Existence of prime ideal lying over $p \implies pS \cap R=p $?

Question

Let $R \subseteq S$ be rings and $p$ be prime ideal in $R$. Suppose there is a prime ideal $q$ in $S$ such that $q\cap R=p$. Then prove that $pS \cap R=p$.

Can one please provide some hint ? I've only noticed extension of $q$ in $S_{R-p}$ is a proper ideal and if $pS \cap R \neq p$ then extension of $pS$ in $S_{R-p}$ is the whole ring $S_{R-p}$. Probably I'm missing something important here, can one please help ?

Answer 1

$q\cap R=p\implies pS\subseteq q\implies pS_{R-p}\subseteq qS_{R-p}$.