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?
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3$\begingroup$ Germs of holomorphic functions at some $z_0\in\mathbb{C}$? $\endgroup$– Kevin VentulloFeb 24, 2011 at 11:19
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8$\begingroup$ More generally, there are lots of discrete valuation rings arising from geometry (local ring of the generic point of a divisor in a smooth variety); I don't suppose they count as "exotic". $\endgroup$– Laurent Moret-BaillyFeb 24, 2011 at 12:46
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$\begingroup$ Dear Laurent, I don't think you can say "more generally" because Kevin's example consists of convergent power series, whereas yours are of algebraic nature. $\endgroup$– Georges ElencwajgFeb 24, 2011 at 14:30
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$\begingroup$ As an aside (that absolutely doesn't answer the question ...) let me recall that between the innocent-looking rings $\mathbb Z$ and $\mathbb Q$ there is a continuum of rings, all of them principal ideal domains, obtained by inverting arbitrary subsets of the prime numbers $\endgroup$– Georges ElencwajgFeb 24, 2011 at 15:39
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3$\begingroup$ Yes, this is covered by "localizations of these." $\endgroup$– Qiaochu YuanFeb 24, 2011 at 16:10
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5 Answers
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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.
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2$\begingroup$ With regards to Bézout domains that are not (elementary submodels of) unltraproducts of PIDs, an example which I think is easier to verify is
$R=\bigcup_nF[x^{1/n}]$, where $F$ is a field. $R$ is Bézout because any its finitely generated subring is contained in some $F[x^{1/n}]$, which is a PID. OTOH, $x$ is not a unit in $R$, but it has no prime divisor. $\endgroup$ Feb 24, 2011 at 15:18 -
$\begingroup$ I see. I suppose all the PIDs I care about are Noetherian anyway, so I am totally okay with taking the perspective that Bezout domains are the more fundamental concept. $\endgroup$ Feb 24, 2011 at 15:30
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1$\begingroup$ All PIDs are Noetherian. A Bezout domain is Noetherian iff it is a PID. $\endgroup$ Feb 24, 2011 at 15:44
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1$\begingroup$ Agh, yes, what I meant to say was "all the Bezout domains I care about are Noetherian anyway." $\endgroup$ Feb 24, 2011 at 16:11
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$\begingroup$ As soon as I saw the word "finicky" I knew this was one of yours, Pete. $\endgroup$– JSEFeb 24, 2011 at 16:42
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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.)
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$\begingroup$ Nice. I suppose I can also take ultraproducts of PIDs to get more exotic ones... $\endgroup$ Feb 24, 2011 at 13:38
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3$\begingroup$ Not really. In almost any nontrivial case, an ultraproduct will contain an infinite chain of divisors, and thus fail to be a UFD. $\endgroup$ Feb 24, 2011 at 13:45
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2$\begingroup$ Oh, I see. Being a PID is not a first-order property. My mistake. $\endgroup$ Feb 24, 2011 at 14:17
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4$\begingroup$ @Qiaochu: right. The corresponding first order property is that every finitely generated ideal is principal -- i.e., Bezout domains. In other words, every ultraproduct of Bezout domains is Bezout. Off the top of my head, I wonder whether every Bezout domain is an ultraproduct of PIDs? $\endgroup$ Feb 24, 2011 at 14:50
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4$\begingroup$ @Pete: No. PIDs have additional first-order properties that do not hold for every Bézout domain, for example: every non-unit is divisible by a prime. BTW, being a PID is not a first-order property, but it's not that bad: it's expressible in $L_{\omega_1\omega}$. $\endgroup$ Feb 24, 2011 at 14:57
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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).
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1$\begingroup$ Care to provide a definition or a reference? $\endgroup$ Feb 24, 2011 at 20:25
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3$\begingroup$ Dear Qiaochu, for a careful definition, you can have a look at this set of notes of Brinon and Conrad: math.stanford.edu/~conrad/papers/notes.pdf But beware that the definitions are rather technical and for many purposes, it's enough to know the main properties of the period rings. For an overview, have a look at David Savitt's notes from the POSTECH winter school: dl.dropbox.com/u/1164264/korea_savitt.pdf (probably not a very permanent link) $\endgroup$– Alex B.Feb 25, 2011 at 0:26
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3$\begingroup$ I would like to add: this fact was truly an absolute shock to specialists in $p$-adic Hodge theory (including Fontaine, who invented the theory!), and the developments that it has led to are indeed very nice (and very recent, with hopefully more to come!). $\endgroup$– EmertonFeb 25, 2011 at 1:08
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1$\begingroup$ For example, starting with this PID, one can produce a Dedekind scheme $X$ over $\mathbb Q_p$ which is infinite type, but which otherwise behaves as if it were proper, and for which $\Gamma(X,\mathcal O_X) = \mathbb Q_p$! So it really look an enormous infinite type version of $\mathbb P^1$ over $\mathbb Q_p$. And it is huge (!): a typical residue field of a closed point is $\mathbb C_p$. $\endgroup$– EmertonFeb 25, 2011 at 1:11
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$\begingroup$ "really look an enormous" --> "really looks like an enormous" $\endgroup$– EmertonFeb 25, 2011 at 1:12
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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.).
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$\begingroup$ I love how a little topology can go a long way in algebra. $\endgroup$ Jan 10, 2019 at 20:12
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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....