dual isogeny

Given an isogeny $f:E\to E^{\prime }$ of elliptic curves of degree $n$, the dual isogeny is an isogeny $\widehat{f}:E^{\prime }\to E$ of the same degree such that $f\circ \widehat{f}=[n]$. Here $[n]$ denotes the multiplication-by-$n$ isogeny $e\mapsto ne$ which has degree $n^2$.

Often only the existence of a dual isogeny is needed, but the construction is explicit as

$$E^{\prime }\to {\mathrm{Div}}^0(E^{\prime })\stackrel{f^{*}}{\to }{\mathrm{Div}}^0(E)\to E$$

where ${\mathrm{Div}}^0$ is the group of divisors of degree 0.To do this, we need maps $E\to {\mathrm{Div}}^0(E)$ given by $P\mapsto P-O$ where $O$ is the neutral point of $E$ and ${\mathrm{Div}}^0(E)\to E$ given by $\sum n_{P}P\mapsto \sum n_{P}P$.

To see that $f\circ \widehat{f}=[n]$, note that the original isogeny $f$ can be written as a composite

$$E\to {\mathrm{Div}}^0(E)\stackrel{f_{*}}{\to }{\mathrm{Div}}^0(E^{\prime })\to E^{\prime }$$

and that since $f$ is finite of degree $n$, $f_{*}{f}^{*}$ is multiplication by $n$ on ${\mathrm{Div}}^0(E^{\prime })$.

Alternatively, we can use the smaller Picard group ${\mathrm{Pic}}^0$, a quotient of ${\mathrm{Div}}^0$. The map $E\to {\mathrm{Div}}^0(E)$ descends to an isomorphism, $E\stackrel{\sim }{\to }{\mathrm{Pic}}^0(E)$. The dual isogeny is

$$E^{\prime }\stackrel{\sim }{\to }{\mathrm{Pic}}^0(E^{\prime })\stackrel{f^{*}}{\to }{\mathrm{Pic}}^0(E)\stackrel{\sim }{\to }E$$

Note that the relation $f\circ \widehat{f}=[n]$ also implies the conjugate relation $\widehat{f}\circ f=[n]$. Indeed, let $\varphi =\widehat{f}\circ f$. Then $\varphi \circ \widehat{f}=\widehat{f}\circ [n]=[n]\circ \widehat{f}$. But $\widehat{f}$ is surjective, so we must have $\varphi =[n]$.