Let $K = \mathbb{Q}_5(\sqrt{5})$. As I learned from my last post, it is possible to write $K$ as the completion of $(k,v)$ where $k$ is a number field (i.e. a finite extension of $\mathbb{Q}$) and a valuation $v$ on $k$ defined by some prime ideal $\mathfrak{p}$ in $\mathcal{O}_k$. While I am still digesting the answer of mentioned post, I think it makes sense to choose $k = \mathbb{Q}(\sqrt{5})$. Now the question remains on how to choose the prime ideal $\mathfrak{p}$ (which is I think not shown in the answer).
I know that $\mathbb{Q}_5$ is the completion of $\mathbb{Q}$ wrt. the $5$-adic value which comes from the prime ideal $(5) \subseteq \mathcal{O}_\mathbb{Q} = \mathbb{Z}$ (cf. also this post of mine). Now I think that the $\mathfrak{p}$ we are looking for has something to do with $(5)$ but I was not able to advance further from this point. Question: How to explicitly describe $\mathfrak{p}$ (say, by finitely many explicit generators)?
$$\Bbb{Z}_5[\sqrt5]= \varprojlim_{n\to \infty} \Bbb{Z}[\sqrt5]/( \sqrt5^n)= \varprojlim_{n\to \infty} \Bbb{Z}[\sqrt5]/(5^n)$$ Well $O_{\Bbb{Q}(\sqrt5)}=\Bbb{Z}[\frac{1+\sqrt5}2]$ but we don't really care here.
The valuation extended to $\Bbb{Q}(\sqrt5)$ is $$v(a)= n\text{ if } \ a\in ( \sqrt5^n),\not \in ( \sqrt5^{n+1}), \qquad v(a/b)=v(a)-v(b), v(0)=\infty$$
$\Bbb{Q}_5(\sqrt5)=\Bbb{Z}_5[\sqrt5][5^{-1}]$ is the completion of $\Bbb{Q}(\sqrt5)$ for $v$.