CONSTRUCTION OF TOTALLY REFLEXIVE MODULES FROM AN EXACT PAIR OF ZERO DIVISORS

Transcription

1 CONSTRUCTION OF TOTALLY REFLEXIVE MODULES FROM AN EXACT PAIR OF ZERO DIVISORS HENRIK HOLM Abstract Let A be a local ring which admits an exact pair x, y of zero divisors as defined by Henriques and Şega Assuming that this pair is regular and that there exists a regular element on the A-module A/x, y, we explicitly construct an infinite family of non-isomorphic indecomposable totally reflexive A-modules with periodic minimal free resolutions In this setting, our construction provides an answer to a question by Christensen, Piepmeyer, Striuli, and Takahashi Furthermore, we compute the module of homomorphisms between any two given modules from the infinite family mentioned above 1 Introduction Throughout this paper, A is a commutative noetherian local ring As indicated by the title, this paper is concerned with explicit constructions of totally reflexive A-modules, as defined by Auslander [1] in 1967 the terminology totally reflexive was introduced in 2002 by Avramov and Martsinkovsky [4] Totally reflexive A-modules always exist, indeed, A itself is totally reflexive However, in general, there need not exist non-free totally reflexive A-modules If A is Gorenstein then the totally reflexive A-modules are exactly the maximal Cohen- Macaulay modules, and their representation theory is a classical field of study A main result of Christensen, Piepmeyer, Striuli, and Takahashi [6, Theorem B] asserts that if A is not Gorenstein, then existence of one non-free totally reflexive A-module implies the existence of infinitely many non-isomorphic indecomposable totally reflexive A-modules Unfortunately, the known proof of this interesting result is not constructive, which is why the authors of [6] raise the following Question [6, 48] Assume that A is not Gorenstein, and that there exists a non-free totally reflexive A-module Are there constructions that produce infinite families of non-isomorphic indecomposable totally reflexive A-modules? Under certain assumptions, we give in this paper exactly such a construction In fact, the modules produced by our construction even have periodic minimal free resolutions We begin mentioning a couple of related results from the literature: Assume that A is complete or has an uncountable residue field Assume furthermore that there exist a prime ideal p in A with grade p > 0 and dim A/p > 1, and a totally reflexive A-module M such that M p is not A p -free Then Takahashi [11] proves the existence of uncountably many non-isomorphic indecomposable totally reflexive A-modules The proof is not constructive, however, in [11, Example 2000 Mathematics Subject Classification 13C13, 13H99 Key words and phrases Computation of module of homomorphisms; exact pair of zero divisors; indecomposability; isomorphism class; periodic minimal free resolution; totally reflexive module 1

2 2 HENRIK HOLM 43] one does find an example A Q[[x, y, z]]/x 2, where Q is a complete local domain which is not a field where uncountably many non-isomorphic indecomposable totally reflexive A-modules are actually constructed Assume that A has an embedded deformation of codimension c 2, ie there is a local ring Q, q, k and a Q-regular sequence x in q 2 of length c such that A Q/x Avramov, Gasharov, and Peeva [2] construct a non-free totally reflexive A-module G, from which infinitely many non-isomorphic indecomposable totally reflexive A- modules can be constructed by using results of Avramov and Iyengar [3]: Consider the graded ring R k[χ 1,, χ c ] where χ i has degree 2 From [3, 41] and [3, proof of 741] it follows that given a closed subset X Spec R, there is a module M X Thick A G A DA with cohomological support Supp M X, k equal to X As M X Thick A G A some syzygy G X of M X is totally reflexive, and by [3, 42] one also has Supp G X, k X Clearly, if X Y then G X and G Y are not isomorphic Furthermore, if X is irreducible, it follows by [3, 362] that Supp G X, k X for some indecomposable summand G X of G X In the author s opinion, the