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Suppose $R$ is a finite ring and the only nilpotent element is zero. Prove that $$R\simeq F_{1}\times F_{2}\times \cdots\times F_{n}$$ where $F_{i}$ are fields.

Because $R$ is finite it is left artinian, then $J(R)$ is a nilpotent ideal, and because there is no nilpotent element except zero thus $J(R)={0}$, so $R$ is semisimple and by Artin-Wedderburn theorem we have: $$R\simeq M_{n_{1}}(D_{1})\times \cdots\times M_{n_{k}}(D_{k})$$ where $D_{i}$s are division rings. Now how I must show that $ M_{n_{i}}(D_{i})$ are fields?

kpax
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2 Answers2

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Hint. If $n_i\ge 2$, then find a non-zero nilpotent matrix in $ M_{n_{i}}(D_{i})$ (eventually of size $2\times 2$ and fill in the other entries by $0$). Then use that the finite division rings are...

  • thank you again!I am telling myself that I must find it out,but I couldn't,I don't know if you have same experience or not,but what is your advice to me to make improvement? – kpax Dec 25 '13 at 11:39
  • @kpax Let me tell you that in both answers to your questions today I've started to think (and made it clear in my answers) at the problem by considering smaller cases: here $n_i=2$ and there $M^2=0$. This is very useful in order to understand the general case! –  Dec 25 '13 at 12:17
  • I have learned lot of things today,thank you!I hope that I will be able to use them in my final exam! – kpax Dec 25 '13 at 14:21
  • @kpax The day is not over yet! Look at the answer I've posted here. –  Dec 25 '13 at 14:28
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Let $\frak{m}_1$, $\ldots$, $m_k$ be the maximal ideals of $R$. We have a surjection $$R \to \prod_{i=1}^k R/\frak{m}_i$$ with kernel $\frak{m}_1 \cap \ldots \cap\frak{m}_k = \frak{m}_1 \cdot \ldots \cdot\frak{m}_k$. So we only need to show that $J \colon = \frak{m}_1 \cdot \ldots \cdot\frak{m}_k=0$. Now the sequence $(J^n)_{n\ge 1}$ stabilizes, so there exists $n$ such that $J^n = J^{n+1} = J \cdot J^n$. Now Nakayama's lemma shows that $J^n=0$, and so $J=0$. Therefore, the above map is an isomorphism $$R \simeq \prod_{i=1}^k A/\frak{m}_i$$

orangeskid
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