⊂ Pn for some i by Lemma 14.7. But as Qi is maximal we have Qi = Pn , and multiplying Q−1 = Pn−1 to the identity we obtain i P1 · · · Pn−1 = Q1 · · · Qi−1 Qi+1 · · · Qm . Hence we conclude the uniqueness by induction. For arbitrary I ∈ FA , write I = (a)−1 (aI), aI ⊂ A for some a ∈ A \ {0}, and we obtain the decomposition of I from that of (a) and aI. The uniqueness also follows from the ideal case by multiplying the factors with negative exponents. (ii): Let P be a prime ideal appearing in the decomposition of I. For other prime v (I) ideals Q 6= P we have QP = AP , hence if PP = (π) we obtain IP = PPP = (π vP (I) ). (iii): ⇐ is clear. ⇒: If I ⊂ J then IP ⊂ JP , but if there is a P ∈ Spec(A) with v (I) v (J) vP (I) < vP (J), then IP = PPP ⊃ PPP = JP and IP 6= JP , a contradiction. ¤ Definition 14.9. For a Dedekind domain A and an I ∈ FA , the vP (I) of Theorem 14.8 is called the P -adic valuation of I. For K := Frac(A), the map vP : K × 3 x 7→ vP ((x)) ∈ Z is a group homomorphism, and is called the P -adic valuation of K. Proposition 14.10. A : PID, not a field ⇐⇒ A : noetherian UFD of dimension 1. Proof. ⇒: Theorem 1.27 and Exercise 11.8. ⇐: By Proposition 10.5, A is a Dedekind domain. For any nonzero prime ideal P of A, if we take an a ∈ P then P appears in the decomposition of (a) by Theorem 14.8(iii) as (a) ⊂ P . If we take the prime (element) decomposition a = p1 · · · pn , then (a) = (p1 ) · · · (pn ), (pi ) ∈ Spec(A), but by uniqueness of the prime ideal decomposition P = (pi ) for some i, therefore P ∈ HA . As all I ∈ FA are products of primes, FA = HA . ¤ Theorem 14.11. (Fundamental theorem of algebraic number theory) In the ring of integers of an algebraic number field, all nonzero ideals are uniquely decomposed into a product of prime ideals. Definition 14.12. For a Dedekind domain A, by Proposition 14.6, FA is an abelian group with A as an identity. This group is called the ideal group of A. The quotient group FA /HA is called the ideal class group of A, and denoted by Cl(A).

Part 4. Arithmetic Galois Theory — Theory of Hilbert 15. Modules over PID 15.1. Free modules. Let A be a ring and M, Mi be A-modules.


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Definition 15.1. For A-modules M1 , . . . , Mn , the direct product M1 × · · · × Mn becomes an A-module by defining the addition and the action of A componentwise: (x1 , . . . , xn ) + (y1 , . . . , yn ) = (x1 + y1 , . . . , xn + yn ), a(x1 , . . . , xn ) = (ax1 , . . . , axn ). L It is called the direct sum of {Mi }, and denoted by ni=1 Mi = M1 ⊕ · · · ⊕ Mn . Ln n 0 The direct sum i=1 M of n copies of M is denoted by M . Set M = 0. This definition generalizes Definition h.1 for vector spaces. Exercise 15.2. If Ni be an A-submodule of Mi for each i, there is an isomorphism: n n n .M M M ∼ = Mi Ni 3 (x1 , . . . , xn ) 7−→ (x1 , . . . , xn ) ∈ (Mi /Ni ). i=1

i=1

i=1

Exercise 15.3. Let M a finitely generated free A-module. As the generators (finitely many of them) are linear combination of a basis involving finite number of its elements, a basis has a finite cardinality. If {x1 , . . . , xn } is a basis of M , then M ∼ = An by the isomorphism: n X n A 3 (a1 , . . . , an ) 7−→ ai xi ∈ M. i=1

Definition 15.4. For an A-module M and an ideal I ⊂ A, we denote by IM the A-submodule of M generated by the set {ax | a ∈ I, x ∈ M }. If B is an A-algebra, then IB is an ideal of B, generated by I as a B-module. Lemma 15.5. For an A-module M and an ideal I ⊂ A, M/IM is naturally an A/Imodule, and if M is free and M ∼ = An , then M/IM ∼ = (A/I)n as A/I-modules. Proof. Because for all a ∈ I and x ∈ M/IM we have ax = 0, the action of A on M/IM is naturally considered as an action of A/I. As IAn = I n , the isomorphism of Exercise 15.3 restricts to I n ∼ = IM , hence by Exercise 15.2 we have an isomorphism (A/I)n ∼ = M/IM of A-modules, which is naturally an isomorphism of A/I-modules. ¤ Proposition 15.6. For a finitely generated free module M , all bases of M have the same number of elements. Proof. We reduce to Theorem c.14, i.e. the case when A is a field. For arbitrary ring A, take a P ∈ m-Spec(A) by Proposition 8.5. Then A/P is a field (Corollary 2.12(ii)), and by Lemma 15.5, M/P M is a vector space over A/P and the cardinality of a basis of M is equal to dimA/P (M/P M ), which is independent of the choice of the basis. ¤ Definition 15.7. The cardinality of a basis of M is called the rank of M and denoted by rankA M . Remark 15.8. We called it the dimension of M when A is a field, but if M is a finite A-algebra, we use the term rank to distinguish from its Krull dimension as a ring.

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15.2. Modules over PID. Now we let A be a domain and M be an A-module. Definition 15.9. If for any a ∈ A \ {0} and x ∈ M we have ax = 0 =⇒ x = 0, we call M a torsion-free module. Exercise 15.10. Free modules are torsion free. Proposition 15.11. Let M be a finitely generated torsion-free module over a domain A. Then M is isomorphic to an A-submodule of a finitely generated free A-module. Proof. Let {x1 , . . . , xk } be a generating set of M , and by suitably changing the indices, take its linearly independent subset {x1 , . . . , xn } such that {x1 , . . . , xn , xi } is linearly dependent for all i: 1 ≤ ∀i ≤ k ∃ci ∈ A \ {0}, ci xi =

n X

bij xj (bij ∈ A).

j=1

Now let c = c1 · · · ck and c0i = c−1 let N be i c for each i. As A is a domain, c 6= 0. If we Pk the A-submodule of M generated by {x1 , . . . , xn }, then for all x ∈ M , if x = i=1 ai xi : cx =

k X i=1

cai xi =

k X

(c0i ai )ci xi =

i=1

k X i=1

c0i ai

n ³X

´ bij xj ∈ N,

j=1

hence we have an A-homomorphism f : M 3 x 7−→ cx ∈ N . As M is torsion free, f is injective, hence M ∼ ¤ = f (M ) ⊂ N , and N is free with a basis {x1 , . . . , xn }. Proposition 15.12. Let A be a PID. If N is an A-submodule of an A-module M : M : finitely generated, free =⇒ N : finitely generated, free. Proof. We prove by induction on the rank n of M . If n = 1, an A-submodule of M∼ = A is an ideal of A, hence is free, being principal (Exercise 9.3(ii)). For n > 1, we let M = An , and f be the projection to the n-th component: f : M = An 3 (a1 , . . . , an ) 7−→ an ∈ A. Then f (N ) is an ideal of A, hence f (N ) = (a) for an a ∈ A. As N ∩ Ker f is an A-submodule of Ker f = {(a1 , . . . , an−1 , 0) ∈ An } ∼ = An−1 , it is finitely generated ∼ and free by the inductive hypothesis: N ∩ Ker f = Am . Now take y ∈ N such that f (y) = a. Then for any x ∈ N , if f (x) = ab then x − by ∈ N ∩ Ker f , and we define an A-homomorphism: ∼ Am+1 . g : N 3 x 7−→ (x − by, b) ∈ (N ∩ Ker f ) ⊕ A = For any (x, b) ∈ (N ∩ Ker f ) ⊕ A, we have g(x + by) = (x, b), hence g is surjective, and as x = (x − by) + by, is injective as well. Hence N ∼ ¤ = Am+1 . Theorem 15.13. All finitely generated torsion-free modules over a PID are free. Proof. Proposition 15.11, Proposition 15.12.

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Remark 15.14. As a module over a field K is always torsion-free by a ∈ K × , ax = 0 =⇒ x = 1x = a−1 ax = 0, and a field is a PID, the existence of the basis for vector spaces (Theorem c.11) is a special case of Theorem 15.13. Theorem 15.13 shows that the ring of integers of an algebraic number field is a free Z-module. In general: Proposition 15.15. Let A be a PID, K = Frac(A), L/K a finite extension, B the integral closure of A in L, and assume that B is a finite A-algebra. Then B is a free A-module of rank [L : K]. For P ∈ m-Spec(A), the quotient ring B/P B is a finite k(P )-algebra of dimension [L : K]. Proof. As B is a domain containing A, it is torsion free as an A-module. Hence Theorem 15.13 shows that it is free. Taking any basis of B over A, it is linearly independent over A, hence linear independent over K (clearing the denominators of a linear relation). By Proposition 10.7(ii), this basis generates L over K, hence is a basis of L over K. Therefore rankA B = dimK L. The latter part follows from Lemma 15.5. ¤ 16. Equations over Rings of Integers — I. Decomposition of Primes In this section, let A be a Dedekind domain, K = Frac(A), L/K a finite extension, and B the integral closure of A in L. We know that L = Frac(B) (Proposition 10.7(ii)). We assume that B is a finite A-algebra, and hence is a Dedekind domain. This assumption is satisfied if L/K is separable, by Theorem 11.18. Example 16.1. If A = OK is the ring of integers of an algebraic number field K, any finite extension L/K is separable, and the integral closure of OK in L is B = OL , as an element of L is integral over OK if and only if it is integral over Z (Lemma 10.6). Exercise 16.2. More generally, let K 0 be an intermediate field of L/K and A0 be the integral closure of A in K 0 . Then B is also the integral closure of A0 in L by Lemma 10.6, and the finiteness of B/A is equivalent to that for B/A0 and A0 /A (for A0 /A, use Proposition 11.5; for the converse use Lemma 9.14). 16.1. Decomposition of primes, ramification index and residual degree. We fix a P ∈ Spec(A) \ {0} = m-Spec(A) and consider its decomposition in B. Definition 16.3. We write Q | P if Q ∈ Spec(B) lies above P (i.e. Q ∩ A = P ). Suppose that the prime decomposition (Theorem 14.8) of the ideal P B of B generQ e ated by the elements of P is P B = gi=1 Qei i = Qe11 · · · Qgg (ei = vQi (P B)). Then: vQ (P B) > 0 ⇐⇒ Q ∈ {Q1 , . . . , Qg } ⇐⇒ P B ⊂ Q ⇐⇒ P ⊂ Q ⇐⇒ P ⊂ Q ∩ A ⇐⇒ Q | P. Definition 16.4. The field k(P ) := A/P is called the residue field at P . (We will see in Proposition 16.10(ii) that it is the residue field of AP as in Definition 12.10.)

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Proposition 16.5. If Q | P , then k(Q) = B/Q is a finite extension of k(P ) = A/P . Proof. If we let f : A → B/Q be the composite of the inclusion A → B and the canonical surjection B → B/Q, then Ker f = P by Q ∩ A = P . Hence B/Q ⊃ Im f ∼ = A/Ker f = A/P by Theorem 2.8. As B is a finite A-algebra (Theorem 11.18), take the generator {ai } of B as an A-module. Then B/Q is generated by {ai } as an A/P -module, hence is a finite extension. ¤ Definition 16.6. Keep the above notations, and let Q | P . (i) (ii) (iii) (iv)

vQ (P B) is called the ramification index of Q/P , and denoted by e(Q/P ). [k(Q) : k(P )] is called the residue degree of Q/P , and denoted by f (Q/P ). P is called ramified in L if e(Q/P ) > 1 for some Q | P . P is called unramified in L if e(Q/P ) = 1 for all Q | P .

