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$\newcommand{\F}{\mathbb{F}} \newcommand{\End}{\mathrm{End}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\Z}{\mathbb{Z}}$ I would like to know if the following is true:

Proposition A : Let $A_1, A_2$ be two abelian varieties over a finite field $k$. If $\End_{\overline k}(A_1) \otimes_{\Z} \Q$ and $\End_{\overline k}(A_2) \otimes_{\Z} \Q$ are isomorphic (as $\Q$-algebras), then $A_1, A_2$ are isogenous over $\overline k$.

The converse holds: let $\phi : A_1 \to A_2$ be an isogeny of degree $m$. There is an isogeny $\psi : A_2 \to A_1$ such that $psi \circ \phi = [m]$ (see Poonen "Lectures on rational points", Proposition 4.1.19 ; this is not the dual isogeny!). Define a map $\End_{\overline k}(A_1) \otimes_{\Z} \Q \to \End_{\overline k}(A_2) \otimes_{\Z} \Q$ via $f \longmapsto \dfrac{1}{m} \phi \circ f \circ \psi $. It is an algebra isomorphism.

The result holds over $\mathbb C$, see here or Prop. 1.2.17. It holds for elliptic curves over a finite field: the supersingular case can be proved "by hand", using Tate isogeny theorem, while the ordinary case follows from Deuring's work on CM (lifting, etc.). Note that the curves do not need to be isogenous over the base field $k$. I skimmed through the paper Endomorphisms of Abelian Varieties over Finite Fields of Tate, but I did not find such a statement.

Maybe one could use Serre–Tate theory instead of Deuring, but it seems to be only available for ordinary abelian varieties. I am not aware of the details here anyway. If the claim does not hold, is there a suitable hypothesis to make it true?

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  • $\begingroup$ Did you mean to assume $A_i$ have CM? $\endgroup$ – Wojowu May 23 at 11:33
  • $\begingroup$ @Wojowu : Tate proved that any abelian variety over a finite field is a CM abelian variety (Prop. 16.27 in Moonen--Geer notes on Abelian Varieties). $\endgroup$ – Watson May 23 at 11:41
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    $\begingroup$ There are examples of abelian varieties with isomorphic endomorphism algebras but different $p$-ranks (see e.g. arxiv.org/pdf/2003.05380.pdf, §5.6) and $p$-ranks are isogenous invariants. If they still have the same endomorphism algebras over $\bar{k}$, they will provide counter-examples. $\endgroup$ – HYL May 23 at 13:09
  • $\begingroup$ @HYL : thanks! In fact, Example 5.19 therein gives an example of two geometrically simple abelian varieties of dimension 3 over $\Bbb F_2$ (lmfdb.org/Variety/Abelian/Fq/3/2/a_c_c, lmfdb.org/Variety/Abelian/Fq/3/2/d_f_h), which are not isogenous, but according to the LMFDB "All geometric endomorphisms are defined over $\Bbb F_{2}$" (namely the geometric endomorphism algebra is a CM number field of degree 6, lmfdb.org/NumberField/6.0.679024.1) $\endgroup$ – Watson May 24 at 19:37
  • $\begingroup$ Update: this question and the answer below gave rise to the following preprint arxiv.org/abs/2107.06432 :-) $\endgroup$ – Watson Jul 15 at 5:55
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Here is a counterexample to Proposition A provided by 8-dimensional abelian varieties $A_1$ and $A_2$ over a finite field $F_{p^2}$ where $p$ is any prime that is congruent to $1$ modulo $17$ (e.g., $p=103$). The corresponding endomorphism algebra is the $17$th cyclotomic field $E=Q(\zeta_{17})$. The congruence condition means that $p$ splits completely in $E$. It is known that $E/Q$ is a cyclic extension, the class number of $E$ is $1$ (see ``Introduction to cyclotomic fields" by Larry Washington) and any proper subfield of $E$ is totally real (because $17$ is a Fermat prime).

Let $O_E$ be the ring of integers in $E$ and $\iota: E \to E$ be the complex conjugation, which is the only element of order 2 in the cyclic Galois group $G:=Gal(E/Q)$ of order 16=2^4. (In particular, every nontrivial subgroup of $G$ contains $\iota$.)

Let S be the 16-element set of maximal ideals $\mathfrak{P}$ of integers $O_E$ with residual characteristic $p$. The group $G$ acts freely transitively on the set $S$ of maximal ideals in $O_E$ and $\Pi_{\mathfrak{P}\in S}\mathfrak{P}=p \cdot O_E$. Let H be the set of ideals $\mathfrak{B}$ of $O_E$ such that $\mathfrak{B}\cdot \iota(\mathfrak{B})=p \cdot O_E$. The set $H$ has $2^8=16^2$ elements. I claim that the natural action of $16$-element $G$ on $H$ is free. Indeed, if it's not free then there is $\mathfrak{B}\in H$ such that $\iota(\mathfrak{B})=\mathfrak{B}$ and therefore $$p \cdot O_E=\mathfrak{B}\cdot \iota(\mathfrak{B})=\mathfrak{B}^2,$$ which implies that $p$ is ramified in $E$, which is not the case. So, the action is free and therefore $H$ consists of $16$ orbits of $G$.

