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Home , Dedekind domain, Integrally closed domain, Principal ideal domain

Dedekind

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 with quotient field F , then the integral closure of A in K is also a . As we will see in the proof, we need various results from theory and field theory. We first recall some basic definitions and facts. A Dedekind domain is an B for which every nonzero 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 B is a Dedekind domain if and only if BM is a discrete ring for every M of B. In particular, a Dedekind domain that is a is a discrete , and vice-versa. We start by mentioning two examples of Dedekind domains.

Example 1. The ring of Z is a Dedekind domain. In fact, any domain is a Dedekind domain since a 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 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 in the number field F . The study of the arithmetic of A is the heart of . 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 (X/k) is equal to the localization of Γ(X/k) at M. Since each OP local ring (X/k) is a , Γ(X/k) is a Dedekind domain. OP Lemma 3. Let A be an 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 ∈ minimal 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 ∈ 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 ∈ 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 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 ∈ i ∈ 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]. ∩ ∈ 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 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. ∈ Therefore, x B. Moreover, since dim(A) = 1 and B/A is integral, dim(B) = 1 by the ∈ 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-. 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 ∈ the b B for the following reason. Given any basis x , . . . , x , we can write x = c /d with i ∈ 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 ∈ 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 ∈ 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 ∈ i i i i ∈ 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 A∈. This shows that B Ay , so B lies in a finitely j ∈ ∩ i ∈P ⊆ 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 ∞ that ap F for all a K. For simplicity, write q = [K : F ]. Let M be the splitting field ∈ ∈ over F of all of the form xq α with α F . Then K M. Let C be the − ∈ ⊆ integral closure of A in M. Note that the map ϕ : M F defined by ϕ(x) = xq 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 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 ∈ ∈ q 1 q d F by the definition of M. Let e = b − d K. Then b e = 1. The elements d i ∈ Pi i i ∈ 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). Thus, we have proven that I(B : I) = B, so I is invertible, and so B is a i ∈ 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 be a discrete valuation ring of O F , and let B be the integral closure of in K. Then the discrete valuation rings of F that O contain are the localizations B at maximal ideals of B. O M Proof. Let B be the integral closure of in K. Then B is a Dedekind domain by Theorem O 4. Thus, if M is a maximal ideal of B, then B is a discrete valuation ring. Since B M O ⊆ because is integrally closed, B . Conversely, let V be a valuation ring of K that O O ⊆ M contains . If b B, then b is integral over . Thus, b is also integral over V . Therefore, O ∈ O b V since V is integrally closed. Therefore, B V . Let P be the maximal ideal of V . ∈ ⊆ Then P B = M is a of B. We have B V = V , and since V = K, the ideal ∩ M ⊆ P 6 M is nonzero. Thus, since every nonzero prime ideal of B is maximal, M is maximal, and

since BM is a discrete valuation ring, BM = V since discrete valuation rings are maximal of their quotient fields. Thus, the valuation rings of K containing are precisely O the localizations BM 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 . Then t is the maximal ideal of . Consider the ideal O O O tB of B. Then this ideal factors into a product of prime ideals since B is a Dedekind domain. e1 eg Thus, we may write tB = M Mg for some integers e 1 and distinct maximal ideals 1 ··· i ≥ M ,...,M . We claim that M ,...,M is exactly the set of maximal ideals of B. These 1 g { 1 g} are maximal ideals since they are nonzero prime ideals. Conversely, let M be a maximal ideal of B. Then B is a discrete valuation ring of K, and B contains since B contains M M O . Thus, MBM = t because MBM is the maximal ideal of BM . Since MBM B = M, O ∩O O e1 eg ∩ we see that t M. Then tB M. So, M Mg M. Because M is prime, M M ∈ ⊆ 1 ··· ⊆ i ⊆ for some i. However, Mi is maximal, which forces M = Mi. We have thus proved that M ,...,M is the set of maximal ideals of B, and so B ,...,B are all the discrete { 1 g} M1 Mg 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 degree 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 M1,...,Mg be the maximal ideals of B. By the prime avoidance theorem, M1 is not contained in M 2 M M . Therefore, there is a t M with t / M 2, and t / M 1 ∪ 2 ∪ · · · ∪ g ∈ 1 ∈ 1 ∈ i for each i 2. Consider the ideal tB. This factors as a product of maximal ideals. Thus, ≥ e1 eg ei there are integers e 0 with tB = M Mg . For i 2, if e 1, then tB M M , i ≥ 1 ··· ≥ i ≥ ⊆ i ⊆ i which says t M . Since this is false, e = 0 for i 2. Therefore, tB = M e1 . Since t / M 2, ∈ i i ≥ 1 ∈ 1 we have ei  2. So, e1 = 1, and so tB = M1. This proves that M1 is principal. Similarly, each Mi 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 M1,...,Mg. Let vi be a valuation on the quotient field F of B whose valuation ring is BMi . If x B and xB factors e1 eg ei ∈ as xB = M Mg , then xB = M B , and so v (x) = e . Thus, we can recover the 1 ··· Mi i Mi i i values of x in terms of the into prime ideals of the ideal xB.

