TITLE: Commutative artinian ring is noetherian
QUESTION [8 upvotes]: Suppose R is a commutative Artinian ring then R is Noetherian.
I am aware of the proof which uses the idea of filtration. But I would like to prove this fact without that idea but haven't got far enough. Is there any reference for such a proof? or is anyone aware of one?

REPLY [6 votes]: Here is a direct proof that the Jacobson radical $J$ of an Artinian ring is nilpotent.
We know that $J^n = J^{n+1}$ for some $n \geq 1$. Let $X = \{a \in A : aJ^n = 0\}$; it is enough to show $X = A$ as then $1 \in X$ and $J^n = 0$.
Now if $X < A$ then we can find a minimal non-zero submodule $Y/X$ of $A/X$ because $A$ is Artinian. Then $Y/X$ is a simple $A$-module, so the definition of the Jacobson radical as the simultaneous annihilator of all simple modules implies that $YJ \subseteq X$. Hence
$$YJ^n = YJ^{n+1} = YJ.J^n \subseteq XJ^n = 0$$
and hence $Y \subseteq X$. This contradicts the assumption that $Y/X$ is non-zero. Thus $X = A$ as required.
This proof also works in the non-commutative case, and shows that a right Artinian ring is right Noetherian.