Proposition 1

Let $h : A \to A'$ be a ring homomorphism such that $A'$ is integral over $A$, $\mathfrak{p}'$ a prime ideal of $A'$ and $\mathfrak{p} = h^{-1}(\mathfrak{p}')$. For $\mathfrak{p}$ to be maximal, it is necessary and sufficient that $\mathfrak{p}'$ be so.

Let us write $B = A/\mathfrak{p}$, $B' = A'/\mathfrak{p}'$ and let $h_1 : B \to B'$ be the homomorphism derived from $h$ by taking quotients; $B$ and $B'$ are integral domains and $B'$ is integral over $B$ ($§ 1$, no. 1, Proposition 2). To say that $\mathfrak{p}$ (resp. $\mathfrak{p}'$) is maximal means that $B$ (resp. $B'$) is a field. The proposition then follows from the following lemma: