Contraction of extension of prime ideals in an integral ring extension

Question

This is probably an easy question, but I don't know the answer:

Let $R \subset S$ be an integral ring extension, and let $p$ be a prime ideal of $R$. Then $p^{ec} = pS \cap R$ can be different from $p$.

We always have $p \subseteq pS \cap R$, so if $\dim(R)=0$ (or if $R$ is Dedekind; the case $p=(0)$ being trivial), then we have equality. So, what would be a counter-example?

Answer 1

Thanks to user26857's helpful comment above, I think that I'm able to provide an answer.

By the lying-over property, there is a prime ideal $q$ of $S$ such that $q^c=p$. Then: $$p^{ec} = q^{cec}=q^c=p.$$

This is wrong is $S$ is not integral over $R$, for instance $p=2\Bbb Z \triangleleft R=\Bbb Z \subset S=\Bbb Q$.