Totally Reflexive Modules and Poincar\'{e} Series

TL;DR

This paper explores the structure of Cohen-Macaulay non-Gorenstein local rings by analyzing Poincaré series through totally reflexive modules, extending Yoshino's results to higher dimensions and constructing diverse modules.

Contribution

It provides a new description of Poincaré series using totally reflexive modules and constructs large families of indecomposable modules in higher-dimensional rings.

Findings

Generalizes Yoshino's result to higher dimensions

Provides a method to compute Poincaré series via reflexive modules

Constructs infinitely many indecomposable modules with large minimal generators

Abstract

We study Cohen-Macaulay non-Gorenstein local rings $(R,\mathfrak{m},k)$ admitting certain totally reflexive modules. More precisely, we give a description of the Poincar\'{e} series of $k$ by using the Poincar\'{e} series of a non-zero totally reflexive module with minimal multiplicity. Our results generalize a result of Yoshino to higher-dimensional Cohen-Macaulay local rings. Moreover, from a quasi-Gorenstein ideal satisfying some conditions, we construct a family of non-isomorphic indecomposable totally reflexive modules having an arbitrarily large minimal number of generators.

Equations

$$P_k^R(t)=\frac{P_M^R(t)(1+t{)}^d}{{\nu }_R(M)(1-rt)}.$$

$$\frac{P_M^R(t)}{{\nu }_R(M)}=\frac{P_{M^{\prime }}^R(t)}{{\nu }_R(M^{\prime })}.$$

$$\frac{P_N^R(t)}{1-rt}\preccurlyeq \frac{C\cdot P_M^R(t)}{1-rt},\text{ where }C=\frac{r\cdot {\ell }_R(N/\mathfrak{q}N)-{\ell }_R(N^{\ast }/\mathfrak{q}{N}^{\ast })}{(r^2-1){\nu }_R(M)}.$$

$$\{0\to (R/I{)}^n\to M_{n,u}\to R/(0{:}_{R}I)\to 0{\}}_{u\in k}$$

$$P_M^R(t)=\sum _{i=0}^{\infty }{\beta }_n^R(M)t^n(I_R^M(t)=\sum _{i=0}^{\infty }{\mu }_R^n(M)t^n).$$

$${\mathrm{cx}}_R(M)=\inf \{d\in \mathbb{N}\mid \exists r\in \mathbb{R}suchthat{\beta }_n^R(M)\le r{n}^{d-1}forn\gg 0\}.$$

$$0\to G_n\to \cdots \to G_0\to M\to 0,$$

$$\operatorname{Ext}^{i}_{R}(R/I,R)\cong\left\{\begin{array}[]{c l}R/I&\text{ \ \ $i=\operatorname{grade}I$,}\\ 0&\text{ \ \ $\textrm{otherwise}$.}\end{array}\right.$$

$${\lambda }_{\ast }^{\overline{R}}:{\wedge }_{\ast }^{\overline{R}}{H}_1(K)⟶H_{\ast }(K)$$

$$I_R^R(t)=t^d{I}_{R/\mathfrak{q}}^{R/\mathfrak{q}}(t)=t^d\cdot \frac{r-t}{1-rt}.$$

$$bs{P}_k^R(t)=P_{\overline{M}{\otimes }_R\Omega \omega }^R(t)=P_{\overline{M}}^R(t)P_{\Omega \omega }^R(t);$$

$$P_{\overline{M}}^R(t)=P_M^R(t)(1+t{)}^d.$$

$$P_{\Omega \omega }^R(t)=\frac{1}{t}(P_{\omega }^R(t)-r)=\frac{1}{t}\left(\frac{1}{t^d}{I}_R^R(t)-r\right)=\frac{r^2-1}{1-rt},$$

$$P_L^R(t)=\frac{r^2-1}{1-rt}{P}_N^R(t),$$

$$\frac{P_N^R(t)}{1-rt}\preccurlyeq \frac{C\cdot P_M^R(t)}{1-rt},\text{ where }C=\frac{{\ell }_R(L)}{(r^2-1){\nu }_R(M)}.$$

$${\ell }_R(L)={\ell }_R(L^{\vee })={\ell }_R({\mathrm{Hom}}_R(N,(\Omega \omega {)}^{\vee }))={\ell }_R((N^{\vee }{)}^{\oplus r})-{\ell }_R(N^{\ast })=r\cdot {\ell }_R(N)-{\ell }_R(N^{\ast }),$$

$$C=\frac{r\cdot {\ell }_R(N)-{\ell }_R(N^{\ast })}{(r^2-1){\nu }_R(M)}.$$

$$P_k^R(t)\preccurlyeq \frac{U}{1-rt}.$$

$$C=\frac{r\cdot e(N)-e(N^{\ast })}{(r^2-1){\nu }_R(M)}.$$

$$C=\frac{{\ell }_R(N)}{(r+1){\nu }_R(M)}.$$

$$\frac{{\beta }_i^R(k)}{r^i}\le \frac{1}{{\nu }_R(M)}\left({\nu }_R(M)+u\sum _{h=1}^i\frac{h^{c-1}}{r^h}\right)$$

$${\mathrm{curv}}_R(k)=\underset{i\to \infty }{\mathrm{limsup}}\sqrt[i]{{\beta }_i^R(k)}\le \left(\underset{i\to \infty }{\mathrm{limsup}}\sqrt[i]{U}\right)\cdot r=\left(\underset{i\to \infty }{\lim }\sqrt[i]{U}\right)\cdot r=1\cdot r=r.$$

$$I_R^C(t)=\frac{t^d(r-t)}{1-rt},\text{ and }{P}_k^R(t)=\frac{P_M^R(t)(1+t{)}^d}{{\nu }_R(M)(1-rt)}.$$

$$e(M)={\nu }_R(\mathfrak{m}M)-(d-1){\nu }_R(M),$$

$$\mathfrak{qm}M\subseteq \left\{\begin{array}{c}{\mathfrak{m}}^{2}M\\ \mathfrak{q}M\end{array}\right\}\subseteq \mathfrak{m}M\subseteq M$$

$$M^{\left(\genfrac{}{}{0ex}{}{d}{2}\right)}\stackrel{A}{\to }{M}^d\to M\to M/\mathfrak{q}M\to 0,$$

$${\ell }_R({\mathfrak{m}}^{2}M/\mathfrak{qm}M)=e(M)+(d-1){\nu }_R(M)-{\nu }_R(\mathfrak{m}M).$$

$$\frac{P_k^R(t)}{1-rt}\preccurlyeq \frac{P_X^R(t)W(t)}{1-rt}\preccurlyeq \frac{C\cdot P_M^R(t)W(t)}{1-rt}.$$

$$\sum _{i=0}^l\frac{c_i}{r^i}\le \left(c_0+BE\sum _{i=1}^l\frac{i^{b+e-1}}{r^i}\right)\le c_0+BET,$$

$${G-dim}_{\overline{R}}(\overline{R}/{\overline{\mathfrak{m}}}^2)={G-dim}_R(R/{\mathfrak{m}}^2)-d=0$$

Full text

2010 Mathematics Subject Classification. 13C13, 13D40

Key words and phrases. totally reflexive module, quasi-Gorenstein ideal, quasi-complete intersection ideal, G-regular local ring, Poincaré series, large homomorphisms.

Takahashi was partly supported by JSPS Grant-in-Aid for Scientific Research 16K05098 and JSPS Fund for the Promotion of Joint International Research 16KK0099

1. Introduction

In the 1960s, Auslander and Bridger [2, 3] introduced the notion of Gorenstein dimension, G-dimension for short, for finitely generated modules as a generalization of projective dimension. Over Gorenstein rings, every finitely generated module has finite G-dimension, similarly to the fact that over regular rings every module has finite projective dimension, and the class of maximal Cohen-Macaulay modules coincides with the class of modules with G-dimension zero, which are called totally reflexive modules. Over rings that are not G-regular in the sense of [30] (see Definition 2.7), there exist modules of finite G-dimension and infinite projective dimension, which is equivalent to the existence of non-free totally reflexive modules. Yoshino [33, Theorem 3.1] proved the following remarkable result for Artinian short local rings possessing non-free totally reflexive modules.

Theorem** (Yoshino).**

Let $(R,\mathfrak{m},k)$ be a non-Gorenstein local ring of type $r$ such that $\mathfrak{m}^{3}=0\neq\mathfrak{m}^{2}$. Assume that $R$ contains $k$, and that there exists a non-free totally reflexive $R$-module $M$. Then the following statements hold.

