# Surjectivity of the adelic Galois Representation associated to a Drinfeld
module of prime rank

Chien-Hua Chen

## Abstract

In this paper, let $\phi$ be the Drinfeld module over $\mathbb{F}_{q}(T)$ of
prime rank $r$ defined by

$\phi_{T}=T+\tau^{r-1}+T^{q-1}\tau^{r}.$

We prove that under certain condition on $\mathbb{F}_{q}$, the adelic Galois
representation

${\rho}_{\phi}:{\rm{Gal}}(\mathbb{F}_{q}(T)^{{\rm{sep}}}/\mathbb{F}_{q}(T))\longrightarrow\varprojlim_{\mathfrak{a}}{\rm{Aut}}(\phi[\mathfrak{a}])\cong{\rm{GL_{r}}}(\widehat{A})$

is surjective.

##  1 Introduction

In [Ser72], Serre proved his famous “Open Image Theorem” for elliptic curves
over number field without complex multiplication. Restricted to elliptic
curves over $\mathbb{Q}$, the theorem says

######  Theorem ([Ser72]).

If $E$ is an elliptic curve over $\mathbb{Q}$ without complex multiplication,
then its associated adelic Galois representation

$\rho_{E}:{\rm{Gal}}(\bar{\mathbb{Q}}/\mathbb{Q})\rightarrow\varprojlim_{{m}}{\rm{Aut}}(E[m])\cong{\rm{GL}}_{2}(\hat{\mathbb{Z}})$

has open image in ${\rm{GL}}_{2}(\hat{\mathbb{Z}})$

The following is then a natural question:

######  Question.

Is it possible to have an elliptic curve $E$ over $\mathbb{Q}$ such that
$\rho_{E}(\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q}))={\rm{GL}}_{2}(\hat{\mathbb{Z}})$?

Serre [Ser72] showed that the answer to the above question is negative, the
main reason being the existence of a non-trivial relation between the
$2$-torsion $E[2](\bar{\mathbb{Q}})$ and some $m$-torsion
$E[m](\bar{\mathbb{Q}})$. Although Serre proved the adelic surjectivity
problem is negative for elliptic curves over $\mathbb{Q}$, Greicious [Gre10]
proved it is true for elliptic curves over a “good enough” number field.
Several mathematicians studied the generalization of adelic surjectivity
problem for abelian varieties of dimension $\geqslant 2$ over a number field,
cf. [AD20], [Hal11], [Zyw15], and [LSTX20]. Currently, it is known that there
is a genus $3$ hyperelliptic curve over $\mathbb{Q}$ whose Jacobian variety
satisfies the adelic surjectivity. Not much is known for adelic surjectivity
of higher dimensional abelian varieties.

Let $A=\mathbb{F}_{q}[T]$, $F=\mathbb{F}_{q}(T)$ and
$\hat{A}=\varprojlim_{\mathfrak{a}}A/\mathfrak{a}$. Pink and Rütsche [PR09a]
proved the Drinfeld module analogue of Serre’s open image theorem:

######  Theorem ([PR09a]).

Let $\phi$ be a Drinfeld $A$-module over $F$ of rank $r$ without complex
multiplication, then its associated adelic Galois representation

$\rho_{\phi}:{\rm{Gal}}(F^{\rm{sep}}/F)\rightarrow\varprojlim_{\mathfrak{a}}{\rm{Aut}}(\phi[\mathfrak{a}])\cong{\rm{GL}}_{r}(\widehat{A})$

has open image in ${\rm{GL}}_{r}(\hat{A})$

Therefore, it is reasonable to study the adelic surjectivity problem for
Drinfeld modules over $F$ of rank $r$. The answer for rank $r=1$ is positive,
which follows from Hayes’ work [Hay74] on the function field analogue of class
field theory. More precisely, the Carlitz module $C_{T}=T+\tau$ has surjective
adelic Galois image. Moreover, the answer for rank $r=2$ is also positive,
assuming that $q=p^{e}\geqslant 5$ is an odd prime power. This was proved by
Zywina [Zyw11] for the rank-$2$ Drinfeld module
$\phi_{T}=T+\tau-T^{q-1}\tau^{2}$. Remark that for the case $q=2$ and $r=2$,
the author [Che21] proved there is no rank-$2$ Drinfeld $A$-module over $F$
with surjective adelic Galois representation due to a similar reason as
Serre’s arguments for elliptic curves over $\mathbb{Q}$. Therefore, some
restrictions on $q$ is necessary.

In [Che20], we have proved the adelic sujectivity of Galois representation
associated to a Drinfeld module of rank $3$ defined by
$\varphi_{T}=T+\tau^{2}+T^{q-1}\tau^{3}$. Thus for the case of Drinfeld
modules of rank $r\geqslant 4$, one may expect the same result will hold for
Drinfeld module $\phi$ over $F$ defined by
$\phi_{T}=T+\tau^{r-1}+(-1)^{r-1}T^{q-1}\tau^{r}$. Our main result is the
following theorem that shows the expectation is true for prime rank Drinfeld
modules:

######  Main Theorem.

Let $q=p^{e}$ be a prime power, $A=\mathbb{F}_{q}[T]$, and
$F=\mathbb{F}_{q}(T)$. Assume $r\geqslant 3$ is a prime number and $q\equiv
1\mod r$, there is a constant $c=c(r)\in\mathbb{N}$ depending only on $r$ such
that for $p>c(r)$ the following statement is true ${\rm{:}}$

Let $\phi$ be a Drinfeld $A$-module over $F$ defined by
$\phi_{T}=T+\tau^{r-1}+T^{q-1}\tau^{r}$. Then the adelic Galois representation

${\rho}_{\phi}:{\rm{Gal}}(\mathbb{F}_{q}(T)^{{\rm{sep}}}/\mathbb{F}_{q}(T))\longrightarrow\varprojlim_{\mathfrak{a}}{\rm{Aut}}(\phi[\mathfrak{a}])\cong{\rm{GL_{r}}}(\widehat{A})$

is surjective.

The general idea comes from the proof for $r=3$. However, when we try to adapt
the strategy of the proof for $r=3$ to $r\geqslant 3$, several technical
problems arise. To explain these problems we briefly recall the idea of the
proof for $r=3$. The problems happened when we try to prove the mod
$\mathfrak{l}$ Galois representations are irreducible and surjective for all
prime ideals $\mathfrak{l}$ of $A$.

To prove irreducibility in the rank-3 case, we aim for a contradiction by
assuming the mod $\mathfrak{l}$ representation
$\bar{\rho}_{\varphi,\mathfrak{l}}$ is reducible. Hence the characteristic
polynomial $\bar{P}_{\varphi,\mathfrak{p}}(x)\in\mathbb{F}_{\mathfrak{l}}[x]$
of Frobenius elements
$\bar{\rho}_{\varphi,\mathfrak{l}}({\rm{Frob}}_{\mathfrak{p}})$ must contain a
linear factor for all prime $\mathfrak{p}\neq(T)\ {\rm{or}}\ \mathfrak{l}$.
Since the degree is $3$, we can explicitly write down a linear factor of
$\bar{P}_{\varphi,\mathfrak{p}}(x)$ when $\deg_{T}(\mathfrak{p})=1$.
Furthermore, we can also determine the characteristic polynomials concretely.
Hence a contradiction can be deduced by comparing coefficients of
$\bar{P}_{\varphi,\mathfrak{p}}(x)$ and its factorization. In the case
$r\geqslant 4$, the assumption of reducibility does not imply the
characteristic polynomial of a Frobenius element would contain a linear
factor.

Following the strategy for the rank-3 case, one might approach the problem of
surjectivity by contradiction. We assume the image
${\rm{Im}}\bar{\rho}_{\varphi,\mathfrak{l}}(G_{F})$ is a proper subgroup of
${\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}})$. Thus
${\rm{Im}}\bar{\rho}_{\varphi,\mathfrak{l}}(G_{F})$ must be contained in some
maximal subgroup of ${\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}})$. By our
knowledge of ${\rm{Im}}\bar{\rho}_{\varphi,\mathfrak{l}}(I_{T})$ (see Lemma
3.2), we can see that $|{\rm{Im}}\bar{\rho}_{\varphi,\mathfrak{l}}(G_{F})|$ is
divisible by certain power of $|A/\mathfrak{l}|$. Hence we can use this
property to rule out possible maximal subgroups. In rank-$3$ case, we can rule
out all possible maximal subgroups of
${\rm{GL}}_{3}(\mathbb{F}_{\mathfrak{l}})$. As the rank increases, we have
some maximal subgroups that can not be ruled out by merely considering their
sizes. The reason is that the growth rate of
$|{\rm{Im}}\bar{\rho}_{\varphi,\mathfrak{l}}(I_{T})|$ is much slower than the
growth rate of $p$-power component of
$|{\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}})|$ when $r$ increases.

Another harder problem is the classification of maximal subgroups in
${\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}})$. From [BHRD13] Theorem 2.2.19, we
can see that those maximal subgroups are divided into $9$ classes in
Aschbacher’s theorem. The first $8$ classes (geometric classes) have general
description, but the ninth class (special class) doesn’t have a known
description for arbitrary $r$ so far. (This is also the reason why the authors
could only describe all the maximal subgroups for low dimensional finite
classical groups.)

Fortunately, if we restrict to the case where $r$ is a prime number, then we
can combine Pink’s work (see [PR09a], section 3) on surjectivity of mod
$\mathfrak{l}$ representations

$\bar{\rho}_{\phi,\mathfrak{l}}:G_{F}\longrightarrow{\rm{Aut}}(\phi[\mathfrak{l}])\cong{\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}})$

with Aschbacher’s theorem ([BHRD13], Theorem 2.1.5) to prove “irreducibility
of mod $\mathfrak{l}$ Galois representations” toward “surjectivity of mod
$\mathfrak{l}$ Galois representations” assuming the characteristic of $F$ is
large enough. This procedure is described in section 3 and 4.

In section 5, we prove the irreducibility of the Galois representation. On the
other hand, we prove the surjectivity of mod $(T)$ Galois representation
directly using a result of Abhyankar.

The proof toward $\mathfrak{l}$-adic surjectivity is similar to the rank-$3$
case and is proved in section 6. In section 7, we prove the adelic
surjectivity under a further assumption $q\equiv 1\mod r$. The proof for
adelic surjectivity is similar to the rank-$3$ case as well.

##  2 Preliminaries

###  2.1 Notation

  * •

$q=p^{e}$ is a prime power with $p\geqslant 5$

  * •

$A=\mathbb{F}_{q}[T]$

  * •

$F=\mathbb{F}_{q}(T)$

  * •

$F^{{\rm{sep}}}$= separable closure of $F$

  * •

$F^{{\rm{alg}}}$= algebraic closure of $F$

  * •

$G_{F}={\rm{Gal}}(F^{{\rm{sep}}}/F)$

  * •

$\mathfrak{l}=(l)$ a prime ideal of $A$, and define
$\deg_{T}(\mathfrak{l})=\deg_{T}(l)$

  * •

$A_{\mathfrak{p}}=$ completion of A at the nonzero prime ideal
$\mathfrak{p}\triangleleft A$

  * •

$\widehat{A}=\underset{\mathfrak{a}\triangleleft
A}{\varprojlim}A/\mathfrak{a}$

  * •

$F_{\mathfrak{p}}=$ fraction field of $A_{\mathfrak{p}}$

  * •

$\mathbb{F}_{\mathfrak{p}}=A/\mathfrak{p}$

###  2.2 Drinfeld module over a field

Let $K$ be a field, we call $K$ an $A$-field if $K$ is equipped with a
homomorphism $\gamma:A\rightarrow K$. Let $K\\{\tau\\}$ be the ring of skew
polynomials satisfying the commutation rule $c^{q}\cdot\tau=\tau\cdot c$.
A Drinfeld $A$-module over $K$ of rank $r\geqslant 1$ is a ring homomorphism

$\displaystyle\phi$ $\displaystyle:A\longrightarrow K\\{\tau\\}$
$\displaystyle\ \ \
a\mapsto\phi_{a}=\gamma(a)+\sum^{r\cdot{\rm{deg}}(a)}_{i=1}g_{i}(a)\tau^{i}.$

It is uniquely determined by
$\phi_{T}=\gamma(T)+\sum^{r}_{i=1}g_{i}(T)\tau^{i}$, where $g_{r}(T)\neq 0$.
$\ker(\gamma)$ is called the $A$-characteristic of $K$, and we say $K$ has
generic characteristic if $\ker(\gamma)=0$.

######  Proposition 2.1.

Let $\phi$ be a Drinfeld module over $K$ of rank $r$ with nonzero
$A$-characteristic $\mathfrak{p}$. For each $a\in A$, we may write $\phi$ as
$\phi_{a}=c(a)\tau^{m(a)}+\cdots+C(a)\tau^{M(a)}$. There is a unique integer
$0<h\leqslant r$ such that $m(a)=hv_{\mathfrak{p}}(a)$ for all nonzero $a\in
A$, where $v_{\mathfrak{p}}$ is the $\mathfrak{p}$-adic valuation of $F$.

This integer $h$ is called the height of $\phi$.

###### Proof.

See [Gos96] Proposition 4.5.7. ∎

An isogeny from a Drinfeld module $\phi$ to another Drinfeld module $\psi$
over $K$ is an element $u\in K\\{\tau\\}$ such that
$u\cdot\phi_{a}=\psi_{a}\cdot u\ \ \forall\ a\in A$. Hence the endomorphism
ring of $\phi$ over $K$ is defined as

${\rm{End}}_{K}(\phi)=\\{u\in K\\{\tau\\}\ |\ u\cdot\phi_{T}=\psi_{T}\cdot
u\\}.$

The Drinfeld module $\phi$ gives $K^{{\rm{alg}}}$ an $A$-module structure,
where $a\in A$ acts on $K^{{\rm{alg}}}$ via $\phi_{a}$. We use the notation
${}^{\phi}K^{{\rm{alg}}}$ to emphasize the action of $A$ on $K^{{\rm{alg}}}$.
The $a$-torsion $\phi[a]=\\{{\rm{zeros}}\ {\rm{of}}\
\phi_{a}(x)=\gamma(a)x+\sum^{r\cdot{\rm{deg}}(a)}_{i=1}g_{i}(a)x^{q^{i}}\\}\subseteq
K^{alg}$. The action $b\cdot\alpha=\phi_{b}(\alpha)\ \forall\ b\in
A,\forall\alpha\in\phi[a]$ also gives $\phi[a]$ an $A$-module structure.

######  Proposition 2.2.

Let $\phi$ be a rank $r$ Drinfeld module over $K$ and $\mathfrak{a}$ an ideal
of $A$,

  1. 1.

If $\phi$ has $A$-characteristic prime to $\mathfrak{a}$, then the
$A/\mathfrak{a}$-module $\phi[\mathfrak{a}]$ is free of rank $r$

  2. 2.

If $\phi$ has nonzero $A$-characteristic $\mathfrak{p}$, let $h$ be the height
of $\phi$, then the $A/\mathfrak{p}$-module $\phi[\mathfrak{p}^{e}]$ is free
of rank $r-h$ for all $e\in\mathbb{Z}_{\geqslant 1}$.

###### Proof.

See [Gos96] Proposition 4.5.7. ∎

######  Note 2.3.

From now on, we consider $K=F$ and $\gamma:A\rightarrow F$ is the natural
injection map.

Let $\phi$ be a rank $r$ Drinfeld module over $F$ of generic characteristic,
then $\phi_{a}(x)$ is separable, so we have $\phi[a]\subseteq F^{{\rm{sep}}}$.
This implies that $\phi[a]$ has a $G_{F}$-module structure. Given a nonzero
prime ideal $\mathfrak{l}$ of $A$, we can consider the $G_{F}$-module
$\phi[\mathfrak{l}]$. We obtain the so-called mod $\mathfrak{l}$ Galois
representation

$\bar{\rho}_{\phi,\mathfrak{l}}:G_{F}\longrightarrow{\rm{Aut}}(\phi[\mathfrak{l}])\cong
GL_{r}(\mathbb{F}_{\mathfrak{l}}).$

Taking inverse limit with respect to $\mathfrak{l}^{i}$, we have the
$\mathfrak{l}$-adic Galois representation

${\rho}_{\phi,\mathfrak{l}}:G_{F}\longrightarrow\varprojlim_{i}{\rm{Aut}}(\phi[\mathfrak{l^{i}}])\cong{\rm{GL_{r}}}(A_{\mathfrak{l}}).$

Combining all representations together, we get the adelic Galois
representation

${\rho}_{\phi}:G_{F}\longrightarrow\varprojlim_{\mathfrak{a}}{\rm{Aut}}(\phi[\mathfrak{a}])\cong{\rm{GL_{r}}}(\widehat{A}).$

###  2.3 Carlitz module

The Carlitz module is the Drinfeld module $C:A\longrightarrow F\\{\tau\\}$ of
rank $1$ defined by $C_{T}=T+\tau$.

######  Proposition 2.4.

(Hayes [Hay74]) For a nonzero ideal $\mathfrak{a}$ of $A$, the Galois
representation

$\bar{\rho}_{C,\mathfrak{a}}:G_{F}\longrightarrow{\rm{Aut}}(C[\mathfrak{a}])\cong(A/\mathfrak{a})^{*}$

is surjective. This implies the adelic Galois representation of Carlitz module
is surjective. Moreover, for prime ideals $\mathfrak{p}$ of $A$ such that
$\mathfrak{p}\not|\;\mathfrak{a}$, we have
$\bar{\rho}_{C,\mathfrak{a}}({{Frob}}_{\mathfrak{p}})\equiv\mathfrak{p}\mod\mathfrak{a}$.

###  2.4 Reduction of Drinfeld modules

Let $K$ be a local field with uniformizer $\pi$, valuation ring $R$, unique
maximal ideal $\mathfrak{p}:=(\pi)$, normalized valuation $v$ and residue
field $\mathbb{F}_{\mathfrak{p}}$. Let $\phi:A\longrightarrow K\\{\tau\\}$ be
a Drinfeld module of rank $r$. We say that $\phi$ has stable reduction if
there is a Drinfeld module $\phi^{\prime}:A\longrightarrow R\\{\tau\\}$ such
that

  1. 1.

$\phi^{\prime}$ is isomorphic to $\phi$ over $K$;

  2. 2.

$\phi^{\prime}$ mod $\mathfrak{p}$ is still a Drinfeld module (i.e.
$\phi^{\prime}_{T}$ mod $\mathfrak{p}$ has ${\rm{deg}}_{\tau}\geqslant 1$ ).

$\phi$ is said to have stable reduction of rank $r_{1}$ if $\phi$ has stable
reduction and $\phi$ mod $\mathfrak{p}$ has rank $r_{1}$.
$\phi$ is said to have good reduction if $\phi$ has stable reduction and
$\phi$ mod $\mathfrak{p}$ has rank $r$.

######  Remark 2.5.

We sometimes denote $\phi\mod\mathfrak{p}$ by
$\phi\otimes\mathbb{F}_{\mathfrak{p}}$.

The Drinfeld module analogue of Nèron-Ogg-Shafarevich is the following:

######  Proposition 2.6.

([Tak82], Theorem 1) Let $\phi:A\longrightarrow K\\{\tau\\}$ be a Drinfeld
module and $\mathfrak{l}$ be a nonzero prime ideal different from the
$A$-characteristic of $\phi\otimes\mathbb{F}_{\mathfrak{p}}$. Then $\phi$ has
good reduction if and only if the $\mathfrak{l}$-adic Galois representation is
unramified at $\mathfrak{p}$. In other words,
$\rho_{\phi,\mathfrak{l}}(I_{\mathfrak{p}})=1$ where $I_{\mathfrak{p}}$ is the
inertia subgroup of $G_{K}$.

Let $u:\phi\rightarrow\psi$ be an isogeny between Drinfeld modules over $K$.
We study the reduction type of the isogenous Drinfeld module. The isogeny $u$
induces an $G_{K}$-equivariant isomorphism between rational Tate modules

$u:V_{\mathfrak{l}}(\phi)\rightarrow V_{\mathfrak{l}}(\psi),$

where $V_{\mathfrak{l}}(\phi):=T_{\mathfrak{l}}(\phi)\otimes
F_{\mathfrak{l}}$. Suppose $\mathfrak{l}$ is different from the
$A$-characteristic of $\phi\otimes\mathbb{F}_{\mathfrak{p}}$, then the
Drinfeld module analogue of Nèron-Ogg-Shafarevich implies $\psi$ has good
reduction at $\mathfrak{p}$. As a result, isogenous Drinfeld modules over a
local field either both have good reduction or bad reduction.

On the other hand, we prove that isogenous Drinfeld modules also preserve
stable bad reduction under a condition on the inertia action.

