# Maximal Cohen-Macaulay tensor products

Olgur Celikbas  Department of Mathematics
West Virginia University
Morgantown, WV 26506 U.S.A<EMAIL_ADDRESS>and  Arash Sadeghi
School of Mathematics, Institute for Research in Fundamental Sciences (IPM),
P.O. Box:19395-5746, Tehran, Iran<EMAIL_ADDRESS>
###### Abstract.

In this paper we are concerned with the following question: if the tensor
product of finitely generated modules $M$ and $N$ over a local complete
intersection domain is maximal Cohen-Macaulay, then must $M$ or $N$ be a
maximal Cohen-Macaulay? Celebrated results of Auslander, Lichtenbaum, and
Huneke and Wiegand, yield affirmative answers to the question when the ring
considered has codimension zero or one, but the question is very much open for
complete intersection domains that have codimension at least two, even open
for those that are one-dimensional, or isolated singularities. Our argument
exploits Tor-rigidity and proves the following, which seems to give a new
perspective to the aforementioned question: if $R$ is a complete intersection
ring which is an isolated singularity such that
$\dim(R)>\operatorname{codim}(R)$, and the tensor product $M\otimes_{R}N$ is
maximal Cohen-Macaulay, then $M$ is maximal Cohen-Macaulay if and only if $N$
is maximal Cohen-Macaulay.

###### Key words and phrases:

Maximal Cohen-Macaulay modules, tensor products, torsion, vanishing of Ext and
Tor.

###### 2000 Mathematics Subject Classification:

13D07, 13C14, 13C15

Sadeghi’s research was supported by a grant from the Institute for Research in
Fundamental Sciences (IPM)

##  1\. Introduction

Throughout $R$ denotes a commutative Noetherian local ring with unique maximal
ideal $\mathfrak{m}$ and residue field $k$. Furthermore
$\operatorname{\operatorname{\mathsf{mod}}}R$ denotes the category of all
finitely generated $R$-modules. For the standard, unexplained notations and
basic terminology, we refer the reader to [3, 5, 10].

It is well-known that the tensor product $M\otimes_{R}N$ of two nonzero
modules $M,N\in\operatorname{\operatorname{\mathsf{mod}}}R$ tends to have
torsion in general, even if $M$ and $N$ are ideals, or maximal Cohen-Macaulay
modules. Therefore one expects that the modules $M$ and $N$ should enjoy
strong homological properties in the rare case that their tensor product has
no torsion. The study of such situation was initiated by Auslander [2] and
Lichtenbaum [26] over regular local rings, and continued by Huneke and Wiegand
[21] over hypersurface rings. Their results easily lead one to raise the
following question, which has an affirmative answer when $c=0$ and $c=1$ by
[2, 3.1] and [21, 2.7] 111 This applies since the ring in Question 1.1 is a
domain; see the erratum [23]., respectively; see also [13, 14, 15].

######  Question 1.1.

Let $R$ be a local complete intersection domain of codimension $c$. If
$M,N\in\operatorname{\operatorname{\mathsf{mod}}}R$ are nonzero modules such
that $M\otimes_{R}N$ satisfies _Serre’s condition_ $(S_{c+1})$, then must $M$
or $N$ satisfy $(S_{c+1})$?

Recall, in our context, a module
$M\in\operatorname{\operatorname{\mathsf{mod}}}R$ satisfies $(S_{n})$ if it is
an $n$th syzygy module; see [20, 3.8]. In this paper we consider Question 1.1
for the case where the ring is an isolated singularity and the tensor product
satisfies Serre’s condition $(S_{n})$ for each nonnegative integer $n$. In
other words, we consider the case where $R$ is a complete intersection domain
such that $R_{\mathfrak{p}}$ is regular for each prime ideal
$\mathfrak{p}\in\operatorname{Spec}(R)-\\{\mathfrak{m}\\}$, and
$M\otimes_{R}N$ is maximal Cohen-Macaulay. To our best knowledge, Question
1.1, even under these very strong hypotheses, is open. Motivated by this fact,
we propose:

######  Conjecture 1.2.

Let $R$ be a local complete intersection domain, and let
$M,N\in\operatorname{\operatorname{\mathsf{mod}}}R$. Assume $R$ is an isolated
singularity. If $M\otimes_{R}N$ is maximal Cohen-Macaulay, then $M$ or $N$ is
maximal Cohen-Macaulay.

Conjecture 1.2, similar to Question 1.1, holds when the codimension of the
ring is at most one; see [2, 3.1] and [21, 3.1]. It is worth pointing out,
when the ring in question has dimension one, we have the following basic
problem, which, quite interestingly, still remains open in general; see [17,
2.10].

######  Conjecture 1.3.

Let $R$ be a one-dimensional local complete intersection domain, and let
$M,N\in\operatorname{\operatorname{\mathsf{mod}}}R$. If $M\otimes_{R}N$ is
torsion-free, then $M$ or $N$ is torsion-free.

Conjecture 1.3 is an easy consequence of a conjecture of Dao [19, 4.2] which,
in particular, claims that all finitely generated modules over one-dimensional
local complete intersection domains are Tor-rigid [2]. We should also mention
that Conjectures 1.2 and 1.3 fail if the ring considered is not assumed to be
a complete intersection. Constapel [18] constructed a one-dimensional
Gorenstein local domain $R$ that is not a complete intersection, and
$M,N\in\operatorname{\operatorname{\mathsf{mod}}}R$ with $M\otimes_{R}N$ is
torsion-free, but both $M$ and $N$ have torsion.

