INDEPENDENCE OF THE TOTAL REFLEXIVITY CONDITIONS FOR MODULES

DAVID A. JORGENSEN AND LIANA M. ŞEGA

Abstract. We show that the conditions defining total reflexivity for modules are independent. In particular, we construct a commutative Noetherian local ring R and a reflexive R-module M such that Ext_{R}^{i}(M, R) = 0 for all i > 0, but Ext_{R}^{i}(M^{∗}, R) ≠ 0 for all i > 0.

introduction

Let R be a commutative Noetherian ring. For any R-module M we set M^{∗} = Hom_{R}(M, R). An R-module M is said to be reflexive if it is finite and the canonical map M → M^{∗∗} is bijective. A finite R-module M is said to be totally reflexive if it satisfies the following conditions:

(i) M is reflexive (ii) Ext_{R}^{i}(M, R) = 0 for all i > 0 (iii) Ext_{R}^{i}(M^{∗}, R) = 0 for all i > 0. This notion is due to Auslander and Bridger [1]: the totally reflexive modules

are precisely the modules of G-dimension zero. The G-dimension of a module is one of the best studied non-classical homological dimensions, and is defined in terms of the length of a resolution of the module by totally reflexive modules.

Given any homological dimension, a serious concern is whether its defining conditions can be verified effectively. For example, the projective dimension of a finite R-module M is zero if and only if Ext_{R}^{1}(M, N) = 0 for all finite R-modules N. Furthermore, when R is local with maximal ideal 𝔪, one only needs to check vanishing for N = R/𝔪. In the same spirit, it is natural to ask whether the set of conditions defining total reflexivity is overdetermined (cf. [4, §2]) and in particular, whether total reflexivity for a module can be established by verifying vanishing of only finitely many Ext modules.

When R is a local Gorenstein ring, (ii) implies the other two conditions above, and it is equivalent to M being maximal Cohen-Macaulay. Recently, Yoshino [9] studied other situations when (ii) alone implies total reflexivity, and raised the question whether this is always the case.

In the present paper we give an example of a local Artinian ring R which admits modules whose total reflexivity conditions are independent, in that (ii) implies neither (i) nor (iii); (i) and (ii) do not imply (iii), equivalently, (i) and (iii) do not imply (ii). More precisely, we prove the following result as Theorem 1.7: Theorem. There exists a local Artinian ring R, and a family {M_{s}}_{s⩾1} of reflexive R-modules such that

(1) Ext_{R}^{i}(M_{s}, R) = 0 if and only if 1 ≤ i ≤ s − 1;

Date: October 22, 2004.

(2) Ext_{R}^{i}(M_{s}^{∗}, R) = 0 for all i > 0. Moreover, there exists a non-reflexive R-module L such that

(1′) Ext_{R}^{i}(L, R) = 0 for all i > 0; (2′) Ext_{R}^{i}(L^{∗}, R) ≠ 0 for all i > 0. By taking N_{s} = M_{s}^{∗} for all s ≥ 1, we get a statement dual to that of the

first part above: there exits a family {N_{s}}_{s⩾1} of reflexive R-modules such that (1) Ext_{R}^{i}(N_{s}^{∗}, R) = 0 if and only if 1 ≤ i ≤ s − 1; (2) Ext_{R}^{i}(N_{s}, R) = 0 for all i > 0.

This theorem shows that in order to check whether or not a module M is totally reflexive — even for a local Artinian ring — one needs to check vanishing of Ext_{R}^{i}(M, R) and Ext_{R}^{i}(M^{∗}, R) for infinitely many values of i. In Section 2, however, we point out that when R is Artinian and standard graded, in the sense that R = ⨁_{i=0}^{∞} R_{i} with R_{0} = k, a field, and R = R_{0}[R_{1}], one may skip checking finitely many values of i of the same parity.

In our example, R is a standard graded Koszul algebra and has Hilbert series ∑_{i≥0} rank_{k}(R_{i})t^{i} = 1 + 4t + 3t^{2}. The ring R is thus local, and its maximal ideal 𝔪 satisfies 𝔪^{3} = 0. Note that our example is minimal in the following sense: if 𝔪^{2} = 0, then any finite R-module M which satisfies Ext_{R}^{i}(M, R) = 0 for some i > 1 is totally reflexive, hence (ii) alone implies total reflexivity. (See 1.1 below.)

