# How does one actually compute the dual isogeny?

Given a non-constant isogeny $f : E_1 \rightarrow E_2$ of degree $n$ between elliptic curves, I'm under the impression that there always exists a unique isogeny $g : E_2 \rightarrow E_1$ satisfying $$g \circ f = [n]_{E_1}, \qquad f \circ g = [n]_{E_2},$$ which is called the dual isogeny of $f$. This seems very amazing and mysterious, so:

How does one actually compute the dual isogeny, assuming the elliptic curves under question are described by Weierstrass equations?

• The thesis by Daniel Shumow at SageMath, "Isogenies of Elliptic Curves:A Computational Approach" gives some theory and examples (see page 78) using Sage's EllipticCurveIsogeny class, and refers the Reader to Sage documentation for more details. – hardmath Dec 30 '17 at 12:57

$\DeclareMathOperator{Pic}{Pic}$The dual isogeny can be shown to exist abstractly, but it can also be defined explicitly using the picard group of degree zero divisors, see Silverman's Arithmetic of Elliptic Curves III.6, this goes as follows: If $$\phi \colon E_1 \to E_2$$ is an isogeny then $$\hat\phi \colon E_2 \to \Pic^0(E_2) \to \Pic^0(E_1) \to E_1$$ is the dual isogeny.

What are all the maps here, well $E \to \Pic^0(E)$ is simply $P \mapsto (P) - (0)$ which has inverse $(P) \mapsto P$, we use the group structure on both sides to find the images of other divisors. The map $\Pic^0(E_2) \to \Pic^0(E_1)$ is $\phi^*$ the pullback which sends $(P) \mapsto \sum_{Q \in \phi^{-1}(P)} (Q)$.

So this is a pretty explicit recipe, to find what happens to $P\in E_2$ we just have to find the preimages of it under $\phi$ and of $0$ under $\phi$ and sum them all in $E_1$.

Let's do an example, if I take a look at the lmfdb I find some not too bad looking curves with a 3-isogeny between them http://www.lmfdb.org/EllipticCurve/Q/14/a/ The curves are $$E_1 = 14.a5\colon y^2 + x y + y = x^{3} - x$$ and $$E_2 = 14.a6\colon y^2 + x y + y = x^{3} + 4 x - 6$$

The 3-isogeny $\phi \colon E_1 \to E_2$ we'll use is given by $$\left(\frac{x^3 - x + 1}{x^2},\frac{x^3y + x^2 + xy - x - 2y - 1}{x^3}\right)$$ (I used SageMath to get the equation for this)

Let's find what $\hat\phi$ of the point $(1,-1) \in E_2(\mathbf{Q})$ is. First we need to find $\phi^{-1}(1,-1)$ so we have to solve $$\frac{x^3 - x + 1}{x^2} = 1$$$$\frac{x^3y + x^2 + xy - x - 2y - 1}{x^3} = -1$$ simplifies to $$x^3 - x^2 - x + 1 = 0 \text{ or } x^3 = x^2 + x - 1$$$$x^3y + x^3 + x^2 + xy - x - 2y - 1 = y(x^2 + 2x - 3) + (2x^2 - 2) = 0$$ the first polynomial factors as $(x-1)^2(x+1)$ so the $x$ coordinates of the preimage are $1,1,-1$, labelling these roots $\alpha_i$ the points are $$\left(\alpha_i, -\frac{2\alpha_i^2 - 2}{\alpha_i^2 + 2\alpha_i -3}\right)=\left(\alpha_i, -\frac{2\alpha_i + 2}{\alpha_i+3}\right) \text{ for }\alpha_i \ne 1$$ when $x = 1$ the second equation above degenerates so to find the $y$ coordinates we have to look at the original curve equation $y^2 + 1\cdot y + y = 1^3 -1 =0$ so we have $y=0,-2$ giving $(1,0),(1,-2),(-1,0)$ as the preimages.

Then the non-zero preimages of $0$ are the points where $x = 0$ so the rational function defining $\phi$ gives $\infty$ so we have the two points on $E_1$ where $x = 0$, they are solutions of $y^2 + y = 0$ i.e. $0, -1$, we also have the point at infinity on $E_1$.

So $\phi^*((P) - (0)) = ((1,0)) + ((1,-2)) + ((-1,0)) - (0) - ((0,0)) - ((0,-1))$ and hence the point $\hat \phi (P)$ is just this sum evaluated on $E_1$, this is simply is $(-1, 0)$ as points with the same $x$ coordinate are each others negation so everything else cancels.

Most of the steps here were just fiddling around with functions, even addition at the end is just complicated rational functions, but its still a little surprising that one can in fact find nice enough rational functions for the dual that do this all for us. In this case SageMath tells me that $$\hat \phi = \left((3 x + 1)^{-2} \cdot (x - 2) \cdot (x^{2} + 2 x + 13), (3 x + 1)^{-3} \cdot (x + 5) \cdot (- x^{3} + x^{2} y - 8 x^{2} - 4 x y + 9 x + 11 y + 8)\right)$$

One final point is that the fundamental property of the dual isogeny allows us to do away with much of this, if $\hat\phi\circ\phi = [\deg \phi]$ then to find $\hat \phi (P)$ we can find $Q$ with $\phi(Q) = P$ so that $[\deg \phi]Q = (\hat\phi \circ \phi)( Q) = \hat \phi (P)$ as required. In the example above this means we could have stopped once we found the point $(-1,0)\in E_1(\mathbf Q)$ and tripled it to find the answer instead as we have a degree 3 isogeny, as letting $x= -1$ in the equation for $E_1$ reduces to $y^2 = 0$ the line $x= -1$ is tangent to $E_1$ and $(-1,0)$ is 2-torsion so $3\cdot(-1,0) = (-1,0)$ just as before.