Remark (2)

Let $h : A \to A'$ be a ring homomorphism such that $A'$ is integral over $A$; then the associated continuous mapping $^a h : \mathrm{Spec}(A') \to \mathrm{Spec}(A)$ is closed. For every ideal $a'$ of $A'$, $A'/a'$ is integral over $A'$, hence also over $A$ ($§ 1$, no. 1, Proposition 6) and $\mathrm{Spec}(A'/a')$ is identified with a closed subspace $V(a')$ of $\mathrm{Spec}(A')$; to show that $^a h$ is closed, we see then (replacing $A'$ by $A'/a'$) that it is sufficient to prove that the image of $\mathrm{Spec}(A)'$ under $^a h$ is a closed subset of $\mathrm{Spec}(A)$; now it follows from Theorem 1 that this image is just the set of prime ideals of $A$ containing the ideal $\mathrm{Ker}(h)$ and this set is closed by definition of the topology on $\mathrm{Spec}(A)$.