Consider a finite field $\mathbb{F}\_p$ such that $p \equiv 1 \ (\mathrm{mod} \ 3)$ and its element $\zeta \neq 1$, $\zeta^3 = 1$. 


Also, let $E\!: y^2 = x^3 + b$ be an elliptic curve of $j$-invariant $0$, where $b \in \mathbb{F}\_p^\* \setminus (\mathbb{F}\_p^\*)^3.$ This curve has the order $3$ automorphism $$[\zeta]\!: (x,y) \mapsto (\zeta x, y).$$


Consider the diagonal matrices
$$A := \mathrm{diag}(1, \zeta, \zeta^2),\qquad B := \mathrm{diag}(\zeta^2, \zeta, 1).$$
They generate the subgroup $$G := \langle A, B\rangle \subset \mathrm{SL}(3, \mathbb{Z}[\zeta]),$$ which is isomorphic to $(\mathbb{Z}/3)^2$. Any element $\mathrm{diag}(\alpha,\beta,\gamma) \in G$ naturally acts on the abelian threefold $E^3$: 
$$
(P, Q, R) \mapsto ([\alpha]P, [\beta]Q, [\gamma]R).
$$


What is the geometric quotient $E^3\!/G$? Could you explicitly write a (affine) model for this variety?



What is the geometric quotient of the abelian threefold?
Let $\mathbb P^1$ have projective coordinates $(y:z)$, so $(\mathbb P^1)^3$ has projective coordinates $(y\_1:z\_1), (y\_2:z\_2), (y\_3:z\_3)$. 


On $(\mathbb P^1)^3$, the line bundle $\mathcal O(1,1,1)$ has sections which are homogeneous functions of tridegree $(1,1,1)$ in these coordinates. Let $\alpha$ denote the value of this function.


Inside this line bundle, we can consider the vanishing locus of the equation $$ \alpha^3= (y\_1^2 + bz\_1^2) (y\_2^2 + bz\_2^2) (y\_3^2+bz\_3^2)z\_1z\_2z\_3$$ as the right side is homogenous of degree $(1,1,1)$. This is your desired geometric quotient.


This is simply because $E$ is the cover defined by adjoining a cube root of $(y^2+bz^2)z$, so $E^3$ is defined by adjoining the cube roots of $$ (y\_1^2 + bz\_1^2)z\_1,$$ $$ (y\_2^2 + bz\_2^2)z\_2,$$ and $$ (y\_3^2+bz\_3^2)z\_3,$$ and then we take the quotient by exactly the symmetries that preserve the product of the three cube roots.


