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Let $R$ be a ring with finitely many maximal ideals such that $R_{\mathfrak m}$ ($\mathfrak m$ maximal ideal) is noetherian ring for all $\mathfrak m$. Is $R$ noetherian?

I think $R$ has to be noetherian. Let $p_1 \subset p_2 \subset \cdots \subset p_n \subset \cdots$ be an infinite ascending chain of prime ideals in $R$, then I claim that there exist a maximal ideal $m$ which will contain all the prime ideals in this chain (from finiteness of the maximal ideals), this will give a chain of prime ideals in $R$ localised at $m$, but since that has to be finite (the local ring $R_m$ is noetherian) so the chain pulled back will terminate in $R$.

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    Noetherian means every ascending chain of ***ideals*** terminates and not necessarily prime ideals.2012-12-22
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    @BenjaLim, I believe that they are equivalent.2012-12-22
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    @Sanchez They are not. Consider the ring $R = k[\epsilon_i]_{i\in\mathbb{Z}_+}$ where $\epsilon_i^2 = 0.$ The ring $R$ has dimension zero but is not Noetherian.2012-12-22
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    @Sanchez: no, ACC on prime ideals is weaker than ACC on all ideals. For instance an infinite Boolean ring satisfies one condition but not the other. You may be thinking of: a ring is Noetherian iff all prime ideals are finitely generated.2012-12-22
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    @PeteL.Clark, Ah, thanks! That was what I was thinking of indeed.2012-12-22

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