$\newcommand{\F}{\mathbb{F}} 
\newcommand{\End}{\mathrm{End}} 
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\Z}{\mathbb{Z}}$
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](https://mathoverflow.net/questions/158410/abelian-varieties-with-the-same-cm-type-are-isogenous) or [Prop. 1.2.17](http://iml.univ-mrs.fr/%7Ekohel/phd/thesis_wilson.pdf). 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?



Isomorphic endomorphism algebras implies isogenous (for abelian varieties over finite fields)?
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 $E$.