1. (1)

The ring $R$ is a Koszul algebra. One has the Poincaré series $P^{R}_{k}(t)=\frac{1}{(1-t)(1-rt)}$, the Bass series $I^{R}_{R}(t)=\frac{r-t}{1-rt}$, and the Hilbert series $H_{R}(t)=(1+t)(1+rt)$. 2. (2)

The Poincaré series of $M$ is $P^{R}_{M}(t)=\frac{\nu_{R}(M)}{1-t}$.

Here, $\nu_{R}(M)$ stands for the minimal number of generators of $M$.

Later, Christensen and Veliche [13] generalized Yoshino’s theorem in terms of acyclic complexes over a non-Gorenstein local ring $(R,\mathfrak{m})$ with $\mathfrak{m}^{3}=0\neq\mathfrak{m}^{2}$; see [13, Theorem A].

In celebration of Yoshino’s theorem, Golod and Pogudin [18] call a ring satisfying the assumption of the theorem, a Yoshino algebra. It turns out that over Yoshino algebras, every non-free indecomposable totally reflexive module $M$ satisfies $\mathfrak{m}^{2}M=0$. Gerko [16] showed that an Artinian non-Gorenstein local ring $(R,\mathfrak{m})$ of type $r$ admitting a non-free totally reflexive module $M$ with $\mathfrak{m}^{2}M=0$ has Bass series $I^{R}_{R}(t)=\frac{r-t}{1-rt}$. This naturally leads us to the following question.

Question**.**

Let $(R,\mathfrak{m},k)$ be an Artinian non-Gorenstein local ring admitting a non-zero totally reflexive module $M$ with $\mathfrak{m}^{2}M=0$. Then what can one say about the Poincaré series of $k$?

In this paper we answer the above question, including even the higher-dimensional case. More explicitly, our main result is the following. We denote by $(-)^{*}$ the algebraic dual functor $\operatorname{Hom}_{R}(-,R)$.

Theorem 1.1**.**

Let $(R,\mathfrak{m},k)$ be a $d$-dimensional Cohen-Macaulay non-Gorenstein local ring of type $r$. Let $M\neq 0$ be a totally reflexive $R$-module with $\mathfrak{m}^{2}M\subseteq\mathfrak{q}M$ for some parameter ideal $\mathfrak{q}$ of $R$. Then the following hold.

1. (1)

One has

[TABLE]

In particular, if $M^{\prime}\neq 0$ is another totally reflexive $R$-module with $\mathfrak{m}^{2}M^{\prime}\subseteq\mathfrak{q}M^{\prime}$, then

[TABLE] 2. (2)

For each totally reflexive $R$-module $N$ there is an inequality of power series

[TABLE]

Here $\ell_{R}(X)$ stands for the length of an $R$-module $X$.

Levin [23] introduced the notion of a large homomorphism of local rings: a surjective homomorphism $f:R\rightarrow S$ of local rings is called large if the induced map $f_{*}:\operatorname{Tor}^{R}_{*}(k,k)\rightarrow\operatorname{Tor}^{S}_{*}(k,k)$ of graded algebras is surjective. It has been shown that $f:R\rightarrow S$ is large if and only if $P^{R}_{M}(t)=P^{S}_{M}(t)P^{R}_{S}(t)$ for every finitely generated $S$-module $M$. There are few examples of large homomorphisms in the literature; see [23, Theorem 2.2 and 2.4]. Also, in [7, Theorem 6.2] it is shown that the natural homomorphism $R\rightarrow R/I$, where $I$ is a quasi-complete intersection ideal with $I\cap\mathfrak{m}^{2}=I\mathfrak{m}$, is large. As an application of Theorem 1.1, we give another example of large homomorphisms.

Theorem 1.2**.**

Let $(R,\mathfrak{m})$ be a Cohen-Macaulay non-Gorenstein local ring. Let $I$ be a proper ideal of $R$ with $\mathfrak{m}^{2}\subseteq I$ and $\operatorname{G-dim}_{R}(R/I)<\infty$. Then

1. (1)

The homomorphism $R\rightarrow R/I$ is large. 2. (2)

If $R$ is a standard graded ring, then $R$ is a Koszul ring if and only if $R/I$ is a Koszul $R$-module.

There are ideals satisfying in the hypothesis of Theorem 1.2 which are not quasi-complete intersection; see Example 4.11(2). Moreover, we show that Theorem 1.2 fails if $\operatorname{G-dim}_{R}(R/I)=\infty$; see Example 3.12.

Our second aim in this paper is to give a construction of infinitely many non-isomorphic indecomposable totally reflexive modules from given ideals of arbitrary local rings. Construction of indecomposable totally reflexive modules with specific properties is widely investigated recently; see [1, 12, 22, 10, 29, 18]. Here, all the references except [18] use exact pairs of zero-divisors for their construction. There exist non-Gorenstein rings that do not admit an exact pair of zero-divisors but admit a non-free totally reflexive module; see for example [12, Propositions 9.1 and 9.2]. Our construction uses quasi-Gorenstein ideals, which are the generalization of ideals generated by exact zero-divisors, and recovers the results of [10, 29]; see Examples 4.7 and 4.11. More precisely, we prove the following in section 4.

Theorem 1.3**.**

Let $(R,\mathfrak{m},k)$ be a local ring. Let $I$ be a quasi-Gorenstein ideal of $R$ with $\operatorname{grade}I=0$ and $\dim_{R}I/I^{2}\geq 2$. Suppose that either of the following holds.

1. (a)

$(0:_{R}I^{2})=(0:_{R}I)$, or 2. (b)

$R/I$* is G-regular.*

Then for every integer $n\geq 1$, there exists a family of exact sequences of $R$-modules

[TABLE]

such that the modules $M_{{n,}u}$ are non-free and pairwise non-isomorphic indecomposable totally reflexive $R$-modules with $\nu_{R}(M_{n,u})=n+1$. In particular, if $k$ is infinite, then for each $n\geq 1$ there exist infinitely many such modules $M_{{n,}u}$.

Finally, we give examples of our construction over rings that satisfy the conditions of Theorem 1.1; see Examples 4.7 and 4.11(3).

2. Preliminaries

Throughout $R$ is a commutative Noetherian local ring with the maximal ideal $\mathfrak{m}$. For an $R$-module $M$ denote $\beta^{R}_{n}(M)$ ($\mu^{n}_{R}(M)$) the $n$-th Betti number ($n$-th Bass number) of $M$. The Poincaré series (Bass series) of $M$ over $R$ is defined by

[TABLE]

The complexity of $M$ over $R$ is defined by

[TABLE]

The curvature of $M$ over $R$ is defined by $\operatorname{curv}_{R}(M)=\underset{n\rightarrow\infty}{\limsup}\sqrt[n]{\beta^{R}_{n}(M)}$.

Definition 2.1**.**

A finitely generated $R$-module $M$ is called totally reflexive (or of $G$-dimension zero) if the evaluation homomorphism $M\to M^{**}$ is an isomorphism and $\operatorname{Ext}_{R}^{i}(M,R)=0=\operatorname{Ext}_{R}^{i}(M^{*},R)$ for all $i>0$.

Recall that the infimum non-negative integer $n$ for which there exists an exact sequence

[TABLE]

such that each $G_{i}$ is a totally reflexive $R$-module, is called the Gorenstein dimension of $M$. If $M$ has Gorenstein dimension $n$, we write $\operatorname{G-dim}_{R}(M)=n$.

Definition 2.2**.**

A proper ideal $I$ of $R$ is called quasi-Gorenstein in sense of [6] if

[TABLE]

Remark 2.3**.**

Let $I$ be a quasi-Gorenstein ideal. Then the following hold.

1. (1)

The ideal $I$ is a $G$-perfect ideal in the sense of Golod [17], that is, $\operatorname{G-dim}_{R}(R/I)=\operatorname{grade}I$. In particular, $\Omega^{\operatorname{grade}I}(R/I)$ is a totally reflexive $R$-module. 2. (2)

Suppose that $\operatorname{grade}I=0$, i.e., $(0:_{R}I)\neq 0$. One then has

$(0:_{R}I)\cong(R/I)^{*}\cong R/I$,  $R/(0:_{R}I)\cong I^{*}$, and   $(0:_{R}(0:_{R}I))=I$.

In particular, $(0:_{R}I)$ is a principal ideal, and $R/I$ and $R/(0:_{R}I)$ are totally reflexive $R$-modules.

We recall the definition of a quasi-complete intersection ideal from [7].

