# The minimal ramification problem for
rational function fields over finite fields

Lior Bary-Soroker  Raymond and Beverly Sackler School of Mathematical
Sciences, Tel Aviv University, Tel Aviv 69978, Israel
<EMAIL_ADDRESS>,  Alexei Entin  Raymond and Beverly Sackler School
of Mathematical Sciences, Tel Aviv University, Tel Aviv 69978, Israel
<EMAIL_ADDRESS>and  Arno Fehm  Technische Universität Dresden,
Fakultät Mathematik, Institut für Algebra, 01062 Dresden, Germany
<EMAIL_ADDRESS>
###### Abstract.

We study the minimal number of ramified primes in Galois extensions of
rational function fields over finite fields with prescribed finite Galois
group. In particular, we obtain a general conjecture in analogy with the well
studied case of number fields, which we establish for abelian, symmetric and
alternating groups in many cases.

MSC2010: 11T55 (primary), 11E25, 12F10 (secondary)

##  1\. Introduction

Let $G$ be a nontrivial finite group. We consider the minimal number of
ramified primes in Galois extensions $L/K$ with Galois group $G$, in various
situations:

In the case $K=\mathbb{Q}$, the primes are as usual equivalence classes of
absolute values, corresponding to the prime numbers and the usual absolute
value as the prime at infinity. Let $r_{K}(G)$ denote the minimal number of
ramified primes111The infinite prime is said to ramify in $L/\mathbb{Q}$ if
$L$ is not totally real. in a Galois extension $L/K$ with Galois group $G$.
Based on class field theory, Boston and Markin came to the following
conjecture:

######  Conjecture 1.1 (Boston–Markin [BM09]).

For every nontrivial finite group $G$,

$r_{\mathbb{Q}}(G)=\max\left\\{d(G^{\rm ab}),1\right\\},$

where $d(G)$ is the smallest cardinality of a set of generators of $G$.

While this was verified by Boston and Markin for abelian groups and for groups
of order up to 32, and since then was proven for certain solvable groups
[Pla04, KS06, KS10, KNS10], it is widely open for most other groups. For
example, for the symmetric group $G=S_{n}$, $n>1$, this conjecture predicts
realizations with only one ramified prime. While this can of course be
verified for small $n$, the best general bound to date is the recent result of
Bary-Soroker and Schlank [BS20], who obtain $r_{\mathbb{Q}}(S_{n})\leq 4$ for
all $n$.

In the case where $K=k(T)$ is a rational function field over a field $k$, the
primes are equivalence classes of ultrametric absolute values on $K$ trivial
on $k$, which correspond to the monic irreducible polynomials $P\in k[T]$ and
the degree valuation as the prime at infinity. As before we let $r_{K}(G)$ be
the minimal number of ramified primes in Galois extensions $L/K$ with Galois
group $G$, where in addition we require that $L/K$ is geometric, i.e. $k$ is
algebraically closed in $L$. The geometric situation $k=\mathbb{C}$ is well
understood:

######  Theorem 1.2 (Riemann Existence Theorem).

For every nontrivial finite group $G$,

$r_{\mathbb{C}(T)}(G)=d(G)+1.$

In positive characteristic, wild ramification leads to smaller bounds:

######  Theorem 1.3 (Abhyankar conjecture for the line, [Ray94, Har94]).

For every prime number $p$ and every nontrivial finite group $G$,

$r_{\overline{\mathbb{F}}_{p}(T)}(G)=d(G/p(G))+1,$

where $p(G)$ is the subgroup of $G$ generated by the $p$-Sylow subgroups of
$G$.

For the global field $K=\mathbb{F}_{q}(T)$, in analogy with Conjecture 1.1, we
thus conjecture the following:

######  Conjecture 1.4.

For every prime power $q=p^{\nu}$ and every nontrivial finite group $G$,

$r_{\mathbb{F}_{q}(T)}(G)=\max\left\\{d((G/p(G))^{\rm ab}),1\right\\}.$

We note that $(G/p(G))^{\rm ab}=G^{\rm ab}/p(G^{\rm ab})$, and that the close
analogy with the case of $K=\mathbb{Q}$ builds in an essential way on our
restriction that the extension $L$ of $K=\mathbb{F}_{q}(T)$ with Galois group
$G$ is geometric. In Section 3 we prove this conjecture for abelian groups,
thereby showing that the right hand side is always a lower bound for
$r_{\mathbb{F}_{q}(T)}(G)$. There we also discuss the contribution of De Witt
[DeW14], who proves Conjecture 1.4 for certain solvable groups.

######  Remark 1.5.

Another potential constraint on $r_{\mathbb{F}_{q}(T)}(G)$ was pointed out to
us by Joachim König. For a finite geometric Galois extension $L$ of
$K=\mathbb{F}_{q}(T)$, the conjugacy classes of inertia groups generate ${\rm
Gal}(L/K)$, and each inertia group is cyclic-by-$p$, i.e. an extension of a
cyclic group by a $p$-group (see [Ser79, Corollary IV.2.4]). Thus Conjecture
1.4 is possibly true only if any finite group $G$ is generated by
$\max\\{d((G/p(G))^{\rm ab}),1\\}$ conjugacy classes of cyclic-by-$p$ groups.
We give the proof of this elementary but nontrivial group-theoretic fact also
in Section 3. A similar remark applies to Conjecture 1.1, where Lemma 3.1
ensures there is no group-theoretic obstruction.

In the main part of this work (Sections 4 and 5) we prove various results for
certain non-solvable groups, both conditional and unconditional, with a focus
on symmetric and alternating groups. Note that Conjecture 1.4 predicts

$r_{\mathbb{F}_{q}(T)}(S_{n})=1,\quad r_{\mathbb{F}_{q}(T)}(A_{n})=1$

for every $n>2$ and every prime power $q=p^{\nu}$. For example, we obtain the
following for $S_{n}$ and $A_{n}$:

######  Theorem 1.6.

Let $n\geq 2$ and $q=p^{\nu}$ a prime power. Then

  1. (1)

$r_{\mathbb{F}_{q}(T)}(S_{n})\leq 2$, and

  2. (2)

$r_{\mathbb{F}_{q}(T)}(S_{n})=1$ in each of the following cases:

    1. (a)

$p<n-1$ or $p=n$ or $p=2$

    2. (b)

$q\equiv 1\mbox{ mod }4$

    3. (c)

$q>(2n-3)^{2}$

    4. (d)

The function field analogue of Schinzel’s hypothesis H (Conjecture 4.13) holds
for $\mathbb{F}_{q}(T)$.

######  Theorem 1.7.

Let $n\geq 3$ and $q=p^{\nu}$ a prime power.

  1. (1)

If $p>2$ or $\mathbb{F}_{q}\supseteq\mathbb{F}_{4}$, then
$r_{\mathbb{F}_{q}(T)}(A_{n})\leq 2$, and

  2. (2)

$r_{\mathbb{F}_{q}(T)}(A_{n})=1$ in each of the following cases:

    1. (a)

$2<p<n-1$ or $p=n$ or $p=n-1,\mathbb{F}_{q}\supseteq\mathbb{F}_{p^{2}}$ or
$p=3,n=4$

    2. (b)

$p=2$ and $\mathbb{F}_{q}\supseteq\mathbb{F}_{4}$ or $10\neq n\geq 8,n\equiv
0,1,2,6,7\pmod{8}$

    3. (c)

$n=10$, $n=11$ or $n\geq 13$, and the function field analogue of Schinzel’s
hypothesis H (Conjecture 4.13) holds for $\mathbb{F}_{q}(T)$.

