Minimal prime ideals and nilpotents Let $R$ be a commutative ring. Let $P\subset R$ be a minimal prime ideal. Let $S=R-P$. Let $x\in P$ and $s \in S$. Is there any property saying that if $sx=0$, then $x=0$?
Should anyone helps me, I will be very grateful. 
Thank you for all the comments. Maybe I was not asking what I wanted to ask. I am studying a proposition:

Let $R$ be a commutative ring. Let $P\subset R$ be a minimal prime ideal. Then every $x\in P$ is nilpotent. 

Its proof is using localization and it is like this:
Let $x\in P$ be nonzero. There is only one prime ideal in $R_P$, namely $PR_P$, and $x/1$ is a nonzero element in it. Indeed, $x$ is nilpotent if and only if $x/1$ is nilpotent...
So here is the problem I don't understand. How do we know $x/1$ is nonzero in $R_P$ if $x$ is nonzero? 
This proposition is on page 194 of Rotman's An Introduction to Homological Algebra. A proposition is proved beforehand:
If $S⊆R$ is a multiplicative set and $h: R → S^{-1}R$ is the localization map, then $\ker h = \{ r∈R : sr=0 \text{ for some } s∈S \}$.
So I was trying to know whether $P$ consists of all the zero divisors. 
 A: There is something wrong with this proposition in Rotman.  It is false whenever $R$ has more than one minimal prime, for example.
In general, the nilradical of a ring is the intersection of all prime ideals.  It follows that a prime ideal can only be contained in the nilradical if there is a unique minimal prime ideal.
The text attempts to get around this by localization, but, as you have observed, the localization will not generally be injective, and so elements may be nilpotent in $R_\mathfrak{p}$ without being nilpotent in $R$.

What we should conclude instead is that every element $x\in\mathfrak{p}$ of a minimal prime ideal is a zero divisor.  We can do that as follows:
First, observe that $x/1$ is nilpotent in $R_\mathfrak{p}$.  Rotman's argument shows this, and it also follows from the fact that $\mathfrak{p}R_\mathfrak{p}$ is the unique prime, hence the nilradical, of $R_\mathfrak{p}$.
So $x^n / 1 = 0$ for some $n$, in other words $x^n$ is in the kernel of the localization map $R\to R_\mathfrak{p}$.  As you mentioned, this means that there is some $s\in R\setminus\mathfrak{p}$ such that $sx^n = 0$.  Since $s\neq 0$, we can conclude that $x$ is a zero divisor.
A: It would even be true for any subset of $R$ if $R$ were an integral domain.  If $R$ is not an integral domain it's not true.  For a counterexample take $\mathbb{Z}/6$; the ideal generated by $3$ is prime (since the quotient by this ideal is an integral domain) and minimal (there are only 2 elements).  Now $2 \notin (3)$ but $2 \cdot 3 = 0$ 
