Dedekind Domains
Total Page：16
File Type：pdf, Size：1020Kb
Dedekind Domains Mathematics 601 In this note we prove several facts about Dedekind domains that we will use in the course of proving the Riemann-Roch theorem. The main theorem shows that if K=F is a ﬁnite extension and A is a Dedekind domain with quotient ﬁeld F , then the integral closure of A in K is also a Dedekind domain. As we will see in the proof, we need various results from ring theory and ﬁeld theory. We ﬁrst recall some basic deﬁnitions and facts. A Dedekind domain is an integral domain B for which every nonzero ideal can be written uniquely as a product of prime ideals. Perhaps the main theorem about Dedekind domains is that a domain B is a Dedekind domain if and only if B is Noetherian, integrally closed, and dim(B) = 1. Without fully deﬁning dimension, to say that a ring has dimension 1 says nothing more than nonzero prime ideals are maximal. Moreover, a Noetherian ring B is a Dedekind domain if and only if BM is a discrete valuation ring for every maximal ideal M of B. In particular, a Dedekind domain that is a local ring is a discrete valuation ring, and vice-versa. We start by mentioning two examples of Dedekind domains. Example 1. The ring of integers Z is a Dedekind domain. In fact, any principal ideal domain is a Dedekind domain since a principal ideal domain is Noetherian integrally closed, and nonzero prime ideals are maximal. Alternatively, it is easy to prove that in a principal ideal domain, every nonzero ideal factors uniquely into prime ideals. Let F be an algebraic number ﬁeld. That is, F a ﬁnite dimensional ﬁeld extension of Q. If A is the integral closure of Z in F , then Theorem 4 below shows that A is a Dedekind domain. The ring A is called the ring of integers in the number ﬁeld F . The study of the arithmetic of A is the heart of algebraic number theory. Example 2. Let X be a nonsingular aﬃne curve over a ﬁeld k. The coordinate ring Γ(X=k) is Noetherian, since it is a quotient of the Noetherian ring k[x; y]. It can be shown that for any maximal ideal M of Γ(X=k), there is a point P X with M = f Γ(X=k): f(P ) = 0 . 2 f 2 g Moreover, the local ring (X=k) is equal to the localization of Γ(X=k) at M. Since each OP local ring (X=k) is a discrete valuation ring, Γ(X=k) is a Dedekind domain. OP Lemma 3. Let A be an integrally closed domain with quotient ﬁeld F , and let K be an extension ﬁeld of F . If b K is algebraic over F , then b is integral over A if and only if the 2 minimal polynomial of b over F is in A[x]. 1 Proof. One direction is easy. Let p(x) be the (monic) minimal polynomial of b over F . If p(x) A[x], then b is integral over A. Conversely, suppose that b is integral over A. Let 2 L be the normal closure of F (b)=F , and let C be the integral closure of A in L. Since b is integral over A, there is a monic g(x) A[x] with g(b) = 0. Then p divides g, and since 2 L=F is normal, p splits in L (recall that if M=F is a normal extension and a M, then 2 the minimal polynomial of a over F splits in M). Any root of g is integral over A, and any root of p is a root of g by the divisibility relation. Therefore, all the roots of p lie in C. Say p(x) = (x c ) (x c ) L[x]. Since each c C, the coeﬃcients of p(x) also lie in − 1 ··· − n 2 i 2 C since they are sums of products of the c . Thus, the coeﬃcients lie in C F . This ring i \ contains A, is contained in F , and is integral over A. Since A is integrally closed, we see that C F = A, so the coeﬃcients of p lie in A. Thus, p(x) A[x]. \ 2 Theorem 4. Let A be a Dedekind domain with quotient ﬁeld F , let K be a ﬁnite extension of F , and let B be the integral closure of A in K. Then B is a Dedekind domain. Proof. We ﬁrst assume that K=F is separable. Recall that a Dedekind domain is a Noethe- rian, integrally closed domain of Krull dimension 1. The ring B is integrally closed because if x K is integral over B, then x is integral over A since integrality is a transitive property. 