One surprising prediction of Conjecture 1.1 is that for an arbitrary product
$G=A_{n_{1}}\times\ldots\times A_{n_{m}},n_{i}\geq 5$, there exists a Galois
extension $L/\mathbb{Q}$ with ${\rm Gal}(L/\mathbb{Q})=G$ which is ramified
over a single prime. This is currently wide open even in the case
$G=A_{5}\times\ldots\times A_{5}$. We will establish Conjecture 1.4 for
arbitrary products of the groups appearing in the statements of Theorems
1.6(2a), 1.7(2a) and 1.7(2b):

######  Theorem 1.8.

Let $G=G_{1}\times\ldots\times G_{m}$ be a product with each $G_{i}$ being one
of the following groups:

  1. (1)

$S_{n},n\geq p$ with $n\neq p+1$ or $p=2$.

  2. (2)

$A_{n},n\geq p>2$ with $n\neq p+1$ or
$\mathbb{F}_{q}\supseteq\mathbb{F}_{p^{2}}$ or $p=3$.

  3. (3)

$A_{n}$ if $p=2$ with either $\mathbb{F}_{q}\supseteq\mathbb{F}_{4}$ or
$10\neq n\geq 8,n\equiv 0,1,2,6,7\pmod{8}.$

Then $r_{\mathbb{F}_{q}(T)}(G)=\max\\{d((G/p(G))^{\rm ab}),1\\}$, i.e.
Conjecture 1.4 holds for $G$.

En route to proving Theorem 1.8 we will obtain evidence for the following
conjecture of Abhyankar (see [Abh01, §16],[HOPS18, §5], a weaker form was
stated in [Abh95, Conjecture 9.2(C)]) which is an analogue over
$\mathbb{F}_{q}$ of the Abhyankar conjecture for the affine line (which is the
case $G=p(G)$ of Theorem 1.3):

######  Conjecture 1.9 (Abhyankar’s arithmetic conjecture for the affine
line).

Let $G$ be a finite group which is cyclic-by-quasi-$p$, meaning that $G/p(G)$
is cyclic (see Definition 5.2 below). Then for every power $q$ of $p$ there
exists a Galois extension $L/\mathbb{F}_{q}(T)$ (not necessarily geometric)
ramified only over the infinite prime such that ${\rm
Gal}(L/\mathbb{F}_{q}(T))=G$.

We will show that Conjecture 1.9 holds for the alternating group $A_{n}$
provided $n\geq p>2,n\neq p+1$ (note that for $n>3$ the group $A_{n}/p(A_{n})$
is cyclic only if $n\geq p$).

######  Theorem 1.10.

Assume $n\geq p>2$. If $n=p+1$ assume additionally that
$\mathbb{F}_{q}\supseteq\mathbb{F}_{p^{2}}$. Then there exists a Galois
extension $L/\mathbb{F}_{q}(T)$ ramified only over the infinite prime with
${\rm Gal}(L/\mathbb{F}_{q}(T))=A_{n}$.

Theorem 1.10 will follow from Theorem 5.7 below. For some values of $q,n$ the
theorem above is implicit in the work of Abhyankar [Abh92, Abh93], but not in
the above generality.

To obtain these results we apply a variety of tools and recent results from
Galois theory, finite group theory (including the Classification of Finite
Simple Groups for some of the results; see Remark 6.4 for the list of cases
which can be proved without the CFSG), analytic number theory and number
theory over function fields, and we also employed the help of a computer to
eliminate some small exceptional cases. See the summary in Section 6 for the
proof of Theorems 1.6 and 1.7, and the corresponding sections for the full
strength and full generality of the results. Using our results for
$\mathbb{F}_{q}(T)$ we also obtain a conditional result for $\mathbb{Q}$
(Theorem 6.6):

######  Theorem 1.11.

Schinzel’s hypothesis H (Conjecture 4.12) implies that
$r_{\mathbb{Q}}(S_{n})=1$ for every $n>1$, i.e. Conjecture 1.1 holds for all
symmetric groups.

This improves upon a result of Plans (see Remark 6.7 below), which shows that
under Schinzel’s hypothesis H there exists a Galois extension $L/\mathbb{Q}$
with Galois group $S_{n}$ and ramified over a single finite prime and the
infinite prime, whereas our construction has no ramification at the infinite
prime.

##  2\. Preliminaries and notation

We start by collecting a few definitions and statements that we will use
throughout the paper.

###  2.1. Function fields

For a thorough introduction to the subject see [Ros02, Sti09]. Here we only
recall some terminology, notation and a few basic facts. Let $k$ be a base
field. A function field (of one variable) over $k$ is a finite extension of
the rational function field $k(T)$. A _global function field_ is a function
field over a finite base field $k=\mathbb{F}_{q}$ (this will be our main case
of interest).

A function field $K/k$ is regular if $K$ is linearly disjoint over $k$ from
the algebraic closure $\bar{k}$ of $k$, which in case that $k$ is perfect
simply means that $k$ is relatively algebraically closed in $K$. If $K/k$ is a
regular function field, there exists a unique geometrically integral non-
singular projective curve $C/k$ such that $K$ is the function field of $C$.
Conversely the function field of a curve as above is a regular function field
over $k$.

Let $K/k$ be a regular function field and $C$ the corresponding curve. A
_prime_ $P$ of $K/k$ is an equivalence class of nontrivial absolute values on
$K$ trivial on $k$. Each such absolute value corresponds to a discrete
valuation $v$, and it corresponds to a prime divisor on $C$. If $P$ is a prime
of $K$, the residue field $k(P)$ at $P$ is a finite extension of $k$ and $\deg
P=[k(P):k]$ is called the _degree_ of $P$. By a _point_ or _geometric point_
of $C$ we will always mean a closed point of $C\times_{k}\bar{k}$. For an
extension $L/k$ we denote by $C(L)$ the set of $L$-rational points on $C$.

The rational function field $k(T)$ corresponds to the projective line
$\mathbb{P}^{1}_{k}$ and has primes of two types: the _finite primes_ of the
form $\mathrm{div}(f)_{0}$ (the zero divisor of a rational function $f$) where
$f\in k[X]$ is an irreducible polynomial (such a prime has degree $\deg f$)
and the _infinite prime_ $\infty=\mathrm{div}(1/T)_{0}$ which has degree 1.

Let $L/K$ be a finite extension of function fields over $k$. It is called
_regular_ , or _geometric_ , if $L$ is also a regular function field over $k$.
Regular extensions correspond to finite covers of the associated curves. If
$P$ is a prime of $K$ and $Q$ a prime of $L$ lying over $P$, we denote by
$e(Q/P)$ the ramification index of $Q$ over $P$. We say that $L/K$ is ramified
(resp. wildly ramified) at $Q$ if $e(Q/P)>1$ (resp.
$\mathrm{char}(k)|e(Q/P)$). We say that $L/K$ is ramified (resp. wildly
ramified) _over_ $P$ if there exists a prime $Q$ of $L$ lying over $P$ at
which the extension is ramified (resp. wildly ramified). Ramification which is
not wild is said to be _tame_. If $L/K$ is tamely ramified over at least one
prime, then $L/K$ is separable.

If $C_{L}$ and $C_{K}$ are the underlying curves then the extension $L/K$
corresponds to a finite morphism $w\colon C_{L}\to C_{K}$. This correspondence
defines an equivalence of categories between regular function fields over $k$
and absolutely irreducible non-singular projective curves defined over $k$. If
$P$ is a prime divisor of $C_{L}$, and $Q=w(P)$ (a prime divisor of $C_{K}$),
we say that $w$ is ramified (resp. tamely ramified) at $P$ whenever the
extension $L/K$ is ramified (resp. tamely ramified) at the corresponding prime
of $L$ which we identify with $P$. In this case we also call the geometric
points corresponding to $P$ ramification points of $w$, and the geometric
points corresponding to $Q$ branch points of $w$.

