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For an exercise I have to show that the number of non-isomorphic degree $2$ extensions of the field of $p$-adic numbers $\mathbb{Q}_p$ equals $3$. I was able to show that $$\mathbb{Q}_p^\times/\left(\mathbb{Q}_p^\times\right)^2 \cong \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$$ where $K^\times$ denotes the group of units in the field $K$. If I could prove that $\mathbb{Q}_p(\sqrt{d}) \cong \mathbb{Q}_p(\sqrt{e})$ if and only if $d = u^2e$ for $p$-adic numbers $d,e,u$, then I would be done.

I recalled the proof that $\mathbb{Q}(\sqrt{2})$ is not isomorphic to $\mathbb{Q}(\sqrt{3})$, where it is shown that (well, at least in the proof I saw) that if there would have been an isomorphism, then $\sqrt{2}$ would be an element of $\mathbb{Q}(\sqrt{3})$, but this proof uses that the isomorphism would fix the rational numbers. I do not immediately see how to generalize this to the $p$-adic case.

Any hints or would my approach be completely wrong?

Answer using hint We might consider integers. Let $d,e$ be integers such that there is no $p$-adic number $u$ satisfying $d = u^2e$. Suppose however that $\mathbb{Q}_p(\sqrt{d}) \cong \mathbb{Q}_p(\sqrt{e})$ by some isomorphism $\phi$. Note that $\phi(1) = 1$, hence we have that $$d = 1 + 1 + \ldots + 1 = \phi(1)+ \ldots + \phi(1) = \phi(d) = \phi(\sqrt{d}^2) = \phi(\sqrt{d})^2$$ hence $\phi(\sqrt{d}) = \pm \sqrt{d}$. This implies that $\sqrt{d} \in \mathbb{Q}_p\sqrt{e})$. Since $\mathbb{Q}_p(\sqrt{e})$ is a vectorspace over $\mathbb{Q}_p$ of dimension $2$ we can pick a basis $\{1, \sqrt{e}\}$, from which we find that $$\sqrt{d} = a + b\sqrt{e}$$ with $a,b \in \mathbb{Q}_p$. However, this implies that $d = a^2 + 2ab\sqrt{e} + b^2e$ and this results in $$\begin{cases} d &= a^2 + b^2e\\ 0&= 2ab \end{cases}$$ The last equations implies that either $a$ or $b$ is zero. But this implies that $d = b^2e$ respectively $d = a^2$. The former is not possible by assumption, while the latter does not give an extension of degree $2$. Hence these extensions can not be isomorphic.

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    $\begingroup$ $p$ is odd right? $\endgroup$ May 2, 2017 at 16:45
  • $\begingroup$ @BrunoJoyal: yes, forgot to mention that. (second exercise was to prove that there are $7$ in the case that $p=2$.) $\endgroup$
    – Student
    May 2, 2017 at 17:16
  • $\begingroup$ "If I could prove that $Q_p(\sqrt d)≅Q_p(\sqrt e)$ if and only if $d=u^2 e $ for p-adic numbers d,e,u, then I would be done" : this is a particular case of Kummer theory, which is valid for any base field of characteristic $\neq 2$ . The general proof usually resorts to Hilbert's theorem 90, see any text book on Galois theory. $\endgroup$ May 2, 2017 at 22:26
  • $\begingroup$ @nguyenquangdo it was something we have never seen in any previous class (kummer theory)... Anyway, I will look it up, thank you! $\endgroup$
    – Student
    May 3, 2017 at 5:40
  • $\begingroup$ For quadratic extensions, a direct proof is easy, following Hilbert 90 i.e., here, -1 is the quotient of 2 conjugates. Concerning the number of cyclic extensions of $Q_p$, perhaps you could be interested by math.stackexchange.com/a/2249315/300700 $\endgroup$ May 3, 2017 at 7:13

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If you knew that the isomorphism must fix $ \mathbf Q_p $, then I imagine you would be done. To do this, you need to show that an isomorphism $ f : K \to L $ of $ p $-adic fields must be continuous. It suffices to show that $ f $ preserves $ p $-adic units, that is, the isomorphism $ f : K \to L $ restricts to a group isomorphism $ \bar f : \mathcal O_K^{\times} \to \mathcal O_L^{\times} $, where $ \mathcal O_K^{\times} = \{ x \in K : v(x) = 0 \} $, where $ v $ is the unique valuation on $ K $ extending the valuation on $ \mathbf Q_p $. We now use the fact that an element $ x \in K, L $ is a $ p $-adic unit iff it can be written as an $ n $th power for arbitrarily large $ n $. Indeed, it's obvious that any such element must be a unit, and any unit $ u $ can be written in that way by picking arbitrarily large $ n $ coprime to $ p(k-1) $ where $ k $ is the cardinality of the residue field, and using Hensel's lemma on $ X^n - u $. Since being an $ n $th power for arbitrarily large $ n $ is a property preserved under field isomorphism, it follows that any isomorphism of $ p $-adic fields preserves $ p $-adic valuations, and is therefore continuous.

Now, simply note that if such an isomorphism is continuous, then it fixes $ \mathbf Q $ since it is a field isomorphism, and it fixes $ \mathbf Q_p $ since $ \mathbf Q $ is dense in $ \mathbf Q_p $.

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  • $\begingroup$ thank you for your answer. We did not see the extensions of valuations, so I will wait to accept your answer to see if there are other approaches which fit more to what we saw in class (we only had one chapter about $p$-adic numbers and in it we saw the definition, some results which followed quite easily from this definition and your last step about $Q$ being dense in $Q_p$). $\endgroup$
    – Student
    May 2, 2017 at 16:38
  • $\begingroup$ @Student You can also do this if you know that all quadratic extensions of $ \mathbf Q_p $ are necessarily obtained by adjoining square roots of integers, and use the result that the isomorphism $ f $ must fix integers. Indeed, if $ p $ is odd and you take an integer $ z $ such that $ z $ is not a square modulo $ p $, then the three quadratic extensions of $ \mathbf Q_p $ are $ \mathbf Q_p(\sqrt{z}), \mathbf Q_p(\sqrt{zp}), \mathbf Q_p(\sqrt{p}) $. $\endgroup$
    – Ege Erdil
    May 2, 2017 at 16:44
  • $\begingroup$ How can we be sure that it must be square roots of integers? (I have tried using an arbitary $p$-adic number, but I could not repeat the argument as for $\mathbb{\sqrt{2}}$ since we are dealing with infinite series of integers...) Moreover, I had to show that there are exactly seven non-isomorphic extensions for $p=2$... (which I did, using the statement I still need to prove) $\endgroup$
    – Student
    May 2, 2017 at 16:47
  • $\begingroup$ Never mind, of course there are seven of them. You show that it must be because $ \mathbf Q_p^{\times} / {\mathbf Q_p^{\times}}^2 \cong V_4 $ shows that there are at most three nonisomorphic quadratic extensions when $ p $ is odd, so it suffices to exhibit three nonisomorphic quadratic extensions to finish the proof that there are exactly three. $\endgroup$
    – Ege Erdil
    May 2, 2017 at 16:49
  • $\begingroup$ yes, so I can just pick regular integers and then use the same kind of reasoning done for $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$? $\endgroup$
    – Student
    May 2, 2017 at 16:52

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