construction of M X given in [3, proof of 41] is not as explicit and accessible as one could hope for Under suitable assumptions, this paper offers a construction of infinitely many non-isomorphic indecomposable totally reflexive A-modules, which is different from the ones mentioned above Our construction has the advantage of being fairly elementary; it applies to large classes of examples, cf Example 24 and Lemma 25; and it makes explicit computations of relevant Hom-modules possible As already mentioned, the totally reflexive modules coming from our construction even have periodic minimal free resolutions Given an exact pair of zero divisors x, y in A, as defined by Henriques and Şega [9], it is easily seen that the A-modules A/x and A/y are non-free and totally reflexive Thus, if A is not Gorenstein, one should by the main result in [6] be able to construct infinitely many non-isomorphic indecomposable totally reflexive A-modules In this paper, we define for each a in A two totally reflexive A-modules G a and H a as cokernels of certain 2 2 matrices with entries in A A special case of our main result Corollary 55 reads as follows Theorem Let x, y A be a regular exact pair of zero divisors, and let a A If a is regular on the A-module A/x, y then G a, G a 2, G a 3, G a 4,, H a, H a 2, H a 3, H a 4, is a double infinite family of non-isomorphic, indecomposable, non-free, totally reflexive A-modules Furthermore, one has the following identities, Hom A H a m, G a n Hom A H a n, G a m G a m+n, Hom A G a n, H a m Hom A G a m, H a n H a m+n, Hom A G a m, G a n Hom A H a n, H a m H a m n for m > n A for m n G a n m for m < n The paper is organized as follows: In Section 2, we introduce exact pairs of zero divisors and discuss regularity In Section 3, the modules G a and H a are defined In the technical Section 4, we compute the module of homomorphisms between various combinations of G a and H b The final Section 5 contains our main results In this section, we study the isomorphism classes and indecomposability of the modules G a and H a

3 CONSTRUCTION OF TOTALLY REFLEXIVE MODULES 3 2 Exact pairs of zero divisors In this section, we consider an exact pair of zero divisors as defined by Henriques and Şega [9] We introduce a notion of regularity for such a pair and give examples 21 Definition Two non-units x, y A are called an exact pair of zero divisors if Ann A x y and Ann A y x Let M be a finitely generated A-module, and let a A be an element Recall that a is weakly regular on M if multiplication by a on M is a monomorphism If, in addition, the element a is not a unit, then a is called regular on M A weakly regular element on the A-module M A is simply referred to as a weakly regular element in the ring A 22 Lemma Let x, y A be an exact pair of zero divisors Then the following conditions are equivalent: i x is regular on A/y ii y is regular on A/x iii x y 0 Proof By symmetry, it suffices to prove i iii First assume i, and let z be in x y As z x we have z ax for some a As z y it follows that x[a] y [z] y [0] y in A/y, and since multiplication by x on A/y is a monomorphism, we conclude that [a] y [0] y, that is, a y Consequently, z ax 0 since, in particular, yx 0 Next assume iii We must argue that multiplication by x on A/y is a monomorphism Thus, let [a] y in A/y satisfy x[a] y [0] y, that is, ax y Then ax x y 0, and hence a Ann A x y Thus [a] y [0] y in A/y 23 Definition An exact pair of zero divisors x, y A that satisfies the equivalent conditions in Lemma 22 is called a regular exact pair of zero divisors 24 Example Let Q be a commutative noetherian local ring, and let f, g be regular elements in Q Set A Q/fg and let x, y A denote the cosets of f, g with respect to the ideal fg Then x, y is an exact pair of zero divisors in A; and this pair is regular if and only if f g fg in Q In particular, if f and g are non-zero and non-units in a noetherian local UFD Q, then x, y is an exact pair of zero divisors in A; and this pair is regular if and only if f and g are relatively prime Note that local UFDs come in all shapes and sizes