###  2.2. Ramification theory

Let $K$ be a field. For a separable polynomial $f\in K[X]$ of degree $n$ we
define its Galois group ${\rm Gal}(f/K)$ to be the Galois group of its
splitting field over $K$. We will always interpret the Galois group ${\rm
Gal}(f/K)$ as a subgroup of the symmetric group $S_{n}$ via its action on the
roots of $f$. The embedding into $S_{n}$ is well-defined up to conjugation.

######  Lemma 2.1.

Let $K$ be a function field over an algebraically closed field $k$ of
characteristic $p\geq 0$, let $f\in K[X]$ be a separable irreducible
polynomial and let $L=K(\alpha)$ where $\alpha$ is a root of $f$. Let $P$ be a
prime (finite or infinite) of $K$ and let $Q_{1},\ldots,Q_{r}$ be the primes
of $L$ lying over $P$. Let $e_{i}=e(Q_{i}/P)$ be the ramification indices.

  1. (i)

Assume that $p\nmid e_{i}$ for all $i$. Then ${\rm Gal}(f/K)$ contains a
permutation which is a product of $r$ disjoint cycles of lengths
$e_{1},\ldots,e_{r}$.

  2. (ii)

Assume that $p\nmid e_{i}$ and $(e_{i},e_{1})=1$ for $2\leq i\leq r$, and
either $e_{1}=p$ or $p\nmid e_{1}$. Then ${\rm Gal}(f/K)$ contains a cycle of
length $e_{1}$.

###### Proof.

See [Abh93, §3, Generalized Cycle Lemma].∎

######  Lemma 2.2 (Abhyankar’s Lemma).

Let $k$ be perfect, $K/k$ a function field, and $L,M$ finite separable
extensions of $K$. Let $P$ be a prime of $K$, $Q$ a prime of $L$ lying over
$P$, $P^{\prime}$ a prime of $M$ lying over $P$. Assume that

  1. (1)

$Q$ is tamely ramified over $P$, and

  2. (2)

$e(Q/P)$ divides $e(P^{\prime}/P)$.

Then every prime $Q^{\prime}$ of $LM$ lying over both $Q$ and $P^{\prime}$ is
unramified over $M$, i.e. $e(Q^{\prime}/P^{\prime})=1$.

###### Proof.

This follows from [Sti09, Theorem 3.9.1] by the multiplicativity of the
ramification index. ∎

###  2.3. Group theory

We denote by $C_{n}$, $D_{n}$, $S_{n}$, $A_{n}$ the cyclic, dihedral,
symmetric respectively alternating group of degree $n$. We will need the
following group-theoretic result of Jones, which relies on the Classification
of Finite Simple Groups (CFSG) for its proof.

######  Theorem 2.3 (Jones).

Let $G\leqslant S_{n}$ be a primitive permutation group containing a cycle of
length $\ell>1$.

  1. (i)

If $\ell\leq n-3$ then $G\geqslant A_{n}$.

  2. (ii)

If $\ell=n-2$ is prime then either $G\geqslant A_{n}$ or $\ell=2^{k}-1$ is a
Mersenne prime and ${PGL}_{2}(2^{k})\leqslant G\leqslant{P\Gamma
L}_{2}(2^{k})$ with its standard action on
$\mathbb{P}^{1}(\mathbb{F}_{2^{k}})$.

###### Proof.

This is a direct consequence of [Jon14, Corollary 1.3].∎

######  Remark 2.4.

In the case that $\ell$ is prime Theorem 2.3(i) is a classical (and
elementary) result of Jordan [DM96, Theorem 3.3E]. Most (though not all) of
the applications in Section 5 only require this weaker version. In Section 4
Jones’s theorem is used in its full form.

We will also make use of the following elementary fact:

######  Lemma 2.5.

Let $G\leqslant S_{n}$ be a primitive group containing a transposition. Then
$G=S_{n}$.

###### Proof.

See [Wie64, Theorem 13.3].∎

###  2.4. Discriminants

Let $K$ be a field. For a polynomial $f\in K[X]$ of degree $n$ we denote its
leading coefficient by $\mathrm{lc}(f)$, and if
$\alpha_{1},\dots,\alpha_{n}\in\overline{K}$ are the zeros of $f$ (listed with
multiplicity), the discriminant of $f$ is defined by

(2.1)
$\Delta(f)\;=\;\mathrm{lc}(f)^{2n-2}\prod_{i<j}(\alpha_{i}-\alpha_{j})^{2}\in
K.$

Sometimes we will use the notation $\Delta_{X}(f)$ to indicate the variable
with respect to which the discriminant is taken.

If $0\neq f,g\in K[X]$ have roots $\alpha_{1},\ldots,\alpha_{m}$ respectively
$\beta_{1},\ldots,\beta_{n}$, their resultant is defined by

(2.2) $\mathrm{Res}(f,g)\;=\;\mathrm{lc}(f)^{{\rm deg}(g)}\mathrm{lc}(g)^{{\rm
deg}(f)}\prod_{i=1}^{m}\prod_{j=1}^{n}(\alpha_{i}-\beta_{j})\;=\;\mathrm{lc}(f)^{{\rm
deg}(g)}\prod_{i=1}^{m}g(\alpha_{i})\in K.$

We will now summarize some formulae which are useful for computing
discriminants, valid in any characteristic. See [CCG08, §3] for the proofs
(but note that there $\Delta(f)$ is defined without the factor
$\mathrm{lc}(f)^{2n-2}$). First we have

(2.3) $\displaystyle\Delta(f)$ $\displaystyle=$ $\displaystyle(-1)^{\frac{\deg
f\cdot(\deg f-1)}{2}+\deg f\cdot\deg f^{\prime}}\mathrm{lc}(f)^{\deg f-\deg
f^{\prime}-2}\mathrm{Res}(f^{\prime},f)$ (2.4) $\displaystyle=$
$\displaystyle(-1)^{\frac{\deg f\cdot(\deg f-1)}{2}+\deg f\cdot\deg
f^{\prime}}\mathrm{lc}(f)^{\deg f-\deg
f^{\prime}-2}\mathrm{lc}(f^{\prime})^{\deg
f}\prod_{f^{\prime}(\rho)=0}f(\rho)$

provided $f^{\prime}\neq 0$. Here $f^{\prime}$ is the formal derivative of $f$
and the product is over all roots of $f^{\prime}$ counted with multiplicity.
Now let $f,g,h\in K[X]$. From (2.1) and (2.2) we see that

(2.5) $\Delta(fg)=\Delta(f)\Delta(g)\mathrm{Res}(f,g)^{2}$

The resultant is bi-multiplicative, i.e.

(2.6)
$\mathrm{Res}(f,gh)\;=\;\mathrm{Res}(f,g)\mathrm{Res}(f,h),\quad\mathrm{Res}(fg,h)\;=\;\mathrm{Res}(f,h)\mathrm{Res}(g,h)$

and satisfies a symmetry property

(2.7) $\mathrm{Res}(f,g)=(-1)^{\deg f\cdot\deg g}\mathrm{Res}(g,f).$

Finally if $g\equiv h\pmod{f}$ then

(2.8) $\mathrm{Res}(f,g)=\mathrm{lc}(f)^{\deg g-\deg h}\mathrm{Res}(f,h).$

######  Lemma 2.6.

If $\mathrm{char}(K)\neq 2$ and $f\in K[X]$ is a separable polynomial of
degree $n$ then ${\rm Gal}(f/K)\subseteq A_{n}$ if and only if $\Delta(f)$ is
a square in $K$. In particular, ${\rm Gal}(f/K(\sqrt{\Delta(f)}))={\rm
Gal}(f/K)\cap A_{n}$.