Remark 16.7. (i) Only finitely many prime ideals ramify in a given L. (ii) It is customary to require the separability of k(Q)/k(P ) for all Q | P for the definition of unramifiedness, but we will not need it. Exercise 16.8. Let K 0 be an intermediate field of L/K, and A0 be the integral closure of A in K 0 (Exercise 16.2). If Q | P is as above and P 0 = Q ∩ A0 ∈ Spec(A0 ), then Q | P 0 , P 0 | P and e(Q/P ) = e(Q/P 0 )e(P 0 /P ), f (Q/P ) = f (Q/P 0 )f (P 0 /P ). 16.2. Localization. Recall that, for P ∈ Spec(A) \ {0}, the localization AP of A at P is a DVR (Proposition 14.6). Setting S = A \ P , the integral closure of AP in L is S −1 B by Proposition 12.7. We write BP = S −1 B, and IP = S −1 I for ideals I ⊂ B. Therefore we have the following situation: AP is a DVR, K = Frac(A), L/K a finite extension, BP is the integral closure of AP in L, and L = Frac(BP ). As BP is generated as an AP -module by the generators of the A-module B, by our assumption it is a finite AP -algebra. This is a special case of our general situation. e

Now consider the decomposition P B = Qe11 · · · Qgg . By Proposition 12.5, we have Spec(BP ) = {QP | Q ∈ Spec(B), Q∩S = ∅}. As Q∩S = ∅ ⇐⇒ Q∩A ⊂ P ⇐⇒ Q = 0 or Q | P , we have Spec(BP ) = {0, (Q1 )P , . . . , (Qg )P }. First observe the following: Lemma 16.9. Let A be a Dedekind domain and P ∈ m-Spec(A). If S ⊂ A is a multiplicative subset and S ∩ P = ∅, then for all n ≥ 1, we have A/P n ∼ = S −1 A/(S −1 P )n . Proof. Consider the composite f : A → S −1 A/(S −1 P )n of the inclusion A → S −1 A and the canonical surjection S −1 A → S −1 A/(S −1 P )n . Note that (S −1 P )i = S −1 (P i ) for all i ∈ N (Proposition 13.9(i)). As S −1 (P n )∩A contains P n , its prime decomposition in A is P i for some i ≤ n (Theorem 14.8 (iii)). Then S −1 (P n ) = S −1 (P i ) by Proposition 12.3(i), hence the uniqueness of prime decomposition in S −1 A shows that i = n. Hence Ker f = (S −1 P )n ∩ A = P n .


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For any as ∈ S −1 A, there is no prime ideal of A containing both s ∈ / P and P n , hence n n (s) + P = A. Therefore there exists b ∈ A with a − sb ∈ P , and as a a − sb −b= ∈ S −1 (P n ) = (S −1 P )n , s s f (b) = as , i.e. f is surjective, hence S −1 A/(S −1 P )n = Im f ∼ = A/Ker f = A/P n by Theorem 2.8. ¤ Proposition 16.10. (i) e(QP /PP ) = e(Q/P ). ∼ (ii) k(PP ) ∼ k(P ), k(Q = P ) = k(Q). In particular f (QP /PP ) = f (Q/P ). e

e

Proof. (i): P B = Qe11 · · · Qgg and Proposition 13.9(i) gives (P B)P = (Q1 )eP1 · · · (Qg )Pg . (ii): AP /PP ∼ ¤ = A/P and BP /QP ∼ = B/Q by Lemma 16.9. By Proposition 16.10, as long as ramification indices and residue degrees are concerned, we can replace A by AP . Hence now we assume that A is DVR. By Proposition 15.15, we have rankk(P ) (B/P B) = [L : K]. 16.3. The quotient ring B/P B. 16.3.1. Chinese remainder theorem and its consequence. Definition 16.11. Let A be a ring and {Xi }i∈Λ be a family of A-algebras, indexed by the elements of a set Λ. The direct product ring (Definition 8.17) is an A-algebra by considering it as a direct sum of A-modules (Definition 15.1). This is called the direct product of A-algebras. Proposition 16.12. (Chinese remainder theorem) Let A be a Dedekind domain, P1 , . . . , Pn distinct non-zero prime ideals of A, and e1 , . . . , en ∈ N. Then: Q (i) The canonical mapQ f : A 3 a 7−→ (a, . . . , a) ∈ ni=1 A/Piei is a surjection. (ii) For the ideal I = i Piei , this f is the composite of the canonical surjection ∼ Q = A → A/I and the isomorphism A/I 3 a 7−→ (a, . . . , a) ∈ i A/Piei . Proof. (i): Let Ii = Piei . If Pi 6= Pj then there is no prime ideal containing both e Piei , Pj j , hence Ii + Ij = A. If we fix one i, then i 6= j ⇒ ∃aj ∈ Ii , bj ∈ Ij , aj + bj = 1. Q Q Now expanding j6=i (aj + bj ) = 1, let xi = bj and let yi be the sum of the rest of the terms. Then xi ∈ Ij for all j 6= i, andT1 − xi = yQ i ∈ Ii , i.e. f (xi ) = (0, . . . , 1, . . . , 0), hence f is a surjection. (ii): As Ker f = ni=1 Ii =Q ni=1 Ii = I, Theorem 2.8 gives the canonical isomorphism A/I = A/Ker f ∼ ¤ = Im f = A/Ii . Q e Applying (ii) to our decomposition P B = gi=1 Qei i = Qe11 · · · Qgg of P ∈ m-Spec(A): Qg e Proposition 16.13. We have B/P B ∼ = i=1 B/Qi i as k(P )-algebras.

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Qg Qg e e Proof. The ring isomorphism B/P B ∼ = B/( i=1 Qi i ) ∼ = i=1 B/Qi i of Proposition ei 16.12(ii) is a k(P )-algebra morphism (note that each B/Qi are naturally a vector space over A/P , as the action of P is 0 by P B ⊂ Qei i ). ¤ 16.3.2. π-adic expansion and its consequence. Lemma 16.14. Let A be a Dedekind domain and P its nonzero prime ideal. Fix a set X of representatives of A/P in A and π ∈ P \ P 2 . Then any x ∈ A/P n is uniquely P ¯ i with ai ∈ X. written as x = n−1 i=0 ai π Proof. By Lemma 16.9 we have A/P n ∼ = AP /(PP )n , and as AP is a DVR and π ∈ PP \ 2 (PP ) , we have PP = (π). Moreover X is also a set of representatives of AP /PP ∼ = A/P (Proposition 16.10(ii)) in AP . Hence any x ∈ A is written as x = a0 + yπ (a0 ∈ X, y ∈ AP ) uniquely, and writing y = a1 + zπ (a1 ∈ X, z ∈ AP ), we have x = a0 + a1 π + zπ 2 P i n n uniquely. Repeating, we obtain ai ∈ X with x − n−1 i=0 ai π ∈ (π ) = (PP ) in AP . ¤ Lemma 16.15. If e = e(Q/P ) and f = f (Q/P ), we have rankk(P ) B/Qe = ef . Proof. By Lemma 16.14, fixing a set X of representatives of B/Q in B, e

(B/Q) 3 (a0 , a1 , . . . , ae−1 ) 7−→ x =

e−1 X

ai π ¯ i ∈ B/Qe (ai ∈ X)

i=0

∼ gives the isomorphism = (B/Q)e as B/Q-vector spaces, therefore as A/P -vector e ∼ spaces we have B/Q = (B/Q)e ∼ ¤ = ((A/P )f )e = (A/P )ef . B/Qe

Applying the above lemma, Proposition 15.15 and Proposition 16.13 to the decompoe sition P B = Q1e1 · · · Qgg (and removing the assumption that A is DVR by Proposition 16.10), we obtain: Theorem 16.16. Let A be a Dedekind domain, K = Frac(A), L/K a finite extension and B the integral closure of A in L. Assume that B is a finite A-algebra. For P ∈ m-Spec(A), let Q1 , . . . , Qg be the prime ideals of B lying above P . Then: [L : K] =

g X

e(Qi /P )f (Qi /P ).

i=1

17. Equations over Rings of Integers — II. Theory of Hilbert We keep the assumptions of the last section: let A be a Dedekind domain, K = Frac(A), L/K a finite extension, and B the integral closure of A in L. In this section we assume in addition:


83

(i) For every P ∈ m-Spec(A), the residue field k(P ) is a perfect field. (ii) L/K is a Galois extension (hence B is a finite A-algebra), and let G = Gal(L/K). 17.1. Decomposition group and inertia group. 17.1.1. Galois group and prime ideals. We fix a nonzero prime ideal P of A, and let Q1 , . . . , Qg be the prime ideals lying over P . Proposition 17.1. (i) Every element of G permutes the set {Q1 , . . . , Qg }. (ii) For all i, j, there is σ ∈ G with σ(Qi ) = Qj . (iii) e = e(Qi /P ) and f = f (Qi /P ) are both independent of i. (iv) |G| = [L : K] = ef g. Proof. (i): A restriction σ|B of σ ∈ G maps B into B, because if x ∈ L is integral over A, then σ(x) is a root of the same monic with coefficients in A, hence integral over A. Therefore σ|B is a ring isomorphism of B into B (with inverse σ −1 |B ), and fixes the elements of A (i.e. an A-algebra automorphism of B). Hence if Q ∈ m-Spec(B) then σ(Q) = Q0 ∈ m-Spec(B) and Q ∩ A = Q0 ∩ A. (ii): Fix i. If there is Qj | P such that σ(Qi ) 6= Qj for all σ ∈ G, then by Chinese remainder theorem (Proposition 16.12(i)), Q there is x ∈ Qj with x ∈ / σ(Qi ) for all σ ∈ G. Then σ∈G σ(x) is invariant under the action of G, hence belongs to K, i.e. to Qj ∩ A = P because B ∩ K = A as A is integrally / Qi for all σ ∈ G and Qi is prime, we Q closed. On the other hand, as σ(x) ∈ have σ(x) ∈ / Qi , which contradicts Qi ∩ A = P . (iii): As an A-algebra automorphism of B maps Qi to Qj , we must have e(Qi /P ) = e(Qj /P ) and f (Qi /P ) = f (Qj /P ). (iv): Theorem 16.16. ¤ 17.1.2. Decomposition group and inertia group. By the above proposition we have P B = Qe1 · · · Qeg = (Q1 · · · Qg )e , and we fix the notation e, f, g as in the proposition. Now we fix a Q | P . As we assumed k(P ) to be perfect, k(Q)/k(P ) is a separable extension of degree f . Definition 17.2. The subgroup Z = {σ ∈ G | σ(Q) = Q} of G is called the decomposition group of Q. For σ ∈ Z and x, y ∈ B, writing x = x mod Q ∈ k(Q), x = y =⇒ x − y ∈ Q =⇒ σ(x) − σ(y) = σ(x − y) ∈ Q =⇒ σ(x) = σ(y), defines a field automorphism σ of k(Q) by σ(x) = σ(x). As σ fixes the elements of A, the map σ fixes the elements of k(P ), hence we have a group homomorphism Z 3 σ 7−→ σ = (σ|B mod Q) ∈ Autk(P ) (k(Q)). The subgroup T = Ker (σ 7→ σ) = {σ ∈ G | ∀x ∈ B σ(x) − x ∈ Q} of Z is called the inertia group of Q (σ = id ⇐⇒ σ(x) = x (∀x ∈ B) ⇐⇒ σ(x) − x ∈ Q (∀x ∈ B)). Proposition 17.3. (i) |G|/|Z| = g, hence |Z| = ef . (ii) k(Q)/k(P ) is Galois and Z 3 σ 7−→ σ ∈ Gal(k(Q)/k(P )) is surjective. (iii) Z B T, Z/T ∼ = Gal(k(Q)/k(P )). Hence |Z/T | = f, |T | = e.

