LECTURES ON ALGEBRAIC NUMBER THEORY. Yichao TIAN


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1 LECTURES ON ALGEBRAIC NUMBER THEORY Yichao TIAN
2 Yichao TIAN Morningside Center of Mathematics, 55 Zhong Guan Cun East Road, Beijing, , China.
3 LECTURES ON ALGEBRAIC NUMBER THEORY Yichao TIAN
4
5 CONTENTS 1. Number fields and Algebraic Integers Algebraic integers Traces and norms Discriminants and integral basis Cyclotomic fields Dedekind Domains Preliminaries on Noetherian rings Dedekind domains Localization Decomposition of Primes in Number Fields Norms of ideals Decomposition of primes in extension of number fields Relative different and discriminant Decomposition of primes in Galois extensions Prime decompositions in cyclotomic fields Finiteness Theorems Finiteness of class numbers Dirichlet s unit theorem Binary Quadratic Forms and Class Number Binary quadratic forms Representation of integers by binary quadratic forms Ideal class groups and binary quadratic forms Distribution of Ideals and Dedekind Zeta Functions Distribution of ideals in a number field Residue formula of Dedekind Zeta functions Dirichlet LFunctions and Arithmetic Applications
6 6 CONTENTS 7.1. Dirichlet characters Factorization of Dedekind zeta functions of abelian number fields Density of primes in arithmetic progressions Values of L(χ, 1) and class number formula Class number formula for quadratic fields Nonarchmedean Valuation Fields The introduction of padic fields Absolute values and completion Structure of complete discrete valuation fields Hensel s Lemma Extensions of valuations Krasner s Lemma and applications Finite Extensions of Complete Discrete Valuation Fields Generalities Unramified extensions Different, discriminant and ramification Galois extension of complete discrete valuation fields Applications of Local Methods to Number Fields Norms and places on number fields Tensor product and decomposition of primes Product formula Comparison of local and global Galois groups Local and global different HermiteMinkowski s finitness theorem Bibliography
7 CHAPTER 1 NUMBER FIELDS AND ALGEBRAIC INTEGERS 1.1. Algebraic integers All the rings in this section are supposed to be commutative. Definition Let A B be an extension of rings. We say an element x B is integral over A if there exists a monic polynomial f(t ) = T n + a 1 T n a n A[T ] such that f(x) = 0. We say B is integral over A, if every x B is integral over A. Example (1) Z[i] is integral over Z. (2) Let L/K be an extension of fields. Then L is integral over K if and only if L/K is an algebraic extension. Proposition Let A B be an extension of rings, x B. Then the following statements are equivalent: 1. x is integral over A. 2. the subring A[x] B is a finite generated Amodule. 3. x belongs to a subring B B such that B is finitely generated as an Amodule. Proof. (1) = (2) = (3) is trivial. We prove now (3) = (1). Choose generators α 1,, α n of the Amodule B. Since xb B, there exists a U M n n (A) such that x(α 1,, α n ) = (α 1,, α n )U (α 1,, α n )(xi n U) = 0. Let V be the cofactor matrix of xi n U. Then one has (α 1,, α n )(xi n U)V = (α 1,, α n ) det(xi n U) = 0. As 1 B is a linear combination of α i s, we get det(xi n U) = x n +a 1 x n 1 + +a n = 0. Corollary Let A B be extensions of rings. Then the elements of B which are integral over A form a subring of B. Proof. Given x, y B integral over A, we need to show that x + y and xy are also integral over A. Actually, one sees easily that A[x, y] is a finitely generated Amodule, and concludes using Proposition 1.1.3(3).