###### Proof.

This standard fact is easy to deduce from (2.1).∎

######  Lemma 2.7.

Let $K=k(T)$ and $f\in k[T,X]$ irreducible and monic in $X$. If $P\in k[T]$ is
a finite prime which is ramified in the splitting field of $f$ over $K$, then
$P|\Delta(f)$. Moreover if $P$ divides $\Delta(f)$ exactly once, then in the
extension $K(\alpha)/K$, where $\alpha$ is a root of $f$, there is exactly one
ramified prime lying over $P$, and its ramification index is 2.

###### Proof.

This standard fact can be easily deduced from the properties of the different
and its connection to the discriminant, see e.g. [Ros02, Proposition 7.9 and
Corollary 7.10.2]. ∎

###  2.5. Critical points and critical values

Let $k$ be a perfect field and let $w\in k(X)$ be a non-constant rational
function. One can view $w$ also as a morphism
$w\colon\mathbb{P}^{1}\to\mathbb{P}^{1}$ defined over $k$. For convenience we
will use $X$ for the variable in the domain of $w$, which we denote
$\mathbb{P}^{1}_{X}$ and $U$ in the range of $w$, which we denote
$\mathbb{P}^{1}_{U}$ and we adopt a similar convention for the affine line
$\mathbb{A}^{1}=\mathbb{P}^{1}\setminus\\{\infty\\}$. A point
$\alpha\in\mathbb{P}^{1}_{X}(\bar{k})$ is called a _critical point_ of $w$ if
$w$ is ramified at $\alpha$, i.e. if the corresponding prime of $\bar{k}(X)$
is ramified in the extension of function fields $\bar{k}(X)/\bar{k}(U)$
induced by $w$. A point $\beta\in\mathbb{P}^{1}_{U}(\bar{k})$ is called a
_critical value_ for $w$ if there exists a critical point
$\alpha\in\mathbb{P}^{1}_{X}(\bar{k})$ such that $\beta=w(\alpha)$, or
equivalently, if $w$ is branched over $\beta$.

The following lemma gives a useful description of the critical points and
values of a rational function in terms of derivatives and discriminants.

######  Lemma 2.8.

Let $w=\frac{f}{c}$ with $f,c\in k[X]$, $(f,c)=1$, be a non-constant rational
function.

  1. (i)

Let $\alpha\in\bar{k}\setminus w^{-1}(\infty)$, let $e$ be the ramification
index of $w$ at $\alpha$ and $\nu$ the order of the derivative $w^{\prime}$ at
$\alpha$. Then $e=\nu+1$ if the ramification at $\alpha$ is tame, and
$e\leq\nu$ otherwise. In particular, $\alpha$ is a critical point of $w$ iff
$\alpha$ is a root of $f^{\prime}c-fc^{\prime}$.

  2. (ii)

Assume additionally that $\deg f>\deg c$, $f^{\prime}\neq 0$, and that $c$ is
separable. Then the finite critical values of $w$ are the roots of
$\Delta_{X}(f-Uc)$, more precisely

$\Delta_{X}(f(X)-Uc(X))=a\cdot\prod_{(f^{\prime}c-fc^{\prime})(\alpha)=0}(U-w(\alpha)),$

with $a\in k^{\times}$ and the product over the roots of
$f^{\prime}c-fc^{\prime}$ in $\bar{k}$ with multiplicity.

###### Proof.

(i). We identify $U=w(X)$, viewing $U$ as an element of $k(X)$. This gives the
extension of function fields $k(X)/k(U)$ corresponding to the rational map
$w$. Let $\alpha\in\bar{k}$ be such that $c(\alpha)\neq 0$ and denote
$\beta=w(\alpha)$. Then $X-\alpha$ is a local parameter for
$\mathbb{P}^{1}_{X}$ at the point $\alpha$ and $U-\beta$ is a local parameter
for $\mathbb{P}^{1}_{U}$ at the point $\beta$. We have
$U-\beta=w(X)-\beta=(X-\alpha)^{e}u(X)$ with $u\in k(X)$ which has no pole at
$\alpha$. Taking derivatives we obtain

$w^{\prime}(X)=e(X-\alpha)^{e-1}u(X)+(X-\alpha)^{e}u^{\prime}(X).$

As $\alpha$ is neither a pole nor a zero of $u$, and hence also not a pole of
$u^{\prime}$, the multiplicity $\nu$ of the zero $\alpha$ of $w^{\prime}$ is
$\nu=e-1$ if ${\rm char}(k)\nmid e$, and $\nu\geq e$ if ${\rm char}(k)\mid e$.

(ii). Denote $h=(f^{\prime},c^{\prime})\in\bar{k}[X]$ and write
$f^{\prime}=hu,c^{\prime}=hv,u,v\in\bar{k}[X]$. As $f^{\prime}\neq 0$, also
$u\neq 0$. If $\deg c=0$ the assertion of (ii) is immediate from (i) and
(2.4), so let us assume $\deg c>0$ and then $c^{\prime},v\neq 0$ by our
assumption that $c$ is separable. Using the properties of resultants and
discriminants (2.2),(2.3),(2.6),(2.7),(2.8) and using $\sim$ to denote
equality up to a non-zero multiplicative constant (in $k$), we calculate:

$\displaystyle\Delta_{X}(f-Uc)$
$\displaystyle\stackrel{{\scriptstyle(\ref{eq:disc})}}{{\sim}}$
$\displaystyle\mathrm{Res}(f^{\prime}-Uc^{\prime},f-Uc)$
$\displaystyle\stackrel{{\scriptstyle(\ref{eq:resbimult})}}{{=}}$
$\displaystyle\mathrm{Res}(h,f-Uc)\cdot\mathrm{Res}(u-Uv,f-Uc)$
$\displaystyle\stackrel{{\scriptstyle(\ref{eq:resbimult})}}{{=}}$
$\displaystyle\mathrm{Res}(h,f-Uc)\cdot\frac{\mathrm{Res}(u-Uv,fv-
Ucv)}{\mathrm{Res}(u-Uv,v)}$
$\displaystyle\stackrel{{\scriptstyle(\ref{eq:ressym}),(\ref{eq:resalt})}}{{\sim}}$
$\displaystyle\mathrm{Res}(h,f-Uc)\cdot\frac{\mathrm{Res}(u-Uv,fv-
uc)}{\mathrm{Res}(u,v)}$
$\displaystyle\stackrel{{\scriptstyle(\ref{eq:ressym})}}{{\sim}}$
$\displaystyle\mathrm{Res}(h,f-Uc)\cdot\mathrm{Res}(uc-fv,u-Uv)$
$\displaystyle\stackrel{{\scriptstyle(\ref{eq:resdefinition})}}{{\sim}}$
$\displaystyle\prod_{h(\alpha)=0}\left(c(\alpha)U-f(\alpha)\right)\prod_{(uc-
fv)(\alpha)=0}(u(\alpha)-Uv(\alpha))$ $\displaystyle\sim$
$\displaystyle\prod_{h(\alpha)=0}\left(U-\frac{f(\alpha)}{c(\alpha)}\right)\prod_{(uc-
fv)(\alpha)=0}\left(U-\frac{f^{\prime}(\alpha)}{c^{\prime}(\alpha)}\right)$
$\displaystyle\sim$
$\displaystyle\prod_{(f^{\prime}c-fc^{\prime})(\alpha)=0}(U-w(\alpha)).$

