# Dimension of finitely generated reflexive module over Noetherian local ring

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.

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 $$\dim M=\dim R$$.