Line bundles on Ind Schemes - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T19:04:08Z http://mathoverflow.net/feeds/question/75888 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75888/line-bundles-on-ind-schemes Line bundles on Ind Schemes solbap 2011-09-19T21:43:34Z 2011-09-20T00:38:35Z <p>I've learned when you have a integral smooth scheme line bundles are the same as Cartier divisors are the same as Weil divisors. My question is to what extent does this continue to hold (if at all) when you are talking about ind objects. Since things become more subtle in positive characteristic lets say we are working over an algebraically closed field of characteristic 0. </p> <p>Background</p> <p>I think the situation can be quite different because smoothness, or algebraic smoothness is more complicated. In the non ind scheme case you can define smoothness of a point $p \in X$ as $S^q(m/m^2) \to m^q/m^{q+1}$ being an isomorphism for all $q\ge 0$, where $m$ is the maximal ideal of the local ring at $p$. In the ind case you have varieties $(X_n)_{n \ge 0}$ and $p \in X_n$ for all sufficiently large $n$ so you have $S_n^q(m_n/m_n^2) \to m^q_n/m^{q+1}_n$ for all sufficiently large $n$. Algebraic smoothness at $p$ means that the map</p> <p>$\varprojlim S_n^q(m_n/m_n^2) \to \varprojlim m^q_n/m^{q+1}_n$</p> <p>is an isomorphism for all $q\ge 0$</p> <p>Thoughts</p> <p>It is true that if $X$ is the union of $X_n$ and each $X_n$ is smooth then $X$ is algebraically smooth. However algebraic smoothness is not inductive in the sense that there are examples of $p \in X$ being algebraically smooth even when $p \in X_n$ is a singular point for all $X_n$ containing $p$. However line bundles are inductive in the sense that a line bundles on $X$ determines line bundles on each $X_n$ compatible under pull back and vice versa.</p> <p>This suggests to me that you could have a collection of compatible Weil divisors in each $X_n$ which are not the zero section of any section of any line bundle. But I don't have an example.</p> http://mathoverflow.net/questions/75888/line-bundles-on-ind-schemes/75898#75898 Answer by Dave Anderson for Line bundles on Ind Schemes Dave Anderson 2011-09-20T00:38:35Z 2011-09-20T00:38:35Z <p>I don't know the general story, but I think the correspondence between Weil divisors and line bundles breaks down already for $\mathbf{A}^\infty$, which is pretty much the smoothest ind-variety out there. Using the standard ind-structure on $\mathbf{A}^\infty$, the example I have in mind is the union of all hyperplanes $x_n-x_1=0$. This is a codimension one closed ind-subscheme, but it's not the zero locus of any polynomial in $\mathcal{O}_{\mathbf{A}^\infty}$.</p>