In the last line we used the fact that for a root $\alpha$ of
$(f^{\prime}c-fc^{\prime})/h$ we have
$f^{\prime}(\alpha)/c^{\prime}(\alpha)=w(\alpha)$. ∎

###  2.6. Monodromy of rational functions

The definitions and results in this subsection will be used only in Section
4.2. Let $k$ be a field of characteristic $p\geq 0$. The (arithmetic)
monodromy $\mathrm{Mon}(w)$ of a non-constant rational function $w\in k(X)$
for which the extension $k(X)/k(w)$ is separable is the Galois group of the
Galois closure of $k(X)/k(w)$. When ambiguity about the base field may arise
we write $\mathrm{Mon}_{k}(w)$. We usually view $\mathrm{Mon}(w)$ as a
subgroup of $S_{n}$ with $n=\deg w$ via its action on the generic fiber of the
corresponding morphism $w\colon\mathbb{P}^{1}\to\mathbb{P}^{1}$. A rational
function $w\in k(X)$ is called _indecomposable_ if it cannot be written as a
composition $w=u\circ v$ with $u,v\in k(X)$ where $\deg u,\deg v>1$.

######  Lemma 2.9.

Let $w\in k(X)$ with $k(X)/k(w)$ separable. Then $w$ is indecomposable if and
only if $\mathrm{Mon}(w)\leqslant S_{n}$ is primitive.

###### Proof.

If $w=u\circ v$ with $\deg u>1$, $\deg v>1$, and
$\alpha_{1},\dots,\alpha_{m}\in\overline{k(T)}$ are the roots of $u-T$, then
$v^{-1}(\alpha_{1}),\dots,v^{-1}(\alpha_{m})$ is a nontrivial partition of the
roots of $w-T$, which is preserved by $\mathrm{Mon}(w)$. The converse
implication is proven for example in [Fri70, Lemma 2] in the case of
polynomials, and the case of rational functions can be proven similarly using
Lüroth’s theorem. ∎

######  Definition 2.10.

Let $w\colon\mathbb{P}^{1}\to\mathbb{P}^{1}$ be a tamely ramified rational
function defined over an algebraically closed field $k$. The _ramification
type_ of $w$ (called _passport_ in [AZ15]) is the unordered tuple of
partitions $(\lambda_{1},\ldots,\lambda_{r})$, where $b_{1},\ldots,b_{r}$ are
the distinct branch points of $w$ and $\lambda_{i}$ is the partition of
$n=\deg w$ given by the ramification indices over $b_{i}$. We write partitions
of $n$ as $n_{1}^{e_{1}}\dots n_{r}^{e_{r}}$ with $0\leq n_{1}<\dots<n_{r}$
and $\sum_{i=1}^{r}e_{i}n_{i}=n$.

We recall that the inertia subgroup of $\mathrm{Mon}(w)\leqslant S_{n}$
corresponding to a branch point $b_{i}$ of the tamely ramified rational
function $w$ is a cyclic subgroup of $S_{n}$ with generator of cycle type
precisely the partition of $n$ given by the ramification indices over $b_{i}$.
Moreover, $\mathrm{Mon}(w)$ is generated by the inertia subgroups.

######  Lemma 2.11.

Let $w\colon\mathbb{P}_{k}^{1}\to\mathbb{P}_{k}^{1}$ be a tamely ramified
rational function defined over an algebraically closed field $k$ with
monodromy group $G\leqslant S_{n}$ and ramification type
$\Pi=(\lambda_{1},\ldots,\lambda_{r})$. Then there exists a rational function
$W\colon\mathbb{P}^{1}_{\mathbb{C}}\to\mathbb{P}^{1}_{\mathbb{C}}$ with the
same monodromy and ramification type.

###### Proof.

This is essentially a direct consequence of the comparison theorem for the
étale fundamental group [GR71, §XIII, Corollaire 2.12]. We use the terminology
and notation of [GR71, §XIII]. Let $b_{1},\ldots,b_{r}\in\mathbb{P}^{1}(k)$ be
the branch points of $w$ and denote
$U=\mathbb{P}^{1}\setminus\\{b_{1},\ldots,b_{r}\\}$. By [GR71, §XIII,
Corollaire 2.12] there exist elements $x_{i}\in\pi_{1}^{t}(U),1\leq i\leq r$
such that $x_{1},\ldots,x_{r}$ generate $\pi_{1}^{t}(U)$,
$\prod_{i=1}^{r}x_{i}=1$ and each $x_{i}$ generates a tame inertia group at
$b_{i}$.

The map $w$ induces a finite étale cover $V=w^{-1}(U)\to U$ and therefore
(using the assumption that $w$ is tamely ramified) a surjection
$\delta\colon\pi_{1}^{t}(U)\twoheadrightarrow G$. Denote
$\sigma_{i}=\delta(x_{i})$. The elements $\sigma_{1},\ldots,\sigma_{r}$
generate $G$, satisfy $\prod_{i=1}^{r}\sigma_{i}=1$ and each $\sigma_{i}\in
G\leqslant S_{n}$ has cycle structure $\lambda_{i}$. Now by the Riemann
Existence Theorem (see e.g. [Völ96, Corollary 5.11]) there exists a finite
irreducible cover $W\colon Y\to\mathbb{P}^{1}_{\mathbb{C}}$ with monodromy
group $G$ and ramification type $\Pi$. The genus of $Y$ can be determined by
the Riemann-Hurwitz formula from its ramification type and is therefore 0
(since the same calculation gives the genus of the domain of $w$, since $w$ is
tamely ramified). ∎

######  Lemma 2.12.

Let $w\colon\mathbb{P}^{1}\rightarrow\mathbb{P}^{1}$ be a rational function of
degree $n>1$ over the algebraically closed field $k$ with monodromy group
$\mathrm{Mon}(w)\leqslant S_{n}$. Assume that $w$ has at most two critical
values $a,b$, and that $w^{-1}(a)$ contains only one critical point
$a^{\prime}$. If the ramification index of $a^{\prime}$ is prime, and all
ramification over $b$ is tame, then $\mathrm{Mon}(w)$ is primitive.

###### Proof.

Let $L$ be the Galois closure of the extension $k(X)/k(T)$ induced by $w$. Let
$k(T)\subseteq M\subsetneq k(X)$ be an intermediate extension. Since there is
only one ramified prime of $k(X)$ over $T=a$ and it has prime ramification
index, the extension $M/k(T)$ has to be unramified over $T=a$. However, it is
also tamely ramified over $T=b$ and unramified elsewhere, hence $M=k(T)$ by
the Riemann-Hurwitz formula. So $k(X)/k(T)$ is a minimal proper extension,
hence by Galois theory ${\rm Gal}(L/k(X))$ is maximal in ${\rm Gal}(L/k(T))$,
which means that $\mathrm{Mon}(w)={\rm Gal}(L/k(T))\leqslant S_{n}$ is
primitive. ∎

##  3\. Abelian groups

In this section we first prove the group-theoretic fact mentioned in Remark
1.5 and then verify Conjecture 1.4 for abelian groups.

If $G$ is a group and $a\in G$ is an element we denote by $a^{G}$ the
conjugacy class of $a$. For a subset $S\subseteq G$ we denote
$S^{G}=\bigcup_{a\in S}a^{G}$. Recall that for a finite group $G$ we denote by
$p(G)$ the subgroup generated by its $p$-Sylow subgroups. A _cyclic-by- $p$_
group is an extension of a cyclic group by a $p$-group.

