I asked this on math.SE a few days ago with no reply, so I'm reposting it here. Hope this is not considered too elementary for MO (feel free to close if so).

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.

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    $\begingroup$ I read this question and thought it was a nice question, and was wondering to myself whether this argument something I knew back in the day... And then I was a bit startled when I clicked through the link. $\endgroup$ Commented Mar 23, 2016 at 19:32
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    $\begingroup$ At any rate, I certainly can't vouch for whether my 21 year old self knew how to deal with this issue. My guess is I overlooked this point. $\endgroup$ Commented Mar 23, 2016 at 19:35
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    $\begingroup$ If memory serves, this computation is tricky! By global reciprocity you reduce immediately to computing $\text{Frob}_{\mathfrak{q}}(\pi)/\pi$, but this is not as easy at it might seem. You can have a look at section 5.100 of math.harvard.edu/~rabinoff/math223a/notes.pdf and the problem set math.harvard.edu/~rabinoff/math223a/hw_a11.pdf. $\endgroup$
    – user84144
    Commented Mar 23, 2016 at 19:40
  • $\begingroup$ Did you look at the exercises in Cassels-Froehlich? My memory was that something was said there about how cubic reciprocity follows from the general machine set up in the book, but I don't know if your explicit question is dealt with. $\endgroup$
    – znt
    Commented Mar 23, 2016 at 19:59
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    $\begingroup$ @znt: The exercise in C-F, 2.14, basically asks one to assemble the previous exercises rather than saying how the argument goes. But the key input is Exercise 2.10, for which the indicated product over $S(a) \cap S(b) = S$ in the notation there collapses to 1 when $a, b \equiv 1 \bmod 3$. This is essentially David Speyer's solution below. $\endgroup$
    – nfdc23
    Commented Mar 23, 2016 at 23:01

1 Answer 1


I was just thinking about this the other day! Here is one solution.

As in the original post, let $K = \mathbb{Q}(\sqrt{-3})$ and let $\theta$ and $\pi$ be distinct primary primes. Also, set $L = K(\sqrt[3]{\theta \pi})$ and $M = K(\sqrt[3]{\theta}, \sqrt[3]{\pi})$.

We note that $M/L$ is unramified. The only primes which might ramify are those lying over $(\sqrt{-3})$, $(\pi)$ and $(\theta)$ in $K$. We can check that there is no ramification at these places using, respectively, a $3$-adic computation and the fact that $\theta$ and $\pi$ are primary, the presentation $M = L[T]/(T^3-\theta)$ and the presentation $M = L[U]/(U^3-\pi)$.

Let $\pi'$ be the prime of $L$ lying over $\pi$, and let $\theta'$ be the prime of $L$ lying over $\theta$. Now, $\theta' \pi'$ is the principal ideal generated by $\sqrt[3]{\theta \pi}$. So Artin reciprocity gives the equation $$\mathrm{Frob}_{M/L}(\theta') \mathrm{Frob}_{M/L}(\pi') = \mathrm{Id} \qquad (\ast)$$ in $\mathrm{Gal}(M/L)$.

Now, $(\sqrt[3]{\pi})^{N(\pi')} = (\sqrt[3]{\pi})^{N(\pi)} \equiv \left[ \frac{\pi}{\theta} \right]_3 \sqrt[3]{\pi} \bmod \theta$. So $$\mathrm{Frob}_{M/L}(\theta')(\sqrt[3]{\pi}) = \left[ \frac{\pi}{\theta} \right]_3 \sqrt[3]{\pi}.$$ Since $\sqrt[3]{\pi} \sqrt[3]{\theta} \in L$, it is fixed by $\mathrm{Frob}_{M/L}(\theta')$, so we compute $$\mathrm{Frob}_{M/L}(\theta')(\sqrt[3]{\theta}) = \left[ \frac{\pi}{\theta} \right]_3^{-1} \sqrt[3]{\theta}.$$

Similarly, $$\mathrm{Frob}_{M/L}(\pi')(\sqrt[3]{\theta}) = \left[ \frac{\theta}{\pi} \right]_3 \sqrt[3]{\theta} \quad \mathrm{Frob}_{M/L}(\pi')(\sqrt[3]{\pi}) = \left[ \frac{\theta}{\pi} \right]_3^{-1} \sqrt[3]{\pi} .$$

Plugging into $(\ast)$ and acting on $\sqrt[3]{\pi}$ we get $$ \left[ \frac{\pi}{\theta} \right]_3 \left[ \frac{\theta}{\pi} \right]_3^{-1} = 1$$ as desired.

On a linguistic note, I recently realized that "reciprocal" is a word, like "moot" or "cleave", which can denote two opposite things:

Life cannot subsist in society but by reciprocal concessions. (Samuel Johnson)

involves the exact opposite meaning from

...the pressures and expansions [volumes] to be in reciprocal relation. (Boyle, p. 60)

The higher reciprocity laws use reciprocal in the former sense.

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    $\begingroup$ I'm afraid I do not believe your claim about the lack of ramification above $3$. Using SageMath I compute that the relative discriminant of $M/L$ for $\pi=1+3\omega,\theta=1+6\omega$ is (the ideal generated by) $9$ (and indeed computing a prime lying above $\sqrt{-3}$ in $L$ and factoring it in $M$ reveals repeated factors). Code I've used: K.<w> = CyclotomicField(3); R.<x> = PolynomialRing(K); pi1 = 1+3*w; pi2 = 1+6*w; L.<a> = NumberField(x^3-pi1*pi2); S.<y> = PolynomialRing(L); M.<b> = NumberField(y^3-pi1); M.relative_discriminant() $\endgroup$
    – Wojowu
    Commented Aug 30, 2019 at 11:13
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    $\begingroup$ I believe this can be salvaged by computing the exponent of this ramified prime in the conductor and showing that, since $\pi,\theta$ are primary, that $\sqrt[3]{\pi\theta}$ is $1$ modulo the conductor. I'm not entirely sure how to perform this calculation though. $\endgroup$
    – Wojowu
    Commented Aug 30, 2019 at 11:16

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