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We know that the fraction field $F:=\operatorname{Frac}(\mathbb Z_p[[X]])$ is strictly contained in the field of Laurent power series $\mathbb Q_p((X))$, thanks to this result of Gilmer. So my question is:

Is it possible to describe explicitly the elements of $F$?

Some similar questions have been already asked here or on Mathoverflow. Maybe the most relevant is this one regarding the explicit computation of the fraction field of $\mathbb Z[[X]]$. Someone suggests in the comments of the linked question that the problem with $\mathbb Z_p$ (instead of $\mathbb Z$) should be easier.

Some general necessary conditions are given here when the coefficients of the power series lie in any domain, but I'd like to find some sufficient conditions in the particular case of $\mathbb Z_p$.

Many thanks in advance

manifold
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2 Answers2

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I wanted to post this as a comment, but I don't have enough reputation. The accepted answer is wrong. It is not true that a nonzero element of $\mathbb{Z}_p[[X]]$ is of the form $X^np^m\sum_k b_kX^k$ with $b_0\notin (p)$. For example, $p+X$ is not of this form. And indeed the inverse of $p+X$ is not in $\mathbb{Z}_p[[X]][X^{-1},p^{-1}]$: $$(p+X)^{-1}=\sum_k (-1)^k\frac{1}{p^{k+1}}X^k .$$ As in this answer to one of the questions you linked, by the Weierstrass preparation theorem it is enough to invert $p$ and all monic polynomials. However, this doesn't give a description of the fraction field in terms of the coefficients of the Laurent power series. I don't know if such a description exists.

fred2015
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    $v(\sum_{n\ge 0} a_n x^n)= \inf_n v_p(a_n)$ is a discrete valuation, it extends to the fraction field $F$. It seems natural to consider the completion $F_v$ which may have a simpler description. The residue field is $\Bbb{F}_p((x))$ and uniformizer $p$ so it's probably $F_v=W(\Bbb{F}_p((x)))$ – reuns Mar 10 '23 at 23:46
  • The completion $F_v$ is the ring of formal series $\sum_{n=-\infty}^\infty a_n x^n$ with $\lim_{n\to \infty} |a_n|p<\infty,\lim{n\to -\infty} |a_n|_p=0$ – reuns Mar 11 '23 at 00:05
  • You're absolutely right, I don't know how I missed this (although I had no memory of answering this question :D ) – Maxime Ramzi Mar 14 '23 at 18:59
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$\newcommand{\Z}{\mathbb{Z}}$ Every element $f\in\mathbb{Z}_p[[x]]$ can be written as $$ f(x) = p^e u(x) q(x) $$ with $e \geq 0$, $u(x)\in \mathbb{Z}_p[[x]]^*$ and $q(x)\in \mathbb{Z}_p[x]$ monic.

This follows from the Weierstrass preparation theorem.

Therefore every element $F \in \text{Frac}\left( \Z_p[[x]] \right)$ can be written as $$ F = \frac{f(x)}{q(x)} $$ with $f(x)\in \Z_p[[x]]$ and $q(x) \in \Z_p[x]$.

Inverting $q(x)$ shows what kinds of bounds we can have on the coefficients $F_n$ of $F$, in particular $\limsup_{n \to \infty}\frac{v_p(F_n)}{n} < \infty$.

Another way to see this is to write $$ F=\frac{1}{x^a}\frac{f(x)}{g(x)}, $$ with $f(x)\in \Z_p[[x]]$, and $x \nmid g(x)$, and $$ g(x) = \sum g_i x^i = p^{e} + g_1x + g_2x^2\ldots $$ with $c \geq 0$. Then $$ p^{-e}g(x) = 1 + \frac{g_1}{p^r}x + \frac{g_2}{p^r}x^2 + \ldots $$ and therefore $$ h(x) = \frac{1}{p^{-r} g(x) } \in \Z_p[[\tfrac{x}{p^r}]]. $$ So $F = \frac{1}{p^e}\frac{1}{x^a} f(x) h(x)$ with $$ h(x) \in \bigcup_{e=0}^\infty {\large{\Z_p}}\big[\big[\tfrac{x}{p^e}\big]\big]. $$

Kolja
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