The main purpose of this paper is to make progress pertaining to Conjecture
1.2. Our argument yields:

######  Theorem 1.4.

Let $R$ be a local complete intersection with
$\dim(R)>\operatorname{codim}(R)$, and let
$M,N\in\operatorname{\operatorname{\mathsf{mod}}}R$. Assume $R$ is an isolated
singularity. If $M\otimes_{R}N$ is maximal Cohen-Macaulay, then $M$ is maximal
Cohen-Macaulay if and only if $N$ is maximal Cohen-Macaulay. In other words,
$M\otimes_{R}N$ cannot be maximal Cohen-Macaulay in case $M$ is maximal Cohen-
Macaulay but $N$ is not maximal Cohen-Macaulay.

It is easy to see that the condition $\dim(R)>\operatorname{codim}(R)$ is
necessary for Theorem 1.4: if $R=k[[x,y]]/(xy)$, $M=R/xR$ and $N=R/(x^{2})R$,
then $M\otimes_{R}N\cong M$ is maximal Cohen-Macaulay, but $N$ is not.

In section 2, we state our main result, give a proof for Theorem 1.4 and
discuss several consequences of our argument. Section 3 is devoted to various
preliminary results and a proof of our main result; see Theorems 2.1 and 3.7.

##  2\. Statement of the Main Result and corollaries

In the following, for a given module
$M\in\operatorname{\operatorname{\mathsf{mod}}}R$,
$\operatorname{\textnormal{G-dim}}-dim_{R}(M)$ and
$\operatorname{\textnormal{CI-dim}}_{R}(M)$ denote the _Gorenstein_ dimension
[3] and the _complete intersection_ dimension [7] of $M$, respectively.
Moreover, $\operatorname{Tr}M$ denotes the Auslander _transpose_ of $M$; see
[3].

The next theorem is our main result. In this section we discuss some of its
consequences and defer its proof to section 3; see Theorem 3.7.

######  Theorem 2.1.

Let $R$ be a local complete intersection ring and let
$M,N\in\operatorname{\operatorname{\mathsf{mod}}}R$. Assume:

  1. (i)

$\operatorname{depth}_{R}(M\otimes_{R}N)\geq\operatorname{depth}_{R}(N)$.

  2. (ii)

$\operatorname{length}(\operatorname{Tor}_{i}^{R}(M,N))<\infty$ for all $i\gg
0$.

  3. (iii)

$M_{\mathfrak{p}}$ is maximal Cohen-Macaulay for all
$\mathfrak{p}\in\operatorname{Supp}_{R}(M\otimes_{R}N)-\\{\mathfrak{m}\\}$.

  4. (iv)

If $\operatorname{depth}_{R}(N)\leq\operatorname{codim}(R)$, assume further
$\operatorname{Tor}_{i}^{R}(M,N)=0$ for all
$i=1,\ldots,\operatorname{codim}(R)-\operatorname{depth}_{R}(N)+1$.

Then $\operatorname{Tor}_{i}^{R}(M,N)=0$ for all $i\geq 1$.

Theorem 1.4, stated in the introduction, is a special case of the following
consequence of Theorem 2.1.

######  Corollary 2.2.

Let $R$ be a local complete intersection ring and let
$M,N\in\operatorname{\operatorname{\mathsf{mod}}}R$. Assume:

  1. (a)

$M\otimes_{R}N$ is maximal Cohen-Macaulay.

  2. (b)

$\operatorname{length}(\operatorname{Tor}_{i}^{R}(M,N))<\infty$ for all $i\gg
0$.

  3. (c)

If $\dim(R)\leq\operatorname{codim}(R)$, assume further
$\operatorname{Tor}_{i}^{R}(M,N)=0$ for all
$i=1,\ldots,\operatorname{codim}(R)-\dim(R)+1$.

Then the following conditions are equivalent:

  1. (i)

$M$ is maximal Cohen-Macaulay.

  2. (ii)

$N$ is maximal Cohen-Macaulay.

  3. (iii)

$\operatorname{Tor}_{i}^{R}(M,N)=0$ for all $i\geq 1$.

  4. (iv)

$\operatorname{Ext}^{i}_{R}(M,N)=0$ for all $i\geq 1$.

  5. (v)

$\operatorname{Ext}^{i}_{R}(N,M)=0$ for all $i\geq 1$.

Prior giving a proof of Corollary 2.2, we record:

######  2.3.

([1, 3.5]) If $R$ is a local ring and
$M,N\in\operatorname{\operatorname{\mathsf{mod}}}R$ are nonzero such that
$\operatorname{Tor}_{i}^{R}(M,N)=0$ for all $i\geq 1$ and
$\operatorname{\textnormal{CI-dim}}_{R}(M)<\infty$, then
$\operatorname{depth}_{R}(M)+\operatorname{depth}_{R}(N)=\operatorname{depth}(R)+\operatorname{depth}_{R}(M\otimes_{R}N)$.

######  2.4.

If $R$ is a local ring and $M,N\in\operatorname{\operatorname{\mathsf{mod}}}R$
such that $\operatorname{\textnormal{CI-
dim}}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}})=0$ for each
$\mathfrak{p}\in\operatorname{Spec}R-\\{\mathfrak{m}\\}$, then
$\operatorname{length}(\operatorname{Tor}_{i}^{R}(M,N))<\infty$ for all $i\gg
0$ if and only if
$\operatorname{length}(\operatorname{Tor}_{i}^{R}(M,N))<\infty$ for all $i\geq
1$; see [6, 4.9].

