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Suppose $S$ is an algebraic surface (possibly projective) over an algebraically closed field $k$. Suppose $D_i$ are irreducible smooth curves (rational, if you want) with negative self-intersection and simple normal crossings such that there is a morphism $f:S\rightarrow S'$ which collapses them to a point.

Can we form an effective divisor $D=\sum d_i D_i$ such that $D\cdot D_i<0$ for all $i$? Note that if I had said $\leq$ rather than $\lt$ this is the content of the Proposition in page 83 of Reid's Park City lectures. I ask for strict inequality.

My belief is that this is possible, but I don't manage to give a proof nor find a reference or even to find a counter-example... This would be very useful for many arguments in birational geometry of surfaces.

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The intersection matrix $(D_i\cdot D_j)$ is negative definite and hence invertible. So the system of equations: $\sum_i d_i D_i\cdot D_j=\alpha_j$ is solvable for any set of $\alpha_j$'s, in particular for negative ones. It is an easy consequence of the negative definite property that if $\alpha_j<0$ for all $j$, then $d_i>0$ for all $i$.

A really slick way to do this (in any dimension) if $S$ is quasi-projective, $f$ is projective, and $S'$ is $\mathbb Q$-factorial the following: Let $H$ be an effective (reduced if you want) ample divisor on $S$ and consider $f_*H$ (as a subscheme). Since $S'$ is $\mathbb Q$-factorial, this is a $\mathbb Q$-Cartier divisor, so $mf_*H$ is Cartier for some $m$. Therefore $f^*(mf_*H)-mH$ is effective, supported on the exceptional divisor and (strictly) negative on every curve contracted by $f$.

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See the proof ON ISOLATED RATIONAL SINGULARITIES OF SURFACES (Artin) Propostiion 2 (i).

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  • $\begingroup$ I knew that paper, although I didn't check it when I wrote the post since I thought Reid's notes were a rewrite of this. I checked it now and it says $D\cdot D_i\leq 0$ not $D\cdot D_i<0$, as I thought. However the proof claims that this is indeed the case although it does not explain why. Sandor does it below. I guess I should accept his answer, if anything because it is more complete but I am also thankful to this post. $\endgroup$
    – Jesus Martinez Garcia
    Commented Jan 23, 2012 at 11:18

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