Does every nonsingular cubic curve $C \subset \mathbb{P}^2$ over a finite field $\mathbb{F}_q$ have a flex over $\mathbb{F}_{q^3}$?

Question

I cannot prove the following statement.

Does every nonsingular cubic curve $C \subset \mathbb{P}^2$ over a finite field $\mathbb{F}_q$ have a flex over $\mathbb{F}_{q^3}$?

Is there an elegant proof (i.e., without the full brute force of possible Frobenius actions on the Hesse configuration of 9 flexes of $C$) or at least a reference to some source?

Thank you in advance.

Answer 1

The question linked in the comments provides a reasonable strategy. Suppose we have a smooth cubic over $\Bbb F_q$ given by the equation $f(X,Y,Z)=0$. If we can put the curve in to Weierstrass form $Y^2Z+a_1XYZ+a_3YZ^2 = X^3+a_2X^2Z+a_4XZ^2+a_6Z^3$ over $\Bbb F_{q^3}$, then we will be done: if we have such a Weierstrass form for our curve, then the point $[0:1:0]$ is on our curve, and the tangent line is given by $Z$, which intersects our cubic with multiplicity 3 at $[0:1:0]$, giving a flex.

Lemma. Given any homogeneous cubic defined over $\Bbb F_q$, we may put the cubic in to Weierstrass form over $\Bbb F_{q^3}$ via a projective change of coordinates.

Proof: By a standard argument, any smooth cubic with a $K$-rational point may be put in to a Weierstrass form over $K$ via a projective change of coordinates (here, or any elliptic curve book). We claim that a smooth cubic defined over $\Bbb F_q$ always has a rational point over $\Bbb F_{q^3}$, which will show the claim.

Intersect our cubic $C$ with any $\Bbb F_q$-rational line (for instance, the $Z$-axis). By Bezout, this intersection is of degree three over $\Bbb F_q$. Thus either it contains a $\Bbb F_q$-rational point or it contains a point defined over a cubic extension of $\Bbb F_q$. But $\Bbb F_q$ has only one cubic extension (up to isomorphism): $\Bbb F_{q^3}$. So $C$ has a rational point over $\Bbb F_{q^3}$ and we're done. $\blacksquare$

One could also just get one's hands dirty with the equation of the curve: there it will be obvious that you need access to $\Bbb F_{q^3}$-coefficients (and no other extensions) to make the relevant coordinate changes. This is very low-tech, which has it's advantages and disadvantages.