Gorenstein, non-gorenstein etc, in fact, by Heitmann [8, Theorem 8] every complete local ring of depth >1, in which no integer is a zero divisor, is the completion of a local UFD Several results in this paper refer to a weakly regular element a on the A-module A/x, y For example, a could be a unit in A However, only in the case where a is regular on the A-module A/x, y do we get interesting applications of our results For the existence of regular elements on A/x, y, we give the following 25 Lemma Let Q be a commutative noetherian local ring, and let f, g be regular elements in Q such that f g fg Consider the regular exact pair x, y of zero divisors in A Q/fg constructed in Example 24 If depth Q > 2 then there exists regular elements on the A-module A/x, y

4 4 HENRIK HOLM Proof We must prove that the A-module A/x, y has depth > 0, equivalently, that the ring A/x, y has depth > 0 By the ring isomorphism A/x, y Q/f, g, we are required to show that Q/f, g has depth > 0 To finish the proof it suffices to argue that f, g is a Q-regular sequence, because then depth Q/f, g depth Q 2 by Bruns and Herzog [5, Proposition 1210d] However, it is given that f is regular on Q, and that g is regular on Q/f follows from the assumption f g fg and the regularity of g on Q 3 Two families of totally reflexive modules Assuming that the ring A admits an exact pair of zero divisors, we introduce in this section two families G a a A and H a a A of totally reflexive A-modules We begin with the following definition due to Auslander [1, 322] Although Auslander himself did not use the terminology totally reflexive module, this usage was introduced by Avramov and Martsinkovsky [4, 2] and has grown to be standard in several later papers on the subject 31 Definition A finitely generated A-module G is totally reflexive if it satisfies the following three conditions: 1 Ext i AG, A 0 for all i > 0 2 Ext i AHom A G, A, A 0 for all i > 0 3 The biduality map G Hom A Hom A G, A, A is an isomorphism 32 Definition Let x, y A be an exact pair of zero divisors, and let a A We define two 2 2 matrices, x a y a γ a and η 0 y a 0 x Furthermore, considering γ a and η a as A-linear maps acting on column vectors by multiplication from the left, we define two finitely generated A-modules, G a Coker γ a and H a Coker η a 33 Remark Let x, y A be an exact pair of zero divisors, let a be any element in A, and let u be a unit in A It is straightforward to see that there are isomorphisms, G ua Ga and H ua Ha In particular, G a Ga and H a Ha 34 Observation Of course, the modules G a and H a do not only depend on the ring element a, but also on the exact pair of zero divisors x, y A more precise notation would therefore be G x,y a and Ha x,y Note that if x, y is a regular exact pair of zero divisors, then so is y, x By Definition 32 and Remark 33, it follows that G y,x a Ha x,y and Ha y,x G x,y a 35 Theorem Let x, y A be an exact pair of zero divisors, and let a A The A-modules G a and H a are totally reflexive, in fact, they are cokernels of some of the differentials in the following periodic and totally acyclic complex of free A-modules, γa ηa γa ηa Furthermore, G a and H a are dual, ie G a HomA H a, A and H a HomA G a, A, and they satisfy the following conditions: a Ext i AH b, G a Ext i AH a, G b for every i Z

5 CONSTRUCTION OF TOTALLY REFLEXIVE MODULES 5 b Ext i AG a, H b Ext i AG b, H a for every i Z c Ext i AG a, G b Ext i AH b, H a for every i Z Proof Denote by F the sequence displayed in the proposition As γ a η a η a γ a 0, we conclude that F is a complex To show that F is acyclic exact, we must argue that Ker γ a Im η a and Ker η a Im γ a We only prove the first inclusion, as the other one is proved analogously To this end, assume that b 2 γ a b1 b 2 b1 x + b 2 a b 2 y 0 0 As b 2 y 0 we get that b 2 c 2 x for some c 2 in A Hence b 1 +c 2 ax b 1 x+b 2 a 0, so there exists c 1 in A with b 1 + c 2 a c 1 y Consequently, b1 c1 y c 2 a c1 η c 2 x a Im η c a 2 There is an isomorphism of complexes Hom A F, A F given by, η t a γ a γ t a η t a γ t a η a γ a η a, where t means