###### A proof of Corollary 2.2.

It suffices to assume (i) and prove (ii) and (iii) hold; see [6, 4.8, 6.1 and
6.3].

Assume (i). Then we have
$\operatorname{depth}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}})+\operatorname{depth}_{R_{\mathfrak{p}}}(N_{\mathfrak{p}})=\operatorname{depth}R_{\mathfrak{p}}+\operatorname{depth}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}}\otimes_{R_{\mathfrak{p}}}N_{\mathfrak{p}})$;
see 2.3 and 2.4. Consequently $N_{\mathfrak{p}}$ is maximal Cohen-Macaulay for
all
$\mathfrak{p}\in\operatorname{Supp}(M)\cap\operatorname{Supp}(N)-\\{\mathfrak{m}\\}$.
Now, by swaping the role of $N$ with that of $M$, we see from Theorem 2.1 that
$\operatorname{Tor}_{i}^{R}(M,N)=0$ for all $i\geq 1$. Thus 2.3 shows that $N$
is maximal Cohen-Macaulay. ∎

The next two corollaries are reformulations of Corollary 2.2; concerning the
first one, see also [24, 1.2].

######  Corollary 2.5.

Let $R$ be a local complete intersection. Assume
$\dim(R)>\operatorname{codim}(R)$ and $R$ is an isolated singularity. If
$M\in\operatorname{\operatorname{\mathsf{mod}}}R$ with
$\operatorname{pd}_{R}(M)=\infty$, then
$\Omega^{i}_{R}(M)\otimes_{R}\Omega^{j}_{R}(M)$ is not maximal Cohen-Macaulay
for each $i\geq 0$ and each $j\geq\dim(R)$.

######  Corollary 2.6.

Let $R$ be a local complete intersection with
$\dim(R)>\operatorname{codim}(R)$. If $J\subseteq I$ are ideals of $R$ such
that $\operatorname{length}(\operatorname{Tor}_{i}^{R}(R/I,R/J))<\infty$ for
all $i\gg 0$, and $R/I$ is maximal Cohen-Macaulay, then $R/J$ is maximal
Cohen-Macaulay. In particular, if $M$ is a maximal Cohen-Macaulay $R$-module
such that $M_{\mathfrak{p}}$ is free over $R_{\mathfrak{p}}$ for all
$\mathfrak{p}\in\operatorname{Spec}(R)-\\{\mathfrak{m}\\}$, then $R/J$ is
maximal Cohen-Macaulay for all ideals $J$ of $R$ with $JM=0$.

The example mentioned in the introduction also shows that the hypothesis
$\dim(R)>\operatorname{codim}(R)$ in Corollary 2.5 and 2.6 is necessary: if
$R=k[[x,y]]/(xy)$, $M=R/I$ and $N=R/J$, where $I=xR$ and $J=(x^{2})R$, then
$\operatorname{pd}_{R}(M)=\infty$ but
$\Omega^{1}_{R}(M)\otimes_{R}\Omega^{1}_{R}(M)$ is maximal Cohen-Macaulay.

We can also show that the hypothesis concerning the dimension and codimension
in Corollary 2.2 is necessary. Let us first recall two elementary facts:

######  2.7.

([3, 2.8]) Let $M\in\operatorname{\operatorname{\mathsf{mod}}}R$ and let
$n\geq 0$ be an integer. Then there is an exact sequences of functors:

$0\longrightarrow\operatorname{Ext}^{1}_{R}(\operatorname{Tr}\Omega^{n}M,-)\longrightarrow\operatorname{Tor}_{n}^{R}(M,-)\longrightarrow\operatorname{Hom}_{R}(\operatorname{Ext}^{n}_{R}(M,R),-)\longrightarrow\operatorname{Ext}^{2}_{R}(\operatorname{Tr}\Omega^{n+1}M,-).$

When $M\in\operatorname{\operatorname{\mathsf{mod}}}R$ and
$\operatorname{\textnormal{G-dim}}-dim_{R}(M)<\infty$, we denote by
$\widehat{\operatorname{Ext}}^{n}_{R}(M,N)$ (respectively, by
$\widehat{\operatorname{Tor}}^{n}_{R}(M,N)$) the $n$th _Tate cohomology_
(respectively, the $n$th _Tate homology_) of $M$ and $N$; see, for example,
[8].

######  2.8.

([11, 4.7]) If $R$ is a a local complete intersection ring with
$\operatorname{codim}(R)=\dim(R)\geq 1$, and $M$ and $N$ are maximal Cohen-
Macaulay modules such that $\operatorname{length}(M\otimes_{R}N)<\infty$, then
$\operatorname{Tor}_{i}^{R}(M,N)\neq 0$ if and only if $i$ is a nonnegative
even integer.

######  Example 2.9.

Let $R=\mathbb{C}[\\![x_{1},x_{2},y_{1},y_{2}]\\!]/(x_{1}y_{1},x_{2}y_{2})$ be
the formal power series ring, $M=R/(x_{1},x_{2})$ and $N=R/(y_{1},y_{2})$.
Then $R$ is a complete intersection with $\operatorname{codim}(R)=\dim(R)=2$.
Notice $M$ and $N$ are maximal Cohen-Macaulay with $M\cong M^{\ast}$,
$M\otimes_{R}N\cong k$ and
$\widehat{\operatorname{Ext}}^{0}_{R}(M,N)=\widehat{\operatorname{Ext}}^{1}_{R}(M,N)=0$;
see [9, 3.17]. Consider a maximal Cohen-Macaulay $R$-module $L$ with
$M\cong\Omega^{1}_{R}(L)$. Then, for each $i\geq 1$, one has:

