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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. – user38268 Dec 22 '12 at 7:47
@BenjaLim, I believe that they are equivalent. – user27126 Dec 22 '12 at 7:52
@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. – jspecter Dec 22 '12 at 8:27
@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. – Pete L. Clark Dec 22 '12 at 8:36
@YACP:Thanks for editing,and i think in a commutative ring it is enough to show that any ascending chain of prime ideal terminates to show that the ring is noetherian . – Sarjbak Dec 22 '12 at 15:30

A classical result of Nagata:

Let $R$ be a commutative ring with the following properties: (i) $R_{\mathfrak m}$ is noetherian for all $\mathfrak m\in\operatorname{Max}(R)$; (ii) every non-zero element of $R$ belongs to finitely many maximal ideals. Then $R$ is noetherian.

Proof. Take an ascending chain of ideals $0\neq I_1\subseteq I_2\subseteq\cdots$. Then $I_1$ is contained in a finite number of maximal ideals $\mathfrak m_1,\dots,\mathfrak m_r$ and we get that there exists $n\ge 1$ such that $$I_nR_{\mathfrak m_i}=I_{n+1}R_{\mathfrak m_i}=\cdots$$ for all $i=1,\dots,r$. Obviously $$I_nR_{\mathfrak m}=I_{n+1}R_{\mathfrak m}=\cdots=R_{\mathfrak m}$$ holds true for maximal ideals $\mathfrak m\neq\mathfrak m_i$ for all $i=1,\dots,r$. Thus we get $$I_nR_{\mathfrak m}=I_{n+1}R_{\mathfrak m}=\cdots$$ for any $\mathfrak m\in\operatorname{Max}(R)$. This is enough to show that $I_n=I_{n+1}=\cdots$.

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In this argument it what you have used for I_1 sits in finitely many maximal ideals :0 is contained in finitely many maximal ideals so number of maximal ideals are finite . So you can weaken the condition (ii)of the statement by assuming that R has finitely many finitely many maximal ideals. – Sarjbak Dec 22 '12 at 18:47
@Sarjbak Do you really think that assuming $R$ has finitely many maximal ideals means a weakness of the hypothesis in (ii)? Btw, $I_1$ is contained in finitely many maximal ideals since it is non-zero. – user26857 Dec 22 '12 at 19:59
In the argument you are only using that $I_1$ is contained in finitely many maximal ideals(from part 2 of the condition). – Sarjbak Dec 23 '12 at 3:53
@Sarjbak So what? – user26857 May 5 '13 at 14:07

Let $I$ be an ideal of $R$. Take a finite subset of $I$ which generates $I$ at all localizations at maximal ideals. This is possible because there are only finitely many maximal ideals.

Edit What I mean above is if $m$ is a maximal ideal of $R$, then $IR_m$ is generated by finitely many elements (in the image by $R\to R_m$) of $I$. In the particular case when $I\not\subseteq m$, then $IR_m=1$, this means there exists $s\in I\setminus m$. Then $s$ generates $IR_m$ (we don't necessarily take $1$ as generator if $1\notin I$).

Let $J$ be the ideal generated by this finite subset. Then the localization of $I/J$ at every maximal ideal is zero. This is then a classical result that $I/J=0$ (consider the annihilator ideal of $I/J$ and show it is not contained in any maximal ideal). So $I=J$ is finitely generated and $R$ is noetherian.

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what is meant by generates I at localisation at a maximal ideal ,do you mean the extension of the ideal I but that might contain 1 in local ring(s) . – Sarjbak Dec 22 '12 at 15:42
@QiL Got your solution thanks :).. – Sarjbak Dec 23 '12 at 4:17

Since $R$ has only finitely many maximal ideals, then we may assume the ascending chain of ideals is such that the first ideal belong to maximal ideal $A_{1}$, the second belong to maximal ideal $A_{2}$, etc. If at some point $\forall j\ge n$, $I_{j}\in A_{n}$ then we will be done. Suppose this does not happen, then the infinitely many ascending ideals are distributed in a finite set of maximal ideals, and there must be at least one maximal ideal containing infinitely many original ideals. Since we know passing to the quotient the part of the ideals outside of the maximal ideal is finitely generated, we only need to prove every ascending chain within the maximal ideal is finitely generated.

But we have an example of a local ring which is not noetherian. So your claim cannot hold.

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This would come into the second category that we have an ascending chain of ideals within a maximal ideal, and we have the desired counter example. – Bombyx mori Dec 22 '12 at 13:25

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