Stability of locally CMFPD homologies under duality

Transcription

1 Journal of Algebra Contents lists available at ScienceDirect Journal of Algebra Stability of locally CMFPD homologies under duality Satya Mandal a,,1, Sarang Sane b a University of Kansas, Lawrence, KS 66045, United States b IIT Madras, Sardar Patel Road, Adyar, Chennai , India a r t i c l e i n f o a b s t r a c t Article history: Received 16 December 2013 Available online xxxx Communicated by Steven Dale Cutkosky MSC: primary 13C10, 11E81 secondary 19G12, 13D We consider bounded complexes P of finitely generated proective A-modules whose homologies have finite proective dimension and are locally Cohen Macaulay. We give a necessary and sufficient condition so that its dual P also has the same property Published by Elsevier Inc. Keywords: Complexes Dual complexes Homologies Cohen Macaulay modules Proective modules Proective dimension 1. Introduction Throughout this paper A will denote a Cohen Macaulay CM ring with dim A m = d m MaxA. Throughout, CM abbreviates Cohen Macaulay and FPD abbreviates finite proective dimension, which clarifies the title of the paper. * Corresponding author. addresses: mandal@ku.edu S. Mandal, sarangsanemath@gmail.com S. Sane. 1 Partially supported by a General Research Grant from University of Kansas / 2015 Published by Elsevier Inc.

2 50 S. Mandal, S. Sane / Journal of Algebra To introduce the main results, in this paper, let M denote a finitely generated A-module with pro dimm = r<. When A is local, then M is Cohen Macaulay if and only if gradem = r. In this case, Ext i M, A = 0 i r. However, even in the non-local case, grade of M is defined as gradem := min{i : Ext i M, A 0} see [6]. Let B denote the category of such finitely generated A-modules M, with pro dimm = gradem. Then, M B, Ext i M, A = 0 i pro dimm. Now suppose P is an obect in the category Ch b BPA of finite complexes of finitely generated proective A-modules, with homologies in B. In this paper, we give a necessary and sufficient condition, for such a complex P Ch b BPA, so that its dual P is also in Ch b BPA. To further describe this condition let the homology H t P 0, at degree t and ρ t = pro dim H t P. The homomorphism H t P P t B t, where B t = t+1 P t+1, induces a homomorphism ι t : Ext ρ t Pt Ext ρ t H t P B t The main Theorem 3.6 states that the dual P Ch b BPA if and only if ι t is an isomorphism, whenever H t P 0. The theorem immediately applies to complexes P Ch b PA whose homologies are locally CMFPD see Corollary 3.7. The theorem also applies to a number of interesting subcategories of B, for which ι t is already an isomorphism, for all such t. Among them are the categories Bn = {M B: pro dimm = n}. Note, A := Bd is the category of modules of finite length and finite proective dimension. The stability of P Ch b APA, under duality is a theorem in [4]. We underscore that this paper is part of a wider study [4,5] of duality of subcategories of derived categories and their Witt groups. While we have particular interest in the setting of singular varieties and we also provide further insight into nonsingular varieties in these articles. Our interest in Witt theory stems from the introduction of Chow Witt groups, [1] and developed by Fasel [3], as obstruction groups for proective modules to split off a free direct summand. The readers are referred to [4] for further introductory comments. To be more specific, one of the primary motivations behind this study has been to address the Witt theory, for non-regular Cohen Macaulay schemes X, with dim X = d, while this article addresses the duality aspect of the same. Let V X denote the category of locally free sheaves on X and D b V X denote the derived category of finite complexes of locally free sheaves. Let MX, d denote the subcategory of CohX with finite length and finite V X-dimension. For unexplained notations, readers are referred to [5,4]. As a consequence of Theorem 3.6, it follows that DMX,d b V X is closed under the usual duality induced by E HomE, O X. This allows us [5,4] to give a definition of shifted Witt groups W r DMX,d b X of Db MX,d X, while Db MX,d X fails to inherit a triangulated structure. Also note that MX, d has a duality sending F Ext d F, O X and hence a Witt group W MX, d is defined. It was established

