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.

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