8 8 CHAPTER 1. NUMBER FIELDS AND ALGEBRAIC INTEGERS Corollary Let A B C be extensions of rings. Then C is integral over A if and only if C is integral over B and B is integral over A. Proof. The only if part is easy. Prove now that the inverse implication holds. Let x C. Since C is assumed integral over B, we have f(x) = x n + b 1 x n b n = 0 for some b 1,, b n B. Note that b 1,, b n are all integral over A by assumption. One proves easily by induction that A[b 1,, b i ] is a finitely generated Amodule for all 1 i n. Then A[b 1,, b n, x] = A[b 1,, b n ][T ]/(f(t )) is also finitely generated as Amodule. One concludes with Proposition 1.1.3(3). Definition (1) Let A R be an extension of rings. Define the integral closure of A in R to be the subring of R consisting of all integral elements over A. (2) If the integral closure of A in R is A, we say A is integrally closed in R. (3) Assume A is an integral domain. We say A is integrally closed if A is integrally closed in its fraction field. Example (1) Z is integrally closed. Indeed, let x = a b Q with gcd(a, b) = 1 and b > 0. If x is integral over Z, then there exist some c 1,, c n Z such that x n + c 1 x n c n = 0 a n + c 1 a n 1 b + + c n b n = 0. If b 1, let p denote a prime dividing b. Then the equality above implies that p a n, hence p a. This contradicts with gcd(a, b) = 1. (2) Let ω = e 2πi 3. Then Z[ω] is integral over Z and integrally closed by similarly arguments as above. Actually, every principal ideal domain is integrally closed. Definition (1) An element x C is an algebraic number (resp. an algebraic integer) if x is integral over Q (resp. over Z). (2) A number field is a finite extension of Q. For a number field K/Q, we define O K to be the integral closure of Z in K, and call it the ring of integers of K. Example The ring of Gauss integers Z[i] is the ring of integers of Q(i), and Z[ω] with ω = is the ring of integers of Q( 3). Proposition Let x C be an algebraic number, and f(t ) = T n + a 1 T n a n Q[T ] be its minimal polynomial. Then x in an algebraic integer if and only if f(t ) Z[T ]. Proof. One side implication is clear. Assume now that x is an algebraic integer. Let {x = x 1,, x n } be the set of complex roots of f(t ). We claim that each x i is also an algebraic integer. Indeed, for each x i, there exists an embedding of fields ι : K = Q(x) C such that ι(x) = x i. Therefore, if x satisfies g(x) = 0 for some monic polynomial g(t ) Z[T ], then g(x i ) = ι(g(x)) = 0. This proves the claim. But each a i is a symmetric function of x j s. It follows that a i O K Q = Z.
9 1.2. TRACES AND NORMS 9 Example Let K = Q( D) with an integer D 1 square free. Then we have O K = Z + Zω D, where { 1+ D ω D = 2 if D 1 mod 4 D if D 2, 3 mod 4. It is easy to check that ω D is an integer. It remains to show that if x = a + b D with a, b 0 Q is an integer, then x Z + Zω D. Indeed, the minimal polynomial of x over Q is T 2 2aT + (a 2 b 2 D) = 0. By the previous Proposition, for x to be an integer, one must have 2a, a 2 b 2 D Z. If a Z, then b must be in Z. Otherwise, if a is a halfinteger, then b has to be an halfinteger as well and D 1 mod Traces and norms Definition Let L/K be a finite extension of fields, and x L. We view L as a finite dimensional Kvector space, and denote by φ x : L L the Klinear endomorphism on L defined by the multiplication by x. We have φ x End K (L). We put Tr L/K (x) = Tr(φ x ), and call it the trace of x (relative to L/K); put N L/K (x) = det(φ x ), and call it the norm of x (relative to L/K). Lemma Let L/K be a finite extension of fields, and x L. 1. One has Tr L/K (x) = [L : K(x)]Tr K(x)/K (x) and N L/K (x) = N K(x)/K (x) [L:K(x)]. 2. If f(t ) = T n + a 1 T n 1 + a n K[T ] is the minimal polynomial of x over K, then Tr K(x)/K (x) = a 1 and N K(x)/K (x) = ( 1) n a n. Proof. Exercise. From now on, we assume that all the fields encountered are number fields. Proposition Let L/K be a finite separable extension of fields, and n = [L : K]. Fix an algebraically closed field Ω, and an embedding τ : K Ω. Then 1. there exists exactly n distinct embeddings σ 1,, σ n : L Ω such that σ i K = τ for 1 i n; 2. the n embeddings σ 1,, σ n are linearly independent over Ω. Proof. (1) By induction on n, one reduces to the case where L = K(x) for some x L. In this case, let f(t ) = T n + a 1 T n 1 + a n be the minimal polynomial of x over K so that L = K[T ]/(f(t )). Put f τ (T ) = T n + τ(a 1 )T n τ(a n ) Ω[T ], and let α 1,, α n Ω be the roots of f τ (T ). Then the α i s must be distinct (because f(t ) is separable). For each α i, there exists a unique embedding σ i : L Ω extending τ such that σ i (x) = α i. Conversely, if σ : L Ω is an extension of τ, then it must send x to some α i, hence it must coincide with one of the σ i s.
