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Let $R$ be a commutative ring such that $R/J(R)$ is von Neumann regular, where $J(R)$ is the Jacobson radical of $R$. I conjecture that the idempotents of $R/J(R)$ lift modulo $J(R)$. Is it a fact?

Indeed, this is equivalent to the statement that every direct summand of the $R$-module $R/J(R)$ would have a projective cover by Proposition 27.4 of "Rings and Categories of Modules" by Fuller & Anderson. Thanks for any help!

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No, this conjecture is not true.

You can semi localize $\mathbb Z$ at two maximal primes, so that the ring mod the radical is a product of two fields.

But the localization is a domain, so obviously there aren't enough idempotents to lift to.

These are studied under the name semiregular rings, by the way. It's well covered in Rings Related to Stable Range Conditions by Chen, and in several works of Tuganbaev, I'm sure.

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  • $\begingroup$ Thanks for your cooperation! What do you mean by "... semi localize $\mathbb Z$ at two maximal primes"? Principally, what does mean "to semi localize"? $\endgroup$ – karparvar Dec 23 '16 at 17:06
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    $\begingroup$ @karparvar I think is pretty clear from the answer what this means: localize a ring $R$ at $R\setminus(p_1\cup p_2)$ for two prime ideals $p_1,p_2$. $\endgroup$ – user26857 Dec 24 '16 at 0:13
  • $\begingroup$ @karparvar yes, user26857 is describing what I mean. It appears in this query in the DaRT $\endgroup$ – rschwieb Dec 24 '16 at 1:19

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