3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 :) $\endgroup$ – Berber Jul 24 '18 at 19:23
  • $\begingroup$ @bert Artinian means "every nonempty collection of ideals has a minimal element". Nothing to do with Zorn. $\endgroup$ – Lord Shark the Unknown Jul 24 '18 at 19:25
  • $\begingroup$ 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? $\endgroup$ – Berber Jul 24 '18 at 19:31
  • $\begingroup$ @Bert You don't need Zorn. $\endgroup$ – Lord Shark the Unknown Jul 24 '18 at 19:32
  • 1
    $\begingroup$ @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. $\endgroup$ – Lord Shark the Unknown Jul 24 '18 at 19:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.