[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.


1 Answer 1


This is only a partial answer, but it's too long to fit in the comments.

For any (commutative) ring $R$, there is a canonical inclusion $\text{Cl}(R) \longrightarrow \text{Pic}(R)$, where $\text{Cl}(R)$ is the ideal class group of $R$, that is, the group of all invertible fractional ideals of $R$ modulo its subgroup of principal regular fractional ideals of $R$.

If $R$ is an integral domain or a Noetherian ring, or more generally if $R$ is a ring with few zerodivisors, then the canonical inclusion $\text{Cl}(R) \longrightarrow \text{Pic}(R)$ is an isomorphism.

Also, if $R$ is a regularly Bezout ring, that is, if every finitely generated regular ideal of $R$ is principal, then clearly the ideal class group $\text{Cl}(R)$ of all invertible ideals of $R$ modulo principal ideals of $R$ is trivial. In fact, a regularly Bezout ring is equivalently a Prufer ring with trivial ideal class group. Prufer rings (with or without zerodivisors) are a well-studied class of rings. The proper invariant to study Prufer rings is the ideal class group rather than the Picard group.

It follows from the comments above any regularly Bezout (or Bezout) ring with nontrivial Picard group cannot be a ring with few zerodivisors (and therefore cannot be Noetherian or a domain). So, to find an example, you might start by looking at the construction of such rings in the literature. It seems like one should be able to make many examples that are Bezout, but then finding one with nontrivial Picard group could be hard.

Marot rings are a generalization of rings with few zerodivisors, and there are also many constructions of rings that are not Marot in the literature.

  • $\begingroup$ Thanks for the comments, Jesse! Another straightforward way to see that (regular) Bezout rings with few zero divisors have trivial picard groups is by noting that $T(R)$ will be semi-local (hence $Pic(T(R)) = 0$), and so again every invertible module is isomorphic to a dense finitely generated ideal of $R$ which will in turn be regular (since the zero-divisors of $R$ are contained in finitely many prime ideals), and thus principal by assumption. $\endgroup$ Nov 20, 2019 at 19:20
  • $\begingroup$ You write "there is a canonical inclusion $\text{Cl}(R) \longrightarrow \text{Pic}(R)$...". This is not true: there is no canonical map $\text{Cl}(R) \longrightarrow \text{Pic}(R)$, let alone an injective one. Hartshorne shows on page 134 of his Algebraic Geometry that for the ring $R$ of regular functions on the the affine cone $xy-z^2=0$ in affine $3$-space $\mathbb A^3_k$ over a field $k$ we have $\text{Cl}(R)=\mathbb Z/2 \mathbb Z$ but $\text{Pic}(R)=0$. $\endgroup$
    – lefuneste
    Oct 22, 2020 at 18:30
  • $\begingroup$ My example above shows that the assertion in your following sentence according to which $\text{Cl}(R)$ and $\text{Pic}(R)$ are isomorphic for integral domains or noetherian rings is false too. $\endgroup$
    – lefuneste
    Oct 22, 2020 at 18:39
  • 1
    $\begingroup$ This isn't correct. "Cl X" in Hartshorne refers to the Weil divisor class group. As I noted, I was referring to the ideal class group of invertible fractional ideals modulo regular principal fractional ideals. $\endgroup$ Oct 23, 2020 at 2:18
  • $\begingroup$ lefuneste, did you verify what I claimed, that the definition of Cl(X) in Hartshorne is not the same as my Cl(R)? I still stand by my answer. $\endgroup$ Oct 27, 2020 at 10:11

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