Given an elliptic curve with a Weierstrass equation, is there any easy way to see whether it has got an isogeny of low degree?
How to know whether an elliptic curve has a low-degree isogeny?
-
0Yes, for a start. – 2012-03-07
1 Answers
You can use the so-called classical modular polynomials $\Phi_n(x,y)$. Suppose we have an elliptic curve $E$ with $j$-invariant $j(E)$, and suppose there is an isogeny \phi:E\to E', to a second elliptic curve E' with $j$-invariant j(E'), and the isogeny has degree $n$. Then, \Phi_n(j(E),j(E'))=0.
So, if you have an elliptic curve $E$ and you know its $j$-invariant $j(E)$, then you can study the polynomials $p_{E,n}(y)= \Phi_n(j(E),y),$ for each $n\geq 1$, since the roots of these polynomials are $j$-invariants of curves isogenous to $E$.
For instance, let $E/\mathbb{Q}: y^2=x^3+x^2+x$, with $j(E)=2048/3$. We know that $E$ is $2$-isogenous to E':y^2=x^3-2x^2-3x, via $\phi(x,y)=(y^2/x^2,y(1-x^2)/x^2).$ Let us see that we could retrieve E' from the classical modular polynomials $\Phi_2(x,y)$. This polynomial is given by $\Phi_2(x,y)=x^3 - x^2y^2 + 1488x^2y - 162000x^2 + 1488xy^2 + 40773375xy + 8748000000x + y^3 - 162000y^2 + 8748000000y - 157464000000000.$ When we evaluate $x=2048/3$ we obtain: $p_{E,2}(y)=y^3 + 3489968y^2/9 + 111828246784y/3 - 4092314705809408/27,$ and $p_{E,2}(y)$ factors as: $p_{E,2}(y)=(y-35152/9)(y^2 + 391680y + 116417691904/3).$ Thus, $E$ is isogenous to a curve E' with $j$-invariant j(E')=35152/9. Now you can check that the curve E' given above has this $j$-invariant. There are $2$ other $j$-invariants of curves that are $2$-isogenous to $E$, but those are defined over quadratic extensions of $\mathbb{Q}$.
-
0Or here http://www.itd.umich.edu/scs/magma/text1091.htm – 2012-03-08