Is $M\otimes E(R/m)$ an indecomposable module?

Question

In abstract algebra, a module is indecomposable if it is non-zero and can not be written as a direct sum of two non-zero submodules. In other words an $R$-module $M$ is indecomposable, if $M=A\oplus B$ which $A,B$ are submodules of $M$, we conclude that $A=0$ or $B=0$. Let $(R,\frak m)$ denote a commutative Gorenstein local ring and $M$ a finitely generated indecomposable $R$-module. Suppose that $E(R/\frak m)$ denotes the injective hull of the field $(R/\frak m)$. I guess $M\otimes_RE(R/\frak m)$ is an indecomposable $R$-module. But I cant prove it. If my guess is not true, may we take a little condition on M (such as M is maximal Cohen-Macaulay), to $M\otimes_RE(R/\frak m)$ be indecomposable?

Answer 1

This is true if $M$ is reflexive, which is the case if $M$ is maximal Cohen-Macaulay, as you assume $R$ to be Gorenstein. Indeed, without loss of generality, we may pass to the completion to suppose $R$ is complete. By Matlis duality, $M \otimes_R E(R/\mathfrak{m})$ is indecomposable if and only if $\operatorname{Hom}_R(M \otimes_R E(R/\mathfrak{m}),E(R/\mathfrak{m})) \cong \operatorname{Hom}_R(M,R)$ is so. Noting that $\operatorname{Hom}_R(M,R)$ satisfies Serre's condition $(S_2)$ and is thus unmixed, any proper decomposition of $\operatorname{Hom}_R(M,R)$ provides a proper decomposition of $\operatorname{Hom}_R(\operatorname{Hom}_R(M,R),R)) \cong M$ as each component has maximal dimension, establishing the claim.

This need not be true if $M$ is not reflexive. E.g., take $R=k[\![x,y]\!]$ and let $M$ be any indecomposable $R$-module of rank $>1$. Then $\operatorname{Hom}_R(M,R)$ is a second syzygy and is thus free over $R$ (as $R$ has global dimension $2$). Since the rank of $M$ is greater than $1$, it follows that $\operatorname{Hom}_R(M,R) \cong R^{\oplus \operatorname{rank}(M)}$ is not indecomposable. If one wants a concrete $M$, take for instance $M=\operatorname{im} \begin{pmatrix} y & x & 0 \\ 0 & y & x \\ \end{pmatrix}$