VANISHING OF EXT AND TOR OVER SOME COHEN-MACAULAY LOCAL RINGS

CRAIG HUNEKE, LIANA M. ŞEGA, AND ADELA N. VRACIU

Abstract. We discuss vanishing of cohomology of finite modules over Cohen-Macaulay local rings (R, 𝔪). Special attention is given to the case when the modules are annihilated by 𝔪^{2}. (Note that if 𝔪^{3} = 0, then we can assume the modules satisfy this condition.) In this case we obtain effective versions of conjectures of Auslander-Reiten and Tachikawa.

Introduction

In recent years, there has been growing interest in understanding the vanishing of Ext and Tor over Noetherian local rings, especially in the case when the ring is Artinian. One motivation for this interest comes from a conjecture of Auslander and Reiten [2], which in the case of commutative local rings can be stated as follows:

Conjecture (Auslander-Reiten). Let (R, 𝔪) be a commutative Noetherian local ring, and M a finitely generated R-module. If Ext_{R}^{i}(M, M ⊕ R) = 0 for all i > 0, then M is free.

This conjecture was initially stated for Artin algebras, and Auslander, Ding and Solberg [1] widened the context to algebras over commutative local rings. A recent result of Huneke and Leuschke [13] establishes the conjecture in the case when R is a quotient by a regular sequence of an excellent Cohen-Macaulay normal domain containing the rational numbers.

To prove the Auslander-Reiten Conjecture for Cohen-Macaulay rings, it suffices to consider Artinian rings. Indeed, if R is assumed to be Cohen-Macaulay, then one can first replace M by a high syzygy in a free resolution of M (see [2]) to assume that M is maximal Cohen-Macaulay. Now, if x_{1}, ..., x_{d} is a maximal M- and R-regular sequence, and I is the ideal generated by it, then the vanishing of Ext_{R}(M, M ⊕ R) passes to Ext_{R/I}(M/IM, M/IM ⊕ R/I); also, M has finite projective dimension (and it is thus free) if and only if M/IM has. Thus, replacing R by R/I and M by M/IM, one can assume without loss of generality that R is Artinian.

In this paper we concentrate on the commutative Artinian case. If 𝔪^{2} = 0, then the first syzygy in a minimal free resolution of any non-free R-module is annihilated by the maximal ideal, and the Auslander-Reiten conjecture follows trivially. The first interesting open case is when 𝔪^{3} = 0.

Rings in which 𝔪^{3} = 0 were systematically studied by Lescot [15]. In particular, his results give the Poincaré series of finitely generated modules none of whose syzygies has a copy of the residue field as a direct summand. Only such modules could provide counterexamples to the Auslander-Reiten conjecture. For if Ext_{R}^{i}(M, M ⊕ R) = 0 for all i > 0, and M is not free, it is clear by shifting degree that no syzygy of M can have the residue field as a direct summand.

Date: September 22, 2003.

C.H. and A.N.V. were partially supported by NSF grant DMS-0098654.

L.M.Ş. was supported by Clay Mathematics Institute during the Summer 2002.

One of our main results proves the Auslander-Reiten conjecture for rings with 𝔪^{3} = 0. Note that the statement gives an effective bound on the required number of vanishing Ext modules.

4.1. Theorem. Let (R, 𝔪) be a commutative Artinian local ring with 𝔪^{3} = 0 and M a finitely generated R-module.

(1) If Ext_{R}^{i}(M, M ⊕ R) = 0 for four consecutive values of i with i ≥ 2, then M is free.

(2) If R is Gorenstein and Ext_{R}^{i}(M, M) = 0 for some i > 0, then M is free.

The case i = 1 in the second part is known: Hoshino [12] proved, more generally, that if R is a (possibly non-commutative) finite dimensional self-injective local algebra with radical cube zero and Ext_{R}^{1}(M, M) = 0, then M is free.

As mentioned above, in the context of the Auslander-Reiten conjecture one can replace the original module M by an arbitrary syzygy in a minimal free resolution of M. Now, if 𝔪^{3} = 0 and N is such a syzygy, then 𝔪^{2}N = 0. A closer look at modules annihilated by 𝔪^{2} shows that the Auslander-Reiten conjecture holds for any such module over an arbitrary Artinian local ring. We also give a bound on the required number of vanishing Exts, in terms of the minimal number of generators, denoted ν(−), of certain modules.

4.2. Theorem. Let (R, 𝔪) be a commutative Artinian local ring and M a finitely generated R-module with 𝔪^{2}M = 0.

If Ext_{R}^{i}(M, M ⊕ R) = 0 for all i with 0 < i ≤ max{3, ν(M), ν(𝔪M)}, then M is free.

We prove our results by considering more generally the vanishing of Ext_{R}(M, N) where M and N are two finitely generated modules over an Artinian ring R. Since the Matlis dual of such Ext modules are Tor modules, we often find it more convenient to work with the vanishing properties of Tor. Our arguments suggest that the vanishing of Tor_{i}^{R}(M, N) for all positive i places restrictions which relate the annihilators of M, N and R. Specifically, we propose the following:

5.1. Conjecture. Let (R, 𝔪) be a commutative Artinian local ring and let M, N be nonzero finitely generated R-modules with 𝔪^{2}M = 𝔪^{2}N = 0. If Tor_{i}^{R}(M, N) = 0 for all i > 0, then 𝔪^{3} = 0.

One can ask a more general question: Let p, q be positive integers and assume that M, N are nonzero modules with 𝔪^{p}M = 𝔪^{q}N = 0. If Tor_{i}^{R}(M, N) = 0 for all i > 0, does it follow that 𝔪^{p+q−1} = 0?

The hypothesis of Conjecture 5.1 imposes a strong condition on the Poincaré series of the residue field of R, cf. Lemma 1.8. This allows us to verify the conjecture for several classes of rings, including complete intersections of codimension greater than 2, Koszul rings, Golod rings, etc. In Theorem 5.4 we prove the conjecture when the ring R is standard graded.

Another conjecture which has received attention recently is a conjecture of Tachikawa. A commutative version of this conjecture for Cohen-Macaulay local rings is the following, cf. Avramov, Buchweitz and Şega [6], and also Hanes and Huneke [11]:

Conjecture (Tachikawa). Let (R, 𝔪) be a Cohen-Macaulay local ring. If R has a canonical module ω and Ext_{R}^{i}(ω, R) = 0 for all i > 0, then R is Gorenstein, i.e. ω is free.

This version of the Tachikawa conjecture is subsumed by the Auslander-Reiten conjecture, since the condition Ext_{R}^{i}(ω, ω) = 0 for all i > 0 holds automatically when R is Cohen-Macaulay.

Assuming that 𝔪^{3} = 0 and R is a finite dimensional local algebra over a field, this conjecture was proved by Asashiba [3] under the weaker assumption that Ext_{R}^{1}(ω, R) = 0.

Recall that Ext_{R}^{i}(ω, R) is Matlis dual to Tor_{i}^{R}(ω, ω) when R is Artinian. Thus, the theorem below is equivalent to the one that appears in Section 2 under the same number.

2.11. Theorem. Let (R, 𝔪) be a commutative Artinian local ring with 𝔪^{3} = 0. The following statements are equivalent:

(1) R is Gorenstein.

(2) Ext_{R}^{1}(ω, R) = 0.

(3) Ext_{R}^{2}(ω, R) = Ext_{R}^{3}(ω, R) = 0.

(4) Ext_{R}^{j}(ω, R) = Ext_{R}^{j+1}(ω, R) = Ext_{R}^{j+2}(ω, R) = 0 for some j ≥ 0.

A different proof of the equivalence (1) ⟺ (2), along the lines of [4], is given in [6].

Section 1 contains a number of lemmas we will use throughout the paper, concerning the growth of Betti numbers of modules under certain conditions on the vanishing of Tor.

Section 2 contains our work on rings with 𝔪^{3} = 0. An important technical result is Theorem 2.5, which gives detailed information comparing the Betti numbers of two modules with three consecutive vanishing Tors. In this section we prove Theorem 2.11.

In Section 3 we show that if sufficiently many Tors vanish for two modules M,N and 𝔪^{2}M = 0, then either M or N is free. However, for this result we impose stringent conditions on the ring.

Section 4 contains the proof of Theorems 4.1 and 4.2.

