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The book I'm reading doesn't provide the definition of degree of an isogeny and I failed to google it. Can anyone tell me?

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An isogeny $f:A \rightarrow A'$ induces a map $f^*:\overline{k}(A') \rightarrow \overline{k}(A)$ on function fields which is necessarily injective. We define the degree of $f$ to be the degree of the extension $\overline{k}(A)/\overline{k}(A').$

Equivalently, if $x$ is any point of $A',$ we can pull back $x$ along $f,$ and obtain a divisor $$\sum_{P \in f^{-1}(x)} e_P(f)P.$$ The degree of $f$ is the degree of this divisor. Because $f$ is a homomorphism this degree is independent of $x.$

If $f$ is separable (which occurs when $char(k)\not|Deg(f)),$ then $f$ is etale and so $Deg(f) = |f^{-1}(0_{A'})| = |ker(f)|.$

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Your answer is too advanced to me. I've just learned about complex tori and $\varphi$ is an isogeny between two complex tori...Yes indeed the book says $ker(\varphi)$ is the degree of $\varphi$.. – hxhxhx88 Jan 7 '13 at 1:33

Let E be an elliptic curve with a p-torsion point which is denoted by P. If you define ϕ:E→E/⟨P⟩ then if ϕ is separable, then the degree of isogeny is the order of the kernel which is given p=#⟨P⟩.

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I have not learned elliptic curve yet, but thanks for your answer! – hxhxhx88 Jan 7 '13 at 1:34

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