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Let $K=\mathbb{Q}(\zeta_p)$. Then if $q\nmid p$ is any odd prime, we know that the Frobenius map $(q,K/\mathbb{Q})$ is just the map

$(q,K/\mathbb{Q})(\zeta_p)=\zeta_p^q$.

Now let $d=(-1)^{(p-1)/2}p$, so that $\mathbb{Q}(\sqrt{d})$ is the unique quadratic field contained in $K$. I'm essentially interested in the canonical restriction map

$(\mathbb{Z}/p\mathbb{Z})^\times=Gal(K/\mathbb{Q})\to Gal(\mathbb{Q}(\sqrt{d})/\mathbb{Q})=\{\pm 1\}$.

In Milne's notes on algebraic number theory on p. 138 available at

http://www.jmilne.org/math/CourseNotes/ANT.pdf

there's this comment that "$(q,K/\mathbb{Q})$ restricts to the identity element of $Gal(\mathbb{Q}(\sqrt{d})/\mathbb{Q})$ if and only if $q$ is a square in $(\mathbb{Z}/p\mathbb{Z})^\times$. My question is why? There must be something very silly that I'm missing. At least any square gets mapped to a square, so a square is taken to $1$. I can't seem to figure out the converse.

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up vote 1 down vote accepted

The squares inside $(\mathbb{Z}/p\mathbb{Z})^\times$ form a subgroup of index 2. If any non-square got sent to the identity, the restriction map would be the trivial map, which would contradict the Fundamental Theorem of Galois Theory.

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Alright, thanks for confirming that I'm stupid. :) –  dstt Apr 12 '11 at 21:58
    
@dstt, don't worry, happens to all of us :) –  Zev Chonoles Apr 12 '11 at 21:59
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