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S Proof. (i): If we let GQ0 = {σ ∈ G | σ(Q) = Q0 }, then G = Q0 |P GQ0 , and for any τ ∈ GQ0 there is a bijection GQ0 3 σ 7−→ τ −1 σ ∈ Z, hence |GQ0 | = |Z|. (ii): Choose x ∈ k(Q) such that k(Q) = k(P )(x) by Theorem 6.4, and by Chinese remainder theorem (PropositionQ16.12(i)) we can choose x ∈ B such that x ∈ σ −1 (Q) for all σ ∈ / Z. Now let P (X) = σ∈G (X −σ(x)), the coefficients of P (X) are the symmetric polynomials of σ(x) ∈ B, hence in B and also in K (being invariant under the action of G), therefore in A, as A is integrally closed. Therefore P (X) ∈ k(P )[X], and having x as a root, P (X) is divisible by the minimal polynomial Px of x. Therefore all the roots of Px can be written as σ(x) = σ(x) ∈ k(Q), which shows that k(Q)/k(P ) is Galois. Now by the choice of x, all the nonzero roots of P (X) are of the form σ(x) with σ ∈ Z. Hence for all τ ∈ Gal(k(Q)/k(P )), there is a σ ∈ Z such that τ (x) = σ(x), and k(Q) = k(P )(x) shows that τ = σ. (iii): As T = Ker (σ 7→ σ), it follows from (ii) and Exercise 2.9. ¤ 17.1.3. Decomposition in intermediate fields. We can track the decomposition of prime ideals in the intermediate fields, as they decompose, inert (residue fields extend) and ramify. Definition 17.4. The intermediate field KZ (resp. KT ) of L/K which corresponds to Z (resp. T ) by Galois theory (Theorem 4.14) is called the decomposition field (resp. inertia field) of Q. We have K ⊂ KZ ⊂ KT ⊂ L. Proposition 17.5. (i) |G|/|Z| = [KZ : K] = g, |Z| = [L : KZ ] = ef . (ii) If we let P 0 = Q ∩ KZ , then P 0 B = Qe , f (Q/P 0 ) = f . (iii) |Z/T | = [KT : KZ ] = f, |T | = [L : KT ] = e. (iv) If we let P 00 = Q ∩ KT , then P 00 B = Qe , f (Q/P 00 ) = 1. Proof. (i),(iii): Proposition 17.3(i),(iii) and the Galois theory. (ii): The decomposition group of Q for the extension L/KZ is again Z, and for this extension we have G = Z, hence g = 1 by (i). Therefore Q is the unique prime ideal lying above P 0 , and if we let e0 = e(Q/P 0 ), f 0 = f (Q/P 0 ) then e0 f 0 = ef by (i), and moreover e0 | e, f 0 | f by P 0 | P (Exercise 16.8), hence e0 = e, f 0 = f . (iv): The inertia group of Q for the extension L/KT is also T , and for this extension G = T . Hence f (Q/P 00 ) = 1, e = e(Q/P 00 ) by (iii). ¤ 17.2. Frobenius automorphism and abelian extensions. 17.2.1. Frobenius automorphism. If P is unramified, we have T = {id} for all Q, and Z = Z/T ∼ = Gal(k(Q)/k(P )). If moreover k(P ) is a finite field Fq , by Theorem 5.28, Gal(k(Q)/k(P )) is a cyclic group generated by the Frobenius map Frq : x 7→ xq . Definition 17.6. The element FrQ ∈ Z such that FrQ = Frq is called the Frobenius automorphism of Q. Then Z ∼ = hFrQ i ∼ = Z/f Z and f is the order of FrQ in G. By definition, we have FrQ (x) = FrQ (x) = (x)q = xq for all x ∈ B, hence FrQ is characterized by the property FrQ (x) − xq ∈ Q for all x ∈ B.


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Exercise 17.7. In the situation of 16.8, if K 0 /K is Galois then FrQ |K 0 = FrP 0 . 17.2.2. The case of abelian extension. For a Q | P , let Z, T, FrQ as above, and for another Q0 | P and take σ ∈ G such that σ(Q) = Q0 by Proposition 17.1(ii). Proposition 17.8. The decomposition/inertia group and Frobenius automorphism of Q0 are given by: Z 0 = σZσ −1 , T 0 = σT σ −1 , FrQ0 = σ FrQ σ −1 . Proof. Compute as follows: τ ∈ Z ⇐⇒ τ (Q) = Q

⇐⇒

⇐⇒ στ σ −1 (Q0 ) = Q0

τ σ −1 (Q0 ) = σ −1 (Q0 ) ⇐⇒

στ σ −1 ∈ Z 0 ,

τ ∈ T ⇐⇒ τ (x) − x ∈ Q (∀x ∈ B) ⇐⇒ στ (x) − σ(x) ∈ Q0 (∀x ∈ B) ⇐⇒ στ σ −1 (x) − x ∈ Q0 (∀x ∈ B) ⇐⇒ στ σ −1 ∈ T 0 . For Frobenius automorphisms, FrQ (x) − xq ∈ Q, hence σ FrQ (x) − σ(x)q ∈ Q0 for all x ∈ B, which shows that σ FrQ σ −1 (x)−xq ∈ Q0 for all x ∈ B, i.e. σ FrQ σ −1 = FrQ0 . ¤ Corollary 17.9. If G is abelian, then the decompostion/inertia group and the Frobenius automorphism depend only on P . We call them the decomposition group, inertia group and Frobenius automorphism FrP of P .

18. Decomposition laws of primes 18.1. Cyclotomic Extensions of Q. Let A be a Dedekind domain and K = Frac(A). Fix n ≥ 1 such that (char K, n) = 1 and consider the cyclotomic extension L = K(µn ) of K (Definition 5.3). We choose a primitive n-th root of unity ζ ∈ µn , so that L = K(ζ). Let B be the integral closure of A in L. As the elements of µn , being the roots of X n − 1, are integral over A, we have µn ⊂ B. Q k Lemma 18.1. For a primitive n-th root of unity ζ, we have n−1 k=1 (1 − ζ ) = n. Proof. Let X = 1 in

Qn−1

k=1 (X

− ζk) =

X n −1 X−1

= X n−1 + X n−2 + · · · + X + 1.

¤

Lemma 18.2. (Good reduction of µn ) Let P ∈ m-Spec(A), assume (char k(P ), n) = 1, and let Q ∈ m-Spec(B) with Q | P . If ζ i − ζ j ∈ Q, then i ≡ j mod n, i.e. ζ i = ζ j . Proof. As ζ ∈ B × and ζ i − ζ j ∈ Q =⇒ ζ i−j − 1 ∈ Q, it suffices to show ζ i − 1 ∈ / Q for Qn−1 1 ≤ i ≤ n − 1. Reducing Lemma 18.1 mod Q, we see that i=1 (1 − ζ i ) 6= 0 in the field ¤ k(Q) = B/Q, because n ∈ / Q. Hence ζ i − 1 6= 0 in k(Q). Note that K(µn )/K is abelian (Theorem 5.22), and Corollary 17.9 applies.

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Proposition 18.3. Let P ∈ m-Spec(A), assume (char k(P ), n) = 1 and k(P ) ∼ = Fq . Recall the injection of Theorem 5.22: Gal(K(µn )/K) 3 (ζ 7→ ζ a ) 7−→ a ∈ (Z/(n))× . Then P is unramified in L = K(µn ), and its Frobenius automorphism FrP is (ζ 7→ ζ q ). Proof. (i): Let Z be the decomposition group of P , choose a prime Q | P of B, and consider the surjection of Proposition 17.3(ii): Z 3 σ = (ζ 7→ ζ a ) 7−→ (σ|B mod Q) ∈ Gal(k(Q)/k(P )). If ζ 7→ ζ a is in the kernel of this map, then ζ − ζ a ∈ Q, therefore a = 1 ∈ (Z/(n))× by Lemma 18.2. Hence the inertia group is trivial and P is unramified in L. Also, the Frobenius automorphism FrP ∈ Gal(k(Q)/k(P )) is equal to ζ 7→ ζ q , because it maps ζ q to ζ = ζ q and ζ a = ζ q implies a = q ∈ (Z/(n))× by Lemma 18.2. ¤ Now let A = Z, K = Q and P = (p). Theorem 18.4. Let n ≥ 1. Recall the injection of Theorem 5.22 for K = Q: Gal(Q(µn )/Q) 3 (ζ 7→ ζ a ) 7−→ a ∈ (Z/(n))× . (i) Any prime (p) of Z with (p, n) = 1 is unramified in Q(µn ), and its Frobenius automorphism Fr(p) is (ζ 7→ ζ p ). (ii) (Irreducibility of cyclotomic polynomials) The above map is an isomorphism: Gal(Q(µn )/Q) ∼ = (Z/(n))× , hence [Q(µn ) : Q] = |(Z/(n))× | = ϕ(n). Proof. (i) follows from the above proposition. (ii): As Fr(p) maps to p ∈ (Z/(n))× for all p prime to n and (Z/(n))× is generated by p for such p, the map is surjective. ¤ As the residue degree f (Q/(p)) = |Z| is the order of Fr(p) in the group Gal(Q(µn )/Q) ∼ = × (Z/(n)) of order ϕ(n), we obtain: Corollary 18.5. Let p be a prime not dividing n. If f is the minimal integer satisfying pf ≡ 1 mod n and f g = ϕ(n), then the prime ideal decomposition of (p) in the ring of integers of Q(µn ) is of the form: (p) = P1 P2 · · · Pg . In particular, p decomposes completely (e = f = 1) in Q(µn ) if and only if p ≡ 1 mod n. Exercise 18.6. Find the decomposition law of unramifed primes in Q(µ12 )/Q. Finally we identify the ramified primes in Q(µn ). Note that µm ⊂ µn for m | n. As Q(µ2n ) = Q(µn ) for odd n, we can assume n is either odd or divisible by 4. Proposition 18.7. Assume n is either odd or divisible by 4. All primes p dividing n ramify in Q(µn ).


87

Proof. First assume p is odd. As Q(µp ) ⊂ Q(µn ), it suffices to show that (p) ramifies in Q i Q(µp ), by Exercise 16.8. For a primitive p-th root of unity ζ, we have p = p−1 i=1 (1 − ζ ) by Lemma 18.1. For each 1 ≤ i ≤ p − 1, if we take j such that ij ≡ 1 mod p, then: 1 − ζ i = (1 − ζ)(1 + ζ + ζ 2 + · · · + ζ i−1 ), 1 − ζ = 1 − ζ ij = (1 − ζ i )(1 + ζ i + ζ 2i + · · · + ζ (j−1)i ), hence 1 − ζ and 1 − ζ i are associate in the ring of integers A of Q(µp ). Hence (p) = (1 − ζ)p−1 as ideals of A, therefore (p) ramifies because p − 1 > 1. (Actually this is the prime ideal decomposition of (p), by [Q(µp ) : Q] = ϕ(p) = p − 1 and Theorem √ 16.16.) For p = 2, it suffices to√show that (2) ramifies in Q(µ4 ) =√Q( −1) because Q(µ4 ) ⊂ Q(µn ), but (2) = (1 − −1)2 in the ring of integers of Q( −1). ¤ 18.2. Quadratic reciprocity law. 18.2.1. Quadratic fields and quadratic residues. Definition 18.8. Let p be an odd prime. An element a = a mod p ∈ F× p is called × 2 a quadratic residue mod p if there is an x ∈ Fp such that x = a. The set of all quadratic residues Qp is a subgroup of F× p of order (p − 1)/2. We define the Legendre symbol by: ( ³a´ 1 (a ∈ Qp ), = p −1 (a ∈ / Qp ). Exercise 18.9. For a, b ∈ F× p , we have

³ ab ´ p

=

³ a ´³ b ´ p

p

.