Our construction involves a minimal acyclic complex C of finite free R-modules such that the sequence {rank_{R}(C_{i})}_{i⩾0} is strictly increasing and has exponential growth, while the sequence {rank_{R}(C_{−i})}_{i⩾0} is constant. In the last section we raise several related questions.

1. Independence

Let R be a Noetherian commutative ring and M a finite R-module. Suppose that ϕ: G → F is a homomorphism of finite free R-modules with M = Coker ϕ. The R-module Tr(M) := Coker ϕ^{∗} is called the transpose of M, and it is unique up to projective equivalence; it is thus well-defined in the stable category of R. The R-modules Tr(Tr(M)) and M are isomorphic up to projective summands and there is an exact sequence

0 → Ext_{R}^{1}(Tr(M), R) → M → M^{∗∗} → Ext_{R}^{2}(Tr(M), R) → 0

where the map in the middle is the natural map. In particular, M is reflexive if and only if Ext_{R}^{i}(Tr(M), R) = 0 for i = 1, 2. Also note that M^{∗} is a second syzygy of Tr(M), hence Ext_{R}^{i}(M^{∗}, R) ≅ Ext_{R}^{i+2}(Tr(M), R) for all i > 0. Thus the definition of M being totally reflexive can be recast as follows:

Ext_{R}^{i}(M, R) = 0 = Ext_{R}^{i}(Tr(M), R) for all i > 0 .

It is convenient to have a uniform notation for the conditions above: let i ∈ ℤ, i ≠ 0. We say that M satisfies condition (TR_{i}) provided

(TR_{i}) : \begin{cases}Ext_{R}^{i}(M, R) = 0 & if i ≥ 1 \\ Ext_{R}^{−i}(Tr(M), R) = 0 & if i ≤ −1. \end{cases}

Thus M is totally reflexive if and only if M satisfies (TR_{i}) for all i ≠ 0.

Note that we did not define a condition (TR_{i}) for i = 0. When referring to these conditions we will assume tacitly that i ≠ 0.

1.1. Remarks. (1) Assume that R is local Gorenstein. The module M is then totally reflexive if and only if (TR_{i}) is satisfied for all i with 0 < i ≤ dim R.

(2) Assume that R is local, with maximal ideal 𝔪 and residue field k. If 𝔪^{2} = 0, then the first syzygy N in a minimal free resolution of M has 𝔪N = 0, hence it is a finite dimensional k-vector space. If Ext_{R}^{i}(M, R) = 0 for some i > 1 then either M is free or Ext_{R}^{i−1}(k, R) = 0. Thus, when 𝔪^{2} = 0 we obtain: M is totally reflexive whenever (TR_{i}) is satisfied for a single value of i with i ≠ ±1.

(3) Assume that R is a commutative Noetherian ring. Yoshino proved in [9] that if the full subcategory of R-modules N with Ext_{R}^{i}(N, R) = 0 for all i > 0 is of finite type, then every module M in this subcategory is totally reflexive.

The remarks above lead to the question: How many conditions (TR_{i}) does one need to check for total reflexivity, without placing extra assumptions on the ring? Is it possible that only finitely many suffice? Is it enough to check (TR_{i}) for all i > 0, or, more generally, for all i > s for some integer s?

Theorem 1.7 (stated also in the introduction) provides negative answers to these questions: in general, one needs to check the conditions (TR_{i}) for infinitely many positive values of i and infinitely many negative values of i.

We now describe the ring of Theorem 1.7. Related rings were used by Gasharov and Peeva in [6] and then by the authors in [7] to disprove various conjectures.

Let k be a field which is not algebraic over a finite field and let α ∈ k be an element of infinite multiplicative order. For the remainder of this section we assume the ring R to be defined as follows.

1.2. Consider the polynomial ring k[V, X, Y, Z] in four variables (each of degree one) and set

R = k[V, X, Y, Z]/I,

where I is the ideal generated by the following quadratic relations:

V^{2}, Z^{2}, XY, V X + αXZ, V Y + Y Z, V X + Y^{2}, V Y − X^{2} .

