There are two great first examples of complete discrete valuation ring with residue field $\mathbb{F}_p = \mathbb{Z}/p$: The $p$-adic integers $\mathbb{Z}_p$, and the ring of formal power series $(\mathbb{Z}/p)[[x]]$. Any complete DVR over $\mathbb{Z}/p$ is a ring structure on left-infinite strings of digits in $\mathbb{Z}/p$. The difference between these two examples is that in the $p$-adic integers, you add the strings of digits with carries. (In any such ring, you can say that a $1$ in the $j$th place times a 1 in the $k$th place is a 1 in the $(j+k)$th place.) At one point I realized that these two examples are not everything: You can also add with carries but move the carry $k$ places to the left instead of one place to the left. The ring that you get can be described as $\mathbb{Z}_p[p^{1/k}]$, or as the $x$-adic completion of $\mathbb{Z}[x]/(x^k - p)$. This sequence has the interesting feature that the terms are made from $\mathbb{Z}_p$ and have characteristic $0$, but the ring structure converges topologically to $(\mathbb{Z}/p)[[x]]$, which has characteristic $p$.

(Edit: Per Mariano's answer, I have in mind a discrete valuation in the old-fashioned sense of taking values in $\mathbb{Z}$, not $\mathbb{Z}^n$.)

I learned from Jonathan Wise in a question on mathoverflow that these examples are still not everything. If $p$ is odd and $\lambda$ is a non-quadratic residue, then the $x$-adic completion of $\mathbb{Z}[x]/(x^2-\lambda p)$ is a different example. You can also call it $\mathbb{Z}_p[\sqrt{\lambda p}]$.

So my question is, is there is a classification or a reasonable moduli space of complete DVRs with residue field $\mathbb{Z}/p$? Or whose residue field is any given finite field? Or if not a classification, an indexed family that includes every example at least once?

I suppose that the question must be related to the Galois theory of $\mathbb{Q}_p$; maybe the best answer would be a relevant sketch of that theory. But part of my interest is in continuous families of DVR structures on the Cantor set of strings of digits in base $p$.

As question authors often say in mathoverflow, I learned stuff from many of the answers and it felt incomplete to only accept one of them. I upvoted several others, though. Following Kevin, the set of pairs $(R,\pi)$, where $R$ is a CDVR and $\pi$ is a uniformizer, is an explicit indexed family that includes every example (of $R$) at least once. The mixed characteristic cases are parametrized by Eisenstein polynomials, and there is only one same-characteristic choice of $R$.

One side issue that was not addressed as much is continuous families of CDVRs. To explain that concern a little better, all CDVRs with finite residue field are homeomorphic (to a Cantor set). For any given finite residue field, there are many homeomorphisms that commute with the valuation. So you could ask what a continuous family of CDVRs can look like, where both the addition and multiplication laws can vary, but the topology and the valuation stay the same. Now that I have been told what the CDVRs are, I suppose that not all that much can happen. It seems key to look at $\nu(p)$, the valuation of the integer element $p$, in a sequence of such structures (say). If $\nu(p) \to \infty$, then the CDVRs have to converge to $k[[x]]$. Otherwise $\nu(p)$ is eventually constant; it eventually equals some $n$. Then it looks like you can convert the convergent sequence of CDVRs to a convergent sequence of Eisenstein polynomials of degree $n$.

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    $\begingroup$ Greg's first question! Too bad they don't give a badge for 10K without asking a question. $\endgroup$ – Noah Snyder Dec 26 '09 at 5:10
  • $\begingroup$ In Firefox 3.5.6, OS X, the unicode \mathbb{Z} in the title doesn't render. $\endgroup$ – Scott Morrison Dec 26 '09 at 21:24
  • $\begingroup$ I'm running Firefox 3.5.6, OS X, and it does render for me. $\endgroup$ – Bjorn Poonen Dec 26 '09 at 22:04
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    $\begingroup$ Greg has added some new stuff at the bottom of his post, and here's a coment on that: Krasner's Lemma implies that if f is an irreducible polynomial in Q_p[x] (or indeed in K[x] for K any non-arch local field) then for any g sufficiently close to f and of the same degree (keeping Greg's nu constant), the field extensions of Q_p (or K) generated by f and g coincide. So if you're just thinking about families of R's generated in this way (by continuously moving f), the isomorphism class of R itself will be locally contant (but pi will move). $\endgroup$ – Kevin Buzzard Dec 29 '09 at 7:21
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    $\begingroup$ Many excellent answers have already been given, and a clear summary is, I think, always better than (just) a reference to a paper; but this question shouldn't go by without a reference to Cohen's classic “On the structure and ideal theory of complete local rings”, TAMS 59 54–106. $\endgroup$ – LSpice Mar 16 '10 at 19:05

Greg, I want to say some basic things, but people are giving quite "high-brow" answers and what I want to say is a bit too big to fit into a comment. So let me leave an "answer" which is not really an answer but which is basically background on some other answers.

