Large Index Theorem for Endomorphism Rings of Drinfeld Modules
speaker: Sumita Garai (Penn State University)abstract: The theory of Drinfeld modules runs parallel to the theory of Elliptic curves, and our result was motivated by a similar result for Elliptic curves. Let $A=\mathbb{F}qT$ be the polynomial ring over $\mathbb{F}q$, and $F$ be the field of fractions of $A$. Let $\phi$ be a Drinfeld $A$-module of rank $r$ over $F$. For all but finitely many primes $\mathfrak{p}\lhd A$, one can reduce $\phi$ modulo $\mathfrak{p}$ to obtain a Drinfeld $A$-module $\phi\otimes\mathbb{F}{\mathfrak{p}}$ of rank $r$ over $\mathbb{F}{\mathfrak{p}}=A\mathfrak{p}$. It is known that the endomorphism ring $\mathcal{E}{\mathfrak{p}}=\text{End} {\mathbb{F}{\mathfrak{p}}}( \phi\otimes\mathbb{F}{\mathfrak{p}})$ 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 $\pi\mathfrak{p}\in \mathcal{E}\mathfrak{p}$ be the Frobenius endomorphism of $\phi\otimes\mathbb{F}\mathfrak{p}$. Then we have the inclusion of orders $A\pi_\mathfrak{p}\subset \mathcal{E}\mathfrak{p}\subset \mathcal{O}\mathfrak{p}$ in $K$. In a joint work with my advisor, Mihran Papikian, we showed that if $\phi$ is a Drinfeld Module without complex multiplication, 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 $\chi(\mathcal{E}{\mathfrak{p}}A\pi_\mathfrak{p})$ and $\mathfrak{m}$ divides the index $\chi(\mathcal{O}\mathfrak{p}\mathcal{E}\mathfrak{p})$. We also give an algorithm to compute $\mathcal{E}{\mathfrak{p}}$ in the rank-2 case.
timetable:
Tue 6 Nov, 14:30 - 15:00, Aula Dini
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