3 S. Mandal, S. Sane / Journal of Algebra that W MX, d W dim X DMX,d b X is an isomorphism [5,4]. When X is regular, this would be a result of Balmer and Walter [2], while in the non-regular case, the right side would be meaningful only in the light of Theorem 3.6. Note, when X is regular MX, d is subcategory of schemes of finite length. We would like to thank Sankar P. Dutta for many helpful discussions. The second named author is thankful to the University of Kansas and the Robert D. Adams Trust. 2. Preliminaries We will borrow some notations from [4]. Notations 2.1. Throughout this article, A will denote a noetherian commutative ring, such that dim A m = d m MaxA. All modules will be assumed to be finitely generated, unless stated. Most often, A will be assumed to be Cohen Macaulay. We will also use the abbreviation CM for Cohen Macaulay. We set up the notations: 1. ModA will denote the category of finitely generated A-modules. 2. MFPDA will denote the category of finitely generated A-modules with finite proective dimension. 3. For integers r 0, denote and Br :=BAr :={M MFPDA :pro dimm =gradem =r} B := BA := r 0 Br Note, when A is local and CM, then B is the category of finitely generated Cohen Macaulay A-modules with finite proective dimension see 2.7. In the non-local case, there are modules M MFPDA which are locally CM, with M / B, e.g. A = k[x, Y ],M = k[x, Y ] X, Y k[x, Y ] X PA: category of finitely generated proective A-modules. 5. Ch b PA will denote the category of bounded complexes of finitely generated proective A-modules. Also, Ch b BPA will denote the category of complexes P in Ch b PA such that all the homologies H i P are in B. 6. We will denote complexes P in Ch b PA by: 0 P m m P m 1 P n 0 7. For a complex P of proective A-modules P will denote the usual dual induced by Hom, A.

4 52 S. Mandal, S. Sane / Journal of Algebra Let B r = B r P := r+1 P r+1 P r denote the module of r-boundaries and Z r = Z r P := ker r P r denote the module of r-cycles or the r th syzygy. 9. The r th -homology of P will be denoted by H r = H r P := Z r B r. So, the r th -homology of the dual is H r P = ker r 1 image r. 10. By width, we mean the number h l where P h 0, P l 0and P k = 0 for all r/ [l, h]. Definition 2.2. A complex P as in 2.1, is called indecomposable, if P = U V = U =0 or V =0. We say that P is decomposable, if P is not indecomposable. Lemma 2.3. Let P be as in 2.1. If B r 1 is proective, then P is a direct sum of the following two complexes P r+1 Z r B r 1 P r 1 P r 2 P r 3 Example 2.4. Given an indecomposable proective module P, there is a trivial indecomposable complex which is quasi-isomorphic to 0. Lemma 2.5. Let 0 P P 0 P P r+1 P r P r 1 be an indecomposable complex of proective modules. Assume, for all r, H r is CM and has finite proective dimension i.e. H r B. If dim A = 0, then widthp = 0or P is a trivial indecomposable complex as in 2.4. If dim A = 1, then widthp = 1. Proof. First, assume d =0. Since, P is a bounded complex, we can assume P r =0 for all r<0and P 0 0. In this case, H 0 =0or pro dim H 0 =0. If H 0 0then P 0 = B 0 H 0. So, P has a direct summand H : 0 H 0 0.

5 S. Mandal, S. Sane / Journal of Algebra Since, it is indecomposable, P = H. So, width is zero. If H 0 =0then P 1 P0 Z 1. Since P is indecomposable, Z 1 =0and P 1 P0. So, P is given by 0 P 1 P 0 0. This is isomorphic to 0 P 0 P 0 0. and P 0 is also indecomposable. Now assume d = 1. We assume P 0 0and P r =0for all r <0. If H 0 P = 0, by the same argument as above P is isomorphic to 0 P 0 P 0 0. and P 0 is also indecomposable. So, assume H 0 0. Then, we have an exact sequence 0 B 0 P 0 H 0 0. Then, pro dim H 0 =0or 1. In either case, it follows B 0 is proective. So, P 1 = Z 1 B 0. Note B 0 P 0. So, P splits into two complexes: Q : 0 B 0 P 0 0 P : P 2 Z and hence P = Q. So the proof is complete CM modules, grade and finite proective dimension Readers are referred to [6] for a definition of gradem and for a proof of the following lemma. Lemma 2.6. Suppose A, m is a Cohen Macaulay local ring and dim A = d. Suppose M is a Cohen Macaulay A-module with finite proective dimension. 1. Then, pro dim M +dimm = d and pro dim M = gradem. 2. Let r = pro dim M. Then, i r, Ext i M, A = 0, and pro dim Ext r M, A = r.