######  Proposition 2.7.

Let $u:\phi\rightarrow\psi$ be an isogeny between Drinfeld modules over $K$.
Suppose there is a prime ideal $\mathfrak{l}$ of $A$, and $\mathfrak{l}$ is
different from the $A$-characteristic of
$\phi\otimes\mathbb{F}_{\mathfrak{p}}$. Assume further that $\text{the inertia
group $I_{\mathfrak{p}}$ acts on $V_{\mathfrak{l}}(\phi)$ via a group of
unipotent matrices},$ then the isogenous Drinfeld module $\psi$ has stable
reduction at $\mathfrak{p}$.

###### Proof.

We may assume $\psi$ is defined over the valuation ring $R$ after replacing
$\psi$ by an isomorphic copy. Moreover, the assumption that $I_{\mathfrak{p}}$
acts on $V_{\mathfrak{l}}(\phi)$ by unipotent matrices implies that the action
of $I_{\mathfrak{p}}$ on $V_{\mathfrak{l}}(\psi)$ is also unipotent.
Therefore, $I_{\mathfrak{p}}$ acts by unipotent matrices on
$\psi[\mathfrak{l}]$, which means the ramification index of
$K(\psi[\mathfrak{l}])/K$ is a power on $p$.

Suppose $\psi$ does not have stable reduction over $K$, then any extension
$L/K$ over which $\psi$ has stable reduction has ramification index divisible
by some prime not equal to $p$. Therefore, if $\psi$ does not have stable
reduction over $K$, then $\psi$ does not have stable reduction over
$K(\psi[\mathfrak{l}])$ either, so we may assume that $\psi[\mathfrak{l}]$ is
rational over $K$.

On the other hand, consider the Newton Polygon ${\rm
NP}(\psi_{\mathfrak{l}}(x)/x)$ of the polynomial $\psi_{\mathfrak{l}}(x)/x$.
If $\psi[\mathfrak{l}]$ is rational over K, then the roots of
$\psi_{\mathfrak{l}}(x)$ have integer valuations, so the slope of the first
line segment of ${\rm NP}(\psi_{\mathfrak{l}}(x)/x)$ is an integer. Otherwise,
we have the slope of the first line segment of ${\rm
NP}(\psi_{\mathfrak{l}}(x)/x)$ is a simplified fraction $\frac{c}{d}$ with
denominator $d\neq 1$. Thus there are roots of $\psi_{\mathfrak{l}}(x)/x$ with
valuation equal to $\frac{1}{d}$, which is a contradiction.

Now we write

$\psi_{\mathfrak{l}}(x)=\sum_{i=0}^{r\cdot\deg_{T}(l)}g_{i}(l)x^{q^{i}},\text{
with }g_{0}(l)=l.$

The first line segment of ${\rm NP}(\psi_{\mathfrak{l}}(x)/x)$ has endpoints
$(0,0)$ and $(q^{m}-1,v(g_{m}(l)))$. The integrality of slope implies
$q^{m}-1\mid v(g_{m}(l))$. Hence after taking a suitable isomorphic copy of
$\psi$, we may assume $v(g_{m}(l))=0$. This implies one of the coefficient of
$\psi_{T}(x)$ other than $Tx$ must be a unit. Hence we deduce that $\psi_{T}$
has stable reduction over $K$, which is a contradiction.

∎

######  Remark 2.8.

Unlike abelian varieties (Corollaire 3.8 in chapter IX of [SGA72]), stable bad
reduction of Drinfeld module over local field does not imply the inertia group
acts on Tate module via unipotent matrices. The following is a counterexample:

Let $\mathfrak{p}=(T)$ and $\phi_{T}=T+\tau+T\tau^{2}$ be the Drinfeld
$A$-module defined over $F_{\mathfrak{p}}$. It’s clear that $\phi$ has stable
bad reduction of rank $1$. For any prime $\mathfrak{l}=(T-c)\neq\mathfrak{p}$,
the Tate uniformization shows that the inertia group $I_{\mathfrak{p}}$ acts
on $T_{\mathfrak{l}}(\phi)$ via matrices of the form

$\left(\begin{array}[]{c|c}1&*\\\ \hline\cr 0&c\end{array}\right).$

We know
$\det\circ\rho_{\phi,\mathfrak{l}}(I_{\mathfrak{p}})=\rho_{\psi,\mathfrak{l}}(I_{\mathfrak{p}})$,
where $\psi$ is the rank-$1$ Drinfeld module $\psi_{T}=T-T\tau$ by Proposition
7.1 in [vdH04]. Now we claim that $\rho_{\psi,\mathfrak{l}}(I_{T})$ is
nontrivial. We observe the Newton polygon of $\psi_{\mathfrak{l}}(x)/x$ with
respect to the valuation $v_{\mathfrak{p}}$, the polygon is a single line with
slope equal to $\frac{1}{q-1}$. Hence the Galois extension
$F_{\mathfrak{p}}(\psi[\mathfrak{l}])/F_{\mathfrak{p}}$ ramifies. This implies
the inertia group $I_{T}$ acts nontrivially on $\psi[\mathfrak{l}]$, hence
acts nontrivially on $T_{\mathfrak{l}}(\psi)$ as well. Therefore,
$\det\circ\rho_{\phi,\mathfrak{l}}(I_{\mathfrak{p}})=\rho_{\psi,\mathfrak{l}}(I_{\mathfrak{p}})$
is nontrivial, so $c\neq 1$.

###  2.5 Determinant of $\rho_{\phi}$

Let $\phi$ be a Drinfeld module over $F$ defined by

$\phi_{T}=T+g_{1}\tau+g_{2}\tau^{2}+\cdots+g_{r}\tau^{r}.$

Let $\mathfrak{p}\neq\mathfrak{l}$ be a prime of good reduction of $\phi$, the
$\mathfrak{l}$-adic Galois representation ${\rho}_{\phi,\mathfrak{l}}$ is
unramified at $\mathfrak{p}$ by Proposition 2.6. Therefore, the matrix
${\rho}_{\phi,\mathfrak{l}}({\rm{Frob}}_{\mathfrak{p}})\in{\rm{GL}}_{r}(A_{\mathfrak{l}})$
is well-defined up to conjugation, so we can consider the characteristic
polynomial
$P_{\phi,\mathfrak{p}}(x)=\det(xI-{\rho}_{\phi,\mathfrak{l}}({\rm{Frob}}_{\mathfrak{p}}))$
of the Frobenius element ${\rm{Frob}_{\mathfrak{p}}}$.
The polynomial $P_{\phi,\mathfrak{p}}(x)$ has coefficients in $A$ which are
independent of the choice of $\mathfrak{l}$. Moreover,
$P_{\phi,\mathfrak{p}}(x)$ is equal to the characteristic polynomial of
Frobenius endomorphism of $\phi\otimes\mathbb{F}_{\mathfrak{p}}$ acting on
$T_{\mathfrak{l}}(\phi\otimes\mathbb{F}_{\mathfrak{p}})$. We may write
$P_{\phi,\mathfrak{p}}(x)$ as follows:

$P_{\phi,\mathfrak{p}}(x)=a_{r}+a_{r-1}x+a_{r-2}x^{2}+\cdots+a_{1}x^{r-1}+x^{r}\in
A[x].$

The constant term $a_{r}$ is equal to
$(-1)^{r}\det\circ\bar{\rho}_{\phi,\mathfrak{a}}({\rm{Frob}}_{\mathfrak{p}}).$

######  Proposition 2.9.

([Yu95], Theorem 1) For $1\leqslant i\leqslant r$, we have
$\deg(a_{i})\leqslant\frac{i\cdot\deg(\mathfrak{p})}{r}$.

######  Proposition 2.10.

([HY00], p.268)

$a_{r}=\epsilon(\phi)\cdot\mathfrak{p}.$

Here
$\epsilon(\phi)=(-1)^{r}(-1)^{{{\rm{deg}}_{T}\mathfrak{p}}(r+1)}{\rm{Nr}}_{\mathbb{F}_{\mathfrak{p}}/\mathbb{F}_{q}}(g_{r})^{-1}$,
which belongs to $\mathbb{F}_{q}^{*}$ and is independent of $\mathfrak{a}$.

Let $\phi$ be the Drinfeld module over $F$ defined by
$\phi_{T}=T+\tau^{r-1}+T^{q-1}\tau^{r}$, where $r$ is an odd prime. By [vdH04]
Proposition 7.1, we have

$\det\circ\bar{\rho}_{\phi,\mathfrak{a}}=\bar{\rho}_{\psi,\mathfrak{a}},$

where $\psi$ is defined by $\psi_{T}=T+T^{q-1}\tau$. It’s clear that $\psi$ is
isomorphic over $F$ to the Carlitz module $C_{T}=T+\tau$, so we have the
following commutative diagram:

$\begin{array}[]{ccc}G_{F}&\xrightarrow{\bar{\rho}_{\phi,\mathfrak{l}}}&{\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}})\\\
\|&&\downarrow\det\\\
G_{F}&\xrightarrow{\bar{\rho}_{C,\mathfrak{l}}}&(A/\mathfrak{l})^{*}\end{array}.$

Hence we can deduce the following Corollary from Proposition 2.4.

######  Corollary 2.11.

For prime ideals $\mathfrak{p}$ of $A$ such that $\mathfrak{p}\not|\
\mathfrak{a}$, we have
$\det\circ\bar{\rho}_{\phi,\mathfrak{a}}({\rm{Frob}}_{\mathfrak{p}})\equiv\mathfrak{p}\mod\mathfrak{a}$.

###  2.6 Drinfeld-Tate uniformization

Let $\phi:A\longrightarrow A_{\mathfrak{p}}\\{\tau\\}$ be a Drinfeld module. A
$\phi$-lattice is a finitely generated free $A$-submodule of
${}^{\phi}F_{\mathfrak{p}}^{{\rm{sep}}}$ and stable under
$G_{F_{\mathfrak{p}}}$-action. Here the discreteness is with respect to the
topology of the local field $F_{\mathfrak{p}}^{{\rm{sep}}}$.
A Tate datum over $A_{\mathfrak{p}}$ is a pair $(\phi,\Gamma)$ where $\phi$ is
a Drinfeld module over $A_{\mathfrak{p}}$ and $\Gamma$ is a $\phi$-lattice.
Two pairs $(\phi,\Gamma)$ and $(\phi^{\prime},\Gamma^{\prime})$ of Tate datum
are isomorphic if there is an isomophism from $\phi$ to $\phi^{\prime}$ such
that the induced $A$-module homomorphism
${}^{\phi}F_{\mathfrak{p}}^{{\rm{sep}}}\longrightarrow{{}^{\phi^{\prime}}F_{\mathfrak{p}}^{{\rm{sep}}}}$
gives an $A$-module isomorphism $\Gamma\longrightarrow\Gamma^{\prime}$.

######  Proposition 2.12.

(Drinfeld) Let $r_{1}$, $r_{2}$ be two positive integers. There is a one-to-
one correspondence between two sets:

  1. 1.

The set of $F_{\mathfrak{p}}-$isomorphism classes of Drinfeld modules $\phi$
over $F_{\mathfrak{p}}$ of rank $r:=r_{1}+r_{2}$ with stable reduction of rank
$r_{1}$

  2. 2.

The set of $F_{\mathfrak{p}}-$isomorphism classes of Tate datum
$(\psi,\Gamma)$ where $\psi$ is a Drinfeld module over $A_{\mathfrak{p}}$ of
rank $r_{1}$ with good reduction, and $\Gamma$ is a $\psi$-lattice of rank
$r_{2}$.

###### Proof.

See chapter 4 in [Leh09]. ∎

######  Remark 2.13.

Fix $a\in A-\mathbb{F}_{q}$. From the proof of Proposition 2.12, we have the
following two properties :

  1. (i)

There is a $G_{F_{\mathfrak{p}}}$-equivariant short exact sequence of
$A$-modules:

$0\longrightarrow\psi[a]\longrightarrow\phi[a]\xrightarrow{\psi_{a}}\Gamma/a\Gamma\longrightarrow
0.$

  2. (ii)

There is an element $u\in{A_{\mathfrak{p}}\\{\\{\tau\\}\\}}$ such that
$u\psi_{a}=\phi_{a}u$, here ${A_{\mathfrak{p}}\\{\\{\tau\\}\\}}$ is the set of
power series in $\tau$ with coefficients in $A_{\mathfrak{p}}$. This element
$u$ induces an isomorphism of $A[G_{F_{\mathfrak{p}}}]$-modules from
$\psi_{a}^{-1}(\Gamma)/\Gamma$ to $\phi[a]$ by mapping $z+\Gamma$ to $u(z)$.

Moreover, the function $u$ can be expressed in the following ways

    1. 1.

$u(x)=x+u_{1}x^{q}+u_{2}x^{q^{2}}+\cdots+u_{i}x^{q^{i}}+\cdots$, and $u_{i}$
belongs to the maximal ideal of $A_{\mathfrak{p}}$ for all $i$.

    2. 2.

When $r_{2}=1$, $u(x)=x\cdot\displaystyle\prod_{0\neq a\in
A}(1-\frac{x}{\phi_{a}(\gamma)})$ where $\gamma$ is a generator of $\Gamma$.

##  3 Image of $\bar{\rho}_{\phi,\mathfrak{l}}$ and Aschbaher’s Theorem

From now on, we work under the following assumptions:

Let $r$ be a prime number, $A=\mathbb{F}_{q}[T]$, and $F=\mathbb{F}_{q}(T)$,
where $q=p^{e}$ and $p>r!$. Let $\phi$ be a Drinfeld $A$-module over $F$ of
rank $r$ with generic characteristic defined by
$\phi_{T}=T+\tau^{r-1}+T^{q-1}\tau^{r}$. Let $\mathfrak{l}$ be a place of $F$
where $\phi$ has good reduction at $\mathfrak{l}$. We denote
$\bar{\rho}_{\phi,\mathfrak{l}}(G_{F})$ by $\Gamma_{\mathfrak{l}}$, so
$\Gamma_{\mathfrak{l}}$ is a subgroup of
${\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}})$

By Aschbacher’s Theorem(Theorem B.1 in Appendix B), $\Gamma_{\mathfrak{l}}$
lies in one of the Aschbacher classes. In this subsection, we’ll show that
classes $\mathcal{C}_{2}$, $\mathcal{C}_{3}$, $\mathcal{C}_{4}$,
$\mathcal{C}_{7}$, and $\mathcal{C}_{8}$ can be ruled out.

######  Lemma 3.1.

There is a basis of $\phi[\mathfrak{l}]$ such that

$\bar{\rho}_{\phi,\mathfrak{l}}(I_{T})\subseteq\left\\{\left(\begin{array}[]{cccc}1&&&b_{1}\\\
&\ddots&&\vdots\\\ &&\ddots&b_{r-1}\\\ &&&1\end{array}\right),\
b_{i}\in\mathbb{F}_{\mathfrak{l}}\ \forall\ 1\leqslant\ i\leqslant\
r-1\right\\}.$

###### Proof.

As $\phi_{T}=T+\tau^{r-1}-T^{q-1}\tau^{r}$ has stable reduction at $(T)$ of
rank $r-1$, we may use Tate uniformization to obtain a Tate datum
$(\psi,\Gamma)$. Here $\psi$ has good reduction of rank $r-1$ and $\Gamma$ has
rank $1$. The Drinfeld module $\psi:A\rightarrow A_{(T)}\\{\tau\\}$ has rank
$r-1$ of good reduction, so the Galois representation
$\bar{\rho}_{\psi,\mathfrak{l}}:G_{F_{(T)}}\rightarrow{\rm{Aut}}(\psi[\mathfrak{l}])$
is unramified. Thus there is a basis $\\{w_{1},w_{2},\cdots,w_{r-1}\\}$ of
$\psi[\mathfrak{l}]$ such that $\sigma(w_{i})=w_{i}\ \forall\sigma\in I_{T},\
\forall\ 1\leqslant i\leqslant r-1$.
Now since $\Gamma$ is a free $A$-module of rank $1$, we may fix a generator
$\gamma$ of $\Gamma$. Choose $z\in F_{(T)}^{\rm{sep}}$ such that
$\psi_{\mathfrak{l}}(z)=\gamma$. The fact that $\Gamma$ is stable under the
Galois action implies that there is a character
$\chi_{\Gamma}:G_{F_{(T)}}\rightarrow\mathbb{F}_{q}^{*}$ such that
$\sigma(\gamma)=\chi_{\Gamma}(\sigma)\gamma,\ \forall\sigma\in I_{T}$. By
Remark 2.13(ii), we have

$\psi_{\mathfrak{l}}(\sigma(z))=\sigma(\psi_{\mathfrak{l}}(z))=\sigma(\gamma)=\chi_{\Gamma}(\sigma)\gamma=\chi_{\Gamma}(\sigma)\psi_{\mathfrak{l}}(z)=\psi_{\mathfrak{l}}(\chi_{\Gamma}(\sigma)z)$

Thus $\sigma(z)-\chi_{\Gamma}(\sigma)z\in\psi[\mathfrak{l}]$, therefore there
are some elements $b_{\sigma,1},b_{\sigma,2},\cdots,b_{\sigma,r-1}$ in
$\mathbb{F}_{\mathfrak{l}}$ such that

$\sigma(z)=b_{\sigma,1}w_{1}+b_{\sigma,2}w_{2}+\cdots+b_{\sigma,r-1}w_{r-1}+\chi_{\Gamma}(\sigma)z.$

Therefore, the action of $\sigma\in I_{T}$ on
$\psi_{\mathfrak{l}}^{-1}(\Gamma)/\Gamma$ with respect to the basis
$\\{w_{1}+\Gamma,w_{2}+\Gamma,\cdots,w_{r-1}+\Gamma,z+\Gamma\\}$ is of the
form $\left(\begin{array}[]{cccc}1&&&b_{\sigma,1}\\\ &\ddots&&\vdots\\\
&&1&b_{\sigma,r-1}\\\ &&&\chi_{\Gamma}(\sigma)\end{array}\right)$ .

Because of our choice of $\phi_{T}$, we have the determinant of mod
$\mathfrak{l}$ Galois representation $\det\circ\bar{\rho}_{\phi,\mathfrak{l}}$
is equal to the mod $\mathfrak{l}$ Galois representation of the Carlitz module
$\bar{\rho}_{C,\mathfrak{l}}$. Hence $\chi_{\Gamma}(\sigma)=1$ for all
$\sigma\in I_{T}$ since the Carlitz module has good reduction at $(T)$.
Therefore, we have deduced

$\bar{\rho}_{\phi,\mathfrak{l}}(I_{T})\subseteq\left\\{\left(\begin{array}[]{cccc}1&&&b_{1}\\\
&\ddots&&\vdots\\\ &&\ddots&b_{r-1}\\\ &&&1\end{array}\right),\
b_{i}\in\mathbb{F}_{\mathfrak{l}}\ \text{ for all }1\leqslant\ i\leqslant\
r-1\right\\}$

with respect to the basis
$\\{w_{1}+\Gamma,w_{2}+\Gamma,\cdots,w_{r-1}+\Gamma,z+\Gamma\\}$.

∎

######  Lemma 3.2.

The inclusion in Lemma 3.1 is an equality. In particular,
$\bar{\rho}_{\phi,\mathfrak{l}}(I_{T})$ has order equal to
$|\mathbb{F}_{\mathfrak{l}}|^{r-1}$.

###### Proof.

So far we have from Lemma 3.1 that

$\bar{\rho}_{\phi,\mathfrak{l}}(I_{T})\subseteq\left\\{\left(\begin{array}[]{cccc}1&&&b_{1}\\\
&\ddots&&\vdots\\\ &&\ddots&b_{r-1}\\\ &&&1\end{array}\right),\
b_{i}\in\mathbb{F}_{\mathfrak{l}}\ \forall\ 1\leqslant\ i\leqslant\
r-1\right\\}.$

Let $F^{un}_{(T)}$ be the maximal unramified extension of $F_{(T)}$ in
$F^{sep}_{(T)}$. Since $I_{T}={\rm{Gal}}(F^{sep}_{(T)}/F^{un}_{(T)})$ by
definition, we have
$\bar{\rho}_{\phi,\mathfrak{l}}(I_{T})\cong{\rm{Gal}}(F^{un}_{(T)}(\phi[\mathfrak{l}])/F^{un}_{(T)})$.
By Remark 2.13 and Lemma 3.1, we can derive that
$F^{un}_{(T)}(\phi[\mathfrak{l}])=F^{un}_{(T)}(w_{1},w_{2},\cdots,w_{r-1},z)$.
Furthermore, $w_{i}$ belongs to $F^{un}_{(T)}$ because
$\bar{\rho}_{\psi,\mathfrak{l}}:G_{F_{(T)}}\rightarrow{\rm{Aut}}(\psi[\mathfrak{l}])$
is unramified for all $1\leqslant i\leqslant r-1$. Therefore,
$F^{un}_{(T)}(\phi[\mathfrak{l}])=F^{un}_{(T)}(z)$ and its ramification index
$e[F^{un}_{(T)}(z):F^{un}_{(T)}]$ at least the order of $v(z)$ in
$\mathbb{Q}/\mathbb{Z}$ where $v$ is the normalized valuation of $F_{(T)}$.

