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If $X\to Y$ is a map of varieties that is Zariski-locally isomorphic to a projection $U\times P\to U$ with $P$ (smooth) proper, I think the pullback $A_{\bullet}(Y)\to A_{\bullet}(X)$ is supposed to be injective.

Is this true and why?


(I keep thinking this should be able to be proven using rigidity lemma and/or Noetherian induction, but I think I'm just being slow.)

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    $\begingroup$ I think that this is probably not always true, even for a product. However, if $U$ has a rational point then the map has a rational section and the map given by composing the pullback with intersection with the class of the closure of the rational section and then pushing forward is the identity (by the projection formula). The same argument works if you have a class in $A_{\bullet}(X)$ which maps to $[Y]$ in $A_{\bullet}(Y)$. $\endgroup$
    – naf
    Commented Jul 22, 2018 at 5:28
  • $\begingroup$ To elaborate on ulrich's comment: if $P$ is defined over a field $k$ and $L$ is a finite extension of $k$ such that the extension of scalar map on Chow groups is not injective (this kernel is always torsion, but it doesn't have to be zero), then $U = Spec(L)$ is a counterexample. $\endgroup$ Commented Jul 22, 2018 at 20:54

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First for a positive result.This is true if $P$ is cellular (or more generally when the motive of $P$ splits into a sum of Tate motives), $X$ is smooth, and when the domain has degree larger than the dimension of $Z=Y\setminus U$.

Proof. The claim is true when $Z$ is empty (arbitrary degree):

Since $P$ is cellular, the motive of $U\times P$ decomposes into a product of the shifts of the motive of $U$, and the pullback of $U\times P\rightarrow U$ is the inclusion in appropriate codimension.

The claim is true in general:

From the localization sequences (with commuting squares and exact rows) $$\begin{matrix} \mathrm{CH}_i(Z)&\rightarrow& \mathrm{CH}_i(Y)&\rightarrow& \mathrm{CH}_i(U)&\rightarrow & 0 \\ \downarrow && \downarrow && \downarrow\\ \mathrm{CH}_{i+d}(X\times_Y Z) &\rightarrow &\mathrm{CH}_{i+d}(X)& \rightarrow & \mathrm{CH}_{i+d}(U\times P)&\rightarrow & 0\end{matrix}$$ one can check (by a diagram chase), for $i>\dim(Z)$ the map $\mathrm{CH}_i(Y)\rightarrow \mathrm{CH}_{i+d}(X)$ is an injection. $\square$

A counterexample is explained in the comments. E.g., one can take a Severi-Brauer variety for $Y$ satisfying the property that it's Chow group has torsion. Then a Galois splitting field for $Y$, say $L$, gives a variety $X=Y_L$ such that $\mathrm{CH}(Y)\rightarrow \mathrm{CH}(Y_L)$ is not injective.

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