In Section 5 we deal with Conjecture 5.1, and give the proof of Theorem 5.4. In Section 6 we prove Tachikawa’s conjecture for Cohen-Macaulay rings of type

two.

1. Betti numbers

In this paper (R, 𝔪, k) denotes a commutative Noetherian local ring with maximal ideal 𝔪 and residue field k. We consider finitely generated R-modules M, N.

The number ν(M) denotes the minimal number of generators of M and λ(M) denotes the length of M.

For every i ≥ 0 we let M_{i} denote the i-th syzygy of M in a minimal free resolution

… ⟶ R^{b_{i+1}(M)} ⟶^{δ_{i}} R^{b_{i}(M)} ⟶ … ⟶^{δ_{0}} R^{b_{0}(M)} ⟶ M ⟶ 0

The number b_{i}(M) is called the i-th Betti number of M over R. The Poincaré series of M over R is the formal power series

P_{M}^{R}(t) = ∑_{i=0}^{∞} b_{i}(M)t^{i}

In this section we describe several restrictions on the Betti numbers of M,N that are imposed under the assumption that Tor_{i}^{R}(M, N) = 0 for certain values of i.

1.1. Remark. If Tor_{i}^{R}(M, N) = 0 for some i > 0 and N has infinite projective dimension, then k is not a direct summand of any of the R-modules M_{0}, …, M_{i−1}. Indeed, we have Tor_{i−j}^{R}(M_{j}, N) ≅ Tor_{i}^{R}(M, N) = 0 for all j < i. If k is a direct summand of M_{j} for some j < i, then Tor_{i−j}^{R}(k, N) = 0, contradicting the assumption on N.

Let P (t) = a_{0} + a_{1}t + ⋯ + a_{i}t^{i} + … be a formal power series. For each n ≥ 0 we denote [P (t)]_{≤n} the polynomial a_{0} + a_{1}t + ⋯ + a_{n}t^{n}.

The next result is a slightly modified version of a technique in [17, 1.1]:

1.2. Lemma. Let n be a positive integer. If Tor_{i}^{R}(M, N) = 0 for all i ∈ [1, n], then

[P_{M⊗_{R}N}^{R}(t)]_{≤n} = [P_{M}^{R}(t)P_{N}^{R}(t)]_{≤n}

Proof. Let F, respectively G, be a minimal free resolution of M, respectively N, over R. The hypothesis implies that H_{i}(F ⊗_{R} G) = 0 for all i ∈ [1, n]. Since F ⊗_{R} G is a minimal complex with H_{0}(F ⊗_{R} G) = M ⊗_{R} N, we note that (F ⊗_{R} G)_{≤n} is the beginning of a minimal free resolution of M ⊗_{R} N. We have thus:

[P_{M⊗_{R}N}^{R}]_{≤n} = ∑_{i=0}^{n} rank(F ⊗_{R} G)_{i}t^{i} = [P_{M}^{R}(t)P_{N}^{R}(t)]_{≤n}  □

1.3. For each nonzero R-module M of finite length we set

γ(M) = \frac{λ(M)}{ν(M)} − 1

Note that γ(M) is also equal to \frac{λ(𝔪M)}{ν(M)}. It is thus a rational number in the

interval [0, λ(R) − 1]. The extreme values on this interval are attained as follows: γ(M) = 0 if and only if 𝔪M = 0 and γ(M) = λ(R) − 1 if and only if M is free.

We will often use the definition of γ(M) in length computations as follows:

(1.3.1)  λ(M) = ν(M)(γ(M) + 1)

1.4. Lemma. Let R be an Artinian ring and let M, N be finiteley generated Rmodules such that M is not zero and N is not free.

(1) If Tor_{i}^{R}(M, N) = 0 for some i > 0, then

(γ(M ⊗_{R} N_{i}) + 1)b_{i}(N) = (γ(M) − γ(M ⊗_{R} N_{i−1}))b_{i−1}(N)

In particular, there is an inequality b_{i}(N) ≤ γ(M)b_{i−1}(N).

(2) If 𝔪^{2}M = 0 and Tor_{i}^{R}(M, N) = 0 for some i > 0, then 𝔪(M ⊗_{R} N_{i}) = 0 and

b_{i}(N) = (γ(N) − γ(M ⊗_{R} N_{i−1}))b_{i−1}(N)

(3) If 𝔪^{2}M = 0 and Tor_{i}^{R}(M, N) = Tor_{i−1}^{R}(M, N) = 0 for some i > 1, then

b_{i}(N) = γ(M)b_{i−1}(N)

Proof. (1) Consider the short exact sequence

0 ⟶ N_{i} ⟶ R^{b_{i−1}(N)} ⟶ N_{i−1} ⟶ 0

The hypothesis implies that the sequence remains exact when tensored with M:

(1.4.1)  0 ⟶ M ⊗_{R} N_{i} ⟶ M ⊗_{R} R^{b_{i−1}(N)} ⟶ M ⊗_{R} N_{i−1} ⟶ 0

For any j we use (1.3.1) to obtain

λ(M ⊗_{R} N_{j}) = ν(M ⊗_{R} N_{j})(γ(M ⊗_{R} N_{j}) + 1) = ν(M)b_{j}(N)(γ(M ⊗_{R} N_{j}) + 1)

λ(M ⊗_{R} R^{b_{j}(N)}) = b_{j}(N)λ(M) = b_{j}(N)ν(M)(γ(M) + 1)

Using these expressions, a length count in (1.4.1) leads to the desired conclusion.

(2) As above, we have a short exact sequence (1.4.1). The image of N_{i} in R^{b_{i−1}(N)} is contained in 𝔪R^{b_{i−1}(N)}, hence the image of M ⊗_{R} N_{i} in M ⊗_{R} R^{b_{i−1}(N)} is contained in 𝔪(M ⊗_{R} R^{b_{i−1}(N)}), and the latter is annihilated by 𝔪. We have then γ(M ⊗_{R} N_{i}) = 0 and the relation follows from (1).

(3) By (2) we have 𝔪(M ⊗_{R} N_{i−1}) = 𝔪(M ⊗_{R} N_{i}) = 0 and therefore γ(M ⊗_{R} N_{i−1}) = γ(M ⊗_{R} N_{i}) = 0. □

1.5. Lemma. Let R be an Artinian ring and let M, N be finitely generated Rmodules such that M is not zero and N is not free.

(1) If Tor_{i}^{R}(M, N) = 0 for all i ∈ [1, ν(N)], then γ(M) ≥ 1.

(2) If 𝔪^{2}M = 0 and Tor_{i}^{R}(M, N) = 0 for all i ∈ [1, log_{2} b_{1}(N) + 2], then γ(M) is an integer.

Proof. (1) Assume that γ(M) < 1. By Lemma 1.4(1) we have then

b_{i}(N) ≤ γ(M)b_{i−1}(N) < b_{i−1}(N) for all i ∈ [1, ν(N)]

Since b_{0}(N) = ν(N) we conclude that b_{ν(N)}(N) = 0, hence N has finite projective dimension and it is thus free, contradicting the hypothesis.

(2) Using Lemma 1.4(3) we have:

b_{i+1}(N) = (γ(M))^{i}b_{1}(N) for all i ∈ [1, log_{2} b_{1}(N) + 1]

Let u, v be relatively prime positive integers such that γ(M) = uv^{−1}. It follows that v^{i} divides b_{1}(N) for all i ∈ [1, log_{2} b_{1}(N) + 1]. If v ≥ 2, then b_{1}(N) ≥ 2^{i} for all such i, a contradiction. □

Let e denote the minimal number of generators of 𝔪.

1.6. Lemma. Let R be an Artinian ring, and let M, N be non-free finitely generated R-modules. If 𝔪^{2}M = 0 and Tor_{2}^{R}(M, N) = Tor_{1}^{R}(M, N) = 0, then the following hold:

(1) b_{1}(M) = (e − γ(M))b_{0}(M).

(2) 𝔪M_{1} = 𝔪^{2}R^{b_{0}(M)}.