As a vector space over k, it has a basis consisting of the following 8 elements:

1, v, x, y, z, vx, vy, vz,

where v, x, y, z denote the residue classes of the variables modulo I. In particular, R has Hilbert series H_{R}(t) = 1 + 4t + 3t^{2}. 1.3. Remark. One may check that the generators for I listed above form a Gröbner basis for I. Therefore by [5, Section 4], the ring R is Koszul, and it follows that the Poincaré series ∑_{i} rank_{k} Tor_{i}^{R}(k, k)t^{i} of the residue field is equal to (1−4t+3t^{2})^{−1}.

For each integer i ≤ 0 we let d_{i}: R^{2} → R^{2} denote the map given with respect to the standard basis of R^{2} by the matrix

(\begin{matrix}v & α^{−i}x \\ y & z \end{matrix}) .

Also, let d_{1} : R^{3} → R^{2} denote the map represented by the matrix

(\begin{matrix}v & α^{−1}x & yz \\ y & z & 0 \end{matrix} )

and let d_{2} : R^{7} → R^{3} be represented by the matrix

(\begin{matrix}v & α^{−2}x & −y & 0 & 0 & 0 & 0 \\ y & z & αx & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & v & x & y & z \end{matrix}) .

Consider a minimal free resolution of Coker d_{2} with d_{2} as the first differential:

⋯ → R^{b_{i}} ⟶^{d_{i}} R^{b_{i−1}} → ⋯ → R^{b_{3}} ⟶^{d_{3}} R^{7} ⟶^{d_{2}} R^{3},

where for each i ≥ 3 the map d_{i}: R^{b_{i}} → R^{b_{i−1}} denotes the (i − 1)st differential in this resolution. 1.4. Lemma. The sequence of homomorphisms:

C : ⋯ → R^{b_{3}} ⟶^{d_{3}} R^{7} ⟶^{d_{2}} R^{3} ⟶^{d_{1}} R^{2} ⟶^{d_{0}} R^{2} ⟶^{d_{−1}} R^{2} ⟶^{d_{−2}} R^{2} ⟶^{d_{−3}} R^{2} → ⋯

is a doubly infinite complex with H_{i}(C) = 0 for all i ∈ ℤ.

Proof. (1). The defining equations of R guarantee that d_{i−1}d_{i} = 0 for all i ≤ 2. For i ≥ 2 the maps d_{i} are differentials in a free resolution, hence the equality holds for all i ≥ 3, as well. We conclude that C is a complex and H_{i}(C) = 0 for all i ≥ 2.

We let (a, b) denote an element of R^{2} written in the standard basis of R^{2} as a free R-module. For each i ≤ 0 the k-vector space Im d_{i} is generated by the elements:

d_{i}(1, 0) = (v, y)  d_{i}(z, 0) = (vz, −vy)
d_{i}(0, 1) = (α^{−i}x, z)  d_{i}(0, v) = (α^{−i}vx, vz)
d_{i}(v, 0) = (0, vy)  d_{i}(0, x) = (α^{−i}vy, −α^{−1}vx)
d_{i}(x, 0) = (vx, 0)  d_{i}(0, y) = (0, −vy)
d_{i}(y, 0) = (vy, −vx)  d_{i}(0, z) = (−α^{−i−1}vx, 0)

It is clear that rank_{k}(Im d_{i}) = 8 for all i ≤ 0. Since rank_{k}(R^{2}) = 16, this implies that rank_{k}(Ker d_{i}) = 8 for all i ≤ 0, showing that H_{i}(C) = 0 for all i ≤ −1.

Notice that for i = 1 the elements above give 7 linearly independent elements in Im d_{1}, and the 8th can be taken to be ε(0, 0, 1) = (yz, 0). (Here (a, b, c) denotes an element of R^{3} in its standard basis as a free R-module.) Thus rank_{k}(Im d_{1}) ≥ 8, and so rank_{k}(Ker d_{1}) ≤ 16. In particular, we obtain H_{0}(C) = 0.