######  Lemma 3.1.

Let $G$ be a finite group. There exist $x_{1},\ldots,x_{r}\in G$ with
$r=\max\\{d(G^{\mathrm{ab}}),1\\}$ such that $G=\langle
x_{1}^{G},\ldots,x_{r}^{G}\rangle$.

###### Proof.

This is a special case of the main theorem in [Kut76], see also [Bae64]. ∎

######  Proposition 3.2.

Let $G$ be a finite group and $p$ a prime number. There exist subgroups
$Q_{1},\ldots,Q_{r}$ of $G$ such that
$r=\max\\{d((G/p(G))^{\mathrm{ab}}),1\\}$, each $Q_{i}$ is cyclic-by-$p$ and
$G=\langle Q_{1}^{G},\ldots,Q_{r}^{G}\rangle$.

###### Proof.

Let $H=G/p(G)$, so that $r=\max\\{d(H^{\mathrm{ab}}),1\\}$. By Lemma 3.1 there
exist $y_{1},\ldots,y_{r}\in H$ with $H=\langle
y_{1}^{H},\ldots,y_{r}^{H}\rangle$. Let $P\leqslant G$ be a $p$-Sylow
subgroup. If $P=1$, i.e. $(|G|,p)=1$, then the claim holds with $Q_{i}=\langle
y_{i}\rangle$, so we may assume $P\neq 1$. By Frattini’s argument [Gor80,
Theorem 3.7] we have $p(G)N_{G}(P)=G$, so we can choose $x_{1},\ldots,x_{r}\in
N_{G}(P)$ such that $y_{i}=p(G)x_{i}$. Denote $Q_{i}=\langle x_{i}\rangle P$.
Each $Q_{i}$ is a subgroup of $G$ since $x_{i}\in N_{G}(P)$, it is cyclic-
by-$p$ since it is an extension of $\langle x_{i}\rangle/(\langle
x_{i}\rangle\cap P)$ by $P$, and finally since $p(G)=\langle P^{G}\rangle$ and
$G=\langle p(G),x_{1}^{G},\ldots,x_{r}^{G}\rangle$ we have $G=\langle
P^{G},x_{1}^{G},\ldots,x_{r}^{G}\rangle=\langle Q_{1}^{G},\ldots
Q_{r}^{G}\rangle$, as required. ∎

######  Theorem 3.3.

Conjecture 1.4 holds for $G$ abelian.

###### Proof.

Let $q=p^{\nu}$ and let $G$ be a nontrivial finite abelian group. The claim is
that $r_{\mathbb{F}_{q}(T)}(G)=1$ if $G$ is a $p$-group and
$r_{\mathbb{F}_{q}(T)}(G)=d(G/p(G))$ otherwise.

We first prove the lower bound. Let $L/\mathbb{F}_{q}(T)$ be an abelian Galois
extension with Galois group $G$ and $\mathbb{F}_{q}$ algebraically closed in
$L$. If $G$ is a $p$-group, the lower bound $r_{\mathbb{F}_{q}(T)}(G)\geq 1$
follows from the Riemann-Hurwitz formula. If $G$ is not a $p$-group, then
$G/p(G)$ is nontrivial, and since $r_{\mathbb{F}_{q}(T)}(G)\geq
r_{\mathbb{F}_{q}(T)}(G/p(G))$ we can assume that $p(G)=1$. Let
$P_{1},\ldots,P_{k}$ be all the finite primes of $\mathbb{F}_{q}(T)$ ramified
in $L$, and let $I_{1},\ldots,I_{k}$ be the corresponding inertia groups (note
that $I_{i}$ are well defined since the extension is abelian). Then each
$I_{i}$ is cyclic, as $P_{i}$ is tamely ramified in $L$. As also the infinite
prime of $\mathbb{F}_{q}(T)$ is tamely ramified in $L$,
$G=\left<I_{1},\ldots,I_{k}\right>$, cf. [Ser08, Proposition 4.4.6]. In
particular, $r_{\mathbb{F}_{q}(T)}(G)\geq k\geq d(G)$.

To show that the lower bound is tight, we construct a suitable Galois
extension. Write $G=G/p(G)\times p(G)$, let $k$ and $p^{\alpha}$ be the
exponents of $G/p(G)$ respectively $p(G)$, and let $d=\max\\{d(G/p(G)),1\\}$.
Choose distinct primes $P_{1},\ldots,P_{d}$ such that $k(q-1)\mid q^{\deg
P_{i}}-1$ (i.e. such that the multiplicative order of $q$ modulo $k(q-1)$
divides $\deg P_{i}$). Let $R_{i}$ be the completion of $\mathbb{F}_{q}[T]$ at
$P_{i}$ and $E_{i}$ its fraction field. By [Neu99, Proposition II.5.7],
$E_{i}^{\times}=\left<\pi\right>\times(\mathbb{F}_{q^{\deg
P_{i}}})^{\times}\times G_{i}$, where $\pi$ is a uniformizer and

$G_{i}=1+\pi\mathbb{F}_{q^{\deg
P_{i}}}[[\pi]]\cong\mathbb{Z}_{p}^{\mathbb{N}}.$

In particular, $R_{i}^{\times}=(\mathbb{F}_{q^{\deg P_{i}}})^{\times}\times
G_{i}$. By class field theory (take inverse limit in [Hay74, Proposition 2.2,
Theorem 2.3, and Theorem 3.2]) the maximal abelian extension of
$\mathbb{F}_{q}(T)$ which is regular over $\mathbb{F}_{q}$, tamely ramified at
infinity, and unramified outside $\\{P_{i},\infty\\}$ has Galois group
$R_{i}^{\times}$ and the tame inertia group at $\infty$ corresponds to
$\mathbb{F}_{q}^{\times}$. Hence $R_{i}^{\times}/\mathbb{F}_{q}^{\times}$ is
the Galois group of the maximal abelian extension that is regular over
$\mathbb{F}_{q}$ and unramified outside $\\{P_{i}\\}$. As $\mathbb{F}_{q}(T)$
has no unramified extensions regular over $\mathbb{F}_{q}$, those extensions
for different $i$ are linearly disjoint, so the Galois group of the maximal
abelian extension that is regular over $\mathbb{F}_{q}$ and unramified outside
$\\{P_{1},\ldots,P_{d}\\}$ is isomorphic to

$\prod_{i=1}^{d}R_{i}^{\times}/\mathbb{F}_{q}^{\times}\cong\prod_{i=1}^{d}C_{(q^{\deg
P_{i}}-1)/(q-1)}\times\mathbb{Z}_{p}^{\mathbb{N}}.$

By the assumption that $k\mid(q^{\deg P_{i}}-1)/(q-1)$, we get that $G/p(G)$
is a quotient of $\prod_{i}C_{(q^{\deg P_{i}}-1)/(q-1)}$ (this is vacantly
true if $G/p(G)=1$). Obviously, $p(G)$ is a quotient of
$\mathbb{Z}_{p}^{\mathbb{N}}$. Hence $G=G/p(G)\times p(G)$ is a quotient of
$\prod_{i}R_{i}^{\times}/\mathbb{F}_{q}^{\times}$. This implies we can realize
$G$ as the Galois group of a geometric Galois extension of $\mathbb{F}_{q}(T)$
unramified outside $\\{P_{1},\ldots,P_{d}\\}$, as needed. ∎

######  Corollary 3.4.

For every prime power $q=p^{\nu}$ and every nontrivial finite group $G$,

$r_{\mathbb{F}_{q}(T)}(G)\geq\max\left\\{d((G/p(G))^{\rm ab}),1\right\\}.$

######  Remark 3.5.

For $K=\mathbb{F}_{p}(T)$, a conjecture was posed in [DeW14, Conjecture 1.1]
and claimed to be proven for abelian finite groups, namely that

$r_{\mathbb{F}_{p}(T)}(G)=\begin{cases}d((G/p(G))^{\rm ab})+1,&\mbox{if
}p\,|\,|G^{\rm ab}|\\\ \max\\{d((G/p(G))^{\rm
ab}),1\\},&\mbox{otherwise}\end{cases}.$

However, already in the case $G=\mathbb{Z}/2p\mathbb{Z}$, $p>2$, the number
obtained there is too big: For example, $G=\mathbb{Z}/6\mathbb{Z}$ can be
realized over $\mathbb{F}_{3}(T)$ with only one ramified prime of degree $2$,
for example by $\mathbb{F}_{3}(T,x,y)$ with $x^{2}=y^{3}-y=(T^{2}+1)^{-1}$.
The incorrect lower bound in the case $p||G^{\rm ab}|$ in [DeW14] presumably
comes from the false assumption there that an extension with such group $G$
must be wildly ramified at infinity.

