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. $\square$

## Comments (2)

Comment #3422 by Jonas Ehrhard on

1. Suggestion: Write $J \subsetneqq J'$ instead of $J' \neq J, J \subset J'$.
2. We use the fact, that the set ${J' \subset R ideal | J \subset J'}$ has a minimal element. This seems similar to the analogous statement for noetherian rings. Should this be proved?

Comment #3484 by on

Dear Jonas, your first suggestion is a style question which I am going to ignore. For your second point: a nonempty ordered set with dcc has a minimal element.

A more general remark is that the Stacks project isn't exactly building everything from scratch. The reader should have some basic knowledge of ring theory, fields, topology, set theory, combinatorics, etc. For example the discussion on Noetherian rings simply states some properties and then starts to do more interesting stuff. This makes the choices we've made in the exposition of earlier stuff in the Stacks project a bit arbitrary at times (the proof here included), but I don't see how to avoid this.

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