6 54 S. Mandal, S. Sane / Journal of Algebra Corollary 2.7. Suppose A, m is a Cohen Macaulay local ring with dim A = d. Suppose M is a finitely generated A-module with finite proective dimension. Then, M is Cohen Macaulay if and only if pro dim M = gradem =heightannm. Proof. Suppose M is Cohen Macaulay. Write I = annm. Then depthm =dimm = dim A I = d heighti. Therefore pro dimm = heighti. Also, gradem = depth I A = heighti see[6, p. 108]. Now suppose pro dim M = gradem = heighti. Now, heighti = d dim M. So, pro dim M +dimm = d. Hence dim M = depthm. So, M is Cohen Macaulay. The following lemma will be useful for our subsequent discussions. Lemma 2.8. Let A be a Cohen Macaulay ring with dim A = d non-local but as in 2.1 and M 0 be in B. Let 0, 1 SuppM and 0 1. Then, pro dimm 0 = pro dimm 1. We state two lemmas without proof.the proof of the second lemma can be found in [4, Proposition 2.8]. Lemma 2.9. Let A be a CM ring and M, N be finitely generated A-modules such that M has finite proective dimension. Let r 0 be an integer. Then heightannext r M, N r. In this case, gradeext r M, N r. Lemma Let P be a bounded complex of proective modules such that all the homologies H i have finite proective dimension. Then so do all the boundaries B i, kernels Z i and the cokernels P i B i. In fact, pro dimb r dim A 1, pro dimz r dim A 2, r. Lemma Let A be a CM ring and P be a complex in Ch b BPA. Let t 0, t 1,... be the list of all the degrees so that H ti P 0and ρ i = pro dimh ti P. Then, Proof. To begin with, note that P t k by 2.9, grade Ext ρ Ptk k = ρ k. has finite proective dimension by 2.10 and hence, grade Ext ρ Ptk k ρ k.

7 S. Mandal, S. Sane / Journal of Algebra Now, consider the exact sequence 0 H tk P tk 1 0 and the corresponding five term exact sequence 0 Ext ρ k 1 Ext ρ Ptk k Ext ρ k H tk Ext ρ k+1 1 Ext ρ k+1 P tk 0 Choose SpecA with height = ρ k, such that Ext ρ k H tk 0. Applying 2.9, we get Ext ρk+1 1 =0. Hence, Ext ρ Ptk k 0. Hence, grade Ext ρ Ptk k = ρ k. The proof is complete. Remark Let s := t t 1. Note in the previous proof that if ρ k 1, then the 5-term exact sequence can be rewritten as: 0 Ext ρ k+s Ptk 1 k 1 Ext ρ Ptk k Ext ρ k H tk Ext ρ k+1+s Ptk 1 k 1 Ext ρ k+1 P tk 0 and hence it follows that for SpecA with height < ρ k + s k, we have 3. Dualizable complexes Ext ρ k Ptk = Ext ρ k H tk B. tk In this section, we classify complexes P Ch b BPA whose dual is also in Ch b BPA. The following lemma will be useful subsequently.

8 56 S. Mandal, S. Sane / Journal of Algebra Lemma 3.1. Suppose P : P 2 d 2 P 1 d 2 P 0 is a complex of proective A-modules. Then, there is an exact sequence 0 P 0 B 0 P 0 P 1 B 1 H 1 P 0 Proof. We have the diagrams: P 2 d 2 P 1 d 1 P 0 P 0 d 1 P 1 d 2 P 2 B 1 B 0 So, kerd 2={λ P 1 : λ B1 =0} = P1 B 1. Also, Imaged 1 = {λ B0 : λ P 0 } Let K = {λ P0 : λ B0 =0} = Then we have an exact sequence P0 B 0. 0 K P 0 Imaged 1 0 So, 0 P 0 B 0 P 0 Imaged 1 0 This completes the proof. The following is a key tool to compute the homologies of the dual. Proposition 3.2. Suppose P : P s+1 P s P s 1 P 1 P 0

9 S. Mandal, S. Sane / Journal of Algebra is a complex of proective A-modules with s > 0 such that H i P = 0for i = 1,..., s 1 and H s P B. Then, 0 <t<s H t P =Ext 1 Pt 1 = Ext t P0. B t 1 B 0 If pro dimh s 1, then the above statement also holds for t = s. Proof. For 0 < t < sthe statement follows because the complex is exact between 0 and s. For the last statement, we have an exact sequence 0 H s P s B s B s 1 0. Since pro dimh s 1and H s B, we conclude that Hs =0and hence, Bs 1 P. = s B s Therefore, in conunction with 3.1, we have a commutative diagram of exact sequences: 0 Ext 0 P s 1 B s 1 Ext 0 P s 1 Ext 0 B s 1 Ext 1 P s 1 B s Ext 0 P s 1 B s 1 Ext 0 P s 1 Ext 0 P s B s H s P 0 Theorem 3.3. Let P be a complex in Ch b BPA. Let t 0 <t 1 <t 2 <... be the list of all the degrees so that H ti P 0. Write ρ i = pro dimh ti P. Assume ρ 1 for all. Then, the following are equivalent: 1. For all k 1 and all, the module Ext k P t is in B. 2. For all, the module Ext ρ Pt is in B. 3. The natural homomorphism ι : Ext ρ Pt Ext ρ H t is an isomorphism. In this case, for all, Ext 0 P t has finite proective dimension. Proof. Clearly, 1 = 2. We will prove 3 = 1 = 2 = 3. 3 = 1: Suppose ι is an isomorphism for all. Since the complex is exact before t 0, it splits and hence, B t0 1 is proective. Hence, P t0 = Zt0 B t0 1. So, Ext k Pt0 = Ext k Zt0 Ext k B t0 1 B t0 B = t0 0 k 0,ρ 0 Ext ρ 0 H t0, k = ρ 0. Bt 0 1 k =0

