Surjectivity of a homomorphism between Picard groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T19:46:39Z http://mathoverflow.net/feeds/question/57405 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57405/surjectivity-of-a-homomorphism-between-picard-groups Surjectivity of a homomorphism between Picard groups AGMAN 2011-03-04T22:01:27Z 2011-03-04T22:32:10Z <p>Let $X$ be a one-dimensional Noetherian scheme over an algebraically closed field $k$. Suppose $X$ is reduced and let <code>$X=\bigcup X_i$</code> be the composition of $X$ into irreducible components. Then, is the following homomorphism surjective?</p> <p><code>$\mathrm{Pic} X\to \bigoplus \mathrm{Pic} X_i$</code>.</p> http://mathoverflow.net/questions/57405/surjectivity-of-a-homomorphism-between-picard-groups/57411#57411 Answer by Brian for Surjectivity of a homomorphism between Picard groups Brian 2011-03-04T22:32:10Z 2011-03-04T22:32:10Z <p>Yes! This can be shown using the isomorphism <code>$H^1(X,\mathcal{O}_X) \cong \mathrm{Pic} X$</code>. First, look at the short exact sequence:</p> <p><code>$0 \to \mathcal{O}_X^* \to \bigoplus \mathcal{O}_{X_i}^* \to \mathcal{C} \to 0$</code></p> <p>From the long exact sequence of cohomological groups associated to the short exact sequence, it suffices to show that <code>$H^1(X,\mathcal{C})\cong 0$</code>. However, this is clear since from the short exact sequence above, we can see that the support of $\mathcal{C}$ is a finite number of points (points that belong to more than just 1 irreducible component) and hence, of dimension 0. Now, use Grothendieck's vanishing theorem and we are done.</p>