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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 \begin{equation} \mathfrak{R}^n = (0) \; , \end{equation} 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.

user26857
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Berber
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1 Answers1

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Assume that $\newcommand{\R}{\mathfrak{R}}\R^{n+1}=\R^n\ne0$, and let $P = \R^n$. Then, there exist left ideals $I$ which are contained in $P$ for which $IP\neq 0$, since $P^2 = P \neq 0$.

There is (by the Artinian condition) a minimal left ideal with $IP \ne0$ and $I\subseteq P$. There must exist some $x\in I$ for which $Px\neq 0$ so $I = Ax$ is principal by minimality. Choose $z\in P$ for which $zx =x$. Then $(1-z)x=0$, but $1-z$ is invertible, so $x=0$ which is a contradiction.

Pedro
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Angina Seng
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  • Why is there such an I by the Artinian condition? I thought this was purely Zorn's Lemma.

  • I don't see how xIR^n≠0 is implied by xR^n≠0

  • 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. – Angina Seng 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. – Angina Seng Jul 24 '18 at 19:32
  • I really don't see how the d.c.c implies every nonempty set of ideals contains a minimal element without zorn's lemma. in fact, that exactly is zorns lemma: "Lemma — Suppose a partially ordered set P has the property that every chain in P has an upper bound in P. Then the set P contains at least one maximal element. " (from wikipedia https://en.wikipedia.org/w/index.php?title=Zorn%27s_lemma&oldid=846327376) – Berber Jul 24 '18 at 19:38
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    @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. – Angina Seng Jul 24 '18 at 19:41
  • Sorry for the "discussion", you are right. – Berber Jul 24 '18 at 19:49