10 10 CHAPTER 1. NUMBER FIELDS AND ALGEBRAIC INTEGERS (2) The statement is trivial if n = 1. Suppose now n 2 and in contrary that σ 1,, σ n are linearly independent over Ω. Up to renumbering, we may assume that d i=1 c iσ i = 0 is a linearly relation with c i Ω such that d 2 is minimal. Thus for any x L, we have d i=1 c iσ i (x) = 0. By dividing c 1, we may assume that c 1 = 1. Choose y L such that σ 2 (y) σ 1 (y). This is possible since σ 1 and σ 2 are distinct. Then d i=1 c iσ i (xy) = d i=1 c iσ i (y)σ i (x) = 0 for all x L. Therefore, one obtains d c i (σ i (y) σ 1 (y))σ i (x) = 0, x L. i=2 This is a nonzero linear relation among σ i s of length at most d 1, whose existence contradicts with the minimality of d. Theorem Let L/K be a finite separable extension of fields, τ and σ i for 1 i n be the embeddings as in Proposition We identify K with its image in Ω via τ. Then one has n n Tr L/K (x) = σ i (x) and N L/K (x) = σ i (x), for all x L. i=1 Moreover, the Kbilinear form L L K given by (x, y) Tr L/K (xy) i=1 for all x, y L is nondegenerate, i.e. if x L such that Tr L/K (xy) = 0 for all y L, then x = 0. Proof. By the construction of σ i s in Proposition 1.2.3, Tr L/K (x) and N L/K (x) are easily verified if L = K(x). The general case follows from Lemma 1.2.2(1). The nondegeneracy of the Kbilinear form Tr L/K (xy) follows immediately from Proposition 1.2.3(2). Remark If L/K is inseparable, then the pairing Tr L/K is no longer nondegenerate. For instance, if K = F p (x) and L = F p (x 1/p ), then Tr L/K (x) = 0 for all x L. Corollary Let α 1,, α n be n = [L : K] elements of L. Then (α 1,, α n ) is a Kbasis of L if and only if det(tr L/K (α i α j )) 0. Proof. Consider the morphisms: K n φ L ψ K n given respectively by φ : (x i ) 1 i n = i x iα i and ψ : x (Tr L/K (xα i )) 1 i n. Then the matrix of ψ φ under the natural basis of K n is (Tr L/K (α i α j )) 1 i,j n. If (α i ) 1 i n is a basis of L, then φ is an isomorphism by definition, and ψ is injective (hence bijective) by the nondegeneracy of Tr L/K (xy). It follows that ψ φ is an isomorphism, thus det(tr L/K (α i α j )) 0. Conversely, if det(tr L/K (α i α j )) 0, then ψ φ is an isomorphism. It follows that φ is injective, hence bijective since L has the same Kdimension as K n.
