[I asked this on stackexchange here a few weeks ago to no response]

A ring is called *Bézout* when its finitely generated ideals are principal.

Q: Is there a nice example of a Bézout ring $R$ with $\operatorname{Pic}(R) \not= 0$ and an explicit description of a (non-free) rank $1$ projective over $R$?

Below are some thoughts and motivation:

Until recently I had assumed (and thought I had a proof in mind) that any Bézout ring would have trivial Picard group. But then I came across the paper Finitely Generated Modules over Bézout Rings of Wiegand and Wiegand, in which **Theorem 2.1 implies that any Hermite ring $R$ which is not an elementary divisor ring contains an element $d$ such that $R/(d)$ has nontrivial Picard group**.

Since I've been carrying around this apparent misconception about Picard groups of Bézout rings for quite a while, I'd love to have an explicit example to sink my teeth into.

Part of my problem seems to be that some common additional properties of Bézout rings *do* ensure trivial Picard group. For example, if the Bézout ring is an elementary divisor ring or if it has compact minimal prime spectrum (with respect to the Zariski topology).

Most of the still-viable candidate Bézout rings I know occur as rings of continuous real-valued functions on the remainder of certain Stone-Čech compactifications, and I find it hard to work with such rings under construction. For example, if we take $X$ to be the union of the positive x-axis in $\mathbb{R}^2$ and the positive half of the $\sin$ curve, and take $R = C(\beta X \setminus X)$ (the ring of associated real-valued continuous functions), then $R$ is Hermite but not an elementary divisor ring (cf example 4.11 here), so the above-cited Theorem 2.1 implies that $R/(d)$ would provide me an example for some $d \in R$. Yet I have no idea how to locate such an element $d$ or, having done that, what this projective module would look like.