In any case, at least for $q=p$, [DeW14, Theorem 2.6] proves Conjecture 1.4
for abelian groups of order prime to $p$ (and thus proves a special case of
Theorem 3.3). Moreover, [DeW14, Theorem 3.6] proves Conjecture 1.4 for all
semiabelian $\ell$-groups $G$, with $\ell$ a prime number different from $p$
(building on results from [KS06] and [KNS10]), and [DeW14, Corollary 6.4]
proves Conjecture 1.4 for all $\ell$-groups with $\ell$ a prime number
dividing $q-1$ (both results are stated for $q=p$, but the proofs go through
also for general $q$).

We note however that the proof of [DeW14, Theorem 6.9], which claims to prove
the above conjecture for all nilpotent groups, seems to be incomplete
regarding both $p$-groups ([DeW14, Theorem 2.5]) and $\ell$-groups with $\ell$
coprime to $p-1$ ([DeW14, Corollary 6.8]).

##  4\. $S_{n}$ and $A_{n}$ for $p>n$

In this section we obtain Galois extensions of group $S_{n}$ and $A_{n}$ with
ramification over at most two primes in the case where $n$ is smaller than the
characteristic $p$. We start with splitting fields of polynomials of the form
$f(X)-T$ in Section 4.1 and $f(X)-Tc(X)$ in Section 4.2, which produce
extensions with one ramified finite prime and tame ramification over infinity.
In Section 4.3 we then explain how to eliminate the tame ramification at
infinity in many cases. Finally, in Section 4.4 we present yet another
approach to obtain extensions with two ramified primes, by working with
trinomials and applying recent results on small gaps between primes in
function fields.

We recall that if $f\in\mathbb{F}_{q}[T,X]$ is monic in $X$, then the only
primes of $K=\mathbb{F}_{q}(T)$ that possibly ramify in the splitting field
$L$ of $f$ over $K$ are the divisors of the discriminant $\Delta_{X}(f)$, and
possibly the infinite prime of $K$ (Lemma 2.7). In particular, if
$\Delta_{X}(f)$ is irreducible or, more generally, a prime power, at most two
primes of $K$ ramify in $L$.

###  4.1. Two ramified primes via Morse polynomials

We start with splitting fields of polynomials of the simple from $f(X)-T$,
where in some cases the classical theory of Morse polynomials produces
suitable extensions of group $S_{n}$ with two ramified primes. Let $k$ be a
field of characteristic $p\neq 2$, and recall that a polynomial $f\in k[X]$ of
degree $n$ not divisible by $p$ is Morse if it has exactly $n-1$ distinct
critical values, i.e. the roots $\alpha_{1},\dots,\alpha_{n-1}$ of
$f^{\prime}$ in $\bar{k}$ are simple and $f(\alpha_{i})\neq f(\alpha_{j})$ for
$i\neq j$. If $f$ is Morse, then ${\rm Gal}(f(X)-T/\overline{k}(T))=S_{n}$ by
[Ser08, Theorem 4.4.5], so the splitting field of $f(X)-T$ over $k(T)$ is
geometric with Galois group $S_{n}$.

######  Lemma 4.1.

Let $f\in k[X]$ be Morse of degree $n$ with $p\nmid n$. Then $f^{\prime}$ is
irreducible if and only if $D(T):=\Delta_{X}(f(X)-T)\in k[T]$ is irreducible.

###### Proof.

Let $A$ be the set of roots of $f^{\prime}$, and $G={\rm Gal}(k(A)/k)$. By
(2.4) or Lemma 2.8, $D\sim\prod_{a\in A}(T-f(a))$ (where as before $\sim$
denotes equality up to a non-zero constant). The fact that $f$ is Morse
implies that $f$ is injective on $A$, so $f$ induces an isomorphism of
$G$-sets $A\rightarrow f(A)$. In particular, $k(f(a))=k(a)$ for every $a\in
A$, hence $D\sim\prod_{a\in A}(T-f(a))$ is irreducible if and only if
$f^{\prime}\sim\prod_{a\in A}(X-a)$ is. ∎

######  Proposition 4.2.

Let $n\geq 3$ and $q=p^{\nu}$ with $p>n$. There are
$\frac{q^{n}}{n-1}+O_{n}(q^{n-1})$ many monic Morse $f\in\mathbb{F}_{q}[X]$ of
degree $n$ such that the splitting field of $f(X)-T$ over $\mathbb{F}_{q}(T)$
is geometric with Galois group $S_{n}$, and $\Delta_{X}(f(X)-T)$ is
irreducible.

###### Proof.

Let $M_{n}\cong\mathbb{A}^{n}$ be the space of monic polynomials of degree $n$
(identified with the $n$-tuple of non-leading coefficients) considered as a
variety over $\mathbb{F}_{q}$. The subset $\mathcal{M}\subseteq M_{n}$ of
monic Morse polynomials of degree $n$ is Zariski-open and dense with
complement $M_{n}\setminus\mathcal{M}$ the zero set of a polynomial in the
coefficients $a_{0},\dots,a_{n-1}$ of degree $O_{n}(1)$, see [Gey99,
Proposition 4.3 and (2) in its proof]. Thus the elementary estimate [LW54,
Lemma 1] gives that the set $\mathcal{M}(\mathbb{F}_{q})$ of $f\in
M_{n}(\mathbb{F}_{q})$ that are Morse satisfies
$|\mathcal{M}(\mathbb{F}_{q})|=q^{n}+O_{n}(q^{n-1})$.

Let $\mathcal{P}$ denote the set of $f\in M_{n}(\mathbb{F}_{q})$ for which
$f^{\prime}$ is irreducible. As $p>n$, the map
$M_{n}(\mathbb{F}_{q})\rightarrow M_{n-1}(\mathbb{F}_{q})$,
$f\mapsto\frac{1}{n}f^{\prime}$ is surjective with fibers of size $q$, so by
the Prime Polynomial Theorem,
$|\mathcal{P}|=\frac{q^{n}}{n-1}+O_{n}(q^{(n-1)/2})$. It follows that

$|\mathcal{M}(\mathbb{F}_{q})\cap\mathcal{P}|=|\mathcal{P}\setminus(M_{n}(\mathbb{F}_{q})\setminus\mathcal{M}(\mathbb{F}_{q}))|=\frac{q^{n}}{n-1}+O_{n}(q^{n-1}),$

since by the above
$|M_{n}(\mathbb{F}_{q})\setminus\mathcal{M}(\mathbb{F}_{q})|=O_{n}(q^{n-1})$.
Now for every $f\in\mathcal{M}(\mathbb{F}_{q})$, the splitting field of $f-T$
is geometric with Galois group $S_{n}$ (see above), and $\Delta_{X}(f-T)$ is
irreducible if and only if $f\in\mathcal{P}$ (Lemma 4.1), so the claim
follows. ∎

######  Lemma 4.3.

Let $f\in\mathbb{F}_{q}[X]$ of degree $n$ where $q=p^{\nu}$, $p\nmid n$ and
$n-1$ is prime. If $f^{\prime}$ is irreducible, then $f$ is Morse.

