Take a finite field $\mathbb{F}_{\!q}$ such that $q \equiv 1 \pmod 3$, i.e., $\omega \mathrel{:=} \sqrt[3]{1} \in \mathbb{F}_{\!q}$, $\omega \neq 1$. Also, for $i \in \{0,1,2\}$ consider the elliptic curves $E_i\!: y^2_i = b^ix_i^3 - b$, where $b \in \mathbb{F}_{\!q}^* \setminus (\mathbb{F}_{\!q}^*)^3$. There is on $E_i$ the order $3$ automorphism $[\omega]\!: (x_i,y_i) \mapsto (\omega x_i, y_i)$.

Look at the quotient $T \mathrel{:=} (E_0 \!\times\! E_1 \!\times\! E_2)/[\omega]^{\times 3}$, which is a Calabi–Yau threefold according to Oguiso and Truong - Explicit examples of rational and Calabi–Yau threefolds with primitive automorphisms of positive entropy. It is easily seen that it has the affine model $$ T\!: \begin{cases} y_1^2 + b = b(y_0^2 + b)t_1^3, \\ y_2^2 + b = b^2(y_0^2 + b)t_2^3 \end{cases} \quad \subset \quad \mathbb{A}^{\!5}_{(y_0,y_1,y_2,t_1,t_2)}, $$ where $t_1 \mathrel{:=} x_1/x_0$, $t_2 \mathrel{:=} x_2/x_0$.

Although $T$ is a quite classical quotient, I cannot find a rational $\mathbb{F}_{\!q}$-curve on it. In my opinion, this is a sufficiently interesting algebraic geometry task. Can you help me please? I can explain the origin of this task if it is necessary.


1 Answer 1


Further intersecting $T$, which is $3$-dimensional in its affine model inside $\Bbb A^5$, with a generic variety of codimension two should lead to a curve. (It may be that i do not catch the point of the question. So i am inserting also an example.)

For instance, after multiplying the equations of $E_0, E_1,E_2$ by $1,b^2,b^4$ we obtain isomorphic curves given by $$ \begin{aligned} Y_0^2 + b^1&= X_0^3\ ,\\ Y_1^2 + b^3&= X_1^3\ ,\\ Y_2^2 + b^5&= X_2^3\ . \end{aligned} $$ Each change of coordinates is linear, so the action of $[\omega]$ translates also as a multiplication with $\omega$ on the $X_i$-components, $i$ being $0,1,2$. Then the model of the cartesian product of the three curves modulo $$ (X_0,X_1,X_2,Y_0,Y_1,Y_2)\sim (\omega X_0,\omega X_1,\omega X_2,Y_0,Y_1,Y_2) $$ is in a similar manner given by the equations: $$ \begin{aligned} \frac{Y_1^2-b^3}{Y_0^2-b} & = \left(\frac {X_1}{X_0}\right)^3=: u_1^3 \ ,\\ \frac{Y_2^2-b^5}{Y_0^2-b} & = \left(\frac {X_2}{X_0}\right)^3=: u_2^3 \ . \end{aligned} $$ (Omit the $X$-occurrences and consider the equations in $\Bbb A^5_{(u_1,u_2;Y_0,Y_1,Y_2)}$.)

Now we can intersect with the codimension two variety given (in an affine model) by $$ Y_1=Y_0^3\ ,\ Y_2=Y_0^5\ .$$ We obtain a curve (of higher degree) parametrized by $Y_0$ as follows: $$ \left\{ \begin{aligned} Y_1 &= Y_0^3\ ,\\ Y_2 &= Y_0^5\ ,\\ u_1^3 &= Y_0^4 + bY_0^2+b^2\ ,\\ u_2^5 &= Y_0^8 + bY_0^6 + b^2Y_0^4 + b^3Y_0 + b^4\ . \end{aligned} \right. $$ (Depending on the direction of research this may be useful or not.)

  • $\begingroup$ Well, "rational" is indeed ambiguous here, but I guess the OP means of genus $0$. $\endgroup$
    – abx
    Commented Jul 27, 2020 at 14:09
  • $\begingroup$ Yes, I mean an $\mathbb{F}_{\!q}$-curve of geometric genus $0$. $\endgroup$ Commented Jul 28, 2020 at 14:14

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