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?
abstract-algebraartiniancommutative-algebra
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?
Best Answer
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.