# $R\subseteq S$ integral extension and $S$ Noetherian implies $R$ Noetherian?

The problem is as follows:

Let $R\subseteq S$ be an integral extension and $S$ a Noetherian ring.

1. Then show that for each $\mathfrak p\in \operatorname{Spec}R$, there are only finitely many $P\in \operatorname{Spec}S$ such that $P\cap R = \mathfrak p$.

2. Is $R$ also Noetherian?

I am able to show the first part, but unable to show the second part. Any help is welcome.

1. Consider the field $$\widetilde{\mathbb{Q}}=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5},\ldots)=\mathbb{Q}(\sqrt{p}\mid\;p\mbox{ is positive prime integer}).$$ The rings $R=\mathbb{Q}+X\widetilde{\mathbb{Q}}[X]$ and $S=\widetilde{\mathbb{Q}}[X]$ provide a counterexample to your question.
• It is unclear for me that $\tilde{\mathbb Q}[X]$ is noetherian. Maybe you can replace $\tilde{\mathbb Q}$ by $\overline{\mathbb Q}$ (an algebraic closure of $\mathbb Q$) ? – Cantlog Nov 1 '13 at 18:28
• @Cantlog $\widetilde{\Bbb Q}$ is a field. The key of this example is that the field extension $\Bbb Q\subset\widetilde{\Bbb Q}$ is not finite. (In general, if $A\subset B$ is a ring extension, then $A+xB[x]$ is Noetherian iff $A$ is Noetherian and $A\subset B$ is finite, and this remark provides plenty of counterexamples.) – user89712 Nov 1 '13 at 18:47
• @user: yes you are right. Note for the OP: $R$ is noetherian if $S$ is finite over $R$ (Eakin's theorem, see Matsumura: Commutative Algebra, p. 263). – Cantlog Nov 1 '13 at 21:00
• @Cantlog In his subsequent book Matsumura repaired the omission of Nagata's name, calling that theorem Eakin-Nagata. Although Nagata proved the theorem one year later than Eakin, his proof was significantly simpler. Nowadays we know Formanek's proof which is probably the simplest possible one. (Btw, the result I've mentioned above about the rings $A+xB[x]$ uses Eakin-Nagata theorem for the if part.) – user89712 Nov 1 '13 at 22:25
• Dear @user89712, could you please give a proof or reference for your impressive criterion of noetherianness of $A+xB[x]$ ? – Georges Elencwajg Mar 16 '18 at 23:31
Remark. If $S$ is not noetherian, then the property can fail; see here.