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Does there exist a smooth irreducible variety $X/\mathbb{C}$ such that $\mathrm{Pic}(X)$ is finite and non-zero?

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Yes: take $\mathbb{P}^2_\mathbb{C}-V(f)$, with $f$ degree $d$. Then, this is smooth and irreducible and has Picard group $\mathbb{Z}/d\mathbb{Z}$.

This is not true if you require $X$ to be projective, then $\mathrm{Pic}(X)$ contains $\mathbb{Z}$ if it's non-zero.

EDIT(Explaining what happens if $X$ is projective): Suppose that $X$ is not affine, and so not $\text{Spec}(\mathbb{C})$ (I'm assuming $X$ is integral, just for convenience). Then, there is a very ample line bundle $\mathcal{O}(1)\in\mathrm{Pic}(X)$. But, any power of $\mathcal{O}(1)$ is still very ample, and since $X$ is not $\text{Spec}(\mathbb{C})$, $\mathcal{O}_X$ can't be very ample. So, $\mathcal{O}(1)$ can't be torsion. So, we see that $\mathbb{Z}\subseteq\mathrm{Pic}(X)$.

In fact, you can say more. I'm going to restrict to the case of $X$ a (smooth projective integral) curve, purely for convenience again. Then, we know that $\text{Pic}^0(X)$ are the $\mathbb{C}$-points of an abelian variety (the Jacobian) of dimension $g=\text{genus}(X)$. So, if $X\ne\mathbb{P}^1$, then $\mathrm{Pic}(X)$ is uncountable.

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  • $\begingroup$ thanks! can you say why $\mathrm{Pic}(X)\neq 0$ needs to contain $\mathbb{Z}$ in the projective case? if the answer is too long, I can post as a separate question. $\endgroup$
    – adrido
    Nov 18, 2014 at 5:17
  • $\begingroup$ @adrido I've edited my response. $\endgroup$ Nov 18, 2014 at 6:53
  • $\begingroup$ Excellent answer, dear Alex. Maybe for the sake of clarity you could make explicit that your Edit addresses the projective case. $\endgroup$ Nov 18, 2014 at 8:20
  • $\begingroup$ @GeorgesElencwajg Thanks :) I've edited for clarity. $\endgroup$ Nov 18, 2014 at 8:59
  • $\begingroup$ Dear Alex, thanks for the clear explanation. $\endgroup$
    – adrido
    Nov 18, 2014 at 20:16

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