Definition 2.4**.**

Let $I$ be an ideal of a local ring $R$. Set $\overline{R}=R/I$ and let $K$ be the Koszul complex with respect to a minimal generating set of $I$. Then $I$ is said to be quasi-complete intersection if $\operatorname{H}_{1}(K)$ is free $\overline{R}$-module, and the natural homomorphism

[TABLE]

of graded $\overline{R}$-algebras with $\lambda^{\overline{R}}_{1}=\mathrm{id}_{\operatorname{H}_{1}(K)}$, is bijective.

Remark 2.5**.**

Every quasi-complete intersection ideal is quasi-Gorenstein; see [7, Theorem 2.5].

Lemma 2.6**.**

Let $I$ be a quasi-Gorenstein ideal of $R$. Then for any finitely generated $R/I$-module $M$, one has $\operatorname{G-dim}_{R}(M)=\operatorname{G-dim}_{R/I}{(}M{)}+\operatorname{grade}I$. In particular, if $\operatorname{grade}I=0$, then $M$ is a totally reflexive $R$-module if and only if $M$ is a totally reflexive $R/I$-module.

Proof.

This follows from [17, Theorem 5] by setting $K=R$. ∎

Definition 2.7**.**

A ring $R$ is called G-regular if all totally reflexive $R$-modules are free.

Every regular ring is G-regular. More generally, a Cohen-Macaulay local ring with minimal multiplicity is G-regular; see Corollary 3.6. The following proposition, stated in [30, Proposition 4.6], provides examples of G-regular local rings. Other examples of G-regular local rings are found in [8, Example 3.5], [30, Examples 5.2–5.5] and [26, Corollary 4.8]; see also [27, Corollary 6.6].

Proposition 2.8**.**

Let $(R,\mathfrak{m})$ be a G-regular local ring and $x\in\mathfrak{m}$ an $R$-regular element. Then $R/(x)$ is G-regular if and only if $x\notin\mathfrak{m}^{2}$.

Definition 2.9**.**

Let $R=\oplus^{\infty}_{i=0}R_{i}$ be a graded ring with $R_{0}=k$ a field. A graded $R$-module $M$ is called Koszul or linear, if $\beta^{R}_{i,j}(M)=0$ for all $j-i\neq r$ and some $r\in\mathbb{Z}$. More precisely, Koszul modules are those modules that have linear resolutions. The ring $R$ is said to be Koszul if $k$ is a Koszul $R$-module. For more details, see [15, 20].

3. Totally reflexive modules with minimal multiplicity

In this section we give the proof of our main result, Theorem 1.1. Before, we bring a lemma.

Lemma 3.1**.**

Let $(R,\mathfrak{m},k)$ be an Artinian local ring of type $r\geq 2$. Suppose there exists a non-zero totally reflexive $R$-module $M$ such that $\mathfrak{m}^{2}M=0$. Then the following statements hold.

1. (1)

There exists an exact sequence $0\rightarrow k^{br}\rightarrow M\rightarrow k^{b}\rightarrow 0$ where $b=\nu_{R}(M)$. 2. (2)

One has $I^{R}_{R}(t)=\frac{r-t}{1-rt}$.

Proof.

It follows from [16, Propositions 5.1 and 5.2] by setting $K=R$. ∎

Proof of Theorem 1.1.

(1) First we show that $I^{R}_{R}(t)=\frac{t^{d}(r-t)}{1-rt}$. Lemma 3.1 settles the case $d=0$. In the case where $d{>0}$, $\mathfrak{q}$ is generated by a regular sequence of length $d$. Therefore there is an isomorphism $\operatorname{Ext}^{i}_{R/\mathfrak{q}}(k,R/\mathfrak{q})\cong\operatorname{Ext}^{i+d}_{R}(k,R)$, which gives $\mu^{i}(R/\mathfrak{q})=\mu^{i+d}(R)$ for $i\in\mathbb{Z}$. Hence

[TABLE]

Set$\overline{(-)}=(-)\otimes_{R}R/\mathfrak{q}$ and let $\nu_{R}(M)={b}$. Replacing $R$ with its completion, we may assume that $R$ admits the canonical module $\omega$. As $\overline{M}$ is a totally reflexive $\overline{R}$-module, $\operatorname{G-dim}_{R}({\overline{M}})<\infty$ by Lemma 2.6. Hence we get $\operatorname{Tor}^{R}_{>0}({\overline{M}},\omega)=0$ by [11, Theorems (3.1.10) and (3.4.6)]. Consider the exact sequence $0\rightarrow\Omega\omega\rightarrow F\rightarrow\omega\rightarrow 0$ where $F$ is a free $R$-module and $\Omega\omega\subseteq\mathfrak{m}F$. As $\operatorname{Tor}^{R}_{1}({\overline{M}},\omega)=0$, applying ${\overline{M}}\otimes_{R}-$, we get an injection ${\overline{M}}\otimes_{R}\Omega\omega\hookrightarrow{\overline{M}}\otimes_{R}F$ whose image is a subset of $\mathfrak{m}({\overline{M}}\otimes_{R}F)$. Since $\mathfrak{m}^{2}({\overline{M}})=0$, it follows that ${\overline{M}}\otimes_{R}\Omega\omega$ is a $k$-vector space, that is, ${\overline{M}}\otimes_{R}\Omega\omega\cong k^{{b}s}$, where $s=\nu_{R}(\Omega\omega)$. Putting this together with $\operatorname{Tor}^{R}_{>0}({\overline{M}},\Omega\omega)=0$, one gets

[TABLE]

see [11, Theorem (A.7.6)]. Since $M$ is maximal Cohen-Macaulay R-module,

[TABLE]

We have

[TABLE]

which especially says $s=r^{2}-1$. The result now follows.

If $M^{\prime}\neq 0$ is another totally reflexive $R$-module such that $\mathfrak{m}^{2}\subseteq\mathfrak{q}M^{\prime}$ then by the last argument $\displaystyle\frac{P^{R}_{M}(t)(1+t)^{d}}{\nu_{R}(M)(1-rt)}=P^{R}_{k}(t)=\frac{P^{R}_{M^{\prime}}(t)(1+t)^{d}}{\nu_{R}(M^{\prime})(1-rt)}$. Thus $\displaystyle\frac{P^{R}_{M}(t)}{\nu_{R}(M)}=\frac{P^{R}_{M^{\prime}}(t)}{\nu_{R}(M^{\prime})}$.

(2) Note that $\overline{M}$ and $\overline{N}$ are totally reflexive $\overline{R}$-modules with $P^{\overline{R}}_{\overline{M}}(t)=P^{R}_{M}(t)$ and $P^{\overline{R}}_{\overline{N}}(t)=P^{R}_{N}(t)$. It follows from [9, Proposition 3.3.3(a)] that $\overline{N^{*}}\cong\operatorname{Hom}_{\overline{R}}(\overline{N},\overline{R})$. Thus, replacing $R$ with $R/\mathfrak{q}$, we may assume $d=0$. The same argument as in part (1) shows that

[TABLE]

where $L:=N\otimes_{R}\Omega\omega$. It is easy to observe by induction on $\ell_{R}(L)$ that $P^{R}_{L}(t)\preccurlyeq\ell_{R}(L)\cdot P^{R}_{k}(t)$. Using (1), we obtain

[TABLE]

There is an exact sequence $0\to\Omega\omega\to R^{\oplus r}\to\omega\to 0$. Taking the Matlis dual $(-)^{\vee}=\operatorname{Hom}_{R}(-,\omega)$, we get an exact sequence $0\to R\to\omega^{\oplus r}\to(\Omega\omega)^{\vee}\to 0$. Applying the functor $\operatorname{Hom}_{R}(N,-)$ to this and noting that $N$ is totally reflexive, we obtain an exact sequence $0\to N^{*}\to(N^{\vee})^{\oplus r}\to\operatorname{Hom}_{R}(N,(\Omega\omega)^{\vee})\to 0$. It follows that

[TABLE]

which yields the equality

[TABLE]

Thus (2) follows. ∎

Corollary 3.2**.**

Keep the notation of Theorem 1.1.

1. (1)

If $M$ has finite complexity, then there exists a real number $U>1$ such that

[TABLE]

In particular, one has $\operatorname{curv}_{R}(k)=r$. 2. (2)

If $\mathfrak{q}$ is a reduction of $\mathfrak{m}$, then $C$ is described in terms of multiplicities:

[TABLE] 3. (3)

Assume $d=0$ and $R$ is positively graded over a field, and $N$ is graded. Then

[TABLE]

Proof.

