# Ordinary abelian varieties over a finite field

Let $q$ be a power of a prime $p$. Deligne's paper "Variétés abéliennes ordinaires sur un corps fini" seems to describe an equivalence of categories between

1. ordinary abelian varieties over a finite field $\mathbf F_q$,
2. complex abelian varieties equipped with an endomorphism $\pi$ which is a $q$-Weil number.

Is that true? The paper actually states the equivalence of categories between (1) and a category of free Z-modules. It sketches the relation with complex abelian varieties, but does it remain an equivalence of categories?

This question has been discussed a bit here: Canonical lifts from $\mathbb F_q$ and CM-theory. The answer seems to be "yes". I still ask this question because of the following, which looks like a contradiction to me:

Fix a $q$-Weil number $\pi$, let $K = \mathbf Q(\pi)$, and $\mathcal O_K$ the maximal order in $K$. Then, $\mathrm{Cl}(K)$ acts freely and transitively on the set of isomorphism classes of complex abelian varieties with endomorphism ring $\mathcal O_K$. On the other hand, this action of $\mathrm{Cl}(K)$ on the abelian varieties over $\mathbf F_q$ with endomorphism ring $\mathcal O_K$ is still free but not necessarily transitive (Waterhouse, Abelian varieties over finite fields, Theorem 5.3: the number of orbits is $2^s$ where $s$ is the number of primes factors of $p$ in $K_0$ staying prime in $K$, and $K_0$ is the real subfield of $K$).

Did I misinterpret Deligne's paper? Or is this $s$ actually always 0?

• For elliptic curves with CM by $K$, p splits if it has ordinary good reduction, and is inert if it has supersingular reduction. This suggests that the answer might be that $s = 0$. – Jesse Silliman Aug 4 '16 at 12:50
• Yes, this means that $s = 0$ for elliptic curves. But I am mostly interested in the general case. Does it generalise in a way that ensures that $s$ is always $0$ when the variety in ordinary? – Calodeon Aug 4 '16 at 12:56

Because the abelian variety is ordinary, $\pi$ has the property that it's $p$-adic valuation at every place is either $0$ or $1$ (this follows easily from condition (IV) on the first page of Deligne's paper.) That means no prime $p'$of $\mathbb Q(\pi)$ lying over $p$ is equal to its complex conjugate, because if the $p'$-adic valuation of $\pi$ is $1$ then the $\overline{p}'$-adic valuation of $\pi$ is $0$, and vice versa.
On the other hand, if a prime of $K_0$ remained a prime of $K$ then it would equal its complex conjugate.
Hence $s=0$.