If I have a quasiprojective variety $X$, and a subscheme $Z$, then the blowup $$f:Y = Bl(X,Z)\rightarrow X$$ is projective over $X$, since it is constructed by a relative Proj construction. Can I find a relatively ample bundle on $Y$ that is trivial on $$f^{-1}(X\backslash Z)?$$

At first I thought the construction of $O(1)$ in the Proj construction guarantees this property. However, if I have a projective small contraction $Y \rightarrow X$, then it is a blowup map, but then certainly an ample bundle can't be trivial on the birational locus, since it would be trivial on $Y$.

So under what conditions can I find such a relatively ample bundle?

Edit: Karl's answer explains away my confusion: I have to blow up more than the image of the exceptional locus in case of a small contraction. It's still reasonable to ask when it's trivial on the birational locus, in which case Sandor's answer applies.

  • $\begingroup$ anon, I'm confused about the "assumption of generation in degree 1" bit. What do you mean exactly? The blow-up of an ideal sheaf is always locally generated in degree 1 by construction. $\endgroup$ – Karl Schwede Jul 26 '11 at 13:21
  • $\begingroup$ No, the confusion was mine - I'll edit out the error... $\endgroup$ – anon Jul 26 '11 at 14:09

This can be done if $X$ has $\mathbb Q$-factorial singularities (but this is not a necessary condition!): Let $H$ be a relative ample effective divisor (not a bundle, divisor!). Then $f_*H$ is a Weil divisor on $X$ and if $X$ has $\mathbb Q$-factorial singularities, then some multiple of $f_*H$ is Cartier. Replacing $H$ with the appropriate multiple we may assume that $f_*H$ is Cartier. Then it can be pulled back and $f^*(f_*H)$ is relatively numerically trivial, hence $H-f^*f_*H$ is still relatively ample. Notice that $f^*f_*H-H$ is an effective exceptional divisor, hence trivial when restricted to the complement of the exceptional set.

Also notice that indeed if $f$ is a small contraction, then one cannot find such a divisor, but also, if $X$ has $\mathbb Q$-factorial singularities, then the exceptional set of $f$ has pure codimension $1$.

| cite | improve this answer | |
  • $\begingroup$ Nice argument - thanks! What are some other sufficient conditions for this to hold? For instance, it seems to me that no matter how singular $X$ is, blowups at smooth centers will have this property (by embedding inside the blowup of projective space at that center). $\endgroup$ – anon Jul 26 '11 at 11:25
  • $\begingroup$ In fact, maybe this embedding inside projective space trick works in general. Won't the blowup of X at Z be the proper transform of X inside the blowup of projective space at Z? In which case the ample bundle Sandor's argument provides (since projective space is Q-factorial) would work for X as well? Not sure why this wouldn't apply in the small resolution setting - are those always non-projective? $\endgroup$ – anon Jul 26 '11 at 12:07
  • $\begingroup$ anon, no small resolutions will still be projective in most cases you write down. You are right, that the blow-up of $Z$ in $\mathbb{P}^n$ will have the strict transform you want. The problem is that if blowing up $Z$ gives a small resolution, then $Z$ had divisorial components, and wasn't just the singular locus. $\endgroup$ – Karl Schwede Jul 26 '11 at 13:11

anon, is your definition of relatively ample local over the base? (There are some different definitions out there, but most people seem to use the one that is local over the base now, compare for instance Ravi Vakil's notes with Hartshorne).

If your definition of relatively ample is local over the base, then there is nothing to show. If you blow-up $Z$, say with ideal sheaf $I_Z$, then $I_Z \cdot O_Y$ is relatively ample.

With respect to a small contraction $\pi : Y \to X$ (at least with $X$ normal), you can NEVER get one of these just by blowing up a scheme supported on the locus where $\pi$ is an isomorphism. In general, the way you get a small resolution, is by blowing up a Weil divisor that is not Cartier. Of course, on a normal variety a Weil divisor is Cartier except on a small close set, and blowing up a Cartier divisor obviously doesn't do anything.

For fun, I'd suggest that you try blowing up the ideal $(x, u)$ on the scheme $\text{Spec} k[x,y,u,v]/(xy-uv)$. $(x,u)$ is a non-Cartier divisor, but it's still a divisor (it's also not $\mathbb{Q}$-Cartier, as Sándor effectively points out above). Blowing it up gives a small resolution.

| cite | improve this answer | |
  • $\begingroup$ Aha - this is great! Totally explains away my confusion... $\endgroup$ – anon Jul 26 '11 at 14:39
  • $\begingroup$ @Karl When you blow up divisors that are not Cartier, under which conditions one can be sure that the resulting small resolution is projective? $\endgroup$ – JME Jul 30 '11 at 20:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.