A quadratic extension is always Galois in charactersitic zero (Exercise 4.12(i)), hence by Proposition 17.1(iv) a prime decomposes as either e = 2, f = 2 or g = 2. Proposition 18.10. √Let d ∈ Z be square free. A √ prime p, which does not divide 2d and unramified in Q( d), decomposes in K = Q( d) as follows: ³d´ ³d´ (p) = P1 P2 ⇐⇒ = 1, (p) = P ⇐⇒ = −1. p p √ √ x2 = d, then p | x2 − d = (x + d)(x − d) in OK , Proof. ⇐: If there is an x ∈ F× p with √ √ but p does not divide either of x + d, x − d in OK , hence is not a prime element of OK and (p) is not a prime ideal of OK (Exercise 1.18(iii)). √ 2 ⇒: If there is no x ∈ F× d mod Q ∈ OK /Q, for a prime Q of p with x = d, then OK lying above (p), does not belong to Fp , hence f (Q/(p)) > 1. ¤

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Exercise 18.11. Comparing the above proposition with Corollary 18.5 for Q(µ4 ) = √ Q( −1), we obtain the first complementary law: If p is an odd prime, then: ( ³ −1 ´ p−1 1 (p ≡ 1 (mod4)), = (−1) 2 = p −1 (p ≡ 3 (mod4)). √ Exercise 18.12. If ζ ∈ Q(µ ) is a primitive 8-th root of unity, then 2 = ζ +ζ 7 , hence 8 √ ∼ Q( 2) corresponds to the subgroup {1, 7} of Gal(Q(µ8 )/Q) = (Z/(8))× = {1, 3, 5, 7}. Using this and Exercise 17.7, and comparing Proposition 18.10 with Corollary 18.5, we obtain the second complementary law: If p is an odd prime, then: ( ³2´ p2 −1 1 (p ≡ 1, 7 (mod8)), = (−1) 8 = p −1 (p ≡ 3, 5 (mod8)). 18.2.2. Gauss sums. Let ` be an odd prime. As Gal(Q(µ` )/Q) ∼ = (Z/(`))× = F× ` is a cyclic group by Proposition 5.26(ii), its subgroup of order (` − 1)/2 is unique and must coincide with Q` ⊂ F× ` by Exercise 5.7(ii). Let K` be the quadratic field corresponding to Q` by Galois theory. Then Corollary 18.5 and Exercise 17.7 shows that: Proposition 18.13. In the unique quadratic field K` contained in Q(µ` ), a prime p 6= ` decomposes as: ³p´ ³p´ (p) = P1 P2 ⇐⇒ = 1, (p) = P ⇐⇒ = −1. ` ` ( √ Q( `) (` ≡ 1 (mod4)), √ Lemma 18.14. K` = Q( −`) (` ≡ 3 (mod4)). P Proof. Let ζ ∈ Q(µ` ) be a primitive `-th root of unity. Setting α = a∈Q` ζ a and β = P P`−1 k b k=1 ζ = −1 and α, β ∈ K` . The nontrivial b∈Q / ` ζ (Gauss sum), we have α+β = × ∼ element σ of Gal(K` /Q) = {id, σ} = F` /Q` interchanges α and β, hence αβ ∈ Q. P P Expanding αβ, we obtain a sum of (` − 1)2 /4 terms (a,b) ζ a+b = k∈Z/(`) ck ζ k = P c0 + k∈F× ck ζ k (ck ≥ 0). This is invariant under the action of Gal(Q(µ` )/Q), hence `

2 all ck for k ∈ F× ` are identical, therefore c0 + (` − 1)c1 = (` − 1) /4, αβ = c0 − c1 . If ` ≡ 1 (mod4), then a ∈ Q` ⇔ −a ∈ Q` (Exercise 18.9, 18.11), hence c0 = 0, c1 = (` − 1)/4, αβ = −(` − 1)/4. If ` ≡ 3 (mod4), then similarly a ∈ Q` ⇔ −a ∈ / Q` , hence c0 = (` − 1)/2, c1 = (` − 3)/4, αβ = (` + 1)/4. Therefore, ( ` (` ≡ 1 (mod4)), (β − α)2 = (α + β)2 − 4αβ = −` (` ≡ 3 (mod4)). √ √ Accordingly ±` ∈ K` , hence Q( ±`) = K` . ¤

Comparing Proposition 18.13 and Proposition 18.10 for K` , and using Exercise 18.9, 18.11 when ` ≡ 3 (mod4), we obtain:


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Corollary 18.15. (Quadratic reciprocity law) For odd primes p, `: ³ ´ ³ ´ p `   = (p or ` ≡ 1 (mod4)), ³ p ´³ ` ´ p−1 `−1 ` p = (−1) 2 · 2 , i.e., ³p´ ³`´  ` p  =− (p ≡ ` ≡ 3 (mod4)). ` p

Preliminaries II: Categories and Functors i. Additive Categories Recall the definitions from §e.4. When a particular isomorphism between two objects X, Y ∈ C is designated, we often say that X, Y are canonically isomorphic, and call the designated isomorphism the canonical isomorphism. i.1. Direct products and direct sums. Definition i.1. Let C be a category. For X, Y ∈ C , a triple (X × Y, pX , pY ) with X × Y ∈ C and pX ∈ HomC (X × Y, X), pY ∈ HomC (X × Y, Y ) is called a direct product of X and Y if HomC (T, X × Y ) 3 f 7−→ (pX ◦ f, pY ◦ f ) ∈ HomC (T, X) × HomC (T, Y ) Q is a bijection for all T ∈ C . The direct product X1 × · · · × Xn is denoted by ni=1 Xi . Exercise i.2. (i) This definition can be restated as follows: for any fX , fY as below, there exists a unique morphism f which makes the diagram commutative: pX pY / X ×O Y X ocGG ;Y GG ww w GG f w G ww ww fY fX GGG w w T

(ii) The direct product of X an Y , if exists, then they are all canonically isomorphic to each other. (We say that they are unique up to canonical isomorphism.) (iii) In the category Sets, the direct product of sets and the projection maps give the direct sum in the above sense: X × Y := {(x, y) | x ∈ X, y ∈ Y }, pX : X × Y 3 (x, y) 7→ x ∈ X,

pY : X × Y 3 (x, y) 7→ y ∈ Y.

(iv) The direct products of Groups, Rings (Definition 8.17), A-Alg (Definition 16.11), and the direct sums in VectK (Definition h.1), A-Mod (Definition 15.1) give direct products in the above sense in each of the categories. In each of these examples, the underlying set of the direct product is the direct product set, and the morphisms (pX , pY ) are the projection maps.

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Definition i.3. For X, Y ∈ C , a triple (X q Y, iX , iY ) with X q Y ∈ C and iX ∈ HomC (X, X q Y ), iY ∈ HomC (Y, X q Y ) is called a direct sum of X and Y if: HomC (X q Y, T ) 3 f 7−→ (f ◦ iX , f ◦ iY ) ∈ HomC (X, T ) × HomC (Y, T ) ` is a bijection for all T ∈ C . The direct sum X1 q · · · q Xn is denoted by ni=1 Xi . Exercise i.4. (i) This definition can be restated as follows: for any fX , fY as below, there exists a unique morphism f which makes the diagram commutative: iX

/ X q Y o iY Y w GG ww GG f www G GG ² ww fY fX # {w

X GG

T

(ii) The direct sum of X an Y , if exists, is unique up to a canonical isomorphism. (iii) In the category Sets, the disjoint union of sets and the projection maps give the direct sum in the above sense: X q Y := {(x, i) | i ∈ {X, Y }, x ∈ i}, iX : X 3 x 7→ (x, X) ∈ X q Y,

iY : Y 3 y 7→ (y, Y ) ∈ X q Y.

(iv) The direct sums in VectK (Definition h.1), A-Mod (Definition 15.1) give direct sums in the above sense in each of the categories. The underlying set of the direct sum is the direct product set, and the morphisms (iX , iY ) are given by: iX : X 3 x 7→ (x, 0) ∈ X ⊕ Y,

iY : Y 3 y 7→ (0, y) ∈ X ⊕ Y.

We can extend the above definition to not necessarily finite family of objects. Definition i.5. Let Λ be a set, and {Xi }i∈Λ be a family of objects in C , indexed by the elements of Λ. ¡Q ¢ Q Q (i) A pair i∈Λ Xi , {pi }i∈Λ with i∈Λ Xi ∈ C and pi ∈ HomC ( i∈Λ Xi , Xi ) is called a direct product of {Xi } if the following map is bijective for all T ∈ C : ³ Y ´ Y HomC T, Xi 3 f 7−→ (pi ◦ f )i∈Λ ∈ HomC (T, Xi ) i∈Λ

¡`

¢

i∈Λ

`

` (ii) A pair i∈Λ Xi ∈ C and ii ∈ HomC (Xi , i∈Λ Xi ) is i∈Λ Xi , {ii }i∈Λ with called a direct sum) of {Xi } if the following map is bijective for all T ∈ C : ³a ´ Y HomC Xi , T 3 f 7−→ (f ◦ ii )i∈Λ ∈ HomC (Xi , T ) i∈Λ

i∈Λ

Exercise i.6. In the category Sets, direct products and direct sums (disjoint unions) are constructed in the similar way: Y Y Xi := {(xi )i∈Λ | xi ∈ Xi }, pi : Xi 3 (xi )i 7→ xi ∈ Xi , a i∈Λ

i∈Λ

Xi := {(x, i) | i ∈ Λ, x ∈ Xi },

i∈Λ

ii : Xi 3 x 7→ (x, i) ∈

a i∈Λ

Xi .


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i.2. Additive categories. Definition i.7. An object X ∈ C is called a zero object of C , denoted by X = 0, if it satisfies |HomC (X, T )| = |HomC (T, X)| = 1 for ∀T ∈ C . Exercise i.8. (i) An object isomorphic to a zero object is also a zero object. (ii) If a zero object exists, it is unique up to a canonical isomorphism. (iii) The 0 in VectK (Example b.22(i)), A-Mod (Example b.27(i)) give zero objects in each of the categories. When there is a zero object in C , we denote the unique element of HomC (0, X), HomC (X, 0) by 0. More generally, for any X, Y ∈ C , we denote the image 0 ◦ 0 of HomC (X, 0) × HomC (0, Y ) −→ HomC (X, Y ) by 0 ∈ HomC (X, Y ). This element is independent of the choice of zero object, and it is called the zero morphism from X to Y . Exercise i.9. For X ∈ C , we have idX = 0 ∈ HomC (X, X) ⇐⇒ X = 0. Definition i.10. A category C is called an additive category if: (i) The zero object 0 ∈ C exists; (ii) For any X, Y ∈ C , the set HomC (X, Y ) has an additive group structure with the zero morphism 0 as the zero element; (iii) g ◦ (f1 + f2 ) = g ◦ f1 + g ◦ f2 ; (iv) (g1 + g2 ) ◦ f = g1 ◦ f + g2 ◦ f ; (v) The direct sum X q Y exists for all X, Y ∈ C . We write X ⊕ Y for X q Y . Exercise i.11. For a ring A, the category A-Mod (in particular VectK = K-Mod when K is a field) is an additive category, because if we define the addition of Ahomomorphisms by (f1 + f2 )(x) := f1 (x) + f2 (x), then it is distributive under composition and HomA (X, Y ) is an additive group with the zero map 0 (defined as 0(x) = 0) as the zero element. The most basic example is the category of abelian groups Ab = Z-Mod. Proposition i.12. Let C be an additive category. For X, Y ∈ C , the object X ⊕Y ∈ C and the diagram Xo

iX pX

/

X ⊕Y

o iY / Y pY

gives the direct sum of X, Y if and only if the following identities hold: pX ◦ iX = idX , pY ◦ iY = idY , pX ◦ iY = 0, pY ◦ iX = 0, iX ◦ pX + iY ◦ pY = idX⊕Y . Proof. If it is the direct sum, by the bijection: ϕ : HomC (X ⊕ Y, T ) 3 f 7−→ (f ◦ iX , f ◦ iY ) ∈ HomC (X, T ) × HomC (Y, T ),

