Extending a vector bundle to a torsion free sheaf - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T12:13:26Z http://mathoverflow.net/feeds/question/75964 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75964/extending-a-vector-bundle-to-a-torsion-free-sheaf Extending a vector bundle to a torsion free sheaf Ian Shipman 2011-09-20T15:49:55Z 2011-09-20T16:55:36Z <p>Let's say that $X$ is an integral scheme of finite type over a field and $Y \subset X$ is a closed subscheme. Given a vector bundle $E$ on $Y$, is $E$ the restriction to $Y$ of a vector bundle on a neighborhood $U$ of $Y$ in $X$? </p> http://mathoverflow.net/questions/75964/extending-a-vector-bundle-to-a-torsion-free-sheaf/75969#75969 Answer by ulrich for Extending a vector bundle to a torsion free sheaf ulrich 2011-09-20T16:11:27Z 2011-09-20T16:55:36Z <p>No, this is not even true for line bundles.</p> <p>For an example, let $X = \mathbb{P}^2$ and $Y$ a smooth curve in $X$ of genus $>0$ (over an algebraically closed field). Since $X$ is smooth, any line bundle on an open set $U$ extends to a line bundle on $X$ so the map $\operatorname{Pic}(X) \to \operatorname{Pic}(U)$ is surjective. Since $\operatorname{Pic}(X) \cong \mathbb{Z}$, it follows that the image of $\operatorname{Pic}(U)$ in $\operatorname{Pic}(Y)$ is of rank $1$ and is independent of $U \supset Y$. Since $\operatorname{Pic}(Y)$ is not even finitely generated we see that there exist (many) line bundles on $Y$ which do not extend to any open $U \supset Y$.</p>