Proof. (1) Lemma 1.4(2) shows that the R-module M ⊗_{R}N_{1} is a finite direct sum of copies of k, hence its first Betti number is eb_{0}(M)b_{1}(N). On the other hand, since Tor_{1}(M, N_{1}) = 0, Lemma 1.2 gives b_{1}(M ⊗_{R} N_{1}) = b_{0}(M)b_{2}(N)+b_{1}(M)b_{1}(N). We also have b_{2}(N) = γ(M)b_{1}(N) by Lemma 1.4(3), hence

eb_{0}(M)b_{1}(N) = b_{0}(M)γ(M)b_{1}(N) + b_{1}(M)b_{1}(N)

(2) A length count in the short exact sequence

0 ⟶ M_{1} ⟶ R^{b_{0}(M)} ⟶ M ⟶ 0

using (1.3.1) gives

λ(M_{1}) = λ(R)b_{0}(M) − λ(M)
= (1 + e + λ(𝔪^{2}))b_{0}(M) − b_{0}(M)(γ(M) + 1)
= (e + λ(𝔪^{2}) − γ(M))b_{0}(M)

We next use (1) to obtain

λ(𝔪M_{1}) = λ(M_{1}) − ν(M_{1}) = λ(M_{1}) − b_{1}(M)
= (e + λ(𝔪^{2}) − γ(M))b_{0}(M) − (e − γ(M))b_{0}(M)
= λ(𝔪^{2})b_{0}(M)

Since 𝔪M_{1} is contained in 𝔪^{2}R^{b_{0}(M)} and both modules have the same length, it follows that they are equal. □

1.7. Lemma. Let R be an Artinian ring and let M, N be non-free finite R-modules with 𝔪^{2}M = 𝔪^{2}N = 0.

If Tor_{2}^{R}(M, N) = Tor_{1}^{R}(M, N) = 0, then γ(M) + γ(N) − γ(M ⊗_{R} N) = e

Proof. Compare the relations

b_{1}(M) = (γ(N) − γ(M ⊗_{R} N))b_{0}(M) and b_{1}(M) = (e − γ(M))b_{0}(M)

given by Lemma 1.4(2), respectively Lemma 1.6(1). □

1.8. Lemma. Let R be an Artinian ring and let M, N be finitely generated non-free R-modules with 𝔪^{2}M = 𝔪^{2}N = 0.

If Tor_{i}^{R}(M, N) = 0 for all i > 0, then

P_{k}^{R}(t) = \frac{1 − γ(M ⊗_{R} N)t}{(1 − γ(M)t)(1 − γ(N)t)}

Proof. By Lemma 1.4 we have

P_{N_{1}}^{R}(t) = \frac{b_{1}(N)}{(1 − γ(M)t)} and P_{M}^{R}(t) = b_{0}(M) + \frac{b_{0}(M)(γ(N) − γ(M ⊗_{R} N))t}{(1 − γ(N)t)}

Lemma 1.2 then yields

P_{M⊗_{R}N_{1}}^{R}(t) = P_{M}^{R}(t)P_{N_{1}}^{R}(t) = b_{0}(M)b_{1}(N)\frac{1 − γ(M ⊗_{R} N)t}{(1 − γ(M)t)(1 − γ(N)t)}

The desired conclusion about P_{k}^{R}(t) is then obtained using the fact that 𝔪(M ⊗_{R} N_{1}) = 0, cf. Lemma 1.4(2). □

2. Rings with 𝔪^{3} = 0

In this section, unless otherwise stated, we assume that (R, 𝔪, k) is an Artinian local ring with 𝔪^{3} = 0. We set e = ν(𝔪) and a = dim_{k} Soc(R).

When 𝔪^{2} = 0, vanishing of homology is not at all mysterious:

2.1. Remark. If 𝔪^{2} = 0 and Tor_{i}^{R}(M, N) = 0 for some i > 1, then M or N is free.

Indeed, assume that M is not free. The module M_{1} is contained in 𝔪R^{b_{0}(M)}, hence 𝔪M_{1} = 0. It is thus a finite sum of copies of k. Since Tor_{i−1}^{R}(M_{1}, N) = 0, we conclude that N has finite projective dimension, hence it is free.

The behavior of Betti numbers of finitely generated R-modules was studied by Lescot [15]. The following results are collected from the proof of [15, 3.3].

2.2. Assume M is not free. For any i ≥ 0 the following hold:

(1) There is an inequality b_{i+1}(M) ≥ eb_{i}(M) − ν(𝔪M_{i}). Equality holds if and only if k is not a direct summand of M_{i+1}.

(2) If i > 1 and k is not a direct summand of M_{i}, then ν(𝔪M_{i}) = ab_{i−1}(M).

2.3. Remark. If 𝔪^{2}M = 0 and k is not a direct summand of M, then Soc(M) = 𝔪M. (This statement holds for all local rings R, not only for those with 𝔪^{3} = 0.)

2.4. Remark. If Tor_{i}^{R}(M, N) = 0 for some i ≥ 3 and M, N are not free, then Soc(R) = 𝔪^{2}. Indeed, it is enough to show that Soc(R) ⊆ 𝔪^{2}, the other inclusion being obvious. Note that Soc(M_{i−1}) = Soc(R^{b_{i−2}(M)}). On the other hand, Remark 1.1 shows that k is not a direct summand in M_{i−1}, hence Soc(M_{i−1}) = 𝔪M_{i−1} ⊆ 𝔪^{2}R^{b_{i−2}(M)} by Remark 2.3, and the conclusion follows.

2.5. Theorem. Let (R, 𝔪) be an Artinian local ring with 𝔪^{3} = 0, and let M,N be non-free R-modules satisfying 𝔪^{2}M = 𝔪^{2}N = 0.

If there exists an integer j > 0 such that

Tor_{j}^{R}(M, N) = Tor_{j+1}^{R}(M, N) = Tor_{j+2}^{R}(M, N) = 0

then the following hold:

(1) γ(M) and γ(N) are positive integers.

(2) \frac{b_{i+1}(M)}{b_{i}(M)} = γ(N) and \frac{b_{i+1}(N)}{b_{i}(N)} = γ(M) for all i with 0 ≤ i ≤ j + 1.

(3) γ(M) = γ(M_{i}) and γ(N) = γ(N_{i}) for all i with 0 ≤ i ≤ j.

(4) γ(M) + γ(N) = e and γ(M)γ(N) = a.

2.6. Remark. It is not true that the vanishing of Tor_{i}^{R}(M, N) for all i ≫ 0 implies Tor_{i}^{R}(M, N) = 0 for all i > 0. Recently, Jorgensen and Şega [14] have constructed a local ring R with 𝔪^{3} = 0 and an R-module M such that for every integer n there exists an R-module N_{n} satisfying Tor_{i}^{R}(M, N_{n}) = 0 if and only if i ≠ 0, n, n + 1.

Proof of Theorem 2.5. (1) We will show that γ(M), γ(N) satisfy the equation γ^{2} − eγ + a = 0. As γ(M), γ(N) are positive rational numbers, this implies that they are integers. The statement is symmetric in M and N, hence it suffices to prove it for γ(M).

By Lemma 1.4 we have:

(2.6.1) b_{j}(N) ≤ γ(M)b_{j−1}(N) and b_{i+1}(N) = γ(M)b_{i}(N) for i = j, j + 1

The hypothesis and Remark 1.1 imply that k is not a direct summand of M_{i} for any i with 0 ≤ i ≤ j + 1 and then 2.2 gives the following relations:

(2.6.2) b_{j+1}(N) = eb_{j}(N) − ab_{j−1}(N) and b_{j+2}(N) ≥ eb_{j+1}(N) − ab_{j}(N)

Combining the second relations of (2.6.1) and (2.6.2) we obtain

γ(M)^{2}b_{j}(N) ≥ eγ(M)b_{j}(N) − ab_{j}(N)

Canceling b_{j}(N) we get γ(M)^{2} ≥ eγ(M) − a.

On the other hand, using the first relation of (2.6.2) and (2.6.1) we have:

γ(M)b_{j}(N) = eb_{j}(N) − ab_{j−1}(N) ≤ eb_{j}(N) − a\frac{b_{j}(N)}{γ(M)}

Canceling b_{j}(N) and multiplying both sides by γ(M) we obtain γ(M)^{2} ≤ eγ(M)−a. We conclude:

(2.6.3)  γ(M)^{2} − eγ(M) + a = 0 and γ(N)^{2} − eγ(N) + a = 0

(2) We show by induction on j + 1 − i that b_{i+1}(N) = γ(M)b_{i}(N) for all i with 0 ≤ i ≤ j + 1. By (2.6.1), the relation holds for i = j, j + 1. Assuming it holds for i = l + 1, with 0 ≤ l < j, we prove that it holds for i = l. By 2.2 we have b_{l+2}(N) = eb_{l+1}(N) − ab_{l}(N), hence, using the inductive hypothesis and (2.6.3) we obtain

ab_{l}(N) = (e − γ(M))b_{l+1}(N) = aγ(M)^{−1}b_{l+1}(N)

and the conclusion follows.

