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.

$\begingroup$ For Ktheory 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 $RMod$ 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

1$\begingroup$ An infinite projective resolution does not naturally "live in" Ktheory. The comparison between Ktheory 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
If $f$ is a nonzerodivisor 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 nonzerodivisor. 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 nonzerodivisor. Now Quillen's resolution theorem applies to show that that the Ktheory of the category of finitely generated $R$modules is the same as the Ktheory of the category of finitely generated $R$modules on which $f$ acts as a nonzerodivisor.