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 intral 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 intral 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, intrally 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 B M 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 inters Z is a Dedekind domain. In fact, any principal ideal domain is a Dedekind domain since a principal ideal domain is Noetherian intrally 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 intral closure of Z in F, then Theorem 4 below shows that A is a Dedekind domain. The ring A is called the ring of inters 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}. Moreover, the local ring O P (X/k) is equal to the localization of Γ(X/k) at M. Since each local ring O P (X/k) is a discrete valuation ring, Γ(X/k) is a Dedekind domain. Lemma 3. Let A be an intrally 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 intral over A if and only if the 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 intral over A. Conversely, suppose that b is intral over A. Let L be the normal closure of F (b)/f, and let C be the intral closure of A in L. Since b is intral over A, there is a monic g(x) A[x] with g(b) = 0. Then p divides g, and since L/F is normal, p splits in L (recall that if M/F is a normal extension and a M, then the minimal polynomial of a over F splits in M). Any root of g is intral 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 1 ) (x c n ) L[x]. Since each c i C, the coefficients of p(x) also lie in C since they are sums of products of the c i. Thus, the coefficients lie in C F. This ring contains A, is contained in F, and is intral over A. Since A is intrally closed, we see that C F = A, so the coefficients of p lie in A. Thus, p(x) A[x]. 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 intral 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 Noetherian, intrally closed domain of Krull dimension 1. The ring B is intrally closed because if x K is intral over B, then x is intral over A since intrality is a transitive property. Therefore, x B. Moreover, since dim(a) = 1 and B/A is intral, dim(b) = 1 by the incomparability theorem: If Q 1 Q 2 are prime ideals of B with Q 1 A = Q 2 A, then Q 1 = Q 2. 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 intral 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 1,..., b n B be an F -basis for K. We can choose the b i B for the following reason. Given any basis x 1,..., x n, we can write x i = c i /d i with each c i, d i B. By using common denominators, we can then write each x i in the form b i /d with b i, d B. The b i then form a basis for K. Now, recall that the trace map T : K F is nonzero. Therefore, the bilinear map K K F given by (a, b) T (ab) is nondenerate. Given the basis b 1,..., b n, there is a dual basis y 1,..., y n with T (b i y j ) = 1 if i = j and 0 otherwise. Let b B, and write b = i α iy i with each α i F. Then bb j = i α iy i b j, so T (bb j ) = α j. However, T (bb j ) = σ Gal(K/F ) σ(bb j), and σ(b) = B for each σ. Therefore, T (bb j ) B F = A, so each α i A. This shows that B i Ay i, so B lies in a finitely generated A-module. This module is Noetherian since A is a Noetherian ring, so B is also a 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 B 0 be the intral closure of 2
A in S. By what we have just proven, B 0 is a Dedekind domain, and note that B is the intral closure of B 0 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 ] = p n, and that a pn F for all a K. For simplicity, write q = [K : F ]. Let M be the splitting field over F of all polynomials of the form x q α with α F. Then K M. Let C be the intral closure of A in M. Note that the map ϕ : M F defined by ϕ(x) = x q is injective since char(f ) = p and q is a power of p, and ϕ is surjective by the definition of M. Thus, ϕ is a field isomorphism. Moreover, ϕ(c) = A. The inclusion A ϕ(c) follows from the definition of intrality, 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 are b i I and d i (C : IC) with i b id i = 1. From this we get i bq i dq i = 1. Moreover, d q i F by the definition of M. Let e i = b q 1 i d q i K. Then i b ie i = 1. The elements d i satisfy d i I C by definition, so e i I C. Intersecting with K gives e i I C K = B, so e i (B : I). Thus, we have proven that I(B : I) = B, so I is invertible, and so B is a Dedekind domain. An important fact we need is to determine the discrete valuation rings of a finite extension F of the rational function field k(x) that contain k[x] (x). Proposition 5. Let K/F be a finite field extension, let O be a discrete valuation ring of F, and let B be the intral closure of O in K. Then the discrete valuation rings of F that contain O are the localizations B M at maximal ideals of B. Proof. Let B be the intral closure of O in K. Then B is a Dedekind domain by Theorem 4. Thus, if M is a maximal ideal of B, then B M is a discrete valuation ring. Since O B because O is intrally closed, O B M. Conversely, let V be a valuation ring of K that contains O. If b B, then b is intral over O. Thus, b is also intral over V. Therefore, b V since V is intrally closed. Therefore, B V. Let P be the maximal ideal of V. Then P B = M is a prime ideal of B. We have B M V P = V, and since V K, the ideal M is nonzero. Thus, since every nonzero prime ideal of B is maximal, M is maximal, and since B M is a discrete valuation ring, B M = V since discrete valuation rings are maximal subrings of their quotient fields. Thus, the valuation rings of K containing O are precisely the localizations B M at maximal ideals of M. This proposition tells us how to obtain the discrete valuation rings of K that contain O. We do not yet know how many there are. In particular, we have not yet shown that there are only finitely many such valuation rings. We do this now. Proposition 6. With notation in the previous proposition, there are only finitely many maximal ideals of B, and so there are only finitely many discrete valuation rings of K that contain O. 3
