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Every ring integral over a Jacobson ring is Jacobson. After quotienting by a prime, an integral equation for a nonzero element has nonzero constant term; choose a maximal ideal of the base avoiding that term and apply the Lying-over theorem.

For an integral extension $A\subseteq B$ of commutative rings,Contraction sends maximal ideals of $B$ to maximal ideals of $A$, while the Lying-over theorem puts a maximal ideal of $B$ over every maximal ideal of $A$.

For a commutative ring $R$, the Jacobson radical isLet $A\subseteq B$ be integral. If $\mathfrak{n}$ is maximal in $B$, then its contraction $\mathfrak{n}\cap A$ is maximal in $A$. Therefore every $a\in J(A)$ belongs to every $\mathfrak{n}$, andConversely, the Lying-over theorem puts a maximal ideal $\mathfrak{n}$ of $B$ above every maximal ideal $\mathfrak{m}$ of $A$. Hence an element of $J(B)\cap A$ lies in every $\mathfrak{m}$, provingThis is the Jacobson radical under an integral extension formula.

Let $d=\mathrm{deg}f$ and let $f_d$ be the nonzero highest homogeneous part. Choose one coordinate, after a permutation, such thatis not the zero polynomial. A polynomial of degree at most $d$ in each variable cannot vanish on the entire grid $\{0,1,\dots ,d{\}}^{n-1}$, by induction on the number of variables. Hence there are $a_i\in \{0,1,\dots ,d\}$ such that

SetThis is given, up to the initial coordinate permutation, by an integer matrix $M$ with determinant $\pm 1$ andIn the inverse coordinates $t_i=y_i+a_i{y}_n$, the coefficient of $y_n^d$ in $f$ is the nonzero real number $f_d(a_1,\dots ,a_{n-1},1)$. Dividing by it makes the defining equation monic in $y_n$. Thus $A$ is integral over $\mathbb{R}[y_1,\dots ,y_{n-1}]$ by linear Noether normalization for a hypersurface. The Lying-over theorem now makessurjective.

It is enough to prove that every prime ideal $\mathfrak{q}$ of $S$ is the intersection of the maximal ideals containing it. Replace $R\subseteq S$ byThe new extension is integral, both rings are domains, and the base remains a Jacobson ring.

Let $0\ne s\in S$. Choose an integral equation of least degreeAs above, $a_0\ne 0$. Since the zero ideal of $R$ is the intersection of its maximal ideals, choose a maximal ideal $\mathfrak{m}$ with $a_0\notin \mathfrak{m}$. By the Lying-over theorem, some maximal ideal $\mathfrak{n}$ of $S$ contracts to $\mathfrak{m}$. If $s\in \mathfrak{n}$, the integral equation would imply $a_0\in \mathfrak{n}\cap R=\mathfrak{m}$, a contradiction. Thus every nonzero $s$ is omitted by some maximal ideal, so their intersection is zero. Therefore $S$ is Jacobson, proving Integral extension of a Jacobson ring.