(3) Let i be as in the statement and set l = j + 1 − i. The hypothesis implies Tor_{l}^{R}(M_{i}, N) = Tor_{l+1}^{R}(M_{i}, N) = 0, hence b_{l+1}(N) = γ(M_{i})b_{l}(N) by Lemma 1.4(3). By (1), we also have b_{l+1}(N) = γ(M)b_{l}(N), hence γ(M_{i}) = γ(M).

(4) By Lemma 1.4(2) we have 𝔪(M ⊗_{R} N_{j}) = 0. The hypothesis implies Tor_{1}(M, N_{j}) = Tor_{2}(M, N_{j}) = 0, hence, by Lemma 1.7 we get γ(M) + γ(N_{j}) = e + γ(M ⊗_{R} N_{j}) = e. Using (2) we have then γ(M) + γ(N) = e. Recall from (2.6.3) that γ(M) and γ(N) are roots for the equation γ^{2} − eγ + a = 0. We obtain:

2γ(M)γ(N) = (γ(M) + γ(N))^{2} − γ(M)^{2} − γ(N)^{2}
= e^{2} − (eγ(M) − a) − (eγ(N) − a)
= e^{2} − e(γ(M) + γ(N)) + 2a = 2a

and this finishes the proof of the theorem. □

Remark. Let M, N, j be as in the statement of Theorem 2.5. If l ≥ j + 3 and k is not a direct summand of M_{i} for all i < l (In view of Lemma 1.1 this happens, for example, when Tor_{l}^{R}(M, N) = 0), then

\frac{b_{i+1}(M)}{b_{i}(M)} = γ(N) and \frac{b_{i+1}(N)}{b_{i}(N)} = γ(M) for all i with 0 ≤ i ≤ l − 1

Indeed, by 2.2 we have b_{i+1}(N) = eb_{i}(N)−ab_{i−1}(N) for all i ≤ l−2. We proceed by induction on i, as in the proof of Theorem 2.5. By this theorem, the statement is true for all i ≤ j + 1. Assuming that i ≤ l − 1 and b_{i}(N) = γ(N)b_{i−1}(N), we then have:

b_{i+1}(N) = eb_{i}(N) − a\frac{b_{i}(N)}{γ(M)} = b_{i}(N)(e − \frac{a}{γ(M)}) = b_{i}(N)γ(M)

where the last inequality is due to (2.6.3).

Since R is Artinian, it has a dualizing module ω. In the remaining part of the section we present results that are obtained when one of the modules M, N is equal to ω. An important part in our arguments is played by Matlis duality. We recall below some basic facts.

2.7. Let R be an Artinian ring (not necessarily with 𝔪^{3} = 0) and let ω denote its dualizing module. Matlis duality then gives: ν(ω) = dim_{k} Soc(R), dim_{k} Soc(ω) = 1 and λ(ω) = λ(R).

2.7.1. For every R-module M we set M^{∨} = Hom_{R}(M, ω). For any R-modules M, N and any i there are isomorphisms:

Tor_{i}^{R}(M, N^{∨}) ≅ Ext_{R}^{i}(M, N)^{∨}

2.7.2. We set M^{∗} = Hom_{R}(M, R). Note that M ⊗_{R} ω ≅ (M^{∗})^{∨}. Indeed, this is given by Matlis duality and the isomorphisms below:

(M ⊗_{R} ω)^{∨} ≅ Hom_{R}(M ⊗_{R} ω, ω) ≅ Hom_{R} (M, Hom_{R}(ω, ω)) ≅ M^{∗}

In particular, we obtain that R is Gorenstein if and only if ω is isomorphic to ω^{∗∗}. Indeed, if R is Gorenstein, then the relation holds trivially. Conversely, if ω^{∗∗} ≅ ω, then, taking M = ω^{∗} above, we obtain

ω^{∗} ⊗_{R} ω ≅ (ω^{∗∗})^{∨} ≅ ω^{∨} ≅ R

and it follows that ω is cyclic, hence R is Gorenstein.

We now return to the case of interest, when 𝔪^{3} = 0.

2.8. Assume that 𝔪^{2} ≠ 0. By the above, we have ν(ω) = a. Since 𝔪^{2}ω is not zero and is contained in Soc(ω), we also have ν(𝔪^{2}ω) = 1. Setting N = ω_{1} and r = ν(𝔪^{2}), we can make then the following computations:

(1) λ(ω) = λ(R) = 1 + r + e.

(2) ν(𝔪ω) = λ(ω) − ν(𝔪^{2}ω) − ν(ω) = 1 + r + e − 1 − a = e + r − a.

(3) λ(N) = (a − 1)(1 + r + e). This follows from a length count in the short exact sequence

0 ⟶ N ⟶ R^{a} ⟶ ω ⟶ 0

(4) If k is not a direct summand of N and a = r, then 2.2(1) and (2) give

ν(N) = eν(ω) − ν(𝔪ω) = ea − e = e(a − 1)

γ(N) = \frac{λ(N)}{ν(N)} − 1 = \frac{(a − 1)(1 + a + e)}{(a − 1)e} − 1 = \frac{1 + a}{e}

2.9. Proposition. Let (R, 𝔪) be an non-Gorenstein Artinian ring with 𝔪^{3} = 0 and M a non-free finitely generated R-module with 𝔪^{2}M = 0.

If there exists an integer j ≥ 2 such that

Tor_{j}^{R}(M, ω) = Tor_{j+1}^{R}(M, ω) = Tor_{j+2}^{R}(M, ω) = 0

then e = a + 1, γ(ω_{1}) = 1, γ(M) = a and b_{0}(M) = b_{1}(M) = ⋯ = b_{j+2}(M).

Proof. Set N = ω_{1}. By Remark 2.4 we have Soc(R) = 𝔪^{2}. Also, k is not a direct summand of N by Remark 1.1, hence 2.8(4) gives γ(N) = (1 + a)/e.

By Theorem 2.5(4), γ(N) is a solution of the equation γ^{2} −eγ +a = 0. It follows that (a + 1)^{2} = e^{2}, hence a + 1 = e. In particular, γ(N) = 1, and Theorem 2.5(4) implies γ(M) = a. The conclusion about the Betti numbers follows from Theorem 2.5(2). □

Proposition 2.9 is related to a result of Yoshino [21, 3.1].

There are examples when the situation in Proposition 2.9 holds, M is not free, and j can be chosen to be arbitrarily large. Such an example is provided by Avramov, Gasharov and Peeva [7, 2.2], as described below.

If F is a complex, then F^{∗} denotes the complex Hom_{R}(F, R), with induced differentials.

2.10. Example. Let l be a field, let X = {X_{1}, X_{2}, X_{3}, X_{4}} be a set of indeterminates over l and set A = l[X]_{(X)}. Let I be the ideal of A generated by elements

X_{1}^{2}, X_{1}X_{2} − X_{3}X_{4}, X_{1}X_{2} − X_{4}^{2}, X_{1}X_{3} − X_{2}X_{4}, X_{1}X_{4} − X_{2}^{2},
X_{1}X_{4} − X_{2}X_{3}, X_{1}X_{4} − X_{3}^{2}

and set R = A/I. The ring R is then local and has 𝔪^{3} = 0.

Let x_{i} denote the image of X_{i} in R for i = 1, …, 4 and consider the sequence of homomorphisms of free R-modules:

F = … ⟶^{ψ} R^{2} ⟶^{φ} R^{2} ⟶^{ψ} R^{2} ⟶^{φ} …

where

φ = (\begin{matrix}x_{3} & x_{1} \\ x_{4} & x_{2} \end{matrix}) ψ = (\begin{matrix}x_{2} & −x_{1} \\ −x_{4} & x_{3} \end{matrix})

Set M = Coker φ. By [7, (2.2)(i)] the complex F is exact. As noted by Veliche [20], a computation similar to one in [7, Section 3] shows that the complex F^{∗} is exact. This yields Ext_{R}^{i}(M, R) = 0 for all i > 0, or equivalently, cf. 2.7.1, Tor_{i}^{R}(M, ω) = 0 for all i > 0.