11 1.3. DISCRIMINANTS AND INTEGRAL BASIS 11 Given a basis (α i ) 1 i n of L over K. Let C = (c ij ) 1 i,j n denote the inverse matrix of (Tr L/K (α i α j )) 1 i,j n, and put αi = n k=1 α kc ki for 1 i n. Then one checks easily that { Tr L/K (α i αj 1 if i = j ) = 0 otherwise. We call (α i ) 1 i n the dual basis of (α i ) 1 i n with respect to Tr L/K. For any x L, if we write x = i x iα i, then x i = Tr L/K (xα i ); similarly if we write x = i y iα i, then y i = Tr L/K (xα i ) Discriminants and integral basis We apply the theory of previous section to the case of number fields. In this section, let K denote a number field of degree n = [K : Q], and O K be its ring of integers. For α 1,, α n K, we put and call it the discriminant of α 1,, α n. Disc(α 1,, α n ) = det(tr K/Q (α i α j )), Lemma (1) The elements α 1,, α n form a basis of K over Q if and only if Disc(α 1,, α n ) 0. (2) If σ 1, σ n denote the n distinct complex embeddings of K given by Proposition 1.2.3, then Disc(α 1,, α n ) = det(σ i (α j )) 2. (3) If C M n n (Q) and (β 1,, β n ) = (α 1,, α n )C, then Disc(β 1,, β n ) = Disc(α 1,, α n ) det(c) 2. Proof. Statement (1) is Corollary For (2), one deduces from Theorem that Tr K/Q (α i α j ) = n k=1 σ k (α i α j ) = k σ k (α i )σ k (α j ). Hence, if A denotes the matrix (Tr K/Q (α i α j )) 1 i,j n and U = (σ i (α j )) 1 i,j n, then A = U T U. Statement (2) follows immediately. Let B denote (Tr K/Q (β i β j )) 1 i,j n. Then one has B = C T A C, and hence (3) follows. Proposition Let α be an arbitrary element of K, and f(t ) Q[T ] be its minimal polynomial. Then one has { 0 if deg(f) < n, Disc(1, α,, α n 1 ) = ( 1) n(n 1) 2 N K/Q (f (α)) if deg(f) = n.
12 12 CHAPTER 1. NUMBER FIELDS AND ALGEBRAIC INTEGERS Proof. Since (1, α,, α n 1 ) form a basis of K if and only if deg(f) = n, the first case follows from Lemma 1.3.1(1). Assume now deg(f) = n. Denote by σ 1,, σ n the complex embeddings of K. By Lemma 1.3.1(2), one has Disc(1, α,, α n 1 ) = det(σ i (α j 1 ) 1 i,j n ) 2 = i<j(σ i (α) σ j (α)) 2, where the last equality uses Vandermonde s determinant formula. The Proposition then follows from n n N K/Q (f (α)) = σ(f (α)) = (σ i (α) σ j (α)). i=1 i=1 j i Theorem The ring of integers O K is a free abelian group of rank n. Proof. Choose a basis (α i ) 1 i n of K over Q. Up to multiplying by an integer, we may assume that α i O K. Consider the abelian subgroup M O K generated by the α i s. Let (αi ) 1 i n be the dual basis of (α i ) 1 i n with respect to Tr K/Q, and put M = n i=1 Zα i as a abelian subgroup of K. It is easy to see that M = {x K Tr K/Q (xy) Z, y M}. Thus M O K M, and one checks easily that M /M is finite with cardinality Disc(α 1,, α n ). Since M is a Zmodule free of rank n, the Theorem follows immediately. Definition A basis (α 1,, α n ) of K over Q is called an integral basis if it is a basis of O K over Z. Proposition Let (α 1,, α n ) be an integral basis of K, and (β 1,, β n ) be an arbitrary ntuple of elements in O K which form a basis of K/Q. Then Disc(β 1,, β n ) equals to Disc(α 1,, α n ) times a square integer. In particular, if (β 1,, β n ) is also an integral basis, if and only if Disc(β 1,, β n ) = Disc(α 1,, α n ). Proof. Write each β i as a Zlinear combination of α j s, then there exists a matrix C M n n (Z) with det(c) 0 and (β 1,, β n ) = (α 1,, α n ) C. Note that det(c) Z, then the Proposition follows from Lemma Definition The discriminant of K, denoted by K Z, is the discriminant of an integral basis of K. By the previous Lemma, this definition does not depend on the choice of the integral basis. For instance, if K = Q( D) where D 0, 1 is a square free integer, then (1, ω D ) considered in Example is an integral basis of K. Thus K = Disc(1, ω D ) which equals to 4D if D 2, 3 mod 4, and to D if D 1 mod 4. We now give a practical criterion for nelements of O K to be an integral basis.
