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Let $X$ be a regular scheme and consider Grothendieck's $\gamma$-filtration $F^nK(X)$ on $K(X)$. For the graded pieces, one has $Gr^0K(X) = CH^0(X)$ and $Gr^1K(X) = \mathrm{Pic}(X) = CH^1(X)$. Does this continue to hold, i.e., do we have $Gr^pK(X) = CH^p(X)$?

I found that for $X/k$ smooth quasi-projective, $CH^q(X,p) \otimes \mathbf{Q} = K_p(X)^{(q)} \otimes \mathbf{Q}$, so this holds after rationalising.

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Note that I've rewritten the second paragraph of my answer, which originally contained too strong a statement. – Steven Landsburg Jun 10 '13 at 19:53
up vote 4 down vote accepted

The map between the graded $K$-theory ring on the one hand and the Chow ring on the other is defined via Chern classes and requires denominators. I know of no good reason to expect an integral isomorphism (or even a map), but I'm not aware of an explicit counterexample (though I'm vaguely aware that the experts think the place to look for that counterexample is over a field with large etale cohomological dimension).

On the other hand, if you replace the $\gamma$-filtration with the filtration by codimension of support, then you do get $Gr^p(X)$ as a quotient (with torsion kernel) of $Ch^p(X)=H^p(X,K_p)$ (over the integers) provided $X$ is both regular and of finite type over a field --- though it would follow from Gersten's conjecture that this holds for all regular $X$.

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Thank you. By the way, one has $F^1\gamma K(X)=F^1_{top}K(X)$ if there is an ample sheaf. – Timo Keller Jun 10 '13 at 16:07

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