2.11. Theorem. Let (R, 𝔪) be an Artinian local ring with 𝔪^{3} = 0. The following statements are equivalent:

(1) R is Gorenstein.

(2) Tor_{1}^{R}(ω, ω) = 0.

(3) Tor_{2}^{R}(ω, ω) = Tor_{3}^{R}(ω, ω) = 0.

(4) Tor_{j}^{R}(ω, ω) = Tor_{j+1}^{R}(ω, ω) = Tor_{j+2}^{R}(ω, ω) = 0 for some j ≥ 0.

We recall a result of Asashiba and Hoshino [4, 2.1]:

2.12. Let R be a local ring. If M is a faithful R-module and the sequence

0 ⟶ N ⟶^{φ} R^{2} ⟶^{ψ} M ⟶ 0

is exact, then there exist homomorphisms α and β, and an isomorphism θ making the following diagram commute:

\xymatrix{0 \ar[r] & N \ar[r]^{φ} \ar[d]^{α} & R^{2} \ar[d]^{θ} \ar[r]^{ψ} & M \ar[d]_{β} \ar[d] \ar[r] & 0 \\ 0 \ar[r] & M^{∗} \ar[r]^{ψ^{∗}} & (R^{2})^{∗} \ar[r]^{φ^{∗}} & N^{∗}}

Proof of Theorem 2.11. Set r = ν(𝔪^{2}) and N = ω_{1} and b_{i} = b_{i}(ω).

The implications (1) ⇒ (2), (1) ⇒ (3), (1) ⇒ (4) are obvious.

(2) ⇒ (1) Assume R is not Gorenstein. By 2.8 we have λ(ω) = 1 + e + r and ν(𝔪ω) = e + r − a. We obtain then

γ(ω) = \frac{1 + e + r}{a} − 1 = \frac{1 + e + r − a}{a}

By 2.2(1) we have b_{1} ≥ eb_{0} − ν(𝔪ω) = ea − (e + r − a) and by Lemma 1.4 we have b_{1} ≤ γ(ω)b_{0}. We conclude:

ea − (e + r − a) ≤ \frac{1 + e + r − a}{a} · a = 1 + e + r − a

hence ea − 2e + 2a − 2r ≤ 1, or, equivalently, e(a − 2) + 2(a − r) ≤ 1. Since e > 1 and a ≥ r we conclude that a = 2 = r and ν(N) = b_{1} ≥ e. By 2.8 we have thus

λ(R) = λ(N) = e + 3

The hypothesis implies there is a short exact sequence

0 → N ⊗_{R} ω → ω^{2} → ω ⊗_{R} ω → 0 .

with λ(ω^{2}) = 2(e + 3), λ(N ⊗_{R} ω) = 2ν(N) + ε and λ(ω ⊗_{R} ω) = 4 + η, where ε and η are nonnegative integers. A length count in the short exact sequence then gives:

(2.10.1)  2e + 6 = 2ν(N) + ε + 4 + η ≥ 2e + 4 + ε + η

In particular, it follows that η ≤ 2, hence λ(ω ⊗_{R} ω) = 4 + η ≤ 6. Note that λ(ω^{∗}) = λ(ω ⊗_{R} ω), as we have Hom_{R}(ω ⊗_{R} ω, ω) ≅ ω^{∗}.

For the rest of the proof we will look at the commutative diagram in 2.12, with M = ω. Note that α: N → ω^{∗} is injective. Also, the lower sequence in the diagram is right-exact, by the hypothesis Ext_{R}^{1}(ω, R) = 0. In particular, β: ω → N^{∗} is surjective, or equivalently, the dual map β^{∨}: (N^{∗})^{∨} → R is injective.

Since N is contained in ω^{∗}, we have e + 3 = λ(N) ≤ λ(ω^{∗}) = 4 + η and thus e ≤ η + 1 ≤ 3. In particular, we have η ∈ {1, 2}.

If η = 1, then e = 2, hence λ(N) = λ(ω^{∗}) = 5. It follows that α is an isomorphism. The commutative diagram in 2.12 yields that β is an isomorphism, hence ω ≅ N^{∗} ≅ ω^{∗∗}. We apply then 2.7.2 to conclude that R is Gorenstein, a contradiction.

If η = 2, then the inequality (2.10.1) yields ε = 0 and ν(N) = 3, hence e ≤ 3 and λ(N ⊗_{R} ω) = 6. On the other hand, N ⊗_{R} ω is isomorphic to (N^{∗})^{∨} by 2.7.1, and the latter is contained in R, as seen above. Since λ(R) = e + 3 ≤ 6, it follows that N ⊗_{R} ω ≅ R, hence ω is cyclic, contradicting our assumption that R is not Gorenstein.

(3) ⇒ (1) Assume R is not Gorenstein. By Remark 2.4, we have Soc(R) = 𝔪^{2} and then 2.8 gives λ(ω) = λ(R) = 1 + a + e and ν(𝔪ω) = e.

Since Tor_{1}^{R}(N, ω) = Tor_{2}^{R}(N, ω) = 0, we use Theorem 1.4(3) to obtain:

b_{2} = γ(N)b_{1} = \frac{λ(𝔪N)}{ν(N)}b_{1} = \frac{ν(𝔪N)}{b_{1}}b_{1} = ν(𝔪N)

On the other hand, 2.2(1) gives b_{2} = eb_{1} − ν(𝔪N) = eb_{1} − b_{2}, hence we have:

2b_{2} = eb_{1} and ν(𝔪N) = \frac{eb_{1}}{2}

and we conclude γ(N) = e/2. By Lemma 1.4(1) we then get

b_{1} ≤ γ(N)b_{0} = \frac{e}{2}a

By 2.2(1) we also have b_{1} ≥ eb_{0} − ν(𝔪ω) = e(a − 1). We obtain thus

e(a − 1) ≤ \frac{e}{2}a

and we conclude a ≤ 2. As we assumed R not to be Gorenstein, we have a = 2. Recall from 2.8 that γ(N) = (a+1)/e. Comparing this with the relation γ(N) = e/2 obtained above, we obtain e^{2} = 6, a contradiction.

(4) ⇒ (1) By the previous implications we may assume j ≥ 2. Assume that R is not Gorenstein and set N = ω_{1}. Applying Proposition 2.9 with M = N we obtain γ(N) = 1. We then use Proposition 2.5(4) with M = N and we conclude a = γ(N)^{2} = 1, hence R is Gorenstein, a contradiction. □

3. Rings of large embedding dimension

For any finitely generated module N we set

c(N) = max{4, log_{2}(b_{1}(N)) + 2}

(where log_{2} 0 = −∞). The Loewy length of the ring R, denoted ℓℓ(R), is the smallest integer n for which 𝔪^{n} = 0.

In this section we prove the following:

3.1. Theorem. Let (R, 𝔪) be an Artinian local ring satisfying 2ν(𝔪) ≥ λ(R) − ℓℓ(R) + 4.

(1) If 𝔪^{2}M = 0 and Tor_{i}^{R}(M, N) = 0 for all i ∈ [1, c(N)], then either M or N is free.

(2) If 𝔪^{3} = 0 and Tor_{i}^{R}(M, N) = 0 for three consecutive values of i ≥ 2, then either M or N is free.

3.2. Remark. The ring in Example 2.10 has ν(𝔪) = 4, λ(R) = 8 and ℓℓ(R) = 3; in particular, the inequality in the statement of Theorem 3.1 is (minimally) not satisfied. This shows that the inequality can not be improved.

Proof of Theorem 3.1. Set e = ν(𝔪) and h = ℓℓ(R) − 1. Assume that M and N are not free.

(1) Let i be a positive integer. In the proof of [10, 2.2] Gasharov and Peeva show that the following inequality holds for any local Artinian ring R and any finitely generated R-module N.

b_{i+1}(N) ≥ eb_{i}(N) − (λ(𝔪^{2}) + 2 − h)b_{i−1}(N)

Setting a = λ(𝔪^{2}) + 3 − h, we conclude that for all positive integers i there is a strict inequality

b_{i+1}(N) > eb_{i}(N) − ab_{i−1}(N)

We let then i be any integer such that 2 ≤ i ≤ c(N) − 2. Lemma 1.4(3) gives that b_{j}(N) = γ(M)^{j−1}b_{1}(N), for j = i, i + 1 and we conclude

(3.2.1)  γ(M)^{2} − eγ(M) + a > 0

The roots of the equation γ^{2} − eγ + a = 0 are

γ_{1,2} = \frac{e ± \sqrt{e^{2} − 4a}}{2}

The hypothesis gives a ≤ e − 1, hence e^{2} − 4a ≥ (e − 2)^{2}. Both γ_{1} and γ_{2} are then real. Assume γ_{1} ≤ γ_{2}. We obtain then γ_{1} ≤ 1 and γ_{2} ≥ e − 1.

The strict inequality in (3.2.1) shows that γ(M) is outside the interval [γ_{1}, γ_{2}]. If γ(M) < γ_{1}, then γ(M) < 1 and this contradicts Lemma 1.5. We conclude that γ(M) > γ_{2}, hence γ(M) > e−1. We recall that γ(M) is an integer, cf. Lemma 1.5, and we conclude γ(M) ≥ e. On the other hand, Lemma 1.6(1) implies e > γ(M), a contradiction.

(2) We replace M with M_{1}, if necessary, so that we may assume 𝔪^{2}M = 0. Proceed then as in (1), using Theorem 2.5. □

4. The Auslander-Reiten Conjecture

In this section we prove the conjecture of Auslander and Reiten (stated in the introduction) when the module is annihilated by 𝔪^{2}. More precise statements are obtained when 𝔪^{3} = 0. We state our main results below. The proofs will follow later in the section.

4.1. Theorem. Let (R, 𝔪, k) be an Artinian local ring with 𝔪^{3} = 0 and M a finitely generated R-module.

(1) If Ext_{R}^{i}(M, M ⊕ R) = 0 for four consecutive values of i with i ≥ 2, then M is free.

(2) If R is Gorenstein and Ext_{R}^{i}(M, M) = 0 for some i > 0, then M is free

The second part was inspired by the following statement of Hoshino [12, 3.4]: If R is a finite dimensional self-injective local algebra (possibly non-commutative) over a field and the cube of its radical is zero, then any finitely generated R-module M with Ext_{R}^{1}(M, M) = 0 is free.

4.2. Theorem. Let (R, 𝔪, k) be an Artinian local ring and M a finitely generated R-module with 𝔪^{2}M = 0.

If Ext_{R}^{i}(M, M ⊕ R) = 0 for all i with 0 < i ≤ max{3, ν(M), ν(𝔪M)}, then M is free.

4.3. Remark. Let (R, 𝔪, k) be an Artinian local ring with 𝔪^{2} ≠ 0 and let M be a non-zero finitely-generated R-module with 𝔪^{2}M = 0. If Ext_{R}^{i}(M, M) = 0 for some i > 0, then γ(M^{∨}) = γ(M)^{−1}.

Indeed, 2.7.1 gives Tor_{i}^{R}(M, M^{∨}) = 0. Since 𝔪^{2} is not zero, the modules M^{∨} and M are not free, hence k is not a direct summand in either of them, cf. Remark 1.1. Set r(M) = rank_{k} Soc(M). As noted in Remark 2.3, r(M) = λ(𝔪M) and r(M^{∨}) = λ(𝔪M^{∨}). Using Matlis duality we obtain

γ(M^{∨}) = \frac{λ(𝔪M^{∨})}{ν(M^{∨})} = \frac{r(M^{∨})}{r(M)} = \frac{ν(M)}{λ(𝔪M)} = \frac{1}{γ(M)}

4.4. Proposition. Let (R, 𝔪, k) be an Artinian local ring. Let M be a non-zero finitely generated R-module such that 𝔪^{2}M = 0. If any of the following conditions holds:

(1) Ext_{R}^{i}(M, M) = 0 for all i with 0 < i ≤ max{3, ν(M), ν(𝔪M)}

(2) 𝔪^{3} = 0 and Ext_{R}^{i}(M, M) = 0 for three consecutive values of i > 0, then 𝔪^{2} = 0, and M is either free or injective.

Proof. By 2.7.1 we have Tor_{i}(M, M^{∨}) = 0 for all i as in the statement. If 𝔪^{2} = 0, then Remark 2.1 implies M or M^{∨} is free, hence M is free or injective. From now in we assume 𝔪^{2} ≠ 0. This implies, in particular, that neither M nor

M^{∨} is free. By Remark 1.1, k is not a direct summand in M or M^{∨}. Matlis duality and Remark 2.3 then yield ν(𝔪M) = ν(M^{∨}), while Remark 4.3 gives γ(M^{∨}) = γ(M)^{−1}. Lemma 1.5(1), respectively Theorem 2.5(1), give then γ(M) ≥ 1 and γ(M^{∨}) ≥ 1, and we conclude γ(M) = γ(M^{∨}) = 1.

We use the notation of the previous sections: e = ν(𝔪) and a = dim_{k} Soc(R). Assume that M satisfies (1). By Lemma 1.7 we have 2 − γ(M ⊗_{R} M^{∨}) = e,

hence e ≤ 2.

By Scheja [19, Satz 9] the ring R is then either a complete intersection, or a Golod ring. If it is a complete intersection, then Ext_{R}^{2}(M, M) = 0 implies M is free, by Auslander, Ding, and Solberg [1, (1.8)], a contradiction. If it is Golod, but not a hypersurface, then e = 2 and γ(M ⊗_{R} M^{∨}) = 0. Lemma 1.4(1) yields then b_{i}(M) = ν(M) and b_{i}(M^{∨}) = ν(M^{∨}) for i = 1, 2, 3. Since R is Golod, P_{k}^{R}(t) = (1 + t)(1 − t − lt^{2})^{−1} with l ≥ 1, hence b_{3}(k) = 2 + 3l ≥ 5. Since M ⊗_{R} M^{∨} is a sum of copies of k, we have then

β_{3}(M ⊗_{R} M^{∨}) = b_{3}(k)ν(M)ν(M^{∨}) ≥ 5ν(M)ν(M^{∨})

On the other hand, Lemma 1.2 gives β_{3}(M ⊗_{R} M^{∨}) = 4ν(M)ν(M^{∨}), and this leads to a contradiction.

Assume that M satisfies (2). Using 2.5(4) we obtain a = 1 and e = γ(M) + γ(M^{∨}) = 2. Thus, R is Gorenstein and it follows that it is a complete intersection. Let j be an even integer among the three consecutive integers in the hypothesis. The hypothesis that Ext_{R}^{j}(M, M) = 0 implies M is free, cf. Avramov and Buchweitz [5, 4.2], a contradiction. □

Proof of Theorem 4.2. Assume that M is not free. Proposition 4.4(1) then shows 𝔪^{2} = 0 and M is injective. By 2.7.1, Ext_{R}^{i}(M, R) = 0 implies Tor_{i}^{R}(M, ω) = 0 and Remark 2.1 shows that ω is free, hence R is Gorenstein. In this case, any finiteley generated injective R-module, and in particular M, is also free, a contradiction. □

4.5. Consider a short exact sequence of R-modules

0 ⟶ N → R^{n} → M → 0

Assume that Ext_{R}^{i+1}(M, R) = 0 for some i > 0. A usual computation with the long exact sequences of Ext shows that Ext_{R}^{i}(M, M) = 0 implies Ext_{R}^{i}(N, N) = 0. If, in addition, Ext_{R}^{i}(M, R) = 0 then Ext_{R}^{i}(M, M) = 0 if and only if only if Ext_{R}^{i}(N, N) = 0.

Proof of Theorem 4.1(1). The hypothesis and 4.5 imply that Ext_{R}^{i}(M_{1}, M_{1}) vanishes for three consecutive values of i > 0. Since 𝔪^{2}M_{1} = 0, we proceed as above, using Proposition 4.4(2). □

In order to prove part (2) of Theorem 4.1 we use the fact that for Gorenstein Artinian rings one can define negative Betti numbers and negative syzygies.