13 1.3. DISCRIMINANTS AND INTEGRAL BASIS 13 Lemma Let β 1,, β n be n elements of O K which form a basis of K. Then Disc(β 1,, β n ) is not an integral basis if and only if there exists a rational prime p with p 2 Disc(β 1,, β n ) and some x i {0, 1,, p 1} for 1 i n such that not all of x i are zero and n i=1 x iβ i po K. Proof. Choose an integral basis (α 1,, α n ) and write (β 1,, β n ) = (α 1,, α n ) C for some matrix C M n n (Z) with det(c) 0. Then (β 1,, β n ) is an integral basis if and only if det(c) = ±1. Assume (β 1,, β n ) is not an integral basis. Let p be a prime dividing det(c). Then p 2 Disc(β 1,, β n ) = det(c) 2 K. Denote by C the reduction of C modulo p. Let ( x 1,, x n ) T F n p be a nonzero column vector such that C( x 1,, x n ) T = 0. If x i denotes the unique lift of x i in {0, 1,, p 1}, then we see that i x iβ i po K. Conversely, if such a nonzero i x iβ i po K exists, then 0 ( x 1,, x n ) Ker( C). Hence, det(c) is divisible by p, and (β 1,, β n ) is not an integral basis. Proposition Let α O K such that K = Q(α), and f(t ) Z[T ] be its minimal polynomial. Assume that for each prime p with p 2 Disc(1, α,, α n 1 ), there exists an integer i (which may depend on p) such that f(t + i) is an Eisenstein polynomial for p. Then O K = Z[α]. Here, recall that a polynomial f(t ) = T n + a 1 T n a n is called an Eisenstein polynomial for p if p a i for all 1 i n and p 2 a n. Proof. Note that Z[α] = Z[α i] for all integer i Z. Using replacing α by α i and using Lemma 1.3.7, we need to show that if f(t ) = T n + a 1 T n a n is an Eisentein polynomial for some prime p, then x = 1 n 1 p i=0 x iα i / O K for x i {0, 1,, p 1} not all zero. Put j = min{i x i 0}. Then N K/Q (x) = N K/Q(α) j n 1 p n N K/Q ( x i α i j ). We claim that N K/Q ( n 1 i=j x iα i j ) x n j mod p. But the denominator of aj n p n is divisible by p, since p N K/Q (α) = ( 1) n a n. Therefore, it follows that N K/Q (x) / Z, and hence x / O K. To prove the claim, let σ 1,, σ n denote the complex embeddings of K. Then n 1 N K/Q ( x i α i j ) = i=j i=j n (x j + x j+1 σ k (α) i j + x n 1 σ k (α) n 1 ). k=1 Expanding the product, we see easily that all terms, except for x n j, are divisible by p, since they can be expressed as linearly combinations of ( 1) k a k for k 1, which is kelementary symmetric functions of α 1,, α n s.