So firstly there is this amazing construction of Witt vectors, which takes as input a finite field $k$ of characteristic p (well, it can take a lot more than that but let me stick with finite fields of characteristic p) and spits out a canonical complete DVR $W(k)$ with residue field $k$ and uniformiser $p$. If you feed in $Z/pZ$ it spits out $Z_p$ and if you feed in, say, the field $Z/5Z[\sqrt{2}]$ with 25 elements it spits out something isomorphic to $Z_5[\sqrt{2}]$, and so on. Witt vectors are just a way of formalising the "carry" business---it gets a bit trickier when the residue field isn't Z/pZ. Turns out that any complete DVR $R$ with fraction field of characteristic 0 and residue field $k$ is canonically and uniquely a $W(k)$-algebra. So that's pretty cool. Furthermore, $R$ will be finite and free over $W(k)$, and $R$ is the "ring of integers" (this makes sense) in its fraction field $K$, which is a finite extension of the fraction field of $W(k)$, something which in turn will be finite over $Q_p$ of degree $d$ if $[k:Z/pZ]=d$. [Note in particular that "fraction field of Witt vectors" gives us a whole bunch of extension of $Q_p$---the so-called "unramified" ones---one for each finite extension of $Z/pZ$.]

Conversely, given an arbitrary finite field extension $K$ of $Q_p$, the ring of integers of $K$ (that is, the elements in $K$ satisfying a monic polynomial with coefficients in $Z_p$) will be a complete DVR with residue field some finite field $k$, and then the field of fractions of $W(k)$ embeds into $K$. This subfield of $K$---the "maximal unramified subextension of $K$" is canonical and intrinsic. An extension $K$ of $Q_p$ is "totally ramified" if the residue field of $R$ is $Z/pZ$. A FABULOUS place to read about this stuff is Serre's book "local fields". Everything there, with complete canonical proofs.

So one aspect of your question is whether there is a moduli space of totally ramified extensions of $Q_p$. I wouldn't rule such a gadget out but I've not seen one. This might be just a question in algebra. If $K$ is a finite totally ramified extension of $Q_p$ then $K=Q_p(\pi)$ with $\pi$ a uniformiser of the integers of $K$, and $\pi$ will satisfy a degree $d$ equation if $d=[K:Q_p]$. Furthermore this degree $d$ equation will be monic, with coefficients in $Z_p$, and the constant term will be $p$ times a unit. Conversely any such equation will give a totally ramified extension of $Q_p$ (they will all be irreducible by Eisenstein's criterion).

So now we have a list of all totally ramified extensions of $Q_p$, and hence all complete DVRs with residue characteristic $Z/pZ$ but generic characteristic zero, because we just list all polynomials of this form. Unfortunately each $R$ is in our list infinitely often. So to make the $R$s the $Z/pZ$-points of a moduli space we need to quotient out the set of degree $d$ Eisenstein polynomials by the relation "the extension of $Q_p$ generated by these polynomials are the same". A map $Q_p(\pi_1)\to Q_p(\pi_2)$ is just another polynomial so it looks to me like there is hope that this can be done, but I've not done it.

Finally, everything I said above has a natural generalisation to any finite field, not just $Z/pZ$.

When people talk about class field theory or Lubin-Tate groups above, what they're saying is that there are certain totally ramified extensions of $Q_p$, namely those which are Galois over $Q_p$ with abelian Galois group, which can be constructed explicitly using other techniques (like formal groups or Artin maps or whatever), and these constructions generalise to give all abelian extensions of an arbitrary finite extension of $Q_p$. However if you're looking for general moduli spaces then it seems to me that these notions might not be of too much use to you because they don't give all extensions, just abelian ones.

There. So really that was just a comment but it was visibly too large. Hopefully someone will now quotient out the Eisenstein polynomials by an equivalence relation thus giving you your moduli space, because that seems to me to be the heart of your question.

