6
$\begingroup$

Let $X$ be a either a projective scheme or a compact complex space. Then one has an exact sequence $$ (1) \quad 0 \to \textrm{Pic}(X) \to \textrm{Cl}(X) \to \bigoplus_{x \in \textrm{Sing}(X)} \textrm{Cl}(\mathcal{O}_{X,x}),$$ where $\textrm{Pic}(X)$ and $\textrm{Cl}(X)$ denote the groups of Cartier (resp. Weil) divisors on $X$ up to linear equivalence.

The last arrow is in general not surjective, as can be shown by simple examples.

Now, in the paper by J.Birgener and U. Storch

Zur Berechnung der Divisorenklassengruppen kompletter lokaler Ringe, Nova Acta Leopoldina 52 Nr. 240, 7-63 (1981)

page 11, it is claimed that in the algebraic case one actually has the sequence $$ (2) \quad 0 \to \textrm{Pic}(X) \to \textrm{Cl}(X) \to \bigoplus_{x \in \textrm{Sing}(X)} \textrm{Cl}(\mathcal{O}_{X,x}) \to H^2(X, \mathcal{O}_X^{\ast}) \to 0.$$

My problem is that $(2)$ seems to me not true. For instance, take a smooth cubic threefold $X \subset \mathbb{P}^4$. Then the inclusion $\textrm{Pic}(X) \to \textrm{Cl}(X)$ is an isomorphism, so $(2)$ would imply $ H^2(X, \mathcal{O}_X^{\ast})=0$. However, by the exponential sequence one finds $H^2(X, \mathcal{O}_X^{\ast})=H^3(X, \mathbb{Z})= \mathbb{Z}^{10}$, and this is a contradiction.

So my question is:

am I missing something? If not, can one correct $(2)$ in some way?

Any answer or reference to the existing literature will be appreciated. Thank you!

Improve this question
$\endgroup$

1 Answer 1

Reset to default
11
$\begingroup$

The exponential sequence does not quite work that way for the algebraic cohomology $H^q(X,\mathcal{O}_{X}^*)$ for $q>1$. Of course the analytic cohomology, $H^q(X^{\text{an}},\mathcal{O}_{X^{\text{an}}}^*)$, does satisfy the exponential sequence. However, GAGA does not directly apply for $q>1$. Of course it does apply for $q=1$, because both the analytic and algebraic cohomology groups classify invertible sheaves, and GAGA gives an equivalence between these groups. But for $q=2$, one only has that the torsion subgroup of $H^q(X,\mathcal{O}_X^*)$ (and this group is a torsion group in "typical" cases) equals the torsion subgroup of $H^q(X^{\text{an}},\mathcal{O}_{X^{\text{an}}}^*)$. For a cubic threefold, there is no torsion in $H^3(X^{\text{an}};\mathbb{Z})$. Thus $H^2(X,\mathcal{O}_X^*)$ is zero.

Edit. I should make clear that "algebraic cohomology" above really means etale cohomology.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thank you for your clear answer, it was really helpful! You said that $H^2(X, \mathcal{O}_X^*)$ is a torsion group in "typical cases". May I ask you what "typical" means? For instance, is this always true for threefolds $X \subset \mathbb{P}^4$ with isolated sigularities? Could you please suggest me some references on this topic? $\endgroup$
    – Francesco Polizzi
    Commented Jul 19, 2013 at 13:48
  • $\begingroup$ The standard references are the three exposes, "Le Groupe de Brauer I,II,III" by Grothendieck in "Dix Exposes sur la Cohomologie des Schemas". Both de Jong and Lieblich have more recent surveys of the progress since then (e.g., work of Gabber and de Jong comparing the cohomological Brauer group to the Brauer group). $\endgroup$
    – Jason Starr
    Commented Jul 19, 2013 at 14:12
  • 2
    $\begingroup$ I found a precise reference for when $H^2(X,\mathcal{O}_X^*)$ is torsion. According to Proposition 1.4, p. 71 of "Le Groupe de Brauer, II" in "Dix Exposes ...", if $X$ is a Noetherian scheme and the strict Henselization of every local ring is factorial, then $H^2(X,\mathcal{O}_X^*)$ is torsion. $\endgroup$
    – Jason Starr
    Commented Jul 19, 2013 at 14:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .