This is a quick note to record some thoughts following from Toby Gee’s first lecture of his course at the Arizona Winter School, where he observes that quadratic reciprocity is a completely immediate consequence of basic algebraic number theory. I feel rather silly for never having noticed this before, and hope I don’t insult the reader by providing a post on it.

That quadratic reciprocity follows immediately from class field theory is standard, and for the rational numbers class field theory can be decomposed into the irreducibility of cyclotomic polynomials (reciprocity laws for cyclotomic extensions) and the Kronecker-Weber theorem (cyclotomic extensions fill out all the abelian extensions). Of these, I would only consider the second to be `hard’.

The key point that makes quadratic reciprocity strictly easier than class field theory is that for $p$ an odd prime, the quadratic extensions $K_p := \mathbb{Q}(\sqrt{(-1)^{(p-1)/2} p})/\mathbb{Q}$ are the unique quadratic extensions ramified only at $p$. They therefore obviously satisfy the Kronecker-Weber theorem, since $\mathbb{Q}(\zeta_p)/\mathbb{Q}$ has a degree 2 subextension which is ramified only at $p$ and thus equal to $K_p$.

We can easily make this more explicit. Consider the quadratic character

$\chi: G_\mathbb{Q} \rightarrow Gal(K_p/\mathbb{Q}) = \{ \pm 1 \}$.

By definition (more or less), for $q \not= p$ odd, this character is unramified and $\chi(Frob_q) = \left(\frac{(-1)^{(p-1)/2} p}{q}\right)$.

On the other hand, by our previous observation (“Kronecker Weber” in this special case), and the standard isomorphism between $Gal(\mathbb{Q}(\zeta_p)/\mathbb{Q})$ and $(\mathbb{Z}/p\mathbb{Z})^*$ (“irreducibility of cyclotomic poly”: here just Eisenstein’s theorem), we obtain the factorisation:

$\chi: G_\mathbb{Q} \rightarrow (\mathbb{Z}/p\mathbb{Z})^* \rightarrow \{\pm 1 \}$,

where $Frob_q$ is mapped to the class of $q$ modulo $p$. Equipped with this description (and recalling that $(\mathbb{Z}/p\mathbb{Z})^*$ is cyclic), it is clear that

$\chi(Frob_q) = \left(\frac{q}{p}\right)$.

Comparing the two expressions obtained, we recover the classical quadratic reciprocity law. One can also handle $q=2$ the same way with a small amount of care (over the correct way to interpret the first expression: Hensel’s lemma doesn’t give a direct comparison with a Legendre symbol in this case).