$\begin{array}[]{rl}{}\operatorname{Ext}^{i}_{R}(\operatorname{Tr}L,N)\cong\widehat{\operatorname{Ext}}^{i}_{R}(\operatorname{Tr}L,N)\cong\widehat{\operatorname{Ext}}^{i-2}_{R}(L^{*},N)\cong\widehat{\operatorname{Ext}}^{i-1}_{R}(M,N).\end{array}$

In particular, we have
$\operatorname{Ext}^{1}_{R}(\operatorname{Tr}L,N)=\operatorname{Ext}^{2}_{R}(\operatorname{Tr}L,N)=0$.
This implies $L\otimes_{R}N\cong\operatorname{Hom}_{R}(L^{\ast},N)$ and hence
$L\otimes_{R}N$ is maximal Cohen-Macaulay; see 2.7 and [10, 1.4.18].

We have, since $\operatorname{Tor}_{i}^{R}(M,N)\neq 0$ if and only if $i$ is a
nonnegative even integer, that $\operatorname{Tor}_{2i}^{R}(L,N)=0$ and
$\operatorname{Tor}_{2i+1}^{R}(L,N)\neq 0$ for all $i\geq 1$; see 2.8.
Furthermore, as $M\cong\Omega^{1}_{R}(L)$ and $L\otimes_{R}N$ is maximal
Cohen-Macaulay, it is easy to see, using the depth lemma, that
$\operatorname{Tor}_{1}^{R}(L,N)\neq 0$.

Consequently $(L,N)$ is a pair of maximal Cohen-Macaulay $R$-modules such that
$L\otimes_{R}N$ is maximal Cohen-Macaulay,
$\operatorname{length}(\operatorname{Tor}_{i}^{R}(L,N))<\infty$ for all $i\geq
1$ and $\operatorname{Tor}_{i}^{R}(L,N)=0$ if and only if $i$ is a nonnegative
even integer; cf. Corollary 2.2.

##  3\. A proof of the main result

This section is devoted to a proof of Theorem 2.1, which requires some
preparation. We skip the proof of the next elementary remark as one can prove
it easily following the argument of [16, 2.2].

######  3.1.

Let $M,N\in\operatorname{\operatorname{\mathsf{mod}}}R$ and let $n\geq 1$ be
an integer.

  1. (i)

Assume $n\geq 3$. If
$\operatorname{depth}_{R}(\operatorname{Hom}_{R}(M,N))\geq n$,
$\operatorname{depth}_{R}(N)\geq n-1$ and
$\operatorname{length}(\operatorname{Ext}^{i}_{R}(M,N))<\infty$ for all
$i=1,\ldots,n-2$, then $\operatorname{Ext}^{i}_{R}(M,N)=0$ for all
$i=1,\ldots,n-2$.

  2. (ii)

If $\operatorname{depth}_{R}(M\otimes_{R}N)\geq n$,
$\operatorname{depth}_{R}(N)\geq n-1$ and
$\operatorname{length}(\operatorname{Ext}^{i}_{R}(\operatorname{Tr}M,N))<\infty$
for all $i=1,\ldots,n$, then
$\operatorname{Ext}^{i}_{R}(\operatorname{Tr}M,N)=0$ for all $i=1,\ldots,n$.

  3. (iii)

If $\operatorname{Ext}^{i}_{R}(\operatorname{Tr}M,N)=0$ for all
$i=1,\ldots,n$, then
$\operatorname{depth}_{R}(M\otimes_{R}N)\geq\min\\{n,\operatorname{depth}_{R}(N)\\}$.

######  3.2.

([6, 4.7] and [28, 4.1]) Let
$M,N\in\operatorname{\operatorname{\mathsf{mod}}}R$ be modules. Assume
$\operatorname{\textnormal{G-dim}}-dim_{R}(M)<\infty$. Assume further
$\operatorname{\textnormal{CI-dim}}_{R}(M)<\infty$ or
$\operatorname{\textnormal{CI-dim}}_{R}(N)<\infty$. Then one has:

${}\widehat{\operatorname{Ext}}^{i}_{R}(M,N)=0\text{ for all }i\gg
0\Longleftrightarrow\widehat{\operatorname{Ext}}^{i}_{R}(M,N)=0\text{ for all
}i\ll 0\Longleftrightarrow\widehat{\operatorname{Ext}}^{i}_{R}(M,N)=0\text{
for all }i\in\mathbb{Z}.$

######  3.3.

Let $R$ be a local ring and let
$M,N\in\operatorname{\operatorname{\mathsf{mod}}}R$. Assume
$\operatorname{\textnormal{CI-dim}}_{R}(N)<\infty$. Then,

  1. (i)

If $\operatorname{\textnormal{G-dim}}-dim_{R}(M)<\infty$ and
$\operatorname{Ext}^{i}_{R}(M,N)=0$ for all $i\gg 0$, then
$\operatorname{\textnormal{G-dim}}-dim_{R}(M)=\sup\\{i\in\mathbb{Z}\mid\operatorname{Ext}^{i}_{R}(M,N)\neq
0\\}$.

  2. (ii)

If $M$ is totally reflexive, then
$\operatorname{Ext}^{i}_{R}(\operatorname{Tr}M,N)=0$ for all $i\geq 1$ if and
only if $\operatorname{Tor}_{i}^{R}(M,N)=0$ for all $i\geq 1$.

