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Let R be a domain with finitely many prime ideals such that the localization at each prime, $R_{\mathfrak p}$, is Noetherian. Then is $R$ necessarily Noetherian?

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  • $\begingroup$ You can read here a Lemma of Nagata which generalizes your situation. $\endgroup$ – user26857 Apr 25 '19 at 22:57
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$R$ is Noetherian.

Proof: Suppose, for a contradiction, that $R$ is not Noetherian. Then, consider a strictly ascending chain of ideals in $R$,

$$ I_1 \lt I_2 \lt I_3 \lt \cdots $$

Now let $M_1, ..., M_n$ be the maximal ideals of $R$ (since maximal ideals are prime, there are only finitely many). Then for each $M_i$,

$$ I_1 R_{M_i} \le I_2 R_{M_i} \le I_3 R_{M_i} \le \cdots $$

is an ascending chain of ideals, so eventually becomes stationary, at $I_{N_i} R_{M_i}$, say.

Now let $N := \mathrm{max} \{ N_1, ..., N_n \}$ so that $I_N R_M = I_{N+1} R_M$ for every maximal ideal, $M$. But this implies$^{\dagger}$ that $I_N = I_{N+1}$ which is a contradiction.


$\dagger$: Here we have used the theorem

If $A$, $B$ are ideals of a domain $R$ such that $A R_M \le B R_M$ for every maximal ideal $M$, then $A \le B$.

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