Once again, the question says it all.


My motivation is the article on factorization I am writing. I want to explain (as well as to understand!) why for normal Noetherian domains of dimension greater than one, the obstruction to factoriality is the nonvanishing of the (Weil) divisor class group $\operatorname{Cl}(R)$, not the Picard group $\operatorname{Pic}(R)$ (equivalently, the Cartier divisor class group).


My understanding is that it is equivalent to find a normal Noetherian domain with vanishing Picard group which is not locally factorial. I would be especially happy to see an example among affine domains, i.e., in which the domain is finitely generated as an algebra over some field. 



Seeking Noetherian normal domain with vanishing Picard group but not a UFD
If the group $C\_2$ acts on $\mathbb{C}[x,y]$ by sending $x$ to $-x$ and $y$ to $-y$, then the fixed subalgebra is the domain $\mathbb{C}[x^2,y^2,xy]$. This is Noetherian and normal. A theorem of Nakajima gives the isomorphism type of the divisor class group (in this case, $C\_2$.) A theorem of Kang asserts for such examples (polynomial rings over $\mathbb{C}$, fixed subalgebra of finite group action), the Picard group is *always* trivial, so one can find many examples.


The reference for all of this is Benson's [Polynomial Invariants of Finite Groups](http://books.google.com/books?id=jxIXivzKFdgC&printsec=frontcover&dq=benson+polynomial+invariants+of+finite+groups&source=bl&ots=O7gr6Ujm7n&sig=XyB4HTSvlMJkxkJ8m7EhmGCF1lA&hl=en&ei=VVX6S-XbJoT58AbWx4H-Cg&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBsQ6AEwAA#v=onepage&q&f=false).
Kang's theorem is [Theorem 3.6.1](http://books.google.com/books?id=jxIXivzKFdgC&lpg=PP1&ots=O7gr6Ujm7n&dq=benson%2520polynomial%2520invariants%2520of%2520finite%2520groups&pg=PA33#v=onepage&q&f=false); Nakajima's theorem is [Theorem 3.9.2](http://books.google.com/books?id=jxIXivzKFdgC&lpg=PP1&ots=O7gr6Ujm7n&dq=benson%2520polynomial%2520invariants%2520of%2520finite%2520groups&pg=PA38#v=onepage&q&f=false) (and Corollary 3.9.3).


Steve