To prove H_{1}(C) = 0 we need to show that rank_{k}(Im d_{2}) ≥ 16. Indeed, the following elements in Im d_{2} are linearly independent:

d_{2}(e_{1}) = (v, y, 0)  d_{2}(ye_{4}) = (0, 0, vy)
d_{2}(e_{2}) = (α^{−2}x, z, 0)  d_{2}(ze_{4}) = (0, 0, vz)
d_{2}(e_{3}) = (−y, αx, 0)  d_{2}(xe_{1}) = (vx, 0, 0)
d_{2}(e_{4}) = (0, 0, v)  d_{2}(ye_{1}) = (vy, −vx, 0)
d_{2}(e_{5}) = (0, 0, x)  d_{2}(ze_{1}) = (vz, −vy, 0)
d_{2}(e_{6}) = (0, 0, y)  d_{2}(ve_{1}) = (0, vy, 0)
d_{2}(e_{7}) = (0, 0, z)  d_{2}(ve_{2}) = (α^{−2}vx, vz, 0)
d_{2}(xe_{4}) = (0, 0, vx)  d_{2}(xe_{2}) = (α^{−2}vy, −α^{−1}vx, 0)

where e_{1}, …, e_{7} denote the elements comprising the standard basis of R^{7} as a free R-module. □

If f: M → N is a homomorphism of R-modules, we let f^{∗} represent the induced map Hom_{R}(f, R): Hom_{R}(N, R) → Hom_{R}(M, R). If (D, δ) is a complex of Rmodules, then the complex (D^{∗}, δ^{∗}) has (D^{∗})_{i} = (D_{−i})^{∗} and differentials (δ^{∗})_{i} = (δ_{−i})^{∗}. We write δ_{i}^{∗} for (δ^{∗})_{i}.

Note that, upon identification of R^{∗} with R, the map d_{i}^{∗}: R^{2} → R^{2} for i ≥ 0 is given in the standard basis of R^{2} by the matrix

(\begin{matrix}v & y \\ α^{i}x & z \end{matrix}) .

Similarly, the maps d_{−1}^{∗} and d_{−2}^{∗} are given by the transposes of the matrices defining d_{1} and d_{2}, respectively.

1.5. Lemma. The complex

C^{∗} : ⋯ → R^{2} ⟶^{d_{2}^{∗}} R^{2} ⟶^{d_{1}^{∗}} R^{2} ⟶^{d_{0}^{∗}} R^{2} ⟶^{d_{−1}^{∗}} R^{3} ⟶^{d_{−2}^{∗}} R^{7} → ⋯

satisfies H_{i}(C^{∗}) = 0 if and only if i ≥ 1.

Proof. As a k-vector space, Im d_{i}^{∗} for i ≥ 0 is generated by the following elements:

d_{i}^{∗}(1, 0) = (v, α^{i}x)  d_{i}^{∗}(z, 0) = (vz, −α^{i−1}vx)
d_{i}^{∗}(0, 1) = (y, z)  d_{i}^{∗}(0, v) = (vy, vz)
d_{i}^{∗}(v, 0) = (0, α^{i}vx)  d_{i}^{∗}(0, x) = (0, −α^{−1}vx)
d_{i}^{∗}(x, 0) = (vx, α^{i}vy)  d_{i}^{∗}(0, y) = (−vx, −vy)
d_{i}^{∗}(y, 0) = (vy, 0)  d_{i}^{∗}(0, z) = (−vy, 0)

One can see therefore that rank_{k}(Im d_{i}^{∗}) = 8 if i ≥ 1 and rank_{k}(Im d_{0}^{∗}) = 7. This implies that rank_{k}(Ker d_{i}^{∗}) = 8 if i ≥ 1 and rank_{k}(Ker d_{0}^{∗}) = 9, and it follows that H_{i}(C^{∗}) = 0 for all i ≥ 1 and H_{0}(C^{∗}) ≠ 0.

