Let $(S,\mathfrak{m})$ be a regular local ring, $0\neq f\in \mathfrak{m}$, and $R=S/(f)$.
(a) Suppose $\varphi, \psi$ are $n\times n$ matrices with entries in $\mathfrak{m}$ such that $\varphi \psi=f\cdot \text{Id}_n$. Let $M=\text{coker}\varphi$. Show that (1) $M$ is an $R$-module, (2) $M$ is maximal Cohen-Macaulay R-module ($\text{depth}_R(M)=\text{dim}R$), (3) the minimal number of generators for $M$ over $R$ is $n$.
(b) Conversely, given a maximal Cohen-Macaulay $R$-module $M$ with $n$ minimal generators, show that there exist $\varphi, \psi$ with $\varphi \psi=f\cdot \text{Id}_n$ and $M=\text{coker}\varphi$.
Theorem 2.1.1 of the following seems relevant, but I don't know how to solve this.
For a), you get map $F\stackrel{\phi}{\to} F\to M\to 0$ where $F$ is a free $S$-m0dule of rank $n$ and since $\phi\psi=f \mathrm{Id}_F$ you have $\det\phi\neq 0$. Thus, $\phi$ is injective. The same condition says $fF=\phi\psi (F)\subset\phi(F)$ and so $M$ is a quotient of $F/fF$ and thus an $R$-module. The exact sequence gives by Auslander-Buchsbaum depth of $M=$ depth of $S-1$, which is depth of $R$ and thus $M$ is maximal Cohen-Macaulay. Finally, the entries of $\phi$ are in the maximal ideal says that $M$ is minimally generated by $n$ elements.
For, b), you have a surjection $F\to M$, $F$ as above and use Auslander-Buchsbaum to deduce that the kernel is free and hence isomorphic to $F$, giving your $\phi$. The rest should be clear.