Timeline for Equations for Elliptic Curves
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Apr 19, 2016 at 11:31 | comment | added | znt | 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). | |
| Apr 19, 2016 at 0:32 | comment | added | Noam D. Elkies | (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.) | |
| Apr 19, 2016 at 0:30 | comment | added | Noam D. Elkies | 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 $y^2 = cubic(x)$. | |
| Apr 18, 2016 at 23:20 | comment | added | user78330 | @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. | |
| Apr 18, 2016 at 23:18 | history | edited | user78330 | CC BY-SA 3.0 |
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| Apr 18, 2016 at 22:09 | comment | added | znt | 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. | |
| Apr 18, 2016 at 21:59 | history | asked | user78330 | CC BY-SA 3.0 |