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Let $R$ be a ring and $f\in R$. Is there something like an $f$-regular $K$-theory group of $R$ based on the category of $f$-regular $R$-modules, i.e. modules that do not have any $f$-torsion? If needed, I could assume that $R$ is an integral domain.

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  • $\begingroup$ For K-theory you consider projective modules, which are direct summands of free, hence don't have torsion if the ring is integral. $\endgroup$ – user1688 Jun 19 '15 at 7:47
  • $\begingroup$ I think the $K$-theory of the entire category $R-Mod$ is also considered. By taking projective resolutions, it equals the $K$-theory coming from projective modules. For this latter, I think it is necessary to assume Noetherian. $\endgroup$ – stupidq75 Jun 19 '15 at 8:13
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    $\begingroup$ An infinite projective resolution does not naturally "live in" K-theory. The comparison between K-theory of finitely generated modules and projective modules needs the assumption that $R$ is regular (which translates into existence of finite free resolutions). $\endgroup$ – Matthias Wendt Jun 19 '15 at 8:36
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If $f$ is a non-zero-divisor in $R$ and $R$ is noetherian, then any finitely generated $R$-module has a resolution of length 1 by finitely generated modules upon which $f$ acts as a non-zero-divisor. One sees that by starting the resolution with a free module and then observing that a submodule of a free module has $f$ acting it as a non-zero-divisor. Now Quillen's resolution theorem applies to show that that the K-theory of the category of finitely generated $R$-modules is the same as the K-theory of the category of finitely generated $R$-modules on which $f$ acts as a non-zero-divisor.

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