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 $\pi_*$ 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?


2 Answers 2


Maybe the following works?

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: $$ \text{Pic}(\widetilde X, O_{\widetilde X}) \cong \text{Pic}(X, \pi_* O_{\widetilde X}) \cong H^1(X, (\pi_{*} O_{\widetilde X})^*) \cong H^1(X, \pi_{*} (O_{\widetilde X}^{*})) $$

  • $\begingroup$ Hi Jason, I was trying to avoid $\pi_*$ exactness (since that seemed to be what Just Campbell was worried about)? Instead I was claiming that all this work can be done on a different ringed space on $X$, and we don't need to consider the scheme $(X, O_X)$ at all. $\endgroup$ Jul 17, 2012 at 2:01
  • $\begingroup$ I guess that might be a better way to go about it, to prove directly that $\pi_*$ is exact on arbitrary sheaves of Abelian groups. $\endgroup$ Jul 17, 2012 at 2:11
  • $\begingroup$ @Jason Starr: I had hoped something like this was true. Would you mind elaborating a bit, perhaps in an answer? What property of $\pi$ is responsible for this fact? I know there is a theorem about higher direct images for proper maps that would give me what I want e.g. in the complex topology, but I'm wary of applying this in the Zariski topology. $\endgroup$ Jul 17, 2012 at 3:25
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    $\begingroup$ @Jason: Are finite morphisms really acyclic for the Zariski topology? Perhaps (likely!) I'm mistaken, but there are semi-local schemes (obtained via finite covers of local ones) with non-trivial Zariski cohomology, which I thought would contradict acyclicity (see Example 1.10 of Morel-Voevodsky). Of course, for Justin's question, one may work with the etale/Nisnevich topology, so the acyclicity is fine. $\endgroup$
    – Bhargav
    Jul 17, 2012 at 3:33
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    $\begingroup$ Bhargav is correct, I was wrong: pushforward by a finite morphism is not necessarily acyclic (for the Zariski topology). You can see this quite clearly for a non-local, semilocal domain if you use Exercise 1.19 of Hartshorne with the locally constant sheaf $\mathbb{Z}$. As Bhargav says, you can use the \'etale topology to compute Pic, and pushforward by a finite morphism is acyclic for the \'etale topology. $\endgroup$ Jul 17, 2012 at 12:05

The proof of the isomorphism can be found in Proposition 2.8 of this 2013 paper by Hartshorne and Polini: https://arxiv.org/pdf/1301.3222.pdf


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