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!
On
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.