Proof. Let t be a uniformizer of O. Then to is the maximal ideal of O. Consider the ideal tb of B. Then this ideal factors into a product of prime ideals since B is a Dedekind domain. Thus, we may write tb = M e 1 for some inters e i 1 and distinct maximal ideals M 1,..., M g. We claim that {M 1,..., M g } is exactly the set of maximal ideals of B. These are maximal ideals since they are nonzero prime ideals. Conversely, let M be a maximal ideal of B. Then B M is a discrete valuation ring of K, and B M contains O since B contains O. Thus, MB M O = to because MB M is the maximal ideal of B M. Since MB M B = M, we see that t M. Then tb M. So, M e 1 M. Because M is prime, M i M for some i. However, M i is maximal, which forces M = M i. We have thus proved that {M 1,..., M g } is the set of maximal ideals of B, and so B M1,..., B Mg are all the discrete valuation rings of K that contain O. We finish this note with a technical looking result, which will be the heart of the proof that principal divisors of an algebraic function field have dree 0. We first need two lemmas, which are useful facts about Dedekind domains or discrete valuation rings. Lemma 7. Let B be a Dedekind domain with only finitely many maximal ideals. Then B is a principal ideal domain. Proof. Let M 1,..., M g be the maximal ideals of B. By the prime avoidance theorem, M 1 is not contained in M1 2 M 2 M g. Therefore, there is a t M 1 with t / M1 2, and t / M i for each i 2. Consider the ideal tb. This factors as a product of maximal ideals. Thus, there are inters e i 0 with tb = M e 1 1 M e g g. For i 2, if e i 1, then tb M e i i M i, which says t M i. Since this is false, e i = 0 for i 2. Therefore, tb = M e 1 1. Since t / M1 2, we have e i 2. So, e 1 = 1, and so tb = M 1. This proves that M 1 is principal. Similarly, each M i is principal. Finally, since every nonzero ideal of B is a product of maximal ideals, each ideal is principal. Let B be a Dedekind domain with finitely many maximal ideals M 1,..., M g. Let v i be a valuation on the quotient field F of B whose valuation ring is B Mi. If x B and xb factors as xb = M e 1 1 M e g g, then xb Mi = M e i i B Mi, and so v i (x) = e i. Thus, we can recover the values of x in terms of the factorization into prime ideals of the ideal xb. Lemma 8. Let O be a discrete valuation ring. O-module is free. Then any finitely generated torsion free Proof. Let T be a finitely generated torsion free O-module. Pick n minimal such that T is generated by n elements, and let {t 1,..., t n } be a generating set of T. We will prove that {t 1,..., t n } is a basis. Since this set generates T, we only need to prove that it is linearly independent over O. Suppose that i x it i = 0 with x i O. Let v be a valuation on the quotient field F of O whose valuation ring is O. If some x i 0, then let v(x j ) = min {v(x i ) : 1 i n}. Then x i x 1 j O for each i since v(x i x 1 j ) 0. Then i x 1 j x i t i = 0, and so t j = i j (x 1 j x i )t i is an O-linear combination of {t i : i j}. Therefore, we have 4
a generating set for T with n 1 elements. This contradiction shows that each x i = 0, and so {t 1,..., t n } is independent, and so is a basis for T as an O-module. Therefore, T is a free module. Proposition 9. Let K/F be a finite separable extension of fields, let O be a discrete valuation ring of F with maximal ideal P = to, and let B be the intral closure of O in K. Let M 1,..., M g be the maximal ideals of B, and let f i = [B/M i : O/P ]. If tb = M e 1, then [K : F ] = g i=1 e if i. Proof. We first remark that the proof of Theorem 4 shows that B is contained in a finitely generated O-module. Since O is a discrete valuation ring, any finitely generated torsion free module is free. Since O is a Noetherian ring, any finitely generated O-module is a Noetherian O-module, and so any submodule is finitely generated. Thus, B is finitely generated as an O-module, and so B is free by Lemma 8. Its rank as an O-module is n = [K : F ] since B O F = K; this fact is (almost) really a restatement that the quotient field of K is B. Let k = O/P, a field. Then dim k B/tB = n because B/tB = B O k. By the Chinese remainder theorem, B/tB = B/(M e 1 ) = g i=1 B/M e i i. It then suffices to prove that dim k (B/M e i i ) = e i f i. Set M = M i, e = e i, and f = f i. We will prove this by considering the chain B M M 2 M e. Each quotient M r /M r+1 is a k-vector space because each is an O-module with tm r M r+1, since tb M e. We then have a sequence of k-vector spaces B/M, M/M 2,..., M e 1 /M e. In fact, B/M e is a k-vector space for the same argument that each quotient above is a k-vector space. We prove that dim k (M r /M r+1 ) = f for each r. Since dim k (V/W ) = dim k (V ) dim k (W ) for any k-vector spaces W V, induction shows that dim k (B/M r ) = rf for each r, and so dim k (B/M e ) = ef, as desired. Now, to prove this, we produce a k-vector space isomorphism M r /M r+1 = B/M. The ideal M is principal by the previous lemma; say M = sb. Then M r = s r B. Define ϕ : B M r /M r+1 by ϕ(b) = s r b + M r+1. Then ϕ is an O-module homomorphism with ϕ(m) = 0. Thus, we have an induced map, which we also denote by ϕ, from B/M to M r /M r+1, given by ϕ(b + M) = s r b + M r+1. This is a k-vector space map since it is an O-module map. It is clear that ϕ is surjective. To prove that it is injective, suppose that ϕ(b + M) = 0. Then s r b M r+1 = s r+1 B. So, there is a c B with s r b = s r+1 c. Then b = sc M, so b + M = 0, as desired. This finishes the argument. 5