Vanishing of Tor over fiber products

TL;DR

This paper investigates how the vanishing of Tor modules over fiber product rings influences module properties, extending previous work and providing new insights into homological behavior in this context.

Contribution

It explores the implications of Tor vanishing over fiber product rings, building on Naseh and Sather-Wagstaff's work to deepen understanding of module interactions.

Findings

Vanishing of Tor imposes specific module structure constraints

Results extend known homological properties to fiber product rings

Provides new criteria for module projectivity over fiber products

Abstract

Let $(S,\mathfrak{m},k)$ and $(T,\mathfrak{n},k)$ be local rings, and let $R$ denote their fiber product over their common residue field $k$. Inspired by work of Naseh and Sather-Wagstaff, we explore consequences of vanishing of ${\rm Tor}^R_m(M,N)$ for various values of $m$, where $M$ and $N$ are finitely generated $R$-modules.

Equations

$$R:=S{\times }_{k}T=\{(s,t)\in S\times T\mid {\pi }_S(s)={\pi }_T(t)\}.$$

$$\begin{array}{c}R\\ S\\ \\ T\\ k\end{array}$$

$$\begin{array}{c}I\cong \mathfrak{m}\text{and}J\cong \mathfrak{n}\text{as R-modules};\text{also}\\ S\cong R/J\text{and}T\cong R/I\text{as rings and as R-modules.}\end{array}$$

$${\Omega }_{R}Y\cong J^{\oplus b}\oplus {\Omega }_{S}Y\cong {\mathfrak{n}}^{\oplus b}\oplus {\Omega }_{S}Y.$$

$$0\to Z_1\to T^n\to Z\to 0.$$

$$0\to {\mathrm{Tor}}_1^R(S,Z)\to Z_1/J{Z}_1\to (T/J{)}^n\to Z/JZ\to 0.$$

$$\operatorname{Tor}^{R}_{1}(X,Y)\cong\operatorname{Tor}^{S}_{1}(X,Y)\oplus\big{(}\frac{Y}{\mathfrak{m}Y}\big{)}^{\oplus da}\,.$$

$${\mathrm{Tor}}_{2m-1}^R(X^{\prime },Y^{\prime })=0⟹X^{\prime }=0\text{ or }{Y}^{\prime }=0$$

$${\mathrm{Tor}}_{2m-1}^R(V,W)=0⟹V=0\text{ or }W=0$$

$$0$$

$$={\mathrm{Tor}}_{2m-1}^R({\mathfrak{n}}^{\oplus a}\oplus {\Omega }_{S}X,{\mathfrak{n}}^{\oplus b}\oplus {\Omega }_{S}Y)={\mathrm{Tor}}_{2m-1}^R({\mathfrak{n}}^{\oplus a},{\mathfrak{n}}^{\oplus b})\oplus \boxed{\text{other terms}}.$$

$$0$$

$$={\mathrm{Tor}}_{2m-1}^R({\mathfrak{n}}^{\oplus b},Z)\oplus {\mathrm{Tor}}_{2m-1}^R({\Omega }_{S}Y,Z)⟹0={\mathrm{Tor}}_{2m-1}^R({\mathfrak{n}}^{\oplus b},Z).$$

$$\begin{array}{c}{\mathrm{Tor}}_s^R(X,Y)={\mathrm{Tor}}_{s-1}^R(X,{\mathfrak{n}}^{\oplus b})\oplus {\mathrm{Tor}}_{s-1}^R(X,{\Omega }_{S}Y)\text{and}\\ {\mathrm{Tor}}_s^R(Y,Z)={\mathrm{Tor}}_{s-1}^R(Y,{\mathfrak{m}}^{\oplus c})\oplus {\mathrm{Tor}}_{s-1}^R(Y,{\Omega }_{T}Z).\end{array}$$

$$\mathrm{depth}M+{\mathrm{pd}}_{A}M=\mathrm{depth}A.$$

$${\mathrm{Tor}}_2^R(Y,X)=0={\mathrm{Tor}}_2^R(Y,Z).$$

$${\mathrm{Tor}}_2^R(S,S)={\mathrm{Tor}}_1^R(S,{\Omega }_{R}S)\cong {\mathrm{Tor}}_1^R(S,T)=0,$$

$${\mathrm{Tor}}_4^R(S,S)={\mathrm{Tor}}_2^R({\Omega }_{R}S,{\Omega }_{R}S)\cong {\mathrm{Tor}}_2^R(T,T)=0.$$

