Let $M$ be a finitely generated reflexive module over a Noetherian local ring $R$. Recall that for a module $M$, $\dim M :=\dim\mathrm{Supp}(M)$, so if $M$ is finitely generated, then $\dim M =\dim R/\mathrm{ann}_R (M)$.

My question is: When can we say that $\dim M =\dim R$ ? In particular, if $R$ is Cohen-Macaulay then can we say $\dim M=\dim R$ ? What happens if $R$ is Generalized Cohen-Macaulay?

My thoughts: I know that $M$ can be embedded in a finite free module, so if $R$ is an integral domain, then $M$ is torsion free, hence faithful, so we're done. I'm not sure what happens for Cohen-Macaulay rings.

Please help.


1 Answer 1


The answer is yes, a fortiori, when $R$ is Cohen-Macaulay. In what follows, by unmixed, we mean "has no embedded primes".

Suppose $R$ is a unmixed local ring. Suppose $M$ is an $R$-module and that we have an injection $M \hookrightarrow R^n$ for some $n$. Then $\dim M=\dim R$.

Proof: Let $\mathfrak{p}$ be a minimal prime of $M$. Then $\mathfrak{p} \in \operatorname{Ass}_R(M)$. Since $M \hookrightarrow R^n$, it follows that $\mathfrak{p} \in \operatorname{Ass}_R(R)$. Since $R$ is unmixed, this means $\mathfrak{p}$ is a minimal prime of $R$, and so $\dim M=\dim R$.


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