Let $R$ be a nonzero commutative ring. Must there exist an $R$-algebra $S$ (that is, a commutative ring $S$ containing a quotient of $R$ as a subring) such that $\mathrm{Pic}(S)$ is not the trivial group?
My first thought is that it suffices to consider the case where $R$ is a field, since $R \neq 0$ implies that there is some $R$-algebra which is a field. But I'm not sure how to produce an algebra over a field which has a non-free rank-$1$ projective module.
This isn't a homework question although was inspired by one.
I just realized I do know how to produce such an algebra. If $K$ is a field, then $K[x,y]/(y^2 - x^3)$ has Picard group $K$ (under addition), which is always nontrivial since fields have at least two elements.