From Remark 2.13(ii), we have
$\phi_{\mathfrak{l}}(x)=\mathfrak{l}x\cdot{\displaystyle\prod_{0\neq\gamma^{\prime}\in\psi_{\mathfrak{l}}^{-1}(\Gamma)/\Gamma}}(1-\frac{x}{u(\gamma^{\prime})})$.
We compute the leading coefficients on both sides up to units of $A_{(T)}$:

$T^{(q-1)({\displaystyle\sum_{i=1}^{\deg_{T}(\mathfrak{l})}q^{r(i-1)}})}=\pm\frac{\mathfrak{l}}{{\displaystyle\prod_{0\neq\gamma^{\prime}\in\psi_{\mathfrak{l}}^{-1}(\Gamma)/\Gamma}}u(\gamma^{\prime})}.$
(1)

As $\\{w_{1}+\Gamma,w_{2}+\Gamma,\cdots,w_{r-1}+\Gamma,z+\Gamma\\}$ is a basis
of $\psi_{\mathfrak{l}}^{-1}(\Gamma)/\Gamma$, an element
$\gamma^{\prime}\in\psi_{\mathfrak{l}}^{-1}(\Gamma)/\Gamma$ can be written as
a $\mathbb{F}_{\mathfrak{l}}$\- linear combination of $w_{i}+\Gamma$ and
$z+\Gamma$. Thus we have

$\begin{array}[]{ccl}(q-1)({\displaystyle\sum_{i=1}^{\deg_{T}(\mathfrak{l})}q^{r(i-1)}})&=&v\left(\pm\frac{\mathfrak{l}}{{\displaystyle\prod_{0\neq\gamma^{\prime}\in\psi_{\mathfrak{l}}^{-1}(\Gamma)/\Gamma}}u(\gamma^{\prime})}\right)\\\
&=&0-\left({{\displaystyle\sum_{a_{1},a_{2},\cdots,a_{r-1},b\in\mathbb{F}_{\mathfrak{l}}\
{\rm{not\ all\
zero}}}}}v(u(a_{1}w_{1}+a_{2}w_{2}+\cdots+a_{r-1}w_{r-1}+bz))\right).\end{array}$

Recall from the proof of Lemma 3.1 that we have
$\gamma=\psi_{\mathfrak{l}}(z)$, where $\gamma$ is a generator of the rank $1$
discrete $A$-module $\Gamma$. The discreteness of $\Gamma$ forces
$v(\gamma)<0$, which implies $v(z)<0$ because every coefficient of
$\psi_{\mathfrak{l}}(x)$ has nonnegative valuation. Moreover, the valuations
$v(w_{i})$ are all nonnegative because they are roots of
$\psi_{\mathfrak{l}}(x)$.

The proof of $v(u(w_{i}))=v(w_{i})$ is easy because $v(w_{i})$ are non-
negative and non-constant coefficients of $u(x)$ lie in the maximal ideal of
$A_{(T)}$. For the proof of $v(u(z))=v(z)$, we use the chosen $\gamma$ to
compare coefficients of two expressions of $u(x)$ in Remark 2.13(ii). We have

$v(u_{n})\geqslant-(q^{n}-1)v(\gamma).$

We also know that $v(\gamma)=q^{(r-1)(deg_{T}(\mathfrak{l}))}v(z)$. Hence

$v(u_{n}z^{q^{n}})=v(u_{n})+q^{n}v(z)\geqslant-(q^{n}-1)v(\gamma)+q^{(n-(r-1)(\deg_{T}(\mathfrak{l})))}v(\gamma).$

For $n\geqslant 1$, $v(u_{n}z^{q^{n}})$ is always non-negative. Therefore,
$v(u(z))=v(z)$.

Thus we can compute the valuation of
$u(a_{1}w_{1}+a_{2}w_{2}+\cdots+a_{r-1}w_{r-1}+bz)$ explicitly:

$v(u(a_{1}w_{1}+a_{2}w_{2}+\cdots+a_{r-1}w_{r-1}+bz))=\begin{cases}q^{(r-1)i}v(z),&{\rm{if}}\
b\neq 0,\deg_{T}(b)=i\\\
v(a_{1}w_{1}+a_{2}w_{2}+\cdots+a_{r-1}w_{r-1}),&{\rm{if}}\ b=0\end{cases}.$

Hence we have

$\begin{array}[]{ccl}(q-1)({\displaystyle\sum_{i=1}^{\deg_{T}(\mathfrak{l})}q^{r(i-1)}})&=&-(q^{(r-1)(\deg_{T}(\mathfrak{l}))})[(q-1)({\displaystyle\sum_{i=1}^{\deg_{T}(\mathfrak{l})}q^{r(i-1)}})]v(z)\\\
&&-v\left({{\displaystyle\prod_{a_{1},a_{2},\cdots,a_{r-1}\in\mathbb{F}_{\mathfrak{l}}\
{\rm{not\ all\
zero}}}}}(a_{1}w_{1}+a_{2}w_{2}+\cdots+a_{r-1}w_{r-1})\right).\end{array}$

In fact, ${{\displaystyle\prod_{a_{1},a_{2},\cdots,a_{r-1}\in\mathbb{F}_{q}\
{\rm{not\ all\ zero}}}}}(a_{1}w_{1}+a_{2}w_{2}+\cdots+a_{r-1}w_{r-1})$ is
equal to the constant term $\mathfrak{l}$ of $\psi_{\mathfrak{l}}(x)/x$.
Finally, we are able to compute the valuation
$v(z)=-\frac{1}{q^{(r-1)(\deg_{T}(\mathfrak{l}))}}$, its order in
$\mathbb{Q}/\mathbb{Z}$ is equal to
$q^{(r-1)(\deg_{T}(\mathfrak{l}))}=|A/\mathfrak{l}|^{r-1}$. Therefore,
$|A/\mathfrak{l}|^{r-1}\leqslant e[F^{un}_{(T)}(z):F^{un}_{(T)}]$ and so
$|\bar{\rho}_{\phi,\mathfrak{l}}(I_{T})|\geqslant|A/\mathfrak{l}|^{r-1}$.
Combining with Lemma 3.1, we have
$|\bar{\rho}_{\phi,\mathfrak{l}}(I_{T})|=|A/\mathfrak{l}|^{r-1}$.

∎

Now we can apply the Aschbacher’s theorem (Theorem B.1 in the Appendix) to
rule out certain Aschbacher classes.

  * •

$\Gamma_{\mathfrak{l}}$ does not lie in Class $\mathcal{C}_{2}$.

###### Proof.

Suppose $\Gamma_{\mathfrak{l}}$ lies in $\mathcal{C}_{2}$, then
$\Gamma_{\mathfrak{l}}$ acting on $\mathbb{F}_{\mathfrak{l}}^{r}$ must be of
the type ${\rm{GL}}_{1}(\mathbb{F}_{\mathfrak{l}})\wr S_{r}$, the wreath
product of $\mathrm{GL}_{1}$ and the symmetric group $S_{r}$. Therefore, we
have $|\Gamma_{\mathfrak{l}}|$ divides
$|\mathbb{F}_{\mathfrak{l}}^{*}|\cdot{r!}$. We then get a contradiction from
the fact that $p$ doesn’t divide $|\mathbb{F}_{\mathfrak{l}}^{*}|\cdot{r!}$. ∎

  * •

$\Gamma_{\mathfrak{l}}$ does not lie in Class $\mathcal{C}_{3}$

###### Proof.

Suppose $\Gamma_{\mathfrak{l}}$ lies in $\mathcal{C}_{3}$, then the action of
$\Gamma_{\mathfrak{l}}$ on $\mathbb{F}_{\mathfrak{l}}^{r}$ must be of the type
${\rm{GL}}_{1}(\mathbb{F}_{\mathfrak{l}^{r}})$, here
$\mathbb{F}_{\mathfrak{l}^{r}}$ is a degree-$r$ extension over
$\mathbb{F}_{\mathfrak{l}}$. Thus we have $|\Gamma_{\mathfrak{l}}|$ divides
${\rm{GL}}_{1}(\mathbb{F}_{\mathfrak{l}^{r}})$, which contradicts to the fact
that $p$ divides $|\Gamma_{\mathfrak{l}}|$. ∎

  * •

$\Gamma_{\mathfrak{l}}$ does not lie in Class $\mathcal{C}_{4}$

###### Proof.

This is clear by the primality of $r$. Since $\mathbb{F}_{\mathfrak{l}}^{r}$
cannot have such tensor product decomposition
$\mathbb{F}_{\mathfrak{l}}^{r}=V_{1}\otimes V_{2}$, where $V_{1}$(resp.
$V_{2}$) is a $\mathbb{F}_{\mathfrak{l}}$-subspace of
$\mathbb{F}_{\mathfrak{l}}^{r}$ of dimension $n_{1}$(resp. $n_{2}$) and
$1<n_{1}<\sqrt{r}$. ∎

  * •

$\Gamma_{\mathfrak{l}}$ does not lie in Class $\mathcal{C}_{7}$

###### Proof.

Suppose $\Gamma_{\mathfrak{l}}$ lies in $\mathcal{C}_{2}$, then the action of
$\Gamma_{\mathfrak{l}}$ on $\mathbb{F}_{\mathfrak{l}}^{r}$ must be in a
quotient of the standard wreath product
${\rm{GL}}_{1}(\mathbb{F}_{\mathfrak{l}})\wr S_{r}$. Hence we still have
$|\Gamma_{\mathfrak{l}}|$ divides $|\mathbb{F}_{\mathfrak{l}}^{*}|\cdot{r!}$,
a contradiction.

∎

  * •

$\Gamma_{\mathfrak{l}}$ does not lie in Class $\mathcal{C}_{8}$

###### Proof.

Suppose $\Gamma_{\mathfrak{l}}$ lies in $\mathcal{C}_{8}$, then
$\Gamma_{\mathfrak{l}}$ would preserve a non-degenerate classical form on
$\mathbb{F}_{\mathfrak{l}}^{r}$ up to scalar multiplication. By classical form
we mean symplectic form, unitary form or quadratic form. As $r$ is odd,
$\mathbb{F}_{\mathfrak{l}}^{r}$ can only have unitary form or quadratic form
structure. We refer to section 1.5 of [BHRD13] for the definitions and
properties for classical forms on a vector space.

    * Case1.

$\Gamma_{\mathfrak{l}}$ preserves a non-degenerate unitary form on
$\mathbb{F}_{\mathfrak{l}}^{r}$ up to scalar multiplication.

In this case, we are dealing with unitary form $<\cdot,\cdot>$ on
$\mathbb{F}_{\mathfrak{l}}^{r}$. There is a basis $\mathcal{B}$ such that
$<\cdot,\cdot>$ corresponds to the identity matrix $I_{r}$( see
Proposition1.5.29 in [BHRD13]). Let $M\in\Gamma_{\mathfrak{l}}$ be a matrix
with respect to the basis $\mathcal{B}$, the fact that $M$ preserves
$<\cdot,\cdot>$ up to a scalar multiplication can be interpreted as the
equality:

$M\cdot M^{\sigma\top}=\lambda_{M}\cdot I.$

Where ${\rm{id}}\neq\sigma\in{\rm{Aut}}(\mathbb{F}_{\mathfrak{l}})$ depends
only on the unitary form $<\cdot,\cdot>$, $\sigma^{2}=1$ , and
$\lambda_{M}\in\mathbb{F}_{\mathfrak{l}}^{*}$ depends on $M$.

Therefore, we can compare the characteristic polynomials of such $M$ and
$M^{-1}$. As $\lambda_{M}\cdot M^{-1}=M^{\sigma\top}$, we have

$\det(xI-M^{\sigma})=\frac{-x^{r}}{\det(\lambda_{M}^{-1}\cdot
M)}\det(\frac{1}{x}I-\lambda_{M}^{-1}\cdot
M)=\frac{-x^{r}}{\det(M)}\det(\frac{\lambda_{M}}{x}I-M)$

Now we consider $M=\bar{\rho}_{\phi,\mathfrak{l}}({\rm{Frob}}_{\mathfrak{p}})$
where $\mathfrak{p}=(T-c)\neq(T)\ {\rm{or}}\ \mathfrak{l}$. The characteristic
polynomial of $\bar{\rho}_{\phi,T}({\rm{Frob}}_{(T-c)})$ is congruent to
$P_{\phi,(T-c)}(x)$ modulo $\mathfrak{l}$. By Proposition 2.10, we may write

$P_{\phi,(T-c)}(x)=-(T-c)+a_{r-1}x+a_{r-2}x^{2}+\cdots+a_{1}x^{r-1}+x^{r}\in
A[x].$

Proposition 2.9 then implies all the $a_{i}$’s are belong to $\mathbb{F}_{q}$.
Because $P_{\phi,(T-c)}(x)$ is also the characteristic polynomial of Frobenius
endomorphism of $\phi\otimes\mathbb{F}_{\mathfrak{p}}$ acting on
$T_{\mathfrak{l}}(\phi\otimes\mathbb{F}_{\mathfrak{p}})$, we have

$-(\phi\otimes\mathbb{F}_{\mathfrak{p}})_{T-c}+(\phi\otimes\mathbb{F}_{\mathfrak{p}})_{a_{r-1}}\tau+(\phi\otimes\mathbb{F}_{\mathfrak{p}})_{a_{r-2}}\tau^{2}+\cdots+(\phi\otimes\mathbb{F}_{\mathfrak{p}})_{a_{1}}\tau^{r-1}+\tau^{r}=0.$

As $\phi_{T}=T+\tau^{r-1}+T^{q-1}\tau^{r}$, we have
$(\phi\otimes\mathbb{F}_{\mathfrak{p}})_{T-c}=\tau^{r-1}+\tau^{r}$. Thus

$a_{r-1}=a_{r-2}=\cdots=a_{2}=0\text{ and }a_{1}=1.$

Hence the characteristic polynomial of
$\bar{\rho}_{\phi,\mathfrak{l}}({\rm{Frob}}_{(T-c)})\in\mathbb{F}_{\mathfrak{l}}[x]$
is

$\bar{P}_{\phi,(T-c)}(x)=x^{r}+x^{r-1}-\bar{\mathfrak{p}},\text{ where
$\bar{\mathfrak{p}}$ denotes the reduction of $\mathfrak{p}$ modulo
$\mathfrak{l}$}.$

Therefore, we have

$x^{r}+x^{r-1}-\sigma(\bar{\mathfrak{p}})=\det(xI-M^{\sigma})=\frac{-x^{r}}{\det(M)}\det(\frac{\lambda_{M}}{x}I-M)=x^{r}-\frac{\lambda_{M}^{r-1}}{\bar{\mathfrak{p}}}x-\frac{\lambda_{M}^{r}}{\bar{\mathfrak{p}}}$

This is a contradiction because the polynomials on both sides can not be
equal.

    * Case 2

$\Gamma_{\mathfrak{l}}$ preserves a non-degenerate quadratic form on
$\mathbb{F}_{\mathfrak{l}}^{r}$ up to scalar multiplication

In this case, we are dealing with quadratic form $<\cdot,\cdot>$ on
$\mathbb{F}_{\mathfrak{l}}^{r}$. There is a basis $\mathcal{B}$ such that
$<\cdot,\cdot>$ corresponds to the the $r\times r$ matrix

$A:=\left(\begin{array}[]{cccc}1&0&0&0\\\ 0&\ddots&0&0\\\ 0&0&1&0\\\
0&0&0&d\end{array}\right),$

here $d$ is either $1$ or some nonsquare element in
$\mathbb{F}_{\mathfrak{l}}^{*}$ . Let $M\in\Gamma_{\mathfrak{l}}$ be a matrix
with respect to the basis $\mathcal{B}$, then $M$ preserves $<\cdot,\cdot>$ up
to a scalar multiplication can be interpreted into the following equalities:

$M\cdot A\cdot M^{\top}=\lambda_{M}\cdot A,$

where $\lambda_{M}\in\mathbb{F}_{\mathfrak{l}}^{*}$ depends on $M$. Thus we
have $M^{\top}=\lambda_{M}\cdot A^{-1}M^{-1}A$, which implies $M$ and
$\lambda_{M}\cdot M^{-1}$ have the same characteristic polynomial.

$\det(xI-M)=\frac{-x^{r}}{\det(\lambda_{M}^{-1}\cdot
M)}\det(\frac{1}{x}I-\lambda_{M}^{-1}\cdot
M)=\frac{-x^{r}}{\det(M)}\det(\frac{\lambda_{M}}{x}I-M).$

Consider $M=\bar{\rho}_{\phi,\mathfrak{l}}({\rm{Frob}}_{\mathfrak{p}})$ where
$\mathfrak{p}=(T-c)\neq(T)\ {\rm{or}}\ \mathfrak{l}$, we can deduce

$x^{r}+x^{r-1}-\bar{\mathfrak{p}}=\det(xI-M)=\frac{-x^{r}}{\det(M)}\det(\frac{\lambda_{M}}{x}I-M)=x^{r}-\frac{\lambda_{M}^{r-1}}{\bar{\mathfrak{p}}}x-\frac{\lambda_{M}^{r}}{\bar{\mathfrak{p}}}.$

Hence we have a contradiction.

∎

In summary, we’ve proved the following Proposition:

######  Proposition 3.3.

Let $r$ be a prime number, $A=\mathbb{F}_{q}[T]$, and $F=\mathbb{F}_{q}(T)$,
where $q=p^{e}$ and $p>r!$. Let $\phi$ be a Drinfeld $A$-module over $F$ of
rank $r$ with generic characteristic defined by
$\phi_{T}=T+\tau^{r-1}+T^{q-1}\tau^{r}$. Let $\mathfrak{l}$ be a place of $F$
where $\phi$ has good reduction at $\mathfrak{l}$. Denote
$\bar{\rho}_{\phi,\mathfrak{l}}(G_{F})$ by $\Gamma_{\mathfrak{l}}$. Then
$\Gamma_{\mathfrak{l}}$ can only lie in Aschbacher classes $\mathcal{C}_{1}$,
$\mathcal{C}_{5}$, $\mathcal{C}_{6}$, or $\mathcal{S}$.

##  4 Some algebraic group theory

Under the assumption in Proposition 3.3, we know that $\Gamma_{\mathfrak{l}}$
can only lie in Aschbacher classes $\mathcal{C}_{1}$, $\mathcal{C}_{5}$,
$\mathcal{C}_{6}$, or $\mathcal{S}$. Now we assume further that
$\Gamma_{\mathfrak{l}}$ is irreducible, then $\Gamma_{\mathfrak{l}}$ can not
lie in class $\mathcal{C}_{1}$. Hence $\Gamma_{\mathfrak{l}}$ lies in class
$\mathcal{C}_{5}$, $\mathcal{C}_{6}$, or $\mathcal{S}$, which implies
$\Gamma_{\mathfrak{l}}$ contains a subgroup that acts on
$\mathbb{F}_{\mathfrak{l}}^{r}$ absolutely irreducibly. Hence
$\Gamma_{\mathfrak{l}}$ acts on $\mathbb{F}_{\mathfrak{l}}^{r}$ absolutely
irreducibly.

For most part, the following arguments are the same as in [PR09a],
p.888-p.889.

