Math 221 (Commutative Algebra) Notes, 2nd Half Live [Locally] Free or Die. This is Math 221: Commutative Algebra as taught by Prof. Gaitsgory at Harvard during the Fall Term of 2008, and as understood by yours truly. There are probably typos and mistakes (which are all mine); use at your own risk. If you find the aforementioned mistakes or typos, and are feeling in a generous mood, then you may consider sharing this information with me.1 -S. Gong [email protected] 1 CONTENTS Contents 1 Divisors 2 1.1 Discrete Valuation Rings (DVRs), Krull Dimension 1 . .2 1.2 Dedekind Domains . .6 1.3 Divisors and Fractional Ideals . .8 1.4 Dedekind Domains and Integral Closedness . 11 1.5 The Picard Group . 16 2 Dimension via Hilbert Functions 21 2.1 Filtrations and Gradations . 21 2.2 Dimension of Modules over a Polynomial Algebra . 24 2.3 For an arbitrary finitely generated algebra . 27 3 Dimension via Transcendency Degree 31 3.1 Definitions . 31 3.2 Agreement with Hilbert dimension . 32 4 Dimension via Chains of Primes 34 4.1 Definition via chains of primes and Noether Normalization . 34 4.2 Agreement with Hilbert dimension . 35 5 Kahler Differentials 37 6 Completions 42 6.1 Introduction to inverse limits and completions . 42 6.2 Artin Rees Pattern . 46 7 Local Rings and Other Notions of Dimension 50 7.1 Dimension theory for Local Noetherian Rings . 50 8 Smoothness and Regularity 53 8.1 Regular Local Rings . 53 8.2 Smoothness/Regularity for algebraically closed fields . 55 8.3 A taste of smoothness and regularity for non-algebraically closed fields . 60 9 Appendix 62 1 1 Divisors 1.1 Discrete Valuation Rings (DVRs), Krull Dimension 1 Let A be a Noetherian local domain with maximal ideal m. Definition 1.1. We say that A is a discrete valuation ring, or DVR, if m = (π), that is, if the maximal ideal is principal. 2 Lemma 1.2. If A is a DVR, then dimA=m(m=m ) = 1. Proof. Note that the image of π generates m=m2 over A=m, so the dimension is at most 1. However, by Nakayama's Lemma, m=m2 is not the zero module, so the dimension is 1. ≥0 n Lemma 1.3. For A a DVR, for any a 2 A, there is a unique n 2 Z such that a = π a0 for a0 invertible. We call this n the order of a. Proof. Uniqueness is clear by cancellation, so we seek to establish existence. Let n be the k n maximal power of π that divides a. This is finite because \m = f0g. Then if a = π a0, then a0 62 m, so a0 is invertible. Corollary 1.4. Any DVR is a PID (principal ideal domain). n Proof. For any ideal I, choose an element a 2 I of minimal order. Then if a = π a0 for a0 invertible, then πn divides all elements of I, but πn 2 I, so I = (πn), as desired. The above corollary leads directly to the following one: Corollary 1.5. If A is a DVR with maximal ideal m, then the only prime ideals in A are m and (0). Now why are they called DVRs? Here is the apparent reason: Proposition 1.6. DVRs are in bijection with fields K, equipped with a map ν : K ! Z [ f1g satisfying the following properties: 1. ν(a + b) ≥ min(ν(a); ν(b)) for any a; b 2 K 2. ν(ab) = ν(a) + ν(b) for any a; b 2 K∗. 3. ν(x) = 1 if and only if x = 0 This map is called a valuation. n Proof. Given a discrete valuation ring A, take K = Frac(A) and take ν(π a0) = n where a0 is invertible. Conversely, for K take A to be its ring of integers, i.e, A = fa 2 Kjν(a) ≥ 0g. One may check that this is a DVR with m = fa 2 Kjν(a) > 0g, because it is clear that any element with valuation 0 is invertible in A and therefore any element of valuation 1 generates m. 2 1.1 Discrete Valuation Rings (DVRs), Krull Dimension 1 Definition 1.7. Let M be a module over a domain A. We say that M is torsion-free, if for any nonzero a 2 A, a : M ! M is injective. We say that M is torsion if for any m 2 M, there is nonzero a 2 A such that am = 0. Lemma 1.8. For any module finitely generated module M over a Noetherian domain A, there is a short exact sequence - - - - 0 Mtors M Mtors−free 0 where Mtors is killed by