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?

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

$R$ 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 $A \le B$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.