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Nov 27, 2013 at 16:03 comment added Jason Starr @StevenLandsburg: Okay, now I see what you mean.
Nov 27, 2013 at 14:44 comment added Steven Landsburg @JasonStarr: Yes, given that $X$ and $Y$ are closed, the problem is completely trivial unless $M$ is reducible, which is why I said in my earlier comment that the OP can and should drop the assumption that $M$ is smooth. For $M$ reducible the problem is still very easy but (though I wavered back and forth on this) I thought it was worth pointing out the simple argument that the OP might have missed.
Nov 27, 2013 at 11:49 comment added Jason Starr @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.
Nov 27, 2013 at 6:07 comment added Steven Landsburg @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.
Nov 27, 2013 at 6:05 answer added Steven Landsburg timeline score: 3
Nov 26, 2013 at 23:10 comment added Bryan261 Yes, $ X $ and $ Y $ are closed. Thank you.
Nov 26, 2013 at 23:07 comment added Steven Landsburg 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.
Nov 26, 2013 at 21:59 comment added Jason Starr 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.
Nov 26, 2013 at 21:51 history asked Bryan261 CC BY-SA 3.0