# Cyclotomic field in proof of quadratic reciprocity

I found a proof of quadratic reciprocity in wikipedia, which I don't quite understand. The link is http://en.wikipedia.org/wiki/Proofs_of_quadratic_reciprocity.

On the last line of the Cyclotomic field setup part, it says $\left(\frac{q}{p}\right)=1$ iff $\sigma_q$ is an element of $H$. I only know $H$ is a subgroup of $\operatorname{Gal}(L/Q)$ of order $\frac{p-1}{2}$. But how could I know $\sigma_q$ is in $H$?

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The fact it's using is that $\mathbb{F}_p^*$ is cyclic, say with generator $g$. For $p$ odd, this is a cyclic group of even size, squares mod $p$ are exactly even powers of $g$.

$H$ by construction is these even powers of $g$.

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Ok so there is another side to this setup.

Starting with $\mathbb{F}_p^{\times}$ take $H'$ to be the subgroup of "squares". By elementary theory this subgroup has order $\frac{p-1}{2}$. Then $H$ is the corresponding subgroup of the Galois group and $\mathbb{Q}(\sqrt{\left(\frac{-1}{p}\right)p})$ will turn out to be the corresponding fixed field.

Now $\left(\frac{q}{p}\right)=1$ is the same as saying that $q$ is a quadratic residue mod $p$, i.e. that the class of $q$ mod $p$ lies in $H'$. This means the same as $\sigma_q\in H$ by definition. But this is the same as saying $\sigma_q$ fixes the field above (by definition).

The next step of the proof is to prove that $\sigma_q$ is actually the Frobenius element of $q$ in the cyclotomic extension. The above deduction about this fixing will be equivalent to $p$ splitting completely in the quadratic field $\mathbb{Q}(\sqrt{\left(\frac{-1}{p}\right)p})$, which happens if and only if $\left(\frac{\left(\frac{-1}{p}\right)p}{q}\right) = 1$. This proves quadratic reciprocity.

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