Picard group of a singular projective curve - MathOverflow most recent 30 from http://mathoverflow.net2013-05-24T03:59:28Zhttp://mathoverflow.net/feeds/question/102408http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/102408/picard-group-of-a-singular-projective-curvePicard group of a singular projective curveJustin Campbell2012-07-17T00:07:39Z2012-07-18T21:59:39Z
<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#102414Answer by Karl Schwede for Picard group of a singular projective curveKarl Schwede2012-07-17T01:39:30Z2012-07-18T21:59:39Z<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>