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the morphisms pX , pY are determined by: HomC (X ⊕ Y, X) 3 pX 7−→ (idX , 0) ∈ HomC (X, X) × HomC (Y, X), HomC (X ⊕ Y, Y ) 3 pY 7−→ (0, idY ) ∈ HomC (X, Y ) × HomC (Y, Y ), hence the first 4 indentities. Then, as ϕ(iX ◦ pX + iY ◦ pY ) = (iX , iY ) = ϕ(idX⊕Y ) and ϕ is injective: iX ◦ pX + iY ◦ pY = idX⊕Y . Conversely, if iX , iY , pX , pY satisfying the 5 identities are given, then the inverse map of ϕ is given by: ψ : (fX , fY ) 7−→ fX ◦ pX + fY ◦ pY , hence ϕ is bijective: ϕ ◦ ψ(fX , fY ) = ((fX ◦ pX + fY ◦ pY ) ◦ iX , (fX ◦ pX + fY ◦ pY ) ◦ iY ) = (fX , fY ), ψ ◦ ϕ(f ) = f ◦ iX ◦ pX + f ◦ iY ◦ pY = f ◦ idX⊕Y = f. ¤ Exercise i.13. Via the above map pX , pY , for all T ∈ C there is a bijection: HomC (T, X ⊕ Y ) 3 f 7−→ (pX ◦ f, pY ◦ f ) ∈ HomC (T, X) × HomC (T, Y ). This means that X ⊕ Y is the direct product in C in the sense of Definition i.1. i.3. Example: Modules over non-commutative rings. Here we extend the notion of modules to the non-commutative rings. Let A be a (possibly non-commutative) ring. We need to distinguish the A-action from the left and that from the right. Definition i.14. (i) When an additive group X has an action of A from the left A × X 3 (a, x) 7−→ ax ∈ X and satisfies the following, X is called a left A-module: (a) 1x = x, (b) a(bx) = (ab)x, (c) (a + b)x = ax + bx, (d) a(x + y) = ax + ay. (ii) When an additive group X has an action of A from the right X ×A 3 (x, a) 7−→ xa ∈ X and satisfies the following, X is called a right A-module: (a) x1 = x, (b) (xa)b = x(ab), (c) x(a + b) = xa + xb, (d) (x + y)a = xa + ya. Exercise i.15. When A is commutative, these two notions are equivalent to each other. For left (right) A-modules, A-submodules, quotient A-modules, zero module, direct sum of modules are defined in exactly the same way as before.


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Definition i.16. For left (resp. right) A-modules X, Y , a homomorphism of additive groups f : X → Y is called an A-homomorphism if it satisfies: f (ax) = af (x) (∀a ∈ A, ∀x ∈ X) (resp. f (xa) = f (x)a (∀a ∈ A, ∀x ∈ X)). The homomorphism theorems (Theorem 2.8) for left (right) A-modules are proved in exactly the same way. Let HomA (X, Y ) be the set of all A-homomorphisms from X to Y . There is an identity and composite homomorphisms. The zero module 0 has the property of the zero object, the direct sum of modules has the property of direct sum of objects. Defining the addition in HomA (X, Y ) by (f1 + f2 )(x) = f1 (x) + f2 (x), which is distributive under composition, the HomA (X, Y ) becomes an additive group with the zero map 0 (defined as 0(x) = 0) as the zero element. Therefore: Definition i.17. We have an additive category A-Mod (resp. Mod-A), category of left (resp. right) A-modules, with left (resp. right) A-modules as objects and A-homomorphisms as morphisms. j. Functors j.1. Functors. Definition j.1. A functor F : C 3 X 7−→ F (X) ∈ D from a category C to a category D is a correspondence satisfying the following conditions: (i) For each X ∈ C there is a F (X) ∈ D, (ii) For all X, Y ∈ C , we have a map: F : HomC (X, Y ) 3 f 7−→ F (f ) ∈ HomD (F (X), F (Y )), (iii) F (idX ) = idF (X) , (iv) F (g ◦ f ) = F (g) ◦ F (f ). If instead of conditions (ii),(iv) we have: (ii) For all X, Y ∈ C , we have a map: F : HomC (X, Y ) 3 f 7−→ F (f ) ∈ HomD (F (Y ), F (X)), (iv) F (g ◦ f ) = F (f ) ◦ F (g), then we call F a contravariant functor (the word ‘contravariant’ is often omitted). Example j.2. (i) For two functors F : C → D, G : D → E , we can define the composite functor G ◦ F : C → E . (ii) For any category C , there is the identity functor idC : C → C defined by idC (X) = X, idC (f ) = f .

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(iii) A functor F : C → D is called an isomorphism of categories if there is a functor G : D → C such that G ◦ F = idC and F ◦ D = idD . Then F, G are called the inverse of each other. In this case, F : HomC (X, Y ) → HomD (F (X), F (Y )) is a bijection for all X, Y ∈ C , i.e., F is fully faithful). Example j.3. Let C be a category. For any object X ∈ C , there are functors: HomC (X, −) : C 3 T 7−→ HomC (X, T ) ∈ Sets, HomC (−, X) : C 3 T 7−→ HomC (T, X) ∈ Sets. For each f ∈ HomC (S, T ), the corresponding morphisms are given by: f∗ = HomC (X, f ) : HomC (X, S) 3 ϕ 7−→ f ◦ ϕ ∈ HomC (X, T ), f ∗ = HomC (f, X) : HomC (T, X) 3 ϕ 7−→ ϕ ◦ f ∈ HomC (S, X). Example j.4. We can consider groups as sets by forgetting the group structure, and this gives the forgetful functor − : Groups 3 X 7→ X ∈ Sets. Similarly we have forgetful functors − : Rings → Groups, − : A-Alg → A-Mod etc. j.2. Additive functors. Definition j.5. A functor (resp. contravariant functor) F : C → D between additivie categories is called an additive functor (resp. additive contravariant functor) if the map: F : HomC (X, Y ) 3 f 7−→ F (f ) ∈ HomD (F (X), F (Y )) (resp. F : HomC (X, Y ) 3 f 7−→ F (f ) ∈ HomD (F (Y ), F (X))) is a homomorphism of additive groups for all X, Y ∈ C . Proposition j.6. (i) Any additive functor F sends the zero object to the zero object: F (0) = 0. (ii) Any additive functor F sends direct sums to direct sums, i.e., canonically: F (X) ⊕ F (Y ) ∼ = F (X ⊕ Y ). Proof. (i): Exercise i.9. (ii) The five identities of Proposition i.12 are preserved by an additive functor. ¤ Example j.7. For an additive category C and all X ∈ C , HomC (X, −) : C 3 T 7−→ HomC (X, T ) ∈ Z-Mod, HomC (−, X) : C 3 T 7−→ HomC (T, X) ∈ Z-Mod are additive functors. (Note the distrubutive law for composite morphisms.) j.3. Morphism of functors. Let C , D be categories. Definition j.8. For two functors F, G from C to D, a morphism of functors α : F → G is a correspondence such that: (i) For each X ∈ C , there is α(X) ∈ HomD (F (X), G(X)).


95

(ii) For each f ∈ HomC (X, Y ), we have α(Y )◦F (f ) = G(f )◦α(X), i.e. the following diagram in D is commutative: F (X) α(X)

²

G(X)

F (f )

G(f )

/ F (Y ) ²

α(Y )

/ G(Y )

Exercise j.9. (i) For two morphisms α : F → G, β : G → H of functors, the composite β ◦ α : F → H is defined by (β ◦ α)(X) = β(X) ◦ α(X). (ii) For any functor F : C → D, there is the identity idF : F → F is defined by idF (X) = idF (X) . For any α : F → G, we have α ◦ idF = idG ◦ α = α. (iii) When D is an additive category, for F : C → D and all n ∈ Z, the multiplication by n of F is defined by n = n · idF : F → F . Definition j.10. A morphism of functors α : F → G is called an isomorphism of functors if β ◦ α = idF , α ◦ β = idG for some β : G → F . In this case the functors F, G are called isomorphic. Remark j.11. When we can specify a particular isomorphism, we say F, G are canonically isomorphic and the specified isomorphism is called a canonical isomorphism. Exercise j.12. (Yoneda’s lemma) For a morphism f ∈ HomC (X, Y ) in C , we have morphisms between the functors (Example j.3, j.7): α = f∗ : HomC (−, X) −→ HomC (−, Y ) β = f ∗ : HomC (Y, −) −→ HomC (X, −) by defining f∗ (T ) = f∗ , f ∗ (T ) = f ∗ for all T ∈ C . Conversely, all morphisms α, β between these functors are of this form for morphisms f = α(idX ), f = β(idY ), i.e., α = α(idX )∗ , β = β(idY )∗ . Definition j.13. A functor (resp. contravariant functor) F : C → D is called representable if it is isomorphic to the functor HomC (X, −) (resp. HomC (−, X)) for an X ∈ C (Example j.3, j.7). We say that F is represented by the object X, which is unique up to isomorphism by Yoneda’s lemma (Exercise j.12). Exercise j.14. (i) A direct product X × Y represents the contravariant functor HomC (−, X) × HomC (−, Y ) : C −→ Sets, and a direct sum X q Y represents the functor HomC (X, −) × HomC (Y, −) : C −→ Sets. (ii) For a field K and an irreducible P ∈ K[X], the field KP := K[X]/(P ) represents the functor F 7−→ RootP (F ) from the category of finite extensions of K to Sets (Proposition 4.9). (This suggests the interpretation of HomK-Alg (KP , F ) for any F ∈ K-Alg as “the set of roots of P in F ”.) j.4. Adjoint functors. Definition j.15. For functors F : C → D, G : D → C , if there are bijections of sets: τXY : HomD (F (X), Y ) −→ HomC (X, G(Y ))

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for all X ∈ C and Y ∈ D, which give isomorphisms of functors D −→ Sets: τX : HomD (F (X), −) −→ HomC (X, G(−)) for all X ∈ C , and isomorphisms of functors C −→ Sets: τY : HomD (F (−), Y ) −→ HomC (−, G(Y )) for all Y ∈ D, then F, G are called the adjoint of each other. We say F is the left adjoint functor of G and G is the right adjoint functor of F . Example j.16. (i) Consider the forgetful functor from the category Fields of fields to the category Dom of integral domains (the morphisms are ring homomorphisms in both cases). Then its left adjoint is given by Frac : Dom 3 A 7−→ Frac(A) ∈ Fields (see Exercise 10.3). (ii) Let A beP a ring. For a set X, consider the set of all formal linear combinations hXi := { ni=1 ai xi | xi ∈ X, n ∈ N} and define the obvious A-module structure on it. This A-module hXi is called the free A-module generated by X, contains X as a subset and is a free A-module generated by X in the sense of Definition 9.1. Then Sets 3 X 7−→ hXi ∈ A-Mod is the left adjoint of the forgetful functor − : A-Mod → Sets. (iii) Let A be a ring. For a set S, consider the ring A[S] of polynomials of elements in S considered as variables, with coefficients in A (see proof of Theorem 8.11). Then Sets 3 S 7−→ A[S] ∈ A-Alg is the left adjoint of the forgetful functor − : A-Alg → Sets. Example j.17. For two categories C , D, define their product C × D as a category of pairs (X, Y ), where X ∈ C and Y ∈ D, as objects, and ¡ ¢ HomC ×D (X, Y ), (X 0 , Y 0 ) := HomC (X, X 0 ) × HomD (Y, Y 0 ). This comes with the projection functors PC : C × D 3 (X, Y ) 7−→ X ∈ C ,

PD : C × D 3 (X, Y ) 7−→ Y ∈ D.

If we define a functor (−, −) : C 3 T 7−→ (T, T ) ∈ C × C (we send a morphism f to (f, f )), then the isomorphisms ¡ ¢ ∼ = HomC (X q Y, T ) −→ HomC ×C (X, Y ), (T, T ) , ¡ ¢ ∼ = HomC (T, X × Y ) −→ HomC ×C (T, T ), (X, Y ) , show that the direct sum q : C × C 3 (X, Y ) 7−→ X q Y ∈ C is the left adjoint of (−, −), and the direct product × : C × C 3 (X, Y ) 7−→ X × Y ∈ C is the right adjoint of (−, −). Composing with the projection functors, we get morphisms of functors HomC (X q Y, −) −→ HomC (X, −), HomC (X q Y, −) −→ HomC (Y, −), HomC (−, X × Y ) −→ HomC (−, X), HomC (−, X × Y ) −→ HomC (−, Y ), which give the canonical morphisms iX , iY , pX , pY by Yoneda’s lemma (Exercise j.12).


