One knows that every commutative von Neumann regular ring with a finite Boolean algebra of idempotents is a product of fields.

Give an example of a commutative von Neumann regular ring which is not a product of fields.


closed as off-topic by user21820, Jean-Claude Arbaut, Adrian Keister, Jendrik Stelzner, José Carlos Santos May 16 at 18:28

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    $\begingroup$ @preethi It would have been nice to give a reference for the result you mentioned before the highlighted question. $\endgroup$ – user26857 May 31 '16 at 7:43

Every subring of a boolean ring is von Neumann regular, so you can pick a subring that isn't a product of fields.

For example, choose $F$ to be the field of two elements, and let $R$ be the subring of $\prod_{i=1}^\infty F$ generated by the identity and $\oplus_{i=1}^\infty F$.

So $R$ is a von Neumann regular ring, but it is not a product of fields. An infinite product of fields must have uncountable cardinality, but this ring is countably infinite.

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    $\begingroup$ It's not true that every subring of a product of fields is also von Neumann regular -- that would imply that every domain (a subring of a field) is von Neumann regular! But a subring of a product of fields is von Neumann regular if it is closed under the weak inverse operation. Your example works, but e.g. the subring generated by $1$ and $\oplus^\infty \mathbb Q$ in $\prod^\infty \mathbb Q$ is not von Neumann regular -- there are no weak inverses for the multiples of $1$. $\endgroup$ – tcamps Sep 13 '18 at 16:37
  • $\begingroup$ @tcamps Odd, I can't believe I claimed this. I would have raised the same objection if i were in your position. You're right, of course. Was I thinking of subrings of boolean rings? $\endgroup$ – rschwieb Sep 13 '18 at 16:54

Let $p \in \mathbb{Z}$ be a prime integer and consider $\mathbb{Z}_p$, the ring of $p$-adic integers. Then $R$ has prime ideals $(0)$ and $(p)$, and hence is a discrete valuation ring. It is easy to check that the localization of $\mathbb{Z}_p$ at each prime ideal is a regular local ring, verifying that $\mathbb{Z}_p$ is a regular commutative ring. Since $\mathbb{Z}_p$ is the ring of integers to the field $\mathbb{Q}_p$ it is clearly not a direct product of fields and you are done.

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    $\begingroup$ I'm afraid the question is about Von Neumann Regular rings. (Otherwise the question is trivial: every local regular ring which is not a field does the job.) $\endgroup$ – user26857 May 30 '16 at 21:36
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    $\begingroup$ I mean von Neumann regular $\endgroup$ – preethi May 31 '16 at 10:21

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