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?



Are there varieties with non finitely generated Picard group and vanishing irregularity?
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 $\textrm{Pic}\,X$.