97

Exercise j.18. (i) The functoriality of τXY is equivalent to the commutativity of the following diagram for all f ∈ HomC (X, X 0 ) and all g ∈ HomD (Y, Y 0 ): HomD (F (X 0 ), Y ) τX 0 Y

²

F (f )∗

∼ =

τXY

HomC (X 0 , G(Y ))

f∗

g∗

/ HomD (F (X), Y ) ²

τXY 0 ∼ =

∼ =

/ HomC (X, G(Y ))

/ HomD (F (X), Y 0 )

G(g)∗

² / HomC (X, G(Y 0 ))

(ii) If F : C → D is an isomorphism of categories with an inverse G, then G : HomD (F (X), Y ) −→ HomC (X, G(Y )) is a bijection because of the fullyfaithfulness of G (Example j.2(iii)), and is functorial by the definition of functors, hence F, G are adjoint of each other. In this case F, G are symmetric. Proposition j.19. (i) An adjoint functor of a given functor is unique up to canonical isomorphism if it exists. (ii) If F : C → D (resp. F 0 : D → E ) is the left adjoint of G : D → C (resp. G0 : E → D), then F 0 ◦ F : C → E is the left adjoint of G ◦ G0 : E → C . Proof. (i) We prove for the right adjoint; the other case is proven in the same way. Let G, G0 be both right adjoint functors of F , and for each Y ∈ D, let τY , τY0 be the isomorphisms defined by G, G0 . Then by Exercise j.12, there is a morphism α(Y ) ∈ HomC (G(Y ), G0 (Y )) such that τY0 ◦ τY−1 = α(Y )∗ as morphisms of functors (the composite of upper horizontal arrows of the below diagram). Now in the following commutative diagram obtained by Exercise j.18(i), starting from id ∈ HomC (G(Y ), G(Y )), we have G0 (f ) ◦ α(Y ) = α(Y 0 ) ◦ G(f ), hence we get a morphism of functors α : G → G0 : HomC (−, G(Y )) o G(f )∗

²

HomC (−, G(Y 0 )) o

∼ = τY ∼ = τY 0

HomD (F (−), Y ) ²

∼ = τY0

/ HomC (−, G0 (Y ))

f∗

HomD (F (−), Y 0 )

∼ = τY0 0

²

G0 (f )∗

/ HomC (−, G0 (Y 0 ))

As τY , τY0 are isomorphisms for each Y , by interchanging τ, τ 0 and doing the same construction, we get the inverse morphism of α. Therefore G ∼ = G0 . (ii) By the functorial isomorphisms defining the adjointness: τ

τ

XY YZ HomD (F (X), Y ) −→ HomC (X, G(Y )), HomE (F 0 (Y ), Z) −→ HomD (Y, G0 (Z))

we obtain the functorial isomorphism τXZ : τF (X)Z

τXG0 (Z)

HomE (F 0 ◦ F (X), Z) −→ HomD (F (X), G0 (Z)) −→ HomC (X, G ◦ G0 (Z)), which shows that F 0 ◦ F is the left adjoint of G ◦ G0 .

¤

Exercise j.20. When C , D are additive categories and F is an additive functor, then G is also an additive functor, and τXY are isomorphisms of additive groups, i.e. for each X and each Y , they give isomorphisms between functors to Z-Mod.

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k. Direct Limits and Inverse Limits k.1. Direct and inverse limits. Definition k.1. An ordered set Λ is called a directed set if it satisfies: (iv) ∀x, y ∈ Λ ∃z ∈ Λ, x ≤ z, y ≤ z. Definition k.2. Let Λ be a directed set. In a category C , let {Ai }i∈Λ be a family of objects indexed by the elements of Λ. (i) If a morphism ϕji ∈ HomC (Ai , Aj ) is defined for all i, j ∈ Λ with satisfies ϕkj ◦ ϕji = ϕki for any triple i ≤ j ≤ k, the pair {Ai , ϕji } of objects and morphisms is called a direct system. (ii) If a morphism ϕij ∈ HomC (Aj , Ai ) is defined for all i, j ∈ Λ with satisfies ϕij ◦ ϕjk = ϕik for any triple i ≤ j ≤ k, the pair {Ai , ϕij } of objects and morphisms is called an inverse system.

i ≤ j and of families i ≤ j and of families

They are often denoted simply by {Ai }, omitting the family of morphisms. The directed set Λ is called the index set of a direct/inverse system. Example k.3. An inverse system with index set N is given is for each i ∈ N we have an object Ai and a morphism Ai+1 → Ai . The family {Z/(pi )}i∈N\{0} is an inverse system in Rings, with the canonical surjection Z/(pi+1 ) → Z/(pi ). Definition k.4. For two direct systems {Ai , ϕji }, {Bi , ψji } (resp. inverse systems {Ai , ϕij }, {Bi , ψij }) in C with the same index set Λ, a morphism from {Ai } to {Bi } is a family {fi } of morphisms fi ∈ HomC (Ai , Bi ) such that ψji ◦ fi = fj ◦ ϕji

(resp. ψij ◦ fj = fi ◦ ϕij )

for all i, j ∈ Λ with i ≤ j. We define {fi } ◦ {gi } = {fi ◦ gi }. We denote the category of direct (resp. inverse) systems in C with the index set Λ by DirΛ (C ) (resp. InvΛ (C )). If C is an additive category, then DirΛ (C ) and InvΛ (C ) are also additive categories with the zero object {0}, {Ai } ⊕ {Bi } := {Ai ⊕ Bi }, and {fi } + {gi } := {fi + gi }. Usually we fix Λ and omit Λ from the notation. k.2. Direct limits and inverse limits. For B ∈ C , we define a direct (resp. inverse) system {B} by Bi = B, ϕji = idB (resp. ϕij = idB ) for all i, j. Then B 7→ {B} gives a functor from C to Dir(C ) (resp. Inv(C )), which is additive if C is additive. Definition k.5. (i) If there exists a left adjoint of {−} : C 3 B 7−→ {B} ∈ Dir(C ), it is called the direct limit and denoted by: lim : Dir(C ) 3 {Ai } 7−→ lim Ai ∈ C . −→

−→

(ii) If there exists a right adjoint of {−} : C 3 B 7−→ {B} ∈ Inv(C ), it is called the inverse limit and denoted by lim : Inv(C ) 3 {Ai } 7−→ lim Ai ∈ C . ←−

←−


99

These functors are unique up to canonical isomorphism of functors by Exercise j.19. They are additive if C is additive. Remark k.6. Direct/inverse systems are sometimes called inductive/projective systems, and their limits inductive/projective limits. The canonical morphisms {ϕi } are defined as follows, using the definition of adjoint functors: HomC (lim Ai , lim Ai ) 3 id 7−→ {ϕi } ∈ HomDir(C ) ({Ai }, {lim Ai }), −→

−→

−→

HomC (lim Ai , lim Ai ) 3 id 7−→ {ϕi } ∈ HomInv(C ) ({lim Ai }, {Ai }). ←−

←−

←−

Using these, the isomorphisms defining the adjoint functors are written as follows: HomC (lim Ai , B) 3 f 7−→ {f ◦ ϕi } ∈ HomDir(C ) ({Ai }, {B}), −→

HomC (B, lim Ai ) 3 f 7−→ {ϕi ◦ f } ∈ HomInv(C ) ({B}, {Ai }). ←−

For a morphism {fi } : {Ai } → {Bi } between direct (resp. inverse) systems, the morphism between the limits: f := lim fi : lim Ai −→ lim Bi , −→

−→

−→

(resp. f := lim fi : lim Ai −→ lim Bi ), ←−

←−

←−

B B A are characterized by the property f ◦ ϕA i = ϕi ◦ fi (resp. ϕi ◦ f = fi ◦ ϕi ) for all i ∈ Λ, A B where ϕi , ϕi are the canonical morphisms.

Exercise k.7. (i) For {Xi` } ∈ Dir(Sets), consider the quotient set X := U/ ∼ of the disjoint union U = i∈Λ Xi by an equivalence relation U defined as (x, i) ∼ (y, j) ⇐⇒ ∃k ∈ Λ, i ≤ k, j ≤ k, ϕki (x) = ϕkj (y). Then as Λ is directed X is a set. For each i ∈ Λ, the composite of the inclusion ii : Xi → U and the canonical surjection U → X gives the canonical morphism ϕi : Xi → X. This pair (X, {ϕi }) gives the direct limit lim Xi of {Xi }. −→ Q (ii) For {Yi } ∈ Inv(Sets), consider the direct product V = i∈Λ Yi , and take its subset defined as Y := {(yi ) ∈ V | i ≤ j =⇒ ϕij (yj ) = yi }. For each i ∈ Λ, the composite of the inclusion Y → V and the projection pi : V → Yi gives the canonical morphism ϕi : Y → Yi . This pair (Y, {ϕi }) gives the inverse limit lim Yi of {Yi }. ←−

(iii) For a direct (resp. inverse) system {Xi } in Groups, A-Mod, A-Alg, its direct (resp. inverse) limit as sets is naturally an object of the category, and is the direct (resp. inverse) limit in the category. (iv) If {Xi } is a direct (resp. inverse) system in Rings, then {Xi× } is a direct (resp. inverse) system in Ab, and canonically (lim Xi )× ∼ = lim Xi× (resp. (lim Xi )× ∼ = lim Xi× ). (See Exercise d.17(v).) ←−

−→

−→

←−

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TERUYOSHI YOSHIDA

Exercise k.8. If {Ai } ∈ Dir(C ) (resp. Inv(C )) and B ∈ C , then {HomC (Ai , B)} (resp. {HomC (B, Ai )}) is in Inv(Sets), and canonically ∼ =

HomDir(C ) ({Ai }, {B}) −→ lim HomC (Ai , B), ←−

∼ =

(resp. HomInv(C ) ({B}, {Ai }) −→ lim HomC (B, Ai )). ←−

Therefore, lim Ai ∈ C (resp. lim Ai ∈ C ), if exists, represents the functor lim HomC (Ai , −) −→

←−

←−

(resp. the contravariant functor lim HomC (−, Ai )). ←−

Proposition k.9. Assume that direct sums (resp. products) exist in C . For {Ai }, {Bi } ∈ Dir(C ) (resp. Inv(C )), the following canonical morphism is an isomorphism: ∼ =

lim Ai q lim Bi −→ lim(Ai q Bi ), −→

−→

−→

∼ =

(resp. lim Ai × lim Bi ←− lim(Ai × Bi )). ←−

←−

←−

Proof. First note that there is an obvious isomorphism of categories: F : Dir(C × C ) 3 {(Ai , Bi )} 7−→ ({Ai }, {Bi }) ∈ Dir(C ) × Dir(C ). Now we have a commutative diagram of functors: C

{−}

/ Dir(C )

RRR RRR(−,−) RRR RRR R) ² {−} F / / Dir(C × C ) ∼ Dir(C ) × Dir(C ) C ×C

(−,−)

=

By Proposition j.19(ii) and Example j.17, the left (resp. right) adjont of ({−}, {−}) = (−, −)◦{−} is lim(−q−) (resp. lim(−×−)). Considering ({−}, {−}) = F ◦{(−, −)} = −→

←−

F ◦ {−} ◦ (−, −), its left (resp. right) adjoint is lim(−) q lim(−) (resp. lim(−) × lim(−)) −→

−→

←−

(use Exercise j.18(ii), hence the claim follows from Proposition j.19(i).