transposition of matrices It follows that Hom A F, A is exact, and hence F is a totally acyclic By [1, prop 8 in 322], the modules G a and H a are totally reflexive That G a and H a are each others dual is immediate from the isomorphism Hom A F, A F It remains to show a, b, and c Since these assertions have similar proofs, we only prove part a This is done below As H a is totally reflexive, Ext i AH a, A 0 for i > 0 Since Hom A H a, A G a, it follows that RHom A H a, A G a in the derived category DA, cf Weibel [12, Chapter 10] Consequently, one has the following isomorphisms in DA, RHom A H b, G a RHom A H b, RHom A H a, A RHom A H a, RHom A H b, A RHom A H a, G b, where the second isomorphism is the so-called swap-isomorphism The isomorphism of A-modules in part a is now obtained by taking homology Under additional assumptions on the ring element a, more can be said about the modules G a and H a from Definition Proposition Let x, y A be an exact pair of zero divisors, and let a A a If a is weakly regular on A/y then G a is isomorphic to the ideal y, a b If a is weakly regular on A/x then H a is isomorphic to the ideal x, a c If a is a unit then G a Ha A d If a is not a unit then neither G a nor H a is free

6 6 HENRIK HOLM Proof a: Since G a Coker γ a, it suffices to prove exactness of the sequence, γ a y a y, a 0 Clearly, the image of y a is all of y, a, and its composite with γ a is zero To prove Kery a Im γ a, assume that b 1, b 2 satisfies b1 y a b b 1 y b 2 a 0 2 As b 2 a b 1 y y and since a is weakly regular on A/y, it follows that b 2 y, ie b 2 c 2 y for some c 2 Now b 1 c 2 ay b 1 y b 2 a 0, and as Ann A y x we conclude that b 1 c 2 a c 1 x for some c 1 Consequently, b1 b 2 c1 x + c 2 a c 2 y x a 0 y c1 c 2 γ a c1 c 2 Im γ a b: Similar to the proof of part a c: Immediate from parts a and b d: If a is not a unit then all the entries in γ a and η a belong to the unique maximal ideal m in A Consequently, if k A/m denotes the residue field of A, the homomorphisms k A γ a and k A η a are zero, and it follows that γa ηa γa G a 0 is an augmented minimal free resolution of G a Hence G a has infinite projective dimension Analogously one sees that H a has infinite projective dimension 4 Computation of Hom-modules In this section, we will explicitly compute certain Hom-modules For example, we will prove that Hom A G a, H b is isomorphic to H ab In order to carry out such computations, we need a concrete way to represent the module of homomorphisms between two given finitely generated A-modules The representation we will use can be found in eg Greuel and Pfister [7, Example 2126]; we give a recap in Remark Let ρ i be an m i n i matrix with entries in A, where i 1, 2 Then Q ρ1,ρ 2 { ψ M m2 m 1 A ψρ 1 ρ 2 ξ for some ξ M n2 n 1 A } is an A-submodule of M m2 m 1 A which contains ρ 2 M n2 m 1 A as a submodule Furthermore, there is an isomorphism of A-modules, Hom A Coker ρ 1, Coker ρ 2 Q ρ1,ρ 2 /ρ 2 M n2 m 1 A 42 Lemma Let x, y A be a regular exact pair of zero divisors Furthermore, let a, b A be elements such that a or b is weakly regular on the A-module A/x, y Then the A-submodule, Q ηb,γ a { ψ M 2 A ψηb γ a ξ for some ξ M 2 A } of M 2 A is generated by the following five matrices, ψ 1, ψ 0 0 2, ψ x b 3, ψ 0 y 4 x 0, ψ a 0 y 0

7 CONSTRUCTION OF TOTALLY REFLEXIVE MODULES 7 Proof First note that if we define, ξ 1, ξ 0 0 2, ξ , ξ b 0 0, ξ 0 0 5, y b then ψ i η b γ a ξ i, and hence ψ i belongs to Q ηb,γ a Next we show that ψ 1,, ψ 5 generate all of Q ηb,γ a To this end, let ψ b ij be any matrix in Q ηb,γ a, that is, there exists ξ c ij such that ψη b γ a ξ; ie b11 y b 11 b + b 12 x c11 x + c 21 a c 12 x + c 22 a b 21 y b 21 b + b 22 x c 21 y c 22 y We must prove the existence of f 1,, f 5 in A such that ψ 5 i1 f iψ i, that is, b11 b 12 f4 x + f 5 a f 1 b 21 b 22 f 2 x + f 5 y f 2 b + f 3 y Of course, if we define f 1 b 12 then entry 1, 2 in holds Also note that from entry 2, 1 in we get b 21 c 21 y 0, and hence b 21 c 21 qx for some q Below, we construct f 2 and f 5 such that entry 2, 1 in holds, that is, b 21 f 2 x + f 5 y However, first