Let $R$ be a commutative ring with zero Jacobson radical such that each maximal ideal of $R$ is idempotent. Does it guarantee that each ideal is idempotent?

I know only that if each maximal ideal is generated by an idempotent element then $R$ turns out to be semisimple Artinian. I think this fact is associated with my question, at least if one could show that any maximal ideal is generated by an idempotent element.
Thanks for any suggestion!

  • $\begingroup$ As for the question in the first paragraph, I hope you're aware that a ring satisfying the conclusion is necessarily Von Neumann regular. Commutative VNR rings (Artinian or not) are consistent with your first question. Atheist ion becomes one about idempotent maximal ideals implying the ring is Von Neumann regular. Seems unlikely, but maybe... $\endgroup$ – rschwieb Mar 2 '16 at 11:16
  • $\begingroup$ @rschwieb What do you mean by "Atheist ion becomes ..."? $\endgroup$ – karparvar Mar 2 '16 at 18:14
  • $\begingroup$ Sorry, didn't notice that autocorrection. It was "The question". $\endgroup$ – rschwieb Mar 2 '16 at 19:27

In a commutative ring $R$ every ideal is idempotent (iff every ideal is radical) iff $R$ is VNR.

Then the question asks if a commutative ring $R$ with $J(R)=0$ and $\mathfrak m^2=\mathfrak m$ for every maximal ideal $\mathfrak m$ is VNR.

The answer is negative: the ring of continuous functions $R=\mathcal C[0,1]$ satisfies both conditions and it's not VNR.

  • $\begingroup$ This was my first thought as well, but I couldn't immediately see why the maximal ideals are idempotent. What's a good way to see that? $\endgroup$ – rschwieb Mar 10 '16 at 18:19
  • $\begingroup$ @rschwieb The maximal ideals are of the form $M_a=\{f\in C[0,1]:f(a)=0\}$. Moreover, any real function $f$ can be written as $(\sqrt[3]f)^3$, so $M_a=M_a^3$. $\endgroup$ – user26857 Mar 10 '16 at 21:22
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    $\begingroup$ Ahh, that is the trick! I had moved on to $C([0,1],\Bbb C)$ to try to get square roots of everything, but this is simpler. Bravo! I'm glad this line of thought works out. Rings like this (continuous functions on compact connected spaces) are pretty much the most conspicuous Jacobson-semisimple non-VNR rings that exist. $\endgroup$ – rschwieb Mar 10 '16 at 21:31
  • $\begingroup$ @user26857 Thank you very much for the answer. Why $R$ is not VNR? $\endgroup$ – karparvar Mar 11 '16 at 14:05
  • $\begingroup$ If $R$ is VNR then $R_{M_a}$ is a field, so $M_aR_{M_a}=0$. In particular, for the map $x-a\in M_a$ there is $s\in R-M_a$ such that $s(x)(x-a)=0$ for all $x\in[0,1]$. Then $s(x)=0$ for $x\ne a$ and since $s$ is continuous we get $s=0$, a contradiction. $\endgroup$ – user26857 Mar 11 '16 at 14:58

The book "Algebra" by "Hungerford", page 437, Corollary 3.5: If $I$ is an ideal in a semisimple left Artinian ring $R$, then $I = Re$, where $e$ is an idempotent which is in the center of $R$.

Edition by OP comment.
Assuming every maximal ideal is finitely generated, there is a positive answer to the question; see here

  • $\begingroup$ Thanks for the answer. If any maximal ideal is generated by an idempotent element then the ring is semisimple Artinian. But, my hypothesis assumes that any maximal ideal is idempotent ( and $J(S)=0$). Now, why is it generated by an idempotent element? $\endgroup$ – karparvar Mar 2 '16 at 4:02
  • $\begingroup$ @karparvar I think the problem is that the way you wrote the question, the second paragraph sounds like you want to study the special case when maximal ideals are generated by idempotents. If you really want to ask the question in the first paragraph, then this solution currently doesn't address the question. $\endgroup$ – rschwieb Mar 2 '16 at 19:28
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    $\begingroup$ @rschwieb Yes, I just mean what I asked in the first paragraph and edited the question. $\endgroup$ – karparvar Mar 3 '16 at 3:29
  • $\begingroup$ @karparvar: Assuming every maximal ideal is f.g., there is an answer to your comment here $\endgroup$ – user 1 Mar 4 '16 at 7:08

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