←−

¤

k.3. Cofinality. Example k.10. If Λ has a maximal element k, i.e. i ≤ k for all i ∈ Λ, then for any {Ai } with the index set Λ, we have lim Ai ∼ = Ak (resp. lim Ai ∼ = Ak ). −→

←−

Definition k.11. Let Λ0 be a ordered subset of a directed set Λ. If for all i ∈ Λ there is an j ∈ Λ0 with i ≤ j, then Λ0 is called cofinal with Λ. the set Example k.12. For k ≥ 1, the ordered set kN = {kj | j ∈ N} is cofinal with N. In this case Λ0 is directed, and by restriction of index sets we have functors: F : DirΛ (C ) 3 {Ai }Λ 7−→ {Aj }Λ0 ∈ DirΛ0 (C ), F : InvΛ (C ) 3 {Ai }Λ 7−→ {Aj }Λ0 ∈ InvΛ0 (C )


101

Proposition k.13. Let Λ0 be a cofinal ordered subset of Λ, and suppose that direct/inverse limits with index set Λ, Λ0 exist in C . Then, if F is as above, we have canonical isomorphisms of functors lim ◦F ∼ = lim, lim ◦F ∼ = lim. −→

−→

←−

←−

Proof. (For lim): For all {Ai }Λ ∈ DirΛ (C ) and all B ∈ C , the map induced by F : −→

HomDirΛ (C ) ({Ai }Λ , {B}) 3 {fi }Λ 7−→ {fj }Λ0 ∈ HomDirΛ0 (C ) ({Aj }Λ0 , {B}) is a bijection, because for any i ∈ Λ, there is a j ∈ Λ0 such that i ≤ j, and for all {fj }Λ0 ∈ HomDirΛ0 (C ) ({Aj }Λ0 , {B}), defining {fi }Λ by: i ≤ j ∈ Λ0 =⇒ fi = fj ◦ ϕji gives the inverse map. Therefore canonically: HomC (lim Ai , B) ∼ = HomDirΛ (C ) ({Ai }Λ , {B}) ∼ = HomDir

Λ0 (C )

−→

({Aj }Λ0 , {B}),

and the proposition follows from the uniqueness of adjoint functors (Proposition j.19(i)). (For lim): For all {Ai }Λ ∈ InvΛ (C ) and all B ∈ C , the map induced by F : ←−

HomInvΛ (C ) ({B}, {Ai }Λ ) 3 {fi }Λ 7−→ {fj }Λ0 ∈ HomInvΛ0 (C ) ({B}, {Aj }Λ0 ) is a bijection, because for any i ∈ Λ, there is a j ∈ Λ0 such that i ≤ j, and for all {fj }Λ0 ∈ HomInvΛ0 (C ) ({B}, {Aj }Λ0 ), defining {fi }Λ by: i ≤ j ∈ Λ0 =⇒ fi = ϕij ◦ fj gives the inverse map. Therefore canonically: HomC (B, lim Ai ) ∼ = HomInvΛ (C ) ({B}, {Ai }Λ ) ∼ = HomInvΛ0 (C ) ({B}, {Aj }Λ0 ) ←−

and the proposition follows from the uniqueness of adjoint functors (Proposition j.19(i)). ¤ l. ⊗ and Hom for Modules l.1. The functor Hom. Let A, B be two (possibly non-commutative) rings. Definition l.1. An additive group W is called an (A, B)-bimodule if it is a left A-module and right B-module at the same time, and satisfies a(xb) = (ax)b (∀a ∈ A, ∀b ∈ B, ∀x ∈ W ). Exercise l.2. (i) Every left A-module is an (A, Z)-bimodule, and every right Bmodule is a (Z, B)-bimodule. (ii) If A is commutative, then every A-module is an (A, A)-bimodule. Now let W be an (A, B)-bimodule. For X ∈ A-Mod, we make the additive group HomA (W, X) into a left B-module by defining (bϕ)(w) = ϕ(wb) (∀w ∈ W )

102

TERUYOSHI YOSHIDA

for ϕ ∈ HomA (W, X) and b ∈ B. If X, Y ∈ A-Mod, for f ∈ HomA (X, Y ), f∗ : HomA (W, X) 3 ϕ 7−→ f ◦ ϕ ∈ HomA (W, Y ) is a B-homomorphism and preserves identity and composition. Similarly, if X ∈ Mod-B, then HomB (W, X) is a right A-module by defining (ϕa)(w) = ϕ(aw) (∀w ∈ W ) for ϕ ∈ HomB (W, X) and a ∈ A, and if X, Y ∈ Mod-B, for f ∈ HomB (X, Y ), f∗ : HomB (W, X) 3 ϕ 7−→ f ◦ ϕ ∈ HomB (W, Y ) is an A-homomorphism and preserves identity and composition. Therefore we obtain the following generalization of Example j.7: Definition l.3. We have the following additive functors: HomA (W, −) : A-Mod 3 X 7−→ HomA (W, X) ∈ B-Mod, HomB (W, −) : Mod-B 3 X 7−→ HomB (W, X) ∈ Mod-A. l.2. Construction of ⊗. Let W be an (A, B)-bimodule, and X a (B, C)-bimodule. Let hW ×Xi be a free Z-module generated by the product set W ×X (Example j.16(ii)), and let F its Z-submodule generated by the set of all elements of the form (x1 + x2 , y) − (x1 , y) − (x2 , y), (x, y1 + y2 ) − (x, y1 ) − (x, y2 ), (xb, y) − (x, by)

(b ∈ B).

Denote the quotient Z-module hW × Xi/F by W ⊗B X. The residue class of (x, y) ∈ W × X in W ⊗B X is denoted by x ⊗ y. We have: (x1 + x2 ) ⊗ y = x1 ⊗ y + x2 ⊗ y, x ⊗ (y1 + y2 ) = x ⊗ y1 + x ⊗ y2 , (xb) ⊗ y = x ⊗ (by) (b ∈ B) by definition. Lastly, we define the structure of (A, C)-bimodule on W ⊗B X as follows: a(x ⊗ y) = (ax) ⊗ y

(a ∈ A),

(x ⊗ y)c = x ⊗ (yc)

(c ∈ C).

Definition l.4. This (A, C)-bimodule W ⊗B X is called the tensor product of W and X over B. Now fix an (A, B)-bimodule W . For X ∈ B-Mod, we have W ⊗B X ∈ A-Mod, and if X, Y ∈ B-Mod, for f ∈ HomB (X, Y ), f] : W ⊗B X 3 w ⊗ x 7−→ w ⊗ f (x) ∈ W ⊗B Y is a left A-homomorphism and preserves identity and composition. Similarly, if X ∈ Mod-A, then X ⊗A W ∈ Mod-B, and if X, Y ∈ Mod-A, for f ∈ HomA (X, Y ), f] : X ⊗A W 3 x ⊗ w 7−→ f (x) ⊗ w ∈ Y ⊗A W


103

is a right B-homomorphism and preserves identity and composition. Therefore: Definition l.5. We have the following additive functors: W ⊗B − : B-Mod 3 X − 7 → W ⊗B X ∈ A-Mod, − ⊗A W : Mod-A 3 X − 7 → X ⊗A W ∈ Mod-B. Theorem l.6. Let W be an (A, B)-bimodule. (i) W ⊗B −, HomA (W, −) are the adjoint of each other. (ii) − ⊗A W, HomB (W, −) are the adjoint of each other. Proof. (i): The following morphisms in Z-Mod are the inverse of each other: HomA (W ⊗B X, Y ) 3 f 7−→ (x 7→ (w 7→ f (w ⊗ x))) ∈ HomB (X, HomA (W, Y )), HomB (X, HomA (W, Y )) 3 g 7−→ (w ⊗ x 7→ g(x)(w)) ∈ HomA (W ⊗B X, Y ), which give the functorial bijections of adjointness. (ii): The following morphisms in Z-Mod are the inverse of each other: HomB (X ⊗A W, Y ) 3 f 7−→ (x 7→ (w 7→ f (x ⊗ w))) ∈ HomA (X, HomB (W, Y )), HomA (X, HomB (W, Y )) 3 g 7−→ (x ⊗ w 7→ g(x)(w)) ∈ HomB (X ⊗A W, Y ), which give the functorial bijections of adjointness.

¤

l.3. Basic properties of ⊗. When a ring B is a left (resp. right) A-module, any left (resp. right) B-module X naturally is a left (resp. right) A-module by: ∀a ∈ A, ∀x ∈ X

ax = (a · 1B )x

(resp. xa = x(1B · a)),

and we have the forgetful functors − : B-Mod → A-Mod, − : Mod-B → Mod-A. Proposition l.7. If the ring B is a left (resp. right) A-module, we have a canonical isomorphism of functors B ⊗B − ∼ = − (resp. − ⊗B B ∼ = −). Proof. For every left (resp. right) B-module X, we have functorial isomorphisms that give inverse of each other: X 3 x 7−→ 1 ⊗ x ∈ B ⊗B X, X 3 x 7−→ x ⊗ 1 ∈ X ⊗B B, B ⊗B X 3 b ⊗ x 7−→ bx ∈ X, X ⊗B B 3 x ⊗ b 7−→ xb ∈ X. ¤ Proposition l.8. If V is an (A, B)-bimodule and W is a (B, C)-bimodule, there are canonical isomorphisms: (i) V ⊗B (W ⊗C −) ∼ = (V ⊗B W ) ⊗C − of functors C-Mod → A-Mod. ∼ (ii) (− ⊗A V ) ⊗B W = − ⊗A (V ⊗B W ) of functors Mod-A → Mod-C. (iii) For v ∈ V, w ∈ W, x ∈ X ∈ C-Mod (resp. Mod-A), above are given by: v ⊗ (w ⊗ x) 7−→ (v ⊗ w) ⊗ x, (x ⊗ v) ⊗ w 7−→ x ⊗ (v ⊗ w).

104

TERUYOSHI YOSHIDA

Proof. (i): By the following canonical isomorphism they are the left adjoints of the same functor, hence Proposition j.19(i) proves the claim: HomA (V ⊗B (W ⊗C −), Y ) ∼ = HomB (W ⊗C −, HomA (V, Y )) ∼ = HomC (−, HomB (W, HomA (V, Y ))), HomA ((V ⊗B W ) ⊗C −, Y ) ∼ = HomC (−, HomA (V ⊗B W, Y )) ∼ HomC (−, HomB (W, HomA (V, Y ))). = (ii): Similarly they are the left adjoints of the same functor: ∼ HomB (− ⊗A V, HomC (W, Y )) HomC ((− ⊗A V ) ⊗B W, Y ) = ∼ = HomA (−, HomB (V, HomC (W, Y ))), HomC (− ⊗A (V ⊗B W ), Y ) ∼ = HomA (−, HomC (V ⊗B W, Y )) ∼ = HomA (−, HomB (V, HomC (W, Y ))). (iii): Clear from the proof.

¤

l.4. Tensor Products of Algebras. Definition l.9. Let A be a commutative ring. For X, Y ∈ A-Alg, we can define a ring structure on the tensor product X ⊗A Y as A-modules by: (x1 ⊗ y1 )(x2 ⊗ y2 ) = x1 x2 ⊗ y1 y2 . This is well-defined and makes X ⊗A Y into an A-algebra. Exercise l.10.

(i) The canonical morphisms in A-Alg:

iX : X 3 x 7−→ x ⊗ 1 ∈ X ⊗A Y,

iY : Y 3 y 7−→ 1 ⊗ y ∈ X ⊗A Y

make X ⊗A Y into an X-algebra and a Y -algebra. This coincides with the module structure on X ⊗A Y , if we consider X as an (X, A)-bimodule (resp. Y as an (A, Y )-bimodule). In particular, we have a functor B ⊗A − : A-Alg −→ B-Alg, which coincides with B ⊗A − : A-Mod −→ B-Mod when we forget the ring structure. It is the left adjoint of − : B-Alg → A-Alg. (ii) The following canonical isomorphisms as A-modules are clearly ring homomorphisms, hence are isomorphisms as A-algebras (or X-algebras, etc): (a) A ⊗A X 3 1 ⊗ x 7−→ x ∈ X (Proposition l.7). (b) X ⊗A Y 3 x ⊗ y 7−→ y ⊗ x ∈ Y ⊗A X. (c) (X ⊗A Y ) ⊗B Z 3 (x ⊗ y) ⊗ z 7−→ x ⊗ (y ⊗ z) ∈ X ⊗A (Y ⊗B Z) (Proposition l.8). Proposition l.11. The tensor product over A gives the direct sum in A-Alg. Proof. The maps of Exercise l.10(i) give canonical morphisms, and for any fX : X → T, fY : Y → T , the map f : X ⊗ Y 3 x ⊗ y 7−→ fX (x)fY (y) ∈ T gives the unique morphism satisfying f ◦ iX = fX , f ◦ iY = fY .

