Is a local ring Artinian, if the maximal ideal is nilpotent?

Question

This answer suggests the idea, that a local ring $(R, \mathfrak{m})$ whose maximal ideal is nilpotent is in fact an Artinian ring.

Is this true? If so, how is it proven?

Answer 1

You need that $R$ is noetherian, else there are counterexamples.

E.g., take $R = K[x_i]_{i \in \mathbb{N}}/(x_i | i \in \mathbb{N})^2$.

If $R$ is Noetherian, this is a special case of Lemma 10.59.4 here which states that a Noetherian ring of dimension $0$ is Artinian.

Answer 2

There is a version of this theorem even for noncommutative rings. It is called the Hopkins-Levitzki theorem.

It says, in a nutshell, that if $R/J(R)$ is semisimple and $J(R)$ is nilpotent, then $R$ is right Artinian iff right Noetherian. Yours is a special case where $R/J(R)$ is a field.

Here is the DaRT search which currently yields two examples of local, non-Artinian rings with nilpotent Jacobson radicals.