What is completion of $ \Bbb{Q}( \sqrt{-1})$ at prime ideal $(3)$?

Question

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).

Answer 1

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.

Answer 2

Denote $K=\mathbb{Q}, L=\mathbb{Q}(i), \mathfrak{p}=(3)_K, \mathfrak{P}=(3)_L$.

Denote the completion at $\mathfrak{p}=(3)_K$ resp. $\mathfrak{P}=(3)_L$ by $\hat{K}=\mathbb{Q}_3$ resp. $\hat{L}$.

As you already know $[L:K]=2$ and $\mathfrak{p}$ is inert in $L$.

By e.g. Fröhlich/Taylor, "Algebraic Number Theory", pp. 109 ff., we have that $\hat{L}$ is a finite extension of $\hat{K}=\mathbb{Q}_3$. Since $\mathfrak{p}$ does not split we have $[\hat{L}:\hat{K}]=[L:K]=2$. We have $i\in L\subset\hat{L}$. If we had $i\in\hat{K}$, we could combine it with $K=\mathbb{Q}\subset\mathbb{Q}_3=\hat{K}$ to get $L=K(i)\subset\hat{K}$. So already $\hat{K}$ would be a complete field containing $L$, which means $\hat{L}\subset\hat{K}$ which would contradict $[\hat{L}:\hat{K}]=2$. Therefore $i\not\in\hat{K}$. From this we see that $\hat{K}\subsetneq\hat{K}(i)\subset\hat{L}$ and for degree reason we conclude: $\hat{K}(i)=\hat{L}$.

Alternatively you could construct explicitly $\hat{L}=\text{Frac}\left(\lim\limits_{\stackrel{\leftarrow}{n}}\mathfrak{o}_L/\mathfrak{P}^n\right)$, where $\mathfrak{o}_L=\mathbb{Z}[i]$ is the ring of integers of $L$. This is a rather cumbersome process, in the end the representation $\hat{L}=\mathbb{Q}_3(i)$ is much more useful.