Projectivity of blowups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T20:31:37Z http://mathoverflow.net/feeds/question/71264 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71264/projectivity-of-blowups Projectivity of blowups anon 2011-07-25T22:00:51Z 2011-07-26T14:11:13Z <p>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 $$f^{-1}(X\backslash Z)?$$</p> <p>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 $Y$.</p> <p>So under what conditions can I find such a relatively ample bundle?</p> <p>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.</p> http://mathoverflow.net/questions/71264/projectivity-of-blowups/71274#71274 Answer by Sándor Kovács for Projectivity of blowups Sándor Kovács 2011-07-25T23:56:00Z 2011-07-26T02:21:19Z <p>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 <code>$f_*H$</code> is a Weil divisor on $X$ and if $X$ has $\mathbb Q$-factorial singularities, then some multiple of <code>$f_*H$</code> is Cartier. Replacing $H$ with the appropriate multiple we may assume that <code>$f_*H$</code> is Cartier. Then it can be pulled back and $f^*(f_*H)$ is relatively numerically trivial, hence <code>$H-f^*f_*H$</code> 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.</p> <p>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 $1$.</p> http://mathoverflow.net/questions/71264/projectivity-of-blowups/71313#71313 Answer by Karl Schwede for Projectivity of blowups Karl Schwede 2011-07-26T13:29:39Z 2011-07-26T13:43:16Z <p>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). </p> <p>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. </p> <p>With respect to a small contraction $\pi : Y \to X$ (at least with $X$ normal), you can <em>NEVER</em> 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 <em>not</em> 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.</p> <p>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.</p>