I saw an exercise which says the following:
For a Cohen-Macaulay ring $R$ and a maximal Cohen-Macaulay module $M$, an $R$-regular sequence is also $M$-regular. Why is this true?
The only thing I know about MCM is that $\operatorname{dim}(R) = \operatorname{dim}(M)$, so $\operatorname{depth}(R) = \operatorname{depth}(M)$. How should I proceed?
One way of looking at this problem is that the set of zero divisors for $M$ is the union of its associated primes. When $R$ is Cohen-Macaulay and $M$ is maximal Cohen-Macaulay, we know the associated primes of both $R$ and $M$ are minimal primes of $R$. But $R$ is fully supported, so if $x$ is a non-zerodivisor on $R$, thus avoiding every minimal prime of $R$, it must also avoid every minimal prime of $M$. Induction then extends this from a single element to showing an $R$-regular sequence must be $M$-regular.