Are there any examples known of an affine variety $A$ over an algebraically closed field such some Chow group (say, of codimension at least $2$) of $A$ with coefficients in $\mathbb{Z}/n\mathbb{Z}$ ($n$ is prime to the residue field characteristic) is non-zero?
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asked Sep 27, 2011 at 8:43
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5$\begingroup$ By a result of Jouanaolou ("Jouanolou's trick", see his article in LNM 341), every quasi-projective variety $X$ over a field (say) has a vector bundle $V_0$ and a torsor $V$ under $V_0$, such that $V$ is an affine scheme. If $X$ is regular, then the pull-back from $X$ to $V$ will induce an isomorphism ${\rm CH}(X)_{\bf Q}\simeq {\rm CH}(V)_{\bf Q}$, so this construction should provide you with a lot of examples. $\endgroup$– Damian RösslerCommented Sep 27, 2011 at 10:28
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$\begingroup$ @Damian: Fantastic! $\endgroup$– WandererCommented Sep 27, 2011 at 10:49
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$\begingroup$ Great! So, I can start with a projective space.:) $\endgroup$– Mikhail BondarkoCommented Oct 4, 2011 at 14:47
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I think there will be many such varieties.
For example, let $Q$ be a smooth $4$-dimensional projective quadric, $Q'$ a hyperplane section of $Q$ and $A = Q \backslash Q'$. For any $i$, we have the localisation sequence
$$ CH^{i-1}(Q') \to CH^i(Q) \to CH^i(A) \to 0 \ . $$
Since $CH^1(Q') = \mathbb{Z}/n\mathbb{Z}$ and $CH^2(Q) = (\mathbb{Z}/n\mathbb{Z})^{\oplus 2}$ (with $\mathbb{Z}/n\mathbb{Z}$ coefficients) it follows that $CH^2(A)$ is nonzero.
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$\begingroup$ Thanks! Still, could you provide me with some references for this example? Also, are there any (natural) generalizations for it? From mine (motivic) point of view, this question should depend on the ground field; does it (for large $n$)? $\endgroup$ Commented Sep 28, 2011 at 20:21
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$\begingroup$ Quadrics (over an algebraically closed field) have a cell decomposition which is what allows one to compute the Chow groups. There is a single cell in each dimension except for $d/2$ if the dimension $d$ is even, so this gives what I said. One can replace quadrics by Grassmannians to get similar examples. Of course, many more can be constructed using Damian Roessler's comment. $\endgroup$– nafCommented Sep 29, 2011 at 6:13