Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I found a proof of quadratic reciprocity in wikipedia, which I don't quite understand. The link is

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 $H$?

share|cite|improve this question

2 Answers 2

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 $g$.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.