I learned algebraic field $K$ and prime ideal $P$ of integer ring of $K$ is given, we can complete $K$ with respect to $P$. $P$'s ideal norm defines metric and complete $K$ by that metric.
I understand the definition, but I cannot calculate nontrivial examples.
For example,
What is completion of $ \Bbb{Q}( \sqrt{-1})$ at prime ideal $(3)$ ?
Some extension of $ \Bbb{Q}_3$ ? I want to know strategy to calculate this (not just result).
Thank you for your kind help.
P.S
　Accurate definition is,
Let $F$ be a number field and let $\mathfrak{p}\in \mathsf{Spec} \: \mathcal{O}_F$. We have a non Archimedean valuation $\nu_\mathfrak{p}\colon F\longrightarrow\mathbb{R}_{\geq 0}$, given by $\nu_p(x):=\mathsf{card}(\mathcal{O}_F/\mathfrak{p})^{\mathsf{ord}_\mathfrak{p}(x)}$.
$F_\mathfrak{p}$ is completion with respect to valuation $\nu_p(x)$.
I want to know strategy to calculate this (not just result).
Best Answer
As observed in @ThomasPreu's answer, it is not efficient to construct a completion directly from the definitions. Rather, we know/realize/show that a finite-dimensional field extension of $\mathbb Q_p$ has a unique topological vector space (over $\mathbb Q_p$) structure. Thus, the essential point is to understand whether an algebraic element $\alpha$ over $\mathbb Q$ extends $\mathbb Q_p$, or not, or how far.
Very often, Hensel's lemma and solvability of equations mod $p$ is the essential starting point.
In the case at hand, since $\mathbb F_3\approx \mathbb Z/3$ has no $\sqrt{-1}$, there is no $\sqrt{-1}$ in $\mathbb Q_3$. Thus, adjoining $\sqrt{-1}$ gives a degree-two extension... whose topology is that of a two-dimensional vector space over $\mathbb Q_3$. The latter is (provably) unambiguous.