Consider the following exact sequence $0\rightarrow
I_{\mathfrak{l}}^{p}\rightarrow I_{\mathfrak{l}}\rightarrow
I_{\mathfrak{l}}^{t}\rightarrow 0$ of inertia group $I_{\mathfrak{l}}$. Here
$I_{\mathfrak{l}}^{p}$ is the wild inertia group at $\mathfrak{l}$ and
$I_{\mathfrak{l}}^{t}$ is the tame inertia group at $\mathfrak{l}$. Let
$h_{\mathfrak{l}}$ be the height of $\phi\otimes\mathbb{F}_{\mathfrak{l}}$,
and let $\mathbb{F}_{n}$ be an extension of $\mathbb{F}_{\mathfrak{l}}$ in
$\bar{\mathbb{F}}_{\mathfrak{l}}$ with
$n:=|\mathbb{F}_{\mathfrak{l}}|^{h_{\mathfrak{l}}}$ elements. By [PR09b]
Proposition 2.7, we have up to conjugation

$\bar{\rho}_{\phi,\mathfrak{l}}(I_{\mathfrak{l}}^{t})=\left(\begin{array}[]{cc}\mathbb{F}_{n}^{*}&0\\\
0&1\end{array}\right)\subseteq\Gamma_{\mathfrak{l}},$

written in block matrices with size $h_{\mathfrak{l}}$, $r-h_{\mathfrak{l}}$.
Since $\mathbb{F}_{\mathfrak{l}}^{*}\neq\\{1\\}$, the centralizer of
$\bar{\rho}_{\phi,\mathfrak{l}}(I_{\mathfrak{l}}^{t})$ in
${\rm{GL}}_{r,\mathbb{F}_{\mathfrak{l}}}$ is

$\left(\begin{array}[]{c|c}T_{\mathfrak{l}}&0\\\ \hline\cr
0&{\rm{GL}}_{r-h_{\mathfrak{l}},\mathbb{F}_{\mathfrak{l}}}\end{array}\right).$

Here $T_{\mathfrak{l}}$ is the torus
${\rm{Res}}_{\mathbb{F}_{\mathfrak{l}}}^{\mathbb{F}_{n}}\mathbb{G}_{m,\mathbb{F}_{n}}$
in the algebraic group
${\rm{GL}}_{h_{\mathfrak{l}},\mathbb{F}_{\mathfrak{l}}}$. We embed
$T_{\mathfrak{l}}$ into ${\rm{GL}}_{r,\mathbb{F}_{\mathfrak{l}}}$ by setting

$T_{\mathfrak{l}}=\left(\begin{array}[]{c|c}T_{\mathfrak{l}}&0\\\ \hline\cr
0&{\rm{I}}_{r-h_{\mathfrak{l}}}\end{array}\right)\subseteq{\rm{GL}}_{r,\mathbb{F}_{\mathfrak{l}}}.$

The $\Gamma_{\mathfrak{l}}$-conjugacy class of $T_{\mathfrak{l}}$ in
${\rm{GL}}_{r,\mathbb{F}_{\mathfrak{l}}}$ is independent of the choice of
place $\bar{\mathfrak{l}}$ of $\bar{F}$ above $\mathfrak{l}$.

Let $H_{\mathfrak{l}}^{\circ}$ be the connected algebraic subgroup of
${\rm{GL}}_{r,\mathbb{F}_{\mathfrak{l}}}$ generated by $\gamma
T_{\mathfrak{l}}\gamma^{-1}$ for all $\gamma\in\Gamma_{\mathfrak{l}}$( see
[Hum75], Proposition 7.5 ). Let $H_{\mathfrak{l}}$ be the algebraic subgroup
of ${\rm{GL}}_{r,\mathbb{F}_{\mathfrak{l}}}$ generated by
$H_{\mathfrak{l}}^{\circ}$ and $\Gamma_{\mathfrak{l}}$. Thus we have

$H_{\mathfrak{l}}^{\circ}\subseteq
H_{\mathfrak{l}}\subseteq{\rm{GL}}_{r,\mathbb{F}_{\mathfrak{l}}}.$

Since $H_{\mathfrak{l}}$ contains $\Gamma_{\mathfrak{l}}$, $H_{\mathfrak{l}}$
acts on $\mathbb{F}_{\mathfrak{l}}^{r}$ absolutely irreducibly. Now we can
make the following claim

  * Claim:

$H_{\mathfrak{l}}^{\circ}$ acts on $\mathbb{F}_{\mathfrak{l}}^{r}$ absolutely
irreducibly. In other words,
$H_{\mathfrak{l},\bar{\mathbb{F}}_{\mathfrak{l}}}^{\circ}$ acts on
$\bar{\mathbb{F}}_{\mathfrak{l}}^{r}$ irreducibly.

###### Proof of claim.

Let $W$ be a non-trivial
$H_{\mathfrak{l},\bar{\mathbb{F}}_{\mathfrak{l}}}^{\circ}$-invariant subspace
of $\bar{\mathbb{F}}_{\mathfrak{l}}^{r}$ of minimal dimension. As
$H_{\mathfrak{l},\bar{\mathbb{F}}_{\mathfrak{l}}}^{\circ}$ is normalized by
$\Gamma_{\mathfrak{l}}$, we know that $\gamma W$ is also invariant under
$H_{\mathfrak{l},\bar{\mathbb{F}}_{\mathfrak{l}}}^{\circ}$ for all
$\gamma\in\Gamma_{\mathfrak{l}}$. Consider
$V=\sum_{\gamma\in\Gamma_{\mathfrak{l}}}\gamma W$, it is invariant under
$\Gamma_{\mathfrak{l}}$. Thus we have
$\bar{\mathbb{F}}_{\mathfrak{l}}^{r}=\sum_{\gamma\in\Gamma_{\mathfrak{l}}}\gamma
W$ because $\Gamma_{\mathfrak{l}}$ acts on
$\bar{\mathbb{F}}_{\mathfrak{l}}^{r}$ irreducibly. Since each $\gamma W$ is
irreducible under the action of
$H_{\mathfrak{l},\bar{\mathbb{F}}_{\mathfrak{l}}}^{\circ}$, there is a natural
number $s_{\mathfrak{l}}$ and a decomposition

${\bar{\mathbb{F}}_{\mathfrak{l}}}^{r}=W_{1}\oplus W_{2}\oplus\cdots\oplus
W_{s_{\mathfrak{l}}}$

into irreducible
$H_{\mathfrak{l},\bar{\mathbb{F}}_{\mathfrak{l}}}^{\circ}$-subspaces which are
conjugate under $H_{\mathfrak{l},\bar{\mathbb{F}}_{\mathfrak{l}}}.$

Let $t_{\mathfrak{l}}$ be the common dimension of $W_{i}$. Then
$H_{\mathfrak{l},\bar{\mathbb{F}}_{\mathfrak{l}}}^{\circ}$ acts on
$\bar{\mathbb{F}}_{\mathfrak{l}}^{r}$ through matrices in
${\rm{GL}}_{t_{\mathfrak{l}},\bar{\mathbb{F}}_{\mathfrak{l}}}^{s_{\mathfrak{l}}}$.
Thus
$H_{\mathfrak{l},\bar{\mathbb{F}}_{\mathfrak{l}}}^{\circ}\subseteq{\rm{GL}}_{t_{\mathfrak{l}},\bar{\mathbb{F}}_{\mathfrak{l}}}^{s_{\mathfrak{l}}}$.
The algebraic subgroup of ${\rm{GL}}_{r,\bar{\mathbb{F}}_{\mathfrak{l}}}$
mapping a summand $W_{i}$ to some summand $W_{j}$ is isomorphic to
${\rm{GL}}_{t_{\mathfrak{l}},\bar{\mathbb{F}}_{\mathfrak{l}}}^{s_{\mathfrak{l}}}\rtimes
S_{s_{\mathfrak{l}}}$.

######  Lemma 4.1.

$H_{\mathfrak{l},\bar{\mathbb{F}}_{\mathfrak{l}}}\subseteq{\rm{GL}}_{t_{\mathfrak{l}},\bar{\mathbb{F}}_{\mathfrak{l}}}^{s_{\mathfrak{l}}}\rtimes
S_{s_{\mathfrak{l}}}$

###### Proof.

See Lemma 3.5 in [PR09a]. ∎

Combining the result $\Gamma_{\mathfrak{l}}\subseteq
H_{\mathfrak{l},\bar{\mathbb{F}}_{\mathfrak{l}}}\subseteq{\rm{GL}}_{t_{\mathfrak{l}},\bar{\mathbb{F}}_{\mathfrak{l}}}^{s_{\mathfrak{l}}}\rtimes
S_{s_{\mathfrak{l}}}$ with the primality of $r$, we can conclude either
$s_{\mathfrak{l}}=1,\ t_{\mathfrak{l}}=r$ or $s_{\mathfrak{l}}=r,\
t_{\mathfrak{l}}=1$. If $s_{\mathfrak{l}}=r,\ t_{\mathfrak{l}}=1$, then we
have

$\Gamma_{\mathfrak{l}}\subseteq{\rm{GL}}_{1,\bar{\mathbb{F}}_{\mathfrak{l}}}^{r}\rtimes
S_{r}.$

Since we assume $p>r!$, there is no element in
${\rm{GL}}_{1,\bar{\mathbb{F}}_{\mathfrak{l}}}^{r}\rtimes S_{r}$ whose order
is divisible by $p$. Thus we’ve deduced a contradiction because
$\Gamma_{\mathfrak{l}}$ has elements of order divisible by $p$. Therefore we
must have $s_{\mathfrak{l}}=1,\ t_{\mathfrak{l}}=r$. So
$H_{\mathfrak{l},\bar{\mathbb{F}}_{\mathfrak{l}}}^{\circ}$ acts on
${\bar{\mathbb{F}}}_{\mathfrak{l}}^{r}$ irreducibly. ∎

So far we have the following information on $H_{\mathfrak{l}}^{\circ}$:

  * •

$H_{\mathfrak{l}}^{\circ}$ is a connected algebraic group over
$\mathbb{F}_{\mathfrak{l}}$

  * •

$H_{\mathfrak{l}}^{\circ}$ acts on ${\mathbb{F}}_{\mathfrak{l}}^{r}$
absolutely irreducibly.

  * •

$H_{\mathfrak{l},\bar{\mathbb{F}}_{\mathfrak{l}}}^{\circ}$ has a cocharacter
of weight $1$ with multiplicity $1$ and weight $0$ with multiplicity $r-1$(In
the proof of Lemma 4.1)

With these conditions in hand, we can apply Proposition A.3 in [Pin97] and
prove that

$H_{\mathfrak{l}}^{\circ}={\rm{GL}}_{r,\mathbb{F}_{\mathfrak{l}}}$

###  4.1 Image of residual representation

######  Theorem 4.2.

Let $q=p^{e}$ be a prime power, $A=\mathbb{F}_{q}[T]$, and
$F=\mathbb{F}_{q}(T)$. Assume $r\geqslant 3$ is a prime number, there is a
constant $c=c(r)\in\mathbb{N}$ depending only on $r$ such that for $p>c(r)$
the following statement is true ${\rm{:}}$

Let $\phi$ be a Drinfeld $A$-module over $F$ of rank $r$ defined by
$\phi_{T}=T+\tau^{r-1}+T^{q-1}\tau^{r}$. Let $\mathfrak{l}$ be a finite place
of $F$ where $\phi$ has good reduction at $\mathfrak{l}$. Then the mod
$\mathfrak{l}$ representation

$\bar{\rho}_{\phi,\mathfrak{l}}:G_{F}\longrightarrow{\rm{Aut}}(\phi[\mathfrak{l}])\cong{\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}})$

is either reducible or surjective.

###### Proof.

Let $r$ be a prime number, $A=\mathbb{F}_{q}[T]$, and $F=\mathbb{F}_{q}(T)$
where $q=p^{e}$ where $p>r!$. Let $\phi$ be a Drinfeld $A$-module over $F$ of
rank $r$ with generic characteristic defined by
$\phi_{T}=T+\tau^{r-1}+T^{q-1}\tau^{r}$. Let $\mathfrak{l}$ be a place of $F$
where $\phi$ has good reduction at $\mathfrak{l}$. Let
$\Gamma_{\mathfrak{l}}=\bar{\rho}_{\phi,\mathfrak{l}}(G_{F})$ be irreducible.
From the argument above, we have
$H_{\mathfrak{l}}^{\circ}={\rm{GL}}_{r,\mathbb{F}_{\mathfrak{l}}}$.

Now we can apply Lemma 3.12 in [PR09a] for all places $\mathfrak{l}$ of $F$
where $\phi$ has good reduction. Thus there exists a natural number
$\tilde{c}$ depends only on $r$ such that

$[{\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}}):\Gamma_{\mathfrak{l}}]\leqslant\tilde{c}.$

Therefore, we replace $p$ by a prime number
$p>c:={\rm{max}}\\{r!,\tilde{c}!\\}$ and run Lemma 3.12 in [PR09a] again, we
get

$[{\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}}):\Gamma_{\mathfrak{l}}]\leqslant\tilde{c}\text{
and }|{\rm PGL}_{r}(\mathbb{F}_{\mathfrak{l}})|>\tilde{c}!.$

By applying Proposition 2.3 in [PR09a], we have

$\Gamma_{\mathfrak{l}}\supseteq{\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}}).$

By Proposition 2.10 and Chebotarev density theorem,
$\det\circ\bar{\rho_{\phi,\mathfrak{l}}}$ is equal to the mod $\mathfrak{l}$
representation $\bar{\rho_{C,\mathfrak{l}}}$ of Carlitz module. Hence we have
$\det(\Gamma_{\mathfrak{l}})=\mathbb{F}_{\mathfrak{l}}^{*}$, which implies

$\Gamma_{\mathfrak{l}}={\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}}).$

∎

##  5 Irreducibility of the mod $\mathfrak{l}$ representations

We assume $\mathfrak{l}\neq(T)$ from here to subsection 5.2, the case when
$\mathfrak{l}=(T)$ will be dealt with in subsection 5.3.

For $\mathfrak{l}=(T-c)$ a degree-$1$ prime ideal different from $(T)$, the
proof for irreducibility is straightforward.

######  Claim.

$\phi_{T-c}(x)/x=(T-c)+x^{q^{r-1}-1}+T^{q-1}x^{q^{r}-1}$ is irreducible over
$F$.

###### Proof of claim.

First of all, $\phi_{T-c}(x)/x$ is irreducible over $F$ if and only if its
reciprocal polynomial

$x^{q^{r}-1}\phi_{T-c}(x^{-1})/x^{-1}=(T-c)x^{q^{r}-1}+x^{q^{r}-q^{r-1}}+T^{q-1}$

is irreducible over $F$. As $T-c\in F^{*}$, it is enough for us to prove the
polynomial

$x^{q^{r}-1}+\frac{1}{T-c}x^{q^{r}-q^{r-1}}+\frac{T^{q-1}}{T-c}$

is irreducible over $F$. As $F=\mathbb{F}_{q}(T)=\mathbb{F}_{q}(T-c)$, we may
consider the completion $F^{\prime}$ of $F$ at the place $(\frac{1}{T-c})$.
Then the polynomial
$x^{q^{r}-1}+\frac{1}{T-c}x^{q^{r}-q^{r-1}}+\frac{T^{q-1}}{T-c}$ is
irreducible over $F^{\prime}$ by the Eisenstein criterion. Hence
$x^{q^{r}-1}+\frac{1}{T-c}x^{q^{r}-q^{r-1}}+\frac{T^{q-1}}{T-c}$ must be
irreducible over $F$, the proof is now complete.

∎

For general prime ideal $\mathfrak{l}\neq(T)$, we suppose $\phi[\mathfrak{l}]$
viewed as a $\mathbb{F}_{\mathfrak{l}}[G_{F}]$ is reducible, then it contains
a $G_{F}$-invariant proper submodule $X$. Let
${\rm{dim}}_{\mathbb{F}_{\mathfrak{l}}}X=d$, we claim that $d\neq 1\text{ or
}r-1$. This follows the same strategy as in the rank-$3$ case (see section 3
in [Che20]). Firstly, there is a basis such that the action of $G_{F}$ on
$\phi[\mathfrak{l}]$ is of one of the following forms:

  1. (i)

$\left(\begin{array}[]{cc}\chi&*\\\ 0&B\end{array}\right)$ (if $X$ has
dimension $1$),

  2. (ii)

$\left(\begin{array}[]{cc}B&*\\\ 0&\chi\end{array}\right)$ (if $X$ has
codimension $1$),

where $B:G_{F}\longrightarrow{\rm{GL}}_{r-1}(\mathbb{F}_{\mathfrak{l}})$ is a
homomorphism and $\chi:G_{F}\longrightarrow\mathbb{F}_{\mathfrak{l}}^{*}$ is a
character.

Now we consider the following exact sequence

$0\longrightarrow\phi[\mathfrak{l}]^{\circ}\longrightarrow\phi[\mathfrak{l}]\longrightarrow\phi[\mathfrak{l}]^{et}\longrightarrow
0$

of $\mathbb{F}_{\mathfrak{l}}[I_{\mathfrak{l}}]$-modules.

######  Proposition 5.1.

([PR09b], Proposition 2.7)

  1. i.

The inertia group $I_{\mathfrak{l}}$ acts trivially on
$\phi[\mathfrak{l}]^{et}$.

  2. ii.

The $\mathbb{F}_{\mathfrak{l}}$-vector space $\phi[\mathfrak{l}]^{\circ}$
extends uniquely to a one dimensional $\mathbb{F}_{\mathfrak{l}}^{(h)}$-vector
space structure such that the action of $I_{\mathfrak{l}}$ on
$\phi[\mathfrak{l}]^{\circ}$ is given by the fundamental character
$\zeta_{n}$.

  3. iii.

The action of wild inertia group at $\mathfrak{l}$ on
$\phi[\mathfrak{l}]^{\circ}$ is trivial.

######  Corollary 5.2.

$\phi[\mathfrak{l}]^{\circ}$ is an irreducible
$\mathbb{F}_{\mathfrak{l}}[I_{\mathfrak{l}}]$-module.

###### Proof.

This immediately follows from Proposition 5.1(ii). ∎

By Corollary 5.2, we have either (i) $X\cap\phi[\mathfrak{l}]^{\circ}=\\{0\\}$
or (ii) $X\cap\phi[\mathfrak{l}]^{\circ}=\phi[\mathfrak{l}]^{\circ}$.

######  Claim.

Either $\det B$ is unramified at every prime ideal $\mathfrak{p}$ of $A$ or
$\chi$ is unramified at every prime ideal $\mathfrak{p}$ of $A$.

###### Proof.

If $\mathfrak{p}$ is a prime different from $(T)$ and $\mathfrak{l}$, then
$\bar{\rho}_{\phi,\mathfrak{l}}$ is unramified at $\mathfrak{p}$. Thus $\det
B$ and $\chi$ are both unramified at $\mathfrak{p}$. It suffices for us to
prove the cases $\mathfrak{p}=(T)$ and $\mathfrak{p}=\mathfrak{l}$.
For $\mathfrak{p}=(T)$, Lemma 3.1 implies
$\\#\bar{\rho}_{\phi,\mathfrak{l}}(I_{T})$ is a $q$-power. This means the
order of $B(\sigma)$ and $\chi(\sigma)$ are also $q$-powers. Therefore, the
order of $\det B(\sigma)$ and $\chi(\sigma)$ are also $q$-powers for all
$\sigma\in I_{T}$. However, $\det B(\sigma)$ and $\chi(\sigma)$ belong to
$\mathbb{F}_{\mathfrak{l}}^{*}$,which is of order prime to $q$. Hence $\det
B(\sigma)=\chi(\sigma)=1$ for all $\sigma\in I_{T}$.
For $\mathfrak{p}=\mathfrak{l}$, we have two cases.

  1. Case (i):

$X$ has dimension $1$. Then $X$ is an irreducible
$\mathbb{F}_{\mathfrak{l}}[G_{F}]$-module. This means either
$X=\phi[\mathfrak{l}]^{\circ}$ or $X\cap\phi[\mathfrak{l}]^{\circ}=\\{0\\}$.
Let $\sigma\in I_{\mathfrak{l}}.$ If $X=\phi[\mathfrak{l}]^{\circ}$, then the
matrix $B(\sigma)$ describes how $\sigma$ acts on the
$\mathbb{F}_{\mathfrak{l}}[I_{\mathfrak{l}}]$-module
$\phi[\mathfrak{l}]/\phi[\mathfrak{l}]^{\circ}\cong\phi[\mathfrak{l}]^{et}$.
Proposition 5.1 then implies $\det B(\sigma)=1$ for all $\sigma\in
I_{\mathfrak{l}}$. When $X\cap\phi[\mathfrak{l}]^{\circ}=\\{0\\}$, we can view
$X$ as a $\mathbb{F}_{\mathfrak{l}}[I_{\mathfrak{l}}]$-submodule of
$\phi[\mathfrak{l}]/\phi[\mathfrak{l}]^{\circ}\cong\phi[\mathfrak{l}]^{et}$.
Therefore, Proposition 5.1 implies $\chi(\sigma)=1$ for all $\sigma\in
I_{\mathfrak{l}}$.

  2. Case (ii):

$X$ has codimension $1$. As $\phi[\mathfrak{l}]^{\circ}$ is an irreducible
$\mathbb{F}_{\mathfrak{l}}[I_{\mathfrak{l}}]$\- module by Corollary 5.2, we
get either $X\supseteq\phi[\mathfrak{l}]^{\circ}$ or
$X\cap\phi[\mathfrak{l}]^{\circ}=\\{0\\}$. When
$X\supseteq\phi[\mathfrak{l}]^{\circ}$, we have the “modulo $X$” map
$\phi[\mathfrak{l}]^{et}\cong\phi[\mathfrak{l}]/\phi[\mathfrak{l}]^{\circ}\rightarrow\phi[\mathfrak{l}]/X$
as $\mathbb{F}_{\mathfrak{l}}[I_{\mathfrak{l}}]$-modules. Thus
$\chi(\sigma)=1$ for all $\sigma\in I_{\mathfrak{l}}$ by Proposition 5.1. The
argument for the case when $X\cap\phi[\mathfrak{l}]^{\circ}=\\{0\\}$ is
similar to what happened in case (i), so $\det B(\sigma)=1$ for all $\sigma\in
I_{\mathfrak{l}}$.

