I'm trying to understand the proof of cubic reciprocity from Artin reciprocity as outlined in this well-known previous math.SE question and the link KCd mentions there. However, there's one final step that I can't get to work. I suspect that the proofs linked above are in fact incomplete but I'd like to confirm that I'm not missing something. (Reason: if Artin alone worked, I think Cox would have done it in his book $x^2+ny^2$, which he conspicuously does not.)

So let me write out what I've figured out so far. Let $K = \mathbb Q(\sqrt{-3})$, and let $\pi$ be a primary prime in $K$. (For me, primary means $\pi \equiv 1 \pmod 3$.) We want to prove

Cubic Reciprocity: If $\theta$ is a primary prime in $K$ distinct from $\pi$ then $$\left( \frac{\pi}{\theta} \right)_3 = \left( \frac{\theta}{\pi} \right)_3.$$

Now the proof proceeds as follows. The main idea is to consider the diagram enter image description here Here, the top row is the Artin symbol, followed by the ``evaluation'' map $$ \text{ev }_\pi : \sigma \mapsto \frac{1}{\sqrt[3]{\pi}} \sigma(\sqrt[3]{\pi}) $$ so that the composition yields the cubic Legendre symbol (top row). Then Artin reciprocity implies that the Artin map is surjective and factors through $I_K(3\pi)/P_K(3\pi)$, which is isomorphic to $\left( \mathcal O_K/\pi \right)^\times$ by taking any prime ideal and sending it to its primary generator (this is the map in the bottom row). Finally, the rightmost arrow $\left( \mathcal O_K/\pi \right)^\times \to \{1, \omega, \omega^2\}$ is surjective from the rest of the diagram.

Now the claim I don't believe is that this implies the rightmost arrow is $\left( \frac{\bullet}{\pi} \right)_3$, which would imply cubic reciprocity. The argument is that the kernel of the rightmost arrow is an index three subgroup of $(\mathcal O_K/\pi)^\times$, hence it consists of the cubes in $(\mathcal O_K/\pi)^\times$. This means that $$ \left( \frac{\theta}{\pi} \right) = 1 \implies \left( \frac{\pi}{\theta} \right) = 1. $$ However, it doesn't seem to work for the other values, for the reason that there are in fact two different nontrivial homomorphisms $(\mathcal O_K/\pi)^\times \to \{1, \omega, \omega^2\}$, namely $\left( \frac{\bullet}{\pi} \right)_3$ and $\left( \frac{\bullet}{\pi} \right)_3^{-1}$. (This is different from the quadratic case, in which there was only one.) In other words, from this we can only conclude that for a fixed $\pi$, either $$ \left( \frac{\theta}{\pi} \right)_3 = \left( \frac{\pi}{\theta} \right)_3 \quad\text{ or }\quad \left( \frac{\theta}{\pi} \right)_3 = \left( \frac{\pi}{\theta} \right)_3^{-1} \qquad \forall \theta \equiv 1 \pmod 3. $$

Question: how do you prove that we are in the former case and not the latter? One idea I had was to pick a convenient value of $\theta$ and just check it directly, but I haven't been able to find a way to make this work.


Just to mark this as answered, I re-posted this question on MathOverflow and got a fantastic answer from David Speyer there.


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