$\newcommand{\ZZ}{\mathbb{Z}}$ $\newcommand{\dim}{\text{dim }}$

Let me begin by apologizing for the length of this question, but I thought this might be interesting to some of you. This ring isn't exactly contrived and yet it seems sort of mysterious.

This is a follow up to an earlier question What is the etale fundamental group of Spec Z((x))?

where it was proved that $\ZZ((x)) := \mathbb{Z}[[x]][x^{-1}]$ has no nontrivial etale extensions, so its etale fundamental group is trivial.

From basic formulas we know that the units are precisely the laurent series whose leading coefficient (ie, ``first nonzero coefficient'') is a unit in $\mathbb{Z}$ (ie, is $\pm1$). Since $\mathbb{Z}$ is a PID, $\mathbb{Z}((x))$ is a unique factorization domain.

Question 1: Is there a nice way to describe the fraction field of $\mathbb{Z}((x))$?

Note that for any $f(x) = x^{-n}(a_0 + a_1x + a_2x^2+\cdots)\in\mathbb{Z}((x))$, we have $\frac{1}{f(x)}\in\mathbb{Z}[a_0^{-1}]((x))$, so Frac $\mathbb{Z}((x))\ne \mathbb{Q}((x))$ (for example, $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\notin$ Frac $\mathbb{Z}((x))$). On the other hand, a series like $$\sum_{n\ge 0}\frac{x^n}{r^{2^n}}\in\mathbb{Z}[r^{-1}]((x))$$ shouldn't be the inverse of anything in $\mathbb{Z}((x))$, since the denominators grow too fast, and the inverse of $f(x)$ described above has denominators growing roughly like $\frac{1}{r},\frac{1}{r^2},\frac{1}{r^3},\cdots$.

Question 2: What is the dimension of $\mathbb{Z}((x))$?

Clearly $\dim\mathbb{Z} = 1$, and I'm pretty sure $\dim\mathbb{Z}[[x]] = 2$ with all prime ideals having the form $(0),(p)$, or $(p,x)$. Thus I'm similarly pretty sure that $\dim\mathbb{Z}((x)) = 1$ again, but I don't have a proof (partially due to the fact that the localizations of $\mathbb{Z}((x))$ seem hard to describe).

If one is to believe that $\dim\mathbb{Z}((x)) = 1$, then since it's a UFD, it must also be a PID. A natural question is:

Question 3: Are the rings $\mathcal{O}_K((x))$ Dedekind domains? ($\mathcal{O}_K$ is the ring of integers of some number field $K$).

Anyway, back to $\ZZ((x))$. This ring seems to have at least two types of primes ideals. Clearly rational primes $p\in\mathbb{Z}$ give maximal ideals with residue fields $\mathbb{F}_p((x))$. On the other hand, any power series of the form $f(x) = p + a_1x + a_2x^2 + \cdots$ for some prime $p$ is also irreducible, hence prime (any factorization must include a factor of the form $\pm 1 + b_1x + b_2x^2+\cdots$, which is a unit), and is not equivalent to a rational prime $p$ as along as $p\nmid f(x)$ . If one believes that $\dim\mathbb{Z}((x)) = 1$, then these also give maximal ideals. One can prove that any power series $a_0 + a_1x + a_2x^2 + \cdots$ with $a_0$ divisible by at least 2 primes is composite. On the other hand, if $a_0 = p^2$, then for it to be composite, in which case it must factor as $(p + b_1x + \cdots)(p + c_1x + \cdots)$, it is necessary but not sufficient that $p\mid a_1$.

Question 4: What are the residue fields of the form $\ZZ((x))/f(x)$, where $f(x) = p^n + a_1x + a_2x^2 + \cdots$? (assuming $f$ is irreducible and $p\nmid f$)

An easy case is $f(x) = p - x$, in which case it seems pretty clear that the quotient field is $\mathbb{Q}_p$. Another example is $p^2 - x$, in which case it seems like the quotient field is also $\mathbb{Q}_p$. For now, lets restrict to the case $n = 1$, so $f(x) = p + a_1x + a_2x^2 + \cdots$.

If $p\nmid f(x)$, then the residue field must be characteristic 0, so it must be an extension of $\mathbb{Q}$. One may ask, what does $x$ map to? Since $x$ is a unit in $\ZZ((x))$, it can't map to 0. Furthermore, it must map to something such that arbitrary power series in $x$ makes sense, so the image should be some kind of field which is complete w.r.t. a nonarchimedean valuation, in which $x$ gets sent to something of positive valuation, and such that $a_1x + a_2x^2 + \cdots = -p$. The only extensions of $\mathbb{Q}$ that I can think of which fit this description are $\mathbb{Q}((t))$ and $\mathbb{Q}_l$. The first can't be a residue field, since $a_1x + a_2x^2 + \cdots \ne -p$, so lets suppose the residue field is some $\mathbb{Q}_l$. Since $v(a_1x + a_2x^2 + \cdots) = v(a_1x) = v(-p)$, and $v(a_1x)\ge v(x) > 0$, $v(-p) > 0$, so $l = p$. Just by considering $f(x) = p - x^n$, we get all totally ramified extensions of $\mathbb{Q}_p$ as residue fields.