Lemma 8. Let be a discrete valuation ring. Then any finitely generated free O -module is free. O Proof. Let T be a finitely generated torsion free -module. Pick n minimal such that T O is generated by n elements, and let t , . . . , t be a generating set of T . We will prove { 1 n} that t , . . . , t is a basis. Since this set generates T , we only need to prove that it is { 1 n} linearly independent over . Suppose that x t = 0 with x . Let v be a valuation O i i i i ∈ O on the quotient field F of whose valuation ring is . If some xi = 0, then let v(xj) = O 1 P O 1 6 1 min v(xi) : 1 i n . Then xixj− for each i since v(xix−j ) 0. Then i xj− xiti = 0, { ≤ ≤ } 1 ∈ O ≥ and so tj = i=j(xj− xi)ti is an -linear combination of ti : i = j . Therefore, we have − 6 O { 6 } P P 4 a generating set for T with n 1 elements. This contradiction shows that each x = 0, and − i so t , . . . , t is independent, and so is a basis for T as an -module. Therefore, T is a free { 1 n} O module.

Proposition 9. Let K/F be a finite separable extension of fields, let be a discrete valuation O ring of F with maximal ideal P = t , and let B be the integral closure of in K. Let O O e1 eg M1,...,Mg be the maximal ideals of B, and let fi = [B/Mi : /P ]. If tB = M1 Mg , g O ··· then [K : F ] = i=1 eifi. Proof. We firstP remark that the proof of Theorem 4 shows that B is contained in a finitely generated -module. Since is a discrete valuation ring, any finitely generated torsion free O O module is free. Since is a Noetherian ring, any finitely generated -module is a Noetherian O 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 -module is n = [K : F ] since O O B F = K; this fact is (almost) really a restatement that the quotient field of K is B. Let ⊗O k = /P , a field. Then dimk B/tB = n because B/tB = B k. By the Chinese remainder O ∼ ⊗O theorem, B/tB = B/(M e1 M eg ) = g B/M ei . 1 ··· g ∼ ⊕i=1 i ei It then suffices to prove that dimk(B/Mi ) = eifi. Set M = Mi, e = ei, and f = fi. We will prove this by considering the chain B M M 2 M e. Each quotient M r/M r+1 is ⊇ ⊇ ⊇ · · · ⊇ a k- because each is an -module with tM r M r+1, since tB M e. We then O ⊆ ⊆ have a sequence of k-vector spaces

2 e 1 e B/M, M/M ,...,M − /M .

In fact, B/M e is a k-vector space for the same argument that each quotient above is a r r+1 k-vector space. We prove that dimk(M /M ) = f for each r. Since dimk(V/W ) = r dimk(V ) dimk(W ) for any k-vector spaces W V , induction shows that dimk(B/M ) = rf − e ⊆ for each r, and so dimk(B/M ) = ef, as desired. Now, to prove this, we produce a k-vector r r+1 space isomorphism M /M ∼= B/M. The ideal M is principal by the previous lemma; say M = sB. Then M r = srB. Define ϕ : B M r/M r+1 by ϕ(b) = srb + M r+1. Then ϕ is an → -module homomorphism with ϕ(M) = 0. Thus, we have an induced map, which we also O denote by ϕ, from B/M to M r/M r+1, given by ϕ(b + M) = srb + M r+1. This is a k-vector space map since it is an -module map. It is clear that ϕ is surjective. To prove that it is O injective, suppose that ϕ(b + M) = 0. Then srb M r+1 = sr+1B. So, there is a c B with ∈ ∈ srb = sr+1c. Then b = sc M, so b + M = 0, as desired. This finishes the argument. ∈

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