Exotic principal ideal domains

Question

Recently I realized that the only PIDs I know how to write down that aren't fields are $\mathbb{Z}, F[x]$ for $F$ a field, integral closures of these in finite extensions of their fraction fields that happen to have trivial class group, localizations of these, and completions of localizations of these at a prime. Are there more exotic examples? Is there anything like a classification?

Answer 1

No, to the best of my knowledge there is nothing like a general classification of PIDs. Despite their easy definition, they turn out to be rather a finicky class of rings, as for instance Gauss conjectured that there are infinitely many PIDs among rings of integers of real quadratic fields, but more than $200$ years later we have not been able to prove that there are infinitely many PIDs among rings of integers of all number fields. And, as came out in the comments to Emil's answer, the property of being a PID is not first order, so is not very robust in a model-theoretic sense. In that regard, the better class of rings are the Bézout domains, i.e., domains in which every finitely generated ideal is principal. A theorem of Kaplansky which can be used to show that various "big" domains (e.g. $\overline{\mathbb{Z}}$, the ring of all algebraic integers) are Bézout can be found at the end of the section on overrings in these notes. (I am now giving less precise citations to my often-changing commutative algebra notes in the hope that they will take longer to become obsolete.)

There are some interesting papers on construction of PIDs with various properties. The one I want to read next is this 1974 paper of Raymond C. Heitmann: given any countable collection $\mathcal{F}$ of countable fields containing only finitely many fields of any given positive characteristic, Heitmann constructs a countable PID of characteristic $0$ with residue fields precisely the elements of $\mathcal{F}$.

Added: note that $\overline{\mathbb{Z}}$ is also an antimatter domain, i.e., it has no irreducible elements (which specialists in the field tend to call "atoms"). Thus this gives an example of a Bézout domain which is not an ultraproduct of PIDs.

Answer 2

Smith constructed a PID which is a nonstandard model of open induction. That should be exotic enough. (Note that nonstandard models of just slightly stronger theories of arithmetic, such as $IE_1$, are never even UFDs.)

Answer 3

Fontaine's ring $B_{cris}^{\varphi=1}$ is a PID, and no expert in the field would have bet on it in the first place (this led to some very nice recent developments by Fargues and Fontaine).

http://www.math.u-psud.fr/~fargues/Courbe.pdf

Answer 4

A commutative algebra is a PID if and only if it is a UFD and all nonzero prime ideals are maximal. This leads to an interesting method to construct PID's: Let $R$ be a UFD and let $S \subset R$ be a multiplicative set such that, for any prime $\mathfrak{p} \subset R$ of height $\geq 2$, there is some $f \in S$ with $f \in P$. Then $S^{-1} R$ will be a PID, because localizations of UFD's are UFD's and the poset of prime ideals in $S^{-1} R$ is obtained from the poset of prime ideals in $R$ by deleting those ideals containing an element of $S$.

This can be useful for building counterexamples, because $S^{-1} R$ is the forward limit of $f^{-1} R$ over all $f \in S$, and each of the $f^{-1} R$ will be a UFD but not a PID, so one can take counterexamples in UFD's and make them into PID counterexamples by this trick. Speaking vaguely, although $S^{-1} R$ has Krull dimension $1$, it often acts more like a ring of dimension equal to the Krull dimension of $R$.

I learned about this construction from Grayson's paper "$SK_1$ of an interesting principal ideal domain". The PID in question is to take $R = \mathbb{Z}[T]$ and $S = \{ T \} \cup \{ T^n-1 : n > 0 \}$, and the interesting property is that $SL_n(S^{-1} R)$ is not generated by elementary matrices.

I can't resist showing off: After I read Grayson's paper, I come up with the following simpler example. Let $R = \mathbb{R}[x,y]$ and let $S$ be the set of nonzero polynomials in $\mathbb{R}[x^2+y^2]$. Then $S^{-1} R$ is a PID by the above argument. I claim that $M= \left[ \begin{smallmatrix} x/(x^2+y^2) &y/(x^2+y^2) \\ -y&x \end{smallmatrix} \right]$ is not a product of elementary matrices. Suppose that $M=E_1 E_2 \cdots E_n$. Then the denominators of the $E_j$ only contain finitely many elements of $S$, so all the $E_j$ lie in $f(x^2+y^2)^{-1} R$ for some nonzero polynomial $f$. Choose some real number $r$ so that $f(r^2) \neq 0$, then each of the $E_j$ is a well defined continuous function on the circle $x^2+y^2 = r^2$. So $M=E_1 E_2 \cdots E_n$ gives a map from this circle to $SL_2(\mathbb{R})$. Consider the class of this map in $H_1(SL_2(\mathbb{R})) \cong \mathbb{Z}$. Rescaling each off diagonal entry of the $E_j$ by a real number $t$ and sliding $t$ from $1$ to $0$ is a homotopy to the trivial map, so this class is $0$. On the other hand, $\left[ \begin{smallmatrix} x/(x^2+y^2) &y/(x^2+y^2) \\ -y&x \end{smallmatrix} \right]$ represents the generator of $H_1$, a contradiction. The same argument shows that the block matrix $\left[ \begin{smallmatrix} M & \\ & \mathrm{Id}_{n-2} \end{smallmatrix} \right]$ in $SL_n(S^{-1} R)$ is also not a product of elementary matrices (this time we have $H_1(SL_n(\mathbb{R}))\cong H_1(SO_n(\mathbb{R})) \cong \mathbb{Z}/2$, and we need spin groups to compute the class in $H_1$, but I think it still works.).

Answer 5

Dear Qiaochu, if $A$ is a discrete valuation ring and if $B$ is an étale algebra over $A$, then $B$ is a discrete valuation ring. In a related vein, the henselization of a discrete valuation ring $A$ is a discrete valuation ring $A^h$ (however it is not étale over $A$, for example because it is not finitely generated ).If $A$ is the local ring of a point on a curve in the Zariski topology, then $A^h$ is the local ring of that point in the étale topology.

A very concrete example: the henselization of the local ring $A=\mathcal O_{\mathbb A^1,0}$ of the complex affine line at the origin is the subring of the ring of formal series $\mathbb C [[T]]$ consisting of those series that are algebraic over $A$.

These seem to be examples not on your list, but I'll let you be the judge of their exotism....