My intuition says that the residue fields of $\ZZ((x))$ are either $\mathbb{F}_p((x))$ or finite extensions of $\mathbb{Q}_p$. However, it's unclear to me if you can get unramified extensions as well as ramified extensions, and it's unclear why there can't be other weird complete fields that could show up.

  • 3
    $\begingroup$ A1: No. A3 (so A2): Yes, since Spec is complement of a divisor containing all closed points in a regular scheme of dimension 2. A4: The quotient you write is artinian by A3, and you may assume $f$ is irreducible (as its irreducible factors have that form) and some $a_i\ne 0$. It is the fraction field of the 1-dimensional $D=\mathbf{Z}[\![x]\!]/(f)$. By $x$-adic completeness, $D$ is $p$-adically complete, so you can replace $\mathbf{Z}$ with $\mathbf{Z}_p$, and $f$ is irreducible over $\mathbf{Z}_p$ as $D$ a domain. Thus, some $a_i$ is a $p$-adic unit, so Weierstrass Preparation does the rest. $\endgroup$
    – user74230
    Jan 16 '15 at 3:48
  • 1
    $\begingroup$ By the way, I am assuming you are aware that if a noetherian ring $R$ is separated and complete for the topology of an ideal $I$ then $I$ lies in every maximal ideal (ultimately by characterization of the Jacobson radical and the invertibility of $1+t$ for $t \in I$). $\endgroup$
    – user74230
    Jan 16 '15 at 3:53
  • 1
    $\begingroup$ @oxeimon: separated for the $I$-adic topology (so yes, Hausdorff, or equivalently $\cap_{n>0} I^n=0$). $\endgroup$
    – user74230
    Jan 16 '15 at 5:56
  • 1
    $\begingroup$ @oxeimon: There is nothing "exceptionally weird" about Weierstrass Preparation. Push your imagination by playing with irreducible quadratic polynomials whose constant term has higher valuation and intermediate coefficient is also a non-unit. $\endgroup$
    – user74230
    Jan 16 '15 at 7:20
  • 1
    $\begingroup$ @oxeimon: I am using exactly the fact specifically highlighted already about maximal ideals in relation to $I$ when adically separated and complete (so all maximal ideals, visibly of height 2 by inspection or other considerations, contain $x$). $\endgroup$
    – user74230
    Jan 16 '15 at 15:17

Question 1: The fraction field is the same as that of $\mathbb{Z}[[x]]$. It can be gotten by inverting all irreducibles. The irreducibles of the UFD $\mathbb{Z}[[x]]$ are described in Theorem 1.4 of http://arxiv.org/abs/1107.4860

Question 2: You missed a lot of prime ideals of $\mathbb{Z}[[x]]$. (It has uncountably many.) In particular, its height one primes are principal (since it is a UFD) and generated by an arbitrary irreducible power series (as described in Question 1), and none of them are maximal. The maximal ideals are all of height two and of the form $(p,x)$, and they are killed by inverting $x$, so the Krull dimension of $\mathbb{Z}((x))$ is indeed $1$ (hence $\mathbb{Z}((x))$ is a PID).

Question 3: I believe user74230 answered that correctly.

Question 4: This is answered by Theorem 1.2 of http://arxiv.org/pdf/1107.4860v4.pdf In particular, the integral domains $\mathbb{Z}[[x]]/(f)$ for irreducible $f$ (besides $x$ and primes $p$) are precisely $\mathbb{Z}_p[\alpha]$ for $\alpha$ in the unique maximal ideal of the integral closure of $\mathbb{Z}_p$ in $\mathbb{C}_p$ (or equivalently for $\alpha$ in $\overline{\mathbb{Q}_p}$ with positive $p$-adic valuation), where $p$ is any prime. Since $\alpha$ is the image of $x$ in these isomorphisms and localization commutes with quotient rings, the answer to your question is all fields of the form $\mathbb{Q}_p(\alpha) = \mathbb{Z}_p[\alpha][\alpha^{-1}]$, where $\alpha$ is as described earlier.

  • $\begingroup$ That's pretty awesome! For your answer to question 4, it seems like you get every finite extension of $\mathbb{Q}_p$ that way. This is indeed a very rich ring. $\endgroup$
    – Will Chen
    Jan 16 '15 at 22:05
  • $\begingroup$ Indeed. I suppose it's an analogue of the fact that every finite extension of $\mathbb{Q}$ (and likewise every finite extension of $\mathbb{F}_p$) is a reside field of $\mathbb{Z}[x]$. $\endgroup$ Jan 17 '15 at 6:09

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