# Is the localization of such a Noetherian domain at every maximal ideal Artinian?

My question here is inspired by this question here. I am not asking how to prove the problem there, but if an alternative approach is possible.

Suppose $A$ is a Noetherian integral domain in which for every maximal ideal $\mathfrak{m}$, the dimension of $\mathfrak{m}/\mathfrak{m}^2$ as a vector space over $A/\mathfrak{m}$ is $1$. Given any maximal ideal $\mathfrak{m}$ of $A$, we know that $A_\mathfrak{m}$ is local and Noetherian with maximal ideal $I = \mathfrak{m}_\mathfrak{m}$. From here I want to conclude from here that $A_\mathfrak{m}$ is actually an Artinian ring as well. Can I argue along the following lines? I want to prove that there is an $n$ such that $I^n = I^{n+1}$. This is because if $I$ is like this then I can apply a Nakayama Lemma argument to show that $I$ is nilpotent and then show that $A_\mathfrak{m}$ is Artinian as desired.

However how can I translate the condition on $\mathfrak{m}/\mathfrak{m}^2$ having $A/\mathfrak{m}$ - dimension 1 into something concrete in the localisation?

Perhaps it may well be that $A_\mathfrak{m}$ does not need to be Artinian and I am wrong.

Thanks.

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An integral domain (never mind noetherian!) in which every prime ideal is maximal is a field. But it seems your question doesn't have anything to do with the title... –  Zhen Lin Dec 2 '12 at 14:47
@ZhenLin Yes that trivially follows from the fact that the zero ideal is now maximal. I have edited the title of my question as I agree that it was misleading before. –  user38268 Dec 3 '12 at 7:07

## 2 Answers

A ring is Artinian if and only if it is Noetherian and of dimension zero. The dimension of $A_\mathfrak{m}$ is equal to the height of $\mathfrak{m}$. Because $A$ is Noetherian, $A_\mathfrak{m}$ is as well, so $A_\mathfrak{m}$ is Artinian if and only if $\mathfrak{m}$ has height zero. So you would need $\mathfrak{m}$ to be both maximal and minimal. But in a domain (such as $A$), the only minimal prime is $0$, so this can only happen if $\mathfrak{m}=0$, in which case $A$ is a field...but your $A$ is not a field because $\mathfrak{m}/\mathfrak{m}^2\neq 0$.

In fact, your $A$ is a Dedekind domain. Indeed, the $k(\mathfrak{m}):=A/\mathfrak{m}$-vector spaces $\mathfrak{m}/\mathfrak{m}^2$ and $\mathfrak{m}A_\mathfrak{m}/(\mathfrak{m}A_\mathfrak{m})^2$ are canonically isomorphic, so both are $1$-dimensional by assumption. By Krull's principal ideal theorem, $\dim(A_\mathfrak{m})\leq 1$. I've explained why it's not possible that $\dim(A_\mathfrak{m})=0$, so this inequality is an equality for all $\mathfrak{m}$. Thus each $A_\mathfrak{m}$ is a regular local ring of dimension $1$, i.e., a discrete valuation ring. So $A$ is Dedekind.

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Take $A = \mathbb{Z}$ for a counterexample.

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