TITLE: Is noetherianity a local property?
QUESTION [15 upvotes]: 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 $R$.

REPLY [9 votes]: 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.