14 14 CHAPTER 1. NUMBER FIELDS AND ALGEBRAIC INTEGERS Example Let K = Q(α) with α 3 = 2. We see easily that Disc(1, α, α 2 ) = But f(t ) = T 3 2 is Eisenstein for p = 2 and f(t 1) = T 3 3T 2 + 3T 3 is Eisenstein for p = 3. Hence, we get O K = Z[α] by the previous Proposition. We now give another property on the sign of the discriminant K. Let σ : K C be a complex embedding. We say that σ is a real embedding if σ(k) R; otherwise, we say σ is complex. Genuine complex embeddings of K always come in pairs. Actually, the composition of σ with the complex conjugation, denoted by σ, is another complex embedding of K. We denote usually by r 1 the number of real embeddings of K, and by r 2 the number of pairs of genuine complex embeddings so that n = r 1 + 2r 2. Proposition The sign of K is ( 1) r 2. Proof. We label the n embeddings of K into C as σ 1,, σ n such that σ 1,, σ r1 are real, and σ r1 +2i = σ r1 +2i 1 for 1 i r 2. Let (α 1,, α n ) denote an integral basis of K. Then det(σ i (α j )) = det(σ i (α j )) = ( 1) r 2 det(σ i (α j )), because the matrix ( σ i (α j )) 1 i,j n is obtained from (σ i (α j )) 1 i,j n by swiping the r 1 + 2i 1th and r 1 + 2ith rows for all 1 i r 2. Therefore, if r 2 is even then det(σ i (α j )) is real, hence K = (det(σ i (α j ))) 2 is positive; and if r 2 is odd then det(σ i (α j )) is purely imaginary, thus K is negative. Remark By Lemma 1.3.1, the discriminant of any Qbasis of K has the sign as K Cyclotomic fields Let N 3 be an integer, and ζ N C be a primitive nth root of unity. Consider the number field Q(ζ N ). Then for any σ Aut Q (C), σ(ζ N ) must be also a primitive Nth root of unity, hence of the form ζ a N for some a coprime to N. Therefore, Q(ζ N) is a Galois extension over Q, and we have an injective map of groups ϕ : Gal(Q(ζ N )/Q) (Z/NZ). Proposition The homomorphism ϕ is an isomorphism Gal(Q(ζ N )/Q) = (Z/NZ). Proof. To prove the subjectivity of ϕ, it suffices to show that the image of every prime p with p N in (Z/NZ) lies in the image of φ (since such elements generate the group (Z/NZ) ). It is equivalent to showing that ζ p N is a conjugate of ζ N. Let f(t ) Z[T ] denote the minimal polynomial of ζ N, and write T N 1 = f(t )g(t ) with g(t ) Z[T ]. Suppose in contrary that ζ p N is not conjugate to ζ N. Then one has g(ζ p N ) = 0, that is ζ n is a root of g(t p ). Since f(t ) is the minimal polynomial of ζ n, one has f(t ) g(t p ). Let f, ḡ F p [T ] denote the reduction modulo p of f and g respectively. Note that ḡ(t ) p = ḡ(t p ), so we get f(t ) ḡ(t ) p. If α is any root of f(t ) in an algebraic closure F p of F p, then ḡ(α) = 0. Then means that α is a multiple root of F (T ) = f(t )ḡ(t ). But F (α) = Nα N 1 0 in F p, hence F (T ) has no multiple root. This is a contradiction.
15 1.4. CYCLOTOMIC FIELDS 15 Corollary If N, M 2 are integers with gcd(n, M) = 1, then we have Q(ζ N ) Q(ζ M ) = Q. Proof. Note that Q(ζ NM ) = Q(ζ M )Q(ζ N ) as subfields of C. By field theory, one has [Q(ζ MN ) : Q(ζ N )] = [Q(ζ M ) : Q(ζ M ) Q(ζ N )]. Therefore, Q(ζ M ) Q(ζ N ) = Q if and only if [Q(ζ MN ) : Q(ζ N )] = φ(m), where φ(m) := #(Z/MZ) is the Euler function. But this follows from [Q(ζ MN ) : Q(ζ N )] = [Q(ζ MN ) : Q]/[Q(ζ N : Q)] = φ(mn)/φ(n) = φ(m). We manage to compute