Proposition 5

Let $A$ be a ring and $B$ an $A$-algebra integral over $A$. If $A$ is a Jacobson ring, so is $B$.

Replacing $A$ by its canonical image in $B$, we may assume that $A \subset B$. Let $\mathfrak{p}'$ be a prime ideal of $B$ and let $\mathfrak{p} = A \cap \mathfrak{p}'$. There exists by hypothesis a family $(\mathfrak{m}_\lambda)_{\lambda \in L}$ of maximal ideals of $A$ whose intersection is equal top. For all $\lambda \in L$ there exists a maximal ideal $\mathfrak{m}'_\lambda$ of $B$ lying above $\mathfrak{m}_\lambda$ and containing $\mathfrak{p}'$ (§ 2, no. 1, Proposition 1 and Corollary 2 to Theorem 1). If we write $\mathfrak{q}' = \bigcap_{\lambda \in L} \mathfrak{m}'_\lambda$, then $\mathfrak{q}' \cap A = \bigcap_{\lambda \in L} \mathfrak{m}_\lambda = \mathfrak{p}$ and $\mathfrak{q}' \supset \mathfrak{p}'$, whence $\mathfrak{q}' = \mathfrak{p}'$ ($2$, no. 1, Corollary 1 to Proposition 1).