I'm wondering about the following: Let $K$ be a numberfield and $v$ a place of $K$. Is it true that the algebraic closure $\bar{K_v}$ of $K_v$ is the composite of the fields $K_v$ and $\bar{K}$? Thanks in advance!
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Yes.
Let $K=\Bbb Q_p(\alpha)$ be a finite extension of $\Bbb Q_p$. Let $f$ be the minimal polynomial of $\alpha$ over $\Bbb Q_p$.
If the coefficients of $g$ are sufficiently close to those of $\alpha$, then $K=\Bbb Q_p(\beta)$ where $\beta$ is a zero of $g$ (this is essentially a Hensel's lemma argument). We can choose $g$ to have rational coefficients, so that $\beta$ is algebraic over $\Bbb Q$.
Angina Seng
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I need more details on the proof. For example, is it possible to enumerate all the extensions of $\mathbb{Q}_p$ of degree $\le n$ ? – reuns Sep 15 '17 at 18:29
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@reuns Look up Krasner's lemma. See here for a link as well as a discussion why that fails for char $p$ local fields. – Jyrki Lahtonen Sep 16 '17 at 12:30
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I don't see how it is deduced by hensel's lemma. But using krasner's lemma its clear to me. – NeukirchLover Sep 17 '17 at 16:04
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@NeukirchLover You use Hensel's Lemma to prove Krasner's Lemma – Angina Seng Sep 17 '17 at 16:15
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Ok I see, thanks! Just for a reference: the proof I have in mind is in cohomology of number fields, Lemma 8.1.6. – NeukirchLover Sep 18 '17 at 08:21