the discriminant of Q(ζ N ). We put Φ N (T ) = (T ζn) a Z[T ], a (Z/NZ) and call it the Ncyclotomic polynomial. Lemma The discriminant of Q(ζ N ) divides N φ(n). Proof. Since Q(ζN ) Disc(1, ζ N,, ζ φ(n) 1 N ), it suffices to prove the latter divides N φ(n). Write T N 1 = Φ N (T )F (T ) for some F (T ) Z[T ]. Then we get NT N 1 = Φ N(T )F (T ) + Φ N (T )F (T ). Thus N Q(ζN )/Q(Φ (ζ N )) N Q(ζN )/Q(Nζ N 1 N ) = N φ(n). We conclude by Proposition Corollary If p is a prime, then the ring of integers of Q(ζ p n) is Z[ζ p n]. Proof. Indeed, Φ p n(x + 1) is an Eisenstein polynomial for p, and the statement follows from Proposition In order to generalize the previous Corollary to arbitrary Q(ζ N ), we need some preparation. Let K and L be two number fields, and KL be the composite field (inside C). Consider the subring O K O L = {x 1 y x r y r x i O K, y j O L }. We have always O K O L O KL, but they are not equal in general. However, we have the following Proposition Assume that K L = Q, and put d = gcd( K, L ). Then we have O KL 1 d O KO L. Proof. Let (α 1,, α n ) and (β 1,, β m ) be integral basis of K and L respectively. Any x O KL writes as x = x i,j r α iβ j, with m i,j, r Z and gcd(m i,j, r) = 1. i,j
16 16 CHAPTER 1. NUMBER FIELDS AND ALGEBRAIC INTEGERS We have to show that r d. By symmetry, it suffices to prove that r L. Let (α i ) 1 i n be the dual basis of (α i ) 1 i n with respect to Tr K/Q. Then we have Tr KL/L (xα i ) = k,l x k,l r Tr KL/L(α k β l α i ) = l x i,l r β l. On the other hand, we have αi 1 K O K by definition of αi and Cramer s rule. So xαi 1 K O KL, and hence Tr KL/L (xαi ) 1 K Tr KL/L (O KL ) 1 K O L, i.e. K Tr KL/L (xαi ) O L. But (β j ) 1 j m is a basis of O L over Z, thus K xi,j r Z for all i, j, and so r K. Corollary Assume that K L = Q and gcd( K, L ) = 1. O KL = O K O L. Then we have We can now prove Theorem The ring of integers of Q(ζ N ) is Z[ζ N ]. Proof. We prove the statement by induction on the number of prime factors of N. When N is a power of some prime p, then this is proved in Corollary If N contains several prime factors, then write N = nm with n, m > 1 and gcd(n, m) = 1. By Corollary and Lemma 1.4.3, the assumptions of Corollary are satisfied. We conclude by induction hypothesis that O Q(ζN ) = O Q(ζn)O Q(ζm) = Z[ζ n, ζ m ] = Z[ζ N ]. We can also compute the exact value of Q(ζN ) when N = p n, with p a prime. Proposition We have Q(ζp n) = Disc(1, ζ p n,, ζ pn 1 (p 1) 1 p n ) = ±ppn 1 (pn n 1), where we have if p 3 mod 4 or p n = 4, and we have + otherwise. Proof. The statement for the sign follows easily from Remark Compute now Disc(1, ζ p n,, ζ pn 1 (p 1) 1 p n ), which is equal to N Q(ζ p n)/q(φ p n(ζ pn)) by Proposition 1.3.2, where Φ p n(t ) = T pn p 1 1 T pn 1 1 = T pn 1i. If p = 2, then Φ 2 n(ζ 2 n) = 2n 1 ζ 2n n and N Q(ζ 2 n)/q(φ 2 n(ζ 2 n)) = 22n 1 (n 1). If p 3, then Φ (ζ p n) = p n 1 ζ pn 1 p n p 1 i=1 i=0 iζ pn 1 (i 1) p n p 2 = p n 1 ζ pn 1 +(p 2) p n (1 ζp). i i=1 = pn 1 ζ pn 1 p n Φ p(ζ p )
17 1.4. CYCLOTOMIC FIELDS 17 Therefore, N Q(ζp n)/q = p pn 1 (p 1)(n 1) p 2 i=1 N Q(ζ p)/q(ζp i 1) pn 1. But the minimal polynomial of ζp i 1 over Q is ( ) p Φ p (X + 1) = X p 1 + px p X + p. 2 Thus we have N Q(ζp)/Q(ζ i p 1) = p, and the Lemma follows immediately.
