# Bézout ring with non-trivial Picard group?

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

Q: Is there a nice example of a Bézout ring $$R$$ with a non-free rank $$1$$ projective $$R$$-module?

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.

However, I'm pretty sure that it's at least the case that any Bézout ring with compact minimal prime spectrum (with respect to the Zariski topology) will necessarily have a trivial Picard group. Reasoning: first reduce to the case that $$R$$ is a reduced ring, since that affects neither the Picard group nor the compactness of $$\operatorname{minSpec}(R)$$. Then observe that $$R$$ will have Von Neumann Regular total ring of fractions $$T(R)$$ (using Bézout and compactness assumptions). Since Von Neumann Regular rings have trivial Picard groups, deduce that every rank 1 projective of $$R$$ is isomorphic to an ideal of $$R$$ which is invertible in $$T(R)$$.

So in looking for an answer to Q I'm looking in the subset of Bézout rings which have non-compact minimal prime spectrum and which aren't elementary divisor rings. I've been having trouble coming up with anything explicit under these constraints. 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. 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. Since I've apparently been carrying around this misconception about Picard groups of Bézout rings for quite a while, I'd love to have an explicit example to sink my teeth into.