It's one of these standard facts that the Picard group of a normal scheme of finite type over $\mathbb{Q}$, or, more generally, an absolutely finitely generated field of characteristic $0$, is finitely generated. This follows from a compactification argument and Mordell-Weil-Néron-Lang. It's easy to think of a scheme that's $S_2$ and not $R_1$ that has an infinitely generated Picard group: the cuspidal cubic $X=\operatorname{Spec}k[x,y]/(y^2-x^3)$ has Picard group $k^+$ which is not finitely generated if $k$ is an absolutely finitely generated field of characteristic $0$. But I can't think of a scheme that's $R_1$ but not $S_2$ that fails to have a finitely generated Picard group. Is there a known example?
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3$\begingroup$ Let $X$ be the affine scheme obtained by gluing two $k$-points on $\mathbb{A}^2$ to each other. Then $X$ is R1 but not S2 and has $k^\times$ as the Picard group. $\endgroup$– Dori BejleriCommented Aug 30 at 3:34
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$\begingroup$ @DoriBejleri If I glued a point on $\mathbb{A}^2_k$ to an infinitesimal neighborhood (creating a cusp), would that be another example? $\endgroup$– KrillCommented Aug 30 at 3:44
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$\begingroup$ Yes I think so, e.g. if you pinch a tangent direction to a point I think you should get $k^+$ via the same computation as the cusp case. $\endgroup$– Dori BejleriCommented Aug 30 at 21:32
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