Let $X$ and $Y$ be subvarieties of a smooth projective variety $M$ such that $M=X \bigcup Y$. Can you explain to me why $ A_k ( X \bigcap Y ) \to A_k ( X ) \oplus A_k ( Y ) \to A_k ( X \bigcup Y ) \to 0 $, such that $ A_k ( X ) $ is the Chow group of $X$. Can you tell me if you know some references about this subject ? Thanks a lot. :-)
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2$\begingroup$ How are the maps in your sequence defined? "Proper pushforward" requires <I>proper</I> morphisms. I recommend that you read the introductory chapters of a good book on intersection theory, e.g., Fulton's book. $\endgroup$– Jason StarrCommented Nov 26, 2013 at 21:59
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$\begingroup$ Do you mean to assume (per Jason Starr's comment) that $X$ and $Y$ are closed? If so, you can (and should!) drop the assumption that $M$ is smooth. $\endgroup$– Steven LandsburgCommented Nov 26, 2013 at 23:07
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$\begingroup$ Yes, $ X $ and $ Y $ are closed. Thank you. $\endgroup$– Bryan261Commented Nov 26, 2013 at 23:10
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$\begingroup$ @JasonStarr: Actually, Fulton (quite reasonably) states this result without proof, so if the OP doesn't already see how to prove it, Fulton won't help. $\endgroup$– Steven LandsburgCommented Nov 27, 2013 at 6:07
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$\begingroup$ @StevenLandsburg: I guess I am confused. For me a "variety" is an irreducible, finite type scheme over some (fixed) field. So if $M$ is a smooth projective variety, if $X$ and $Y$ are closed subsets, and if $M$ equals $X\cup Y$, then one of $X$ or $Y$ equals $M$, say $X=M$. Then the sequence is ridiculous. My suspicion is that, actually, the OP does not know quite what he wants to ask, he has written something a bit absurd, and the readers are struggling to project some meaning on what is written. $\endgroup$– Jason StarrCommented Nov 27, 2013 at 11:49
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1 Answer
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We can assume $M=X\cup Y$. (And, per the comments above, we assume $X$ and $Y$ are closed.)
Suppose $x$ and $y$ are cycles on $X$ and $Y$ such that $x+y$ becomes trivial in $A_k(M)$. A trivialization of $x+y$ restricts in the obvious way to trivializations of $x$ on $M-Y$ and of $y$ on $M-X$. Thus it restricts to trivializations of $x+z$ and $y+z$ on $X$ and $Y$, for some cycle $z$ supported on $X\cap Y$.
answered Nov 27, 2013 at 6:05