For the proof that H_{i}(C^{∗}) ≠ 0 for i = −1, −2, note that the image of d_{−1}^{∗} is generated as a k-vector space by the following elements.

d_{−1}^{∗}(1, 0) = (v, α^{−1}x, yz)  d_{−1}^{∗}(z, 0) = (vz, −α^{−2}vx, 0)
d_{−1}^{∗}(0, 1) = (y, z, 0)  d_{−1}^{∗}(0, v) = (vy, vz, 0)
d_{−1}^{∗}(v, 0) = (0, α^{−1}vx, 0)  d_{−1}^{∗}(0, x) = (0, −α^{−1}vx, 0)
d_{−1}^{∗}(x, 0) = (vx, α^{−1}vy, 0)  d_{−1}^{∗}(0, y) = (−vx, −vy, 0)
d_{−1}^{∗}(y, 0) = (vy, 0, 0)  d_{−1}^{∗}(0, z) = (−vy, 0, 0)

One sees easily that rank_{k}(Im d_{−1}^{∗}) ≤ 8. Therefore rank_{k}(Ker d_{−1}^{∗}) ≥ 8, and so H_{−1}(C^{∗}) ≠ 0. Clearly rank_{k}(Ker d_{−2}^{∗}) consists of at least nine linearly independent elements, namely the nine quadric elements in R_{2}^{3}. This shows that H_{−2}(C^{∗}) ≠ 0.

Finally, we note from the matrix representing d_{2} that Coker d_{2} ≅ N ⊕k, for some finite R-module N. Therefore H_{i}(C^{∗}) ≅ Ext_{R}^{−i−2}(N ⊕ k, R) ≠ 0 for all i ≤ −3, since R is not Gorenstein. □

1.6. For each integer s ≥ 1 let M_{s} be the cokernel of the map d_{−s}: R^{2} → R^{2}. Using Lemma 1.4, note that

M_{s} = Coker(d_{−s}) ≅ Im(d_{−s−1}) = Ker(d_{−s−2})

and a truncation of the complex C gives the beginning a minimal free resolution of the R-module M_{s}:

⋯ → R^{7} ⟶^{d_{2}} R^{3} ⟶^{d_{1}} R^{2} ⟶^{d_{0}} R^{2} ⟶^{d_{−1}} R^{2} → ⋯ → R^{2} ⟶^{d_{−s+1}} R^{2} ⟶^{d_{−s}} R^{2} → M_{s} → 0

The proof of Lemma 1.4 shows that M_{s} has Hilbert series H_{M_{s}}(t) = 2t + 6t^{2}.

We are now ready to state our main theorem:

1.7. Theorem. For the family of R-modules {M_{s}}_{s≥1} defined above, we have: (1) M_{s} satisfies (TR_{i}) if and only if i < s.

(2) Tr(M_{s}) satisfies (TR_{i}) if and only if i > −s.

Note that this contains the Theorem stated in the introduction: indeed, one can take there the modules M_{s} to be as above and L = Tr(M_{1}).

Proof of Theorem 1.7. The second part of the theorem follows from the first part and the simple observation that a finite R-module N satisfies the condition (TR_{i}) if and only if Tr(N) satisfies (TR_{−i}).

To compute Ext_{R}^{∗}(N, R) for an R-module N we take a minimal free resolution of N, we apply (−)^{∗} to it, and then compute homology of the resulting complex.

Applying (−)^{∗} to the minimal free resolution of M_{s} given in 1.6, and identifying R with R^{∗}, one obtains the complex

R^{2} ⟶^{d_{s}^{∗}} R^{2} ⟶^{d_{s−1}^{∗}} R^{2} → ⋯ → R^{2} ⟶^{d_{1}^{∗}} R^{2} ⟶^{d_{0}^{∗}} R^{2} ⟶^{d_{−1}^{∗}} R^{3} ⟶^{d_{−2}^{∗}} R^{7} → ⋯

Lemma 1.5 shows that Ext_{R}^{i}(M_{s}, R) = 0 for all 1 ≤ i ≤ s − 1, and Ext_{R}^{i}(M_{s}, R) ≠ 0 for i ≥ s.

A minimal free resolution of Tr(M_{s}) is given by

⋯ → R^{2} ⟶^{d_{s+1}^{∗}} R^{2} ⟶^{d_{s}^{∗}} R^{2},

and applying (−)^{∗} we get

R^{2} ⟶^{d_{−s}} R^{2} ⟶^{d_{−s−1}} R^{2} → · · · .

