On condition when the push-forward of coherent sheaf is locally free - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T05:05:19Z http://mathoverflow.net/feeds/question/114238 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114238/on-condition-when-the-push-forward-of-coherent-sheaf-is-locally-free On condition when the push-forward of coherent sheaf is locally free stjc 2012-11-23T14:05:35Z 2012-11-26T14:03:54Z <p>This is a result being widely used in the literature:</p> <p>$f:X\rightarrow Y$ proper morphism between Noetherian schemes. $F\in Coh(X)$ flat over $Y$, if $H^i(X_y,F_y)=const$, $y\in Y$, then $R^if_\ast F$ is locally free.</p> <p>The problem is that I can only find references (EGA or GTM 52, etc) of this result with a condition 'Y is reduced', which seems to be unavoidable if one try to prove it by using the canonical technique of 'Grothendieck complex'.</p> <p>However, the general case is crucial in many arguments.</p> <p>So my question is that: Is the general case true? When can I find the proof of the general case?</p> <p>Thanks!</p> http://mathoverflow.net/questions/114238/on-condition-when-the-push-forward-of-coherent-sheaf-is-locally-free/114289#114289 Answer by Olivier Benoist for On condition when the push-forward of coherent sheaf is locally free Olivier Benoist 2012-11-23T21:09:06Z 2012-11-26T14:03:54Z <p>I believe that this statement is not true. Take $Y=Spec(k[t]/t^2)$ and $X=\mathbb{P}^1_Y$. On $X$, extensions of $\mathcal{O}$ by $\mathcal{O}(-2)$ are parametrized by : $$Ext^1_X(\mathcal{O},\mathcal{O}(-2))=H^1(X,\mathcal{O}(-2))=H^0(X,\mathcal{O})^{\vee}\simeq k[t]/t^2,$$ as a $k[t]/t^2$-module.</p> <p>Let $0\to\mathcal{O}(-2)\to E\to\mathcal{O}\to 0$ be the extension corresponding to $t\in k[t]/t^2$. By construction, the map $H^0(X,\mathcal{O})\to H^1(X,\mathcal{O}(-2))$ in the long exact sequence associated to the above short exact sequence is multiplication by $t$. It follows that $H^0(X,E)$, that is the kernel of this map, is necessarily $(t)\subset k[t]/t^2$. Hence it is isomorphic to $k[t]/t$ as a $k[t]/t^2$-module and it is not free.</p> <p>As $f_*E=H^0(X,E)$ because $Y$ is affine, and $E$ is flat over $Y$ because it is a vector bundle on $X$ that is flat over $Y$, this is a counter-example to the statement.</p>