# Exact sequence for divisor class groups

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!

-

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.
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? – Francesco Polizzi Jul 19 '13 at 13:48
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. – Jason Starr Jul 19 '13 at 14:26