Lemma 1.4 shows that Ext_{R}^{i}(Tr(M_{s}), R) = 0 for all i ≥ 0. This establishes (1), and hence the entire theorem. □

2. Dependence

Theorem 1.7 shows that the conditions (TR_{i}) are, to a large extent, independent. However, in the Artinian graded case, there is some overlap between these conditions.

Let R be a standard graded ring and M a finite graded R-module. As noted by Avramov and Martsinkovsky in [4], the module M is totally reflexive if and only if it satisfies (TR_{i}) for all i, except possibly for i = 1 or i = −1.

This result is based on a formula obtained by Avramov, Buchweitz and Sally in [3]. Buchweitz pointed out to us that the same formula also yields the following. 2.1. Proposition. Assume that R is an Artinian standard graded ring, and M is a finitely generated graded R-module. Let A be a finite set of integers of the same parity. If M satisfies (TR_{i}) for all i ∈ ℤ∖A, then the module M is totally reflexive.

Proof. We may assume 0 ∉ A. Suppose that M satisfies (TR_{i}) for all i ∈ ℤ ∖ A. Since A is a finite set, we conclude that M satisfies (TR_{i}) for all i with |i| ≫ 0. This allows us to use the main formula in [3] which asserts equalities of rational functions

∑_{n∈\{0\}∪A}(−1)^{n}H_{Ext_{R}^{n}(M,R)}(t) = \frac{H_{M}(t^{−1}) · H_{R}(t)}{H_{R}(t^{−1})}  (1)

∑_{−n∈\{0\}∪A}(−1)^{n}H_{Ext_{R}^{n}(M^{∗},R)}(t) = \frac{H_{M^{∗}}(t^{−1}) · H_{R}(t)}{H_{R}(t^{−1})}  (2)

We set

P (t) = ∑_{n∈A} H_{Ext_{R}^{n}(M,R)}(t) and Q(t) = ∑_{−n∈A} H_{Ext_{R}^{n}(M^{∗},R)}(t).

Let σ = 0 if A contains only odd integers and σ = 1 if A contains only even integers. The formulas (1) and (2) give then

H_{M^{∗}}(t) = \frac{H_{M}(t^{−1}) · H_{R}(t)}{H_{R}(t^{−1})} + (−1)^{σ}P (t)

H_{M^{∗∗}}(t) = \frac{H_{M^{∗}}(t^{−1}) · H_{R}(t)}{H_{R}(t^{−1})} + (−1)^{σ}Q(t)

Substituting the formula for H_{M^{∗}}(t^{−1}) given by the first equation into the second equation, we obtain:

H_{M^{∗∗}}(t) = H_{M}(t) + \frac{H_{R}(t)}{H_{R}(t^{−1})} · (−1)^{σ}P (t^{−1}) + (−1)^{σ}Q(t)

and it follows that

H_{R}(t) · P (t^{−1}) + H_{R}(t^{−1}) · Q(t) = (−1)^{σ}H_{R}(t^{−1})(H_{M^{∗∗}}(t) − H_{M}(t)).

Assume that A contains only odd integers. We have then −2 ∉ A, hence (TR_{i}) is satisfied for i = −2. The map M → M^{∗∗} is then surjective, implying a coefficientwise inequality H_{M^{∗∗}}(t) ≤ H_{M}(t). Since σ = 0 in this case and H_{R}(t^{−1}) has positive coefficients, it follows that the Laurent polynomial on the right has nonpositive coefficients.

Both terms of the left-hand side sum are Laurent polynomials with nonnegative coefficients, and it follows that P (t) = 0 and Q(t) = 0, implying that Ext_{R}^{n}(M, R) = 0 for all n ∈ A and Ext_{R}^{n}(M^{∗}, R) = 0 for all n with −n ∈ A. In conclusion, (TR_{i}) is satisfied for all i ≠ −1. Furthermore, we conclude from the formula above that H_{M^{∗∗}}(t) = H_{M}(t). Since the map M → M^{∗∗} is surjective, it follows that it is an isomorphism, hence (TR_{i}) is satisfied for i = −1 as well.

Proceed similarly when A contains only even integers. □

Proposition 2.1 leads to the following question: Question. Let R be a commutative (local) Artinian ring. If a finite R-module M satisfies (TR_{i}) for all but finitely many values of i, does it follow that M is totally reflexive?