we demonstrate how f 3 and f 4 can be constructed from and such that entries 1, 1 and 2, 2 in hold: Existence of f 3 : From entry 2, 2 in we get b 21 b + b 22 x c 22 y Combining this with we obtain b 22 f 2 bx c 22 +f 5 by As the exact pair of zero divisors x, y is regular, see Definition 23, it follows that b 22 f 2 b f 3 y for some f 3 Thus entry 2, 2 in holds Existence of f 4 : From entry 1, 1 in, from the equation b 21 c 21 qx found above, and from, we get the following equalities, b 11 y c 11 x + c 21 a c 11 x + b 21 qxa c 11 + f 2 a qax + f 5 ay Consequently, b 11 f 5 ay x As the pair x, y is regular, see Definition 23, it follows that b 11 f 5 a f 4 x for some f 4 This shows that entry 1, 1 in holds It remains to prove that holds, ie that b 21 belongs to x, y The proof is divided into two cases: Existence of in the case where a is weakly regular on A/x, y: From entry 1, 1 in, we have c 21 a c 11 x + b 11 y x, y As a is weakly regular on A/x, y, it follows that c 21 x, y Combining this with the equation b 21 c 21 qx found above, it follows that b 21 x, y Existence of in the case where b is weakly regular on A/x, y: From entry 2, 2 in, we have b 21 b b 22 x c 22 y x, y As b is weakly regular on A/x, y, it follows that b 21 x, y as desired 43 Theorem Let x, y A be a regular exact pair of zero divisors Furthermore, let a, b A be elements such that a or b is weakly regular on the A-module A/x, y Then there are isomorphisms of A-modules, a Hom A H a, G b Hom A H b, G a G ab b Hom A G a, H b Hom A G b, H a H ab Proof First note that by applying part a to the regular exact pair of zero divisors y, x instead of x, y, then part b follows from Observation 34

8 8 HENRIK HOLM In part a, the first isomorphism is by Theorem 35a To prove the second isomorphism, we first apply Remark 41 to get the representation, Hom A H b, G a Q ηb,γ a /γ a M 2 A Using this identification, we can consider the A-linear map, π s Hom A H b, G a, s[ψ t 1 ] + t[ψ 2 ] Here [ ] denotes coset with respect to the submodule γ a M 2 A of Q ηb,γ a, and ψ i are the matrices introduced in Lemma 42 Recall that G ab Coker γ ab ; thus to prove the theorem, it suffices to show exactness of the sequence, γ ab π Hom A H b, G a 0 To show that π is surjective, it suffices by Lemma 42 to see that each [ψ i ] is in the image of π However, this is clear since [ψ 3 ] a[ψ 1 ] and [ψ 4 ] [ψ 5 ] [0] To see that πγ ab 0, we must show that if s, t γ ab u, v, where u, v, then sψ 1 + tψ 2 is in γ a M 2 A But as s u γ t ab v x ab 0 y u v the desired conclusion follows since, sψ 1 + tψ 2 ux + vab + vy 0 0 x b ux + vab, vy 0 ux + vab 0 u γ 0 vyb a 0 vb It remains to prove that Ker π Im γ ab To this end, assume that s, t satisfies πs, t 0 This means that sψ 1 + tψ 2 belongs to γ a M 2 A, that is, there exists ξ c ij in M 2 A such that sψ 1 + tψ 2 γ a ξ; equivalently, 0 s c11 x + c 21 a c 12 x + c 22 a tx tb c 21 y c 22 y From entry 2, 1 in we get that tx c 21 y Since the exact pair of zero divisors x, y is regular, see Definition 23, it follows that t vy for some v From entry 2, 2 in we get that tb c 22 y Combining this with t vy one gets vb c 22 y 0 Consequently, vb c 22 px for some p Now, inserting c 22 vb px in the equation s c 12 x + c 22 a coming from entry 1, 2 in, we get that s c 12 pax + vab ux + vab, where u c 12 pa Since s ux+vab and t vy, we see from that s, t γ ab u, v Im γ ab It is also desirable to know what the module Hom A G u, G v Hom A H v, H u see Theorem 35c for this isomorphism looks like for every combination of u and v Since the author was not able to figure this out, we restrict ourselves to the case where u is in the ideal generated by v, or vice versa, see Theorem Lemma Let x, y A be a regular exact pair of zero divisors Furthermore, let a A be weakly regular on the A-module A/x, y, and let b A be any element Then the A-submodule, Q γab,γ a { ψ M 2 A ψγ ab γ a ξ for some ξ M 2 A } of M 2 A is generated by the following five matrices, x ψ 1, ψ 0 x 2, ψ 0 b 3, ψ Proof Similar to but easier than the proof of Lemma 42 a 0, ψ y a 0 y

9 CONSTRUCTION OF TOTALLY REFLEXIVE MODULES 9 45 Theorem Let x, y A be a regular exact pair of zero divisors Furthermore, let a A be weakly regular on the A-module A/x, y, and let b A be any element Then there are isomorphisms of A-modules, a Hom A H a, H ab Hom A G ab, G a H b b Hom A G a, G ab Hom A H ab, H a G b Proof First note that by applying part a to the regular exact pair of zero divisors y, x instead of x, y, then part b follows from Observation 34 In part a, the first isomorphism is by Theorem 35c To prove the second isomorphism, we apply Remark 41 to get the representation, Hom A G ab, G a Q γab,γ a /γ a M 2 A Using this identification, we can consider the A-linear map, π s Hom A G ab, G a, s[ψ t 1 ] + t[ψ 2 ] Here [ ] denotes coset with respect to the submodule γ a M 2 A of Q γab,γ a, and ψ i are the matrices introduced in Lemma 44 Recall that H b Coker η b, see Definition 32; thus to prove the result, it suffices to show exactness of the sequence, η b π Hom A G ab, G a 0 To see that π is surjective, it suffices by Lemma 44 to see that each [ψ i ] belongs to Im π However, this is clear since [ψ 3 ] [ψ 4 ] [ψ 5 ] [0] It remains to prove Im η b Ker π The arguments needed to show this are similar to the ones found in the proof of Theorem 43 5 Isomorphism classes and indecomposability In this section, we address two natural questions about the family of totally reflexive modules G a a A H a a A defined in 32, namely: Are the modules in this family indecomposable, and are they pairwise non-isomorphic? In general, both of these questions have a negative answer, see Example 54, Remark 33, and Proposition 36c Under suitable assumptions on a A, we will prove that End A G a and End A H a are both isomorphic to A, in particular, these endomorphism rings are commutative, noetherian and local Hence G a and H a are indecomposable in a quite strong sense We will also give a condition on ring elements a, b A which ensures that, for example, G a and H b are not isomorphic 51 Remark Let M be a reflexive A-module Clearly, M is indecomposable if and only if its dual Hom A M, A is indecomposable It is not hard to see that something stronger holds, namely there is an isomorphism of A-algebras, End A M op EndA Hom A M, A In particular, G a is indecomposable if and only if H a is indecomposable The following result is probably folklore However, since the author was not able to find a reference, a short proof has been included 52 Lemma Let M be a finitely generated A-module If A and Hom A M, M are isomorphic as A-modules, then the canonical homomorphism of A-algebras, given by χ: A End A M, χa a1 M, is also an isomorphism

10 10 HENRIK HOLM Proof Let ζ : A Hom A M, M be an isomorphism of A-modules As ζ is surjective, there exists b in A such that bζ1 ζb 1 M It follows that bm M Furthermore, M 0 since Hom A M, M is non-zero Hence Nakayama s Lemma [10, Theorem 22] implies that b is a unit in A, and thus bζ is also an isomorphism It remains to note that bζ χ since bζa baζ1 aζb a1 M χa 53 Theorem Let x, y A be a regular exact pair of zero divisors, and let a, b be arbitrary elements in A Then one has: a If a is weakly regular on A/x, y then there are isomorphisms of A-algebras, End A G a A End A H a In particular, the A-modules G a and H a are indecomposable b Assume that a or b is weakly regular on A/x, y Assume furthermore that a and b are not both units Then G a H b c If a is weakly regular on A/x, y, and b is not a unit, then G a G ab and H a H ab Proof a: Taking b 1 in Theorem 45a, we get the first two isomorphism of A-modules in the chain Hom A H a, H a Hom A G a, G a H 1 A The last isomorphism is by Proposition 36c Now Lemma 52 completes the proof b: Since a or b is weakly resular on A/x, y, part a gives Hom A G a, G a A or Hom A H b, H b A Thus, to prove that G a and H b are not isomorphic, it suffices to see Hom A G a, H b A By Theorem 43b there is an isomorphism, Hom A G a, H b H ab Since a and b are not both units, ab is not a unit, and hence H ab is not free by Proposition 36d c: We only prove G a G ab, as the proof of H a H ab is similar By