Suppose $A$ is Artinian and commutative with 1. If $J(A)M=M$, then $M=\{0\}$. $J(A)$ is the Jacobson radical of A.

Question

I just want to know if the following proof is valid for the above theorem. (Note: $M$ is a $A$-module)

Sketch: Since $A$ is Artinian we know that $J(A)$ is nilpotent i.e. there exists $k\geq 1, k\in \mathbb{Z}$ such that $J(A)^k=\{0\}$. Hence, ${0}=\{0\}M=J(A)^kM=J(A)^{k-1}M=...=J(A)M=M$. Is this valid? Does it also mean that Nakayama's Lemma holds for non-finitely generated modules of Artinian Rings?

Answer 1

It clearly shows that (this formulation of) Nakayama’s lemma holds for all modules, not just finitely generated ones, over a ring with nilpotent Jacobson radical.

The formulation of Nakayama’s lemma you gave is also still true for noncommutative rings.