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Let $X$ be an integral variety defined over an algebraically closed field $k$ of characteristic 0 with finitely generated Picard group $Pic(X)$ and such that $k[X]^\times=k^\times$ (i.e. the only invertible global sections of the structure sheaf of $X$ are the constants).

To my knowledge, if $X$ is quasi-projective or locally factorial, then it satisfies the following property:

($*$) $Pic(X)$ is generated by effective Cartier divisors (equivalently every Cartier divisor is linearly equivalent to a difference of two effective Cartier divisors).

My question is: does there exist a variety $X$ as above that does not satisfy ($*$)?

EDIT: I recall that the group of Cartier divisors of $X$ is $\text{Div}(X)=H^0(X,\mathcal{K}(X)/\mathcal{O}^*_X)$, where $\mathcal{K}(X)$ is the constant sheaf associated to the function field of $X$ and $\mathcal{O}^*_X$ is the sheaf of units of the structure sheaf of $X$. A Cartier divisor $\{(U_i,f_i)\}_{i\in I}$ is said effective if $f_i\in H^0(U_i,\mathcal{O}_X)$ for all $i\in I$.

I have no experience with singular varieties, any reference related to this problem would be helpful.

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If I understand correctly, you are asking when the natural map $\mathrm{Div}(X)\rightarrow \mathrm{Pic(X)}$ is surjective (by definition $\mathrm{Div}(X)$ is the free group generated by effective Cartier divisors). EGA IV, section 21.3 gives some general conditions under which this holds: in particular, when $X$ is reduced with finitely many irreducible components. So of course $X$ integral suffices.

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    $\begingroup$ why do you say that $\text{Div}(X)$ is the free group generated by effective Cartier divisors? this is not EGA IV, Definition 21.1.1. $\endgroup$ – Marta Jul 6 '14 at 19:34
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    $\begingroup$ According to EGA IV, Definition 21.1.1, the group of Cartier divisors is $\text{Div}(X)=H^0(X,\mathcal{K}(X)/\mathcal{O}^*_X)$, where $\mathcal{K}(X)$ is the (here constant) sheaf associated to the function field of $X$ and $\mathcal{O}^*_X$ is the sheaf of units of the structure sheaf of $X$. I do not know whether $\text{Div}(X)$ is always free and/or generated by effective divisors. Denote by $\text{Div}_{\text{eff}}(X)$ its subgroup of effective divisors. I am asking whether the map $\text{Div}_{\text{eff}}(X)\to\text{Pic}(X)$ is always surjective. I edited my question to make it clearer. $\endgroup$ – Marta Jul 21 '14 at 11:34

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