Let us work in the category of smooth, projective varieties (say, over an algebraically closed field $k$). If $X$ and $X'$ are birational, then do they have the same Chow groups? Is there at least a canonical morphism, between the Chow rings of these?
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7$\begingroup$ Think of what happens to Picard groups when you blow up points in surfaces. $\endgroup$– AngeloNov 21, 2011 at 15:47
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1$\begingroup$ @Angelo: What DOES happen to them? Can something be said in the case where the surface has singularities? I know there is a theorem on divisor class groups (Hartshorne, II.6.5), but I'd like to know if anything can be said specifically about the Picard group. $\endgroup$– topspin1617Nov 21, 2011 at 16:34
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1$\begingroup$ To topspin1617: please don't scream. Do you want to pose another question? $\endgroup$– AngeloNov 21, 2011 at 18:07
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1$\begingroup$ I don't know if there is a general result for the Picard group of a blowup. However, in the simplest cases (say, if you blow up a smooth point on a surface) the answer is known, and is sufficient to address the original question. $\endgroup$– Laurent Moret-BaillyNov 21, 2011 at 19:54
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2$\begingroup$ There is a general result for the blow-up along a regular embedding: in a quasi-compact quasi-separated (e.g. noetherian separated) scheme, if you blow-up a codimension at least 2 regularly embedded connected subscheme, the Picard group of the blow-up will be the original one with an extra copy of $\mathbb Z$. N.B. I assume quasi-compact and quasi-separated because I can only figure out a proof using derived categories of perfect complexes, where these assumptions play a role, but they might be unnecessary. $\endgroup$– Baptiste CalmèsNov 22, 2011 at 12:43
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1 Answer
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Proposition 6.3 of Colliot-Thélène (Jean-Louis) and Coray (Daniel), L'équivalence rationnelle sur les points fermés des surfaces rationnelles fibrées en coniques, Compositio Math. 39 (1979), no. 3, 301–332, says that the Chow group of $0$-cycles of degree$\;0$ is a birational invariant of a smooth projective absolutely connected $k$-variety $X$ in the following two cases :
(i) the field $k$ is perfect and $X$ is a surface,
(ii) the field $k$ has characteristic $0$.
You can read the proof at Numdam (in French) or in Fulton's book on Intersection theory (in American).
answered Nov 21, 2011 at 16:18
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$\begingroup$ In Fulton's Intersection Theory, Example 16.1.11, he proves this under the assumption that $k$ is algebraically closed (of arbitrary characteristic). $\endgroup$ Jun 11, 2016 at 4:13
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2$\begingroup$ @R.vanDobbendeBruyn is too modest, he provided a wonderful explanation of this in an answer to mathoverflow.net/questions/241860/… $\endgroup$– JoachimJun 12, 2016 at 14:50