$${\mathrm{Tor}}_1^R(X\oplus Z,Y\oplus W)={\mathrm{Tor}}_1^R({\Omega }_R^{2}M,{\Omega }_R^{2}N)={\mathrm{Tor}}_5^R(M,N)=0.$$

$${\mathrm{Tor}}_1^R(X,Y)={\mathrm{Tor}}_1^R(X,W)={\mathrm{Tor}}_1^R(Z,Y)={\mathrm{Tor}}_1^R(Z,W)=0.$$

$$\begin{array}{c}X=0\text{or}Y=0;\text{and}\\ Z=0\text{or}W=0.\end{array}$$

$${\mathrm{Tor}}_4^R(M,Y)=0\text{and}{\mathrm{Tor}}_4^R(M,W)=0.$$

$${\Omega }_R^{2}M\cong X\oplus Z\text{ and }{\Omega }_R^{2}N\cong Y\oplus W,$$

$${\mathrm{Tor}}_4^R(M,{\Omega }_R^{2}N)=0={\mathrm{Tor}}_4^R({\Omega }_R^{2}M,N),$$

$${\mathrm{Tor}}_4^R(M,Y)$$

$${\mathrm{Tor}}_4^R(X,N)$$

$${\mathrm{Tor}}_2^R(X\oplus Z,Y)={\mathrm{Tor}}_2^R({\Omega }_R^{2}M,Y)=0,\text{and hence}{\mathrm{Tor}}_2^R(Z,Y)=0.$$

$${\mathrm{Tor}}_5^R({\Omega }_R^{2i+1-5}M,N)={\mathrm{Tor}}_{2i+1}^R(M,N)=0.$$

$${\mathrm{Tor}}_5^R(M,N)=0={\mathrm{Tor}}_{2j}^R(M,N),$$

$$0={\mathrm{Tor}}_{2j-4}^R(S,T)={\mathrm{Tor}}_{2j-5}^R(S,{\Omega }_{R}T)={\mathrm{Tor}}_{2j-5}^R(S,\mathfrak{m}),$$

$$0=$$

$${\mathrm{Tor}}_{2i-3}^R(S^{\oplus a},S^{\oplus b})=0={\mathrm{Tor}}_{2i-3}^R(T^{\oplus c},T^{\oplus d}).$$

Full text

All four authors were partially supported by FAPESP-Brazil 2018/05271-6, 2018/05268-5 and CNPq-Brazil 421440/2016-3. RW was partially supported by Simons Collaboration Grant 426885. SW was partially supported by a UNL Emeriti & Retiree Association Wisherd Award

1. Introduction

Recently there has been renewed interest in the homological properties of fiber product rings. In particular, the results obtained by Nasseh and Sather-Wagstaff on the vanishing of Tor in [6] have inspired us to try to extend their computations. This note should be regarded as an addendum to that paper, or perhaps an advertisement for the utility of the nice results established there.

Setting 1.1**.**

Let $(S,\mathfrak{m},k)$ and $(T,\mathfrak{n},k)$ be commutative local rings. Let $S\stackrel{{\scriptstyle\pi_{S}}}{{\twoheadrightarrow}}k\stackrel{{\scriptstyle\pi_{T}}}{{\twoheadleftarrow}}T$ denote the natural surjections onto the common residue field, and assume that $S\neq k\neq T$. Let $R$ denote the fiber product:

[TABLE]

Then $R$ is a local ring with maximal ideal $\mathfrak{m}\times\mathfrak{n}$, and $R$ is a subring of the usual direct product $S\times T$. Let $\eta_{S}:R\twoheadrightarrow S$ and $\eta_{T}:R\twoheadrightarrow T$ be the projections $(s,t)\mapsto s$ and $(s,t)\mapsto t$, respectively. The maps $\eta_{S}$ and $\eta_{T}$ are surjective, with respective kernels $J:=0\times\mathfrak{n}$ and $I:=\mathfrak{m}\times 0$. Then $R$ is represented as a pullback diagram:

[TABLE]

The maximal ideal $\mathfrak{m}\times\mathfrak{n}$ is decomposable: $\mathfrak{m}\times\mathfrak{n}=I\oplus J$. For future reference we note that

[TABLE]

Whenever we use the symbols $R$, $S$, or $T$, we tacitly assume that $R$ is the fiber product as described here; the notation $\mathfrak{m}$, $\mathfrak{n}$, $I$, and $J$ for the various ideals will be preserved throughout the paper. Moreover, every module over $R$, $S$, or $T$ is assumed to be finitely generated.

The general theme of the paper is to assume the vanishing of $\operatorname{Tor}_{m}^{R}(M,N)$ for certain $R$-modules $M$ and $N$, and certain values of $m$, and then describe properties of the modules that result from this assumption.