10 58 S. Mandal, S. Sane / Journal of Algebra Hence, Ext k P t0 B t0 is in B for all k. Now we do the induction step. So, we assume the statement holds for all 0 and prove it for 1 = Let s = t 1 t 0. Corresponding to the short exact sequence: 0 H t1 P t I, we get a long exact sequence 0 Ext Ext 0 Pt1 Bt 1 0 Ext ρ Ext ρ 1 1 Pt1 Bt 1 0 Ext ρ 1 Bt1 1 ι1 Pt1 Ext ρ 1 Bt 1 Ext ρ 1 Ht1 Ext ρ Ext ρ 1 +1 Pt1 Bt 1 0 Since ι 1 is an isomorphism, Then, Ext ρ 1 Bt1 1=0, Ext ρ Pt1 1 = Ext ρ 1 Ht1, which is in B. 1 Also, Ext ρ = Ext ρ 1 +1 Pt1 1. k ρ 1,k 1 Ext k Pt1 = Ext k 1 1 B = Ext k+s Pt0 t1 0 which are all in B, by induction hypothesis. Also Ext 0 Pt1 = Ext has finite proective dimension. 1 2 = 3: To prove this we assume that ι is not an isomorphism for some. So, m and such that ι m is not an isomorphism. It will be enough to prove that Ext ρ i Pti either has infinite proective dimension or is not CM, for some i. By m replacing A by A m, we assume A is local. Further, we write P = r n=1 P n, where P n are indecomposable complexes in Ch B PA. If H t P n 0, ρ = gradeh t gradeh t P n pro dimh t P n pro dimh t = ρ and so pro dimh t P n = ρ and also, for some, n the homo-

11 morphism ι : Ext ρ S. Mandal, S. Sane / Journal of Algebra P n t B Ext ρ t n H t P n is not an isomorphism. We will replace P by such a P n and assume that P is indecomposable. Let { 1 =min : Ext ρ Pt ι Ext ρ Ht } is not an isomorphism Note that 1 > 0since P t 0 B t0 = H t0 B t0 1. Let 0 = 1 1and s = t 1 t 0. By minimality of 1, since we already proved 3 = 1, it follows that 1 applies to P t 0. B t0 So, Ext 0 Pt0 has finite proective dimension and 0 k 1 Ext k Pt0 0 is in B. It follows that Ext has finite proective dimension and k 1 Ext k 1 1 = Ext k+s Pt0 0 is in B. Corresponding to the short exact sequence I, we have the five term exact sequence 0 Ext ρ 1 +s Pt0 Ext ρ Pt1 ι1 1 Ext ρ 1 Ht1 0 1 Ext ρ 1 +1+s Pt0 0 Ext ρ 1 +1 Pt By 2.11, grade Ext ρ Pt1 1 = ρ 1. 1 For notational convenience, write K = Ext ρ 1 +s Pt0 0 E = Ext ρ Pt1 1, C = imageι 1, 1 H = Ext ρ 1 H1, B = image, K = Ext ρ 1 +1+s Pt0 0 If E has infinite proective dimension then E is not in B. So, assume E has finite proective dimension. All the modules listed above have finite proective dimension. Since ι 1 is not an isomorphism, either K 0or C 0. Assume K 0. We have g := gradek ρ 1 + s. Let SuppK such that height = g. Therefore