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    $\begingroup$ Greg---perhaps a "better" moduli space would be pairs (R,pi) with R a complete DVR and pi a uniformiser. Now it seems to me that you're in much better shape. For example the mixed characteristic ones will be precisely the Eisenstein polys. Of course the stackists amongst us would tell you not to look for a moduli space at all, just to look for a stack. There is still a geometric question here---whether the functor is a stack---it's just better-hidden. $\endgroup$ – Kevin Buzzard Dec 26 '09 at 9:56
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    $\begingroup$ This is just daft. My "answer" above is just a comment, and a review of the basics of the theory of local fields, but it has picked up votes, presumably from people who didn't understand the question but learned something from my comments! Next time someone asks a question, remind me to just list a few basic facts about one of the words in the question rather than answering it ;-) $\endgroup$ – Kevin Buzzard Dec 27 '09 at 18:03
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    $\begingroup$ As you might guess, I'm not a student of local fields. As it turned out, a good summary of the basics is a good answer to the question. $\endgroup$ – Greg Kuperberg Dec 29 '09 at 7:57

The classification of CDVRs with residue field any given perfect field k is discussed in Chapter 2 of Serre's Local Fields. In particular:

Theorem II.2: Let R be a CDVR with residue field k. Suppose R and k have the same characteristic and that k is perfect. Then R is isomorphic to k[[t]].

Theorem II.3: For every perfect field k of characteristic p, there exists a unique CDVR (up to unique isomorphism) which is absolutely unramified [i.e., p is a uniformizing element] and has k as its residue field: namely W(k), the Witt vector ring.

Theorem II.4: Let R be a CDVR of unequal characteristic with perfect residue field k. Let e be its absolute ramification index. Then there exists a unique homomorphism W(k) -> R commuting with reduction modulo the maximal ideal. This is injective, and R is a free W(k)-module of rank e.

Thus the CDVRs with residue field Z/p are: Z/p[[t]] and the valuation ring of a totally ramified extension of Z_p. In particular, the set of such isomorphism classes is countably infinite, and there no moduli in any sense known to me.

  • $\begingroup$ I think this is essentially the same as my answer. $\endgroup$ – Hailong Dao Dec 26 '09 at 7:19
  • $\begingroup$ After reading your answer more carefully, I agree with this assessment. I find my answer to be slightly clearer, but then again I would... $\endgroup$ – Pete L. Clark Dec 26 '09 at 7:39
  • $\begingroup$ Well, we used slightly different languages (: $\endgroup$ – Hailong Dao Dec 26 '09 at 7:41
  • $\begingroup$ Pete: the set of isomorphism classes is countably infinite, which is what one might expect for the Z/p-points of a moduli space over Z/p. So there might yet be a moduli space perhaps? $\endgroup$ – Kevin Buzzard Dec 26 '09 at 9:17
  • $\begingroup$ @buzzard: OK, possibly, if the representing space is not of finite type. But I think Greg wanted some kind of geometric structure on the set of CDVRs with residue field F_p, and this seems unlikely. $\endgroup$ – Pete L. Clark Dec 26 '09 at 9:37

I don't know what all the cdvr's with residue field $F_p$ are, but local class field theory gives you a huge family. Those that are finite extensions of $Z_p$ are the rings of integers in the totally ramified extensions of $Q_p$. For each prime element $\pi$ of $Z_p$, there is a unique totally ramified extension $K_{\pi}$ of $Q_p$ such that (a) $\pi$ is a norm from every finite subextension of $K_{\pi}$ and (b) the maximal abelian extension of $Q_p$ is obtained by composing $K_{\pi}$ with the unramified extensions of $Q_{p}$. For example, if $\pi=p$, then $K_{\pi}$ is obtained by adjoining the $p^{n}$th roots of $1$ to $Q_{p}$ for all $n$. In general, $K_{\pi}$ is given quite explicitly by Lubin-Tate theory. Different $\pi$ give different field extensions, so the $K_{\pi}$'s are parametrized by the units in $Z_p$. See, for example, Chapter I of my notes


I think the answer to your question is Section 29 of Matsumura's book "Commutative ring theory".

A summary: A complete $p$-ring is a complete local ring whose maximal ideal is generated by $p$. For any field $k$ of characteristic $p$, there is a unique complete $p$-ring whose residue field is $k$. In the case of $k=\mathbb Z/(p)$, that ring is $\mathbb Z_p$.

We call a DVR $(R,m,k)$ unramified if $char(R)= char(k)$ or if it is mixed characteristic and $p \notin m^2$. We call it ramified if $p\in m^2$.