(1) Set $c=\operatorname{cx}_{R}(M)$. Then $\operatorname{cx}_{R}(M/\mathfrak{q}M)=c$ by [4, Proposition 4.2.4(5)], and there exist a real number $u>0$ and an integer $e>0$ such that $\beta^{R}_{i}(M/\mathfrak{q}M)\leq u\,i^{c-1}$ for all integers $i\geq e$. Replacing $u$ with $\max_{1\leq i\leq e-1}\{u,\,i\cdot\beta_{i}^{R}(M/\mathfrak{q}M)\}$, we may assume $e=1$. Theorem 1.1(1) shows $P^{R}_{k}(t)=\frac{P_{M}^{R}(t)(1+t)^{d}}{\nu_{R}(M)(1-rt)}=\frac{P^{R}_{M/\mathfrak{q}M}(t)}{\nu_{R}(M)(1-rt)}$, which implies

[TABLE]

for all $i\geq 1$. Since $c>0$ and $r>1$, by d’Alembert’s ratio test, the series $\sum_{h=1}^{\infty}\frac{h^{c-1}}{r^{h}}$ converges to $T$. Setting $U=1+\frac{u\,T}{\nu_{R}(M)}$, we get $\beta^{R}_{i}(k)\leq Ur^{i}$ for all $i\geq 0$. Therefore $P^{R}_{k}(t)\preccurlyeq\frac{U}{1-rt}$, and

[TABLE]

On the other hand, since $\frac{P_{M}^{R}(t)}{\nu_{R}(M)}=\frac{P_{M}^{R}(t)}{\beta^{R}_{0}(M)}=1+(\frac{\beta^{R}_{1}(M)}{\beta^{R}_{0}(M)})t+\cdots\succcurlyeq 1$ and $(1+t)^{d}\succcurlyeq 1$, we have $\frac{P_{M}^{R}(t)(1+t)^{d}}{\nu_{R}(M)}\succcurlyeq 1$. Therefore one has $P^{R}_{k}(t)=\frac{P_{M}^{R}(t)(1+t)^{d}}{\nu_{R}(M)(1-rt)}\succcurlyeq\frac{1}{1-rt}$. Hence $\beta^{R}_{n}(k)\geq r^{n}$ and so $\operatorname{curv}_{R}(k)\geq r$.

(2) It follows from [9, Theorems 4.7.6 and 4.7.10] that $\operatorname{e}(N)=\ell_{R}(N/\mathfrak{q}N)$ and $\operatorname{e}(N^{*})=\ell_{R}(N^{*}/\mathfrak{q}N^{*})$. Thus the assertion follows from Theorem 1.1(2).

(3) Since $\ell_{R}(N)=\ell_{R}(N^{*})$ by [19, Lemma 6.6], Theorem 1.1(2) implies the assertion. ∎

Remark 3.3**.**

A similar argument to the proof of Theorem 1.1(1) actually shows a more general statement:

Let $R$ be a $d$-dimensional Cohen-Macaulay local ring. Let $C$ be a semidualizing $R$-module of type $r{\neq 1}$. Let $M$ be a non-zero totally $C$-reflexive $R$-module with $\mathfrak{m}^{2}M\subseteq\mathfrak{q}M$ for some parameter ideal $\mathfrak{q}$ of $R$. Then

[TABLE]

The following result gives a sufficient condition for the assumption $\mathfrak{m}^{2}M\subseteq\mathfrak{q}M$ used in Theorem 1.1, which no longer involves a parameter ideal $\mathfrak{q}$.

Proposition 3.4**.**

Let $(R,\mathfrak{m},k)$ be a $d$-dimensional Cohen-Macaulay local ring with $k$ infinite. Let $M$ be a maximal Cohen-Macaulay $R$-module satisfying

[TABLE]

Then there exists a parameter ideal $\mathfrak{q}$ of $R$ with $\mathfrak{m}^{2}M\subseteq\mathfrak{q}M$.

Proof.

Since $k$ is an infinite field, one can choose a reduction $\mathfrak{q}$ of $\mathfrak{m}$ which is a parameter ideal of $R$; see [9, Corollary 4.6.10]. Consider the chain

[TABLE]

of submodules of $M$. We claim that $\ell_{R}(\mathfrak{q}M/\mathfrak{q}\mathfrak{m}M)=d\cdot\nu(M)$. In fact, as $\mathfrak{q}$ is generated by a regular sequence on $R$ and $M$, its Koszul complex on $M$ makes an exact sequence

[TABLE]

where $A$ is a $d\times\binom{d}{2}$ matrix with entries in $\mathfrak{q}$. Tensoring with $k$ the induced exact sequence $M^{\binom{d}{2}}\xrightarrow{A}M^{d}\to\mathfrak{q}M\to 0$ shows the claim. Since $\ell_{R}(M/\mathfrak{q}M)=\operatorname{e}(M)$ by [9, Lemma 4.6.5], the above chain of inclusions gives rise to:

[TABLE]

Equality (3.4.1) is nothing but the right-hand side is zero, which implies $\mathfrak{m}^{2}M=\mathfrak{q}\mathfrak{m}M\subseteq\mathfrak{q}M$. ∎

Recall that a Cohen-Macaulay local ring $R$ is said to have minimal multiplicity if (3.4.1) holds for $M=R$; see [9, Exercise 4.6.14]. The proof of Proposition 3.4 shows that if $k$ is infinite then for a maximal Cohen-Macaulay $R$-module $M$ one has $\operatorname{e}(M)\geq\nu_{R}(\mathfrak{m}M)-(d-1)\nu_{R}(M)$. In fact, the last inequality holds even if $k$ is a finite field. To see that, one can use the trick of passing to the flat extension $R[w]_{\mathfrak{m}[w]}$ where $w$ is an indeterminate; see [25, page 114, Remark] for more details. Therefore, it makes sense to give the following definition.

Definition 3.5**.**

Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring of dimension $d$. We say that a maximal Cohen-Macaulay $R$-module $M$ has minimal multiplicity if $\operatorname{e}(M)=\nu_{R}(\mathfrak{m}M)-(d-1)\nu_{R}(M)$.

Theorem 1.1 and Proposition 3.4 immediately recover the following result given in [33, Corollary 2.5] (see also [8, Examples 3.5(2)]).

Corollary 3.6**.**

Let $(R,\mathfrak{m},k)$ be a Cohen-Macaulay non-Gorenstein local ring with minimal multiplicity. Then $R$ is G-regular.

Proof.

Replacing $R$ with the faithfully flat extension $R[w]_{\mathfrak{m}[w]}$ with $w$ an indeterminate, we may assume that $k$ is infinite. Then by Proposition 3.4 there is a parameter ideal $\mathfrak{q}$ such that $\mathfrak{m}^{2}\subseteq\mathfrak{q}$. Therefore if $M$ is any non-zero totally reflexive module, then one has $\mathfrak{m}^{2}M\subseteq\mathfrak{q}M$. Now, Theorem 1.1(1) implies that $\frac{P^{R}_{M}(t)}{\nu_{R}(M)}=\frac{P^{R}_{R}(t)}{\nu_{R}(R)}=1$, i.e. $M$ is a free $R$-module. ∎

In [31, Question 6.6] Takahashi asked if a local ring is Gorenstein when a syzygy of the residue field (with respect to a minimal free resolution) has a non-zero direct summand of finite G-dimension. As another application of Theorem 1.1 we prove the following result which specifically says that [31, Question 6.6] has affirmative answer over Artinian short local rings.

Corollary 3.7**.**

Let $(R,\mathfrak{m},k)$ be a Cohen-Macaulay non-Gorenstein local ring. Suppose that there exists a non-zero totally reflexive $R$-module $M$ with minimal multiplicity. If $\operatorname{cx}_{R}(M)<\infty$, then for each $n\geq 0$, the $n$th syzygy $\Omega^{n}k$ of the $R$-module $k$ does not have a non-zero direct summand of finite G-dimension.

Proof.

Passing to the faithfully flat extension $R[w]_{\mathfrak{m}[w]}$ with $w$ an indeterminate, we may assume that $k$ is infinite. Proposition 3.4 implies $\mathfrak{m}^{2}M\subseteq\mathfrak{q}M$ for some parameter ideal $\mathfrak{q}$ of $R$. Assume that there is an integer $n\geq 0$ such that $\Omega^{n}k\cong X\oplus Y$, where $X$ is a non-zero $R$-module with $\operatorname{G-dim}_{R}X<\infty$. Replacing $n$ with a bigger integer if necessary, we may assume that $X$ is totally reflexive.

