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).