Going mod an element which annihilates a prime ideal (Associated primes)

Question

Suppose $R$ is a Cohen-Macaulay local ring and $P=ann(x)$, an associated prime of $R$. Now consider the ring $S=R/(x)$. Will $PS$ (extension of $P$ to $S$) still consists of zero divisors of $S$? Or will $PS$ have a non zero divisor on S?

Answer 1

It can happen either way. $PS$ may be either the maximal ideal of $S$ or $0$ in the extreme cases.

Note also, since $R$ is a local C-M ring, it is either local Artininian or the maximal ideal is not associated.

For example, let $A$ be a Discrete valuation Ring with maximal ideal $pA$ and let $R=A[[x]]/(px)$. Then the minimal primes $x,p$ of $R$ are each other's annihilators, and $S= R/xR= A$, $pS = pA$ contains the regular element $p$.

On the other hand, if $R$ is a domain (take it to be C-M if you want) that is not a field, then pick a non-unit $r \in R$. You have $P= 0 = Ann(r)$ is the unique associated prime of $R$, and setting $S = R/rR$, also $PS= 0$.

Another example: if $R$ is an Artinian local ring that is not a principal ideal ring, then its unique prime ideal $M$ is associated, say $M = Ann(x)$,and $S = R/xR$ is an Artinian local ring that is not a field. Thus, $MS$ is the maximal ideal of $S$, it is nonzero, and it consists entirely of zero divisors.