Endomorphism ring of simple ordinary abelian variety

Is there an example of an ordinary and simple abelian variety $A$ over an algebraically closed field $K$ (of characteristic $p>0$) such that ${\rm End}(A)$ is not commutative? Note that the answer is no if $K=\overline{\bf F}_p$ (in that case ${\rm End}(A)_{\bf Q}$ is a CM field). My question is for other fields.

• Maybe include the word "simple" in the title for emphasis. – nfdc23 Apr 11 '18 at 15:12
• @nfdc23: quite right. I changed it. – Damian Rössler Apr 11 '18 at 15:42
• Consider a moduli space of fake elliptic curves X over F_p. This carries a universal abelian surface A with an action by (an order in) a quaternion algebra over Q. Then take the generic fiber. – Raju Apr 11 '18 at 18:24
• @Raju: what is a fake elliptic curve? – Laurent Moret-Bailly Apr 12 '18 at 7:17
• @Rössler Here's a silly argument. So, let K be the function field of the Shimura curve (over F_p) and let A_K be the abelian variety. If it's not geom. simple, then it is isogenous to a product of elliptic curves after a finite extension L/K. This would yield family of elliptic curves over a smooth projective curve, contradiction. (I guess I'm using Néron-Ogg-Shafarevich to show that the elliptic curve has good reduction over the curve whose function field is L.) – Raju Apr 12 '18 at 11:37

Let $D$ be a non-split quaternion algebra over $\mathbb{Q}$, split at $p$ and $\infty$. Let $\mathcal{O}$ be a maximal order (I think these are all conjugate). Then $\mathcal{O}^1$, the multiplicative group of norm 1 elements, embeds into $SL_2(\mathbb{R})$ by picking an isomorphism $\iota:D\otimes \mathbb{R}\cong M_{2\times 2}(\mathbb{R})$. In particular, $\mathcal{O}^1$ acts on the upper half plane $\mathbb{H}$. It turns out that this action is properly discontinuous and cocompact; form the quotient $\mathcal{X}=[\mathbb{H}/\mathcal{O}^1]$. This (Deligne-Mumford stack) is the moduli space of principally polarized abelian surfaces with multiplication by $\mathcal{O}$; these abelian surfaces are also sometimes called "fake elliptic curves (with multiplication by $\mathcal{O}$)".
It turns out this Shimura curve has a good integral model, and in particular may be reduced modulo $p$. Let $\pi:A\rightarrow X$ denote the reduction mod $p$ together with the universal fake elliptic curve and let $K:=\mathbb{F}_p(X)$ denote the function field of $X$.
First of all, $A$ is generically ordinary. Here is one way to see this: $A[p^{\infty}]$ has an action by $\mathcal{O}\otimes\mathbb{Z}_p\cong M_{2\times 2}(\mathbb{Z}_p)$. By projecting onto idempotents, we see that $A[p^{\infty}]\cong \mathcal{G}\oplus \mathcal{G}$ where $\mathcal{G}$ is a height 2, dimension 1 $p$-divisible group. Moreover, $\mathcal{G}$ is everywhere versally deformed along $X$ (this is essentially Serre-Tate). On the other hand, there are only two possible Newton Polygons by the numerical constraints on $\mathcal{G}$: ordinary and supersingular. Therefore $\mathcal{G}$ and hence $A$ is generically ordinary.
Now, $A_{K}$ is an ordinary abelian surface over $K$ with multiplication by $\mathcal{O}$. We just need to check that $A_{K}$ is geometrically simple. If it weren't, there would be a finite extension $L/K$ such that $A_L$ is isogenous to $E\times E'$. By Ogg-Shafarevich, $E$ and $E'$ both have good reduction over the smooth projective curve $Y$ corresponding to $L$. This implies that $E$ and $E'$ are isotrivial over $Y$, which is a contradiction (e.g. by monodromy).