Note that as the map $\operatorname{Hom}_{R}(X,k)\rightarrow\operatorname{Hom}_{R}(\Omega^{n}k,k)$ induced from the projection $\Omega^{n}k\to X$ is non-zero, by the proof of [5, Theorem 8] there exists a power series $W(t)=\frac{(1+t)^{\alpha_{1}}\cdots(1+t^{2n-1})^{\alpha_{n}}}{(1-t^{2})^{\beta_{1}}\cdots(1-t^{2n})^{\beta_{n}}}$ such that $W(t)P^{R}_{X}(t)\succcurlyeq P^{R}_{k}(t)$, where $\alpha_{1},\cdots,\alpha_{n}$, and $\beta_{1},\cdots,\beta_{n}$ are positive integers. Let $r$ be the type of $R$. By Theorem 1.1(2) we have $\frac{P^{R}_{X}(t)}{1-rt}\preccurlyeq\frac{C\cdot P^{R}_{M}(t)}{1-rt}$ with $C$ constant. Note that multiplication by a power series with positive coefficients preserves an inequality of formal power series. Therefore by multiplying $W(t)\succcurlyeq 1$, we get

[TABLE]

Since $\frac{P^{R}_{M}(t)(1+t)^{d}}{\nu_{R}(M)}\succcurlyeq 1$, we have $\frac{P^{R}_{k}(t)}{1-rt}=\frac{P^{R}_{M}(t)(1+t)^{d}}{\nu_{R}(M)(1-rt)^{2}}\succcurlyeq\frac{1}{(1-rt)^{2}}$.

Let $\operatorname{cx}_{R}(M)=e<\infty$. Then there exists $E>\beta^{R}_{0}(M)$ such that $\beta_{i}^{R}(M)\leq E\cdot i^{e-1}$ for all $i\geq 1$. Also, note that $W(t)$ has finite complexity, i.e. if $a_{i}$ stands for $i$th coefficient of $W(t)$, there exist $b>0$, and $B>1$ such that $a_{i}<B\cdot i^{b-1}$ for all $i\geq 1$. Set $V(t)=P^{R}_{M}(t)W(t)$. Hence if $c_{i}$ is the $i$th coefficient of $V(t)$, then $c_{i}\leq BE\cdot i^{b+e-1}$ for all $i\geq 1$. Therefore we get

[TABLE]

where $T=\sum_{i=0}^{\infty}\frac{i^{b+e-1}}{r^{i}}$ is a convergent power series as $r\geq 2$. Now by setting $D=C(c_{0}+BET)$ one has $\frac{C\cdot P^{R}_{M}(t)W(t)}{1-rt}\preccurlyeq\frac{D}{1-rt}$. Putting these together with (3.7.1), we get an inequality of formal power series $\frac{1}{(1-rt)^{2}}\preccurlyeq\frac{D}{1-rt}$, which implies that $i\cdot r^{i}\leq D\cdot r^{i}$ for all $i\geq 1$, contradiction. ∎

Lemma 3.8**.**

Let $(R,\mathfrak{m},k)$ be an Artinian non-Gorenstein local ring of type $r$. Let $I$ be a proper ideal of $R$ containing $\mathfrak{m}^{2}$ such that ${\operatorname{G-dim}_{R}}R/I{<\infty}$. Then $I$ is a quasi-Gorenstein ideal, and the following hold.

(1)* $\nu_{R}(\mathfrak{m})=\nu_{R}(I)+r$,  (2) $\nu_{R}(\mathfrak{m}/I)=\ell_{R}(\mathfrak{m}/I)=r$,  (3) $\mathfrak{m}^{2}=I\mathfrak{m}$.*

Proof.

First we show that $I$ is a quasi-Gorenstein ideal. Since $R/I$ is totally reflexive, we only need to show that $(R/I)^{*}\cong R/I$; see Definition 2.2. As $\mathfrak{m}^{2}(R/I)^{*}=0$, by Lemma 3.1(1) there is an exact sequence $0\to k^{br}\to(R/I)^{*}\to k^{b}\to 0$ with $b=\nu_{R}(R/I)^{*}$. Since $\ell_{R}(R/I)=\ell_{R}((R/I)^{*})$ by [16, Proposition 5.2], we get $br+b=\ell_{R}((R/I)^{*})=\ell_{R}(R/I)=r+1$, whence $b=1$.

(1) By Theorem 1.1(1) we have $P^{R}_{k}(t)=\frac{P^{R}_{R/I}(t)}{1-rt}$, which shows $\nu_{R}(\mathfrak{m})=\nu_{R}(I)+r$.

(2) One has $\ell_{R}(R/I)=r+1$ which implies $\ell_{R}(\mathfrak{m}/I)=r$. Since $\mathfrak{m}/I$ is annihilated by $\mathfrak{m}$, we get $\nu_{R}(\mathfrak{m}/I)=\ell_{R}(\mathfrak{m}/I)$.

(3) Consider the chain $I\mathfrak{m}\subseteq\mathfrak{m}^{2}\subseteq I\subseteq\mathfrak{m}$ of ideals of $R$. Computing the lengths of subquotients, we have $\ell_{R}(\mathfrak{m}^{2}/I\mathfrak{m})+\nu_{R}(\mathfrak{m})=\nu_{R}(I)+\ell_{R}(\mathfrak{m}/I)$. It follows from (1) and (2) that $\ell_{R}(\mathfrak{m}^{2}/I\mathfrak{m})=0$, and we obtain $\mathfrak{m}^{2}=I\mathfrak{m}$. ∎

Lemma 3.9**.**

Let $(R,\mathfrak{m},k)$ be a $d$-dimensional Cohen-Macaulay non-Gorenstein local ring with $\mathfrak{m}^{2}\neq 0$ (e.g. ${d}>0$). Then $R/\mathfrak{m}^{2}$ has infinite G-dimension as an $R$-module.

Proof.

Assume that $R/\mathfrak{m}^{2}$ has finite G-dimension. Choose a parameter ideal $\mathfrak{q}$ of $R$ contained in $\mathfrak{m}^{2}$. Setting $\overline{R}=R/\mathfrak{q}$ and $\overline{\mathfrak{m}}=\mathfrak{m}/\mathfrak{q}$, we have

[TABLE]

by [11, (1.5.4) and (1.4.8)]. Using Lemma 3.8(3), we get ${\overline{\mathfrak{m}}}^{2}={\overline{\mathfrak{m}}}^{2}\overline{\mathfrak{m}}$. Nakayama’s lemma implies ${\overline{\mathfrak{m}}}^{2}=0$, whence $\mathfrak{m}^{2}=\mathfrak{q}$. Since $\mathfrak{q}$ is generated by an $R$-sequence, $\mathfrak{m}^{2}$ is a nonzero $R$-module of finite projective dimension. It follows from [24, Theorem 1.1] that $R$ is regular, which contradicts the assumption that $R$ is non-Gorenstein. ∎

Now, we are in a position to prove Theorem 1.2.

Proof of Theorem 1.2.

We use induction on $d=\dim R$. First, assume $d=0$. Lemma 3.8 says $\nu_{R}(\mathfrak{m}/I)=r$, where $r$ is the type of $R$. As $(\mathfrak{m}/I)^{2}=0$, $R/I$ is a Cohen-Macaulay ring with minimal multiplicity, and one has $P^{R/I}_{k}(t)=\frac{1}{1-rt}$; see [1, Example 5.2.8]. Theorem 1.1(1) implies $P^{R}_{k}(t)=\frac{P^{R}_{R/I}(t)}{1-rt}=P^{R}_{R/I}(t)\,P^{R/I}_{k}(t)$. Now by [23, Theorem 1.1] $R\rightarrow R/I$ is a large homomorphism.

Assume $d\geq 1$. Then $I$ strictly contains $\mathfrak{m}^{2}$ by Lemma 3.9, and by the Prime Avoidance Theorem, there exists a regular element $x\in I\setminus\mathfrak{m}^{2}$. Set $\overline{R}=R/(x)$ and $\overline{I}=I/(x)$. Then $\operatorname{G-dim}_{\overline{R}}(R/I)<\infty$. Consider the sequence $R\rightarrow\overline{R}\rightarrow R/I$ of surjective local homomorphisms. By the induction hypothesis $\overline{R}\rightarrow R/I$ is large. Since $R\rightarrow\overline{R}$ is also large by [23, Theorem 2.2], it follows that so is the composition $R\rightarrow R/I$. This proves (1).

For (2), since $\mathfrak{m}^{2}\subseteq I$, $I$ is homogeneous and $R/I$ is a graded Koszul ring. Similarly as above, the graded Poincaré series of $k$ over $R/I$ is $P^{R/I}_{k}(s,t)=\frac{1}{1-r(st)}$. As $R\rightarrow R/I$ is large, we have

[TABLE]

Thus the minimal graded free resolution of $k$ over $R$ is linear if and only if so does that of $R/I$ over $R$. ∎

Corollary 3.10**.**

Let $(R,\mathfrak{m},k)$ be a $d$-dimensional Cohen-Macaulay non-Gorenstein local ring of type $r$. Let $I$ be a quasi-complete intersection ideal of $R$ containing $\mathfrak{m}^{2}$. Then

[TABLE]

Proof.

