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 $R$.
Best Answer
A classical result of Nagata:
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 $I_n=I_{n+1}=\cdots$.