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13. Dedekind Domains 117 13. Dedekind Domains 117 13. Dedekind Domains In the last chapter we have mainly studied 1-dimensional regular local rings, i. e. geometrically the local properties of smooth points on curves. We now want to patch these local results together to obtain global statements about 1-dimensional rings (resp. curves) that are “locally regular”. The corresponding notion is that of a Dedekind domain. Definition 13.1 (Dedekind domains). An integral domain R is called Dedekind domain if it is Noetherian of dimension 1, and for all maximal ideals P E R the localization RP is a regular local ring. Remark 13.2 (Equivalent conditions for Dedekind domains). As a Dedekind domain R is an integral domain of dimension 1, its prime ideals are exactly the zero ideal and all maximal ideals. So every localization RP for a maximal ideal P is a 1-dimensional local ring. As these localizations are also Noetherian by Exercise 7.23, we can replace the requirement in Definition 13.1 that the local rings RP are regular by any of the equivalent conditions in Proposition 12.14. For example, a Dedekind domain is the same as a 1-dimensional Noetherian domain such that all localizations at maximal ideals are discrete valuation rings. This works particularly well for the normality condition as this is a local property and can thus be transferred to the whole ring: Lemma 13.3. A 1-dimensional Noetherian domain is a Dedekind domain if and only if it is normal. Proof. By Remark 13.2 and Proposition 12.14, a 1-dimensional Noetherian domain R is a Dedekind domain if and only if all localizations RP at a maximal ideal P are normal. But by Exercise 9.13 (c) this is equivalent to R being normal. 25 Example 13.4. (a) By Lemma 13.3, any principal ideal domain which is not a field is a Dedekind domain: it is 1-dimensional by Example 11.3 (c), clearly Noetherian, and normal by Example 9.10 since it is a unique factorization domain by Example 8.3 (a). For better visualization, the following diagram shows the implications between various properties of rings for the case of integral domains that are not fields. Rings that are always 1-dimensional and / or local are marked as such. It is true that every regular local ring is a unique factorization domain, but we have not proven this here since this requires more advanced methods — we have only shown in Proposition 11.40 that any regular local ring is an integral domain. dim = 1 13.4 (a) dim = 1 =) 12.14 (b) PID Dedekind 13.3 =) dim = 1 =) = def. DVR ) 8.3 (a) normal =) domain local =) =) 12.14 (e) regular =) UFD 9.10 local (b) Let X be an irreducible curve over an algebraically closed field. Assume that X is smooth, i. e. that all points of X are smooth in the sense of Example 11.37 and Definition 11.38. Then the coordinate ring A(X) is a Dedekind domain: it is an integral domain by Lemma 2.3 (a) since I(X) is a prime ideal by Remark 2.7 (b). It is also 1-dimensional by assumption and 118 Andreas Gathmann Noetherian by Remark 7.15. Moreover, by Hilbert’s Nullstellensatz as in Remark 10.11 the maximal ideals in A(X) are exactly the ideals of points, and so our smoothness assumption is the same as saying that all localizations at maximal ideals are regular. In fact, irreducible smooth curves over algebraically closed fields are the main geometric examples for Dedekind domains. However, there is also a large class of examples in number theory, which explains why the concept of a Dedekind domain is equally important in num- ber theory and geometry: it turns out that the ring of integral elements in a number field, i. e. in a finite field extension of Q, is always a Dedekind domain. Let us prove this now. Proposition 13.5 (Integral elements in number fields). Let Q ⊂ K be a finite field extension, and let R be the integral closure of Z in K. Then R is a Dedekind domain. Proof. As a subring of a field, R is clearly an integral domain. Moreover, by Example 11.3 (c) and Lemma 11.8 we have dimR = dimZ = 1. It is also easy to see that R is normal: if a 2 QuotR ⊂ K is integral over R it is also integral over Z by transitivity as in Lemma 9.6 (b), so it is contained in the integral closure R