 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.