∎

We first assume that $\det B$ is unramified at every prime $\mathfrak{p}$. The
homomorphism $\det B:G_{F}\rightarrow\mathbb{F}_{\mathfrak{l}}^{*}$ factors
through ${\rm{Gal}}(K/\mathbb{F}_{q}(T))$, where $K$ is a finite abelian
extension of $\mathbb{F}_{q}(T)$, unramified at every finite place and tamely
ramified at infinity. Hayes [Hay74] studied the class field theory in function
field analogue and showed that such finite abelian extension is just some
constant extension of $\mathbb{F}_{q}(T)$ (see section 5 and Theorem 7.1 in
[Hay74]), hence we have $K\subseteq\bar{\mathbb{F}}_{q}(T)$. Now we can write
$\det B$ in the following way:

$\det
B:G_{F}\twoheadrightarrow{\rm{Gal}}(\bar{\mathbb{F}}_{q}(T)/\mathbb{F}_{q}(T))\xrightarrow{\sim}{\rm{Gal}}(\bar{\mathbb{F}}_{q}/\mathbb{F}_{q})\rightarrow\mathbb{F}_{\mathfrak{l}}^{*},$

where the first map is the restriction map.

This implies there is some element $\xi\in\mathbb{F}_{\mathfrak{l}}^{*}$ such
that $\det B({\rm{Frob}}_{\mathfrak{p}})=\xi^{\deg\mathfrak{p}}$ for every
prime ideal $\mathfrak{p}$ of $A$ not equal to $(T)$ or $\mathfrak{l}$. Now we
consider the Frobenius element for a degree-$1$ prime
$\mathfrak{p}=(T-c)\neq(T)\text{ or }\mathfrak{l}$. The characteristic
polynomial of ${\rm{Frob}}_{\mathfrak{p}}$ acting on $\phi[\mathfrak{l}]$ is
$\bar{P}_{\phi,\mathfrak{p}}(x)=x^{r}+x^{r-1}-\bar{\mathfrak{p}}$. Thus we
have the following factorization

$x^{r}+x^{r-1}-\bar{\mathfrak{p}}=(x^{r-1}-\alpha_{\mathfrak{p}}x^{r-2}+\cdots+\xi)(x-\xi^{-1}{\bar{\mathfrak{p}}})\in\mathbb{F}_{\mathfrak{l}}[x].$

Hence for distinct prime ideals $\mathfrak{p}_{1}$ and $\mathfrak{p}_{2}$ of
degree 1, we can factorize the characteristic polynomial of
$\bar{\rho}_{\phi,\mathfrak{l}}({\rm{Frob}}_{\mathfrak{p}_{1}})$ and
$\bar{\rho}_{\phi,\mathfrak{l}}({\rm{Frob}}_{\mathfrak{p}_{2}})$,
respectively. By computing their coefficients, we get

$\left\\{\begin{array}[]{ccc}\alpha_{\mathfrak{p}_{1}}+\xi^{-1}\bar{\mathfrak{p}}_{1}&=\alpha_{\mathfrak{p}_{2}}+\xi^{-1}\bar{\mathfrak{p}}_{2}&=-1\\\
\alpha_{\mathfrak{p}_{1}}\xi^{-1}\bar{\mathfrak{p}}_{1}+\xi&=\alpha_{\mathfrak{p}_{2}}\xi^{-1}\bar{\mathfrak{p}}_{2}+\xi&=0\end{array}\right.$

$\Rightarrow\left\\{\begin{array}[]{ccc}\alpha_{\mathfrak{p}_{1}}&=&-1-\xi^{-1}\bar{\mathfrak{p}}_{1}\\\
\alpha_{\mathfrak{p}_{2}}&=&-1-\xi^{-1}\bar{\mathfrak{p}}_{2}\\\
\alpha_{\mathfrak{p}_{1}}\bar{\mathfrak{p}}_{1}&=&\alpha_{\mathfrak{p}_{2}}\bar{\mathfrak{p}}_{2}\end{array}\right.$

$\Rightarrow(-\xi-\bar{\mathfrak{p}}_{1})\bar{\mathfrak{p}}_{1}=(-\xi-\bar{\mathfrak{p}}_{2})\bar{\mathfrak{p}}_{2}$
$\Rightarrow-\xi(\bar{\mathfrak{p}}_{1}-\bar{\mathfrak{p}}_{2})=(\bar{\mathfrak{p}}_{1}^{2}-\bar{\mathfrak{p}}_{2}^{2}).$

So $\xi\equiv-({\mathfrak{p}}_{1}+{\mathfrak{p}}_{2})\ {\rm{mod}}\
\mathfrak{l}.$ As $q>5$, there are at least three different prime ideals of
degree $1$ not equal to $(T)$ or $\mathfrak{l}$. Thus we can derive the above
argument for any two such prime ideals, which implies that all these prime
ideals are congurent to each other modulo $\mathfrak{l}$. This gives us a
contradiction. The arguments for “ If $\chi$ is unramified at every prime
$\mathfrak{p}$” follows the same process as above.

In conclusion, we have shown that if there is an proper $G_{F}$-submodule $X$
of $\phi[\mathfrak{l}]$, then

$2\leqslant{\rm{dim}}_{\mathbb{F}_{\mathfrak{l}}}X=d\leqslant r-2.$

Therefore, we may assume in the subsection 5.1 and 5.2 that

  1. (a)

$2\leqslant{\rm{dim}}_{\mathbb{F}_{\mathfrak{l}}}X=d\leqslant r-2.$

  2. (b)

$\deg_{T}\mathfrak{l}\geqslant 2$

The following subsections will split into the cases (i)
$X\cap\phi[\mathfrak{l}]^{\circ}=\\{0\\}$ and (ii)
$X\cap\phi[\mathfrak{l}]^{\circ}=\phi[\mathfrak{l}]^{\circ}$ to derive
contradictions. Then the irreducibility for mod $\mathfrak{l}$ representations
when $\mathfrak{l}\neq(T)$ is proved.

###  5.1 The case $X\cap\phi[\mathfrak{l}]^{\circ}=\\{0\\}$

If $X\cap\phi[\mathfrak{l}]^{\circ}=\\{0\\}$, then Proposition 5.1 (i) shows
the action of $I_{\mathfrak{l}}$ on $X$ is trivial. Now we consider the action
of $I_{T}$ on $X$. By Lemma 3.2, there is a basis
$\\{w_{1},w_{2},\cdots,w_{r-1},z\\}$ of $\phi[\mathfrak{l}]$ such that
$\bar{\rho}_{\phi,\mathfrak{l}}(I_{T})=\left\\{\left(\begin{array}[]{cccc}1&&&b_{1}\\\
&\ddots&&\vdots\\\ &&\ddots&b_{r-1}\\\ &&&1\end{array}\right),\
b_{i}\in\mathbb{F}_{\mathfrak{l}}\ \forall\ 1\leqslant\ i\leqslant\
r-1\right\\}.$ Since $X$ is a proper submodule of $\phi[\mathfrak{l}]$, we
have $X\subseteq{\rm{span}}\\{w_{1},w_{2},\cdots,w_{r-1}\\}$. Therefore,
$I_{T}$ acts trivially on $X$. For places $\mathfrak{p}\neq(T)\ {\rm{or}}\
\mathfrak{l}$, we know that $I_{\mathfrak{p}}$ acts trivially on
$\phi[\mathfrak{l}]$, hence $I_{\mathfrak{p}}$ acts trivially on $X$.

In conclusion, we have proved the following proposition:

######  Proposition 5.3.

The action of $G_{F}$ on $X$ is unramified at every finite place.

Now we consider the ramification index at infinity of the extension
$F(\phi[\mathfrak{l}])/F$. Let $F_{\infty}=\mathbb{F}_{q}((\frac{1}{T}))$ be
the local field of $F$ at $\infty$ with valuation $v_{\infty}$, and
$|\cdot|_{\infty}$ be its corresponding absolute value. Let $\Lambda$ be the
period lattice of the polynomial
$\phi_{T}(x)=Tx+x^{q^{r-1}}+T^{q-1}x^{q^{r}}.$ From section 1.1 of [Gek19], we
know $\Lambda$ is a discrete free $A$-submodule of $\hat{\bar{F}}_{\infty}$,
the completed algebraic closure of $F_{\infty}$, of rank $r$. We also have

$e_{\Lambda}(Tz)=\phi_{T}(e_{\Lambda}(z))\ \ \ {\rm{for\ all\
}}z\in\hat{\bar{F}}_{\infty},$

where

$e_{\Lambda}(z)=z\prod_{0\neq\lambda\in\Lambda}(1-z/\lambda).$

Let $F_{\infty}(\Lambda)$ be the field extension generated by the period
lattice $\Lambda$. From Proposition 1.2 of [Gek19], we have

$F_{\infty}(\Lambda)=F_{\infty}({\rm{tor}}(\phi)),$

where ${\rm{tor}}(\phi)=\bigcup_{0\neq a\in A}\phi[a]$. Let
$\\{\lambda_{1},\lambda_{2},\cdots,\lambda_{r}\\}$ be a successive minimum
basis of the lattice $\Lambda$ (see [Gek19], 1.3 for the definition). By
Proposition 1.4 (i) of [Gek19], the spectrum
$(|\lambda_{1}|_{\infty},|\lambda_{2}|_{\infty},\cdots,|\lambda_{r}|_{\infty})$
is related to the Newton polygon of $\phi_{T}(x)$ with respect to the
valuation $v_{\infty}$. The Newton polygon of $\phi_{T}(x)$ is a line segment
with slope $\frac{2-q}{q^{r}-1}$. Hence, by Proposition 1.4(ii) in [Gek19], we
have

$|\lambda_{1}|_{\infty}=|\lambda_{2}|_{\infty}=\cdots=|\lambda_{r}|_{\infty}.$

Let $\mu_{i}=e_{\Lambda}(\frac{\lambda_{i}}{T})$, the set
$\\{\mu_{1},\mu_{2},\cdots,\mu_{r}\\}$ is a $\mathbb{F}_{q}$-basis of
$\phi[T]$. By formula 1.3.2 of [Gek19], we have

$|\mu_{i}|_{\infty}=|e_{\Lambda}(\frac{\lambda_{i}}{T})|_{\infty}=|\frac{\lambda_{i}}{T}|_{\infty}.$

From the Newton polygon of $\phi_{T}(x)$, we can compute
$v_{\infty}(\mu_{i})=\frac{q-2}{q^{r}-1}$. Thus we can deduce

$v_{\infty}(\lambda_{i})=v_{\infty}(\mu_{i})-1=\frac{-q^{r}+q-1}{q^{r}-1}.$

Now we can study the extension $F_{\infty}(\Lambda)/F_{\infty}$. We use the
notation set up in [Gek19], section 2.1. In our case, $L=F_{\infty}$,
$\mathbf{B}_{1}=\\{\lambda_{1},\lambda_{2},\cdots,\lambda_{r}\\}$, $\tau=t=1$,
$L_{\tau}=F_{\infty}(\Lambda)$. By applying diagram 2.8.4 of [Gek19] to our
situation, we have the followings:

  1. 1.

$\tilde{L}_{\tau}=L_{0}(V_{\tau})=F_{\infty}(\Lambda)$

  2. 2.

$L_{\tau-1}=L_{0}=F_{\infty}$

  3. 3.

$V_{\tau}=\text{the $\mathbb{F}_{q}$-vector space generated by basis
$e_{0}(B_{1})=\\{\lambda_{1},\cdots,\lambda_{r}\\}$}$ (see [Gek19], 2.4)

  4. 4.

$V_{\tau}$ is pure with precise denominator of weight equal to $q^{r}-1$ by
our computation on $v_{\infty}(\lambda_{i})$, Definition 2.5 and Proposition
2.6 in [Gek19].

  5. 5.

$L^{\prime}_{\tau-1}=\text{a completely ramified separable extension of
$L_{\tau-1}$ of degree $q^{r}-1$}$. (see [Gek19], 2.8)

  6. 6.

$M_{\tau}=L_{\tau-1}(V_{\tau})\cap L^{\prime}_{\tau-1}=F_{\infty}(\Lambda)\cap
L^{\prime}_{\tau-1}$, and
$[M_{\tau}:L_{\tau-1}]=e(\tilde{L}_{\tau}|L_{\tau-1})=e(F_{\infty}(\Lambda)|F_{\infty})$
(see [Gek19], 2.9.1)

Therefore, $[M_{\tau}:L_{\tau-1}]$ must divide $q^{r}-1$. This implies the
ramification index $e(F_{\infty}(\Lambda)|F_{\infty})$ is some number prime to
$p$. Hence the field extension $F_{\infty}({\rm{tor}}(\phi))/F_{\infty}$ is
tamely ramified, which means the field extension $F({\rm{tor}}(\phi))/F$ is
tamely ramified at $\infty$. Since $F(\phi[\mathfrak{l}])$ is an intermediate
field of the extension $F({\rm{tor}}(\phi))/F$, the extension
$F(\phi[\mathfrak{l}])/F$ is tamely ramified at $\infty$. Combining with
Proposition 5.3, we can make the following statement:

  1. ($\ast$).

The action of $G_{F}$ on $X$ is unramified at every finite place and tamely
ramified at the place of infinity.

Finally, we claim that the field extension
$\bar{\mathbb{F}}_{q}F(X)/\bar{\mathbb{F}}_{q}F$ is nontrivial. This leads to
a non-trivial Galois cover of $\mathbb{P}^{1}_{{\bar{\mathbb{F}}_{q}}}$ which
is unramified away from infinity and tamely ramified at infinity, hence is a
contradiction by Hurwitz genus formula.

Suppose the field extension $\bar{\mathbb{F}}_{q}F(X)/\bar{\mathbb{F}}_{q}F$
is trivial. Then for a element $0\neq w\in X\subset\bar{\mathbb{F}}_{q}(T)$,
the valuation $v_{T}(w)$ is an integer. We may write

$w=\sum_{i=-m}^{n}c_{i}T^{i},\text{ where $m,n\in\mathbb{Z}$, and
$c_{i}\in\bar{\mathbb{F}}_{q}$ with
$c_{-m},c_{n}\in\bar{\mathbb{F}}_{q}^{*}$}.$

We show that $w$ must lie in $\bar{\mathbb{F}}_{q}[T]$. Suppose $m\geqslant
1$, we have the valuation $v_{T}(w)=-m$. Now we compute the valuation of
$\phi_{T}(w)=Tw+w^{q^{r-1}}+T^{q-1}w^{q^{r}}$. As
$v_{T}(w^{q^{r-1}})=q^{r-1}\cdot(-m)$ and
$v_{T}(T^{q-1}w^{q^{r}})=q^{r}\cdot(-m)+(q-1)$, we have

$q^{r}\cdot(-m)+(q-1)=v_{T}(\phi_{T}(w))<v_{T}(w)=-m.$

Therefore, we can see that

$v_{T}(\phi_{T^{j}}(w))<v_{T}(\phi_{T^{j-1}}(w))\text{ for any integer
$j\geqslant 1$}.$

Hence the term of $\phi_{T^{\deg_{T}(l)}}(w)$ with smallest valuation does not
appear in $\phi_{T^{j}}(w)$ for any integer $0\leqslant j<\deg_{T}(l)$. We can
deduce from this that the term of $\phi_{\mathfrak{l}}(w)$ with smallest
valuation cannot be eliminated, which contradicts to $w\in
X\subset\phi[\mathfrak{l}]$. Thus $w$ must lie in $\bar{\mathbb{F}}_{q}[T]$.

On the other hand, we can replace $v_{T}$ by $v_{\infty}$ and run the same
process as above to show that $w$ must lie in $\bar{\mathbb{F}}_{q}^{*}$.
Indeed, we assume $v_{\infty}(w)=-m\leqslant-1$. We can deduce that

$q^{r}\cdot(-m)-(q-1)=v_{\infty}(\phi_{T}(w))<v_{\infty}(w)=-m.$

Hence we have $v_{\infty}(\phi_{T^{j}}(w))<v_{\infty}(\phi_{T^{j-1}}(w))\text{
for any integer $j\geqslant 1$}.$ The remains follow the same argument as
above.

Now we have the element $0\neq w\in X\subset\phi[\mathfrak{l}]$ lies in
$\bar{\mathbb{F}}_{q}^{*}$. However, this implies the term of
$\phi_{\mathfrak{l}}(w)$ with largest $T$-degree is the leading term

$T^{(q-1)({\displaystyle\sum_{i=1}^{\deg_{T}(l)}q^{r(i-1)}})}\cdot
w^{q^{r\cdot\deg_{T}(l)}}.$

Moreover, this term in $\phi_{\mathfrak{l}}(w)$ cannot be eliminated since the
$T$-degree of the leading coefficient in $\phi_{\mathfrak{l}}(x)$ is strictly
larger than the $T$-degree of any other coefficient. Thus we have a
contradiction due to the fact that $w\in X\subset\phi[\mathfrak{l}]$. In
conclusion, the field extension
$\bar{\mathbb{F}}_{q}F(X)/\bar{\mathbb{F}}_{q}F$ is nontrivial.

###  5.2 The case $X\cap\phi[\mathfrak{l}]^{\circ}=\phi[\mathfrak{l}]^{\circ}$

If $X\cap\phi[\mathfrak{l}]^{\circ}=\phi[\mathfrak{l}]^{\circ}$, then we have
$\phi[\mathfrak{l}]^{\circ}\subseteq X\subsetneq\phi[\mathfrak{l}]$. We
consider the following isogeny over $F$:

$u:\phi\rightarrow\psi:=\phi/X.$

Now we can study the action of $G_{F}$ on the proper $G_{F}$-submodule
$u(\phi[\mathfrak{l}])$ of $\psi[\mathfrak{l}]$. We claim that the extension
$F(u(\phi[\mathfrak{l}]))/F$ is unramified at every finite place and tamely
ramified at the place of infinity.

The isogeny induces an $G_{F}$-equivariant isomorphism between rational Tate
modules

$u:V_{\mathfrak{l}}(\phi)\rightarrow V_{\mathfrak{l}}(\psi),$

where $V_{\mathfrak{l}}(\phi):=T_{\mathfrak{l}}(\phi)\otimes
F_{\mathfrak{l}}$. For any prime ideal $\mathfrak{p}\neq(T)\text{ or
}\mathfrak{l}$. Since $\phi$ has good reduction at $\mathfrak{p}$ and $u$ is
$G_{F}$-equivariant, we know $\psi$ has good reduction at $\mathfrak{p}$.
Therefore, $G_{F}$ acts on $u(\phi[\mathfrak{l}])$ is unramified at
$\mathfrak{p}\neq(T)\text{ or }\mathfrak{l}$.

For the prime $\mathfrak{p}=\mathfrak{l}$, the action of inertia group
$I_{\mathfrak{l}}$ on $u(\phi[\mathfrak{l}])$ is determined by the action of
$I_{\mathfrak{l}}$ on $\phi[\mathfrak{l}]/X$. Combining the condition
$X\supseteq\phi[\mathfrak{l}]^{\circ}$ and Proposition 5.1 (ii), we deduce
that $I_{\mathfrak{l}}$ acts trivially on $\phi[\mathfrak{l}]/X$. Hence
$I_{\mathfrak{l}}$ acts trivially on $u(\phi[\mathfrak{l}])$ as well.

For the place $\mathfrak{p}=\infty$, we proved the extension
$F({\rm{tor}}(\phi))/F$ is tamely ramified at $\infty$ in the end of previous
subsection. This implies the wild inertia group at $\infty$ acts trivially on
the Tate module $T_{\mathfrak{l}}(\phi)$, hence the wild inertia group acts
trivially on $V_{\mathfrak{l}}(\phi)$ as well. Now the above
$G_{F}$-equivariant isomorphism $V_{\mathfrak{l}}(u)$ implies the wild inertia
at $\infty$ acts trivially on $V_{\mathfrak{l}}(\psi)$. Therefore, we can
deduce that the wild inertia group at $\infty$ acts trivially on
$u(\phi[\mathfrak{l}])$. In other words, the extension
$F(u(\phi[\mathfrak{l}]))/F$ is tamely ramified at infinity.

Finally, for the prime $\mathfrak{p}=(T)$, from Proposition 2.7 and Lemma 3.1
we know that $\psi$ has stable bad reduction at $\mathfrak{p}$.

Once $\psi$ has stable bad reduction at $\mathfrak{p}$, we know that $\psi$
has stable bad reduction at $\mathfrak{p}$ of rank $r-1$. Moreover, both $u$
and $\psi$ can be defined over $A_{\mathfrak{p}}$. As
${\rm{dim}}_{\mathbb{F}_{\mathfrak{l}}}X=d$, then the $\tau$-degree of $u$ is
$d$. We may write

$u=a_{0}+a_{1}\tau+\cdots+a_{d}\tau^{d}\text{ and
}\psi_{T}=T+g_{1}\tau+\cdots+g_{r}\tau^{r},$

where $a_{i}$ and $g_{j}$ are elements in $A_{\mathfrak{p}}$ with $a_{d}$,
$g_{r}\in A_{\mathfrak{p}}^{*}$. By comparing the leading coefficients of
$u\phi_{T}=\psi_{T}u$, we get

$T^{q^{d}(q-1)}=g_{r}a_{d}^{q^{r}-1}.$

Therefore, we can conclude $g_{r}=T^{q^{d}(q-1)}\cdot m$ for some $m\in
A_{\mathfrak{p}}^{*}$.

Now we can follow the same process as in Lemma 3.1 to deduce the following:

There is a basis $\\{w_{1}\cdots,w_{r-1},z\\}$ of $\psi[\mathfrak{l}]$ such
that

$\bar{\rho}_{\psi,\mathfrak{l}}(I_{\mathfrak{p}})\subseteq\left\\{\left(\begin{array}[]{cccc}1&&&b_{1}\\\
&\ddots&&\vdots\\\ &&\ddots&b_{r-1}\\\ &&&1\end{array}\right),\
b_{i}\in\mathbb{F}_{\mathfrak{l}}\ \forall\ 1\leqslant\ i\leqslant\
r-1\right\\}.$

with respect to the basis.

Furthermore, we can give an estimation on the size of
$\bar{\rho}_{\psi,\mathfrak{l}}(I_{\mathfrak{p}})$ by using the same process
as in Lemma 3.2. The computation only involves in the leading coefficient
$g_{r}$ of $\psi_{T}$. Because $g_{r}=T^{q^{d}(q-1)}\cdot m$ for some $m\in
A_{\mathfrak{p}}^{*}$, the left hand side of equation (1) in the proof of
Lemma 3.2 becomes

$T^{(q-1)q^{d}({\displaystyle\sum_{i=1}^{\deg_{T}(\mathfrak{l})}q^{r(i-1)}})}.$

Hence we can deduce the estimation

$|\bar{\rho}_{\psi,\mathfrak{l}}(I_{\mathfrak{p}})|\geqslant
q^{(r-1)\deg_{T}\mathfrak{l}-d}.$

######  Claim.

$I_{\mathfrak{p}}$ acts trivially on
$u(\phi[\mathfrak{l}])\subset\psi[\mathfrak{l}]$.

###### Proof of claim.

We split ${\rm{dim}}_{\mathbb{F}_{\mathfrak{l}}}X=d$ into two cases.

  1. $d\leqslant\deg_{T}(\mathfrak{l})$:

In this case, we can deduce from the estimation
$|\bar{\rho}_{\psi,\mathfrak{l}}(I_{\mathfrak{p}})|\geqslant
q^{(r-1)\deg_{T}\mathfrak{l}-d}$ that

$|\bar{\rho}_{\psi,\mathfrak{l}}(I_{\mathfrak{p}})|\geqslant
q^{(r-2)\deg_{T}\mathfrak{l}}.$

As ${\rm{dim}}_{\mathbb{F}_{\mathfrak{l}}}u(\phi[\mathfrak{l}])=r-d$ and
$1<d<r-1$, we know
${\rm{dim}}_{\mathbb{F}_{\mathfrak{l}}}u(\phi[\mathfrak{l}])\leqslant r-2$.
Now we use the basis $\\{w_{1},\cdots,w_{r-1},z\\}$ to present vectors in the
$\mathbb{F}_{\mathfrak{l}}$-space $u(\phi[\mathfrak{l}])$. Suppose there is an
element $w\in u(\phi[\mathfrak{l}])$ with linear combination
$c_{1}w_{1}+\cdots+c_{r-1}w_{r-1}+cz$ where
$c_{i}\in\mathbb{F}_{\mathfrak{l}}$ and $c\in\mathbb{F}_{\mathfrak{l}}^{*}$.
Since we have

$\bar{\rho}_{\psi,\mathfrak{l}}(I_{\mathfrak{p}})\subseteq\left\\{\left(\begin{array}[]{cccc}1&&&b_{1}\\\
&\ddots&&\vdots\\\ &&\ddots&b_{r-1}\\\ &&&1\end{array}\right),\
b_{i}\in\mathbb{F}_{\mathfrak{l}}\ \forall\ 1\leqslant\ i\leqslant\
r-1\right\\}.$

For every $\sigma=\left(\begin{array}[]{cccc}1&&&b_{1,\sigma}\\\
&\ddots&&\vdots\\\ &&\ddots&b_{r-1,\sigma}\\\
&&&1\end{array}\right)\in\bar{\rho}_{\psi,\mathfrak{l}}(I_{\mathfrak{p}})$, we
consider

$\sigma\cdot w-w=b_{1,\sigma}\cdot w_{1}+\cdots+b_{r-1,\sigma}\cdot w_{r-1}.$

We may start with an element
$\sigma_{1}\in\bar{\rho}_{\psi,\mathfrak{l}}(I_{\mathfrak{p}})$ with
$b_{1,\sigma_{1}},\cdots,b_{r-1,\sigma_{1}}$ not all zero. Hence there are at
most $|\mathbb{F}_{\mathfrak{l}}|=q^{\deg_{T}\mathfrak{l}}$ many elements
$\sigma\in\bar{\rho}_{\psi,\mathfrak{l}}(I_{\mathfrak{p}})$ such that
$\sigma\cdot w-w$ is linearly dependent with $\sigma_{1}\cdot w-w$. From our
estimation $|\bar{\rho}_{\psi,\mathfrak{l}}(I_{\mathfrak{p}})|\geqslant
q^{(r-2)\deg_{T}\mathfrak{l}}$, we can find
$\sigma_{2}\in\bar{\rho}_{\psi,\mathfrak{l}}(I_{\mathfrak{p}})$ such that
$\sigma_{2}\cdot w-w$ is linearly independent with $\sigma_{1}\cdot w-w$. Now
there are at most $|\mathbb{F}_{\mathfrak{l}}|^{2}=q^{2\deg_{T}\mathfrak{l}}$
many elements $\sigma\in\bar{\rho}_{\psi,\mathfrak{l}}(I_{\mathfrak{p}})$ such
that $\sigma\cdot w-w$ is linearly dependent with $\\{\sigma_{1}\cdot
w-w,\sigma_{2}\cdot w-w\\}$. From the estimation again, we can find
$\sigma_{3}\in\bar{\rho}_{\psi,\mathfrak{l}}(I_{\mathfrak{p}})$ such that
$\\{\sigma_{1}\cdot w-w,\sigma_{2}\cdot w-w,\sigma_{3}\cdot w-w\\}$ is a
linearly independent set. By following this procedure, we can pick
$\sigma_{1},\sigma_{2},\cdots,\sigma_{r-2}\in\bar{\rho}_{\psi,\mathfrak{l}}(I_{\mathfrak{p}})$
such that

$\\{\sigma_{i}\cdot w-w\mid 1\leqslant i\leqslant r-2\\}$

is a linearly independent set. Therefore, $\sigma_{i}\cdot w-w$ together with
$w$ produce $r-1$ many linearly independent vectors in
$u(\phi[\mathfrak{l}])$, a contradiction. Thus all elements in
$u(\phi[\mathfrak{l}])$ are linear combinations of the basis vectors
$w_{1},w_{2},\cdots,w_{r-1}$, which implies that $I_{\mathfrak{p}}$ acts
trivially on $u(\phi[\mathfrak{l}])$.

  2. $d>\deg_{T}(\mathfrak{l})$:

In this case, we are in the situation where

$1<\deg_{T}(\mathfrak{l})<d\leqslant r-2.$

Therefore, the estimation
$|\bar{\rho}_{\psi,\mathfrak{l}}(I_{\mathfrak{p}})|\geqslant
q^{(r-1)\deg_{T}\mathfrak{l}-d}$ implies

$|\bar{\rho}_{\psi,\mathfrak{l}}(I_{\mathfrak{p}})|\geqslant
q^{(r-1)\deg_{T}\mathfrak{l}-d}>q^{(r-1-d)\deg_{T}\mathfrak{l}}.$

As ${\rm{dim}}_{\mathbb{F}_{\mathfrak{l}}}u(\phi[\mathfrak{l}])=r-d$, we use
the basis $\\{w_{1},\cdots,w_{r-1},z\\}$ to present vectors in the
$\mathbb{F}_{\mathfrak{l}}$-space $u(\phi[\mathfrak{l}])$ again. Again, we
suppose there is an element $w\in u(\phi[\mathfrak{l}])$ with linear
combination $c_{1}w_{1}+\cdots+c_{r-1}w_{r-1}+cz$ where
$c_{i}\in\mathbb{F}_{\mathfrak{l}}$ and $c\in\mathbb{F}_{\mathfrak{l}}^{*}$.
For every $\sigma\in\bar{\rho}_{\psi,\mathfrak{l}}(I_{\mathfrak{p}})$, we
consider

$\sigma\cdot w-w=b_{1,\sigma}\cdot w_{1}+\cdots+b_{r-1,\sigma}\cdot w_{r-1}.$

By the procedure in the previous case and our estimation
$|\bar{\rho}_{\psi,\mathfrak{l}}(I_{\mathfrak{p}})|>q^{(r-1-d)\deg_{T}\mathfrak{l}}$,
we can find $r-d$ many elements
$\sigma_{1},\cdots,\sigma_{r-d}\in\bar{\rho}_{\psi,\mathfrak{l}}(I_{\mathfrak{p}})$
such that the set

$\\{\sigma_{i}\cdot w-w\mid 1\leqslant i\leqslant r-d\\}$

is linearly independent. Thus $\sigma_{i}\cdot w-w$ together with $w$ produce
$r-d+1$ many linearly independent vectors in $u(\phi[\mathfrak{l}])$, which is
a contradiction because
${\rm{dim}}_{\mathbb{F}_{\mathfrak{l}}}u(\phi[\mathfrak{l}])=r-d$. Therefore,
all elements in $u(\phi[\mathfrak{l}])$ are linear combinations of the basis
vectors $w_{1},w_{2},\cdots,w_{r-1}$. Hence $I_{\mathfrak{p}}$ acts trivially
on $u(\phi[\mathfrak{l}])$.

∎

From our study of the ramifications of $G_{F}$-action on
$u(\phi[\mathfrak{l}])$, we conclude

######  Lemma.

The action of $G_{F}$ on $u(\phi[\mathfrak{l}])$ is unramified at every finite
place and tamely ramified at the place of infinity.

Finally, we prove the field extension
$\bar{\mathbb{F}}_{q}F(u(\phi[\mathfrak{l}]))/\bar{F}_{q}F$ is nontrivial.
This leads to a non-trivial Galois cover of
$\mathbb{P}^{1}_{{\bar{\mathbb{F}}_{q}}}$ which is unramified away from
infinity and tamely ramified at infinity, which is a contradiction by Hurwitz
genus formula.

Suppose the field extension
$\bar{\mathbb{F}}_{q}F(u(\phi[\mathfrak{l}]))/\bar{F}_{q}F$ is trivial, then
for any element $\alpha\in\phi[\mathfrak{l}]$ the valuation
$v_{\infty}(u(\alpha))$ is an integer. On the other hand, we take the minimum
successive basis $\\{\lambda_{1},\cdots,\lambda_{r}\\}$ from the previous
subsection and set $\alpha_{i}=e_{\Lambda}(\frac{\lambda_{i}}{l})$. The set
$\\{\alpha_{1},\cdots,\alpha_{r}\\}$ is a $\mathbb{F}_{\mathfrak{l}}$-basis of
$\phi[\mathfrak{l}]$. Moreover, we have

$v_{\infty}(\alpha_{i})=v_{\infty}(e_{\Lambda}(\frac{\lambda_{i}}{l}))=v_{\infty}(\frac{\lambda_{i}}{l})=\deg_{T}(l)+\frac{-q^{r}+q-1}{q^{r}-1}.$

Now we compute the valuation $v_{\infty}(u(\alpha_{i}))$. As
$u=a_{0}+a_{1}\tau+\cdots+a_{d}\tau^{d}\in F\\{\tau\\}$, we have

$u(\alpha_{i})=a_{0}\alpha_{i}+a_{1}\alpha_{i}^{q}+\cdots+a_{d}\alpha_{i}^{q^{d}}.$

For each nonzero term $a_{j}\alpha_{i}^{q^{j}}$ of $u(\alpha_{i})$ where
$0\leqslant j\leqslant d<r-1$, its valuation is

$v_{\infty}(a_{j}\alpha_{i}^{q^{j}})=v_{\infty}(a_{j})+q^{j}\cdot
v_{\infty}(\alpha_{i})=\text{some
integer}+q^{j}\cdot\frac{-q^{r}+q-1}{q^{r}-1}=\text{some
integer}+\frac{q^{j}\cdot(q-2)}{q^{r}-1}.$

Therefore, the valuation of each nonzero term of $u(\alpha_{i})$ has distinct
fractional part. Thus $v_{\infty}(u(\alpha_{i}))$ is not an integer by the
strong triangle inequality because the valuation of each summand
$v_{\infty}(a_{j}\alpha_{i}^{q^{j}})$ is a distinct fraction. This shows the
field extension $\bar{\mathbb{F}}_{q}F(u(\phi[\mathfrak{l}]))/\bar{F}_{q}F$ is
nontrivial.

In conclusion, we have proved the irreducibility of the mod $\mathfrak{l}$
Galois representations for $\mathfrak{l}\neq(T)$. Combining with Theorem 4.2,
we have the following corollary:

######  Corollary 5.4.

Let $q=p^{e}$ be a prime power, $A=\mathbb{F}_{q}[T]$, and
$F=\mathbb{F}_{q}(T)$. Assume $r\geqslant 3$ is a prime number, there is a
constant $c=c(r)\in\mathbb{N}$ depending only on $r$ such that for $p>c(r)$
the following statement is true ${\rm{:}}$

Let $\phi$ be a Drinfeld $A$-module over $F$ of rank $r$ with generic
characteristic, which is defined by $\phi_{T}=T+\tau^{r-1}+T^{q-1}\tau^{r}$.
Let $\mathfrak{l}\neq(T)$ be a finite place of $F$. Then the mod
$\mathfrak{l}$ representation

$\bar{\rho}_{\phi,\mathfrak{l}}:G_{F}\longrightarrow{\rm{Aut}}(\phi[\mathfrak{l}])\cong{\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}})$

is surjective.

###  5.3 The case when $\mathfrak{l}=(T)$

For the mod $(T)$ Galois representation
$\bar{{\rho}}_{\phi,T}:G_{F}\longrightarrow{\rm{Aut}}(\phi[T])\cong{\rm{GL_{r}}}(\mathbb{F}_{q})$,
we are actually computing the Galois group ${\rm{Gal}}(F(\phi[T])/F)$ of the
field extension obtained by adjoining the roots of
$\phi_{T}(x)=Tx+x^{q^{r-1}}+T^{q-1}x^{q^{r}}$ to $F$. This question has been
studied by Abhyankar [Abh94]. Here we use the same notation as in [Abh94],
Theorem 3.2 (3.2.3).

Let $k_{0}=\mathbb{F}_{q}$, $K=k_{0}(\frac{1}{T})$, and

$V=V(Y)=Y^{q^{r}-1}+\frac{1}{T^{q-1}}Y^{q^{r-1}-1}+\frac{1}{T^{q-2}}.$

We have $\mu=r-1$ and ${\rm{GCD}}(\nu,\tau)=1$, where $\nu=1+q+\cdots+q^{r-1}$
and $\tau=1+q+\cdots+q^{r-2}$. Further, we have $C_{1}=(\frac{1}{T})^{q-1}$
and $C_{r}=(\frac{1}{T})^{q-2}$. Thus $\rho=q-1$ and $\sigma=q-2$. Certainly
we have $\rho\neq\frac{\sigma(\nu-\tau)}{\nu}$. Thus we can apply [Abh94],
Theorem 3.2 (3.2.3). Since ${\rm{GCD}}(\sigma,q-1)=1$, the theorem implies

${\rm{Gal}}({\mathbb{F}}_{q}(T)(\phi[T])/{\mathbb{F}}_{q}(T))={\rm{GL}}_{r}(\mathbb{F}_{q}).$

As a summary of section 5, we have proved the surjectivity of mod
$\mathfrak{l}$ Galois representation $\bar{\rho}_{\phi,\mathfrak{l}}$ for any
prime ideal $\mathfrak{l}$ of $A$ under the assumption on $q$ in Theorem 4.2

##  6 Surjectivity of $\mathfrak{l}$-adic Galois representations

Similar to the rank $3$ case, we wish to apply [PR09a], Proposition 4.1 to
prove surjectivity of $\mathfrak{l}$-adic representations. We separate
$\mathfrak{l}$ into two cases:

  1. Case 1.

$\mathfrak{l}\neq(T)$
Our proof of the equality
$\bar{\rho}_{\phi,\mathfrak{l}}(I_{T})=\left\\{\left(\begin{array}[]{cccc}1&&&b_{1}\\\
&\ddots&&\vdots\\\ &&\ddots&b_{r-1}\\\ &&&1\end{array}\right),\
b_{i}\in\mathbb{F}_{\mathfrak{l}}\ \forall\ 1\leqslant\ i\leqslant\
r-1\right\\}$ only restrict $\mathfrak{l}$ to be prime to $(T)$. Hence we can
prove the above equality for mod $\mathfrak{l}^{2}$ representation. In other
words, we have

$\bar{\rho}_{\phi,\mathfrak{l}^{2}}(I_{T})=\left\\{\left(\begin{array}[]{cccc}1&&&b_{1}\\\
&\ddots&&\vdots\\\ &&\ddots&b_{r-1}\\\ &&&1\end{array}\right),\
b_{i}\in(A/\mathfrak{l}^{2})\ \forall\ 1\leqslant\ i\leqslant\ r-1\right\\}$

Therefore, the mod $\mathfrak{l}^{2}$ representation certainly contains a non-
scalar matrix that becomes identity after modulo $\mathfrak{l}$.

  2. Case 2.

$\mathfrak{l}=(T)$
We consider the following diagram and focus on the representations of
decomposition subgroups.

$\begin{array}[]{cccc}\bar{\rho}_{\phi,\mathfrak{l}^{2}}(G_{F})&\xrightarrow{\rm{modulo}\
\mathfrak{l}}&\bar{\rho}_{\phi,\mathfrak{l}}(G_{F})&={\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}}).\\\
\cup&&\cup&\\\
\bar{\rho}_{\phi,\mathfrak{l}^{2}}(G_{F_{(T)}})&\xrightarrow{\rm{modulo}\
\mathfrak{l}}&\bar{\rho}_{\phi,\mathfrak{l}}(G_{F_{(T)}})&\end{array}$

It’s clear that the modulo $\mathfrak{l}$ homomorphism
$\bar{\rho}_{\phi,\mathfrak{l}^{2}}(G_{F_{(T)}})\xrightarrow{\rm{modulo}\
\mathfrak{l}}\bar{\rho}_{\phi,\mathfrak{l}}(G_{F_{(T)}})$ is surjective.
Suppose it is injective, then we can conclude that the splitting field of
$\phi_{\mathfrak{l}^{2}}(x)$ and of $\phi_{\mathfrak{l}}(x)$ are isomorphic
over $F_{(T)}$. Now we compare the Newton’s polygons of
$\phi_{\mathfrak{l}^{2}}(x)/x$ and $\phi_{\mathfrak{l}}(x)$.

$\begin{array}[]{ccc}\phi_{T^{2}}&=&T^{2}+T(T^{q^{r-1}-1}+1)\tau^{r-1}+T^{q}(T^{q^{r}-1}+1)\tau^{r}+\tau^{2r-2}\\\
&&\ \ \ \ \ \
+T^{q-1}(T^{(q-1)(q^{r-1}-1)}+1)\tau^{2r-1}+T^{(q-1)(q^{r}+1)}\tau^{2r}.\end{array}$

There is some root of $\phi_{\mathfrak{l}^{2}}(x)/x$ has valuation equal to
$-\frac{1}{q^{2r-2}}$, but there is no element in the splitting field of
$\phi_{\mathfrak{l}}(x)$ with the same valuation. Thus the splitting fields of
both polynomials are not isomorphic, which implies
$\bar{\rho}_{\phi,\mathfrak{l}^{2}}(G_{F_{(}T)})\xrightarrow{\rm{modulo}\
\mathfrak{l}}\bar{\rho}_{\phi,\mathfrak{l}}(G_{F_{(T)}})$ has nontrivial
kernel.

