Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
2  
Germs of holomorphic functions at some $z_0\in\mathbb{C}$? –  Kevin Ventullo Feb 24 '11 at 11:19
6  
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". –  Laurent Moret-Bailly Feb 24 '11 at 12:46
    
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. –  Georges Elencwajg Feb 24 '11 at 14:30
    
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 –  Georges Elencwajg Feb 24 '11 at 15:39
3  
Yes, this is covered by "localizations of these." –  Qiaochu Yuan Feb 24 '11 at 16:10

4 Answers 4

up vote 8 down vote accepted

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.

share|improve this answer
2  
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. –  Emil Jeřábek Feb 24 '11 at 15:18
    
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. –  Qiaochu Yuan Feb 24 '11 at 15:30
1  
All PIDs are Noetherian. A Bezout domain is Noetherian iff it is a PID. –  Chris Eagle Feb 24 '11 at 15:44
1  
Agh, yes, what I meant to say was "all the Bezout domains I care about are Noetherian anyway." –  Qiaochu Yuan Feb 24 '11 at 16:11
    
As soon as I saw the word "finicky" I knew this was one of yours, Pete. –  JSE Feb 24 '11 at 16:42

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.)

share|improve this answer
    
Nice. I suppose I can also take ultraproducts of PIDs to get more exotic ones... –  Qiaochu Yuan Feb 24 '11 at 13:38
3  
Not really. In almost any nontrivial case, an ultraproduct will contain an infinite chain of divisors, and thus fail to be a UFD. –  Emil Jeřábek Feb 24 '11 at 13:45
2  
Oh, I see. Being a PID is not a first-order property. My mistake. –  Qiaochu Yuan Feb 24 '11 at 14:17
4  
@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? –  Pete L. Clark Feb 24 '11 at 14:50
4  
@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 $L_{\omega_1\omega}$. –  Emil Jeřábek Feb 24 '11 at 14:57

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

share|improve this answer
1  
Care to provide a definition or a reference? –  Qiaochu Yuan Feb 24 '11 at 20:25
3  
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) –  Alex B. Feb 25 '11 at 0:26
3  
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!). –  Emerton Feb 25 '11 at 1:08
1  
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 $\mathbb C_p$. –  Emerton Feb 25 '11 at 1:11
    
"really look an enormous" --> "really looks like an enormous" –  Emerton Feb 25 '11 at 1:12

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....

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.