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In order to prove that algebraic closure of $\mathbb{Q}_p$ is infinite, I took the polynomial $x^n-p$ with $n>1$ over $\mathbb{Q}_p$ to show that this eqaution has no solution for infinite cases to get the result but I want to make one part of the proof precise.

It's clear that $x^2-p$ has no solution in the field so we adjoin $\sqrt{p}$ to get $\mathbb{Q}_p(\sqrt{p})$. For higher powers of $n$, clearly $p^{1/n}$ is not contained in $\mathbb{Q}_p$ but how do we show that $p^{1/n}$ is not contained in the extensions with smaller $n$? For instance how do we show that $\sqrt[3]{p}\notin \mathbb{Q}_p(\sqrt{p})$?

Addendum: Reading Prof. Conrad's comment is helpful for the discussion.

Spock
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    You might as well make things easier for yourself by looking not at all $p^{1/n}$ but at a subset: all $p^{1/2^m}$. The corresponding fields clearly form a totally ordered chain, and since $[\mathbb Q_p(p^{1/2^m})\colon\mathbb Q_p]=2^m$, you’re home free. – Lubin Apr 11 '14 at 22:17
  • That's a nice argument. – Spock Apr 11 '14 at 22:25
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    @Spock: The question you should have asked is not if an $n$th root $\sqrt[n]{p}$ is in any of the extensions $\mathbf Q_p(\sqrt[m]{p})$ for $m < n$, but whether it is in the field generated by smaller roots: could $\sqrt[n]{p}$ lie in $\mathbf Q_p(\sqrt{p},\sqrt[3]{p},\dots,\sqrt[n-1]{p})$ for some choice of square, cube,..., $(n-1)$th roots? And indeed this can happen, e.g., $\sqrt[6]{p} \in \mathbf Q_p(\sqrt{p},\sqrt[3]{p})$ since $1/6 = 1/2 - 1/3$ (that is, $\sqrt{p}/\sqrt[3]{p}$ is a 6th root of $p$). – KCd May 20 '14 at 21:50
  • @KCd: Thank you Prof. Conrad for your remarks. Indeed that's what I had in my mind. – Spock May 21 '14 at 19:53

2 Answers2

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The $p$-adic valuation extends uniquely to any finite extension of $\mathbf Q_p$.

Remark that $v_p(p^{1/n})=1/n$, so $v_p(\mathbf Q_p(p^{1/n})^\times) = 1/n\cdot\mathbf Z$. In particular, the field $\mathbf Q_p(p^{1/n})$ cannot be embedded in $\mathbf Q_p(p^{1/m})$ unless $n|m$.

If you just want to show that $\overline{\mathbf Q_p}$ is not finite over $\mathbf Q_p$, then Keenan's method is the easiest. But there's a funny way to hit this simple statement very hard with the theory of real closed fields. According to a theorem, if $F$ is a non-algebraically closed field whose algebraic closure $\overline{F}$ is finite over $F$, then $F$ is elementarily equivalent to the real numbers. Since $x^2-p$ splits over $\mathbf R$ but not over $\mathbf Q_p$, it follows that $\overline{\mathbf Q_p}/\mathbf Q_p$ is an infinite extension!

Bruno Joyal
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  • Thank you. could you explain whay $v_p(\mathbf Q_p(p^{1/n})^\times) = 1/n\cdot\mathbf Z$? – Spock Apr 11 '14 at 21:19
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    @Spock Try to apply the ultrametric inequality to a $\mathbf Q_p$-linear combination of powers of $p^{1/n}$, and remember that the inequality is an equality whenever there is one term with valuation smaller than all of the other terms. – Bruno Joyal Apr 11 '14 at 21:26
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I assume you mean that you want to show the algebraic closure is of infinite degree over $\mathbf{Q}_p$. It's then enough to show that $\mathbf{Q}_p$ admits finite extensions of arbitrary degree (if an algebraic closure of $\mathbf{Q}_p$ has finite degree $n$, note that it contains a finite extension of degree larger than $n$, a contradiction). For this it's enough to observe that $x^n-p$ is irreducible over $\mathbf{Q}_p$ for all $n\geq 1$ by Eisenstein's criterion ($p$ is prime in the UFD $\mathbf{Z}_p$).

Keenan Kidwell
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