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Let $f:E_1\rightarrow E_2$ be a nonzero isogeny between elliptic curves.

Take a point $Q \in E_2$.

I am looking for a reference to a proof, or a proof, of the following fact:

$|f^{-1}(Q)|=\text{deg}_s(f)$

where $\text{deg}_s(f)$ is the separable degree of $f$.

In fact, it's enough to show that this is true for all but a finite set of points $Q$. The desired claim would then follow because $f$ is a group homomorphism.

EDIT:

  1. I'm working over an algebraically closed field.
  2. If you prefer to, you may assume that $f$ is a separable isogeny.
Olnek
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  • (1). Isogenies of elliptics curves are always non-zero. (2). You need to work over an algebraically closed field.(3) You can easily reduce to the case $f$ is separable. – Cantlog Jun 03 '14 at 13:46
  • You can check III.4 of Silverman's "The Arithmetic of Elliptic Curves", or II-6.8 of Harthsorne's "Algebraic Geometry" . – DonAntonio Jun 03 '14 at 13:53
  • @DonAntonio: III.4.10 in Silverman refers to II.2.6b in Silverman, which refers to II.6.8 in Hartshone. So, the part of the proof I'm interested in is probably in Hartshone's book. I'll grab it now. – Olnek Jun 03 '14 at 13:59
  • Indeed so, @Olnek...if I get another reference I'll write it down, since Hartshorne's book is, imo, written as if it was assumed the reader already knows almost all... – DonAntonio Jun 03 '14 at 14:01
  • @DonAntonio: Thanks! I indeed don't understand exactly how II.6.8 gives what I need. I guess that the part I need is related to the proof of the fact that $f$ is a finite morphism, is that right? – Olnek Jun 03 '14 at 14:03
  • @DonAntonio: I'm looking for a proof in Milne's notes http://www.jmilne.org/math/CourseNotes/AG.pdf . If it's there, it's probably written well. – Olnek Jun 03 '14 at 14:05
  • @Olnek, I've to go in a little while but I'll really try to come back on this later. I've tens of books on this subject (as my thesis dealt with this...), yet in all the books I've checked so far the proof of this fact always send you back to Silverman's book (yes, up to and including my thesis, I'm sorry to say...), and this sucks big time. Now, I think the proof doesn't have to be that messy so we can try it together later. – DonAntonio Jun 03 '14 at 14:13
  • Th. 10.18 in Milne's book is around the matter here, yet the proof is "to be added"...and it hasn't so far. I think it is enough to prove for $;f^{-1}(0);$ as isogenies are morphisms of these algebraic (elliptic) curves, yet the fact that nobody proves it anywhere (in what I've checked so far) makes me wonder whether this fact is not that easy to prove... – DonAntonio Jun 03 '14 at 14:18
  • @Olnek: using the group structure, it is enough to show that one fiber $f^{-1}(x)$ has $\deg f$ points. Now for a finite separable morphism, this is true for $x$ in a non-empty open subset of the target space. I think there are several proofs of this fact here, see for instance http://math.stackexchange.com/questions/341281/ – Cantlog Jun 03 '14 at 17:15

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