Notation 1.2**.**

For a local ring $(A,\mathfrak{m}_{A})$ and a finitely generated $A$-module $M$, we let $\Omega_{A}^{i}M$ denote the $i^{\text{th}}$ syzygy of $M$ with respect to a minimal $A$-free resolution. We often write $\Omega_{A}M$ for $\Omega^{1}_{A}M$. The $i^{\text{th}}$ Betti number $\beta^{A}_{i}M$ is the minimal number of generators required for the $A$-module $\Omega^{i}_{A}M$.

For an $A$-module $L$ and a non-negative integer $c$, we denote by $L^{\oplus c}$ the direct sum of  $c$   copies of $L$. (When $L$ is an ideal of a ring $A$, this distinguishes the direct sum from the ideal product $L^{c}=L\cdots L$.)

First we focus on vanishing of $\operatorname{Tor}^{R}_{m}(M,N)$, where $M$ and $N$ are $S$-modules or $T$-modules, using two useful lemmas. Lemma 1.3 is due to Nasseh and Takahashi, and Lemma 1.4 is from the paper [6] by Nasseh and Sather-Wagstaff and consists of the first and sixth of eight formulas given in [6, Lemma 2.3].

Lemma 1.3**.**

• [7, Lemma 3.2] Let $Y$ be an $S$-module, and set $b=\beta_{0}^{S}Y$. Then*

[TABLE]

Lemma 1.4**.**

[6, Lemma 2.4]**. Let $X$ and $Y$ be $S$-modules, and let $Z$ be a $T$-module.

1. (1)

$\operatorname{Tor}^{R}_{1}(X,Y)\cong\operatorname{Tor}_{1}^{S}(X,Y)\oplus\left(\frac{Y}{\mathfrak{m}Y}\right)^{\oplus(\beta_{1}^{T}k)(\beta_{0}^{S}X)}$. 2. (2)

$\operatorname{Tor}^{R}_{1}(Y,Z)\cong\operatorname{Tor}_{1}^{S}(Y,k)^{\oplus\beta_{0}^{T}Z}\oplus\left(\frac{Y}{\mathfrak{m}Y}\right)^{\oplus\beta_{1}^{T}Z}$.

Proposition 1.5**.**

Let $Y$ be a non-zero $S$-module and $Z$ a $T$-module.

1. (1)

If $\operatorname{Tor}_{1}^{R}(Y,Z)=0$, then $Z$ is a free $T$-module. 2. (2)

If $Y$ and $Z$ are free over $S$ and $T$, respectively, then $\operatorname{Tor}_{1}^{R}(Y,Z)=0$.

Proof.

If $\operatorname{Tor}_{1}^{R}(Y,Z)=0$, then \big{(}\frac{Y}{\mathfrak{m}Y}\big{)}^{\oplus\beta_{1}^{T}Z}=0, by Lemma 1.4. Since $\frac{Y}{\mathfrak{m}Y}\neq 0$ by Nakayama’s Lemma, we have $\beta^{T}_{1}Z=0$, that is, $Z$ is free as a $T$-module. For the converse, we may assume $Y=S$ and $Z=T$. As in [6, Remark 2.6], one has $\operatorname{Tor}_{1}^{R}(S,T)=\operatorname{Tor}_{1}^{R}(R/J,R/I)\cong\frac{J\cap I}{JI}=0$. ∎

When $Y$ is a free $S$-module, we obtain quantitative information about $\operatorname{Tor}_{1}^{R}(Y,Z)$:

Proposition 1.6**.**

Let $Z$ be a $T$-module and $a\in{\mathbb{N}}$. Then $\operatorname{Tor}^{R}_{1}(S^{\oplus a},Z)$ is a $k$-vector space whose dimension is equal to $a\beta_{1}^{T}Z$.

Proof.

Letting $n=\beta^{T}_{0}Z$ and $Z_{1}=\Omega^{1}_{T}Z$, we get an exact sequence

[TABLE]

For $a=1$, apply the functor S\otimes_{R}-\ \ \big{(}=(R/J)\otimes_{R}-\big{)} to this short exact sequence to get the exact sequence

[TABLE]

The zero on the left is by Proposition 1.5(2). Each of these modules is annihilated by the maximal ideal $I\oplus J$ of $R$, so they are $k$-vector spaces. Moreover the last two non-zero terms have the same $k$-dimension, namely $n$. It follows that the first two terms have the same dimension, and hence that \mathop{\rm dim}_{k}\big{(}\operatorname{Tor}_{1}^{R}(S,Z)\big{)}=\mathop{\rm dim}_{k}(Z_{1}/JZ_{1})=\mathop{\rm dim}_{k}(Z_{1}/\mathfrak{n}Z_{1})=\beta_{0}^{T}Z_{1}=\beta_{1}^{T}Z.

