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! 



Exact sequence for divisor class groups
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. 