4.6. Let R be a Gorenstein Artinian local ring, F a free resolution of M and G a free resolution of M^{∗}. Note that the complex G^{∗} is acyclic, with H_{0}(G^{∗}) = M^{∗∗} ≅ M. Gluing together the complexes F and G^{∗}, we obtain an exact complex P, which is called a complete resolution of M.

Furthermore, if M is a first syzygy in a minimal free resolution of some other module, and the resolutions F and G are chosen to be minimal, then the complex P is minimal. In this case, we have rank(P_{i}) = b_{i}(M) for all i ≥ 0 and M_{i} = Coker ∂_{i}^{P}. In general, the Betti numbers and syzygies of M are defined by setting b_{i}(M) = rank(P_{i}) and M_{i} = Coker ∂_{i}^{P} for all integers i. Note that for any j ≥ i the module M_{j} is a (j − i)’th syzygy of M_{i}; in our notation: M_{j} = (M_{i})_{j−i}.

Proof of Theorem 4.1(2). Assume that M is not free. Since R is Gorenstein, the hypothesis and 4.5 imply Ext_{R}^{i}(M_{j}, M_{j}) = 0 for all j. Replacing M by M_{1}, we may assume 𝔪^{2}M = 0. We can now use the notation of 4.6. The assumption that M is not free implies that both M and M^{∗} have infinite projective dimension, hence M_{j} is not free for any j.

For all j we have thus 𝔪^{2}M_{j} = 0. Since Ext_{R}^{i}(M_{j}, M_{j}) = 0, 2.7.1 then gives Tor_{i}^{R}(M_{j}, M_{j}^{∗}) = 0. (Note that ω ≅ R, hence M_{j}^{∗} ≅ M_{j}^{∨}). In particular, k is not a direct summand in any of the M_{j}’s.

We set e = ν(𝔪) and b_{j} = b_{j}(M) for all j. Since R is Gorenstein, Soc(R) is 1-dimensional. We use Lescot’s results recalled in 2.2 to get:

(4.1.1)  b_{j} + b_{j−1} = eb_{j−2} and ν(𝔪M_{j}) = b_{j−1} for all j .

Recall that γ(M) = λ(𝔪M)/ν(M). Using (4.1.1) and Remark 4.3 we have:

(4.1.2)  γ(M_{j}) = \frac{b_{j−1}}{b_{j}} and γ(M_{j}^{∗}) = \frac{b_{j}}{b_{j−1}} for all j .

Since Tor_{i}^{R}(M_{j}, M_{j}^{∗}) = 0 for all j, Lemma 1.4 yields:

(4.1.3)  b_{i+j} = (γ(M_{j}^{∗}) − γ(M_{j+i−1} ⊗_{R} M_{j}^{∗}))b_{j+i−1}

For all j we obtain:

b_{j+i} ≤ γ(M_{j}^{∗})b_{j+i−1} = \frac{b_{j}}{b_{j−1}} b_{j+i−1}

where the inequality comes from (4.1.3) and the equality from (4.1.2). Equivalently:

\frac{b_{j+i}}{b_{j+i−1}} ≤ \frac{b_{j}}{b_{j−1}} for all j .

Each j = 0, 1, ⋯, i − 1 yields thus a non-increasing sequence (b_{j+ni}/b_{j+ni−1})_{n}. Let L_{j} denote the limit of the j’th sequence.

If L_{j} < 1 for some j = 0, 1, …, i − 1 it follows that there exists an eventually strictly decreasing subsequence of {b_{n}}. This implies that b_{n} = 0 for some n ≫ 0, a contradiction.

Thus, L_{j} ≥ 1 for all j. This implies b_{n} ≤ b_{n+1} for all n. If b_{n_{0}} < b_{n_{0}+1} for some n_{0}, then we obtain

1 < \frac{b_{n_{0}+1}}{b_{n_{0}}} ≤ \frac{b_{n_{0}−i+1}}{b_{n_{0}−i}} ≤ ⋯

hence

b_{n_{0}+1} > b_{n_{0}} ≥ b_{n_{0}−i+1} > b_{n_{0}−i} ≥ b_{n_{0}−2i+1} > b_{n_{0}−2i} ≥ ⋯

It follows that b_{n} = 0 for some n ≪ 0, a contradiction. In conclusion, b_{n} = b_{n+1} for all n. We use then (4.1.1) to obtain e = 2. It follows that R is a complete intersection. Since the Betti numbers of any M_{j} are constant, by a result of Eisenbud [9, 4.1] then M_{j+n} ≅ M_{j+n+2} for all n > 0. We conclude M_{j} ≅ M_{j+2} for all j.

If i is even, then M has finite projective dimension by [5, 4.2], hence it is free, a contradiction.

If i is odd, then M ≅ M_{−i+1}, hence

Ext_{R}^{1}(M, M) ≅ Ext_{R}^{1}(M, M_{−i+1}) ≅ Ext_{R}^{i}(M, M) = 0

Since the Betti numbers of M are constant, (4.1.2) gives γ(M^{∗}) = 1. Taking i = 1 and j = 0 in (4.1.3) we get b_{1} = (1 − γ(M ⊗_{R} M^{∗}))b_{0}, and it follows γ(M ⊗_{R} M^{∗}) = 0. This means that 𝔪(M ⊗_{R} M^{∗}) = 0. However, M ⊗_{R} M^{∗} is the Matlis dual of Hom_{R}(M, M) and the later is annihilated by 𝔪 only when 𝔪M = 0. In view of the hypothesis, this implies that M is free, which provides the desired contradiction. □

5. Vanishing of Tor and Loewy length

In this section, unless otherwise stated, we assume that (R, 𝔪, k) is an Artinian local ring. We recall that the Loewy length of R, denoted ℓℓ(R), is the smallest integer n with 𝔪^{n} = 0.

We propose the following conjecture:

5.1. Conjecture. Assume that M, N are nonzero modules with 𝔪^{2}M = 𝔪^{2}N = 0. If Tor_{i}^{R}(M, N) = 0 for all i > 0, then 𝔪^{3} = 0.

5.2. One can ask a more general question: Assume that M, N are nonzero modules with 𝔪^{p}M = 𝔪^{q}N = 0. If Tor_{i}^{R}(M, N) = 0 for all i > 0, does it follow that 𝔪^{p+q−1} = 0? This is trivially true when p = 1 or q = 1, or when one of p, q is greater than ℓℓ(R). The conjecture takes up the case p = 2 = q.

5.3. In view of Lemma 1.8, the conjecture holds whenever 𝔪^{3} = 0 or P_{k}^{R}(t) ≠ (1 − at)(1 − bt)^{−1}(1 − ct)^{−1} with rational numbers a ≥ 0 and b, c > 0. The class of such rings include: complete intersection rings of codimension different from 2, generalized Golod rings, Koszul rings, rings with irrational Poincaré series and many others. More evidence for the conjecture can also be gathered from the preceding two sections.

Theorem 5.4 below establishes the conjecture in yet another important case. We say that the local ring R is standard graded if it has a decomposition R =

R_{0}⊕R_{1}⊕⋯⊕R_{h} such that R_{i}R_{j} ⊆ R_{i+j} for all i, j ∈ [0, h], R_{0} = k and R = R_{0}[R_{1}] (it is thus generated in degree one).

5.4. Theorem. Let (R, 𝔪) be a standard graded local ring, and let M, N be nonzero finitely generated R-modules satisfying 𝔪^{2}M = 𝔪^{2}N = 0.

If Tor_{i}^{R}(M, N) = 0 for all i > 0, then 𝔪^{3} = 0.

Let L, U be any two R-modules. We set \overline{L} = L/𝔪L and we consider the short exact sequence

0 ⟶ 𝔪L ⟶^{μ_{L}} L ⟶ \overline{L} ⟶ 0

and the induced long exact sequence:

⋯ → Tor_{i+1}^{R}(U, \overline{L}) ⟶^{Δ_{i}(U,L)} Tor_{i}^{R}(U, 𝔪L) ⟶^{Tor_{i}^{R}(U,μ_{L})} Tor_{i}^{R}(U, L) → …

where Δ_{i}(U, L) denote the connecting homomorphisms.

5.5. Lemma. Let (R, 𝔪, k) be a Noetherian local ring and let M be a non-zero finitely generated R-module with 𝔪^{2}M = 0. Let j ≥ 2 be an integer.

