Examples for completions of number fields i'm currently learning for an exam on class field theory.
The first thing i thought about are examples for completions of number fields (here $K$), for example of the field extensions $\mathbb{Q}[\sqrt{p}]$ where $\textit{p}$ is a prime number. The only example i know is the field of $\textit{p}$-adic numbers $\mathbb{Q}_p$ if we take $K = \mathbb{Q}$. I also didn't find good examples on stackexchange (if i searched not good enough, please correct me). Do you know any examples?
The second thing i wanted to ask you is about the numbers in the $p$-adic numbers. As we know from the real case, there are elements like $e$ which we can express by the limit of the cauchy sequence $(1 + \frac{1}{n})_{n \in \mathbb{N}}$. How can one get a Feeling of such elements in $\mathbb{Q}_p$? Is there a good example of a number which does not lie in $\mathbb{Q}$?
Thanks for your help!
 A: For each prime ideal $p\in P\subset O_K$ there is a $p$-adic completion $$K_v= \operatorname{Frac}(\varprojlim O_K/P^n)$$ where $v$ is the discrete valuation $v(a)= n$ if $a\in P^n,\not \in P^{n+1}$.

$K_v$ is the field of limits of sequences of elements of $K$ that converge for the absolute value $|a|_v=p^{-v(a)}$ and $\varprojlim O_K/P^n$ mostly means the same.

From the primitive element theorem $K=\Bbb{Q}[x]/(f(x))$ then $K_v \cong \Bbb{Q}_p[x]/(f_j(x))$ where $f_j$ is one of the $\Bbb{Q}_p$-irreducible factor of $f$.
For a Galois extension $\{ \sigma \in \operatorname{Gal}(K/\Bbb{Q}), \sigma(P)=P\}=\operatorname{Gal}(K_v/\Bbb{Q}_p)$.
Try with $K=\Bbb{Q}(i)$ and $p=2,3,5$ to see how it works.
A: For the first question it is good to know the following observation:
Let $K:\color{red}{\Bbb Q_p}$ be an algebraic extension of fields.
Later edit: Above, it is important to work over $\Bbb Q_p$. (Initially it was $\Bbb Q$, which is deadly wrong. For instance, in case of an extension $\Bbb Q(\alpha):\Bbb Q$ where $f=\operatorname{Irr}_{\Bbb Q}$ splits into two or more factors over $\Bbb Q_p$ the tensor product $\Bbb Q(\alpha)\otimes_{\Bbb Q}\Bbb Q_p$ splits as a product of extensions of $\Bbb Q_p$. Each field component comes then with its own extensions, and the following applies for each such component.)
After passing to the Galois closure, we may and do assume it is a Galois extension.
Let $|\cdot|=|\cdot|_p$ be the $p$-adic norm on $\Bbb Q_p$. (So $p$ is a prime.) Then it has a unique extension to $K$ which is also a multiplicative norm. Moreover, Galois conjugated elements have the same $p$-adic norm, and thus if $f=x^n+a_1x^{n-1}+\dots+a_{n-1}x+a_n$ is the minimal polynomial of some algebraic number $\alpha\in K$, then
$$
\begin{aligned}
|\alpha|
&=(|\alpha|^n)^{1/n}
=\left(\prod_{\sigma\in\text{Gal}(K:\Bbb Q)} |\sigma\alpha|\right)^{1/n}
=\left|\prod_{\sigma\in\text{Gal}(K:\Bbb Q)} \sigma\alpha\right|^{1/n}
\\
&=|a_n|^{1/n}\ .
\end{aligned}
$$
This shows how to take the unique norm on $\Bbb Q_p(\alpha)$.

Now it can be that $f$ is no longer irreducible over $\Bbb Q_p$, this is also a point with interest for the second question, so let us give an example.
Consider the algebraic integer $\sqrt 3$ over $\Bbb Q_p$ with $p=11$. Note that $3$ is a quadratic residue modulo $11$, since $(\pm5)^2=25=3$ modulo $11$. The key word is now Hensel's Lemma. We need only the "cheap" version of it telling us that we can start with any of the two solutions $\pm 5$, and we will start with $5$, and improve it successively to have a solution modulo $11$, then modulo $11^2$, then modulo $11^3$, and so on. An example is shown using sage:
sage: R = Qp(11, prec=7)
sage: sqrt(R(3))
5 + 2*11 + 6*11^2 + 8*11^3 + 11^4 + 9*11^5 + 9*11^6 + O(11^7)

(One can use a higher precision to get a longer part from the series.)
Instead, one can also use the binomial theorem to get an other convergent series. For instance:
$$
\begin{aligned}
3 &= 25\cdot \frac3{25}=25\left(1-\underbrace{\frac {22}{25}\_x\right)\ ,\\
\sqrt 3 &=5(1-x)^{1/2}
\\
&=5\left(1-\binom{1/2}1x+\binom{1/2}2x^2-\binom{1/2}3x^3+\dots\right)\ ,
\end{aligned}
$$
and the series in the parentheses converges. To the same value. Sage check:
sage: s = sqrt(R(3))
sage: t = 5 * sum([binomial(1/2, k)*R(-22/25)^k for k in [0..10]])
sage: s == t
True
sage: s
5 + 2*11 + 6*11^2 + 8*11^3 + 11^4 + 9*11^5 + 9*11^6 + O(11^7)
sage: t
5 + 2*11 + 6*11^2 + 8*11^3 + 11^4 + 9*11^5 + 9*11^6 + O(11^7)

There are many examples of convergent series. (Above there is the binomial series.) The logarithmic and exponential series converge in a ball around zero. Over $\Bbb R$ the $\exp$ is a good friend for the convergence, but in characteristic $p$ the "denominators are bad". See also wiki/P-adic_exponential_function.
For these reasons, there is a "good wish" to have a p-adic world of (over)convergent "objects". And start an industry of analytic explorations as it was the case some centuries ago.