For $a>1$, $\operatorname{Tor}_{1}^{R}(S^{\oplus a},Z)=\operatorname{Tor}_{1}^{R}(S,Z)^{\oplus a}$ and the dimension is $a\beta_{1}^{T}Z$. ∎

Part (1) of Theorem 1.7 is a generalization of a result of Nasseh and Sather-Wagstaff [6, Lemma 2.4] and part (2) generalizes Proposition 1.5(1).

Theorem 1.7**.**

Let $X$ and $Y$ be $S$–modules, let $Z$ be a $T$-module, and let $m$ be a nonnegative integer.

1. (1)

$\operatorname{Tor}^{R}_{2m+1}(X,Y)=0\implies X=0$* or $Y=0$.* 2. (2)

$\operatorname{Tor}^{R}_{2m}(Y,Z)=0\implies Y=0$* or $Z=0$.* 3. (3)

$\operatorname{Tor}^{R}_{2m}(X,Y)=0$* and $X\neq 0\implies Y$ is a free $S$-module.* 4. (4)

*If $T$ is not a discrete valuation domain (DVR), then *

$\operatorname{Tor}^{R}_{2m}(X,Y)=0\implies X=0$* or $Y=0$.* 5. (5)

$\operatorname{Tor}^{R}_{2m+1}(Y,Z)=0$* and $Y\neq 0\implies Z$ is a free $T$-module.* 6. (6)

If $S$ is not a DVR and $m>0$, then $\operatorname{Tor}^{R}_{2m+1}(Y,Z)=0\implies Y=0$ or $Z=0$.

By symmetry, the corresponding statements hold if $X$ and $Y$ are $T$–modules and $Z$ is an $S$-module.

Proof.

Put $a=\beta_{0}^{S}X$, $b=\beta_{0}^{S}Y$, $c=\beta_{0}^{T}Z$, and $d=\beta_{1}^{T}k=\beta^{T}_{0}\mathfrak{n}$. The assumption $T\neq k$ in Setting 1.1 implies that $d\neq 0$.

The case $m=0$ of (1) is [6, Lemma 2.4]; we give their proof for completeness. Thus we assume $\operatorname{Tor}^{R}_{1}(X,Y)=0$ and $X\neq 0$ (so $a\neq 0$). By Lemma 1.4(1),

[TABLE]

Since $da\neq 0$, we have $\frac{Y}{\mathfrak{m}Y}=0$. Hence $Y=0$, and so (1) holds for $m=0$.

For (1) with $m>0$, proceed by induction. Assume $\operatorname{Tor}^{R}_{2m+1}(X,Y)=0$, and that

[TABLE]

whenever $X^{\prime}$ and $Y^{\prime}$ are $S$-modules. By symmetry,

[TABLE]

whenever $V$ and $W$ are $T$-modules. Using Lemma 1.3, we get

[TABLE]

This implies $0=\operatorname{Tor}^{R}_{2m-1}(\mathfrak{n}^{\oplus a},\mathfrak{n}^{\oplus b})$. Since $\mathfrak{n}^{\oplus a}$ and $\mathfrak{n}^{\oplus b}$ are both $T$-modules, the inductive hypothesis implies that $\mathfrak{n}^{\oplus a}=0$ or $\mathfrak{n}^{\oplus b}=0$. But $\mathfrak{n}\neq 0$, and so $a=0$ or $b=0,$ that is, $X=0$ or $Y=0$, as desired for (1).

If $Y\otimes_{R}Z=0$, then $\overline{Y}\otimes_{k}\overline{Z}=0$, where $\overline{Y}$ and $\overline{Z}$ are reductions modulo the maximal ideal of $R$. Therefore either $\overline{Y}=0$ or $\overline{Z}=0$, and Nakayama’s Lemma implies that $Y=0$ or $Z=0$. This gives the case $m=0$ in (2), and the same argument verifies the case $m=0$ in statements (3) and (4).

To see that (2) holds for $m\geq 1$, we use Lemma 1.3:

[TABLE]

By Equation (1.1.2), $\mathfrak{n}^{\oplus b}$ is a $T$-module, as is $Z$. If $Z\neq 0$, then part (1) (applied to $T$) shows $\mathfrak{n}^{\oplus b}=0$. One of the assumptions in Setting 1.1 is that $\mathfrak{n}\neq 0$. Therefore $\beta^{S}_{0}Y=b=0$, that is, $Y=0$. A symmetric argument shows that $Y\neq 0$ implies $Z=0$, and so (2) holds.

