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Is there a subfield $K \subset \Bbb Q_p$ such that the degree $[\Bbb Q_p : K]$ is finite and $>1$? I know it is wrong if we replace $\Bbb Q_p$ by $\Bbb R$, because of Artin–Schreier theorem. I'm not sure about the $p$-adic case, though.

Thank you.

Alphonse
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  • The Wikipedia article p-adic number has this: "$\mathbb Q_p$ contains the $n$-th cyclotomic field $(n>2)$ if and only if $n|p-1$." But I don't know whether the degree is finite or infinite. – TonyK Aug 25 '18 at 09:27
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    @TonyK: The degree is infinite, as any number field is countable, whereas $\Bbb Q_p$ is uncountable. – Torsten Schoeneberg Aug 25 '18 at 09:29
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    I think one can show the nonexistence of such $K$ from the fact that $\Bbb Q_p$ has no nontrivial field automorphism. – Torsten Schoeneberg Aug 25 '18 at 09:37
  • I think that the answer is no, thanks to Torsten Schoeneberg's comment. Assume there is such a $K \subsetneq \Bbb Q_p$ of finite degree. The extension being separable and finite, pick a primitive element $a \in \Bbb Q_p$, and a conjugate $b \neq a$ of $a$ (it exists since the extension is separable and non-trivial). Notice that we also have $\Bbb Q_p =K(b)$... are we sure about this? (contd) – Alphonse Aug 25 '18 at 09:43
  • (contd) We can define a morphism $f : \Bbb Q_p = K(a) \to \Bbb Q_p = K(b)$ by sending $a$ to $b$. It is clearly surjective and non-trivial, contradicting this result. – Alphonse Aug 25 '18 at 09:43
  • @TorstenSchoeneberg : thank you :-) . In my above comment, can we prove that $\Bbb Q_p = K(b)$ ? It probably holds if $\Bbb Q_p / K$ is Galois, at least. – Alphonse Aug 25 '18 at 09:45
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    That's sort of what my idea was, and indeed if $\Bbb Q_p|K$ is Galois, we are done. I do not see a simple argument for that i.e. $K(b) = \Bbb Q_p$ though. Maybe instead one can show with an extension of the usual argument that for any finite extension $L|\Bbb Q_p$, all field automorphisms of $L$ are continuous and hence fix $\Bbb Q_p$ pointwise. Then, in the argument you outline, you could just look at the (necessarily finite) Galois closure $L$ of $\Bbb Q_p|K$ and then get $L =K= \Bbb Q_p$. – Torsten Schoeneberg Aug 25 '18 at 21:31
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    A nice question. Lemme think about it. – Lubin Aug 27 '18 at 00:32
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    @TorstenSchoeneberg I gave details on your idea of passing to Galois closure of $\mathbf Q_p$ in another MSE question: see https://math.stackexchange.com/questions/4660292/subfield-of-a-complete-valuation-field-also-complete. – KCd Mar 19 '23 at 18:37

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Here’s my proof. It depends on a lemma that might not be as well-known as it ought to be.

Lemma. Let $L$ be a finite extension of $\Bbb Q_p$. Then every automorphism of $L$ is continuous.

It follows from this, for instance, that the only automorphism of $\Bbb Q_p$ is the identity. I won’t give a proof here unless requested.

Now let’s take our field “finite beneath” $\Bbb Q_p$, that is, $K\subset\Bbb Q_p$ with $[\Bbb Q_p:K]<\infty$. Let $L$ be the normal closure, or indeed, any finite normal extension of $K$ that contains $\Bbb Q_p$, and consider $\sigma\in\text{Gal}^L_K$. It’s an automorphism of $L$, thus continuous on $L$, and also continuous when restricted to $\Bbb Q_p$, thus identity on $\Bbb Q_p$. But that says that $\Bbb Q_p$ is normal over $K$, and in particular the Galois group consists of only the identity. Consequently, $K=\Bbb Q_p$.

Lubin
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  • Reassuring. Am I correct that one would prove the lemma essentially by the same method as one proves it for $\Bbb Q_p$ itself, namely, an automorphism sends $p$ to itself and stabilises the group of units of the valuation ring, because these can be characterised algebraically e.g. as those elements that have $n$-th roots for infinitely many $n$? – Torsten Schoeneberg Aug 27 '18 at 09:32
  • Yes, precisely, @TorstenSchoeneberg. – Lubin Aug 27 '18 at 13:21
  • Sorry, a stupid question: why is sigma restricted to Qp an automorphism of Qp? I.e. why doesn't it send Qp to some other subfield of L? – R.P. Nov 08 '19 at 05:33
  • Well, @RP_, I guess it’s because it has to be identity on $\Bbb Q$, and by continuity, has to be identity on $\Bbb Q_p$, in particular maps $\Bbb Q_p$ to itself. – Lubin Nov 08 '19 at 12:59