###### Proof.

Part (i) follows from 3.2 and [28, 4.3]. Therefore we proceed to prove part
(ii). Note that $\operatorname{Tr}M$ is totally reflexive. Hence, by part (i)
and 3.2, we have:

(3.3.1) $\operatorname{Ext}^{i}_{R}(\operatorname{Tr}M,N)=0\text{ for all
}i\geq
1\Longleftrightarrow\widehat{\operatorname{Ext}}^{i}_{R}(\operatorname{Tr}M,N)=0\text{
for all }i\in\mathbb{Z}.$

Since $M^{*}\cong\Omega^{2}\operatorname{Tr}M$ up to projectives, we can make
use of [6, 4.4.7] and obtain:

(3.3.2) $\widehat{\operatorname{Tor}}_{i}^{R}(M,N)=0\text{ for all
}i\in\mathbb{Z}\Longleftrightarrow\widehat{\operatorname{Ext}}^{i}_{R}(\operatorname{Tr}M,N)=0\text{
for all }i\in\mathbb{Z}.$

On the other hand, by [6, 4.9], we know:

(3.3.3) $\widehat{\operatorname{Tor}}_{i}^{R}(M,N)=0\text{ for all
}i\in\mathbb{Z}\Longleftrightarrow\operatorname{Tor}_{i}^{R}(M,N)=0\text{ for
all }i\geq 1.$

Now the assertion follows from (3.3.1), (3.3.2) and (3.3.3). ∎

In passing we record a consequence of 3.1 and 3.3, which can also be used to
see the vanishing of $\operatorname{Tor}_{2i}^{R}(L,N)$ in Example 2.9; cf.
2.8.

######  3.4.

If $R$ is a a local complete intersection ring with
$\operatorname{codim}(R)=\dim(R)$, and $M$ and $N$ are maximal Cohen-Macaulay
modules such that $M\otimes_{R}N$ is maximal Cohen-Macaulay and
$\operatorname{length}(\operatorname{Tor}_{i}^{R}(M,N))<\infty$ for $i\gg 0$,
then $\operatorname{Tor}_{2i}^{R}(M,N)=0$ for all $i\geq 1$.

###### Proof.

Set $d=\operatorname{codim}(R)=\dim(R)$. Then it follows from 2.4, 3.1 and 3.3
that $\operatorname{Ext}^{i}_{R}(\operatorname{Tr}M,N)=0$ for all
$i=1,\ldots,d$. Therefore
$\widehat{\operatorname{Tor}}_{i}^{R}(M,N)\cong\widehat{\operatorname{Ext}}^{-i+1}_{R}(\operatorname{Tr}M,N)=0$
for all $i=-d+1,\ldots,0$; see [6, 4.4.7]. Now the assertion follows from [9,
3.9]. ∎

From now on, we assume the following setup throughout the rest of the paper.

######  Setup.

Assume $R=S/(\textbf{x})$, where $(S,\mathfrak{n})$ is a Gorenstein local ring
and $\textbf{x}=x_{1},\ldots,x_{c}$ is an $S$-regular sequence in
$\mathfrak{n}$ for some $c\geq 0$. Assume further
$M,N\in\operatorname{\operatorname{\mathsf{mod}}}R$ are nonzero modules such
that $\operatorname{\textnormal{CI-dim}}_{S}(N)<\infty$ and $N$ is Tor-rigid
over $S$. In particular we have that $\operatorname{\textnormal{CI-
dim}}_{R}(N)<\infty$; see [7, 1.12.2].

The following is a generalization of [29, 4.1], which plays an important role
in the proof of our main result, Theorem 2.1.

######  3.5.

Assume $n\geq 1$ and $v\geq 0$ are integers, and that the following conditions
hold.

  1. (i)

$M$ satisfies $(S_{v})$.

  2. (ii)

$\operatorname{depth}_{R}(N)\leq n+c+v$.

  3. (iii)

$\operatorname{Ext}^{i}_{R}(M,N)=0$ for all $i=n,\ldots,n+c$.

Then
$\operatorname{\textnormal{G-dim}}-dim_{R}(M)=\sup\\{i\in\mathbb{Z}\mid\operatorname{Ext}^{i}_{R}(M,N)\neq
0\\}\leq n-1$.

###### Proof.

We proceed by induction on $v$. Assume first the case where $v=0$ holds, and
suppose $v\geq 1$. Then consider the pushforward of $M$, i.e., an exact
sequence in $\operatorname{\operatorname{\mathsf{mod}}}R$ of the form

(3.5.1) $0\rightarrow M\rightarrow F\rightarrow M_{1}\rightarrow 0,$

where $F$ is free; see, for example [11, page 174]. It follows from (3.5.1)
that $M_{1}$ satisfies $(S_{v-1})$ and

(3.5.2)
$\operatorname{Ext}^{i}_{R}(M,N)\cong\operatorname{Ext}^{i+1}_{R}(M_{1},N)\text{
for all }i\geq 1.$

Therefore $\operatorname{Ext}^{i}_{R}(M_{1},N)=0$ for all
$i=n+1,\ldots,n+c+1$. So, by the induction hypothesis, we conclude that
$\operatorname{Ext}^{i}_{R}(M_{1},N)=0$ for all $i\geq n+1$. Now, in view of
(3.5.2), the claim follows from 3.2 and 3.3(i).

Next we proceed by induction on $c$ to establish the case where $v=0$.

