# Locally Noetherian Domain With Finitely Many Prime Ideals

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?

• You can read here a Lemma of Nagata which generalizes your situation. – 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 $$A \le B$$.