Editing Elliptic curve (section)

===Construction of the dual isogeny===

Often only the existence of a dual isogeny is needed, but it can be explicitly given as the composition
: <math>E' \to \operatorname{Div}^0(E') \to \operatorname{Div}^0(E) \to E,</math>
where  <math>\operatorname{Div}^0</math> is the group of [[Divisor (algebraic geometry)|divisors]] of degree 0. To do this, we need maps <math>E \to \operatorname{Div}^0(E)</math> given by <math>P \to P - O</math> where <math>O</math> is the neutral point of <math>E</math> and <math>\operatorname{Div}^0(E) \to E</math> given by <math>\sum n_P P \to \sum n_P P.</math>

To see that <math>f \circ \hat{f} = [n]</math>, note that the original isogeny <math>f</math> can be written as a composite
: <math>E \to \operatorname{Div}^0(E) \to \operatorname{Div}^0(E') \to E',</math>
and that since <math>f</math> is [[Wikt:finite|finite]] of degree <math>n</math>, <math>f_* f^*</math> is multiplication by <math>n</math> on  <math>\operatorname{Div}^0(E').</math>

Alternatively, we can use the smaller [[Picard group]] <math>\operatorname{Pic}^0</math>, a [[factor group|quotient]] of  <math>\operatorname{Div}^0.</math> The map <math>E \to \operatorname{Div}^0(E)</math> descends to an [[isomorphism]], <math>E \to \operatorname{Pic}^0(E).</math> The dual isogeny is
: <math>E' \to \operatorname{Pic}^0(E') \to \operatorname{Pic}^0(E) \to E.</math>

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