# Are there conditions for an elliptic curve to have a quadratic $\mathbb{F}_q$-cover of the line without ramification $\mathbb{F}_q$-points?

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!

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 $$x\mapsto Q-x$$.
Conversely, any $$Q\in E(k)$$ gives rise to an involution $$\sigma:x\mapsto Q-x$$. 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 $$E(k)$$.