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Let $X$ be a smooth projective variety over an algebraically closed field $k$.

Can it happen that $q(X) := \dim H^1(X,\mathcal O_X) =0$ and $\textrm{Pic} \,X$ is not finitely generated?

Certainly, if $q(X) =0$, then $\textrm{Pic}^0 \,X =0$. But is that enough to conclude that the abelian group $\textrm{Pic} \, X$ is finitely generated?

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    $\begingroup$ Yes, that is enough to conclude. Classically this is the theorem of Lang-Neron, sometimes also called the "Theorem of the Base". An exposition in modern terms is contained in SGA 6, one of the exposes by Steven Kleiman (I do not have my copy of SGA 6 handy) . . . I just looked it up: Expose XIII. $\endgroup$ – Jason Starr Nov 2 '15 at 13:00
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As pointed out by Jason Starr, the answer to your last question is yes, so the answer to the question in your title is no. Let me give a quick (and straightforward) proof in the case $k= \mathbb{C}$.

First of all, any projective variety over $\mathbb{C}$ has the homotopy type of a finite CW-complex, hence all (co)homology groups of $X$ with coefficients in $\mathbb{Z}$ are finitely generated abelian groups.

Next, let us consider the exponential sequence $$0 \to \mathbb{Z} \to \mathcal{O}_X \to \mathcal{O}_X^* \to 0.$$ Passing to cohomology, and using the assumption $q(X)=0$, we deduce that $\textrm{Pic}\,X=H^1(X, \, \mathcal{O}_X^*)$ injects into $H^2(X, \, \mathbb{Z})$.

Since the latter is a finitely generated abelian group, the same holds for $\textrm{Pic}\,X$.

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I will quote Jason Starr's comment:

"Yes, that is enough to conclude. Classically this is the theorem of Lang-Neron, sometimes also called the "Theorem of the Base". An exposition in modern terms is contained in SGA 6, one of the exposes by Steven Kleiman (I do not have my copy of SGA 6 handy) . . . I just looked it up: Expose XIII."

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