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In "Elliptic Curves: Number Theory and Cryptography" we see in the proof of proposition 2.21 that for the separable endomorphism $\alpha : E(\overline{K}) \rightarrow E(\overline{K})$ for every point $(a,b)$ with $(a,b) \neq \infty$ and $a,b \neq 0$ there are exactly $\deg(\alpha)$ points with $\alpha(x_1,y_1) = (a,b)$.

How can we then conclude that the kernel also has $\deg(\alpha)$ points since we excluded $\infty$? Do we look at the limit as $(a,b)$ tends towards $\infty$?

NightRain23
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  • The points mapped to $\infty$ are those in $\ker(\alpha)$. Let $k(E)$ the function field, $\tau_Q(P) = P+Q, G = { (f \in k(E) \mapsto f \circ \tau_Q), Q \in \ker(\alpha)}$, $\alpha^* k(E) = { f \circ \alpha, f \in k(E)}$, $k(E)^G = { f \in k(E), \forall Q \in \ker(\alpha), f = f\circ \tau_Q}$. We have the tower of field extensions $$k(E) \ / \ k(E)^G \ / \ \alpha^* k(E)$$ where $k(E) \ / \ k(E)^G$ is separable and Galois with Galois group $G$ and $k(E)^G \ / \ \alpha^* k(E)$ is purely inseparable. – reuns Apr 06 '19 at 14:33
  • Thank you for your response, is there an explanation that does not require Galois theory? – NightRain23 Apr 06 '19 at 14:33
  • That $\alpha \in \text{End}(E)$ is separable means $|\ker(\alpha)| = \deg(\alpha) \overset{def}= [k(E):\alpha^* k(E)]$ ie. $k(E)^G = \alpha^* k(E)$. Galois theory is involved in the definition of function field, field extension, degree, separable, you can't avoid it. – reuns Apr 06 '19 at 14:34
  • Thanks, as Washington defines $deg(\alpha) = Max${deg $p(x)$, deg $q(x)$} with $\alpha(x,y) = (\frac{p(x)}{q(x)}, r_2(x)y)$ I thought there would be a less technical explanation. – NightRain23 Apr 06 '19 at 15:02

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