Next observe that Proposition 1.5(1) implies that (5) holds for $m=0$. Taking first syzygies over $R$, shifting, and using Lemma 1.3 (and its counterpart over $T$), we obtain, for every $s\geq 2$,

[TABLE]

Setting $s=2$ in the first equation, we deduce the case $m=1$ of (3) from the case $m=0$ of (5). Setting $s=3$ in the second equation, we deduce the case $m=1$ of (5) from the case $m=1$ of (3). Next we set $s=4$ in the first equation and get the case $m=2$ of (3) from the case $m=1$ of (5). Continuing in this way, alternating between the two equations, we obtain (3) and (5) by induction.

For (4), suppose $m>0$ , $X\neq 0$, and $\operatorname{Tor}^{R}_{2m}(X,Y)=0$. Using Lemma 1.3, we learn that $\operatorname{Tor}_{2m-1}^{R}(X,\mathfrak{n}^{\oplus b})=0$. Now (5) implies that $\mathfrak{n}^{\oplus b}$ is a free $T$-module. Since $\mathfrak{n}$ is not $T$-free (as $T$ is not a DVR) we conclude that $b=0$, that is, $Y=0$.

To prove (6), suppose $m>0$, $Y\neq 0$, and $\operatorname{Tor}^{R}_{2m+1}(Y,Z)=0$. Then $\operatorname{Tor}^{R}_{2m}(Y,\mathfrak{m}^{\oplus c})=0$, and now (3) implies that $\mathfrak{m}^{\oplus c}$ is free. As in the proof of (4), it follows that $Z=0$. ∎

We note that item (6) of Theorem 1.7 does not hold when $m=0$, since one always has $\operatorname{Tor}^{R}_{1}(S,T)=0$.

Corollary 1.8**.**

Let $Y$ and $Z$ be non-zero modules over $S$ and $T$ respectively, and let $m\geq 0$. If $\operatorname{Tor}_{m}^{R}(Y,Z)=0$, then $m$ is odd and $Y$ and $Z$ are free.

Proof.

Use (2) and (5) of Theorem 1.7. ∎

Remark 1.9**.**

$\operatorname{depth}R=\min\{\operatorname{depth}S,\operatorname{depth}T,1\}\,.$

Remark 1.9 follows from the work of Lescot [4]. See, for example, [2, (3.2) Remark]. Actually, a low-tech, direct proof is easy: Note first that an element $(s,t)\in R$ is a non-zerodivisor (NZD) of $R$ if and only if $s$ is a NZD of $S$ and $t$ is a NZD of $T$. It follows that $\operatorname{depth}R>0\iff\operatorname{depth}S>0$ and $\operatorname{depth}T>0$. To see that $\operatorname{depth}R\leq 1$, suppose that $(s,t)$ is a NZD of $R$ in $\mathfrak{m}\times\mathfrak{n}$, and let $(u,v)$ be a arbitrary element of $\mathfrak{m}\times\mathfrak{n}$. Then $(s,0)(u,v)=(u,0)(s,t)\in R(s,t)$. Moreover, $(s,0)\notin R(s,t)$: Indeed, if $(s,0)=(a,b)(s,t)$, then $b=0$, as $t$ is a NZD; also, the equation $as=s$ forces $a=1$, a contradiction, since $(1,0)\notin R$. Thus every element of $\mathfrak{m}\times\mathfrak{n}$ is a zero-divisor modulo $R(s,t)$.

Auslander-Buchsbaum Formula 1.10**.**

[5, A.5. Theorem, p. 310] Let $M$ be a finitely generated module of finite projective dimension over a local ring $(A,\mathfrak{m}_{A})$. Then

[TABLE]

From now on, our conclusions are going to be that one of the modules has finite projective dimension over $R$. It is important to realize, however, the following fact:

Fact 1.11**.**

If $M$ is an $R$-module with $\operatorname{pd}_{R}M<\infty$, then $\operatorname{pd}_{R}M\leq 1$.

Proof.

See the Auslander-Buchsbaum Formula 1.10 and Remark 1.9. ∎

We now analyze implications of $\operatorname{Tor}_{\mathfrak{m}}^{R}(M,Y)=0$, where $M$ is an arbitrary $R$-module, that is, not necessarily an $S$-module or a $T$-module. The conditions imposed in the next theorem may appear a bit contrived, but Example 1.14, which follows the proof of the theorem, shows that they are exactly what is needed.