part a there is an isomorphism, Hom A G a, G a A Thus, to prove that G a and G ab are not isomorphic, it is enough to show Hom A G ab, G a A By Theorem 45a one has Hom A G ab, G a H b, which is not free as b is not a unit, see 36d 54 Example For a general ring element a, the conclusion in Theorem 53a fails For example, if a is in x, then G a A/x A/y is not indecomposable For the last assertion in the following corollary, it is useful to keep in mind that H 1 A G1 by Proposition 36c 55 Corollary Let x, y A be a regular exact pair of zero divisors Let b n n1 be a sequence of not necessarily distinct elements in A which are regular on the A-module A/x, y If we define a n b 1 b n then G a1, G a2, G a3, G a4,, H a1, H a2, H a3, H a4, is a double infinite family of indecomposable, non-free, totally reflexive A-modules, which are pairwise non-isomorphic Furthermore, one has the following identities: Hom A H am, G an Hom A H an, G am G ama n, Hom A G an, H am Hom A G am, H an H ama n, { Hom A G am, G an Hom A H an, H am Ham/a n for m n G an/a m for m n

11 CONSTRUCTION OF TOTALLY REFLEXIVE MODULES 11 Proof By Theorem 35, the modules G an and H an are totally reflexive As a n is not a unit, G an and H an are non-free by Proposition 36d Since a n is regular on A/x, y, it follows from Theorem 53a that G an and H an are indecomposable By Theorem 53b, the modules G am and H an are not isomorphic If m < n then a n a m b where b b m+1 b n which is not a unit Thus Theorem 53c gives that G am and G an are not isomorphic, and furthermore that H am and H an are not isomorphic Hence the modules in the given list are pairwise non-isomorphic The Hom-identities are immediate from Theorems 43 and 45 Acknowledgements It is a pleasure to thank the referee for useful comments and suggestions References [1] Maurice Auslander, Anneaux de Gorenstein, et torsion en algèbre commutative, Secrétariat mathématique, Paris, 1967, Séminaire d Algèbre Commutative dirigé par Pierre Samuel, 1966/67 Texte rédigé, d après des exposés de Maurice Auslander, par Marquerite Mangeney, Christian Peskine et Lucien Szpiro École Normale Supérieure de Jeunes Filles Available from MR [2] Luchezar L Avramov, Vesselin N Gasharov, and Irena V Peeva, Complete intersection dimension, Inst Hautes Études Sci Publ Math 1997, no 86, MR [3] Luchezar L Avramov and Srikanth B Iyengar, Constructing modules with prescribed cohomological support, Illinois J Math , no 1, 1 20 MR [4] Luchezar L Avramov and Alex Martsinkovsky, Absolute, relative, and Tate cohomology of modules of finite Gorenstein dimension, Proc London Math Soc , no 2, MR [5] Winfried Bruns and Jürgen Herzog, Cohen-Macaulay rings, Cambridge Studies in Advanced Mathematics, vol 39, Cambridge University Press, Cambridge, 1993 MR [6] Lars Winther Christensen, Greg Piepmeyer, Janet Striuli, and Ryo Takahashi, Finite Gorenstein representation type implies simple singularity, Adv Math , no 4, MR [7] Gert-Martin Greuel and Gerhard Pfister, A singular introduction to commutative algebra, extended ed, Springer, Berlin, 2008, With contributions by Olaf Bachmann, Christoph Lossen and Hans Schönemann, With 1 CD-ROM Windows, Macintosh and UNIX MR j:13001 [8] Raymond C Heitmann, Characterization of completions of unique factorization domains, Trans Amer Math Soc , no 1, MR [9] Inês B Henriques and Liana M Şega, Free resolutions over short Gorenstein local rings, preprint 2009, arxiv: v2 [mathac] [10] Hideyuki Matsumura, Commutative ring theory, second ed, Cambridge Studies in Advanced Mathematics, vol 8, Cambridge University Press, Cambridge, 1989, Translated from the Japanese by M Reid MR [11] Ryo Takahashi, An uncountably infinite number of indecomposable totally reflexive modules, Nagoya Math J , MR [12] Charles A Weibel, An introduction to homological algebra, Cambridge Studies in Advanced Mathematics, vol 38, Cambridge University Press, Cambridge, 1994 MR Department of Basic Sciences and Environment, Faculty of Life Sciences, University of Copenhagen, Thorvaldsensvej 40, DK-1871 Frederiksberg C, Denmark address: hholm@lifekudk URL: