Endomorphism rings of reductions of Drinfeld modules

Abstract

Let $A=\mathbb{F}_q[T]$ be the polynomial ring over $\mathbb{F}_q$, and $F$ be the field of fractions of $A$. Let $ϕ$ be a Drinfeld $A$-module of rank $r\geq 2$ over $F$. For all but finitely many primes $\mathfrak{p}\lhd A$, one can reduce $ϕ$ modulo $\mathfrak{p}$ to obtain a Drinfeld $A$-module $ϕ\otimes\mathbb{F}_\mathfrak{p}$ of rank $r$ over $\mathbb{F}_\mathfrak{p}=A/\mathfrak{p}$. The endomorphism ring $\mathcal{E}_\mathfrak{p}=\mathrm{End}_{\mathbb{F}_\mathfrak{p}}(ϕ\otimes\mathbb{F}_\mathfrak{p})$ is an order in an imaginary field extension $K$ of $F$ of degree $r$. Let $\mathcal{O}_\mathfrak{p}$ be the integral closure of $A$ in $K$, and let $π_\mathfrak{p}\in \mathcal{E}_\mathfrak{p}$ be the Frobenius endomorphism of $ϕ\otimes\mathbb{F}_\mathfrak{p}$. Then we have the inclusion of orders $A[π_\mathfrak{p}]\subset \mathcal{E}_\mathfrak{p}\subset \mathcal{O}_\mathfrak{p}$ in $K$. We prove that if $\mathrm{End}_{F^\mathrm{alg}}(ϕ)=A$, then for arbitrary non-zero ideals $\mathfrak{n}, \mathfrak{m}$ of $A$ there are infinitely many $\mathfrak{p}$ such that $\mathfrak{n}$ divides the index $χ(\mathcal{E}_\mathfrak{p}/A[π_\mathfrak{p}])$ and $\mathfrak{m}$ divides the index $χ(\mathcal{O}_\mathfrak{p}/\mathcal{E}_\mathfrak{p})$. We show that the index $χ(\mathcal{E}_\mathfrak{p}/A[π_\mathfrak{p}])$ is related to a reciprocity law for the extensions of $F$ arising from the division points of $ϕ$. In the rank $r=2$ case we describe an algorithm for computing the orders $A[π_\mathfrak{p}]\subset \mathcal{E}_\mathfrak{p}\subset \mathcal{O}_\mathfrak{p}$, and give some computational data.