Fraction field of $p$-adic power series ring

Question

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.

Answer 1

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 $\mathcal{O}_{\mathcal{E}}[1/p]$.

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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 $L$.