Line bundles vs. Cartier divisors on a non-integral scheme - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T05:28:24Z http://mathoverflow.net/feeds/question/22080 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22080/line-bundles-vs-cartier-divisors-on-a-non-integral-scheme Line bundles vs. Cartier divisors on a non-integral scheme J.C. Ottem 2010-04-21T17:09:04Z 2010-04-21T17:42:37Z <p>It is well-known that if $X$ is an integral scheme, then there is an isomorphism $CaCl(X)\to Pic(X)$ taking $[D]$ to $[\mathcal{O}_X(D)]$. Does anyone know any simple examples where the above map fails to be surjective, i.e., a line bundle on a scheme $X$, not isomorphic to $\mathcal{O}_X(D)$ for any Cartier divisor D? </p> http://mathoverflow.net/questions/22080/line-bundles-vs-cartier-divisors-on-a-non-integral-scheme/22083#22083 Answer by VA for Line bundles vs. Cartier divisors on a non-integral scheme VA 2010-04-21T17:22:50Z 2010-04-21T17:22:50Z <p>From the exact sequence </p> <p>$$1\to O^*\to K^*\to K^* / O^*\to 1$$</p> <p>you see that, for as long as $H^1(K^*)=0$, the map from $H^0( K^*/ O^*)$ (i.e. Cartier divisors) to $H^1( O^*)$ (i.e. line bundles) is surjective.</p> <p>On a Noetherian scheme without embedded primes (for example, reduced), $\mathcal K^*$ is the direct sum of several constant sheaves on the irreducible components, so it has trivial $H^1$.</p> <p>So the example would have to be a scheme with embedded primes with a tricky nonconstant $K^*$ (the sheaf of nonzero divisors). I've seen it but can't remember right now. So this is just some general observations to narrow the search.</p> http://mathoverflow.net/questions/22080/line-bundles-vs-cartier-divisors-on-a-non-integral-scheme/22085#22085 Answer by Hailong Dao for Line bundles vs. Cartier divisors on a non-integral scheme Hailong Dao 2010-04-21T17:42:37Z 2010-04-21T17:42:37Z <p>An example is given in this <a href="http://webs.uvigo.es/martapr/Investigacion/Divisors.pdf" rel="nofollow">note</a> (it was credited to Kleiman). </p>