Theorem 1.12**.**

Let $M$ be an $R$-module and $Y$ a non-zero $S$-module. Assume that $\operatorname{Tor}^{R}_{m}(Y,M)=0$ for some $m\geq 4$ and that at least one of these two conditions holds:

1. (1)

$T$* is not a discrete valuation ring, or* 2. (2)

$Y$* is not a free $S$-module.*

Then $\operatorname{pd}_{R}M\leq 1$.

If, in addition, $S$ or $T$ has depth [math], then $M$ is a free $R$-module.

The proof uses a lemma due to Dress and Krämer.

Lemma 1.13**.**

[3, Bemerkung 3]** Let $M$ be an $R$-module. Then $\Omega^{2}_{R}M$ decomposes as a direct sum: $\Omega^{2}_{R}M=X\oplus Z$, where $X$ is an $S$-module and $Z$ is a $T$-module.

Proof of Theorem 1.12.

Set $N=\Omega_{R}^{m-4}M$. If we show that $\operatorname{pd}_{R}N<\infty$, it will follow that $M$ too has finite projective dimension, and hence projective dimension at most $1$. Moreover, if either $S$ or $T$ has depth [math], then $\operatorname{depth}R=0$ by Remark 1.9, and hence $\operatorname{pd}_{R}M\leq 0$ by Remark 1.10. Thus our goal is to show that $N$ has finite projective dimension if $\operatorname{Tor}_{4}^{R}(N,Y)=0$.

Write $\Omega^{2}_{R}N\cong X\oplus Z$ as in Lemma 1.13, where $X$ is an $S$-module and $Z$ is a $T$-module. Since $\operatorname{Tor}_{2}^{R}(Y,\Omega_{R}^{2}N)=\operatorname{Tor}^{R}_{4}(Y,N)=0$, we have

[TABLE]

Since $Y\neq 0$, Theorem 1.7(2) implies that $Z=0$. If also $X=0$, then $\Omega^{2}_{R}N=0$, and so $\operatorname{pd}N_{R}\leq 1$, and we are done.

If, on the other hand, $X\neq 0$, then Theorem 1.7(3) shows that $Y$ is a free $S$-module. Therefore assumption (2) fails, so (1) must hold, that is, $T$ is not a discrete valuation ring. Now Theorem 1.7(4) yields $Y=0$, a contradiction. ∎

Nasseh and Sather-Wagstaff ask [6, Question 2.14] whether the vanishing of $\operatorname{Tor}^{R}_{4}(M,N)$ (for a fiber product $R$) forces one of the modules to have finite projective dimension. The following example shows that the answer is “no” and justifies the hypotheses imposed in Theorem 1.12. (But see Corollary 1.19 for a result in the positive direction.) The example also shows the need for two vanishing Tors in the hypotheses of [6, Theorem 1.1(b)].

Example 1.14**.**

Let $(S,\mathfrak{m},k)$ and $(T,\mathfrak{n},k)$ be discrete valuation rings, and let $R$ be the fiber product of $S$ and $T$. Then $\Omega_{R}S=\Omega_{R}(R/J)=J\cong\mathfrak{n}\cong T$, since $\mathfrak{n}$ is a principal ideal in the domain $T$. Similarly $\Omega_{R}T=S$. Both $S$ and $T$ have non-zero annihilators and therefore are not free as $R$-modules. It follows, from the syzygy relations above, that both $S$ and $T$ have infinite projective dimension over $R$. However,

[TABLE]

by Proposition 1.5(2). Similarly

[TABLE]

It follows that $\operatorname{Tor}^{R}_{m}(S,S)=0$ and $\operatorname{Tor}^{R}_{m}(T,T)=0$ for every even positive integer, and $\operatorname{Tor}^{R}_{m}(S,T)=0$ for every odd positive integer.

Theorem 1.15**.**

Let $M$ and $N$ be $R$-modules with $\operatorname{Tor}_{5}^{R}(M,N)=0$. Then at least one of the following four things happens:

1. (1)

$\operatorname{pd}_{R}M\leq 1$. 2. (2)

$\operatorname{pd}_{R}N\leq 1$. 3. (3)

$\Omega^{2}_{R}M$* is a free $S$-module and $\Omega^{2}_{R}N$ is a free $T$-module.* 4. (4)

$\Omega^{2}_{R}M$* is a free $T$-module and $\Omega^{2}_{R}N$ is a free $S$-module.*

Proof.