If $c=0$, then $N$ is a Tor-rigid $R$-module and the assertion follows from
[12, 5.8(2)] and [1, 2.5]. Now let $c\geq 1$ and set
$Q=S/(x_{1},\ldots,x_{c-1})$. Therefore $R\cong Q/(x_{c})$. It follows from
[27, 11.66] that there is a long exact sequence of the form:

(3.5.3)
$\cdots\rightarrow\operatorname{Ext}^{i}_{R}(M,N)\rightarrow\operatorname{Ext}^{i}_{Q}(M,N)\rightarrow\operatorname{Ext}^{i-1}_{R}(M,N)\rightarrow\operatorname{Ext}^{i+1}_{R}(M,N)\rightarrow\cdots.$

It follows that $\operatorname{Ext}^{i}_{Q}(M,N)=0$ for all
$i=n+1,\ldots,n+c$, and so, by the induction hypothesis, we conclude that
$\operatorname{\textnormal{G-dim}}-dim_{Q}(M)=\sup\\{i\mid\operatorname{Ext}^{i}_{Q}(M,N)\neq
0\\}\leq n$. Thus, by (3.5.3), we have
$\operatorname{Ext}^{i-1}_{R}(M,N)\cong\operatorname{Ext}^{i+1}_{R}(M,N)$ for
all $i\geq n+1$. Since $c\geq 1$, it is clear that
$\operatorname{Ext}^{i}_{R}(M,N)=0$ for all $i\geq n$. Consequently the claim
follows from 3.2 and 3.3(i). ∎

######  3.6.

Set $t=\operatorname{depth}_{R}(N)$. Assume $t\geq c+1$ and
$\operatorname{Ext}^{i}_{R}(M,N)=0$ for all $i=1,\ldots,t$. Then $M$ is
maximal Cohen-Macaulay, and $\operatorname{Ext}^{i}_{R}(M,N)=0$ for all $i\geq
1$.

###### Proof.

If $t=c+1$, then the assertion follows from 3.5. Hence we assume $t\geq c+2$.
Pick a regular sequence $y_{1},\ldots,y_{t-c-1}$ on $N$, and set
$N_{j}=N/(y_{1},\ldots,y_{j})N$ for each $j=0,1,\ldots,t-c-1$, where
$N_{0}=N$. This yields, for each $j=0,1,\ldots,t-c-1$, a short exact sequence
of the form:

(3.6.1) $0\to N_{j}\overset{y_{j+1}}{\longrightarrow}N_{j}\longrightarrow
N_{j+1}\longrightarrow 0.$

Since $N$ is Tor-rigid over $S$, it follows from (3.6.1) and Nakayama’s lemma
that $N_{j}$ is Tor-rigid over $S$ for each $j=1,\ldots,t-c-1$. Moreover, for
each $j=0,1,\ldots,t-c-1$, (3.6.1) induces a long exact sequence of the form:

(3.6.2)
$\cdots\rightarrow\operatorname{Ext}^{i}_{R}(M,N_{j})\rightarrow\operatorname{Ext}^{i}_{R}(M,N_{j+1})\rightarrow\operatorname{Ext}^{i+1}_{R}(M,N_{j})\rightarrow\cdots.$

Setting $j=0$ and using our assumption, we see from (3.6.2) that
$\operatorname{Ext}^{i}_{R}(M,N_{1})=0$ for all $i=1,\ldots,t-1$. By repeating
this argument, we conclude:

$\operatorname{Ext}^{i}_{R}(M,N_{t-c-1})=0\text{ for all }i=1,\ldots,c+1.$

Notice $\operatorname{depth}_{R}(N_{t-c-1})=c+1$. Therefore, by using 3.5 for
the pair $(M,N_{t-c-1})$ and for the cases where $n=1$ and $v=0$, we conclude
$\operatorname{\textnormal{G-dim}}-dim_{R}(M)=\sup\\{i\mid\operatorname{Ext}^{i}_{R}(M,N_{t-c-1})\neq
0\\}\leq 0$. Thus, since $M$ is not zero, we deduce that $M$ is maximal Cohen-
Macaulay and $\operatorname{Ext}^{i}_{R}(M,N_{t-c-1})=0$ for all $i\geq 1$.
Now, proceding inductively and using Nakayama’s lemma, we see from (3.6.2)
that $\operatorname{Ext}^{i}_{R}(M,N)$ vanishes for all $i\geq 1$. ∎

We are now ready to give a proof of our main result. Note that, since modules
over regular local rings are Tor-rigid [2], the conditions in our setup hold
when $S$ is regular, i.e., when $R$ is a complete intersection ring. In
particular the result below contains Theorem 2.1.

######  Theorem 3.7.

Set $t=\operatorname{depth}_{R}(N)$ and assume the following conditions hold:

  1. (i)

$\operatorname{depth}_{R}(M\otimes_{R}N)\geq t$.

  2. (ii)

$\operatorname{length}(\operatorname{Tor}_{i}^{R}(M,N))<\infty$ for all $i\gg
0$.

  3. (iii)

$M_{\mathfrak{p}}$ is maximal Cohen-Macaulay for all
$\mathfrak{p}\in\operatorname{Supp}_{R}(M\otimes_{R}N)-\\{\mathfrak{m}\\}$.

  4. (iv)

If $t\leq c$, assume further that $\operatorname{Tor}_{i}^{R}(M,N)=0$ for all
$i=1,\ldots,c-t+1$.

Then $\operatorname{Tor}_{i}^{R}(M,N)=0$ for all $i\geq 1$.

