Let $p$ be a prime number and let $R$ be the ring of Cauchy sequences in $(\mathbb{Q}, |\cdot|_{p})$. Let $\mathfrak{m} \subseteq R$ be the ideal of all sequences converging to $0$, forming a maximal ideal of $R$. Also, let $R_{0} = \{x \in R/\mathfrak{m} : |x|_{p} \leq 1\}$.
Prove that $R/\mathfrak{m} \cong R_{0}[\frac{1}{p}]$, in other words, for every $x \in R/\mathfrak{m}$ there is an $n \in \mathbb{Z}_{\geq 0}$ such that $p^{n}x \in R_{0}$. Conclude that $R/\mathfrak{m} \cong Q(R_{0})$.
I have proven earlier that $R$ forms a commutative ring and that $\mathfrak{m}$ is a maximal ideal of $R$.
My attempt: Suppose that $x \in R/\mathfrak{m}$ with $x = \lim_{n \to \infty} x_{n}$. Notice that this sequence does not converge to 0. Furthermore, it holds that $|x|_{p} = \lim_{n \to \infty} |x_{n}|_{p}$. Since $x_{n} = p^{k_{n}}\frac{a_{n}}{b_{n}}$ voor integers $k_{n}, a_{n}, b_{n} \in \mathbb{Z}$ with $\gcd(p, a_{n}b_{n}) = 1$, it follows that $|x_{n}| = p^{-k_{n}}$.
At this point, I would like to prove the existence of a $k \in \mathbb{Z}$ for which $|x|_{p} = p^{-k}$.
We can now make a distinction of cases. If $k < 0$, then $|x|_{p} > 1$, so we can take $n = -k \in \mathbb{Z}_{\geq 0}$. Then, $|p^nx|_{p} = |p^{-k}|_{p}|x|_{p} = p^{k}p^{-k} = 1$, and thus $p^n x \in R_{0}$. If $k \geq 0$, then it directly follows that $|x|_{p} = p^{-k} \leq 1$, so we can take $n = 0$. We conclude that there is always a $n \in \mathbb{Z}_{\geq 0}$ such that $p^{n} x \in R_{0}$.
If anyone has hints or suggestions on how I could justify the existence of $k$ with the sequence, or if someone has suggestions on improvements, I would appreciate that.