an element of A and Mtors−free is torsion-free. Proof. This is because we may take Mtors to be all the elements which are killed by a non- zero element of A. Then this is clearly a sub-module. Since A is Noetherian, it is finitely generated, which means that it can be killed by one element of A (take the product of the elements that kill the generators). Then it is easy to check that the quotient M=Mtors is torsion-free. Lemma 1.9. For R a PID, a module M over R is flat if and only if it is torsion-free. Proof. This is the content of Problem 2 on the Midterm. Using this, we will classify modules over DVRs. Proposition 1.10. let M be a finitely generated module over a DVR R. Then ⊕n M = Mtors ⊕ R ; n i.e, where Mtors can be annihilated by π for some n. Proof. Set Mtors ⊂ M be as in Lemma 1.8 so that M=Mtors is torsion-free. Therefore, by Corollary 1.4 and Lemma 1.9 we see that it is flat. But it is over a local ring, so that means ⊕n ⊕n that it is free. So we have M=Mtors = R for some n. Furthermore, since R is free, it is additionally projective, so the above sequence splits, so ⊕n M = Mtors ⊕ R as desired. There is nothing more to say about the free part, so let us discuss the torsion part in more detail. Lemma 1.11. Any finitely generated torsion module over a DVR is M R/πnR: Before we prove this, let us give two examples: 3 1.1 Discrete Valuation Rings (DVRs), Krull Dimension 1 1. Take R = k[[t]], which is a DVR with maximal ideal (t). Thus, by the lemma, for a finitely generated torsion module M, t : M ! M is a nilpotent operator. However, k[[t]]=tn is a Jordan block so we are exactly saying that linear transformations can be written in Jordan block form. 2. Let R = Zp. Here the lemma implies that finitely generated torsion modules over Zp can be written as a direct sum of p-groups. Miaaaoow?? Now let us proceed with the proof of the lemma. Proof of Lemma 1.11. Let n be the minimal integer such that πn kills M. This means that n M is a module over Rn = R/π R, and also there is an element m 2 M, and an injective n−1 map Rn ,! M, because we may choose m to be an element which is not annihilated by π , and then take the map to be 1 7! m. Proceeding by induction, it suffices to show that the above map Rn ,! M splits. But for this it suffices that Rn is an injective module over itself. This property of rings is called the Frobenius property, and it is very rare. We will write this as a lemma. Lemma 1.12. Rn is injective as a module over itself. Proof of Lemma 1.12. Note that a module M over a ring R is injective if and only if for any ideal I ⊂ R, Ext1(R=I; M) = 0. This was shown on Problem Set 8, Problem 2a. 1 Thus we wish to show that for any ideal I, ExtRn (Rn=I; Rn) = 0. Note that since R is a k DVR, we know that it is a PID, and also any element has the form r = π r0 for some k ≥ 0 k and some r0 invertible. Then all ideals in R are of the form (π ) for some k, so all ideals in Rn are also of this form. Therefore, Rn=I = Rm for some m ≤ n, so it suffices to show that 1 for m ≤ n, ExtRn (Rm;Rn) = 0. But note that we have short exact sequence m - π -· - - 0 Rn−m Rn Rm 0 which gives a corresponding long exact sequence of Exts - - ~- 0 HomRn (Rm;Rn) HomRn (Rn;Rn) HomRn (Rn−m;Rn) - 1 - 1 - ExtRn (Rm;Rn) ExtRn (Rn;Rn) ··· But note that any map of Rn modules, Rn−m ! Rn, must map 1 2 Rn−m to an element which is killed by πn−m, which means it must be a multiple of πm, so say is is πma. Then the map is r 7! πmar; which is the image of the map [r 7! ar] 2 HomRn (Rn;Rn): 1 Thus, ~ is surjective. Also note that Rn is projective over itself, so ExtRn (Rn;Rn) = 0. This, along with the surjectivity of ~ shows that 1 ExtRn (Rm;Rn) = 0 as desired. 4 1.1 Discrete Valuation Rings (DVRs), Krull Dimension 1 As mentioned earlier, this lemma concludes our proof of Lemma 1.11 as well. Definition 1.13. We say that a Noetherian domain A is of Krull dimension 1, if any nonzero prime ideal is maximal.