# The Jacobson radical of an artinian ring is nilpotent

I am looking for a nice proof for the following lemma, which will later help prove that a ring is artinian if and only if it is noetherian and all prime ideals are maximal.

Let $A$ be an artinian ring. Then $\mathfrak{R}$ is nilpotent, id est, there exists an $n \in \mathbb{N}$ such that $$\mathfrak{R}^n = (0) \; ,$$ where $\mathfrak{R} = \operatorname{Jac}A$ denotes the Jacobson radical of $A$.

We shall not use that artinian rings are noetherian, since, as stated above, this is what I want to prove after having shown this statement. My ansatz is the following:

Since $A$ is artinian, we have $\mathfrak{R}^{n+1} = \mathfrak{R}^n$ for some $n\in \mathbb{N}$. Let $x \in \mathfrak{R}^n$. I would like to find a finitely generated $A$-submodule (that is, a finitely generated ideal contained in $\mathfrak{R}^n$) $x \in I \unlhd \mathfrak{R}^n$ such that we have $\mathfrak{R}I = I$, and we could thus use Nakayama's lemma to obtain $I=\{0\}$ and therefore $x = 0$, which in conclusion would show $\mathfrak{R}^n = (0)$. Do we already have $\mathfrak{R}(x) = (x)$? I don't think so. Maybe we need to use Zorn's lemma to find such a submodule.

What I know is that $\operatorname{Max} A$, the set of all maximal ideals of $A$, is finite, and that we have $\operatorname{Spec}A = \operatorname{Max}A$, that is, every prime ideal is maximal. Also, for a short exact sequence of $A$-modules $0 \rightarrow M^\prime \rightarrow M \rightarrow M^{\prime\prime} \rightarrow 0$, $M$ is artinian if and only if $M^\prime$ and $M^{\prime\prime}$ are artinian. I don't think this will help, though. Rather, these facts together with this lemma are useful to show that a ring is artinian if and only if it is noetherian and every prime ideal is maximal.

## 1 Answer

Assume that $\newcommand{\R}{\mathfrak{R}}\R^{n+1}=\R^n\ne0$. There is (by the Artinian condition) a minimal ideal with $I\R^n \ne0$. Then $(I\R^n)\R^n=I\R^{2n}=I\R^n\ne0$ so $I\R^n=I$ by minimality. There is $x\in I$ with $x\R^n\ne0$. Then $xI\R^n\ne0$ so $I=xI=xR$ by minimality. Then $I$ is finitely generated (principal even) and $xI=I$. Now use Nakayama.

This is a standard textbook proof.

• 1. Why is there such an I by the Artinian condition? I thought this was purely Zorn's Lemma. 2. I don't see how xIR^n≠0 is implied by xR^n≠0 3. I hope I don't sound condescending or mean when I say I don't view this proof as a "nice proof", since the resulting contradiction is "ugly" ("ugly" meaning that it is not a simple contraposition). Well that's for me to deal with :) – Berber Jul 24 '18 at 19:23
• @bert Artinian means "every nonempty collection of ideals has a minimal element". Nothing to do with Zorn. – Lord Shark the Unknown Jul 24 '18 at 19:25
• Ah ok. My definition of Artinian is that it satisfies the d.c.c. That is equivalent to "every nonempty set of submodules has a minimal element" if we assume Zorn's lemma, that was my thinking. Or am I wrong? – Berber Jul 24 '18 at 19:31
• @Bert You don't need Zorn. – Lord Shark the Unknown Jul 24 '18 at 19:32
• @Bert The hypothesis that each chain $I_1\supseteq I_2\subseteq\cdots$ always stabilises is rather stronger than that saying that each chain in $P$ has a lower bound. – Lord Shark the Unknown Jul 24 '18 at 19:41