Let |V| be a (incomplete) linear series on a nonsingular projective surface. Hironaka says that there is a resolution of the singularities of |V| along smooth centers. If the base locus of |V| is just a collection of points, does it mean I can acheive this resolution by a series of blow ups at points?
3 Answers
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I may be misunderstanding something but this question does not seem to have anything to do with Hironaka's desingularization. You are asking if you can resolve the indeterminacy of a rational map, right? If this is the question, then you can do it with finitely many blow-ups at points.
Suppose you have a rational map $f$ from a smooth projective surface $S$ to $\mathbb{P}^{n}$ given by some linear system $V \subset |D|$. Without loss of generality you may assume that the image of $f$ is non-degenerate, i.e. is not contained in any hyperplane. In that case we can subtract the fixed component of $V$ to make the base locus of $V$ at most finite. Now take a base point of $V$ and blow it up $g : \widehat{S} \to S$. The pullback $g^{\*}V$ will consist of divisors in $|g^{\*}D|$ which contain a positive multiple $nE$ of the exceptional divisor $E$. So $g^{\*}V$ has a fixed component and therefore is of the form $\widehat{V} + nE$, where $\widehat{V} \subset |g^{\*}D - nE|$. The linear system $\widehat{V}$ gives a rational map $\hat{f}$ from $\widehat{S}$ to $\mathbb{P}^{n}$ which has no fixed component and coincides with the original $f$ on $\widehat{S} - E$. Now set $\widehat{D} = g^{\*}D - nE$. Note that the fact that $\widehat{D}$ has no fixed component implies that $\widehat{D}^{2} = D^{2} - n^{2} < D^{2}$. If $\hat{f}$ has no base point, then we are done. If it does have base points, then we can repeat the process. Since the degree of the linear system decreases on each step the process will terminate.
answered Dec 20, 2009 at 16:11
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1$\begingroup$ I vote your anwer, but there is a point to be corrected, I think. The image being non-degenerate does not guarantee that the base locus is finite (think of f:P²->P² given by (x²,xy,xz)). But he fixed part can be subtracted, just as you do after the blowup, and this is just extending the map to "removable" indeterminacy points. Ah, and in my opinion the blowup resolution is indeed related to Hironaka's, it is just an easy particular case of a usually much more complicated process. $\endgroup$– quimCommented Dec 21, 2009 at 10:10
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$\begingroup$ Yes, of course! Thanks for the comment! I edited my answer to fix this. As for the resolution theorem, I just meant what Valery explicitly said in his answer - Hironaka's theorem is too powerful a tool to use for such a simple question. $\endgroup$ Commented Dec 21, 2009 at 13:01
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$\begingroup$ You and Valery are right, of course! I was just over-responding to your "not having anything to do with" part. Cheers. $\endgroup$– quimCommented Dec 21, 2009 at 13:22
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Not necessarily, I don't think. There are surface singularities (though I can't recall an example easily, but one whose resolution has exceptional divisor an elliptic curve) for which if you blow up at a point you get a singular curve. The example I'm thinking of is in Kollar's Exercises.
In the case I'm thinking of, you have a surface with a single point singularity, you blow it up, and you get a rational curve singularity, which if you blow up (or normalize, either one) gives you an elliptic curve over it.
EDIT: found it, it's exercise 68, do all three parts to see some of the things that can happen.
answered Dec 20, 2009 at 13:30
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$\begingroup$ you mean this is the case even if the original surface is nonsingular? $\endgroup$– user2529Commented Dec 20, 2009 at 13:37
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$\begingroup$ I haven't worked it out, but I expect that it is, because resolving singularities of linear systems and resolving singularities of varieties are extremely similar, and I'm pretty sure that resolving linear systems is more general, so if it can go wrong with varieties, it should be able to go wrong in maps, though I don't have a simple example off the top of my head. $\endgroup$ Commented Dec 20, 2009 at 13:51
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$\begingroup$ I am not sure that resolving linear systems is more general than resolving singularities. What I think is clear is the converse: if you consider the closure of the graph of the rational map induced by the linear system you obtain a variety ( blow-up of the original variety at the base scheme of the linear system ) and to finish resolving the original linear system it suffices to resolve the singularities of this variety. $\endgroup$ Commented Dec 20, 2009 at 16:19
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$\begingroup$ Ahh, yeah, I think I see where my mistake was. I concede to jvp and Tony. $\endgroup$ Commented Dec 20, 2009 at 16:49
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Over $\mathbb C$ at least, If the surface is non-singular then a finite number of blow-ups at points suffices to resolve the linear system. Indeed any birational morphism between smooth surfaces is a finite composition of point blow-ups.
answered Dec 20, 2009 at 15:50