Now let's construct Weil's $p^2$-numbers $\pi_1$ and $\pi_2$, using $\mathfrak{B}_1, \mathfrak{B}_2 \in H$ that belong to different orbits of $G$. Let $z_1, z_2 \in O_E$ be generators of ideals $\mathfrak{B}_1$ and $\mathfrak{B}_2$ respectively. Then both $v_1=z_1 \iota(z_1)$ and $v_2=z_2 \iota(z_2)$ are ``real" (i.e., $\iota$-invariant) generators of $p\cdot O_E$, i.e., there are exist units $u_1,u_2 \in O_E^*$ such that $$v_1=p u_1, \ v_2=p u_2.$$ Clearly, both $u_1$ and $u_2$ are real totally positive. Now let us put $$\pi_1=z_1^2/u_1 \in O_E, \ \pi_2=z_2^2/u_2\in O_E.$$ Clearly, $$\pi_1\cdot \iota(\pi_1)=p^2=\pi_2\cdot \iota(\pi_2)$$ (recall that $\iota(u_1)=u_1$ and $\iota(u_2)=u_2$). Taking into account that $$\pi_1 O_E=\mathfrak{B}_1^2, \pi_2 O_E=\mathfrak{B}_2^2,$$ we conclude that $\pi_1$ and $\pi_2$ are not Galois-conjugate (and the same is true for powers $\pi_1^m$ and $\pi_2^m$ for any positive integer $m$). Clearly, $$\pi_1 \ne \iota(\pi_1), \ \pi_2 \ne \iota(\pi_2)$$ and therefore $$Q(\pi_1)=E= Q(\pi_2).$$ Now if $A_1$ (resp. $A_2$) is a simple abelian variety over $F_{p^2}$ attached (by Honda-Tate) to $\pi_i$ then the center of the division algebra $End(A_i)\otimes Q$ is isomorphic to $E$. Since $\pi_1$ and $\pi_2$ (and even their powers) are not Galois-conjugate then $A_1$ and $A_2$ are not isogenous over $F_{p^2}$ (and even over its algebraic closure). On the other hand, both $A_1$ and $A_2$ are obviously ordinary. Since they are simple, their endomorphism algebras are commutative, i.e., coincide with their centers and therefore both $End(A_1)\otimes Q$ and $End(A_2)\otimes Q$ are isomorphic to $E$, Hence, $$\dim(A_1)=[E:Q]/2=\dim(A_2),$$ i.e., both $A_1$ and $A_2$ are $8$-dimensional and $End(A_1)\otimes Q$ is isomorphic to $End(A_2)\otimes Q$.

Let me stress that both $A_1$ and $A_2$ remain simple over an algebraic closure of $F_{p^2}$ and their endomorphism algebras remain isomorphic to $E$.

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  • $\begingroup$ Thank you very much for this extended answer! I am going to read it, but just a quick question: you are considering the geometric endomorphism algebras, right (and not just the endomorphisms defined over $\mathbb F_{p^2}$)? Oh, I just saw that you mention this at the very end of your post. Because then your construction contradicts the other answer, if I understand correctly... ! $\endgroup$ – Watson May 24 at 16:15
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    $\begingroup$ You are welcome. Yes, you understand it correctly: it certainly contradicts. The source of a mistake in the other answer is an improper use of Tate's theorem. Indeed, this theorem describes the local $v$-adic (or $\mathfrak{P}$-)invariants of the division algebra of endomorphisms. While these invariants lie in $Q/Z$, the other answer tacitly (and erroneously) assumes that they are rational numbers. $\endgroup$ – Yuri Zarhin May 24 at 17:51
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This is true if $A_{1}$ and $A_{2}$ are geometrically simple. We may suppose that $A_{1}$ and $A_{2}$ have all their endomorphisms defined over $k$. Let $\pi_{i}$ be the Frobenius endomorphism of $A_{i}$. Tate's theorem (see his 1968 Bourbaki talk) shows that the numbers $ord_{v}(\pi)/ord_{v}(q)$ are the same for both abelian varieties, and these numbers determine $\pi$ up to a root of $1$ (as an element of a number field) because they determine all of its valuations. Extending the field raises $\pi$ to a power, and so gets rid of a root of $1$. Now apply Tate's (1966) theorem.

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    $\begingroup$ If they're not geometrically simple, the endomorphism algebra splits as a product of matrix algebras over division algebras, and applying your argument to those division algebras seems to do the trick. $\endgroup$ – Will Sawin May 23 at 13:43
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    $\begingroup$ Your answer is not correct (please see Prof. Zarhin's comment below). $\endgroup$ – Watson May 24 at 18:36

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