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Here is the situation: $S$ is a nonsingular complex projective surface, and $C=X\cup_AY\subset S$ is a uninodal curve of compact type: it is obtained by glueing two nonsingular curves $X,Y\subset S$ at a single node $A$.

[If this helps (or is needed), I obtained $C$ as the special fiber of a family of smooth curves $\pi:S\to B$, where $B=\textrm{Spec }\mathbb C[[t]]$ and the generic fiber $S_\eta$ is a nonsingular curve.]

Question. Why is $X.C=0$?

If $X$ was a fiber of $\pi$ as well, we would be done. But it isn't. Moreover, the assertion $X.C=0$ looks strange to me, because (please, correct me here because I feel like I am dramatically wrong) it means that $X$ is linearly equivalent to $X+Y$. But $X-(X+Y)=-Y$ is not equivalent to $0$!
Moreover, how should I interpret the fact that $X.X=-A$? (This follows by the claim because $0=X.C=X.(X+Y)=X.X+X.Y=X.X+A$).

Thank you.

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The point is that it doesn't matter if X is a fiber, only that is contained in a fiber. Since C is a fiber, it's algebraically equivalent to any other fiber C', so X.C=X.C'. But now if C' is different from C, then X and C' are disjoint, so X.C'=0. I don't see why anything in this setup should imply that X and X+Y are linearly (or algebraically) equivalent. –  Asal Beag Dubh Apr 23 '13 at 13:51
    
You are perfectly right. Linear equivalence has nothing to do with the kernel of the intersection pairing. Thank you! –  Brenin Apr 23 '13 at 20:04
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I guess my earlier comment was a little sketchy, since we don't really know that C moves in a family of curves whose base is really a variety over $\mathbf{C}$. Maybe a less sketchy way to argue is that C is the pullback of a line bundle on B (i.e. the ideal sheaf of the closed point), so it must have degree zero on any curve which maps to a point on B, including X. I think that argument is more or less free of funny business, but correct me if I'm wrong... –  Asal Beag Dubh Apr 23 '13 at 20:42

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