of Z in K. Hence by Lemma 13.3 it only remains to show that R is Noetherian — which is in fact the hardest part of the proof. We will show this in three steps. (a) We claim that jR=pRj < ¥ for all prime numbers p 2 Z. Note that R=pR is a vector space over Zp = Z=pZ. It suffices to show that dimZp R=pR ≤ dimQ K since this dimension is finite by assumption. So let a1;:::;an 2 R=pR be linearly independent over Zp. We will show that a1;:::;an 2 K are also independent over Q, so that n ≤ dimQ K. Otherwise there are l1;:::;ln 2 Q not all zero with l1a1 + ··· + lnan = 0. After multiplying these coefficients with a common scalar we may assume that all of them are integers, and not all of them are divisible by p. But then l1 a1 + ··· + ln an = 0 is a non-trivial relation in R=pR with coefficients in Zp, in contradiction to a1;:::;an being independent over Zp. (b) We will show that jR=mRj < ¥ for all m 2 Znf0g. In fact, this follows by induction on the number of prime factors in m: for one prime factor the statement is just that of (a), and for more prime factors it follows from the exact sequence of Abelian groups ·m2 0 −! R=m1R −! R=m1m2R −! R=m2R −! 0; since this means that jR=m1m2Rj = jR=m1Rj · jR=m2Rj < ¥. (c) Now let I E R be any non-zero ideal. We claim that m 2 I for some m 2 Znf0g. Otherwise we would have dimR=I = dimZ=(I \ Z) = dimZ = 1 by Lemma 11.8, since R=I is integral over Z=(I \ Z) by Lemma 9.7 (a). But dimR has to be bigger than dimR=I, since a chain of prime ideals in R=I corresponds to a chain of prime ideals in R containing I, which can always be extended to a longer chain by the zero ideal since R is an integral domain. Hence dimR > 1, a contradiction. Putting everything together, we can choose a non-zero m 2 I \ Z by (c), so that mR E I. Hence jI=mRj ≤ jR=mRj < ¥ by (b), so I=mR = fa1;:::;ang for some a1;:::;an 2 I. But then the ideal I = (a1;:::;an;m) is finitely generated, and hence R is Noetherian. p Example 13.6. Consider againp the ring R = Z[ 5i] of Example 8.3 (b). By Example 9.16, it is the integral closure of Z in Q( 5i). Hence Proposition 13.5 shows that R is a Dedekind domain. We see from this example that a Dedekind domain is in general not a unique factorization domain, as e. g. by Example 8.3 (b) the element 2 is irreducible, but not prime in R, so that it does not have a factorization into prime elements. However, we will prove now that a Dedekind domain always has an analogue of the unique factorization property for ideals, i. e. every non-zero ideal can be written uniquely as a product of non-zero prime ideals (which are then also maximal since Dedekind 13. Dedekind Domains 119 domains are 1-dimensional). In fact, this is the most important property of Dedekind domains in practice. Proposition 13.7 (Prime factorization of ideals in Dedekind domains). Let R be a Dedekind domain. (a) Let P E R be a maximal ideal, and let Q E R be any ideal. Then k Q is P-primary , Q = P for some k 2 N>0: Moreover, the number k is unique in this case. (b) Any non-zero ideal I E R has a “prime factorization” k1 kn I = P1 ····· Pn with k1;:::;kn 2 N>0 and distinct maximal ideals P1;:::;Pn E R. It is unique up to permuta- tion of the factors, and P1;:::;Pn are exactly the associated prime ideals of I. Proof. (a) The implication “(” holds in arbitrary rings by Lemma 8.12 (b), so let us show the opposite e direction “)”. Let Q be P-primary, and consider the localization map R ! RP. Then Q is a non-zero ideal in the localization RP, which is a discrete valuation ring by Remark 13.2. So by Corollary 12.17 we have Qe = (Pe)k for some k, and hence Qe = (Pk)e as extension commutes with products by Exercise 1.19 (c). Contracting this equation now gives Q = Pk by Lemma 8.33, since Q and Pk are both P-primary by Lemma 8.12 (b). The number k is unique since Pk = Pl for k 6= l would imply (Pe)k = (Pe)l by extension, in contradiction to Corollary 12.17. (b) As R is Noetherian, the ideal I has a minimal primary decomposition I = Q1 \···\ Qn by Corollary 8.21. Since I is non-zero, the corresponding associated prime ideals P1;:::;Pn of these primary ideals are distinct and non-zero, and hence maximal as dimR = 1.
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