Let $A,B$ be positive dimensional Abelian varieties over a finite field and $p$ be an arbritrary prime. By Zarhin, Homomorphisms of abelian varieties over finite fields http://www.math.nyu.edu/~tschinke/books/finite-fields/final/10_zarhin.pdf, Theorem 10.2, one has an isomorphism $$\mathrm{Hom}(A,B) \otimes \mathbf{Z}_p \to \mathrm{Hom}(A(p),B(p))$$ with $A(p)$, $B(p)$ the $p$-divisible groups of $A$ and $B$ (a generalisation of Tate's Endomorphisms of Abelian Varieties over Finite Fields to $p = \mathrm{char}(K)$).

Is the Galois action on the $p$-adic Tate module also semisimple as it is for $\ell \neq p$?


1 Answer 1


You're just asking whether Frobenius acts semsimply on the $p$-adic Tate module.

We know from Tate's theorem that Frobenius acts semisimply on the $\ell$-adic Tate module, and hence satisfies some squarefree polynomial (its minimal polynomial). Now because the map from the ring of endomorphisms to the endomorphisms of the Tate module is injective, it also satisfies this squarefree polynomial in the ring of endomorphisms of the abelian variety. Hence it satisfies a squarefree polynomial, and thus is semisimple, when acting on the $p$-adic Tate module as well.

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    $\begingroup$ To prove that the minimal polynomial of the Frobenius action on $\mathbb Q_\ell$ is actually defined over $\mathbb Q$ (this is an essential feature of your argument), note that it equals the radical of the characteristic polynomial, and the latter is defined over $\mathbb Q$. $\endgroup$ Jul 13, 2017 at 23:02
  • $\begingroup$ Actually you don't need Tate's theorem to prove that the Frobenius of an abelian variety A over a finite field acts semi-simply. It is enough to observe that the Frobenius endomorphism in $End(A)\otimes \mathbb{Q}$ is a central element of a semi-simple algebra. $\endgroup$ Dec 12, 2021 at 16:00

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