Suppose $R$ is a commutative ring. Let $q$ be a $p$-primary ideal of $R$ and $S$ be a multiplicative subset of $R$ such that $S\cap p=\emptyset$. Show that $S^{-1}q$ is $S^{-1}p$ primary.
It is easy to show that ${\rm rad}(S^{-1}q)=S^{-1}p$. I'm trying to show that $S^{-1}q$ is primary.
Let $\frac{a_1}{s_1}\cdot\frac{a_2}{s_2}\in S^{-1}q.$ So there is $\frac{r}{s_3}\in S^{-1}q$ such that
$$(a_1a_2s_3-s_1s_2r)s=0 \ \ \text{for some} \ s\in S$$
Let $\frac{a_1}{s_1} \not\in S^{-1}q$. I want to show that there is an integer $m$ such that $(\frac{a_2}{s_2})^m\in S^{-1}q$. Can anyone give me a hint?
Since $(a_1a_2s_3-s_1s_2r)s=0$ and $r\in q ,$ we have $s_1s_2rs=a_1a_2s_3s\in q.$ Notice that $S\cap p=\emptyset.$ So, $s_3s\notin p$ and thus $a_1a_2\in q$.