By Theorem 1.2, the homomorphism $R\rightarrow R/I$ is large. Therefore, as we saw before, $P^{R}_{k}(t)=P^{R}_{R/I}(t)\cdot P^{R/I}_{k}(t)=\frac{P^{R}_{R/I}(t)}{(1-rt)}$. It follows from [7, Theorem 6.2] that $P^{R}_{R/I}(t)=\frac{(1-t^{2})^{d}}{(1-t)^{\nu_{R}(I)}}$. It remains to show that $\nu_{R}(I)=\nu_{R}(\mathfrak{m})-r$. Lemma 3.8(1) establishes the case $d=0$, and the case $d>0$ follows by induction. ∎

The following is an example for Theorem 1.2 and Corollary 3.10.

Example 3.11**.**

Let $R=k[{w},x,y,z]/({w^{2}},x^{2},y^{2},z^{2},z{w})$. Then $R$ is an Artinian local ring with the maximal ideal $\mathfrak{m}=({w},x,y,z)$. One checks $(xyz,xyw)\subseteq(0:_{R}\mathfrak{m})$ and hence $R$ is not Gorenstein. Note that $R\cong P[w,z]/(w^{2},z^{2},wz)$, where $P=k[x,y]/(x^{2},y^{2})$ is a Koszul complete intersection; see [21, 4.5]. Since $P\rightarrow R$ is flat, the ideal $I=(x,y){R}$ is a quasi-complete intersection ideal of $R$ containing $\mathfrak{m}^{2}$; see [7, Lemma 2.1]. By Lemma 3.8(1), the type of $R$ is $\nu_{R}(\mathfrak{m})-\nu_{R}(I)=2$. Therefore by Corollary 3.10 one has $P^{R}_{k}(t)=\frac{1}{(1-t)^{2}(1-2t)}$. Since $k$ has a linear resolution over $P$, $R/I$ has linear free resolution over $R$ and so $R/I$ is a Koszul module. Hence $R$ is a Koszul ring by Theorem 1.2(2).

The following example shows that Theorem 1.2 fails if $\operatorname{G-dim}_{R}(R/I)=\infty$.

Example 3.12**.**

[7, Theorem 3.5] Let $k$ be a field of characteristic different from $2$, and let

[TABLE]

Then $R$ is an Artinian local ring with the maximal ideal $m=(w,x,y,z)$. Since $(y^{2},xz)\subseteq(0:_{R}\mathfrak{m})$, $R$ is not Gorenstein. By [7, Theorem 3.5] $x,y$ is an exact pair of zero-divisors, and therefore $(x)$ is a quasi-complete intersection ideal of $R$. One can easily check that $\mathfrak{m}^{2}\subseteq(x)$. Thus, Lemma 3.8(1) says that type of $R$ is $3$. Hence by Corollary 3.10, one has $P^{R}_{k}(t)=\frac{1}{(1-t)(1-3t)}$. Let $I=(x-z)$. Then by using Macaulay2 one checks $\mathfrak{m}^{2}\subseteq I$, and $(0:_{R}I)=(x-y+z,yw)$. Hence $I$ is not a quasi-Gorenstein ideal and therefore $\operatorname{G-dim}_{R}(R/I)=\infty$ by Lemma 3.8. We have $\beta_{2}(R/I)=\nu_{R}(0:_{R}I)=2$. If $R\rightarrow R/I$ is large then by [23, Theorem 1.1] we have $P^{R}_{k}(t)=P^{R}_{R/I}(t)\cdot P^{R/I}_{k}(t)$, and so $\frac{1}{(1-t)(1-3t)}=P^{R}_{R/I}(t)\cdot\frac{1}{1-3t}$. Hence one gets $P^{R}_{R/I}(t)=\frac{1}{1-t}$. Since $\beta^{R}_{2}(R/I)=2$, this is impossible.

4. Construction

In this section, we generalize results of [10, 29]. Both of these references use exact zero-divisors to construct infinitely many totally reflexive modules. Our ingredients for construction of such modules are quasi-Gorenstein ideals, more general than exact zero-divisors. For example, the ideal $I$ in Example 3.11 is a quasi-Gorenstein ideal, but since $I$ is not principal, $I$ is not generated by an exact zero-divisor; see also Example 4.11.

Lemma 4.1**.**

Let $I$ be an ideal of $R$. If $(0:_{R}I^{2})=(0:_{R}I)$ then for every finitely generated $R/I$-module $N$, $\operatorname{Hom}_{R}(N,R/(0:_{R}I))=0$.

Proof.

Applying $\operatorname{Hom}_{R}(-,R/(0:_{R}I))$ to a surjection $(R/I)^{n}{\twoheadrightarrow}N$, we get an injection $\operatorname{Hom}_{R}(N,R/(0:_{R}I)){\hookrightarrow}\operatorname{Hom}_{R}((R/I)^{n},R/(0:_{R}I))$. Since $\operatorname{Hom}_{R}(R/I,R/(0:_{R}I))\cong(0:_{R}I^{2})/(0:_{R}I)=0$, we are done. ∎

Lemma 4.2**.**

Let $I$ be an ideal of $R$. Assume one of the following hold:

1. (a)

$(0:_{R}I^{2})=(0:_{R}I)$, or 2. (b)

$I$* is quasi-Gorenstein and $R/I$ is a G-regular ring.*

Then every exact sequence

[TABLE]

with $n\geq 0$, as an $R$-complex, has a direct summand isomorphic to an exact sequence $0\to(R/I)^{t}\to N\to R/(0:_{R}I)\to 0$ for some $t$ with $0\leq t\leq n$ such that $N$ is indecomposable. In particular, $M\cong(R/I)^{n-t}\oplus N$.

Proof.

Set $J=(0:_{R}I)$. The assertion clearly holds if $J=0$, so let us assume $J\neq 0$. Hence $\operatorname{grade}I=0$. Assume $M$ is decomposable and write $M\cong X\oplus Y$ for some nonzero $R$-modules $X,Y$. Then we have an exact sequence $0\to(R/I)^{n}\to X\oplus Y\xrightarrow{(f,g)}R/J\to 0$ with $n\geq 1$. Since $R/J$ is a cyclic $R$-module and $(f,g)$ is surjective, one of $f$ or $g$ has to be surjective. Assume that $f$ is surjective. Then the following commutative diagram with exact rows and columns

[TABLE]

shows that $Y$ is an $R/I$-module. Set $Z=\operatorname{Ker}f$.

The assumption (a) and Lemma 4.1 imply that $\operatorname{Hom}_{R}(Y,R/J)=0$. Hence $g=0$, and the map $h:(R/I)^{n}\rightarrow Y$ yields a splitting $Y\rightarrow(R/I)^{n}$. Thus $Y\oplus Z\cong(R/I)^{n}$ and hence $Y$ and $Z$ are free $R/I$-modules.

Assume now that (b) applies. Then both $R/I$ and $R/J$ are totally-reflexive by Remark 2.3(2). Hence $M$ is a totally reflexive module. As $X$ and $Y$ are direct summands of $M$, they are also totally reflexive modules. Then Lemma 2.6 shows that $Y$ is a totally reflexive $R/I$-module, and since $R/I$ is G-regular, it is a free $R/I$-module.

Thus in both cases, we have $Y\oplus Z\cong(R/I)^{n}$ and $Y\cong(R/I)^{r}$ for some $r$ with $0<r\leq n$. Therefore there is an exact sequence $0\to(R/I)^{n-r}\to X\to R/J\to 0$, where $Z\cong(R/I)^{n-r}$. Iterating this procedure, we get the desired exact sequence. ∎

Lemma 4.3**.**

Let $I=(x_{1},\dots x_{r})$ be an ideal, and let $y$ and $a_{1},\dots,a_{n}$ be in $R$. Then for an $R$-module $M$ the following are equivalent.

1. (1)

There exists an exact sequence $0\to(R/I)^{n}\to M\to R/(y)\to 0$ of $R$-modules. 2. (2)

There exists an exact sequence $R^{nr+1}\xrightarrow{T(\underline{x},y,\underline{a})}R^{n+1}\to M\to 0$, where $a_{1},\dots,a_{n}\in(I:_{R}(0:_{R}y))$ and

[TABLE]

Proof.

