$p$ splits completely in $\mathbb{Q}(\zeta_3,\sqrt[3]{2}) \Leftrightarrow p=x^2+27y^2$

Question

I want to show the assertion above. I already know: $p$ splits completely in $\mathbb{Q}(\zeta_3,\sqrt[3]{2}) \Leftrightarrow p\mathcal{O}_{\mathbb{Q}(\zeta_3)}=R \overline{R}$ in the ring of integers $\mathcal{O}_{\mathbb{Q}(\zeta_3)}=\mathbb{Z}[\zeta_3]$ and $R$ is a principal ideal, i.e. $R=(\pi)$ where $\pi\in \mathbb{Z}[\zeta_3]$, and $6\mid (\pi-1)$. (This equivalence uses class fields theory) It is claimed that the second one is equivalent to: $p\mathcal{O}_{\mathbb{Q}(\zeta_3)}=(\pi)(\overline{\pi})$ and $\pi=x+3y\sqrt{-3}$ . Why is it true?

Answer 1

Although the cubic reciprocity law (you alluded to CFT in your OP) will not appear in the proof of your equivalence, let us stress that cubic reciprocity plays a central role in the statement of Euler's conjecture that "$p=x^2+27y^2$ iff $p\equiv 1$ mod $3$ and $2$ is a cubic residue mod $p$", as well as in its later proof by Gauss. Sticking to your notations (but dropping the index $3$ for a primitive $3$-rd root of unity), recall that a prime $\pi \in \mathbf Z[\zeta]$ is called primary if $\pi \equiv \pm 1$ mod $3$ . Now :

1) Suppose that $p=\pi.\bar\pi$, with $\pi=x+y\sqrt {-3}$. Then $p\equiv 1$ mod $3$ because $p$ splits in $\mathbf Z[\zeta]$, $\pi$ is a primary prime and $p=x^2+27y^2$. Replacing $\pi$ by $-\pi$ if necessary, we may assume that $\pi \equiv 1$ mod $3$. Since $\sqrt {-3}=1+2 \zeta , \pi=x+y+6y \zeta$, and we have $\pi\equiv x+3y\equiv x+y \equiv 1$ mod $2$ because $x, y$ have opposite parity. It follows that $\pi\equiv 1$ mod $6$.

2) Conversely, suppose that $\pi\equiv 1$ mod $6$. The congruence $\pi\equiv 1$ mod $3$ implies that the prime $p$ under $\pi$ splits as $p=\pi.\bar\pi$. Let us exploit the congruence $\pi\equiv 1$ mod $2$. Since $\pi$ is primary, we can write $\pi=a+b\zeta$, with $a, b \in \mathbf Z$, so the congruence $\pi\equiv 1$ mod $2$ becomes $a+3b\zeta\equiv 1$ mod $2$, which implies straightforwardly that $a$ is odd and $b$ is even. Then, writing $4p=4\pi\bar\pi=4(a^2-3ab+9b^2)=(2a-3b)^2+27b^2$, the parity of $b$ shows that $p$ is of the form $x^2+27y^2$ ./.

Answer 2

Perhaps the proof in K. Conrad's notes is also helpful, which goes as follows. We know that $p$ splits completely in $\mathbb{Q}(\zeta_3,\sqrt[3]{2})$ if and only if $x^3-2$ splits completely modulo $p$, i.e., has three distinct roots in $\mathbb{F}_p$. Now this is the case if and only if $p\equiv 1\bmod 3$ and and $2^{\frac{p−1}{3}} \equiv 1 \bmod p$, which is that $p$ is represented by $X^2 + 27Y^2$, see Theorem $9.9$ here, using cubic reciprocity. In fact, $p=\pi\overline{\pi}$, and $2$ and $\pi= x+3\sqrt{-3}y$ are primary primes etc.