2 Therefore, x B. Moreover, since dim(A) = 1 and B=A is integral, dim(B) = 1 by the 2 incomparability theorem: If Q Q are prime ideals of B with Q A = Q A, then 1 ⊆ 2 1 \ 2 \ Q1 = Q2. Therefore, a chain of prime ideals of length more than 1 cannot contract to a chain in A of length at most 1, so dim(B) 1. Moreover, if dim(B) = 0, then (0) is a maximal ≤ ideal of B, so B is a ﬁeld. But then A would be a ﬁeld, which is false. Thus, dim(B) = 1. The last step in the case that K=F is separable is to show that B is Noetherian. We prove more by showing that B is a Noetherian A-module. Since ideals of B are A-submodules of B, ascending chain condition on A-submodules is a stronger condition than ascending chain condition on ideals, so B is then a Noetherian ring. We will assume that K=F is Galois, since if N is the normal closure of K=F and if C is the integral closure of A in N, then if C is a Noetherian A-module, then B is also a Noetherian A-module since B is a submodule of C. So, assume K=F is Galois, and let b ; : : : ; b B be an F -basis for K. We can choose 1 n 2 the b B for the following reason. Given any basis x ; : : : ; x , we can write x = c =d with i 2 1 n i i i each c ; d B. By using common denominators, we can then write each x in the form b =d0 i i 2 i i with b ; d0 B. The b then form a basis for K. Now, recall that the trace map T : K F is i 2 i ! nonzero. Therefore, the bilinear map K K F given by (a; b) T (ab) is nondegenerate. × ! ! Given the basis b1; : : : ; bn, there is a dual basis y1; : : : ; yn with T (biyj) = 1 if i = j and 0 otherwise. Let b B, and write b = α y with each α F . Then bb = α y b , so 2 i i i i 2 j i i i j T (bb ) = α . However, T (bb ) = σ(bb ), and σ(B) = B for each σ. Therefore, j j j σ Gal(P K=F ) j P T (bb ) B F = A, so each α A2. This shows that B Ay , so B lies in a ﬁnitely j 2 \ i 2P ⊆ i i generated A-module. This module is Noetherian since A is a Noetherian ring, so B is also a P Noetherian A-module, as we wanted to show. We now remove the assumption that K=F is separable. We may assume that char(F ) = p > 0. Let S be the separable closure of F in K, and let B0 be the integral closure of 2 A in S. By what we have just proven, B0 is a Dedekind domain, and note that B is the integral closure of B0 in K. Thus, by replacing F by S, it suﬃces to assume that K=F purely inseparable. Recall that since [K : F ] < , there is an n with [K : F ] = pn, and n 1 that ap F for all a K. For simplicity, write q = [K : F ]. Let M be the splitting ﬁeld 2 2 over F of all polynomials of the form xq α with α F . Then K M. Let C be the − 2 ⊆ integral closure of A in M. Note that the map ' : M F deﬁned by '(x) = xq is injective ! since char(F ) = p and q is a power of p, and ' is surjective by the deﬁnition of M. Thus, ' is a ﬁeld isomorphism. Moreover, '(C) = A. The inclusion A '(C) follows from the ⊆ deﬁnition of integrality, and the other inclusion is true by the lemma. Therefore, A and C are isomorphic, so C is a Dedekind domain. To ﬁnish the proof, we recall another criterion for a ring to be a Dedekind domain: B is a Dedekind domain if and only if every ideal of B is invertible. Let I be an ideal of B. Then IC is an ideal of C, so IC is invertible. Thus, there q q are bi I and di (C : IC) with i bidi = 1. From this we get i bi di = 1. Moreover, q 2 2 q 1 q d F by the deﬁnition of M. Let e = b − d K. Then b e = 1. The elements d i 2 Pi i i 2 i i Pi i satisfy d I C by deﬁnition, so e I C. Intersecting with K gives e I C K = B, so i ⊆ i ⊆ P i ⊆ \ e (B : I).