= Minimal-right-ideal proof of radical nilpotence
{title2=$J(R)^m=0\text{ for some }m$}
In a <right Artinian ring>, the powers of its <Jacobson radical> stabilize at an ideal $I$ satisfying $I^2=I$. If $I\ne0$, choose a minimal <right ideal> $K$ with $KI\ne0$. Since $(KI)I=KI$, minimality gives $KI=K$. Choosing $k\in K$ with $kI\ne0$ gives $K=kR$ by the same minimality. Consequently $k=kj$ for some $j\in J(R)$. The <unit criterion for the Jacobson radical> makes $1-j$ invertible, forcing $k=0$, a contradiction. This proof avoids incorrectly applying the finitely generated form of <Nakayama lemma> to an arbitrary Artinian module.
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