18
19 CHAPTER 2 DEDEKIND DOMAINS 2.1. Preliminaries on Noetherian rings All rings in this section are commutative. Proposition Let R be a ring, and M be an Rmodule. The following statements are equivalent: 1. Every submodule of M (including M itself) is finitely generated. 2. For any increasing chain of ideals N 1 N 2 N n N n+1, there exists an integer m such that N n = N n+1 for all n m. 3. Every nonempty subset S of submodules of M contains a maximal element N under inclusion, i.e. if N S contains N, then N = N. Proof. We prove first (1) = (2). Given an increasing chain of submodules N 1 N 2 N n, put N = n 1 N n. Write N = (x 1,, x r ). If m 1 is large enough so that all x i N m, then N n = N for all n m. For (2) = (3), we assume that S does not contain any maximal element. Take an arbitrary N 1 S. Since N 1 is not maximal, there exists N 2 S such that N 1 N 2. Continuing this process, we produce an increasing chain of ideals N 1 N 2 N n N n+1, whose existence contradicts with (2). Finally, we prove (3) = (1). It is enough to prove that M is finitely generated, since the same arguments apply with M replaced by any submodule N M. Consider the set S consisting of all finitely generated submodules of M. Then S is nonempty, because (0) S. Let N S be a maximal element. For any x M, N = N + R x is also finitely generated and N N. Then one has N = N by the maximality of N. This implies that x N, i.e. N = M. Definition (1) We say an Rmodule M is Noetherian if it satisfies the equivalent conditions in the previous Proposition. (2) We say a ring R is Noetherian, if R itself is Noetherian as an Rmodule. Proposition Let 0 M 1 M M 2 0 be a short exact sequence of Rmodules. Then M is Noetherian if and only if both M 1 and M 2 are Noetherian.
20 20 CHAPTER 2. DEDEKIND DOMAINS Proof. The only if part is easy and left as an Exercise. Assume now both M 1 and M 2 are Noetherian. Let N be a submodule of M. Put N 1 = M 1 N and N 2 M 2 to be the image of N. By assumption, both N 1 and N 2 are finitely generated. Let N 1 = (x 1,, x r ), and x r+1,, x r+s N be such that their image in N 2 generate N 2. Then we claim that N is generated by x 1,, x r+s. Indeed, for any x N, there exist a r+1,, a r+s R such that the image of x s i=1 a r+ix r+i in N 2 is zero, i.e. x s i=1 a r+ix r+i N 1. Thus there exist a 1,, a r R such that x s i=1 a r+ix r+i = r i=1 a ix i, that is x (x 1,, x r+s ). Corollary Any finitely generated Rmodule is Noetherian. Proof. Indeed, any finitely generated Rmodule is a quotient of R n for some n. Corollary (1) If R is a Noetherian ring, then any quotient of R is also Noetherian. (2) If R 1 and R 2 are Noetherian rings, then so is R 1 R 2. Proof. (1) If R is quotient of R, then any ideal of R is finitely generated as Rmodule, hence as Rmodule. (2) It suffices to note that any ideal of R 1 R 2 is of the form I 1 I 2 where I j is an ideal of R j. Example (1) Principal ideal domains such as Z, Q[X] are Noetherian. (2) The ring Q[X 1, X 2,, X n, ] is nonnoetherian. The ring of all algebraic numbers is also nonnoetherian. Finally, we have famous theorem of Hilbert. Theorem (Hilbert basis theorem). If R is Noetherian, then R[X] is also Noetherian. Proof. Let J R[X] be an ideal. Let I R denote the subset consisting of R such that there exists some f J whose top coefficient is a. Then we see easily that I is an ideal of R. Choose f i = a i,di X d i + + a i,0 J for 1 i r such that I = (a 1,d1,, a r,dr ). Let d = max i {d i }. The polynomials of degree < d contained in J form a finitely generated Rmodule; let {g 1,, g s } denote a set of generators over R. Let f J of degree n = deg(f). We claim that f is generated by {f 1,, f r, g 1,, g s }. If n < d, then f is generated by {g 1,, g s } (even over R). If n d, then there exist b 1,, b r R such that f = f r i=1 b ix n d i f i has degree strict less than n. Repeating the process with f replaced by f, one may finally reduce to the case of degree < d. Combining with Corollary 2.1.5, we have the following Corollary If R is a finitely generated algebra over Z or over a field, then R is Noetherian.
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