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Let $(A,\Theta)$ be a principally polarized abelian variety over an algebraically closed field $k$, and let $f$ be a symmetric endomorphism of $A$ (that is, $f^\dagger=f$ where $\dagger$ denotes the Rosati involution). Now, if $p\neq\mbox{char}(k)$ is a prime, $f$ has a representation as an endomorphism of $(T_p A)\otimes_{\mathbb{Z}_p}\mathbb{Q}_p$, where $T_pA:=\lim_{\leftarrow}A[p^l]$ and $\mathbb{Z}_p$ (resp. $\mathbb{Q}_p$) denotes the $p$-adic integers (resp. $p$-adic numbers). Moreover, the characteristic polynomial of $f$ when seen as an endomorphism here is characterized by $P_f(t)=\deg(f-t)$ for $t\in\mathbb{Z}$ (see Milne's notes).

My question is the following: If the eigenvalues of (the representation of) $f$ are all integers, does it follow that the minimal polynomial of $f$ splits into distinct linear factors (that is, no repeated factors)?

Observation: Over the complex numbers this is true, since the fact that $f$ is symmetric under $\dagger$ means that the analytical representation of $f$ is self-adjoint with respect to the Hermitian form $H=c_1(\mathcal{O}_A(\Theta))$, and so is diagonalizable.

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  • $\begingroup$ I don't understand. "Diagonalizable" doesn't mean "no repeated factors". What about the identity?? $\endgroup$
    – abx
    Commented Nov 29, 2013 at 14:56
  • $\begingroup$ Dear abx, I'm asking about the minimal polynomial splitting into distinct linear factors, not the characteristic polynomial. $\endgroup$
    – rfauffar
    Commented Nov 29, 2013 at 15:50

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Yes. The $\mathbb{Q}$-algebra $B$ generated by $f$ is a finite-dimensional $\mathbb{Q}$-algebra with a positive involution, hence semisimple. It is also commutative, hence a product of fields. The minimum polynomial of the image of $f$ in each of the factors divides the characteristic polynomial of $f$ on the abelian variety, and so has integer roots. We conclude that $B$ is a product of copies of $Q$.

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  • $\begingroup$ What is the positive involution on $B$? $f$ is fixed by $\dagger$. $\endgroup$
    – rfauffar
    Commented Nov 29, 2013 at 1:48
  • $\begingroup$ It's the identity map in your case. $\endgroup$
    – anon
    Commented Nov 29, 2013 at 2:02
  • $\begingroup$ Dear @anon, I'm afraid I don't follow. Why does this assure me that the minimum polynomial has no repeated factors? $\endgroup$
    – rfauffar
    Commented Nov 29, 2013 at 2:53
  • $\begingroup$ The action of $f$ on $V_p(A)$ comes from an action of $B$ on $V_p(A)$, and the minimum polynomial of $f$ in $B$ has distinct roots (even distinct integer roots in your case). $\endgroup$
    – anon
    Commented Nov 29, 2013 at 19:05

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