An elliptic curve $C$ over a field $k$ is a smooth, genus 1 curve defined over $k$ with an associated $k$-rational point. If char$(k) \ne 2$, we can show that $C$ has a model of the form $y^2 = f(x)$ where deg$(f) = 3$. The method I have seen in Silverman and Hartshorne finds this model from a change of coordinates of the general Weierstrass equation which you find via Riemann-Roch. My question involves whether you can see this transformation through a different method.

One can show that $C$ admits a degree 2 map to $\mathbb{P}_k^1$. So, as char($k) \ne 2$, the corresponding extension of function fields is a Kummer extension and hence generated by a square root. Thus, we have an affine curve $y^2 = f(x)$ which is birational to $C$. By Riemann-Hurwitz, there should be 4 ramification points which occur either as roots of $f$ or occur at infinity. Thus, $f$ has degree either 3 or 4.

Is there some way to see that the polynomial $f$ can be chosen to have degree 3? That is, can one see **without explicitly writing our the transformation** that one of the ramification points can be chosen to be at infinity over a **non-algebraically closed** field?

The motivation for this question comes from the case of genus 2 where a hyperelliptic curve of genus 2 has a model of the form $y^2 = f(x)$ where the degree of $f$ is either 5 or 6. One can choose a model where $f$ is of degree 5 only when $f$ has a root over the given field $k$.

Essentially my question boils down to, is there some way to view from the function field side that you can choose a model $y^2 = f(x)$ where deg($f) = 3$? Also, from the function field side, can you see that there should be a difference in the case of genus 2?