12 60 S. Mandal, S. Sane / Journal of Algebra g = gradek pro dimk dima =g, and so pro dimk =g. Since Tor g+1 A,C =0, it follows from the sequence of Tor modules that A pro dime = g. So, pro dime pro dime = g ρ 1 + s > ρ 1. So, E is not CM and hence not in B. Now assume K =0. Then it follows that B 0, otherwise ι 1 would be an isomorphism. Consider the exact sequence 0 E H B 0. Since B K, g := gradeb gradek ρ 1 + s + 1. Let SuppB such that height = g and pro dimb = g. Since pro dimh = ρ 1 g 2, it follows pro dime = pro dimb 1 = g 1 > ρ 1. So, E is not CM and hence not in B. The proof is complete. The following will be needed for our subsequent inductive argument, which follows from the proof of 3.3. Corollary 3.4. Let P be as in 3.3. Further, assume A is local and P is indecomposable. Suppose ι : Ext ρ Pt Ext ρ H t is not an isomorphism for some. Let 0 =min{ : ι is not an isomorphism}. Then, 1. E := Ext ρ Pt0 0 0 is not in B. 2. For all k < 0, Ext ρ Ptk k is in B. 3. For all SuppE with height = ρ 0, E Bfor the ring A. Proof. The proof follows from the proof of 3.3 and We proceed to give an equivalence condition for duality. The following proposition will be useful for our inductive argument. Proposition 3.5. Let P as in 3.3. Further, assume A is local and P is indecomposable. X k = { X = SpecA :height =k} and g k = grade Ext Ptk. Assume, for some r 0 1 and some t i 1, Ext r 0 X g i 1, 1 r 0 Ext Pti 1 1 Pti 1 is not in B, and 1 is in BA.

13 S. Mandal, S. Sane / Journal of Algebra Then 1. either H r P is not in B for some r >t i or there is an r 1 1, such that the following hold: Ext r 1 Pti is not in B g X i, 1 r 1 Ext Pti and is in B. In particular, if t i 1 = t n is the last one, then 1 must hold. Proof. Let s = t i t i 1. If r 0 s then by 3.2, H r0 +sp =Ext r 0 Pti 1 1 is not in B. So, assume r 0 s > 0. The latter part of assertion 2 would follow if Ext P ti had finite proective dimension, which may not always be the case. The long exact sequence of Ext modules corresponding to the usual exact sequence 0 H ti P ti 1 0 gives us that: Ext l 1 There is a 5-term sequence Ext l Pti l ρ i,ρ i Ext ρ i 1 Ext ρ i Pti B 1 ι i Ext ρ i H ti Ext ρ i+1 1 Ext ρ i+1 P ti 0 We break the proof into various cases. Case r 0 ρ i <s: So, 1 r 0 s < ρ i. We have k 1 Ext k 1 = Ext k+s Pti 1. 1

14 62 S. Mandal, S. Sane / Journal of Algebra So, from the long exact sequence above, we extract the following row: 0 Ext r Pti Ext r 0 s P ti 0 With r 1 = r 0 s, our assertion is satisfied. Case ρ i r 0 s: It follows from the long exact sequence of the ext modules 1 <ρ i Ext Pti = Ext +s Pti 1 1 For <ρ i, we have + s ρ i + s < r 0. Therefore, 1 <ρ i, X gi Ext Pti hasfpdandiscm. I By the grade Lemma 2.11 grade Ext ρ Pti i = ρ i II So, gρ i i = ρ i and we have X ρ i Ext ρ i Pti = Ext ρ i H ti B is in B. III. ti We will show that E := Ext ρ i Pti is not in B. If E has infinite proective dimension, then E is not in B. So, we will assume E has finite proective dimension. For convenience, we use notations K = Ext ρ i+s P ti 1 1 E = Ext ρ i Pti,C= imageι, H 1 = Ext ρ i H ti, B = image, K = Ext ρ i+1+s P ti 1 1, E = Ext ρ i+1 P ti. 1. Now assume K 0. So, we have an exact sequence 0 K E C 0 Now g := gradek ρ i + s and hence C 0. Let SuppK and height = g. Since ρ i +s r 0, K has finite proective dimension. It follows that pro dimk = g. So, C also has finite proective dimension. From the sequence of Tor modules, it follows that pro dime = g ρ i + s. So, pro dime ρ i + s > ρ i. Hence E is not in B. With r 1 = ρ i the assertion of the theorem is valid.

15 S. Mandal, S. Sane / Journal of Algebra Assume K =0and K 0. Since K =0, we have r 0 s ρ i +1. Also, E C. The five term exact sequence reduces to 0 E H 1 K E 0 We have g := gradek ρ i +1 + s. Assume B 0: Then, g := gradeb gradek ρ i +1 + s. Now, we have an exact sequence 0 E H 1 B 0 Since E, H 1 have finite proective dimension, so does B. Let SuppB and height = g. Then dima = g and pro dimb = g ρ i +1 + s ρ i +2. Hence it follows that pro dime = g 1 ρ i + s. So, pro dime ρ i + s and hence E is not in B. In this case also, with r 1 = ρ i the assertion of the theorem is valid. Assume B =0: Then we get that E = H 1, K = E and choosing r 1 = r 0 s proves the assertion. 3. Assume K = K =0. In this case, ι : E H 1 is an isomorphism, E =0and hence r 0 s ρ i +2. The assertion of the theorem is satisfied with r 1 = r 0 s. The proof is complete. Theorem 3.6. Let P be a complex in Ch B PA with notations as in 3.3. Then, P is in Ch b BPA if and only if ι : Ext ρ Pt Ext ρ H t P is an isomorphism. Proof. Assume that ι is an isomorphism for all. Then by Theorem 3.3, k 1, Ext k Pt B is in B. So, by 3.2, t <t t +1 H t P =Ext t t Pt B is in B. So, P is in Ch b BPA. Now suppose ι is not an isomorphism for some. Arguing exactly as in the proof of Theorem 3.3, we assume that A is local and P is indecomposable. Now, by Corollary 3.4, there exists r 1and i such that Ext r Pti is not in B, and X g i 1, 1 r Ext Pti is in B.

