TITLE: Exact sequence for divisor class groups
QUESTION [6 upvotes]: 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!

REPLY [11 votes]: 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.