Lemma 10.53.4. Let $R$ be Artinian. The Jacobson radical of $R$ is a nilpotent ideal.

**Proof.**
Let $I \subset R$ be the Jacobson radical. Note that $I \supset I^2 \supset I^3 \supset \ldots $ is a descending sequence. Thus $I^ n = I^{n + 1}$ for some $n$. Set $J = \{ x\in R \mid xI^ n = 0\} $. We have to show $J = R$. If not, choose an ideal $J' \not= J$, $J \subset J'$ minimal (possible by the Artinian property). Then $J' = J + Rx$ for some $x \in R$. By NAK, Lemma 10.20.1, we have $IJ' \subset J$. Hence $xI^{n + 1} \subset xI \cdot I^ n \subset J \cdot I^ n = 0$. Since $I^{n + 1} = I^ n$ we conclude $x\in J$. Contradiction.
$\square$

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