16 64 S. Mandal, S. Sane / Journal of Algebra This is exactly the hypothesis of Proposition 3.5, which guarantees us that either there exists a homology H r P which is not in B or that one obtains r 1 1so that the above hypothesis repeats for i +1. Note that for i = n, the only possibility is that there exists a homology H r P which is not in B. Thus, by induction, the above process repeats until we produce a homology H r P which is not B. So, P is not in Ch b BPA. The proof is complete. The following is a version of 3.6 for complexes with CMFPD homologies. Corollary 3.7. Let P be a complex in Ch b PA such that, for all maximal ideals m and all degrees t, H t P m is a CMFPD module, if nonzero. We denote ρ t m := pro dimh t P m. Then the dual P also has the same property, if and only if, ι t : Ext ρ m Pt Ext ρtm H t P B m is an isomorphism, whenever t m H t P m Computations on dualizable complexes In fact, we can say considerably more about the homologies of P under the above conditions. We first note that another look at the long exact Ext sequence along with induction, gives us the following theorem. Theorem 4.1. Let A be a CM ring. Suppose P is a complex in Ch b BPA such that P Ch b BPA. Then, with notations as in the equivalence Theorem 3.6, 1. t, we have 1 i, =0, 1,...,n Also, Ext 0 P t B t has finite proective dimension. 2. Further, Ext i Pt =0 or are in B. Ext i+s Pt i<ρ s Ext i Pt Ext ρ 1 H t 1 for i = ρ s = B 0 ρ 1 s <i<ρ t Ext ρ H t for i = ρ 0 ρ <i 3. For all t, the homologies H t P are in B. In fact, for t 1 <t t, H t P =Ext t t 1 Pt 1. 1

17 S. Mandal, S. Sane / Journal of Algebra Proof. The equation in 3 follows immediately from Proposition 3.2. So it is enough to prove 1, 2.These statements are clearly true for =0. We assume that 1, 2 are valid upto and prove them +1. By the equivalence Theorem 3.6, ι +1 : Ext ρ +1 Pt+1 Ext ρ +1 H t+1 +1 is an isomorphism. and We have, Ext i Pt +l Ext i+l Pt, i 1,l ρ +1 +l Ext i +l Ext i+l+1 Pt, i 1, 1 l<s +1 which are hence in B by induction. As in the previous proofs, looking at the long exact Ext sequence corresponding to P t+1 0 H t along with induction, gives us all the statements in 1, 2 except the finite proective dimension of Ext 0 P t An inductive argument with the exact sequence 0 Ext 0 Pt +l Bt +l Ext 0 P t +l Ext 0 +l Ext 1 Pt +l Bt +l 0 for 0 l<s +1 and P t +l +l has finite proective dimension. Then +l 1, 0 <l<s +1 gives us that Ext 0 Pt Ext Ext 0 Pt +1 Bt +1 Ext 0 H t+1 Ext Since all the other terms have finite proective dimension, so does Ext 0 Pt+1. The +1 proof is complete. We give some explicit formulas for nonzero ext-modules. Corollary 4.2. Let A, P and all the notations be as in Theorem 4.1. Let 0 be an integer. For r =0,..., let τ r = ρ r s + + s r+1 =ρ r t t r.

