primes represented integrally by a homogeneous cubic form

Solution 1:

The discriminant of $\Bbb Q \subset \Bbb Q(\sqrt[3]2)$ is $-108$, and the Minkowski bound for this extension is $\frac {3!}{3^3} \frac 4 \pi \sqrt {108} \approx 2.940$. So to prove that this number field has class number $1$ we only need to find a way to represent $2$, and $2$ is indeed represented by $(0,1,0)$. Thus $p$ is represented by this norm form if and only if the ideal $(p)$ has an ideal factor of norm $p$, which happens if and only if $2$ is a cube modulo $p$.

If $p \equiv 2 \pmod 3$ then any nonzero element of $\Bbb F_p$ has one cube root in $\Bbb F_p$ and two cube roots in $\Bbb F_p^2$, so $2$ is a cube modulo $p$.

If $p \equiv 1 \pmod 3$ then $(p)$ splits in $\Bbb Q(\zeta_3)$, and $2$ is a cube if and only it further splits in $\Bbb Q(\zeta_3,\sqrt[3]2)$. Since $\Bbb Q(\zeta_3) \subset \Bbb Q(\zeta_3,\sqrt[3]2)$ is an abelian extension, it is a ray class field for some modulus $\mathfrak m$ of $\Bbb Q(\zeta_3)$.

Working modulo $6$, we have $(a+b\zeta_3)^3 = (a^3+b^3) - 3ab^2+3ab(a-b)\zeta_3 = a^3 + b^3-3ab^2 \in \Bbb Z/6 \Bbb Z$, and thus for any $a,b,c \in \Bbb Z[\zeta_3]$, $a^3+2b^3+4c^3-6abc = a^3+2b^3+4c^3 \in \Bbb Z/6\Bbb Z$. So, norms that are coprime to $6$ are units ($\pm 1$) modulo $6$. So $\Bbb Q(\zeta_3,\sqrt[3]2)$ is an extension of the ray class field of modulus $(6)$ for $\Bbb Q(\zeta_3)$.

On the other hand, $G = (\Bbb Z[\zeta_3]/(6))^*/\langle \overline{\zeta_6} \rangle$ is isomorphic to $\Bbb Z/3 \Bbb Z$, which is the Galois group of the extension $\Bbb Q(\zeta_3) \subset \Bbb Q(\zeta_3,\sqrt[3]2)$, so $\mathfrak m = (6)$, and $2$ is a cube modulo $p$ if and onlt if $p \equiv 2 \mod 3$ or $p = a^2-ab+b^2$ where $a+\zeta_3 b$ is congruent modulo $6$ to one of $\{1,1+\zeta_3,\zeta_3,-1,-1- \zeta_3,- \zeta_3\}$.

Each element of $G$ (modulo complex conjugation) corresponds to a class of primitive binary quadratic forms of discriminant $-108$, or a corresponding lattice class (modulo multiplication by a unit and complex conjugation) whose endomorphism ring is $\Bbb Z[3\sqrt{-3}]$:

$\Lambda = \langle 1, 3\sqrt{-3} \rangle$ is a lattice corresponding to the neutral element of $G$ : it contains $(6)$ and the numbers coprime with $(6)$ it meets all fall in the neutral class.
while $\Lambda = \langle 2, \frac {1+3\sqrt{-3}}2 \rangle$ corresponds to the other two classes : it contains $(6)$ and the numbers coprime with $(6)$ it meets all fall in the other two classes.

So if $p \equiv 1 \pmod 3$, then $p$ is represented either as $a^2 + 27b^2$ (when $2$ is a cube) or as $4u^2 \pm 2uv + 7v^2$ (when $2$ is not a cube), and never both at the same time.