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I'm working on the following problem and I'm stuck. Any hints or solutions would be appreciated

Let $R$ be a left Artinian ring with Jacobson radical $J(R)$. If $R \neq J(R)$. show that $R$ is a left Noetherian ring.

Here are my thoughts: I think we have to use the ascending/descending chain definitions for artinian and noetherian since I dont see how we can show that all ideals are finitely generated. Besides that, maybe we can somehow use the condition that $J(R)\neq R$ by considering an element in $R$ that is not in $J(R)$... but I'm not sure how that's useful.

Source: Spring 1996

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The ring $R/J(R)$ is semisimple Artinian. Thus every Artinian left module over $R/J(R)$ is Noetherian and the same is true for every Artinian left $R$-module $M$ such that $JM=0$.

Since $J^n$ is Artinian as a left $R$-module, we can conclude that $J^{n}/J^{n+1}$ is Noetherian.

It remains to show that $J=J(R)$ is nilpotent, say $J^m=0$, because then we can consider the chain $$ 0=J^m\subseteq J^{m-1}\subseteq \dots \subseteq J^2\subseteq J\subseteq R $$ where each factor is Noetherian.

Since $R$ is Artinian, there is $m$ such that $J^k=J^m$, for every $k\ge m$. Suppose $J^m\ne0$. There is a left ideal $I$ such that $J^mI\ne0$, namely $I=J$. So we can pick $I_0$ minimal such that $J^mI_0\ne0$. Let $x\in I_0$ with $J^mx\ne0$; then $J^m(J^mx)=J^{2m}x=J^mx\ne0$. Thus we conclude $J^mx=I_0$, by minimality. In particular, there exists $y\in J^m$ with $yx=x$. However, $y\in J$, so $-y$ is left-quasi regular: there exists $z$ with $zy=z+y$, hence $$ zx=zyx=zx+yx $$ from which $yx=0$: a contradiction. (The proof is from Kaplansy’s “Fields and Rings”.)

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