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

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.