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How to prove that a semilocal ring such that all of its localizations at any maximal ideal are Noetherian is a Noetherian ring?

For example, for the easier example, $\operatorname{Max}(R)=\{m_1,m_2\}$, and $R_{m_1}$ and $R_{m_2}$ are noetherian rings. I believe that the ring $R$ is Noetherian, but I don't know how to prove it.

Thanks in advance.

Community
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Eärendil Elrond Arwen
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1 Answers1

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The general result is this:

Proposition: Let $A$ be a ring such that $A_m$ is Noetherian for every maximal ideal $m$. If every element of $A$ is contained in at most finitely many maximal ideals, then $A$ is Noetherian.

This is a standard exercise, and used in constructing the famous example (due to Nagata) of a Noetherian ring of infinite Krull dimension. Here is a proof sketch:

Let $I_0 \subset I_1 \subset \ldots$ be an ascending chain of ideals. Passing to $A/I_0 =: \overline{A}$, since $\overline{A}$ has finitely many maximal ideals by the assumption, we reduce to the semilocal case. Continuing to denote the ideals by $I_i$, we have that for each maximal ideal $m_i$ of $\overline{A}$, there exists $n_i$ with $(I_{n_i})_{m_i} = (I_{n_i+1})_{m_i}$. As there are finitely many indices, we can take their maximum, to obtain an $n$ with $(I_n)_m = (I_{n+1})_m$ for every maximal ideal $m$ of $\overline{A}$. Then $I_{n+1}/I_n$ is locally $0$ at every maximal ideal, hence is $0$.

zcn
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  • I included a proof sketch of the proposition in my original answer: are there specific points you would like clarified? – zcn Jan 18 '14 at 23:13
  • I believe that was interesting remark that the subset ${\mathcal{M}}0$ of the maximal ideals ${\mathfrak{m}} \in {{\mathcal{M}\mathtt{ax}}{A}}$ where the localization makes the chain trivial ($(0)$ or $A_{\mathfrak{m}}$ or $(0)\subset A_{\mathfrak{m}$})$ can be infinite but the set that can make (as posibility) infinite the lenght of the localized chain, say ${\mathcal{M}}1$ this is the set which is finite. The partition was ${{\mathcal{M}}ax}{A}={{\mathcal{M}}_0}\cup{{\mathcal{M}}_1}$, ${{\mathcal{M}}_0}\cap{{\mathcal{M}}_1}=\emptyset$, $\vert {{\mathcal{M}}_1} \vert \in \mathbb{N}$. – Eärendil Elrond Arwen Jan 30 '14 at 11:43