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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.

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Hajime_Saito
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2 Answers2

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  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.
user26857
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arienda
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  • 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 01 '13 at 18:28
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    @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.) –  Nov 01 '13 at 18:47
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    @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 01 '13 at 21:00
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    @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.) –  Nov 01 '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
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  1. The proof goes similarly to the one given in this answer.

Remark. If $S$ is not noetherian, then the property can fail; see here.

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