Contraction of a maximal ideal under an integral extension
= Contraction of a maximal ideal under an integral extension
If $A\subseteq B$ is integral and $\mathfrak n$ is a maximal ideal of $B$, then $\mathfrak n\cap A$ is maximal in $A$. Indeed, $B/\mathfrak n$ is an integral domain integral over $A/(\mathfrak n\cap A)$, and a subring over which a field is integral is itself a field.