Corollary 1

Let $h : A \to A'$ a ring homomorphism such that $A'$ is integral over $A$, $\mathfrak{p}$ a prime ideal of $A$ and $\mathfrak{p}'$ and $a'$ two ideals of $A'$ lying above $\mathfrak{p}$ such that $\mathfrak{p}' \subset a'$. If $\mathfrak{p}'$ is prime, then $a' = \mathfrak{p}'$.

Let us write $S = A - \mathfrak{p}$; then $S^{-1}A'$ is integral over $S^{-1}A$ ($§ 1$, no. 5, Proposition

16), $S^{-1}p$ is a maximal ideal of $S^{-1}A$ (Chapter 11, § 2, no. 5, Proposition 11), $S^{-1}a'$ and $S^{-1}p'$ are ideals of $S^{-1}A'$ lying above $S^{-1}p$ (Lemma 1) and $S^{-1}a' \supseteq S^{-1}p'$. As $S^{-1}p'$ is prime, it is maximal by Proposition 1 and hence $S^{-1}p' = S^{-1}a'$; therefore $a'$ is contained in the saturation of $p'$ with respect to $h(S)$, which is equal to $p'$ (Chapter II, § 2, no. 5, Proposition 11).