Now we prove such a nontrivial element
$\bar{\rho}_{\phi,\mathfrak{l}^{2}}(\sigma)\in\bar{\rho}_{\phi,\mathfrak{l}^{2}}(G_{F_{(T)}})$
can not be a scalar matrix. Suppose
$\bar{\rho}_{\phi,\mathfrak{l}^{2}}(\sigma)\in\bar{\rho}_{\phi,\mathfrak{l}^{2}}(G_{F_{(T)}})$
is a scalar matrix, then there is some $a\in\mathbb{F}_{q}^{*}$ such that
$\sigma$ maps every root $\alpha$ of $\phi_{T^{2}}(x)/x$ to
$\phi_{1+Ta}(\alpha)$. Furthermore, as $\sigma$ lies in the decomposition
subgroup $G_{F_{(T)}}$, $\sigma$ should preserve the valuation $v_{T}$. We
pick a root $\alpha$ with valuation equal to $-\frac{1}{q^{2r-2}}$, then
compare the valuation of $\sigma$ with $\sigma(\alpha)$.

$-\frac{1}{q^{2r-2}}=v_{T}(\alpha)\neq
v_{T}(\sigma(\alpha))=v_{T}(\phi_{1+Ta}(\alpha))=v_{T}([1+Ta]\alpha+a\alpha^{q^{r-1}}-[aT^{q-1}]\alpha^{q^{r}})=-\frac{1}{q^{r-1}}.$

Therefore $\sigma$ cannot be a scalar matrix, and the mod $(T^{2})$
representation contains a non-scalar matrix that becomes identity after modulo
$\mathfrak{l}$.

Now we can apply [PR09a], Proposition 4.1 to show that the $\mathfrak{l}$-adic
representation is surjective for every prime ideal $\mathfrak{l}$ of $A$.

##  7 Adelic surjectivity of Galois representations

In this section, we make a further assumption that $q\equiv 1\mod r$. The
proof of adelic surjectivity is similar to the proof for rank $2$ and $3$
cases in [Zyw11] and [Che20].

######  Lemma 7.1.

For each finite place $\mathfrak{l}$ of $F$, ${\rm{SL}}_{r}(A_{\mathfrak{l}})$
is equal its commutator subgroup. The only normal subgroup of
${\rm{SL}}_{r}(A_{\mathfrak{l}})$ with simple quotient is

$N:=\left\\{B\in{\rm{SL}}_{r}(A_{\mathfrak{l}})|B\equiv\delta\cdot
I_{r}\mod\mathfrak{l},\ {\rm{where}}\ \delta\in\mathbb{F}_{\mathfrak{l}}\
{\rm{satisfies}}\ \delta^{r}=1\right\\}.$

###### Proof.

Let $H$ be the commutator subgroup of ${\rm{SL}}_{r}(A_{\mathfrak{l}})$. It’s
a closed normal subgroup of ${\rm{SL}}_{r}(A_{\mathfrak{l}})$ and
${\rm{GL}}_{r}(A_{\mathfrak{l}})$. We define
$S^{0}:={\rm{SL}}_{r}(A_{\mathfrak{l}})$ and for $i\geqslant 1$, we set
$S^{i}:=\\{s\in{\rm{SL}}_{r}(A_{\mathfrak{l}})\ |\ s\equiv
1\mod\mathfrak{l}^{i}\\}.$ For $i\geqslant 0$, define $H^{i}=H\cap S^{i}$. For
$i\geqslant 0$, we define $S^{[i]}:=S^{i}/S^{i+1}$ and
$H^{[i]}:=H^{i}/H^{i+1}$. There is a natural injection form $H^{[i]}$ to
$S^{[i]}$ and we claim that $H^{[i]}=S^{[i]}$ for all $i\geqslant 0$.

For $i=0$, modulo $\mathfrak{l}$ induces an isomorphism
$S^{[0]}\xrightarrow{\sim}{\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}})$ and the
image of $H^{[0]}$ under this isomorphism becomes the commutator subgroup of
${\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}})$. It’s well known that
${\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}})$ is quasi-simple, i.e.
${\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}})$ equals to its commutator subgroup
and the quotient of ${\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}})$ by its center
$Z({\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}}))$ is a simple group. Therefore, we
have $H^{[0]}=S^{[0]}$.

Now we fix $i\geqslant 1$, let $\mathfrak{sl}_{r}(\mathbb{F}_{\mathfrak{l}})$
be the additive subgroup in $M_{r}(\mathbb{F}_{\mathfrak{l}})$ consisting of
matrices with trace $0$. We have the following isomorphism:

$\begin{array}[]{ccc}S^{[i]}&\xrightarrow{\sim}&\mathfrak{sl}_{r}(\mathbb{F}_{\mathfrak{l}})\\\
\left[1+l^{i}y\right]&\mapsto&[y]\end{array},$

where $l$ is the monic polynomial in $A$ that generates $\mathfrak{l}$.
Consider ${\rm{GL}}_{r}(A_{\mathfrak{l}})$ acting on both sides via
conjugation action, it factors through
${\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}})$. By [PR09a] Proposition 2.1,
$\mathfrak{sl}_{r}(\mathbb{F}_{\mathfrak{l}})$ is an irreducible
${\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}})$-module (here we used the assumption
$q=p^{e}$ is a prime power with $p\geqslant 5$). On the other hand, $H^{[i]}$
injects into $S^{[i]}$ and $H$ is normal in ${\rm{GL}}_{r}(A_{\mathfrak{l}})$
imply that $H^{[i]}$ is also stable under
${\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}})$-action. Once we can show that
$H^{[i]}$ is nontrivial, we have $H^{[i]}=S^{[i]}$ for all $i\geqslant 0$ and
so $H=S^{0}$. Consider the commutator map $S^{0}\times S^{i}\rightarrow H^{i}$
that maps $(g,h)$ to $ghg^{-1}h^{-1}$. This induces a map $S^{[0]}\times
S^{[i]}\rightarrow H^{[i]}$. Combining with the isomorphism
$S^{[i]}\xrightarrow{\sim}\mathfrak{sl}_{r}(\mathbb{F}_{\mathfrak{l}})$, we
obtain the following map:

${\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}})\times\mathfrak{sl}_{r}(\mathbb{F}_{\mathfrak{l}})\rightarrow\mathfrak{sl}_{r}(\mathbb{F}_{\mathfrak{l}}),\
\ \ \ (s,X)\mapsto sXs^{-1}-X.$

This map is not a zero map, so we have $H^{[i]}=S^{[i]}$ for all $i\geqslant
0$ and $H=S^{0}$.

Now let $N^{\prime}$ be a normal subgroup of ${\rm{SL}}_{r}(A_{\mathfrak{l}})$
with simple quotient. We consider the subgroup
$S^{1}\triangleleft{\rm{SL}}_{r}(A_{\mathfrak{l}})$. Note that $S^{1}$ is a
pro-$p$ group by the definition of $S^{i}$ and
$S^{0}/S^{1}={\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}})$ is quasi-simple. Thus
there is a composition series of ${\rm{SL}}_{r}(A_{\mathfrak{l}})$ and its
composition factors are
${\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}})/Z({\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}})),\
\mathbb{Z}/p\mathbb{Z}$ (comes from the composition factors of $S^{1}$) and
$\mathbb{Z}/r\mathbb{Z}$ (comes from composition factors of
$Z({\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}}))$ if it is nontrivial). As
${\rm{SL}}_{r}(A_{\mathfrak{l}})$ equals to its commutator, it has no abelian
quotient. Thus
${\rm{SL}}_{r}(A_{\mathfrak{l}})/N^{\prime}\cong{\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}})/Z({\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}}))$.
On the other hand, we also have
${\rm{SL}}_{r}(A_{\mathfrak{l}})/N\cong{\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}})/Z({\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}}))$
Therefore, we get
${\rm{SL}}_{r}(A_{\mathfrak{l}})/N\cong{\rm{SL}}_{r}(A_{\mathfrak{l}})/N^{\prime}$
and so $N^{\prime}=N$. Otherwise, we may have $N\subsetneq
NN^{\prime}\triangleleft{\rm{SL}}_{r}(A_{\mathfrak{l}})$. Furthermore, we
prove $NN^{\prime}\neq{\rm SL}_{r}(A_{\mathfrak{l}})$. Suppose
$NN^{\prime}={\rm SL}_{r}(A_{\mathfrak{l}})$, then we have

${\rm PSL}_{r}(\mathbb{F}_{\mathfrak{l}})\cong{\rm
SL}_{r}(A_{\mathfrak{l}})/N^{\prime}\cong NN^{\prime}/N^{\prime}\cong N/{N\cap
N^{\prime}}.$

On the other hand, we look at the composition factors of ${\rm
SL}_{r}(A_{\mathfrak{l}})$. We have
${\rm{SL}}_{r}(A_{\mathfrak{l}})/N\cong{\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}})/Z({\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}}))\cong{\rm
PSL}_{r}(\mathbb{F}_{\mathfrak{l}})$, and the composition factor ${\rm
PSL}_{r}(\mathbb{F}_{\mathfrak{l}})$ appears only once. Therefore, the
composition factors of $N$ are all abelian, either $\mathbb{Z}/p\mathbb{Z}$ or
$\mathbb{Z}/r\mathbb{Z}$. It is impossible to have a quotient $N/N\cap
N^{\prime}$ of $N$ isomorphic to ${\rm PSL}_{r}(\mathbb{F}_{\mathfrak{l}})$.
Thus $NN^{\prime}$ is a proper subgroup sits between ${\rm
SL}_{r}(A_{\mathfrak{l}})$ and $N$, which contradicts to the fact that
${\rm{SL}}_{r}(A_{\mathfrak{l}})/N\cong{\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}})/Z({\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}}))$
is simple. ∎

######  Lemma 7.2.

([Zyw11], Lemma A.4) Let $B_{1}$ and $B_{2}$ be finite groups and suppose that
$H$ is a subgroup of $B_{1}\times B_{2}$ for which the two projections
$p_{1}:H\rightarrow B_{1}$ and $p_{2}:H\rightarrow B_{2}$ are surjective. Let
$N_{1}$ be the kernel of $p_{2}$ and $N_{2}$ be the kernel of $p_{1}$. We may
view $N_{1}$ as a normal subgroup of $B_{1}$ and $N_{2}$ as a normal subgroup
of $B_{2}$. Then the image of $H$ in $B_{1}/N_{1}\times B_{2}/N_{2}$ is the
graph of the isomorphism $B_{1}/N_{1}\xrightarrow{\sim}B_{2}/N_{2}$.

######  Lemma 7.3.

([Zyw11], Lemma A.6) Let $S_{1},S_{2},\cdots,S_{k}$ be finite groups with no
non-trivial abelian quotients. Let $H$ be a subgroup of
$S_{1}\times\cdots\times S_{k}$ such that each projection $H\rightarrow
S_{i}\times S_{j}$ $(1\leqslant i<j\leqslant k)$ is surjective. Then
$H=S_{1}\times\cdots\times S_{k}$.

######  Lemma 7.4.

Let $\mathfrak{l}_{1}$ and $\mathfrak{l}_{2}$ be two prime ideals of $A$, and
set $\mathfrak{a}=\mathfrak{l}_{1}\mathfrak{l}_{2}$. Let $\phi$ be the
Drinfeld module as before, and let
$H=\bar{\rho}_{\phi,\mathfrak{a}}(G_{F})\subseteq{\rm{GL}}_{r}(A/\mathfrak{a})$.
Then $H$ satisfies the following properties:

  1. 1.

$\det(H)=(A/\mathfrak{a})^{*}$;

  2. 2.

the projections
$p_{1}^{\prime}:H^{\prime}\rightarrow{\rm{SL}}_{r}(A/\mathfrak{l}_{1})$ and
$p_{2}^{\prime}:H^{\prime}\rightarrow{\rm{SL}}_{r}(A/\mathfrak{l}_{2})$ are
surjective, where $H^{\prime}=H\cap{\rm{SL}}_{r}(A/\mathfrak{a})$;

  3. 3.

the subring of $A/\mathfrak{a}$ generated by the set

$\mathcal{S}=\\{{\rm{tr}}(h)^{r}/\det(h)\ |\ h\in
H\\}\cup\\{\det(h)/{\rm{tr}}(h)^{r}\ |\ h\in H\ {\rm{with}}\
{\rm{tr}}(h)\in(A/\mathfrak{a})^{*}\\}$

is exactly $A/\mathfrak{a}$.

These three properties will imply $H={\rm{GL}}_{r}(A/\mathfrak{a})$.

###### Proof.

Condition $1$ follows from the fact that
$\det\circ\bar{\rho}_{\phi,\mathfrak{a}}=\bar{\rho}_{C,\mathfrak{a}}$, the mod
$\mathfrak{a}$ Galois representation of the Carlitz module. Condition $2$
follows from our result in section 5 and 6 that
$\bar{\rho}_{\phi,\mathfrak{l}}$ is surjective for any prime ideal
$\mathfrak{l}$. For condition $3$, we may take $c\in\mathbb{F}_{q}^{*}$ such
that $\mathfrak{p}=T-c$ not equal to $\mathfrak{l}_{1}$ or $\mathfrak{l}_{2}$.
By looking at the characteristic polynomial of
$\bar{\rho}_{\phi,\mathfrak{a}}({\rm{Frob}}_{\mathfrak{p}})$, we have

$\frac{\det(\bar{\rho}_{\phi,\mathfrak{a}}({\rm{Frob}}_{\mathfrak{p}}))}{{\rm{tr}}(\bar{\rho}_{\phi,\mathfrak{a}}({\rm{Frob}}_{\mathfrak{p}}))^{r}}\equiv\frac{T-c}{1^{r}}\equiv
T-c\mod\mathfrak{a}.$

The subring generated by these $T-c$ is equal to $A/\mathfrak{a}$.

The following shows how these three properties imply
$H={\rm{GL}}_{r}(A/\mathfrak{a})$. Let $N_{1}^{\prime}$ be the kernel of
$p_{2}^{\prime}$ and $N_{2}^{\prime}$ be the kernel of $p_{1}^{\prime}$. We
may view $N_{i}^{\prime}$ as a normal subgroup of
${\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}})$. Lemma 7.2 then implies the
image of $H^{\prime}$ in
${\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}_{1}})/N_{1}^{\prime}\times{\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}_{2}})/N_{2}^{\prime}$
is the graph of a group isomorphism
${\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}_{1}})/N_{1}^{\prime}\xrightarrow{\sim}{\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}_{2}})/N_{2}^{\prime}$.
If one of $N_{1}^{\prime}$ or $N_{2}^{\prime}$ is the whole group, then the
group isomorphism will force the other one to be the whole group. Therefore,
if $N_{1}^{\prime}={\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}_{1}})$ or
$N_{2}^{\prime}={\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}_{2}})$, we have
$H^{\prime}={\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}_{1}})\times{\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}_{2}})$.
Combining with condition 1, we have $H={\rm{GL}}_{r}(A/\mathfrak{a})$.

Now we suppose both $N_{i}^{\prime}$ are proper normal subgroups of
${\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}})$. By the proof of Lemma 7.1, we
can see that $N_{i}^{\prime}\subseteq
Z({\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}}))$ for $i=1,2$. Since the order
$|Z({\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}}))|$ is either $1$ or $r$, the
order $|N_{i}^{\prime}|$ is either $1$ or $r$. Comparing the order on both
sides of the isomorphism
${\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}_{1}})/N_{1}^{\prime}\xrightarrow{\sim}{\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}_{2}})/N_{2}^{\prime}$,
we can show that
$|\mathbb{F}_{\mathfrak{l}_{1}}|=|\mathbb{F}_{\mathfrak{l}_{2}}|$, i.e.
$\mathbb{F}_{\mathfrak{l}_{1}}$ and $\mathbb{F}_{\mathfrak{l}_{2}}$ are
isomorphic fields. For $i=1$ or $2$, define the projection
$p_{i}:H\rightarrow{\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}})$. Let $N_{1}$
be kernel of $p_{2}$ and $N_{2}$ be kernel of $p_{1}$, we may also view
$N_{i}$ as normal subgroup of ${\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}})$.
Lemma 7.2 then implies the image of $H$ into
${\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{1}})/N_{1}\times{\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{2}})/N_{2}$
is the graph of a group isomorphism
${\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{1}})/N_{1}\xrightarrow{\sim}{\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{2}})/N_{2}$.

As $N_{i}/N_{i}^{\prime}\cong
N_{i}{\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}})/{\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}})\subseteq{\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}})/{\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}})\cong\mathbb{F}_{\mathfrak{l}_{i}}^{*}$,
we see that $N_{i}/N_{i}^{\prime}$ and $N_{i}^{\prime}$ are abelian. Thus
$N_{i}$ is a solvable normal subgroup of
${\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}})$. Furthermore, we see that
$N_{i}\cong
N_{i}Z({\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}}))/Z({\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}}))$
is a solvable normal subgroup of
${\rm{PGL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}})$. Hence we can deduce a normal
series

$\\{1\\}\lhd
N_{i}Z({\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}}))/Z({\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}}))\lhd{\rm{PGL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}}).$

If
${\rm{PSL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}})={\rm{PGL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}})$,
then we have $N_{i}\subseteq Z({\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}}))$
by the simplicity of ${\rm{PSL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}})$. If
${\rm{PSL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}})\neq{\rm{PGL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}})$,
then we consider the comsposition series

$\\{1\\}\lhd{\rm{PSL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}})\lhd{\rm{PGL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}}).$

Its factors are ${\rm{PSL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}})$ and $C_{r}$,
the cyclic group of order $r$. Thus by the Jordan-Hölder theorem, $N_{i}\cong
N_{i}Z({\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}}))/Z({\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}}))\cong
C_{r}$. Let $\delta$ be a generator of
$N_{i}\subset{\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}})$. We know that
$\delta$ satisfies the polynomial $x^{r}-1$. Since $q\equiv 1\mod r$,
$x^{r}-1$ splits over $\mathbb{F}_{{\mathfrak{l}_{i}}}$. The minimal
polynomial of $\delta$ over $\mathbb{F}_{\mathfrak{l}_{i}}$ is a product of
distinct degree one polynomials. Now we prove that the minimal polynomial of
$\delta$ has to be a degree one polynomial. Let $\zeta$ be a generator of
$\mathbb{F}_{q}^{*}$ and $\omega=\zeta^{\frac{q-1}{r}}$. The roots of
$x^{r}-1$ in $\mathbb{F}_{q}$ are $\omega^{i}$ with $0\leqslant i\leqslant
r-1$. Suppose the minimal polynomial of $\delta\in
N_{i}\subset{\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}})$ is not a degree one
polynomial, then $\delta$ can be written as a diagonal matrix with diagonal
entries of the form $\omega^{i}$ but not all the same. Let’s assume there are
two distinct diagonal entries
$\omega_{1},\omega_{2}\in\\{\omega^{i},0\leqslant i\leqslant r-1\\}$ in
$\delta$. The proof for general situation will follow the same process. Now we
may write $\delta$ as

$\delta=\left(\begin{array}[]{cccccc}\omega_{1}&&&&&\\\ &\ddots&&&&\\\
&&\omega_{1}&&&\\\ &&&\omega_{2}&&\\\ &&&&\ddots&\\\
&&&&&\omega_{2}\end{array}\right).$

Now since $N_{i}$ is a normal subgroup of
${\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}})$, we can conjugate $\delta$ by a
suitable permutation matrix to get a matrix in $N_{i}$ where two distinct
diagonal entries interchanged while the other diagonal entries fixed. Namely,
we have

$\left(\begin{array}[]{cccccccc}\omega_{1}&&&&&&&\\\ &\ddots&&&&&&\\\
&&\omega_{1}&&&&&\\\ &&&\omega_{2}&&&&\\\ &&&&\omega_{1}&&&\\\
&&&&&\omega_{2}&&\\\ &&&&&&\ddots&\\\ &&&&&&&\omega_{2}\end{array}\right)\in
N_{i}$

As $N_{i}$ is generated by $\delta$, this matrix is equal to $\delta^{j}$ for
some $1\leqslant j\leqslant r-1$. By comparing the fixed entries in $\delta$
and $\delta^{j}$, we can deduce that $j$ must be equal to $1$ because
$\omega^{i}$ has order equal to $r$ for $1\leqslant i\leqslant r-1$ (here we
use $r$ is a prime number). This implies $\omega_{1}=\omega_{2}$, a
contradiction. Therefore, the minimal polynomial of $\delta$ is a degree one
polynomial. Hence $\delta$ is a scalar matrix and we have $N_{i}\subseteq
Z({\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}}))$ in both cases.

By taking further quotient, the image of $H$ into
${\rm{PGL}}_{r}(\mathbb{F}_{\mathfrak{l}_{1}})\times{\rm{PGL}}_{r}(\mathbb{F}_{\mathfrak{l}_{2}})$
is the graph of a group isomorphism

$\alpha:{\rm{PGL}}_{r}(\mathbb{F}_{\mathfrak{l}_{1}})\xrightarrow{\sim}{\rm{PGL}}_{r}(\mathbb{F}_{\mathfrak{l}_{2}}).$

By [Die80] Theorem 2, $\alpha$ can be lifted to an isomorphism

$\tilde{\alpha}:{\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{1}})\xrightarrow{\sim}{\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{2}}).$

Let
$\sigma:\mathbb{F}_{\mathfrak{l}_{1}}\xrightarrow{\sim}\mathbb{F}_{\mathfrak{l}_{2}}$
be a field isomorphism and
$\chi:{\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{1}})\rightarrow\mathbb{F}_{\mathfrak{l}_{2}}^{*}$
be a homomorphism. Now we are able to create two group homomorphisms
${\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{1}})\xrightarrow{\sim}{\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{2}})$:

  1. (i)

$A\mapsto\chi(A)gA^{\sigma}g^{-1}$,

  2. (ii)

$A\mapsto\chi(A)g((A^{T})^{-1})^{\sigma}g^{-1}$,

where $A\in{\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{1}})$, $A^{\sigma}$ is the
matrix that applies $\sigma$ to each entry of $A$, and
$g\in{\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{2}})$. By [Die80] Theorem 1,
there are $\sigma$, $\chi$, and $g$ such that $\tilde{\alpha}$ is one of the
homomorphisms above.

######  Lemma.

: $\tilde{\alpha}$ must be of the first type.

###### Proof.

Suppose $\tilde{\alpha}$ is of second type, then we choose a degree $1$ prime
ideal $\mathfrak{p}$ of $A$ different from $\mathfrak{l}_{1}$ and
$\mathfrak{l}_{2}$. We consider the image of
$\bar{\rho}_{\phi,\mathfrak{a}}({\rm{Frob}}_{\mathfrak{p}})\in H$ in
${\rm{PGL}}_{r}(\mathbb{F}_{\mathfrak{l}_{1}})\times{\rm{PGL}}_{r}(\mathbb{F}_{\mathfrak{l}_{2}})$
under

$\begin{array}[]{ccc}\ \ \ \ \ \ \
H&\hookrightarrow&{\rm{PGL}}_{r}(\mathbb{F}_{\mathfrak{l}_{1}})\times{\rm{PGL}}_{r}(\mathbb{F}_{\mathfrak{l}_{2}})\\\
\ \ \ \ \ \ \
\bar{\rho}_{\phi,\mathfrak{a}}({\rm{Frob}}_{\mathfrak{p}})&\mapsto&(\bar{\rho}_{\phi,{\mathfrak{l}}_{1}}({\rm{Frob}}_{\mathfrak{p}})\cdot
Z({\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{1}})),\bar{\rho}_{\phi,{\mathfrak{l}_{2}}}({\rm{Frob}}_{\mathfrak{p}})\cdot
Z({\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{2}})))\\\ &&\|\\\
&&(\bar{\rho}_{\phi,{\mathfrak{l}}_{1}}({\rm{Frob}}_{\mathfrak{p}})\cdot
Z({\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{1}})),\tilde{\alpha}(\bar{\rho}_{\phi,{\mathfrak{l}}_{1}}({\rm{Frob}}_{\mathfrak{p}}))\cdot
Z({\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{2}})))\\\ &&\|\\\
&&(\bar{\rho}_{\phi,{\mathfrak{l}}_{1}}({\rm{Frob}}_{\mathfrak{p}})\cdot
Z({\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{1}})),g(((\bar{\rho}_{\phi,{\mathfrak{l}}_{1}}({\rm{Frob}}_{\mathfrak{p}}))^{\sigma})^{T})^{-1}g^{-1}\cdot
Z({\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{2}})))\end{array}$

Therefore,
$\bar{\rho}_{\phi,{\mathfrak{l}_{2}}}({\rm{Frob}}_{\mathfrak{p}})\cdot
Z({\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{2}}))$ and
$g(((\bar{\rho}_{\phi,{\mathfrak{l}}_{1}}({\rm{Frob}}_{\mathfrak{p}}))^{\sigma})^{T})^{-1}g^{-1}\cdot
Z({\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{2}}))$ are the same coset in
${\rm{PGL}}_{r}(\mathbb{F}_{\mathfrak{l}_{2}})$. For each element in these two
cosets, we can compute its trace. All elements in
$\bar{\rho}_{\phi,{\mathfrak{l}_{2}}}({\rm{Frob}}_{\mathfrak{p}})\cdot
Z({\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{2}}))$ has nonzero trace because the
characteristic polynomial of
$\bar{\rho}_{\phi,{\mathfrak{l}_{2}}}({\rm{Frob}}_{\mathfrak{p}})$ is
$x^{r}+x^{r-1}-\mathfrak{p}$ (mod $\mathfrak{l}_{2}$). However, the trace of
all elements in
$g(((\bar{\rho}_{\phi,{\mathfrak{l}}_{1}}({\rm{Frob}}_{\mathfrak{p}}))^{\sigma})^{T})^{-1}g^{-1}\cdot
Z({\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{2}}))$ are equal to zero because the
characteristic polynomial of
$(\bar{\rho}_{\phi,{\mathfrak{l}}_{1}}({\rm{Frob}}_{\mathfrak{p}}))^{-1}$ is
$x^{r}-\frac{1}{\mathfrak{p}}x-\frac{1}{\mathfrak{p}}$ (mod
$\mathfrak{l}_{1}$). Hence we have a contradiction and $\tilde{\alpha}$ must
be of first type. ∎

Writing $\tilde{\alpha}(A)=\chi(A)gA^{\sigma}g^{-1}$ for all
$A\in{\rm{GL}}_{r}(\mathbb{F}_{\mathfrak{l}_{1}})$, we have

$\frac{{\rm{tr}}(\tilde{\alpha}(A))^{r}}{\det(\tilde{\alpha}(A))}=\sigma\left(\frac{{\rm{tr}}(A)^{r}}{\det(A)}\right).$

Therefore, for each element $(h_{1},h_{2})\in H$, we have

$\frac{{\rm{tr}}(h_{2})^{r}}{\det(h_{2})}=\sigma\left(\frac{{\rm{tr}}(h_{1})^{r}}{\det(h_{1})}\right).$

Let $W=\\{(x_{1},x_{2})\ |\ \sigma(x_{1})=x_{2}\\}$ be the subring of
$A/\mathfrak{a}\cong\mathbb{F}_{\mathfrak{l}_{1}}\times\mathbb{F}_{\mathfrak{l}_{2}}$.
We have $\mathcal{S}\subseteq W$. However, $W\neq A/\mathfrak{a}$ by counting
cardinality on both sides. Thus we get a contradiction from the assumption
that $N_{i}^{\prime}$ is a proper normal subgroup of
${\rm{SL}}_{r}(\mathbb{F}_{\mathfrak{l}_{i}})$ for $i=1$ or $2$. The proof is
complete.

∎

######  Lemma 7.5.

Let $\mathfrak{l}_{1}$ and $\mathfrak{l}_{2}$ be two finite places of $F$.
Define

$\rho:G_{F}\rightarrow{\rm{GL}}_{r}(A_{\mathfrak{l}_{1}})\times{\rm{GL}}_{r}(A_{\mathfrak{l}_{2}}),\
\
\sigma\mapsto(\rho_{\phi,\mathfrak{l}_{1}}(\sigma),\rho_{\phi,\mathfrak{l}_{2}}(\sigma)).$

Then
$\rho(G_{F})={\rm{GL}}_{r}(A_{\mathfrak{l}_{1}})\times{\rm{GL}}_{r}(A_{\mathfrak{l}_{2}})$.

###### Proof.

It’s enough to show that for any positive integers $n_{1}$ and $n_{2}$, we
have

$\bar{\rho}_{\phi,\mathfrak{a}}(G_{F^{\rm{ab}}})={\rm{SL}}_{r}(A/\mathfrak{a})$

where $\mathfrak{a}=\mathfrak{l}_{1}^{n_{1}}\mathfrak{l}_{2}^{n_{2}}$.

Suppose the equality doesn’t hold, then we can apply Lemma 7.2 with
$H=\bar{\rho}_{\phi,\mathfrak{a}}(G_{F^{\rm{ab}}})$,
$B_{1}={\rm{SL}}_{r}(A/\mathfrak{l}_{1}^{n_{1}})$ and
$B_{2}={\rm{SL}}_{r}(A/\mathfrak{l}_{2}^{n_{2}})$. The image of $H$ in
${\rm{SL}}_{r}(A/\mathfrak{l}_{1}^{n_{1}})/N_{1}\times{\rm{SL}}_{r}(A/\mathfrak{l}_{2}^{n_{2}})/N_{2}$
is the graph of the isomorphism
${\rm{SL}}_{r}(A/\mathfrak{l}_{1}^{n_{1}})/N_{1}\xrightarrow{\sim}{\rm{SL}}_{r}(A/\mathfrak{l}_{2}^{n_{2}})/N_{2}$.
From the second paragraph in the proof of Lemma 7.4, we know that
${\rm{SL}}_{3}(A/\mathfrak{l}_{i}^{n_{i}})/N_{i}$ are not trivial for $i=1,2$.
By Lemma 7.1, $N_{i}$ is a subgroup of

$\bar{N}:=\left\\{B\in{\rm{SL}}_{r}(A/\mathfrak{l}_{i}^{n_{i}})|B\equiv\delta\cdot
I_{r}\mod\mathfrak{l},\ {\rm{where}}\ \delta\in\mathbb{F}_{\mathfrak{l}}\
{\rm{satisfies}}\ \delta^{r}=1\right\\}.$

Taking further quotient, the image of $H$ in
${\rm{SL}}_{r}(A/\mathfrak{l}_{1})/Z({\rm{SL}}_{r}(A/\mathfrak{l}_{1}))\times{\rm{SL}}_{r}(A/\mathfrak{l}_{2})/Z({\rm{SL}}_{r}(A/\mathfrak{l}_{2}))$
is the graph of an isomorphism
${\rm{SL}}_{r}(A/\mathfrak{l}_{1})/Z({\rm{SL}}_{r}(A/\mathfrak{l}_{1}))\xrightarrow{\sim}{\rm{SL}}_{r}(A/\mathfrak{l}_{2})/Z({\rm{SL}}_{r}(A/\mathfrak{l}_{2}))$.
However, Lemma 7.4 shows the reduction of $H$ modulo
$\mathfrak{l}_{1}\mathfrak{l}_{2}$ is the whole group
${\rm{SL}}_{r}(A/\mathfrak{l}_{1})\times{\rm{SL}}_{r}(A/\mathfrak{l}_{2})$.
Thus the image of $H$ into
${\rm{SL}}_{r}(A/\mathfrak{l}_{1})/Z({\rm{SL}}_{r}(A/\mathfrak{l}_{1}))\times{\rm{SL}}_{r}(A/\mathfrak{l}_{2})/Z({\rm{SL}}_{r}(A/\mathfrak{l}_{2}))$
should be the whole group, this gives a contradiction. ∎

###  7.1 Proof of the main theorem

######  Main Theorem.

Let $q=p^{e}$ be a prime power, $A=\mathbb{F}_{q}[T]$, and
$F=\mathbb{F}_{q}(T)$. Assume $r\geqslant 3$ is a prime number and $q\equiv
1\mod r$, there is a constant $c=c(r)\in\mathbb{N}$ depending only on $r$ such
that for $p>c(r)$ the following statement is true ${\rm{:}}$

Let $\phi$ be a Drinfeld $A$-module over $F$ of rank $r$ with generic
characteristic, which is defined by $\phi_{T}=T+\tau^{r-1}+T^{q-1}\tau^{r}$.
Then the adelic Galois representation

${\rho}_{\phi}:{\rm{Gal}}(\mathbb{F}_{q}(T)^{{\rm{sep}}}/\mathbb{F}_{q}(T))\longrightarrow\varprojlim_{\mathfrak{a}}{\rm{Aut}}(\phi[\mathfrak{a}])\cong{\rm{GL_{r}}}(\widehat{A})$

is surjective.

###### Proof.

Firstly, $\det\circ\rho_{\phi}=\rho_{C}$ is the adelic representation of
Carlitz module. We know that the adelic representation of the Carlitz module
is surjective, so $\det\circ\rho_{\phi}=\widehat{A}^{*}$. It suffices for us
to prove $\rho_{\phi}(G_{F^{\rm{ab}}})={\rm{SL}}_{r}(\widehat{A})$. It’s
equivalent to show that for every nonzero ideal
$\mathfrak{a}=\mathfrak{l}_{1}^{n_{1}}\mathfrak{l}_{2}^{n_{2}}\cdots\mathfrak{l}_{k}^{n_{k}}$
of $A$ , we have

$\bar{\rho}_{\phi,\mathfrak{a}}(G_{F^{\rm{ab}}})={\rm{SL}}_{r}(A/\mathfrak{a})\cong\prod_{i}{\rm{SL}}_{r}(A/\mathfrak{l}_{i}^{n_{i}}).$

By Lemma 7.1, we know that each ${\rm{SL}}_{r}(A/\mathfrak{l}_{i}^{n_{i}})$
has no nontrivial abelian quotient. Thus we can apply Lemma 7.3 once we can
prove that each projection
$\bar{\rho}_{\phi,\mathfrak{a}}(G_{F^{\rm{ab}}})\rightarrow{\rm{SL}}_{r}(A/\mathfrak{l}_{i}^{n_{i}})\times{\rm{SL}}_{r}(A/\mathfrak{l}_{j}^{n_{j}})$
is surjective for $1\leqslant i<j\leqslant k$. The surjectivity of each
projection is proved by Lemma 7.5. Thus Lemma 7.3 implies
$\bar{\rho}_{\phi,\mathfrak{a}}(G_{F^{\rm{ab}}})={\rm{SL}}_{r}(A/\mathfrak{a})$
for every nonzero ideal $\mathfrak{a}$ of $A$, the proof of theorem is
complete.

∎

## Acknowledgment

The author would like to thank his thesis advisor, Professor Mihran Papikian,
for helpful discussions and suggestions on earlier drafts of this paper.

In the Appendices, we restate Aschbacher’s theorem in [BHRD13] and focus on
the classification of subgroups in general linear group over a finite field.

##  Appendix A Notations and Aschbacher classes

###  A.1 Notation

At here, $A,B,G$ are groups, and $a,b,n$ belong to $\mathbb{N}$.

  * •

${\rm{GL}}_{n}(q)={\rm{GL}}_{n}(\mathbb{F}_{q})$

  * •

$Z(G)$ denote the center of $G$.

  * •

$[G,G]$ or $G^{\prime}$ denote the derived subgroup of $G$.

  * •

For $n>1$, $G^{(n)}=[G^{(n-1)},G^{(n-1)}]$. $G^{\infty}=\bigcap_{i\geqslant
0}G^{(i)}$.

  * •

$A.B=$ an extension of $A$ by $B$ with unspecified splittness.

  * •

$A\wr B=$ wreath product of $A$ by a permutation group $B$.

  * •

${\rm{E}}_{p^{n}}$ or just $p^{n}$= elementary abelian group of order $p^{n}$.

  * •

For an elementary abelian group $A$, $A^{m+n}=$ a group with elementary
abelian normal subgroup $A^{m}$ such that the quotient is isomorphic to
$A^{n}$.

###  A.2 Aschbacher classes

Let $H$ be a subgroup of ${\rm{GL}}_{n}(q)$, where $n\geqslant 3$. In this
subsection we give a summary of information when $H$ lies in an Aschbacher
class. For a complete edition of Aschbacher classes and the definition of each
class we refer to chapter 2 in [BHRD13].

  * •

$H$ lies in class $\mathcal{C}_{1}\Rightarrow$ $H$ stabilizes a proper non-
zero subspace of $\mathbb{F}_{q}^{n}$.

  * •

$H$ lies in class $\mathcal{C}_{2}\Rightarrow$ there is a direct sum
decomposition $\mathcal{D}$ of $\mathbb{F}_{q}^{n}$ into $t$ subspaces, each
of dimension $m=n/t$:

$\mathcal{D}:\mathbb{F}_{q}^{n}=V_{1}\oplus V_{2}\oplus\cdots\oplus V_{t},\ \
{\rm{where}}\ t\geqslant 2$

The action of $H$ on $\mathbb{F}_{q}^{n}$ is of the type ${\rm{GL}}_{m}(q)\wr
S_{t}$.

  * •

$H$ lies in class $\mathcal{C}_{3}\Rightarrow$ There is a prime divisor
$s\geqslant 2$ of $n$ and $m=n/s$ such that $\mathbb{F}_{q}^{n}$ has a
$\mathbb{F}_{q^{s}}$-vector space structure and $H$ acts
$\mathbb{F}_{q^{s}}$-semilinear on $\mathbb{F}_{q}^{n}$. The action of $H$ on
$\mathbb{F}_{q}^{n}$ is of the type ${\rm{GL}}_{m}(q^{s})$

  * •

$H$ lies in class $\mathcal{C}_{4}\Rightarrow$ $H$ preserves a tensor product
decomposition $\mathbb{F}_{q}^{n}=V_{1}\otimes V_{2}$, where $V_{1}$(resp.
$V_{2}$) is a $\mathbb{F}_{q}$-subspace of $\mathbb{F}_{q}^{n}$ of dimension
$n_{1}$(resp. $n_{2}$) and $1<n_{1}<\sqrt{n}$. The action of $H$ on
$\mathbb{F}_{q}^{n}$ is of the type
${\rm{GL}}_{n_{1}}(q)\otimes{\rm{GL}}_{n_{2}}(q)$.

  * •

$H$ lies in class $\mathcal{C}_{5}\Rightarrow$ $H$ acts on
$\mathbb{F}_{q}^{n}$ absolutely irreducible and there is a subfield
$\mathbb{F}_{q_{0}}$ of $\mathbb{F}_{q}$ such that a conjugate of $H$ in
${\rm{GL}}_{n}(q)$ is a subgroup of
$<Z({\rm{GL}}_{n}(q)),{\rm{GL}}_{n}(q_{0})>$.

  * •

$H$ lies in class $\mathcal{C}_{6}\Rightarrow$ $n=r^{m}$ with $r$ prime. There
is an absolutely irreducible group $E$ such that $E\trianglelefteq H\leqslant
N_{{\rm{GL}}_{n}(q)}(E)$. Here $E$ is either an extraspecial $r$-group or a
$2$-group of symplectic type. The action of $H$ on $\mathbb{F}_{q}^{n}$ is of
the type ${r^{1+2m}}.{\rm{Sp}}_{2m}(r)$ when $n$ is odd. And $H$ is of the
type ${2^{2+2m}}.{\rm{Sp}}_{2m}(2)$ when $n$ is even.

  * •

$H$ lies in class $\mathcal{C}_{7}\Rightarrow$ $H$ preserves a tensor induced
decomposition $\mathbb{F}_{q}^{n}=V_{1}\otimes V_{2}\otimes\cdots\otimes
V_{t}$ with $t\geqslant 2$, dim$V_{i}=m$ and $n=m^{t}$. The action of $H$ on
$\mathbb{F}_{q}^{n}$ is of the type ${\rm{GL}}_{m}(q)\wr S_{t}$. Here the
wreath product is a tensor wreath product, which is a quotient of standard
wreath product.

  * •

$H$ lies in class $\mathcal{C}_{8}\Rightarrow$ $H$ preserves a non-degenerate
classical form on $\mathbb{F}_{q}^{n}$ up to scalar multiplication. By
classical form we mean symplectic form, unitary form or quadratic form.

  * •

$H$ lies in class $\mathcal{S}\Rightarrow$ $H$ doesn’t contain
${\rm{SL}}_{n}(q)$ and $H^{\infty}$ acts on $\mathbb{F}_{q}^{n}$ absolutely
irreducibly.

##  Appendix B Note on Aschbacher’s theorem

######  Theorem B.1.

(Special case of Aschbacher’s Theorem)

Let $H$ be a subgroup of ${\rm{GL}}_{n}(q)$, then $H$ lies in one of the
Aschbacher classes $\mathcal{C}_{i}$ or $\mathcal{S}$.

In [BHRD13] Theorem 2.2.19, the authors state a detailed version of
Aschbacher’s Theorem and describe the structure of $H\cap{\rm{SL}}_{n}(q)$
when $H$ maximal in ${\rm{GL}}_{n}(q)$ that lies in a class $\mathcal{C}_{i}$
for $1\leqslant i\leqslant 8$.

## Conflict of interest statement

On behalf of the author, the author states that there is no conflict of
interest.

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