Theorem 3

Let $A$ be a Jacobson ring, $B$ a finitely generated $A$-algebra and $\varphi : A \to B$ the canonical homomorphism. Then: (i) $B$ is a Jacobson ring. (ii) For every maximal ideal $m'$ of $B$, $m = \varphi^{-1}(m')$ is a maximal ideal of $A$ and $B/m'$ is a finite algebraic extension of $A/m$.

Let $\mathfrak{p}'$ be a prime ideal of $B$ and $\mathfrak{p} = \varphi^{-1}(\mathfrak{p}')$. Let $v$ be an element $\neq 0$ of $B/\mathfrak{p}'$.

As $B/p'$ is a finitely generated integral (Alp)-algebra and the canonical homomorphism $\phi : A/p \to B/p'$ is injective, there exists an element $u \neq 0$ of $A/p$ such that, for every homomorphism $f$ from $A/p$ to an algebraically closed field $L$ whose kernel does not contain $u$, there exists a homomorphism $g$ from $B/p'$ to $L$ whose kernel does not contain $v$ and for which $f = g \circ \phi$ (no. 1, Corollary 3 to Theorem 1). Since $A$ is a Jacobson ring, there exists a maximal ideal $m$ of $A$ containing $p$ and such that $u \notin m/p$. We take $L$ to be an algebraic closure of $A/m$ and $f$ to be the canonical homomorphism $A/p \to L$; let

be a homomorphism such that $f = g \circ \phi$ and $g(v) \neq 0$. Then

hence $g(B/p')$ is a subfield of $K$ (Algebra, Chapter V, § 3, no. 2, Proposition 3) and the kernel of $g$ is therefore a maximal ideal of $B/p'$ not containing $v$. Thus it is seen that the intersection of the maximal ideals of $B/p'$ is reduced to 0, which proves that $B$ is a Jacobson ring. Moreover, if $p'$ is maximal, $g$ is necessarily injective and hence $p = m$ is maximal; finally $B/p'$ is then a finitely generated algebra over the field $A/m$ and hence is a finite extension of $A/m$ (no. 1, Corollary 2 to Theorem 1).