Strict Transform under Blow-Up along singular Subscheme - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T06:31:27Z http://mathoverflow.net/feeds/question/69933 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69933/strict-transform-under-blow-up-along-singular-subscheme Strict Transform under Blow-Up along singular Subscheme gio 2011-07-10T11:57:35Z 2011-07-11T13:54:21Z <p>My question is if <a href="http://mathoverflow.net/questions/68973/strict-transform-under-blow-up-along-nonsingular-subvariety" rel="nofollow" title="this">this</a> can be generalized as well: Let $\beta:\widetilde{X}\mathrel{\mathop:}=\mathop{\mathrm{Bl}}_Z(X)\to X$ be the blow-up of a smooth algebraic variety $X$ along a irreducible generically smooth subscheme $Z$. Let $E\mathrel{\mathop:}=\beta^{-1}(Z)$ be the exceptional divisor. Now, let us assume I have a divisor $D$ on $X$, with $Z$ closed subscheme of $D$ ($D$ smooth and irreducible, if it helps). Then I ask if $\beta^\ast D \sim \widetilde{D} + \alpha E$, for some integer $\alpha$, where $\widetilde{D}$ denotes the strict transform of $D$ and "$\sim$" is linear equivalence.</p> http://mathoverflow.net/questions/69933/strict-transform-under-blow-up-along-singular-subscheme/70016#70016 Answer by Karl Schwede for Strict Transform under Blow-Up along singular Subscheme Karl Schwede 2011-07-11T13:54:21Z 2011-07-11T13:54:21Z <p>I think the answer is, almost. </p> <p>$E$ might have more than one component. Indeed, since you only assume that $Z$ is generically smooth, $E$ might have many different components. $E = \sum E_i$.</p> <p>You can always write $\beta^* D = \widetilde{D} + \sum_i \alpha_i E_i$, but you can't use the same $\alpha$ for all of them.</p> <p>Let me give an example. Suppose $X = \mathbb{A}^2$, $Z = V(x \cdot (x,y) \cdot (x,y^2)) = V(x^3, x^2y^2, x^2y, xy^3)$.<br> Certainly $Z$ is generically smooth and irreducible (its just not reduced, in higher dimensions, I'm think I can rig reduced examples as well). Set $D = \text{Div}(x^3)$. The blow-up of $X$ has two exceptional divisors.<br> Note $Z$ is a closed subscheme of $D$ as well. By working in higher dimensions, one can rig examples where the support of $Z$ and $D$ are different.</p> <p>Anyway, one chart of the blow-up looks like $k[x/y, y^2/x]$ with the obvious map to $\text{Spec} k[x,y]$. On that chart, the two exceptional divisors are $E_1 = \text{Div}(x/y)$ and $E_2 =\text{Div}(y^2/x)$. </p> <p>On this chart, we compute $\beta^*D$. Note $$x^3 = ((x/y)^2 \cdot (y^2/x))^3$$ and so $$\beta^*D = \widetilde{D} + 6 E_1 + 3 E_2.$$</p>