See [10, Lemmas 3.2 and 3.3], whose proofs are explained just before them. ∎

Definition 4.4**.**

Let $I$ be an ideal of $R$. Define

[TABLE]

Lemma 4.5**.**

Let $I$ be an ideal of $R$ and let $y\in\mathfrak{m}$ be an element. For any integer $n>\operatorname{s}(I+(y))$ and any exact sequence $0\to(R/I)^{n}\to M\to R/(y)\to 0$ of $R$-modules, $M$ has a direct summand isomorphic to $R/I$.

Proof.

The assertion follows from a similar argument to the proof of [10, Proposition 3.5] using Lemma 4.3. ∎

Proposition 4.6**.**

Let $I$ be a quasi-Gorenstein ideal of $R$ of grade zero. Assume either of the following holds.

1. (a)

$(0:_{R}I^{2})=(0:_{R}I)$, or 2. (b)

$R/I$* is G-regular.*

Then for each $0<n<\operatorname{s}(I+(0:_{R}I))$, there exists a family

[TABLE]

of exact sequences of $R$-modules such that each $M_{{n,}u}$ is non-free indecomposable totally reflexive $R$-module, and that if $M_{{n,}u}\cong M_{{m,}v}$ for $u,v\in R$, then $n=m$ and $\overline{u}=\overline{v}$ in $k$.

Proof.

The proof is similar to that of [10, Theorems 3.6 and 3.8], but we bring the proof for convenience of the reader. By Remark 2.3(2) $(0:_{R}I)$ is a principal ideal. Write $(0:_{R}I)=(y)$, for some $y\in R$, and set $\overline{R}=R/(I+(y))$. As $n<\operatorname{s}(I+(y))$, we have $\nu_{\overline{R}}(\overline{a_{1}},\dots,\overline{a_{n}})=n$ for some $a_{1},\dots,a_{n}\in R$. Let $M$ be the cokernel of the $R$-linear map defined by $T(\underline{x},y,\underline{a})$, where $I=(x_{1},\dots,x_{r})$. Since $(I:_{R}(0:_{R}y))=(I:_{R}I)=R$ by Remark 2.3(2), Lemma 4.3 provides an exact sequence

[TABLE]

Suppose the $R$-module $M$ is decomposable. It follows from Lemma 4.2 that there is an exact sequence $0\to(R/I)^{n-1}\to N\to R/(y)\to 0$, which is isomorphic to a direct summand of (4.6.1) as an $R$-complex. Now, Lemma 4.3 shows that there is an exact sequence $R^{(n-1)r+1}\xrightarrow{T(\underline{x},y,b_{1},\dots,b_{n-1})}R^{n}\to N\to 0$. Since $M\cong R/I\oplus N$, it is seen that $M$ has two presentation matrices $T(I,y,a_{1},\dots,a_{n})$ and $T(I,y,0,b_{1},\dots,b_{n-1})$, both of which are $(nr+1)\times(n+1)$ matrices. Taking $nr$-th Fitting invariant, we have equality $I+(y,a_{1},\dots,a_{n})=I+(y,b_{1},\dots,b_{n-1})$ of ideals of $R$, see [9, Page 21]. Taking the image in $\overline{R}$, we see that $\nu_{\overline{R}}(\overline{a_{1}},\dots,\overline{a_{n}})\leq n-1$, which is a contradiction.

Now choose $a_{1},\dots,a_{n},b\in R$ such that $\nu_{\overline{R}}(\overline{a_{1}},\dots,\overline{a_{n}},\overline{b})=n+1$. For each $u\in R$, let $M_{{n,}u}=\operatorname{Coker}(T(\underline{x},y,a_{1}+ub,a_{2},\dots,a_{n}))$. Note that $\nu_{\overline{R}}(\overline{a_{1}+ub},\overline{a_{2}},\dots,\overline{a_{n}})=n$. By the last argument, $M_{{n,}u}$ is an indecomposable $R$-module admitting an exact sequence of $R$-modules of the form $0\to(R/I)^{n}\to M_{{n,}u}\to R/(y)\to 0$. Let $u,v\in R$ be elements such that $M_{{n,}u}\cong M_{{m,}v}$ for some $n$ and $m$. Clearly, one has $m=n$ by Lemma 4.3. Then, taking the $nr$-th Fitting invariants of $M_{{n,}u}$ and $M_{{n,}v}$, we see that $I+(y,a_{1}+ub,a_{2},\dots,a_{n})=I+(y,a_{1}+vb,a_{2},\dots,a_{n})$. This induces an equality $(\overline{a_{1}+ub},\overline{a_{2}},\dots,\overline{a_{n}})=(\overline{a_{1}+vb},\overline{a_{2}},\dots,\overline{a_{n}})$ of ideals of $\overline{R}$. Now by [10, Lemma 3.7] we get $\overline{u}=\overline{v}$ in $k$. ∎

The following example shows that a non-Gorenstein local ring that admits a totally reflexive module with minimal multiplicity, may also admit infinitely many totally reflexive modules with non-minimal multiplicities; see Definition 3.5.

Example 4.7**.**

Let $(R,\mathfrak{m},k)$ and $I$ be same as Example 3.11. Assume $k$ is an infinite field. Note that since $\dim R=0$ and $\mathfrak{m}^{2}(R/I)=0$, $R/I$ has minimal multiplicity, and so $R/I$ is a G-regular ring by Corollary 3.6. Since $(0:_{R}I)=(xy)\subseteq I$, $\operatorname{s}(I+(0:_{R}I))=\operatorname{s}(I)=2$. For each $\lambda\in k$, let $M_{\lambda}$ be an $R$-module with a minimal presentation

[TABLE]

Then by Proposition 4.6, $\{M_{\lambda}\}_{\lambda\in k}$ is an infinite family of non-free indecomposable and pairwise non-isomorphic totally reflexive $R$-modules. By Lemma 4.3 there exists an exact sequence $0\rightarrow R/I\rightarrow M_{\lambda}\rightarrow R/(xy)\rightarrow 0$. Since $\mathfrak{m}^{2}$ is not contained in $(xy)$, the surjection $M_{\lambda}\twoheadrightarrow R/(xy)$ shows that $\mathfrak{m}^{2}M_{\lambda}\neq 0$. Therefore $M_{\lambda}$ does not have minimal multiplicity.

Now we are ready to prove Theorem 1.3.

Proof of Theorem 1.3.

First, we observe that $\dim R/(I+(0:_{R}I))=\dim_{R}I/I^{2}$. In fact,

[TABLE]

Now, the theorem follows from Proposition 4.6, and the fact if $\dim R/(I+(0:_{R}I))\geq 2$ then $\operatorname{s}(I+(0:_{R}I))=\infty$. Note that $\nu_{R}(M_{n,u})=n+1$ follows from Lemma 4.3(2). ∎

Lemma 4.8**.**

Let $Q$ be a local ring. Suppose $x_{1},\dots,x_{r}$ and $y_{1},\dots,y_{r}$ are $Q$-regular sequences and set $I=(x_{1},\dots,x_{r})$ and $J=(y_{1},\dots,y_{r})$. Choose any regular sequence $f_{1},\dots,f_{r}\in IJ$ and set $R=Q/(f_{1},\dots,f_{r})$. Then the image of the ideal $I$ (respectively $J$) in $R$ is a quasi-complete intersection ideal of grade zero.

Proof.

Since $(f_{1},\dots,f_{r})$ and $I$ both are ideals of same grade $r$, $\operatorname{grade}IR=0$. Since $(f_{1},\dots,f_{r})$ and $I$ both are complete intersection ideals and $(f_{1},\dots,f_{r})\subset I$, $I/(f_{1},\dots,f_{r})$ is a quasi-complete intersection ideal of $R$; see [7, 8.9]. ∎

Corollary 4.9**.**

Let $(Q,\mathfrak{n},k)$ be a G-regular local ring, and let $x_{1},\dots,x_{r}\in\mathfrak{n}\setminus\mathfrak{n}^{2}$ be a regular sequence. Assume $\dim Q\geq r+2$. Then for any choice of a regular sequence $f_{1},\dots,f_{r}\in(x_{1},\dots,x_{r})^{2}$ and for any integer $n\geq 1$, there exists a family $\{M_{n,u}\}_{u\in k}$ of pairwise non-isomorphic indecomposable totally reflexive $R=Q/(f_{1},\dots,f_{r})$-modules such that $\nu_{R}(M_{n,u})=n+1$ and $\operatorname{cx}_{R}(M_{n,u})\leq r$.

Proof.

