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Let $[i]$ be the endomorphism such that $[i](x,y) = (-x,iy)$ with $[i]^2+[1]=[0]$. I am trying to prove that the degree of the endomorphism $[a]+[b]\circ[i]$ is equal to $a^2+b^2$. After multiplication in $End(E)$ by $[a]+[-b]\circ[i]$ I get $[a]^2+[b]^2$. I now have two endomorphisms whose composition is $[a^2+b^2]$. Is this enough for me to conclude that the dual of $[a]+[b]\circ[i]$ is equal to $[a]+[-b]\circ[i]$ and that the degree is equal to $a^2+b^2$? Theorems about the existence of the dual seem to start with the assumption that you know the degree already.

user288451
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    Presumably you’re speaking of the elliptic curve $Y^2=X^3-X$ or something close to that. Surely $[a+bi]$ and$[a-bi]$ have the same degree, and the degree is multiplicative, so that, together with the fact that the degree of $[n]$ is $n^2$, seems to be enough to get your desired result. – Lubin Nov 10 '15 at 06:16
  • Thank you. Indeed, that the is the elliptic curve. I just fail to see how changing the curve affects the described approach. Where is it relevant? – user288451 Nov 10 '15 at 10:47
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    For most choices of $a$ and $b$, the curve $Y^2=X^3+aX+b$ does not have an $[i]$-endomorphism. – Lubin Nov 10 '15 at 21:43

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