Using Lemma 1.13, write $\Omega^{2}_{R}M=X\oplus Z$ and $\Omega^{2}_{R}N=Y\oplus W$, where $X$ and $Y$ are $S$-modules and $Z$ and $W$ are $T$-modules. Now

[TABLE]

Hence

[TABLE]

From Theorem 1.7(1), $\operatorname{Tor}^{R}_{1}(X,Y)=0$ and $\operatorname{Tor}^{R}_{1}(Z,W)=0$ imply

[TABLE]

If $X=Z=0$ we get (1), and if $Y=W=0$ we get (2). There are two remaining cases:

(a) $Y=0=Z$ and $X\neq 0\neq W$ .

(b) $X=0=W$ and $Y\neq 0\neq Z$ .

Assume (a) holds. Then $\Omega^{2}_{R}M=X$, which is an $S$-module, and $\Omega^{2}_{R}N=W$, a $T$-module. Now Theorem 1.7 and the equation $\operatorname{Tor}_{1}^{R}(X,W)=0$ yield conclusion (3). Similarly, case (b) leads to conclusion (4). ∎

Now on to $\operatorname{Tor}_{6}$ . Here, as in the result above, both $M$ and $N$ are allowed to be arbitrary $R$-modules (that is, not necessarily annihilated by $I$ or by $J$). In view of Example 1.14, however, we cannot do away with the assumption that $S$ and $T$ are not DVRs.

Theorem 1.16**.**

Let $M$ and $N$ be $R$-modules. Assume that neither $S$ nor $T$ is a discrete valuation domain. If $\operatorname{Tor}_{m}^{R}(M,N)=0$ for some $m\geq 6$, then $\operatorname{pd}_{R}M\leq 1$ or $\operatorname{pd}_{R}N\leq 1$. Therefore $\operatorname{Tor}_{i}^{R}(M,N)=0$ for all $i\geq 2$.

Proof.

If $m>6$ we easily reduce to the case $m=6$ by taking syzygies of one of the modules. Therefore we assume that $m=6$. Assume $\operatorname{pd}_{R}N>1$; we show that $\operatorname{pd}_{R}M\leq 1$. From Lemma 1.13 we have $\Omega^{2}_{R}N\cong Y\oplus W$, where $Y$ is an $S$-module and $W$ is a $T$-module. Moreover, since $\operatorname{pd}_{R}N>1$, at least one of $Y$ and $W$ is non-zero. From the equation $\operatorname{Tor}^{R}_{4}(M,\Omega_{R}N)=0$, we get

[TABLE]

If $Y\neq 0$, Theorem 1.12 and our assumption that $T$ is not a discrete valuation ring imply that $\operatorname{pd}_{R}M\leq 1$. If $Y=0$, then $W\neq 0$, and the assumption that $S$ is not a DVR shows that $\operatorname{pd}_{R}M\leq 1$ in this case. ∎

Theorem 1.17**.**

Let $M$ and $N$ be $R$-modules such that $\operatorname{Tor}_{m}^{R}(M,N)=0$ for some  $m\geq 6$. Using Lemma 1.13, write

[TABLE]

where $Y$ and $X$ are $S$-modules and $W$ and $Z$ are $T$-modules. Assume that at least one of the following conditions holds:

1. (1)

At least one of $X$ and $Y$ is not free as an $S$-module; or 2. (2)

At least one of $Z$ and $W$ is not free as a $T$-module; or 3. (3)

At least one of $\{X,Y\}$ is [math] AND at least one of $\{Z,W\}$ is [math]; or 4. (4)

Neither $S$ nor $T$ is a discrete valuation domain.

Then  $\operatorname{pd}_{R}M\leq 1$ or $\operatorname{pd}_{R}N\leq 1$. Therefore $\operatorname{Tor}_{i}^{R}(M,N)=0$ for all $i\geq 2$.

Proof.

If (4) holds, we quote Theorem 1.16. For the other cases, we may assume, by taking syzygies, that $m=6$. We then have

[TABLE]

and hence

[TABLE]

If $Y$ is not $S$-free, the first equation above and Theorem 1.12 show that $\operatorname{pd}_{R}M\leq 1$. Symmetric arguments give the desired conclusion under either assumption (1) or (2). The remaining case to consider is (3).