###### Proof.

We start by recalling that $\operatorname{\textnormal{CI-dim}}_{R}(N)<\infty$.
It follows from 3.3 that
$\operatorname{length}(\operatorname{Tor}_{i}^{R}(M,N))<\infty$ and
$\operatorname{length}(\operatorname{Ext}^{i}_{R}(\operatorname{Tr}M,N))<\infty$,
for all $i\geq 1$. Hence, by 3.1(ii), we have:

(3.7.1) $\operatorname{Ext}^{i}_{R}(\operatorname{Tr}M,N)=0\text{ for all
}i=1,\ldots,t.$

Suppose $t\geq c+1$. Then (3.7.1) and 3.6 imply $\operatorname{Tr}M$ is
maximal Cohen-Macaulay and
$\operatorname{Ext}^{i}_{R}(\operatorname{Tr}M,N)=0$ for all $i\geq 1$. Thus
$M$ is totally reflexive, and the vanishing of
$\operatorname{Tor}_{i}^{R}(M,N)$ for all $i\geq 1$ follows from 3.3(ii).

Next assume $t\leq c$. Then, by assumption, we have
$\operatorname{Tor}_{i}^{R}(M,N)=0$ for all $i=1,\ldots,n$, where $n=c-t+1$.
We will proceed by induction on $n$.

First suppose $n=1$. We have, by [3, 2.21], a short exact sequence of modules
in $\operatorname{\operatorname{\mathsf{mod}}}R$ of the form

(3.7.2) $0\longrightarrow F\longrightarrow L\longrightarrow M\longrightarrow
0,$

where $F$ is a free $R$-module and $\operatorname{Ext}^{1}_{R}(L,R)=0$. Since
$\operatorname{Tor}_{1}^{R}(M,N)=0$, tensoring (3.7.2) with $N$, we obtain the
exact sequence:

(3.7.3) $0\longrightarrow N\otimes_{R}F\longrightarrow
N\otimes_{R}L\longrightarrow N\otimes_{R}M\longrightarrow 0.$

It follows from (3.7.2) that
$\operatorname{Tor}_{1}^{R}(L,N)=\operatorname{Tor}_{1}^{R}(M,N)=0$ and
$\operatorname{Tor}_{i}^{R}(L,N)=\operatorname{Tor}_{i}^{R}(M,N)$ for all
$i\geq 2$. In particular, by (2.7), we see:

(3.7.4) $\operatorname{Ext}^{1}_{R}(\operatorname{Tr}\Omega L,N)=0.$

It follows from (3.7.2) that $L_{\mathfrak{p}}$ is totally reflexive for all
$\mathfrak{p}\in\operatorname{Supp}_{R}(L\otimes_{R}N)-\\{\mathfrak{m}\\}$.
Therefore, since
$\operatorname{length}(\operatorname{Tor}_{i}^{R}(L,N))<\infty$ for all $i\geq
1$, we observe from 3.3(ii) that
$\operatorname{length}(\operatorname{Ext}^{i}_{R}(\operatorname{Tr}L,N))<\infty$
for all $i\geq 1$. It is easy to see $\operatorname{Tr}L$ is stably isomorphic
to $\Omega\operatorname{Tr}\Omega L$ because
$\operatorname{Ext}^{1}_{R}(L,R)=0$. Furthermore, by (3.7.3), we have
$\operatorname{depth}_{R}(L\otimes_{R}N)\geq\operatorname{depth}_{R}(N)$. Thus
3.1 gives:

(3.7.5) $\operatorname{Ext}^{i+1}_{R}(\operatorname{Tr}\Omega
L,N)\cong\operatorname{Ext}^{i}_{R}(\Omega\operatorname{Tr}\Omega
L,N)\cong\operatorname{Ext}^{i}_{R}(\operatorname{Tr}L,N)=0\text{ for all
}i=1,\ldots t.$

Consequently, by (3.7.4) and (3.7.5), we obtain:

(3.7.6) $\operatorname{Ext}^{i}_{R}(\operatorname{Tr}\Omega L,N)=0\text{ for
all }i=1,\ldots,t+1.$

We are now in a position to conclude that $\operatorname{Tor}_{i}^{R}(N,M)=0$
for all $i\geq 1$ by using 3.5 and 3.3(ii).

Now we consider the case where $n\geq 2$. Set $T=S/(x_{1},\cdots,x_{c-1})$ so
that $R\cong T/(x_{c})$. Consider an exact sequence of modules in
$\operatorname{\operatorname{\mathsf{mod}}}T$ of the form

(3.7.7) $0\rightarrow X\rightarrow P\rightarrow M\rightarrow 0,$

where $P$ is a free $T$-module. Tensoring (3.7.7) with $N$ over $T$, we have
the following exact sequence

(3.7.8) $0\rightarrow\operatorname{Tor}_{1}^{T}(M,N)\rightarrow
X\otimes_{T}N\rightarrow P\otimes_{T}N\rightarrow M\otimes_{T}N\rightarrow 0.$

The change of rings spectral sequence [27, 11.64] yields a long exact sequence
of the form

$\setcounter{MaxMatrixCols}{12}\begin{CD}&&&&&&&&\\\
&&&&\vdots&&\vdots&&\vdots\\\ \ \
&&&&\operatorname{Tor}_{1}^{R}(M,N)@>{}>{}>\operatorname{Tor}_{2}^{T}(M,N)@>{}>{}>\operatorname{Tor}_{2}^{R}(M,N)@>{}>{}>&\\\
\ \
&&&&M\otimes_{R}N@>{}>{}>\operatorname{Tor}_{1}^{T}(M,N<EMAIL_ADDRESS>\end{CD}$

