I want to show that the algebraic closure of $\mathbb{Q}_{p}$ has infinite degree. I did a similar problem on quadratic extension, we used Hensel's Lemma. But here it seems there is no such a polynomial to choose. How should I start this problem?
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2For a rather extreme way to do this, google for Konrad's notes on the Artin-Schreier theorem. – Mariano Suárez-Álvarez Feb 04 '18 at 22:23
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@Watson The link you put in the comment is exactly this page. – finiteness Mar 04 '18 at 20:57
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Maybe this question is related, and this one, also. – Watson Mar 04 '18 at 21:01
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$x^n-p$ is irreducible over $\Bbb Q_p$ by Eistenstein, so $[\Bbb Q_p(\sqrt[n]{p}):\Bbb Q_p]=n$. Thus $[\Bbb Q_p^{alg}:\Bbb Q_p] \geq [\Bbb Q_p(\sqrt[n]{p}):\Bbb Q_p] = n$ for all $n$.
Lukas Heger
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You could try showing that if you let $\zeta_m$ be a $p^m$-th root of unity, then $\Bbb Q_p(\zeta_m)$ has degree $\varphi(p^m)=p^{m-1}(p-1)$ over $\Bbb Q_p$. Then, if the algebraic closure $\Bbb Q_p^{\mathrm{alg}}$ has finite degree $n$ over $\Bbb Q_p$, you could take $m$ with $p^{m}>n$ to obtain inclusions $\Bbb Q_p\subseteq \Bbb Q_p(\zeta_{m+1})\subseteq\Bbb Q_p^{\mathrm{alg}}$, which gives a contradiction by looking at the degrees over $\Bbb Q_p$.
Alex Mathers
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