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Timeline for Chow group and base change

Current License: CC BY-SA 3.0

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Jan 3, 2018 at 2:20 vote accept CommunityBot
Jan 3, 2018 at 2:20 comment added user39380 @JasonStarr Sorry I misunderstood your example, thank you!
Jan 2, 2018 at 22:59 answer added Eoin timeline score: 5
Jan 2, 2018 at 20:28 comment added Jason Starr Typo correction: The appropriate Chow group is $\text{CH}_0(X)$, not $\text{CH}^1(X)$.
Jan 2, 2018 at 19:26 comment added R. van Dobben de Bruyn @Qixiao: Jason's example is about the base change from $K(Y)$ to $\overline{K(Y)}$ (not $\kappa$ to $K(Y)$), so that is a base change to the algebraic closure.
Jan 2, 2018 at 19:16 comment added Jason Starr I do not quite understand the question. It is true that the scheme $X$ over $k$ is the base change of $Y$ by the field extension $k/\kappa$. However, there are birational modifications where that is no longer true, e.g., if we blow up $Y\times_{\text{Spec}\ \kappa} Y$ along a "typical" subvariety that dominates $Y$ with respect to $\text{pr}_1$. For an appropriate blowing up $\mathcal{X}\to Y\times_{\text{Spec}\ \kappa}Y$, for the generic fiber $\widetilde{X}$ over $\text{Spec}\ k$, the smallest "field of definition" of $X$ equals the field $k$.
Jan 2, 2018 at 14:54 comment added Jason Starr The base change homomorphism is not always injective. For a smooth, projective, rationally (chain) connected variety $Y$ over a field $\kappa$, let $k$ be $\kappa(Y)$, and let $X$ be the generic fiber of $$\text{pr}_1:Y\times_{\text{Spec}\ \kappa}Y \to Y.$$ Consider the cycle class of the diagonal, and consider the cycle class of $Y\times\{y_0\}$, where $y_0$ is a $\kappa$-point. These induce zero cycles on $X$ whose images in $\text{CH}^1(X_{\overline{k}})$ are equal. However, these cycles are equal in $\text{CH}^1(X)$ if and only if there is an integral decomposition of the diagonal.
Jan 2, 2018 at 14:48 history edited user39380 CC BY-SA 3.0
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Jan 2, 2018 at 14:43 history asked user39380 CC BY-SA 3.0