Since $\operatorname{Tor}_{i}^{R}(M,N)=0$ for all $i=1,\ldots,n$, it follows
$M\otimes_{R}N\cong\operatorname{Tor}_{1}^{T}(M,N)$ and that
$\operatorname{Tor}_{i}^{T}(M,N)=0$ for all $i=2,\ldots,n$. Hence
$\operatorname{Tor}_{i}^{T}(X,N)=0$ for all $i=1,\ldots,n-1$. Similarly, it is
easy to check that the hypotheses (i), (ii) and (iii) hold for the pair
$(X,N)$ over the ring $T$. Therefore, by the induction hypothesis, we have
$\operatorname{Tor}_{i}^{T}(X,N)=0$ for all $i\geq 1$, i.e.,
$\operatorname{Tor}_{i}^{T}(M,N)=0$ for all $i\geq 2$. This implies
$\operatorname{Tor}_{i}^{R}(M,N)\cong\operatorname{Tor}_{i+2}^{R}(M,N)$ for
all $i\geq 1$. Since, by assumption, $\operatorname{Tor}_{1}^{R}(M,N)$ and
$\operatorname{Tor}_{2}^{R}(M,N)$ vanish, the claim follows. ∎

We finish this section by recording a consequence of Theorem 3.7; cf.
Corollary 2.2. In the following $\operatorname{cx}_{R}(M)$ denotes the
_complexity_ of $M$; see [4].

######  3.8.

Assume, in our setup, the ring $(S,\mathfrak{n})$ is an unramified or equi-
characteristic regular local ring, and the regular sequence
$\textbf{x}=x_{1},\ldots,x_{c}$ is contained in $\mathfrak{n}^{2}$. Assume
further the following conditions hold:

  1. (a)

$M\otimes_{R}N$ is maximal Cohen-Macaulay.

  2. (b)

$\operatorname{length}(\operatorname{Tor}_{i}^{R}(M,N))<\infty$ for all $i\gg
0$.

  3. (c)

$\operatorname{cx}_{R}(M)\neq c$ and $\operatorname{cx}_{R}(N)\neq c$.

  4. (d)

If $\dim(R)<\operatorname{codim}(R)$, assume further
$\operatorname{Tor}_{i}^{R}(M,N)=0$ for all
$i=1,\ldots,\operatorname{codim}(R)-\dim(R)$.

Then the following conditions are equivalent:

  1. (i)

$M$ is maximal Cohen-Macaulay.

  2. (ii)

$N$ is maximal Cohen-Macaulay.

  3. (iii)

$\operatorname{Tor}_{i}^{R}(M,N)=0$ for all $i\geq 1$.

  4. (iv)

$\operatorname{Ext}^{i}_{R}(M,N)=0$ for all $i\geq 1$.

  5. (v)

$\operatorname{Ext}^{i}_{R}(N,M)=0$ for all $i\geq 1$.

###### Proof.

Proceeding as in the proof of Corollary 2.2, it suffices to assume (i) and
prove (ii) and (iii) hold; see [6, 4.8, 6.1 and 6.3]. Hence we assume (i) and
induct on $c$. If $c=1$, then $M$ is free so the result follows. Thus suppose
$c\geq 2$.

Notice, for an indeterminate $z$ over $S$, the regular local ring
$S[z]_{\mathfrak{n}S[z]}$ is unramified or equi-characteristic. Hence
extending the residue field by using the faithfully flat extension
$S\hookrightarrow S[z]_{\mathfrak{n}S[z]}$, we may assume $R$ is complete and
has infinite residue field.

It follows there exists a regular sequence
$\textbf{f}=f_{1},f_{2},\cdots,f_{c}$ on $S$ such that
$(\textbf{f})S=(\textbf{x})S\subseteq\mathfrak{n}^{2}$, $R=R_{1}/(f)$ and
$\operatorname{cx}_{R_{1}}(M)<\operatorname{codim}(R_{1})=c-1$, where
$R_{1}=S/(f_{2},\cdots,f_{c})$ and $f=f_{1}$; see [25, 1.3] and also [11,
2.5]. Repeating this argument, we obtain a regular sequence
$\textbf{y}=y_{1},y_{2},\ldots,y_{c}$ on $S$ such that
$(\textbf{y})S=(\textbf{f})S$, $R=R_{c-1}/(y_{1},\ldots,y_{c-1})$ and
$\operatorname{cx}_{R_{c-1}}(M)<\operatorname{codim}(R_{c-1})=1$, where
$R_{c-1}=S/(y_{c})$. Since $\operatorname{pd}_{R_{c-1}}(M)<\infty$, we have
that $M$ is Tor-rigid as an $R_{c-1}$-module; see [22, 1.9] and [26, Theorem
3]. Therefore, for each
$\mathfrak{p}\in\operatorname{Supp}(M\otimes_{R}N)-\\{\mathfrak{m}\\}$, it
follows that
$\operatorname{depth}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}})+\operatorname{depth}_{R_{\mathfrak{p}}}(N_{\mathfrak{p}})=\operatorname{depth}R_{\mathfrak{p}}+\operatorname{depth}_{R_{\mathfrak{p}}}(M\mathfrak{p}\otimes_{R_{\mathfrak{p}}}N_{\mathfrak{p}})$
and $N_{\mathfrak{p}}$ is maximal Cohen-Macaulay. Consequently, by swaping the
role of $N$ with that of $M$, we conclude from Theorem 3.7 that
$\operatorname{Tor}^{R}_{i}(M,N)=0$ for all $i\geq 1$. ∎

## Acknowledgments

We would like to thank Roger Wiegand for useful discussions related to this
work.

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