3. Minimimal acyclic complexes of free modules

Let S be a commutative Noetherian local ring with maximal ideal 𝔫.

A complex F of free S-modules

⋯ → F_{i+1} ⟶^{ϕ_{i+1}} F_{i} ⟶^{ϕ_{i}} F_{i−1} → ⋯

is said to be minimal if ϕ_{i}(F_{i}) ⊆ 𝔫F_{i−1} for all i. The complex F is acyclic if H_{i}(F) = 0 for all i ∈ ℤ. For any minimal acyclic complex of finite free S-modules F we can consider two sequences:

β_{F}^{+} := {rank_{R} F_{i}}_{i⩾0} and β_{F}^{−} := {rank_{R} F_{−i}}_{i⩾0}

Assuming that F_{i} ≠ 0 for all i, it is then natural to ask whether these two sequences have similar asymptotic behavior.

A sequence {β_{i}}_{i≥0} is said to have exponential growth if there exist numbers 1 < A ≤ B such that inequalities A^{i} ≤ β_{i} ≤ B^{i} hold for all i ≫ 0.

When the maximal ideal of S satisfies 𝔫^{3} = 0, Lescot [8] proved that the Betti numbers of a finitely generated S-module N are either eventually stationary, or they have exponential growth; in the last case they are eventually strictly increasing. It is clear from the Poincaré series given in 1.2 that the Betti numbers of k over our ring R have exponential growth. Furthermore, with d_{2} as defined there, since Coker(d_{2}) has a copy of k as a direct summand, its Betti numbers have exponential growth and are eventually strictly increasing.

In conclusion, the complex C of Lemma 1.4 has the following properties:

(a) β_{C}^{+} has exponential growth and is eventually strictly increasing.

(b) β_{C}^{−} is constant (nonzero).

Several questions arise: Question. Does there exist a ring S as above and a minimal acyclic complex of free nonzero S-modules F such that β_{F}^{−} has exponential growth (or is eventually strictly increasing) and β_{F}^{+} is eventually constant? Question. Do there exist examples of different asymptotic behavior for β_{F}^{−} and β_{F}^{+} if we also require H(F^{∗}) = 0 ? Can such examples exist over a Gorenstein ring?

The last question is equivalent to asking whether the Betti numbers of M and M^{∗} can have different asymptotic behavior when M is totally reflexive, and, in particular, when S is Gorenstein. Theorem 5.6 of [2] shows that the answer to this question is “no” when S is a complete intersection.

Acknowledgements

The authors thank Luchezar Avramov, Ragnar Buchweitz, Lars Christensen, Srikanth Iyengar, and Hamid Rahmati for their useful comments and suggestions.

References

[1] M. Auslander, M. Bridger, Stable Module Theory, Memoirs of the A.M.S. 94 (1969), American Math. Society, Providence, R.I.

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[3] L. L. Avramov, R.-O. Buchweitz, J. D. Sally, Laurent coefficients and Ext of finite graded modules, Math. Ann. 307 (1997), 401–415.

[4] L. Avramov, A. Martsinkovsky, Absolute, relative, and Tate cohomology of modules of finite Gorenstein dimension, Proc. London Math. Soc. 85 (2002), 393-440.

[5] R. Fröberg, Koszul algebras, Advances in commutative ring theory (Fez, 1997), Lecture Notes in Pure and Appl. Math. 205 Dekker, New York 1999; pp. 337–350.

[6] V. Gasharov, I. Peeva, Boundedness versus periodicity over commutative local rings, Trans. Amer. Math. Soc. 320 (1990), 569–580.

[7] D. A. Jorgensen, L. M. Şega, Nonvanishing cohomology and classes of Gorenstein rings, Adv. Math. 188 (2004), 470–490.

[8] J. Lescot, Asymptotic properties of Betti numbers of modules over certain rings, J. Pure Appl. Algebra 38 (1985), 287–298.

[9] Y. Yoshino, A functorial approach to modules of G-dimension zero, preprint, 2004; arXiv:math.AC/0408162.

Department of Mathematics, University of Texas at Arlington, Arlington, TX 76019

  E-mail address: djorgens@math.uta.edu

Department of Mathematics, Michigan State University, East Lansing, MI 48824

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