Is noetherianity a local property?

Question

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 $R$.

Answer 1

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.

Answer 2

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 $I_n=I_{n+1}=\cdots$.

Answer 3

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.