Consider an elliptic curve $E: y^2 = f(x) := x^3 + ax + b$ over a finite field $\mathbb{F}_q$ of characteristic $> 3$. Obviously, the projection to $x$ is a quadratic $\mathbb{F}_q$cover of the line with at least one ramification $\mathbb{F}_q$point, namely the infinity point. Are there (simple) necessary or sufficient conditions for $E$ to have a quadratic $\mathbb{F}_q$cover of the line without ramification $\mathbb{F}_q$points? In other words, is there or not a quartic $\mathbb{F}_q$polynomial $g(x)$ without $\mathbb{F}_q$roots such that $E$ is birationally $\mathbb{F}_q$isomorphic to the curve $y^2 = g(x)$? Thank you!
1 Answer
More generally, let $k$ be a field of characteristic $≠2$. For the analogous question over $k$, a necessary and sufficient condition is that the doubling map $[2]_E:E\to E$ is not surjective on $k$points.
Of course, if $k$ is finite, this in turn is equivalent to "$\#E(k)$ is even".
Indeed, given a polynomial $g$ as in the question, consider the divisor $D$ in $E$ given by zeros of $g$. Over $\overline{k}$, $D$ must be a translate of $E[2]=\ker[2]_E$ by some point $P\in E(\overline{k})$, and the image of $D$ by $[2]_E$ is $Q:=2P$ which must be $k$rational because $D$ is. Then it is immediately checked that the double cover $\pi:E\to\mathbb{P}^1_k$ associated to $g$ is the quotient of $E$ by the involution $x\mapsto Qx$.
Conversely, any $Q\in E(k)$ gives rise to an involution $\sigma:x\mapsto Qx$. The quotient $E/\langle\sigma\rangle$ is smooth of genus zero and has a rational point because $E$ does, hence $E/\langle\sigma\rangle\cong\mathbb{P}^1_k$. The branch divisor of $E\to E/\langle\sigma\rangle$ on $E$ is $[2]_E^{1}(Q)$, so the desired condition is satisfied iff $Q$ is not divisible by $2$ on $E(k)$.

$\begingroup$ Thank you! The question is now solved. $\endgroup$ Commented Jul 3 at 17:29