If Tor_{i}^{R}(M, N) = 0 for all i ∈ [1, j], then Tor_{i}^{R}(k, μ_{N}) = 0 for all i ∈ [0, j − 1].

Proof. (1) We will show, equivalently, that Δ_{i}(k, N) is surjective for all i ∈ [0, j−1]. Of course, this is true when i = 0. We prove the claim by induction. Assume that it is true for i = n and we prove it for i = n + 1, with 0 ≤ n < j − 1.

In the diagram below the horizontal lines are long exact sequences of the type considered above.

\xymatrix{Tor_{n+2}^{R}(M, \overline{N}) \ar[r] \ar[d]^{≅}_{Δ_{n+1}(M,N)} & Tor_{n+2}^{R}(\overline{M}, \overline{N}) \ar[r] \ar[d]_{Δ_{n+1}(\overline{M},N)} & Tor_{n+1}^{R}(𝔪M, \overline{N}) \ar[r] \ar@{->>}[d]_{Δ_{n}(𝔪M,N)} & Tor_{n+1}^{R}(M, \overline{N}) \ar@{^{(}->}[d]_{Δ_{n}(M,N)} \\ Tor_{n+1}^{R}(M, 𝔪N) \ar[r] & Tor_{n+1}^{R}(\overline{M}, 𝔪N) \ar[r] & Tor_{n}^{R}(𝔪M, 𝔪N) \ar[r] & Tor_{n}^{R}(M, 𝔪N)}

By [8, V, 4.1] the exterior squares commutes and the interior one anticommutes. The map Δ_{n+1}(M, N) is bijective because Tor_{n+2}^{R}(M, N) = Tor_{n+1}^{R}(M, N) = 0. Also, the map Δ_{n}(𝔪M, N) is surjective by the induction hypothesis, using the fact that 𝔪M is a finite direct sum of copies of k, and Δ_{n}(M, N) is injective because Tor_{n+1}^{R}(M, N) = 0. By the “Five Lemma” we conclude that the map Δ_{n+1}(\overline{M}, N) is surjective. Since \overline{M} is a finite direct sum of copies of k, we obtain that Δ_{n+1}(k, N) is surjective, and this finishes the induction argument. □

Proof of Theorem 5.4. We may assume that M and N are not free. By Lemma 5.5 we have Tor_{i}^{R}(k, μ_{N_{1}}) = 0 for all i ≥ 0. Choose such an i and set F = R^{b_{0}(N)}. By Lemma 1.6(2) we have 𝔪N_{1} = 𝔪^{2}F. Consider the following commutative diagram, in which the vertical maps are induced by the inclusion N_{1} ↪ 𝔪F:

\xymatrix{Tor_{i}^{R}(k, 𝔪N_{1}) \ar[r]^{Tor_{i}^{R}(k,μ_{N_{1}} )=0} \ar[d]^{≅} & Tor_{i}^{R}(k, N_{1}) \ar[d] \\ Tor_{i}^{R}(k, 𝔪^{2}F) \ar[r]^{Tor_{i}^{R}(k,μ_{𝔪F})} & Tor_{i}^{R}(k, 𝔪F)}

We conclude that Tor_{i}^{R}(k, μ_{𝔪F}) = 0, hence Tor_{i}^{R}(k, μ_{𝔪}) = 0 for all i. Equivalently, the map Ext_{R}^{i}(k, k) → Ext_{R}^{i}(R/𝔪^{2}, k) induced by the projection R/𝔪^{2} → k is zero for all i, hence [18, Corollary 1] implies that the algebra Ext_{R}^{∗}(k, k) (with Yoneda product) is generated by its elements of degree 1. This means that the k-algebra R is Koszul, hence P_{k}^{R}(t) = Hilb_{R}(−t)^{−1}, cf. [16, Theorem 1.2]. Comparing with the relation of Remark 1.8 we conclude that Hilb_{R}(t) is a polynomial of degree at most 2, hence 𝔪^{3} = 0. □

6. Rings of type at most 2

In this section we show that the conjecture of Tachikawa stated in the introduction holds for Cohen-Macaulay rings of type at most 2.

6.1. Theorem. Let R be a Cohen-Macaulay local ring with a canonical module ω and such that type(R) ≤ 2.

(1) If Tor_{2}^{R}(ω, ω) = 0, then R is Gorenstein.

(2) If Ext_{R}^{i}(ω, R) = 0 for i = 1, 2, then R is Gorenstein.

The proof will be given at the end of the section, after discussing some preliminaries. We recall a well-known fact:

6.2. Let R be a commutative local ring and consider a short exact sequence of R-modules:

0 ⟶ N ⟶^{φ} R^{n} ⟶^{ψ} M ⟶ 0

The map φ induces a natural map Λ_{R}^{n}(φ): Λ_{R}^{n}(N) → Λ_{R}^{n}(R^{n}). If a ∈ R is in the image of this map, via the identification of R with Λ_{R}^{n}(R^{n}), then aM = 0.

In particular, if M is faithful, then Λ_{R}^{n}(φ) = 0.

6.3. Proposition. Let (R, 𝔪, k) be a Noetherian local ring and M a finitely generated R-module. Consider a short exact sequence of R-modules

0 ⟶ N ⟶^{φ} R^{2} ⟶^{ψ} M ⟶ 0

If Tor_{2}^{R}(M, M) = 0 and M is faithful, then ν(N) ≤ 1.

Proof. For any module L we define a map ι^{L}: Λ_{R}^{2}(L) → L ⊗_{R} L given by

ι^{L}(x ∧ y) = x ⊗ y − y ⊗ x

The hypothesis implies Tor_{1}^{R}(M, N) = 0. It follows that the induced map

φ ⊗_{R} N: N ⊗_{R} N → R^{2} ⊗_{R} N

is injective. The map φ ⊗_{R} φ: N ⊗_{R} N → R^{2} ⊗_{R} R^{2} is the composition

N ⊗_{R} N ⟶^{φ⊗_{R}N} R^{2} ⊗_{R} N ⟶^{R^{2}⊗_{R}φ} R^{2} ⊗_{R} R^{2}

Both maps are injective, hence so is φ ⊗_{R} φ.

Recall from 6.2 that Λ_{R}^{2}(φ) = 0. The commutative diagram

\xymatrix{Λ_{R}^{2}(N) \ar[r]^{ι^{N}} \ar[d]_{Λ_{R}^{2}(φ)} & N ⊗_{R} N \ar@{^{(}->}[d]^{φ⊗_{R}φ} \\ Λ_{R}^{2}(R^{2}) \ar[r]^{ι^{R}} & R^{2} ⊗_{R} R^{2}}

then yields ι^{N} = 0. In particular, the map ι^{N} ⊗_{R} k is zero. Note that this map can be identified with ι^{N⊗_{R}k}. As N ⊗_{R} k is a finite direct sum of copies of k, the map ι^{N⊗_{R}k} is clearly injective. It follows that Λ_{R}^{2}(N ⊗_{R} k) = 0, hence N ⊗_{R} k ≅ k. Nakayama’s Lemma then shows that N is cyclic. □

Proof of Theorem 6.1. (1) Assume that R is not Gorenstein, hence type(R) = 2. Set N = ω_{1}. We denote e(L) the multiplicity of an R-module L. Since e(ω) = e(R), we use the fact that multiplicity is additive on short exact sequences of maximal Cohen-Macaulay modules to obtain e(N) = e(R). By Proposition 6.3, N is cyclic, hence there is a surjection R → N. If K is the kernel of this map, then e(K) = 0, hence N is free, a contradiction.

(2) By [6, B.4] we have Tor_{i}^{R}(ω, ω) = 0 for i = 1, 2, so we can apply (1). □

Acknowledgements

The authors want to thank Luchezar Avramov, Ragnar Buchweitz, Graham Leuschke, and Roger Wiegand for their comments and suggestions.

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Mathematics Department, University of Kansas, Lawrence, KS 66045

E-mail address: huneke@math.ukans.edu

Mathematical Sciences Research Institute, 1000 Centennial Drive, Berkeley, CA 94720

Current address: Department of Mathematics, Michigan State University, East Lansing, MI 48824

E-mail address: lsega@math.msu.edu

Mathematics Department, University of Kansas, Lawrence, KS 66045 Current address: Department of Mathematics, University of South Carolina, Columbia, SC

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E-mail address: vraciu@math.sc.edu