On regular elements and Maximal Cohen-Macaulay modules

Question

I was reading theorem 3.3.3 in Bruns-Herzog: we have a Cohen-Macaulay local ring $(R,\mathfrak m,k)$, $C$ and $M$ are maximal Cohen-Macaulay modules. (Probably to solve my question some of these hypothesis are superfluous.) Anyway I would like to prove that

if $x\in\mathfrak m$ is regular then $\mathrm{Hom}_R(M/xM,C)=0$.

I see that if I prove $x$ regular on $C$ then I'm done. I know that if you have high enough syzygies then they are MCM, so I was trying to prove that every MCM is an high enough syzygy, I don't even know if that is true. Could you help me please?

Answer 1

In order to prove $\operatorname{Hom}_R(M/xM,C)=0$ it's enough to show that $x$ is $C$-regular; see Bruns and Herzog, Proposition 1.2.3. This holds when $C$ is MCM. (For a proof see Eisenbud, Commutative Algebra, Proposition 21.9.)