Set $I=(x_{1},\dots,x_{r})R$. It follows from Lemma 4.8 and Remark 2.5 that $I$ is a quasi-Gorenstein ideal of $R$ with $\operatorname{grade}I=0$. Also, by Proposition 2.8, $R/I\cong Q/(x_{1},\dots,x_{r})$ is a $G$-regular ring. Note that since $f_{1},\dots,f_{r}\in(x_{1},\dots,x_{r})^{2}$, one has $(0:_{R}I)\subset I$. Therefore ${\dim_{R}I/I^{2}=}\dim R/(I+(0:_{R}I))=\dim R/I\geq 2$. Now the first part follows from Theorem 1.3.

For the last part, consider the exact sequence $0\to(R/I)^{n}\to M_{n,u}\to R/(0:_{R}I)\to 0$. Since $R/I\cong(0:_{R}I)=\Omega R/(0:_{R}I)$, $\Omega^{j}(R/(0:_{R}I))\cong\Omega^{j-1}(R/I)$, for all $j\geq 1$. Hence we have $\operatorname{cx}_{R}(M_{n,u})\leq\operatorname{cx}_{R}(R/I)$. Note that $P^{R}_{R/I}(t)=\frac{1}{(1-t)^{r}}$; see [7, Theorem 6.2]. This implies that $\operatorname{cx}_{R}(R/I)\leq r$. ∎

We need the following lemma for Example 4.11.

Lemma 4.10**.**

Let $S=k[x_{0},\dots,x_{n}]$ be a polynomial ring over a field $k$ with $n\geq 3$. Then $\mathfrak{c}=(x_{i}^{2}-x_{i-1}x_{i+1}\mid i=1,\dots,n-1)$ is a complete intersection ideal.

Proof.

Since $\mathfrak{c}$ is generated by $n-1$ elements, $\operatorname{grade}\mathfrak{c}=\operatorname{ht}\mathfrak{c}\leq n-1$. We have $S/\mathfrak{c}+(x_{0})\cong k[x_{1},\dots,x_{n}]/(x_{1}^{2},x_{2}^{2}-x_{1}x_{3},x_{3}^{2}-x_{2}x_{4},\dots,x_{n-1}^{2}-x_{n-2}x_{n})$, and easily see that in this ring $(x_{i})^{2^{i}}=0$ for all $1\leq i\leq n-1$. Hence $S/\mathfrak{c}+(x_{0},x_{n})$ is Artinian, which shows $\dim S/\mathfrak{c}\leq 2$. Thus $\operatorname{grade}\mathfrak{c}=n-1$. ∎

In the following examples, first, we construct a class of local rings satisfying in the assumptions of Theorem 1.3. Next, in (2) we construct a quasi-Gorenstein ideal containing $\mathfrak{m}^{2}$ which is not a quasi-complete intersection ideal. As a result the base ring satisfy in the assumptions of Theorem 1.2(1) but not Corollary 3.10. Finally, in (3) we give an example of $3$-dimensional local ring that satisfy in the assumptions of Theorems 1.2 and 1.3, and Corollary 3.10.

Example 4.11**.**

Let $Q=k[[x_{0},\dots,x_{n}]]$ be a formal power series ring over a field $k$ with $n\geq 3$. Then $\mathfrak{c}=(x_{i}^{2}-x_{i-1}x_{i+1}\mid i=1,\dots,n-1)$ is a complete intersection ideal of $Q$ by Lemma 4.10. Let $\mathfrak{p}$ be the ideal of $Q$ generated by the $2\times 2$ minors of the matrix $\left(\begin{smallmatrix}x_{0}&\cdots&x_{n-1}\\ x_{1}&\cdots&x_{n}\end{smallmatrix}\right)$. Clearly $\mathfrak{p}$ contains $\mathfrak{c}$, and it is well known that $\mathfrak{p}$ is a Cohen-Macaulay prime ideal of height $n-1$; see [14, Theorem 6.4]. Since $x_{1}\dots x_{n-1}\in(\mathfrak{c}:_{Q}\mathfrak{p})\backslash\mathfrak{p}$ and $\mathfrak{p}$ is prime, $\operatorname{grade}_{Q}(\mathfrak{p}+({\mathfrak{c}}:_{Q}\mathfrak{p}))\geq n$. Hence $\mathfrak{p}$ is geometrically linked to $(\mathfrak{c}:_{Q}\mathfrak{p})$, and therefore the quotient $A=Q/(\mathfrak{p}+(\mathfrak{c}:_{Q}\mathfrak{p}))$ is a $1$-dimensional Gorenstein ring; see [28] and [32, Propositions 1.2 and 1.3].

1. (1)

Then $J=(x_{1},\dots,x_{n})A$. Then $A/J\cong k[[x_{0}]]$ is a discrete valuation ring. In particular, $J$ is a Gorenstein ideal of $A$ of grade zero. Hence $R=A[[y,z,w,v]]/(y^{2},yz,z^{2})$ is a $3$-dimensional Cohen-Macaulay non-Gorenstein local ring, $I=JR=(x_{1},\dots,x_{n})R$ is a quasi-Gorenstein ideal of $R$ of grade zero. Also $R/I$ is G-regular and $I/I^{2}=(J/J^{2})\otimes_{A}R$ has dimension at least $2$. Therefore Theorem 1.3(b) applies for $R$ and $I$. Thus for each integer $s\geq 1$ there exists a family $\{M_{s,u}\}_{u\in k}$ of pairwise non-isomorphic indecomposable totally reflexive $R$-modules such that $\nu_{R}(M_{s,u})=s+1$. 2. (2)

Let $n=4$, and let $\mathfrak{n}=(x_{0},\dots,x_{4})A$ be the maximal ideal of $A$. Since $A$ is Gorenstein, $\mathfrak{n}$ is a Gorenstein ideal of $A$ of grade $1$. Hence $R=A[[y,z]]/(y^{2},yz,z^{2})$ is a $1$-dimensional Cohen-Macaulay local ring of type $2$, and $I=\mathfrak{n}R=(x_{0},\dots,x_{4})R$ is a quasi-Gorenstein ideal of grade $1$ containing $\mathfrak{m}^{2}$, where $\mathfrak{m}=(x_{0},\dots,x_{4},y,z)R$ is the maximal ideal of $R$. We have $\nu_{R}(\mathfrak{m})=7$ and $\nu_{R}(I)=5$. The second coefficient of the formal power series $\frac{1-t^{2}}{(1-t)^{5}(1-2t)}$ is 28 while by using Macaulay2, we get $\beta^{R}_{2}(k)=33$. Hence the equality in Corollary 3.10 does not hold, which implies that $I$ is not a quasi-complete intersection ideal of $R$. This shows that one can not relax the hypothesis of $I$ being a quasi-Gorenstein ideal in Corollary 3.10. 3. (3)

Let $n=3$ and $R=(Q/\mathfrak{c})[[y,z,w]]/(yz,zw,wy)=\displaystyle\frac{k[[x_{0},x_{1},x_{2},x_{3},y,z,w]]}{(x_{1}^{2}-x_{0}x_{2},x_{2}^{2}-x_{1}x_{3},yz,zw,wy)}.$ Then $R$ is a Cohen-Macaulay non-Gorenstein local ring of dimension $3$. One has $\mathfrak{q}=(0:_{R}\mathfrak{p})=(x_{1},x_{2})R$. The ideal $\mathfrak{q}$ is quasi-complete intersection; see Lemma 4.8. We have $\operatorname{grade}\mathfrak{q}=0$, $\mathfrak{p}\cap\mathfrak{q}=0$ and $\dim R/\mathfrak{p}+\mathfrak{q}=2$. Hence Theorem 1.3(a) applies for $R$. Therefore for each integer $s\geq 1$ there exists a family $\{M_{s,u}\}_{u\in k}$ of pairwise non-isomorphic indecomposable totally reflexive $R$-modules such that $\nu_{R}(M_{s,u})=s+1$. Moreover, the ideal $J=(x_{0},x_{1},x_{2},x_{3},y+z+w)R$ contains $\mathfrak{q}$ and $J/\mathfrak{q}$ is a complete-intersection ideal of $R/\mathfrak{q}$. Hence $J$ is also a quasi-complete intersection ideal of $R$. As $J$ contains $\mathfrak{m}^{2}$, the type of $R$ is $\nu_{R}(\mathfrak{m})-\nu_{R}(J)=2$, and $P^{R}_{k}(t)=\frac{(1-t^{2})^{3}}{(1-t)^{5}(1-2t)}$ by Corollary 3.10.

Acknowledgments**.**

The authors thank Olgur Celikbas for reading the article and giving comments. They thank Sean Sather-Wagstaff for suggesting 3.3. Finally, they are very grateful to the referee for many valuable suggestions and helpful comments.