It is enough to prove that every prime ideal $\mathfrak q$ of $S$ is the intersection of the maximal ideals containing it. Replace $R\subseteq S$ by
$$
R/(\mathfrak q\cap R)\subseteq S/\mathfrak q.
$$
The 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 degree
$$
s^r+a_{r-1}s^{r-1}+\cdots+a_0=0.
$$
As above, $a_0\ne0$. 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>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 101 / 4 / iv / Solution

{c}

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>.

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 Incoming links: Integral extension of a Jacobson ring