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Let $A$ be an abelian variety defined over a finite field $k$, and $End(A)$ be it ring of endomorphisms defined over an algebraic closure $\overline{k}$ of $k$. Suppose that for an integer $M$ coprime with the characteristic of $k$ there exist an endomorphism $\phi\in End(A)$ such that

$M=(1+F-2F^2)\circ \phi$

where $F$ is the Frobenius endomorphism.

Do we have that $\phi$ is defined over $k$? If the answer is yes, how can I proove it?

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The eigenvalues of $F$ on $H^1(A,\mathbb Q_l)$ are algebraic integers absolute value $q^{1/2}$ (thus $\ne 1, \ne -1/2$), so the eigenvalues of $1+F-2F^2$ are all non-zero, so $\sigma \in Gal(\overline{k}/k) \Rightarrow (1+F-2F^2)(\phi - \phi^\sigma)=0$ on $H^1(A,\mathbb Q_l) \Rightarrow \phi = \phi^\sigma$, thus $\phi$ is defined over $k$.

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  • $\begingroup$ Thank you for your answer. Do you know a good reference to understand the object $H^1(A,Q_l)$ and why we have to look $F$ on it. $\endgroup$ – A. GM Mar 1 '17 at 15:32
  • $\begingroup$ @A.GM Milne's "Abelian Varieties", also Mumford's "Abelian Varieties". $\endgroup$ – David Lampert Mar 1 '17 at 15:40
  • $\begingroup$ If the variety $A$ is simple over $k$, it's enough to show that the endomorphism $1+F-2F^2$ is not $0$? $\endgroup$ – A. GM Mar 9 '17 at 17:05
  • $\begingroup$ @A.GM True: if $A/k$ is simple then $End_{k}A$ has no zero-divisors. $\endgroup$ – David Lampert Mar 9 '17 at 23:58

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