¤


105

Proposition l.12. (i) Let S be an arbitrary set of variables (Example j.16(iii)) and I be an ideal of A[S]. Then for B ∈ A-Alg, we have: B ⊗A (A[S]/I) ∼ = B[S]/IB[S]. (ii) We have the following canonical isomorphisms in B-Alg, A-Alg: B ⊗A (X × Y ) ∼ = (B ⊗A X) × (B ⊗A Y ). (iii) If B is a finite free A-algebra of rank n, then B ⊗A − ∼ = (−)n as functors A-Mod → A-Mod. In particular, if X ∈ A-Alg, then B ⊗A X is a finite free X-algebra of rank n. (iv) If B is a finite free A-algebra, then in B-Alg: B ⊗A (lim −) ∼ = lim (B ⊗A −). ←−

←−

Proof. (i): By Proposition l.11 and Exercise i.4(ii), it is enough to show that B[S]/IB[S] (and the canonical morphisms from B and A[S]/I) gives the direct sum B q (A[S]/I) in A-Alg. For any fB : B → T and fA[S]/I : A[S]/I → T , we extend fB to a map f 0 : B[S] → T by sending ∀X ∈ S to fA[S]/I (X mod I) ∈ T . As f 0 is a ring homomorphism and the image of I in B[S] is contained in Ker f 0 , we have IB[S] ⊂ Ker f 0 , therefore f 0 factors through f : B[S]/IB[S] → T , which gives fB and fA[S]/I when composed with the canonical morphisms. (ii): As there is a canonical morphism of functors by the definition of direct products, it is enough to show that it is an isomorphism in A-Mod. But direct products of rings are direct sums as A-modules, which commute with any additive functor by Proposition j.6(ii). (iii): As ⊗ is additive, B ⊗A − ∼ = An ⊗A − ∼ = (A ⊗A −)n ∼ = (−)n by Proposition j.6(ii), l.7. (iv) By (iii), the functor B ⊗A − commutes with any additive functor on A-Mod (or Inv(A-Mod)) by Proposition j.6(ii). For the latter claim, as there is a canonical morphism of functors by the definition of lim, it is enough to know that they are isomorphic in A-Mod. ¤ ←−

Index A-algebra, 60 A-algebra homomorphism, 60 A-homomorphism, 14, 93 A-module, 7 A-subalgebra, 60 A-submodule, 8 K-algebra, 7 K-linear map, 11

category of abelian groups, 91 category of groups, 18 category of left A-modules, 93 category of right A-modules, 93 category of rings, 18 category of sets, 18 category of vector spaces, 18 change of bases, 17 characteristic (field), 43 characteristic polynomial, 29 Chinese remainder theorem, 81 cofactor, 25 cofactor marix, 25 cofinal, 100 commutative, 4 commutative diagram, 17 commutative group, 5 commutative ring, 5 composite, 3 composite (morphisms of functors), 95 composite field, 53 composite functor, 93 composition (of morphisms), 17 contravariant functor, 93 coprime, 31 cyclic extension, 51 cyclic group, 43 cyclotomic extension, 42 cyclotomic polynomial, 45

abelian extension, 45 abelian group, 5 absolute Galois group, 56 action, 6 addition, 4 addition (for N), 4 additive category, 91 additive contravariant functor, 94 additive functor, 94 additive group, 5 adjoining a root, 38 adjoining all roots, 42 adjoint (functor), 96 algebra, 60 algebra homomorphism, 60 algebraic closure, 55 algebraic exension field, 54 algebraic integer, 62 algebraic number field, 38 algebraically closed field, 54 alternating multilinear form, 22 antisymmetric law, 54 Artin’s lemma, 66 associate, 30 associativity (of morphisms), 17 automorphism group, 15

decomposition field, 84 decomposition group, 83 Dedekind domain, 65 derivation, 44 determinant (linear transformation), 23 determinant (matrix), 24 diagonal matrix, 29 diagonalizable, 29 dimension, 11 dimension (ring), 63 dimension formula, 13 direct limit, 98 direct product (algebra), 81 direct product (group/ring), 57 direct product (object), 90 direct product (object, finite), 89 direct product (set), 4 direct sum (linear map), 27 direct sum (modules), 77 direct sum (object), 90 direct sum (object, finite), 90

basis, 9 basis (free module), 59 bijection, 4 bimodule, 101 canonical basis (of K n ), 9 canonical isomorphism, 95 canonical isomorphism (object), 89 canonical morphism (limits), 99 canonical surjection, 34 canonically isomorphic, 95 canonically isomorphic (object), 89 category, 17 category of A-algebras, 60 category of A-modules, 18 106


direct sum (pair), 28 direct sum (vector space), 26 direct sum decomposition (linear map), 27 direct sum decomposition (pair), 28 direct sum decomposition (vector space), 27 direct system, 98 directed set, 98 discrete valuation ring, 74 divisible, 30 divisor, 30 proper —, 30 domain, 30 dual basis, 67 dual module, 67 DVR, 74 eigenvalue, 29 eigenvector, 29 elementary pair, 28 elementary subpair, 28 endomorphism ring, 15 entry, 18 equivalence class, 33 equivalence relation, 33 equivariant, 14 euclidean domain, 33 Euler’s function, 43 existence of primitive root, 45 extension, 3, 38 extension degree, 38 extension field, 38 field, 6 finite (algebra), 60 finite extension, 38 finite field, 6 finite group, 43 finite-dimensional, 11 finitely generated (module), 59 finitely generated (vector space), 9 first complementary law, 88 forgetful functor, 94 fraction field, 62 fractional ideal, 72 free (module), 59 free module generated by a set, 96 Frobenius automorphism, 84 Frobenius map, 44, 58 fully faithful, 94 functor, 93 fundamental theorem of algebraic number theory, 76 fundamental theorem of arithmetic, 33

fundamental theorem of Galois theory, 41 Galois closure, 49 Galois extension, 41, 47 Galois extension (infinite), 56 Galois group, 41 Galois group (infinite), 56 Gauss sum, 88 general linear group, 20 generate, 8 generate (module), 59 generated by (field), 40 generated by (finite group), 43 generated by (submodule), 59 generating set, 8 generating set (module), 59 generator (finite group), 43 group, 5 group homomorphism, 14 height, 63 homomorphism, 14 homomorphism (algebra), 60 homomorphism theorem, 35 ideal, 8, 31 ideal class group, 76 ideal group, 76 identity, 4 identity (morphism of functors), 95 identity functor, 93 identity map, 3 identity matrix, 19 identity morphism, 18 image, 3 inclusion map, 3 index set, 98 inductive, 54 inductive limit, 99 inductive system, 99 inertia field, 84 inertia group, 83 infinite extension, 38 infinite-dimensional, 11 injection, 4 integer (in algebraic number field), 62 integer ring, 62 integral (algebra), 60 integral (element over a ring), 60 integral closure, 61 integral domain, 30 integral extension, 60 integrally closed, 62 intermediate field, 38

107

108

TERUYOSHI YOSHIDA

inverse, 5 inverse (functor), 94 inverse image, 3 inverse limit, 98 inverse map, 4 inverse matrix, 19 inverse system, 98 invertible (fractional ideal), 72 invertible (in general), 5 invertible matrix, 19 irreducibility of cyclotomic polynomials, 86 irreducible (element), 30 isomorphic, 11, 14 isomorphic (as extensions), 39 isomorphic (functor), 95 isomorphic (object), 18 isomorphism, 11, 14, 28 isomorphism (algebra), 60 isomorphism (category), 94 isomorphism (functor), 95 isomorphism (object), 18 K-automorphism (of extensions), 39 K-homomorphism (of extensions), 39 K-isomorphism (of extensions), 39 kernel (group), 14 kernel (rings, A-modules), 14 kernel (vector space), 12 Kronecker’s delta, 19 Krull dimension, 63 Krull topology, 58 Kummer extension, 51 left A-module, 92 left adjoint functor, 96 Legendre symbol, 87 lie above, 64 lie below, 64 linear combination, 8 linear combination (module), 59 linear map, 11 linear relation, 8 linear relation (module), 59 linear transformation, 11 linearly dependent, 8 linearly dependent (module), 59 linearly independent, 8 linearly independent (module), 59 local domain, 70 local ring, 70 localization, 70 map, 3 matrix, 18

matrix element, 19 maximal (ideal), 32 maximal abelian extension, 56 maximal cyclotomic extension, 57 maximal element, 54, 100 minimal polynomial, 40, 51 module, 7 monic, 37 morphism, 17 morphism (algebra), 60 morphism (direct/inverse system), 98 morphism (functor), 94 morphism (pair), 28 morphism of functors, 94 multilinear form, 22 multiple, 30 multiple root, 38 multiplication, 4 multiplication (for N), 4 multiplication by n (morphism of functors), 95 multiplicative subset, 68 multiplicity, 38 noetherian, 64 normal (domain), 62 normal subgroup, 35 object, 17 operation, 4 order (cyclic group), 43 order (element of a finite group), 43 ordered set, 54 perfect field, 49 permutation, 22 PID, 32 polynomial ring, 5, 37 prime (element), 30 prime (ideal), 31 prime field, 44 primitive n-th root of unity, 44 primitive root, 45 principal fractional ideal, 72 principal ideal, 31 principal ideal domain (PID), 32 product, 19 product (category), 96 product (fractional ideal), 72 profinite completion (of Z), 58 projection, 89 projection (functor), 96 projective limit, 99 projective system, 99


quadratic reciprocity law, 89 quadratic residue, 87 quotient A-module, 34 quotient group, 34 quotient ring, 34 quotient set, 33

torsion-free, 78 totally ordered set, 54 trace, 66 transcendental exension field, 54 transitive law, 33, 54 trivial (linear relation), 8, 59

radical extension, 53 ramification index, 80 ramify, 80 rank, 77 reflexive law, 33, 54 relatively prime (UFD), 31 representable, 95 representation matrix, 20 representative element, 33 residue class ring, 5 residue degree, 80 residue field (local ring), 70 residue field (prime of Dedekind domain), 79 restriction, 3 right A-module, 92 right adjoint functor, 96 ring, 5 ring homomorphism, 14 ring of fractions, 68 ring of integers (algebraic number field), 62 root, 37 roots of unity, 42

uniformizer (DVR), 74 unique factorization domain (UFD), 31 unique up to canonical isomorphism, 89 unit, 6, 30 unit group, 6 unramified, 80 upper bound, 54

second complementary law, 88 semisimple, 29 separable (extension), 47 separable (polynomial), 47 separable closure, 56 sign (permutation), 22 solvable extension, 52 solvable group, 52 splitting field, 42 square matrix, 19 Steinitz’ theorem, 55 subalgebra, 60 subfield, 38 subgroup, 6 submodule, 8 subpair, 28 subring, 6 subspace, 7 sum (ideals), 32 surjection, 4 symmetric group, 23 symmetric law, 33 tensor product (modules), 102

valuation (Dedekind domain), 76 valuation (DVR), 74 valuation (fractional ideal), 76 vector space, 6 volume form, 23 Yoneda’s lemma, 95 zero divisor, 30 zero element, 4 zero morphism, 91 zero object, 91 zero ring, 5 Zorn’s lemma, 54

109