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Let $R$ be an artinian ring. Then

  • $R$ is semilocal.

  • $(\operatorname{rad}(R))^n=0$ for some $n$.

Proof: Let $I=\text{rad}(R)$. By \hyperref[3.15]{3.15}, $\exists n$ so that we have $\operatorname{ann}_{\overline{R}} \overline{I}=\overline{0}$, where we denote $\overline{I}\subset \overline{R} = R / (0:I^n)$. We claim $\overline{R}=0$. ($\Rightarrow 0:I^n=R \Rightarrow I^n=0$).

Suppose $\overline{R}\neq 0$, then since $\overline{R}$ is artinian, $\exists N$ minimal with the property $N\neq 0$. Now $N$ is simple. Thus $N$ is finite, and $\overline{I}N=0$ or $\overline{I}N=N.$ The first equation is ruled out by \hyperref[3.15]{3.15}, the second one by \hyperref[2.2]{2.2} Nakayama's Lemma, ($I=\text{rad}(R)$), $N$ finite $R$-module.

I am going over my notes from last semester and I am trying to make the distinction of the proof of 1st bullet point and the 2nd (if there is one). Can someone help me with that? I am providing below the results that I am referring to.

3.15: Let $R$ be a ring that is artinian or noetherian. $I$ an $R$-ideal. Then $0:I^n=0:I^{n+1}$ for some $n$. For such $n$, write

$ \overline{R} = \left. R \middle/ (0:I^n) \right.$ and $\overline{I}=IR$.

Then $\operatorname{ann}_{\overline{R}} \hspace{0.1cm} \overline{I}=0$.

2.2: (Nakayama's Lemma) $M$ a finite $R$-module, $I$ an $R$-ideal. If $M=IM$, then $aM=0$ for some $a\in 1+I$. In particular, if $M=IM$ and $I\subset \operatorname{Rad}(I)$, then $M=0$ (since $a\in R$).

Bernard
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Username Unknown
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1 Answers1

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Everything you have written seems to be geared to proving the second point. I see nothing about the first point.

The first point is trivial though, so maybe you didn’t write it down? $R/J(R)$ has Jacobson radical zero, and is artinian since $R$ is. Therefore it’s semisimple, so $R$ is semilocal.

rschwieb
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  • This can also be proved from "first principles" i.e. without knowing about semisimple rings. Such a proof is in Atiyah-Macdonald. – RghtHndSd Jan 05 '19 at 22:38
  • @RghtHndSd Well, if your definition of semilocal is $R/J(R)$ is semisimple” it might be hard to do without semisimple rings , but I suppose in this case his definition might be “finitely many maximal ideals?” Those are the two I’m aware of. Is that what’s in A-M? – rschwieb Jan 06 '19 at 01:52
  • Yes - it's Proposition 8.3 in A-M. Consider the collection of all finite intersections of maximal ideals in $R$. $R$ artin implies that this collection has a minimal element, say $I = m_1 \cap \dots \cap m_k$. For any maximal ideal $m$, $m \cap I = I$ by minimality, and so $I \subset m$. By a straightforward element argument (A-M 1.11b), this implies $m_i \subset m$ for some $i$. That $m_i$ is maximal finishes the argument. – RghtHndSd Jan 06 '19 at 21:18