# Fraction field of $p$-adic power series ring

Let $L$ be a finite extension of $\mathbf{Q}_p$. Write $$\mathcal{O}_{\mathcal{E}} = \left\{ f = \sum_{k \in \mathbf{Z}} a_kT^k \in \mathcal{O}_{L}[[T,T^{-1}]] \mid \lim_{k \to -\infty} a_k = 0\right\}$$

I read somewhere that the fraction field of this ring is $\mathcal{E} = \mathcal{O}_{\mathcal{E}}[1/p]$. Is that true ?

It seems weird to me because for example $p+T$ has inverse $\sum_{k=0}^\infty (-1)^k \dfrac{T^k}{p^{k+1}}$ which doesn't seem to be in $\mathcal{E}$ since it has arbitrarily big coefficients.

• I think that the inverse of $p+T$ is $\sum_{n\geq 0}(-1)^n\frac{p^n}{T^{n+1}}$ that is in $O_E$ Commented Jul 14, 2016 at 17:42
• C Hawkins, $L[[T,T^{-1}]]$ is not a ring. At least I don't see you could define a product there. What would be, for example, the square of $\sum_{k\in\Bbb{Z}}T^k$? That square has infinitely many terms of degree zero, namely $T^k\cdot T^{-k}=1$ for all $k$. The sum of countably infinitely many ones doesn't converge. That extra condition on the limit of coefficients is needed to make the products defined. Commented Jul 14, 2016 at 20:02
• I can sort of see why the claim should be true. By truncating the series $a(T)$ by dropping low degree terms (i.e. negative degree terms) up to the first term of size $p^{-1}$, and then applying Weierstrass preparation theorem you get a kind of an "inverse" $u_n(T)$ such that all the high degree terms from some point on are tiny. Letting $n\to\infty$ I think (?) you get $u(T)=\lim_{n\to\infty}u_n(T)$ such that $u(T)a(T)$ has a maximum degree term. Whatever you then have left also has an inverse in your ring. Commented Jul 14, 2016 at 20:54
• But there are so many fuzzy steps and potential pitfalls that I will wait on the sidelines for now. We have users who are quite knowledgable in this area, so I'm optimistic about somebody capable of answering showing up. Commented Jul 14, 2016 at 20:56
• C Hawkins, do you think the following argument for the case $L=\bf{Q}_p$ works and generalizes? Let $f$ be an element of your ring. Multiply it by a power of $p$ to ensure that all coefficients are p-adic integers and at least one is a unit. Reducing that series modulo $p$ then gives a one-sided series in $S=\Bbb{F}_p[[T]][T^{-1}]$ with a non-zero lowest degree term. That series is invertible in $S$. Multiply the original series with a "lift" of that inverse gives a series with a single unit coefficient at degree zero. Invert that with the formula for a geometric series. Commented Jul 20, 2016 at 16:46

The ring $\mathcal{O}_{\mathcal{E}}$ is equipped with the Gauss norm : $$\| \sum_{n \in \mathbb{Z}} a_n T^n \| = \max_{n \in \mathbb{Z}}(|a_n|),$$ which is multiplicative.

Lemma 1: The ring is complete for this norm.

Proof: Let $f_s(T) = \sum_{n} a_n^{(s)} T^n$ with $s \geq 0$ be a sequence such that $\|f_s(T)\| \xrightarrow{s\to \infty} 0$. For each $n \in \mathbb{Z}$ the series $\sum_{s \geq 0} a_n^{(s)}$ converges to some $b_n$. The power series $g(T) = \sum_{n \in \mathbb{Z}} b_n T^n$ is in the ring for the following reason : let $\varepsilon >0$, and $S>0$ such that $|a_n^{(s)}| \leq \varepsilon$ for all $n \in \mathbb{Z}$ and $s \geq S$. Choose $N$ such that $|a_n^{(s)}| \leq \varepsilon$ for all $s \leq S$ and $n \leq N$. Then $|b_n| \leq \varepsilon$ for all $n \leq N$. The series $\sum_{s\geq 0} f_s(T)$ converges to $g(T)$ which proves that the ring is complete.

Lemma 2: If $f(T)= \sum_{n} a_n^{(s)} T^n$ is such that $\|f(T)\| =1$, then $f$ is invertible.

Proof: Denote $n_0$ the least integer $n$ such that $|a_n|=1$. Write $f(T)=g(T)+h(T)$ where $g(T) = \sum_{n < n_0} a_nT^n$ and $h(T) = \sum_{n \geq n_0} a_nT^n$. Then $\|g(T)\| <1$ and $h(T)$ is invertible, and since the ring is complete it is well known that their sum is invertible.

With Lemma 2 it is easy to conclude that the ring $\mathcal{O}_{\mathcal{E}}$ is a DVR with maximal ideal generated by $\pi_L$, a uniformizer of $L$, hence its quotient field is $\mathcal{O}_{\mathcal{E}}[1/\pi_L]$ which is the same as $\mathcal{O}_{\mathcal{E}}[1/p]$.

Note : Jyrki Lahtonen's arguments works (once you have proven that geometric series converge) and generalizes to $L$ by replacing $p$ by a uniformizer of $L$ and $\mathbb{F}_p$ by the residue field of $L$.

• Nice and simple! Thanks. Commented Jul 25, 2016 at 7:09
• Sorry for not answering earlier, I didn't check the site. Thank you and Jyrki very much ! Commented Jul 25, 2016 at 12:46
• @JyrkiLahtonen In the proof of lemma 2, is it correct that the sum of an invertible element and an element of norm strictly less than $1$ is invertible? Thanks!
– user685167
Commented Jan 22, 2020 at 17:25