Comparing blow-ups with comparable centres - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T18:21:30Z http://mathoverflow.net/feeds/question/88356 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88356/comparing-blow-ups-with-comparable-centres Comparing blow-ups with comparable centres Lierre 2012-02-13T16:44:34Z 2012-02-14T08:07:12Z <p>Let $X$ a variety. Say $X=\operatorname{Spec} A$.</p> <p>Consider two ideals of $A$, say $I$ and $J$, with equal radical ; and consider the blow-ups of X with centre $I$ and $J$, say $Y_I$ and $Y_J$. <strong>How can I decide if the blow-up map $Y_I \to X$ factors through $Y_J$ ?</strong></p> <p>For example, if $X = \operatorname{Spec}k[x,y]$ is the affine plane. Then we can consider the blow-ups with centre $I_1 = (x,y)$, $I_2 = (x,y^2)$, $I_3=(x^2,y^2)$ and $I_4=(x^2,xy,y^2)=I_1^2$. We have $I_4\subset I_3\subset I_2\subset I_1$. Let $Y_i$ denote the corresponding blow-ups.</p> <p>It is well-known that $Y_1$ and $Y_4$ are isomorphic. However $Y_2$ and $Y_3$ are two other different varieties. Using the universal property of the blow-up we see that $Y_1\to X$ factors through $Y_3$ but not through $Y_2$.</p> <p><strong>What kind of property should I look at to predict this kind of factorization, without computing the actual blow-up ?</strong></p> http://mathoverflow.net/questions/88356/comparing-blow-ups-with-comparable-centres/88400#88400 Answer by Karl Schwede for Comparing blow-ups with comparable centres Karl Schwede 2012-02-14T04:39:58Z 2012-02-14T04:39:58Z <p>There are <em>two</em> conditions I can think of.</p> <ol> <li><p>Perhaps $I = J \cdot J'$ for some other ideal $J'$. Indeed, blowing up a product of ideals is the same as blowing up one ideal (say $J$) and then blowing up the total transform of the other (say $J'$).</p></li> <li><p>The other is requiring that $I = \overline{J}$, here this is integral closure of ideals (more generally $I$ and $J$ have the same integral closure and $I \subseteq J$). Indeed, two ideals have the same integral closure if and only if they have the same normalized blow-up and if their total transforms on the common normalized blow-up agree. </p></li> </ol> <p>Let me give an alternate explanation of 2., suppose for example that $I = \overline{J}$. Then the blow-up of $I$ and the blow-up of $J$ have the same normalization <em>and</em> $Y_I$ is just a partial normalization of $Y_J$. </p> <p>So this explains some of your examples. Indeed, the integral closure of $I_3$ is just $I_4$. On the other hand, $I_4 = I_1^2$. So by 1., $I_1$ and $I_4$ have the same blowup but by 2., the blow up of $I_4$ factors through the blowup of $I_3$. However, if you consider $I_1 \cdot I_2$ or $I_3 \cdot I_2$, you will find that their blowups factor through $I_2$.</p> <p>For smooth surfaces, there is a theory of factorization of complete (integrally closed) ideals due to Zariski. This should allow you to answer these sorts of questions even more precisely.</p> http://mathoverflow.net/questions/88356/comparing-blow-ups-with-comparable-centres/88410#88410 Answer by Sasha for Comparing blow-ups with comparable centres Sasha 2012-02-14T08:07:12Z 2012-02-14T08:07:12Z <p>The universal property of the blowup $\pi:Bl_I X\to X$ of an ideal $I$ in $X$ is that the sheaf of ideals $\pi^{-1}I\cdot O_{Bl_I X}$ is invertible. So, if you want to find out whether $Bl_I$ factors through $Bl_J$ you just have to check whether $\pi^{-1}J\cdot O_{Bl_I X}$ is invertible.</p>