Which primes belong to the support of $S^{-1}M$ with respect to $R$

Question

Let $R$ be a noetherian ring, $S$ a multiplicatively closed subset of $R$ and $M$ a finitely generated $R$-module.

The support of $M$, denoted ${\rm Supp}_R(M)$, is the set of prime ideals $p$ of $R$ for which the localization $M_p$ is non-zero. Now $S^{-1}M$ is $S^{-1}R$-module and also an $R$-module (via the natural ring map $R\rightarrow S^{-1}R$).

We already know that $${\rm Supp}_{S^{-1}R}(S^{-1}M)=\{p\in {\rm Supp}_R(M): p\cap S=\emptyset\}$$

The question is which primes belong to ${\rm Supp}_{R}(S^{-1}M)$? In other words, which primes $p$ of $R$ satisfy $(S^{-1}M)_p\neq 0$?

I need the answer particularly when $S$ equals the set of non-zero divisors of $R$.

Answer 1

I'm not sure that the following considerations will be helpful.

I believe that $p$ is a prime ideal in $R$. Also I believe that $(S^{-1}M)_p$ means $C(p)^{-1}M$, where $C(p) = R\setminus p$ (as sets; $C(p)$ is a multiplicatively closed subset due to the fact that $p$ is a prime ideal).

So if $(S^{-1}M)_p\neq 0$ then there is $x\in M$ such $C(p)^{-1}(Rx)\neq 0$. If $C(p)^{-1}(Rx)\neq 0$ then $\textrm{Ann}(x)\subseteq p$ (where $\textrm{Ann}(x)=\{r\in R|rx=0\}$).

So it is possible to say that if $(S^{-1}M)_p\neq 0$ for a prime ideal $p$ then there is $x\in M$ such that $\textrm{Ann}(x)\subseteq p$.