If $X=0=Z$, then $\Omega^{2}_{R}M=0$, and hence $\operatorname{pd}_{R}M\leq 1$. Similarly, if $Y=0=W$, then $\operatorname{pd}_{R}N\leq 1$. The remaining possibilities, under assumption (3), are (a) $X=0=W$ and (b) $Y=0=Z$. By symmetry, we need to consider only possibility (a) $X=0=W$. We have, from the first equation in (1.2),

[TABLE]

By Theorem 1.7(2), either $Y=0$, in which case $\Omega^{2}_{R}N=0$; or $Z=0$, in which case $\Omega^{2}_{R}M=0$. Thus either $\operatorname{pd}_{R}N\leq 1$ or $\operatorname{pd}_{R}M\leq 1$, as desired. ∎

The next theorem is a slight generalization of a result due to Nasseh and Sather-Wagstaff [6, Theorem 1.1 (b)].

Theorem 1.18**.**

Let $M$ and $N$ be finitely generated modules over a fiber product $R$ with $\operatorname{Tor}^{R}_{2i+1}(M,N)=0=\operatorname{Tor}_{2j}^{R}(M,N),$ for some $i\geq 2$ and $j\geq 3$. Then $\operatorname{pd}_{R}M\leq 1$ or $\operatorname{pd}_{R}N\leq 1$.

Proof.

We consider separately the cases where $2i+1<2j$ and $2i+1>2j$.

Case 1: First suppose $2i+1<2j$. If $i>2$, then

[TABLE]

Since it suffices to show either $\operatorname{pd}(\Omega_{R}^{2i-4}M)$ or $\operatorname{pd}N$ is finite, we may suppose that $i=2$; that is,

[TABLE]

where $j\geq 3$.

By Theorem 1.15, at least one of the following four things happens:

1. (1)

$\operatorname{pd}_{R}M\leq 1$. 2. (2)

$\operatorname{pd}_{R}N\leq 1$. 3. (3)

$\Omega^{2}_{R}M$ is a free $S$-module and $\Omega^{2}_{R}N$ is a free $T$-module. 4. (4)

$\Omega^{2}_{R}M$ is a free $T$-module and $\Omega^{2}_{R}N$ is a free $S$-module.

If either of the first two happens, we are done. If not, suppose (3) happens, so that $\Omega^{2}_{R}M=S^{\oplus a}$ and $\Omega^{2}_{R}N=T^{\oplus b}$.

Suppose $a\neq 0\neq b$. Now $0=\operatorname{Tor}_{2j}^{R}(M,N)=\operatorname{Tor}^{R}_{2j-4}(S^{\oplus a},T^{\oplus b})$ implies

[TABLE]

the last equality from (1.1.2). Since both $S$ and $\mathfrak{m}$ are non-zero $S$-modules, this contradicts Theorem 1.7(1). Therefore either $\Omega^{2}_{R}M=S^{\oplus a}=0$ or $\Omega^{2}_{R}N=T^{\oplus b}=0$. Thus $\operatorname{pd}_{R}M\leq 1$ or $\operatorname{pd}_{R}N\leq 1$.

A similar argument works if (4) happens.

Case 2: Assume that $2i+1>2j$. By taking syzygies, we can reduce to the case where $j=3$, and so $\operatorname{Tor}_{6}^{R}(M,N)=0$. As in Theorem 1.17, let $\Omega^{2}_{R}M=X\oplus Z$ and $\Omega^{2}_{R}N=Y\oplus W$. We may take each of the pieces to be free, so that $X=S^{\oplus a},Y=S^{\oplus b},Z=T^{\oplus c},W=T^{\oplus d}.$

Now we also have $0=\operatorname{Tor}_{2i+1}^{R}(M,N)=\operatorname{Tor}_{2i-3}^{R}(\Omega_{R}^{2}M,\Omega_{R}^{2}N)$. Thus

[TABLE]

By Theorem 1.7(1), since $2i-3$ is an odd positive number, we have $a=0$ or $b=0$; AND $c=0$ or $d=0$. Thus condition (3) of Theorem 1.17 holds, and so $\operatorname{pd}_{R}M\leq 1$ or $\operatorname{pd}_{R}N\leq 1$.∎

Recall that a finitely generated module $M$ over a Noetherian ring $A$ is torsionless [1] provided the canonical biduality map $\delta_{M}:M\to M^{**}$ is injective. By mapping a finitely generated free module onto $M^{*}$ and then dualizing, we get an embedding of $M^{**}$ into a free module. It follows that every torsionless module over a local ring is, up to free summands, a syzygy module. Therefore we get the following corollary of Theorem 1.16 by representing each of the two modules as a syzygy and then shifting up two homological degrees.

Corollary 1.19**.**

Assume that neither $S$ nor $T$ is a DVR, and let $M$ and $N$ be torsionless $R$-modules. If $\operatorname{Tor}_{4}^{R}(M,N)=0$, then at least one of $M$ and $N$ has projective dimension at most one. ∎