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Homedual isogeny

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# dual isogeny

Given an isogeny $f:E\rightarrow E^{{\prime}}$ of elliptic curves of degree $n$, the *dual isogeny* is an isogeny $\hat{f}:E^{{\prime}}\rightarrow E$ of the same degree such that $f\circ\hat{f}=[n]$. 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

$E^{{\prime}}\rightarrow\Div^{0}(E^{{\prime}})\stackrel{f^{*}}{\rightarrow}\Div% ^{0}(E)\rightarrow E$ |

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

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

$E\rightarrow\Div^{0}(E)\stackrel{f_{*}}{\rightarrow}\Div^{0}(E^{{\prime}})% \rightarrow E^{{\prime}}$ |

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

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

$E^{{\prime}}\stackrel{\sim}{\rightarrow}\Pic^{0}(E^{{\prime}})\stackrel{f^{*}}% {\rightarrow}\Pic^{0}(E)\stackrel{\sim}{\rightarrow}E$ |

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

## Mathematics Subject Classification

14-00*no label found*

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## Recent Activity

new correction: Error in proof of Proposition 2 by alex2907

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