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

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.