Theorem: An unramified, complete DVR is isomorphic to $k[[x]]$ or is the unique $p$-ring whose residue field is $k$. A ramified one is isomorphic to $A[[x]]/(f)$, here $A$ is said unique $p$-ring and $f = p + g(x)$, here $g \in (n^2)$, with $n=(p,x)$ is the maximal ideal of $A[[x]]$.

So apart from the 2 unramified ones, the ramified ones are parametrized by the elements $g \in (n^2)$.

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    $\begingroup$ Taking g=0 leads to k[[x]]. To me it would make more sense to group k[[x]] with the ramified examples (in some sense, it is infinitely ramified: look at the valuation of p!) $\endgroup$ – Bjorn Poonen Dec 26 '09 at 21:53
  • $\begingroup$ You are right! In the problems I am working on, the increasing order of difficulty is almost always: contains a field < unramified <ramified, so I never thought of equicharacteristic as limit case of ramified. $\endgroup$ – Hailong Dao Dec 26 '09 at 22:19

Let $k$ be a finite field of characteristic $p$. There is only one characteristic-$p$ local field with residue field $k$, namely $k((\pi))$, where $\pi$ is transcendental, and there is a "smallest" characteristic-$0$ local field with residue field $k$, namely the degree-$a$ unramified extension $K$ of $\mathbb{Q}_p$, where $q=p^a$ is the cardinality of $k$.

Every characteristic-$0$ local field with residue field $k$ is a finite totally ramified extension of $K$, and every finite totally ramified extension of $K$ has residue field $k$.

As $K$ has only finitely many extensions of any given degree, we might ask for the number of totally ramified extensions $L|K$ of given degree $[L:K]=n$. Serre's "mass formula" (Comptes rendus 1978), says that there are exactly $n$ such extensions, when counted properly. This formula was also proved by Krasner by his methods.

Let Serre explain:

"Let $K$ be a local field, with finite residue field with $q$ elements. Let $n$ be a positive integer, and let $\Sigma_n$ be the set of all totally ramified extensions of $K$ of degree $n$ contained in a given separable closure of $K$. If $\operatorname{gcd}(n,p)=1$, it is easy to show that $\text{Card}(\Sigma_n)=n$. If $L$ belongs to $\Sigma_n$, put $c(L)=d(L)-n+1$, where $d(L)$ is the valuation of the discriminant of $L/K$. Our 'mass formula' is

$\sum_{L\in \Sigma_n} q^{-c(L)}=n$.

We give two proofs. The first one uses Eisenstein polynomials, while the second one applies the H. Weyl integration formula to the multiplicative group of a division algebra." MR0500361 (80a:12018)

(Note that $k((\pi))$ has many totally ramified extensions $L$, but they are all of the form $L=k((\varpi))$ for some uniformiser $\varpi$ of $L$.)


Greg notes that $(Z/pZ)[[x]]$ is a limit of the integers of a sequence of totally ramified extensions of $Q_p$. At the back of my mind I wonder whether he'd be interested by the Fontaine-Wintenberger theory of "corps de normes" , where local fields of characteristic $p$ are written as limits of $p$-adic fields in more generality. I think that Greg has just noted a special case of this theory, correctly interpreted. Or maybe I've misremembered what this theory is and it's something slightly different---I'm certainly no expert. Worth checking out though. Unfortunately I'm not in a position to give a good web reference or even paper reference but hopefully someone, or google, can.

  • $\begingroup$ I looked for papers that mention Fontaine-Wintenberger. The problem is that the theory addresses constructions that are much more complicated than my question, so I can't tell whether or not it is relevant. $\endgroup$ – Greg Kuperberg Dec 27 '09 at 19:06
  • $\begingroup$ Oh dear. I'll try and dig out some comprehensible summary of what's going on when I get back to work after the holidays. $\endgroup$ – Kevin Buzzard Dec 28 '09 at 17:56
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    $\begingroup$ @KevinBuzzard—I know it's still the holidays, but, 8 years later, any news on a good summary for the relevant Fontaine–Wintenberger theory? :-) $\endgroup$ – LSpice Dec 28 '17 at 19:10

Those of dimension 1 (which I guess are the ones you have in mind?) and of equal characteristic are described in [Sekiguchi, Koji. On the structure of strictly complete valuation rings. Tokyo J. Math. 26 (2003), no. 2, 393--402. MR2020792] and those which are unramified are just the Witt vectors, IIRC.

  • $\begingroup$ Koji's paper seems concerned with higher dimensions. In 1D and equal characteristic, I don't see that there is anything other than $k[[x]]$. projecteuclid.org/… $\endgroup$ – Greg Kuperberg Dec 26 '09 at 6:01

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