I accidentally proved quadratic reciprocity in class today, or at least three quarters of a proof. Can you finish it off? Here’s the proof: start with a real quadratic field \(K\), and the sequence
\(1 \rightarrow \mathcal{O}^{\times}_K \rightarrow K^{\times} \rightarrow K^{\times}/\mathcal{O}^{\times} \rightarrow 1 \)
then take cohomology. If \(P_{K}\) is the group of principal ideals of \(K\), then from Hilbert Theorem 90 you deduce that
\(P^{G}_K/P_{\mathbf{Q}} \simeq H^1(G,\mathcal{O}^{\times}_K)\).
If \(I_K\) is the group of all ideals of \(K\), the left hand side is a subgroup of \(I^G_K/P_{\mathbf{Q}}\) which is a product of groups of order two for each ramified prime. Now if \(K = \mathbf{Q}(\sqrt{pq})\) with \(p \equiv q \equiv 3 \bmod 4\), there is no unit of norm \(-1\) because \((-1/p) = (-1/q) = -1\), so you deduce that the cohomology of the unit group has order \(4\) and so from order considerations that the prime ideals of norm \(p\) and \(q\) are principal. Write \(\mathfrak{p} = (\alpha)\) and \(\mathfrak{q} = (\beta)\). One has \(x\) and \(y\) with
\(x^2 – p q y^2 = N(\alpha)\)
Here \(N(\alpha) = \pm p\). But the right hand side must be a square modulo \(q\), and so \(N(\alpha) = p\) if \((p/q)=+1\) and \(-p\) if \((p/q)=-1\). Equivalently, \(N(\alpha) = (p/q)p\), and similarly \(N(\beta) = (q/p)q\). But \(\mathfrak{p} \mathfrak{q} = (\sqrt{pq})\), so \(\alpha \beta = \varepsilon \sqrt{pq}\) for some unit \(\varepsilon\), and since all units in \(K\) have norm one, it follows that
\( pq(p/q)(q/p) = N(\alpha \beta) = N(\varepsilon) N(\sqrt{p q}) = – p q,\)
which is quadratic reciprocity! There is a similar argument for \(p \equiv 1 \bmod 4\) and \(q \equiv -1 \bmod 4\) (though now using facts about \((2/p)\) rather than \((-1/p)\)). However, it is not so clear if one can prove the case \(p \equiv q \equiv 1 \bmod 4\) using this argument. Is there one? Of course the challenge here is to “only” use Hilbert 90 and unique factorization of ideals, but never to make any arguments about how primes are splitting.