Picard group of a singular projective curve - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T03:59:28Z http://mathoverflow.net/feeds/question/102408 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102408/picard-group-of-a-singular-projective-curve Picard group of a singular projective curve Justin Campbell 2012-07-17T00:07:39Z 2012-07-18T21:59:39Z <p>Let $X$ be a singular irreducible projective curve over an algebraically closed field and $\pi : \widetilde{X} \to X$ the normalization morphism. In the book on Neron models by Bosch et al. (I have slightly simplified the setup), the Picard group of $X$ is computed by taking the long exact sequence associated to $$0 \to \mathcal{O}_X^* \to \pi_*\mathcal{O}_{\widetilde{X}}^* \to \pi_*\mathcal{O}_{\widetilde{X}}^*/\mathcal{O}_X^* \to 0.$$ This seems to depend on the assertion that $\text{Pic}(\widetilde{X}) \cong H^1(X,\pi_*\mathcal{O}_{\widetilde{X}}^*)$, which is not obvious to me. If $\mathcal{O}_{\widetilde{X}}^*$ were coherent, then this would be a consequence of the exactness of <code>$\pi_*$</code> on coherent sheaves ($\pi$ is finite), but of course $\mathcal{O}_{\widetilde{X}}^*$ is not even a sheaf of $\mathcal{O}_{\widetilde{X}}$-modules. How can we resolve this difficulty?</p> http://mathoverflow.net/questions/102408/picard-group-of-a-singular-projective-curve/102414#102414 Answer by Karl Schwede for Picard group of a singular projective curve Karl Schwede 2012-07-17T01:39:30Z 2012-07-18T21:59:39Z <p>Maybe the following works?</p> <p>Let's consider the ringed space $(X, \pi_* O_{\widetilde X})$ and also the ringed space $(\widetilde X, O_{\widetilde X})$. Certainly $\pi_{*}$ gives us an isomorphism $$\text{Pic}(\widetilde X, O_{\widetilde X})) \cong \text{Pic}(X, \pi_* O_{\widetilde X})$$ which sends an $O_X$-line bundle $L$ to the $\pi_* O_X$-module $\pi_* L$, right? (I'm using the fact that $X$ is projective here to show that there exist open trivializing sets containing various finite collections of points). Also observe that $(\pi_* O_{\widetilde X})^{*} = \pi_* (O_{\widetilde X}^{*})$. Thus by Hartshorne Chapter III, Exercise 4.5 (which works on arbitrary ringed spaces), we see that $$\text{Pic}(X, \pi_* O_{\widetilde X}) \cong H^1(X, (\pi_* O_{\widetilde X})^*).$$ Combining with the isomorphisms already written, we obtain: $$\text{Pic}(\widetilde X, O_{\widetilde X}) \cong \text{Pic}(X, \pi_* O_{\widetilde X}) \cong H^1(X, (\pi_{*} O_{\widetilde X})^*) \cong H^1(X, \pi_{*} (O_{\widetilde X}^{*}))$$</p>