# How to know whether an elliptic curve has a low-degree isogeny?

Given an elliptic curve with a Weierstrass equation, is there any easy way to see whether it has got an isogeny of low degree?

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Is this elliptic curve defined over $\mathbb{Q}$? Are you looking for $\mathbb{Q}$-isogenies? – Álvaro Lozano-Robledo Mar 7 '12 at 15:16
Yes, for a start. – Alan Lee Mar 7 '12 at 15:47

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 $\mathbb{Q}$.
That's a different question! It is a polynomial relation between the functions $j(\tau)$ and $j(n\tau)$. – Álvaro Lozano-Robledo Mar 7 '12 at 20:28