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Let $R$ be a Cohen-Macaulay local ring. Let $M$ be a finitely generated $R$-module with $\text{pd }M<\infty$. Let $N$ be a maximal Cohen-Macaulay $R$-module. Then $\text{Tor}_i^R(M,N)=0$ for all $i>0$.

If this statement is true. It suffices to show that if $0\longrightarrow R^n\longrightarrow R^m\longrightarrow L$ is exact, and after applying $\otimes N$ it remains exact. But I don't know how to show this. I also tried to use induction, but I didn't get anything.

Bromelain
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  • Hint: If you are able to reduce to the case where $\operatorname{pd}R(M) \le 1$, then note that $M{\mathfrak{p}}$ is a free $R_{\mathfrak{p}}$-module for any $\mathfrak{p} \in \operatorname{Ass}(R)$. So $\operatorname{Tor}^R_1(M,N)$ is torsion. But what happens if we tensor a free resolution $0 \to R^{\oplus n} \to R^{\oplus m} \to M \to 0$ with $N$ and consider the long exact sequence in $\operatorname{Tor}$? – metalspringpro Oct 11 '23 at 00:08
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    @metalspringpro Thanks. So it suffices to show for every $p\in \text{Ass}(N)\subset \text{Ass}(R)$,$\text{Tor}_1^R(M,N)_p=0$, and this is obvious. – Bromelain Oct 11 '23 at 01:22

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