I read that every finite extension of $\mathbb{Q}_p$ is in fact a completion of a numberfield K with a place of K. I also heard that this is a consequence of Krasner´s Lemma.
Do you have any hint how to prove this?
And how can i prove conversely that every completion of a numberfield is a finite extension of $\mathbb{Q}_p$?
Every hint is strongly appreciated. Thanks!
Maybe Krasner’s Lemma is the most direct route to your first statement. It says, very roughly, that if you take an irreducible polynomial over $\Bbb Q_p$ and jiggle its coefficients just slightly, each root $\rho'$ of the new polynomial will be close to a unique root $\rho$ of the original, and in fact the two will generate the same field over $\Bbb Q_p$. So, if $K=\Bbb Q_p(\alpha)$, take its $\Bbb Q_p$-polynomial $f(X)$ and jiggle it to an $\,\bar f\in\Bbb Q[X]$. Then $\,\bar f$ has a root $\alpha'$ also generating $K$ over $\Bbb Q_p$, but $\alpha'$ is algebraic (over $\Bbb Q$).
In case your extension $K$ is unramified over $\Bbb Q_p$, you don’t need Krasner or anything like him. For, such a $K$ can be gotten by adjoining a root of unity to $\Bbb Q_p$. To get your field that completes to give $K$, adjoin a root of unity of the same order to $\Bbb Q$. Of course the really interesting extensions are the ramified ones, so this argument doesn’t apply.
Let me try to prove my second question:
Let $L$ be a Numberfield together with a place $w$. Restricting $w$ to $\mathbb{Q}$ gives a place $v$ of $\mathbb{Q}$. Claim: The completion $L_w$ of L is equal to the compositum of fields $L\mathbb{Q}_v$. i.e. $L_w = L\mathbb{Q}_v$.
Since $\mathbb{Q}$ and $L$ is in $L_w$, surely $L\mathbb{Q}_v \subset L_w$ holds. Because L is a finite extension of $\mathbb{Q}$, its also a finite extension of $\mathbb{Q}_v$ and since $\mathbb{Q}_v$ is complete, $L\mathbb{Q}_v$ is also (See Neukirch, Algebraic Numbertheory, Theorem 4.8 for the last statement). So $L\mathbb{Q}_v$ is complete and contains L, so the equality $L_w = L\mathbb{Q}_v$ holds.