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i already know how to get the 7 quadratic extensions of $\mathbb{Q}_2$ from hensel's lemma. they are $\mathbb{Q}_2(\sqrt{d})$ for $d = -10, -5, -2, -1, 2, 5, 10$.

question: which of these are unramified?

i looked it up (local fields, cassels) and it says the answer is $d=5$ is unramified and the rest are totally ramified, but his argument uses the discriminant which wasn't covered in the course i'm taking

EDIT: so i can work out that the ones where $d$ is even are totally ramified by using the result that $L/K$ is totally ramified iff $L=K[a]$ where $a$ is a root of an eisenstein polynomial, so that leaves the three odd cases

Batrachotoxin
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One Winged Pterodactyl
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1 Answers1

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The unramified extensions of $\mathbb{Q}_p$ are very simple. There is exactly one unramified extension of $\mathbb{Q}_p$ of degree $f$ for every prime $p$ - it is the one obtained by adjoining the $p^f − 1$ roots of unity. This unramified extension is Galois and has Galois group isomorphic to $\mathbb{Z}/f\mathbb{Z}$ over $\mathbb{Q}_p$.
Among the seven quadratic extensions of $\mathbb{Q}_2$ already six are (totally) ramified, namely for $d=-1,\pm 2, -5,\pm 10$. Hence the unrafimied must be the one with $d=5$. Alternatively, let $\zeta_3=(-1+\sqrt{-3})/2$, then $\mathbb{Q}_2(\zeta_3)=\mathbb{Q}_2(\sqrt{-3})=\mathbb{Q}_2(\sqrt{5})$ is unramified by the above statement.

See also the discussion here.

Community
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Dietrich Burde
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    hmm ok i get it. but how do i check that d = -1 and -5 are "obviously" ramified? do i have to manually check that $\sqrt[3]{1}$ isn't part of the extension? – One Winged Pterodactyl May 27 '15 at 19:20
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    Note that $\mathbb{Q}_2(\sqrt{5})=\mathbb{Q}_2(\sqrt{-3})$ and $\zeta_3=(-1+\sqrt{-3})/2$. – Dietrich Burde May 27 '15 at 19:55
  • @OneWingedPterodactyl: One easy way to see those are obviously ramified is noting that e.g. $N(1+\sqrt{-1})=2$ and $N(1+\sqrt{-5})=6$ in the respective field extensions, which both have $2$-adic valuation $1$, which according to basic principles of extensions of complete DVRs gives the result (e.g. $1+\sqrt{-1}$ must have valuation $\frac12$ then). Note how on the contrary, $v_2(a^2+3b^2) \in 2\mathbb Z$ for all $a,b \in \mathbb Q_2$. – Torsten Schoeneberg Dec 27 '20 at 22:50
  • @DietrichBurde, why is $\mathbb{Q}_2(\sqrt{-3}) = \mathbb{Q}_2(\sqrt{5})$? – Batrachotoxin Nov 25 '23 at 00:46
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    @Batrachotoxin The quadratic extensions of $\Bbb Q_2$ are $\Bbb Q_2(\sqrt a)$ where $a\in\Bbb Q_2$ is not a square, and $\Bbb Q_2(\sqrt a)=\Bbb Q_2(\sqrt b)$ if and only if $a/b$ is a square - see Mahler's book on $$-adic numbers and functions, which has a whole chapter on quadratic extensions of $\Bbb Q_p$. See also the MSE post here. – Dietrich Burde Nov 25 '23 at 10:43