2
$\begingroup$

What is a reference for the following construction?

Let $K$ be a finite extension of $\Bbb F_q(\theta)$ and $M$ a Drinfeld module of rank $r$ defined over $K$. Its lattice $L(M)$ is a free $\Bbb F_q[\theta]$-module of rank $r$ in $\Bbb C_\infty$. Like for the analogous case of abelian varieties, the ratios of elements of a basis of $L(M)$ over $\Bbb F_q[\theta]$ are transcendental, if $M$ is not of complete multiplication, hence from the first sight there is no $Gal(\bar K/K)$-action on $L(M)$.

But we have $L(M)=Hom_{\Bbb C_\infty[T]}(M,Z_1)^\tau$, where $Z_1=\{\sum a_i T^i \mid a_i\in \Bbb C_\infty,\ a_i \to0\}.$

Let $f_*:=(f_1,\dots, f_r)$ be a basis of $M$ over $\Bbb C_\infty[T]$. Let $l\in L(M)$ and $l(f_i)=\sum_{j=0}^\infty a_{ij}T^j$. The coefficients $a_{ij}$ are roots to polynomial equations with coefficients in $K$, hence the $Gal(\bar K/K)$-action on $L(M)$ is defined. It is well-defined, i.e. it does not depend on a choice of $f_*$ (for the same $K$-structure).

Example: the Carlitz module $\mathfrak C$. Its lattice $L(\mathfrak C)$ is the set of roots to $\exp_{\mathfrak C}(x)=0$. We know that $$ \exp_{\mathfrak C}(x)=x+\frac{1}{\theta^q-\theta}x^q+\frac{1}{(\theta^{q^2}-\theta^{q})(\theta^{q^2}-\theta)}x^{q^2}+\cdots $$ A basis element $\pi_q$ of $L(\mathfrak C)$ is transcendental, its ord is $\frac{-q}{q-1}$.

From another side, let $l\in Hom_{\Bbb C_\infty[T]}(M,Z_1)^\tau$ be a basis element, and coefficients $a_j$ are defined by $l(f_1)=\sum_{j=0}^\infty a_{j}T^j$. They are roots to the equations $$ a_0^q+\theta a_0=0; \quad a_1^q+\theta a_1-a_0=0; \quad a_2^q+\theta a_2-a_1=0 \quad \text{etc.} $$ $Gal(\bar K/K)$ acts on them. There exists one (up to a factor from $\Bbb F_q^*$) sequence $a_0, \ a_1,\dots$ such that $ord\ a_n=n-\frac1{q-1}$. We have $\pi_q= \lim_{n\to\infty} \theta^{n+1} a_n$.

$\endgroup$
1
  • $\begingroup$ Galois Theories by Borceux and Janelidze has the underlying theory for this kind of Galois construction somewhere in it, I'm not sure about this specific example though. $\endgroup$
    – Alec Rhea
    Mar 24, 2023 at 6:30

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.