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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?

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    $\begingroup$ You have a k-rational point P so use Riemann-Roch to compute dimensions of spaces of global functions with at worst a pole of order n at P, for n=0,1,2,3,4,5. You should get dimensions 1,1,2,3,4,5. If y has a pole of order 2 and x a pole of order 3 at P then both y^2 and x^3 have poles of order 6 and now it's not hard to see that there's a linear relation between 1,y,y^2,y^3,x,xy,x^2. Now complete the square in y if char(k) isn't 2 and you're done. This is surely in Hartshorne or something, and is a bit easy for this site. $\endgroup$
    – znt
    Commented Apr 18, 2016 at 22:09
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    $\begingroup$ @znt Perhaps my question was not entirely clear. I'm aware of the method you have mentioned, the primary aim of my question is whether you can see this model through a different method. Also, why do you not get an odd degree model in the case of genus 2? I've edited my post so that hopefully it is clearer. $\endgroup$
    – user78330
    Commented Apr 18, 2016 at 23:20
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    $\begingroup$ Would you prefer something along these lines? We know the curve has a model $Y^2 = P(X)$ with $P$ of degree $3$ or $4$ and some rational point. Changing projective coordinates, we may assume this point has $X = \infty$. If $\deg P = 3$ then we're done. Else $\deg P = 4$ and the leading coefficient of $P$ is a square, so at infinity $P = Q^2 + R$ for some $Q,R$ with rational coefficients such that $\deg Q = 2$ and $\deg Q \leq 1$. Now write $Y=Q+x$ and get an equation quadratic in $X$ whose discriminant is cubic in $x$. This gives a birational map from $Y^2 = P(X)$ to $y^2 = cubic(x)$. $\endgroup$
    – Noam D. Elkies
    Commented Apr 19, 2016 at 0:30
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    $\begingroup$ (Inevitably that's not far from the usual Riemann-Roch argument, but is what people actually did in the old days and sometimes still do nowadays.) $\endgroup$
    – Noam D. Elkies
    Commented Apr 19, 2016 at 0:32
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    $\begingroup$ The argument of @Noam D. Elkies is presented in Cassels' little blue book on elliptic curves (where he also explains how to get various other models of genus 1 curves with a rational point into the form y^2=cubic). $\endgroup$
    – znt
    Commented Apr 19, 2016 at 11:31

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