"M is reflexive" implies "M is maximal Cohen-Macaulay". Is the converse true?

Question

Let $R$ be a local integrally closed domain of dimension $2$. Let $M$ be a nonzero finitely generated $R$-module. We know that "$M$ is reflexive" implies "$M$ is maximal Cohen-Macaulay". Is the converse true?

Answer 1

If $R$ is a local normal domain with $\dim R=2$, then every MCM is reflexive.

First prove that $M$ is torsion-free. This shows that $M_{\mathfrak p}$ is free over $R_{\mathfrak p}$ for any prime $\mathfrak p$ of height $\le 1$.

Next, if $\mathfrak p$ is a prime of height $2$ it's obvious that $M_{\mathfrak p}$ satisfies Serre's condition $(S_2)$.

In the end, use Proposition 1.4.1(b) from Bruns and Herzog.

(Maybe this can help you.)