Is there a Prüfer domain that is a UFD but not a PID?

Question

An integral domain $R$ is called a Prüfer domain if every non-zero finitely generated ideal $I$ of $R$ is invertible. Many equivalent characterizations can be found here.

Question. Is there a Prüfer domain $R$ that is also a unique factorization domain (UFD) but not a principal ideal domain (PID)?

I guess that such $R$ exists as the definition says nothing about the ideals that are not finitely generated. It is well-known that a Dedekind domain or a Bézout domain that is a UFD is a PID, so such $R$ must be neither a Dedekind domain nor a Bézout domain. But I already had a hard time to find a Prüfer domain that is neither a Dedekind domain nor a Bézout domain. An example is $\operatorname{Int}(\mathbb{Z})$, the ring of integer-valued polynomials with rational coefficients. This ring is Prüfer (see here) but not noetherian (so not a Dedekind domain). It is not a Bézout domain as the ideal $(2,X)$ is not principal. Unfortunately, it is also not UFD as $X(X-1)=2\cdot\dfrac{X(X-1)}{2}$ is a factorization of $X(X-1)$ into different irreducibles.

Thank you for any help in advance.

@MartinBrandenburg Ah yes thanks, I forgot to mention it. There are no results with the given conditions. — Jianing Song, Nov 29 '24 at 22:47
@JianingSong This is only vaguely connected to your post. In A046015 or discriminants $d$ with class number $h(d)=18$, you gave the 9 discriminants that are isomorphic to $C_3 \times C_6$. In A351674 or discriminants $d$ with class number $h(d)=36$, someone mentioned there are 16 discriminants that are isomorphic to $C_{12} \times C_3$, but did not write them down. Would you happen to know those 16 $d$? — Tito Piezas III, Jan 06 '25 at 08:59
@JianingSong In case you are interested, the question about $C_{12} \times C_3$ is now in this MO post. Feel free to answer if you wish. — Tito Piezas III, Jan 07 '25 at 11:58

score 4 · Accepted Answer · answered Nov 29 '24 at 23:05

4

Let $R$ be a Prüfer domain which is also a UFD. Let $\mathfrak{p}$ be a prime ideal of $R$. Then $R_{\mathfrak{p}}$ is a valuation ring which is also a UFD. As valuation rings are Bezout domains, this implies that $R_{\mathfrak{p}}$ is a PID.

As $$\dim R = \sup_{\mathfrak{p}\in \mathrm{Spec}(R)} \dim R_{\mathfrak{p}}$$

This implies that $R$ has dimension $\leq 1$. Finally, a UFD of dimension at most one is a PID.

answered Nov 29 '24 at 23:05

Lukas Heger

25,763

This result is cool. Thanks! – Jianing Song Nov 30 '24 at 00:59
The software at DaRT is actually able to explain a proof for this, but it's not available for everyone's use yet. The explanation probably boils down to this anyway :) – rschwieb Dec 02 '24 at 14:02

Is there a Prüfer domain that is a UFD but not a PID?

1 Answers1