18 66 S. Mandal, S. Sane / Journal of Algebra Then { Ext i Pt Ext ρ r H tr P if i = τr 1 = 0 if 1 i τr r. Proof. Fix and note that τr >τ r 1 for all r =1,...,. First, for i = τ = ρ, by Theorem 4.1, we have Ext i Pt = Ext ρ H t. For i = τ 1 = ρ 1 s by theorem 4.1, we have Ext i Pt = Ext ρ Pt 1 s = Ext ρ 1 H t 1 Also, by Theorem 4.1, Ext i P t =0for τ 1 <i <τ. Now, we use induction. Assume the theorem is established i τ r+1. We prove it for τ r+1 <i τ r. For τr = ρ r s r s i it follows inductively that Ext i Pt = Ext i+s Pt 1 = Ext i+s r+1+ +s Ptr 1 B tr The proof is complete by application of Theorem 4.1. We compute the homologies of the dual for complexes as in Theorem 4.1. Corollary 4.3. Let P be a complex as in Theorem 4.1. Then, H t P = Proof. We assume t <t t +1. By 3.2, By 4.2, H t P is non-zero, only if That means, if { Ext ρ r H tr P if t = t r + ρ r for some r 0 otherwise. H t P =Ext t t Pt t t = τ r = ρ r s + + s r+1. t = ρ r +[t s r+1 + s r s ] = t r + ρ r.

19 S. Mandal, S. Sane / Journal of Algebra For t = t r + ρ r, we have H t P =Ext ρ r s + +s r+1 Pt = Ext τ Pt r = Ext ρ r H tr. The proof is complete. 5. Some dualizable categories of complexes We consider some special categories of complexes and use the equivalence condition in the previous section to show that they are closed under duality. We also show that these conditions are almost the best we can hope for. We begin with some initial lemmas that will be used in the subsequent proofs. Lemma 5.1. Let A be a noetherian ring and P s P s 1 P 1 P 0 be a complex of proective A-modules. Assume that the complex does not split at degree i = 1,..., s 1. Then, s pro dim P0 B 0 1. Proof. Write ρ = pro dim P0 B 0. We have the exact sequence 0 B s 1 P s 1 P 1 P 0 P 0 B 0 0 If s ρ then B s 1 is proective which contradicts the hypothesis that the complex is not split at i = 1,..., s 1. Hence, the proof is complete. Lemma 5.2. Let A, m be a noetherian local ring. Let P be an indecomposable complex in Ch b BPA. Let t 0 <t 1 <t 2 < be the list of all the degrees so that H ti P 0. Let ρ i = pro dim H ti. Assume Z, ρ ρ +1 t +1 t =: s +1 Then, Pt Z, pro dim H t = pro dim.

20 68 S. Mandal, S. Sane / Journal of Algebra Conversely, for any fixed : Ptk pro dim = ρ k for k =, +1 = ρ ρ +1 s Proof. We assume t 0 =0. Since P is indecomposable, H 0 = P 0 B 0. So the lemma holds trivially in this case. Now assume that the lemma holds upto t. Note that if P t is proective, then so is. By our assumption of indecomposability, 2.3 tells us that this is possible only if P t +1 = and this is the final term in the complex, in which case we are already done by the induction hypothesis there is no t +1. So assume that Pt pro dim 1and we will prove the equality for t +1. Then Inductively, we have Pt pro dim =pro dim 1=ρ 1. pro dim +1 1 =ρ 1 t +1 1 t =ρ s +1 ρ +1. Consider the exact sequence 0 H t+1 P t We write the long exact sequence: 0 Tor ρ+1 +1 Pt+1 +1 /m Tor ρ+1 +1 Bt+1 1/m Tor ρ+1 Ht+1 /m Pt+1 Tor ρ+1 +1 /m Tor ρ+1 Bt+1 1/m Since the top right term vanishes, and the bottom left term is non-zero, the middle term Pt+1 in the second row cannot vanish. It follows that pro dim +1 = ρ +1. For the converse, pro dim Pt Then the above long exact sequence gives us that = ρ and hence pro dim +1 1 = ρ s +1. ρ s +1 = pro dim +1 1 ρ This completes the proof.

21 S. Mandal, S. Sane / Journal of Algebra Theorem 5.3. Let A be a CM ring. Let P be a complex in Ch b BPA. Let t 0 <t 1 < t 2 < <t n be the list of the non-vanishing homologies. Let ρ i = pro dim H ti > 0. Assume Then, P Ch b BPA. ρ ρ +1 <t +1 t =: s +1, =0, 1, 2,...,n 1. Proof. By equivalence Theorem 3.6, we only need to prove the homomorphisms ι are isomorphisms. Note that the condition ρ ρ +1 <t +1 t =: s +1,=0, 1, 2,...,n 1 localizes and restricts to indecomposable summands of the complex for the subsequence of {0, 1, 2,..., n 1} consisting of nonzero homologies, because ρ ρ k <t k t for all <k. Hence, we can assume that A is local and P is indecomposable. Note that H t0 = P t 0 B t0 is an isomorphism and hence, ι 0 : Ext ρ 0 Pt0 Ext ρ 0 H t0 B t0 Pt is an isomorphism. By Lemma 5.2, pro dim = ρ. Hence by Lemma 5.1, we have s +1 = t +1 t ρ 1which also ensures that ρ 2 and pro dim +1 1 = ρ s +1 <ρ +1. Thus, the long exact sequence corresponding to the exact sequence: 0 H t+1 P t gives us that ι +1 : Ext ρ +1 Pt+1 Ext ρ +1 H t Hence, by the equivalence Theorem 3.6, we get that P Ch b BPA. The proof is complete. Corollary 5.4. Let P be a complex as in Theorem 4.1. Suppose u, v be integers and Then H u P 0, H v P 0, and H t P =0 u<t<v. pro dim H v P pro dim H u P <v u.

