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Let $\tilde{X}\to X$ be a blow-up of a variety $X$ (over an algebraically closed field).

Is it true that the Chow group of zero cycles of $\tilde{X}$ is isomorphic to that of $X$? What if $X$ is a smooth variety?

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    $\begingroup$ The accepted answer at mathoverflow.net/questions/241860/… proves birational invariance of $CH_0$ for smooth proper varieties (over any field). $\endgroup$ Commented Mar 2, 2017 at 10:17
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    $\begingroup$ To add one note to that answer: the result fails for some singular varieties $X$. For instance, for a smooth plane curve $Y\subset \mathbb{P}^2_k$ of degree $d\geq 3$, for $X\subset \mathbb{P}^3_k$ a projective cone over $Y$, then $\text{CH})_0(X)\xrightarrow{\text{deg}}\mathbb{Z}$ is an isomorphism, because every point is rationally equivalent to the vertex of the cone. However, for the blowing up of the vertex, $\widetilde{X}$ is a projective bundle over $Y$. Thus, $\text{CH}_0(\widetilde{X})\to \text{CH}_0(Y)$ is an isomorphism, and $\text{CH}_0(Y)$ equals $\text{Pic}(Y)$. $\endgroup$ Commented Mar 2, 2017 at 11:00
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    $\begingroup$ @JasonStarr: I would like to say that Hacon-McKernan's proof of Shokurov's conjecture implies that if $X$ is log terminal, then $CH_0 \tilde{X}$ is still isomorphic to $CH_0 X$. (So your counterexample is somehow as good as one can do.) Maybe I am missing a subtlety. $\endgroup$ Commented Mar 2, 2017 at 11:44
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    $\begingroup$ @potentiallydense. Yes, that is correct (assuming that the characteristic is zero, smiley-face). The class of singularities such that $\text{CH}_0(\widetilde{X})\to \text{CH}_0(X)$ is an isomorphism, even after base change of $k$ to an arbitrary (not necessarily algebraically closed) extension is one of the key ingredients in the work of Voisin, Colliot-Th'el`ene -- Pirutka, etc., on non-stable rationality via specialization of decompositions of the diagonal. $\endgroup$ Commented Mar 2, 2017 at 11:56

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