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Let $X:=\rm{Spec}(A)$ be an integral, noetherian, affine variety, and let $L$ be an invertible sheaf on $X$, I try to find an example where $L$ is not isomorphic to the structure sheaf of $X$.

In the language of commutative algebra, this simply means to find an $A$-module $M$ which is locally free, but $M$ is not isomorphic to $A$ as a module.

Because when $A$ is UFD, the Weil class group is trivial, such example should not exists in this case. Also, I know for affine toric variety, the Picard group is also trivial.

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Take any non-principal ideal in your favorite Dedekind domain with nontrivial class group.

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  • $\begingroup$ To add to this, assuming that $A$ is normal, then such a line bundle will exist if and only if $A$ is not a UFD. Since, in this case $\text{Pic}(A)=\text{Cl}(A)$, and $\text{Cl}(A)=0$ if and only if $A$ is a UFD> $\endgroup$ – Alex Youcis Jul 25 '14 at 1:24
  • $\begingroup$ @Bruno Joyal, sorry, but why this is such example? Do you mean the class group of Dedekind domain is the same as Picard group? $\endgroup$ – Li Yutong Jul 26 '14 at 1:09
  • $\begingroup$ @AlexYoucis I'm not convinced that A normal is enough to have Pic(A) =Cl(A). Please see Hartshorne Chapt. II Remark 6.11.2. $\endgroup$ – Li Yutong Jul 26 '14 at 1:13
  • $\begingroup$ @LiYutong I was talking about for $A$ a dimension one domain, and Noetherian, as in Bruno's example. In that case, normality is the same as regularity. $\endgroup$ – Alex Youcis Jul 26 '14 at 1:22
  • $\begingroup$ @AlexYoucis is Dedekind domain necessarily normal? $\endgroup$ – Li Yutong Jul 26 '14 at 1:31
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Any commutative group yields a counterexample!
Indeed , a mind-boggling theorem proved by Claborn in 1966 states that:

Any abelian group is isomorphic to the class group of some Dedekind ring.

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