22 70 S. Mandal, S. Sane / Journal of Algebra Proof. Note that t r +ρ r t r 1 +ρ r 1 = s r +ρ r ρ r 1 > 0. So, by 4.3, u = t r 1 +ρ r 1, v = t r + ρ r for some r. Also, v u = s r + ρ r ρ r 1. From, 4.3, So, H v P =Ext ρ r H tr, H u P =Ext ρ r 1 H tr 1. Therefore, pro dim H v P =ρ r, and pro dim H u P =ρ r 1 pro dim H v P pro dim H u P =ρ r ρ r 1 <ρ r ρ r 1 + s r = v u. This completes the proof. Putting together all the above theorems, lemmas and corollaries gives us the following theorem: Theorem 5.5. Let A be a CM ring. Let C be the additive subcategory of Ch b BPA generated by complexes P which are satisfy the condition that: ρ ρ +1 <t +1 t =: s +1, =0, 1, 2,...,n 1 where the non-zero homologies of P are in degrees t 0 <t 1 <t 2 < <t n and ρ i = pro dim H ti > 0. Then C is closed under the duality Hom A, A. We give some more easy consequences of the above results. Theorem 5.6. Let A be a CM ring and let Bk ={P Ch b BPA : ih i P =0or pro dimp =k}. Then the category Ch b BkPA Hom A, A. is an exact category closed under the duality Remark 5.7. When k = d, the above category Bd is exactly the category of finite length modules with finite proective dimension and in [4], a dévissage theorem for Bd was proved. We end with a lemma that seems to suggest that the conditions in Theorem 5.5 are almost optimal for duality to hold. Lemma 5.8. Suppose P : P 1 P 0 is a complex of proective A-modules, such that H 0 P, H 1 P B. Let ρ 1 = pro dim H 1 and ρ 0 = pro dim H 0.

23 S. Mandal, S. Sane / Journal of Algebra Assume ρ 0 = ρ 1 +1 and pro dim B 0 ρ 0 1 = ρ 0. Then Ext ρ 1 Macaulay. P 1 B 1 is not Cohen Proof. Consider the exact sequence 0 H 1 P 1 B 1 B 0 0. Since Ext ρ 1 B 0 =Ext ρ 1 1 H 1 =0, from the corresponding long exact sequence of the ext modules, we have an exact sequence 0 Ext ρ 1 B 0 Ext ρ 1 P 1 B 1 Ext ρ 1 H 1 0 Since pro dimb 0 ρ 0 1 = ρ 1, we have pro dimz 0 ρ 0 2. So, form the exact sequence 0 B 0 Z 0 H 0 0 we have Ext ρ 1 B 0 = Ext ρ 0 H 0. Now, pro dim Ext ρ 0 H 0 = ρ 0 = ρ 1 +1, pro dim Ext ρ 1 H 1 = ρ 1 It follows pro dim Ext ρ 1 P1 = ρ 0 = ρ B 1 Also, the height of the annihilator of the middle term is the minimum of that of the other two. So, Therefore, by 2.7, Ext ρ 1 References grade Ext ρ 1 P1 = ρ 0 = ρ 1. B 1 P 1 B 1 is not Cohen Macaulay. The proof is complete. [1] Jean Barge, Fabien Morel, Groupe de Chow des cycles orientés et classe d Euler des fibrés vectoriels The Chow group of oriented cycles and the Euler class of vector bundles, C. R. Acad. Sci. Paris Sér. I Math in French. [2] Paul Balmer, Charles Walter, A Gersten Witt spectral sequence for regular schemes, Ann. Sci. Éc. Norm. Supér [3] Jean Fasel, Groupes de Chow Witt, Mém. Soc. Math. Fr. N.S [4] Satya Mandal, Sarang Sane, On Dévissage for Witt groups, preprint, arxiv: [5] Satya Mandal, Derived Witt group formalism, J. Pure Appl. Algebra [6] Hideyuki Matsumura, Commutative Ring Theory, Cambridge Stud. Adv. Math., vol. 8, Cambridge University Press, Cambridge, 1986, xiv+320 pp.