The proof of unobstructedness of deformations for curves - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T06:05:25Z http://mathoverflow.net/feeds/question/108698 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108698/the-proof-of-unobstructedness-of-deformations-for-curves The proof of unobstructedness of deformations for curves ZhuangXiaobo 2012-10-03T09:12:00Z 2012-10-03T10:29:45Z <p>I am reading Illusie's lecture notes "topics in algebraic geometry", and I have difficulty in following his proof of unobstructedness of deformation of curves. Here is the statement of the proposition:</p> <p><strong>Prop</strong> Let $f_0: X_0 \to Y_0$ be a smooth proper morphism with relative dimension 1, and $i : Y_0 \to Y$ a first-order thickening with ideal $I$. If moreover $Y$ is affine, then there always exists a lifting of $X_0$ over $Y$ .</p> <p><strong>Proof</strong>. First, since $Y_0$ is affine, we note that $H^q(X_0, T_{X_0/Y_0} \otimes f_{0}^{*}I)= \Gamma (Y_0,R^qf_{0*}(T_{X_0/Y_0}\otimes f_{0}^{*}I))$<br> By Zariski’s main theorem, for any $q > 1$,<br> $R^qf_{0*}(T_{X_0/Y_0}\otimes f_{0}^{*}I))=0$ (*)<br> Hence the obstruction $o(f_0, i)\in H^2(X_0, T_{X_0/Y_0}\otimes. f_0^*I)$ vanishes. # </p> <p>And I wonder how to derive (*) by Zariski's main theorem. </p> <p>Thank you!</p> http://mathoverflow.net/questions/108698/the-proof-of-unobstructedness-of-deformations-for-curves/108699#108699 Answer by Francesco Polizzi for The proof of unobstructedness of deformations for curves Francesco Polizzi 2012-10-03T09:36:00Z 2012-10-03T10:29:45Z <p>Probably Illusie wrote "Zariski's Main Theorem", but he intended the Theorem of Formal Functions (which is the key result needed in the modern proof of Zariski's Theorem).</p> <p>In fact, the Theorem of Formal Functions implies the following result, see [Hartshorne, Algebraic Geometry, Corollary 11.2 page 279].</p> <blockquote> <p><strong>Proposition.</strong> Let $f \colon X \to Y$ be a projective morphism of noetherian schemes, and let $r= \textrm{max} \{\dim X_y | y \in Y \}$. Then $R^qf_* (\mathscr{F})=0$ for all $q >r$ and for all coherent sheaves $\mathscr{F}$ on $X$.</p> </blockquote> <p>Now by projection formula we have $$R^qf_{0*}(T_{X_0/Y_0}\otimes f_{0}^{*}I))=R^qf_{0*}(T_{X_0/Y_0}) \otimes I,$$ so you can apply the previous proposition with $r=1$ and $\mathscr{F}=T_{X_0/Y_0}$ in order to get the desired vanishing. </p>