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Consider the Weil group $W$ of $\mathbb{Q}_p$, that is, the subgroup of those elements of $\mathrm{Gal}(\overline{\mathbb{Q}}_p/\mathbb{Q}_p)$ mapping to an integer power of Frobenius. Class field theory tells us that there is an isomorphism $$ \mathbb{Q}_p^\times \stackrel{\sim}{\longrightarrow} W^{\text{ab}}. $$

There are several ways to normalise it, say I pick the one that sends the uniformiser $p$ to a lifting of geometric Frobenius.

Question: what is then the image of $-1 \in \mathbb{Q}_p^\times$ under this map?

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    $\begingroup$ What would count as an answer to this? (I mean, what sort of description of elements of $W$, or of elements of $W^{\mathrm{ab}}$, is more familiar?) $\endgroup$
    – LSpice
    Commented Dec 1, 2018 at 2:24
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    $\begingroup$ Well, I would like for example to see how this element acts on quadratic extensions of $\mathbb{Q}_p$. If $p$ is odd, there are three of them: the unramified one, on which it acts as the identity, $\mathbb{Q}_p(\sqrt{p})$ and $\mathbb{Q}_p(\sqrt{cp})$ where $c$ is not quadratic residue. Does it act as identity on the first and minus identity on the second or is it more complicated? $\endgroup$
    – quas93
    Commented Dec 1, 2018 at 2:39
  • $\begingroup$ For $ [K:\mathbb{Q}_p]= 2,Gal(K/\mathbb{Q}_p) = \{ \sigma^2,\sigma\}$, or $p \equiv 1 \bmod 4 \land -1 = N_{K/\mathbb{Q}_p}(\sqrt{-1}) \land (\frac{-1}{K/\mathbb{Q}_p}) = \sigma^2 $ or $p \equiv 3 \bmod 4 \land K=\mathbb{Q}_p(\sqrt{-1}) \land -1 = N_{K/\mathbb{Q}_p}(\zeta_{2 (p+1)}) \land (\frac{-1}{K/\mathbb{Q}_p}) = \sigma^2$ or $ -1 \not \in N_{K/\mathbb{Q}_p}(K^*) \land (\frac{-1}{K/\mathbb{Q}_p}) = \sigma $ $\endgroup$
    – reuns
    Commented Dec 1, 2018 at 5:26

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