###### Proof.

Suppose that $f^{\prime}$ is irreducible and that there exist $a\neq b$ in
$\overline{\mathbb{F}}_{q}$ with $f^{\prime}(a)=f^{\prime}(b)=0$ and
$f(a)=f(b)$. As $f^{\prime}$ is irreducible of degree $n-1$, we have
$a,b\in\mathbb{F}_{q^{n-1}}$ and there exists $1\neq\sigma\in{\rm
Gal}(\mathbb{F}_{q^{n-1}}/\mathbb{F}_{q})$ with $b=a^{\sigma}$. Thus
$c:=f(a)=f(b)=f(a)^{\sigma}$ is in the fixed field of $\sigma$, which is
$\mathbb{F}_{q}$ due to the assumption that $n-1$ is prime. So
$f-c\in\mathbb{F}_{q}[X]$ has root $a$, hence $f-c=f^{\prime}g$ for some
$g\in\mathbb{F}_{q}[X]$ of degree $1$. Deriving gives
$f^{\prime\prime}g=f^{\prime}(1-g^{\prime})$, so since $f^{\prime}$ is
irreducible and ${\rm deg}(f^{\prime})>{\rm deg}(g)$, we conclude that
$f^{\prime}|f^{\prime\prime}$, which contradicts the separability of
$f^{\prime}$. ∎

######  Proposition 4.4.

Let $p\geq 3$ and $q=p^{\nu}$. Suppose that $n-1$ is prime and
$n\in\\{2,\dots,p-1\\}\cup\\{p+1\\}$. Then there exists
$f\in\mathbb{F}_{q}[X]$ of degree $n$ such that the splitting field of
$f(X)-T$ over $\mathbb{F}_{q}(T)$ is geometric with Galois group $S_{n}$, and
$\Delta_{X}(f(X)-T)$ is irreducible.

###### Proof.

There exists $f\in\mathbb{F}_{q}[X]$ of degree $n$ with $f^{\prime}$
irreducible: If $n<p$ we can choose $f^{\prime}$ to be any irreducible
polynomial of degree $n-1$; if $n=p+1$ we can choose $f^{\prime}$ to be the
minimal polynomial of any $\alpha\in\mathbb{F}_{q^{p}}\setminus\mathbb{F}_{q}$
with ${\rm Tr}_{\mathbb{F}_{q^{p}}/\mathbb{F}_{q}}(\alpha)=0$, of which there
exist $q^{p-1}-q$ many. By Lemma 4.3, $f$ is Morse, and therefore the
splitting field of $f-T$ is geometric with Galois group $S_{n}$, and
$\Delta_{X}(f-T)$ is irreducible by Lemma 4.1. ∎

###  4.2. Two ramified primes via monodromy of rational functions

The main result in this subsection is Theorem 4.11 below, which generalizes
the ideas of the previous subsection by studying polynomials of the form
$f(X)-Tc(X)$ rather than $f(X)-T$. We adopt the notation and terminology of
Section 2.6.

The proof of the following key proposition uses the classification of
monodromy of indecomposable rational functions (in special cases) by Adrianov
and Zvonkin [AZ15, Adr17], which is based on the classification of primitive
permutation groups containing a cycle by Jones [Jon14] combined with the
earlier monodromy classification results of Müller [Mül93, Mül94]. These
results rely heavily on the CFSG, although in many cases more elementary group
theory is sufficient. Recall the definition of the ramification type of a
rational function (Definition 2.10).

######  Proposition 4.5.

Let $k$ be algebraically closed of characteristic $p\geq 0$. Let $f,c\in k[X]$
be polynomials with $\deg f=n,\deg c=m\leq n-2,(f,c)=1$ and $c$ squarefree.
Assume further that $w=\frac{f}{c}\in k(X)$ is indecomposable and $m\not\equiv
n\pmod{p}$ if $p>0$.

  1. (i)

Assume that $p\neq 2$, $m\equiv n\pmod{2}$ and $g=f^{\prime}c-fc^{\prime}$ is
squarefree. Then $\mathrm{Mon}(w)=S_{n}$ or one of the following:

     * •

$n=6,m=2,\mathrm{Mon}(w)=PGL_{2}(5)$ with ram. type
$(1^{2}2^{2},1^{2}2^{2},2^{3},1^{2}4^{1})$.

     * •

$n=8,m=2,\mathrm{Mon}(w)=PGL_{2}(7)$ with ram. type
$(1^{2}2^{3},1^{2}2^{3},1^{2}2^{3},1^{2}6^{1})$.

     * •

$n=9,m=1,\mathrm{Mon}(w)=AGL_{2}(3)$ with ram. type
$(1^{3}2^{3},1^{3}2^{3},1^{3}2^{3},1^{1}8^{1})$.

     * •

$n=10,m=2,\mathrm{Mon}(w)=P\Gamma L_{2}(9)$ with ram. type
$(1^{4}2^{3},1^{4}2^{3},2^{5},1^{2}8^{1})$.

  2. (ii)

Assume that $p\notin\\{2,3\\}$, $m\not\equiv n\pmod{2}$, and
$f^{\prime}c-fc^{\prime}=g^{2}$ where $g\in k[X]$ is squarefree. Then
$\mathrm{Mon}(w)=A_{n}$ or one of the following:

     * •

$n=8,m=1,\mathrm{Mon}(w)=L_{2}(7)$ or $A\Gamma L_{1}(8)$ with ram. type
$(1^{2}3^{2},1^{2}3^{2},1^{1}7^{1})$.

     * •

$n=9,m=0,\mathrm{Mon}(w)=P\Gamma L_{2}(8)$ with ram. type
$(1^{3}3^{2},1^{3}3^{2},9^{1})$.

     * •

$n=12,m=1,\mathrm{Mon}(w)=M_{12}$ with ram. type
$(1^{3}3^{3},1^{3}3^{3},1^{1}11^{1})$.

     * •

$n=24,m=1,\mathrm{Mon}(w)=M_{24}$ with ram. type
$(1^{6}3^{6},1^{6}3^{6},1^{1}23^{1})$.

###### Proof.

The classification of monodromy groups of indecomposable rational functions
with precisely one multiple pole over $\mathbb{C}$ is given in [AZ15, §2.2 and
Theorem 12] (case of 3 critical values) and [Adr17, Theorem 1] (case of 4 or
more critical values). By Lemma 2.11 the possible monodromy and ramification
type pairs for a tamely ramified rational function over an algebraically
closed field in arbitrary characteristic can only be the ones occurring over
$\mathbb{C}$. The lists of possible monodromy groups different from
$A_{n},S_{n},C_{n},D_{n}$ together with the corresponding ramification data
appear in [AZ15, §3] and [Adr17, Table 1]. We will refer to these henceforth
as _the tables_.

The case $C_{n}$ only occurs if $n$ is prime with ram. type $(n^{1},n^{1})$,
and the case $D_{n}$ occurs only when $n>2$ is prime with ram. type
$(1^{1}2^{(n-1)/2},1^{1}2^{(n-1)/2},n^{1})$, see [AZ15, Section 2.2] where
this can be read off from the corresponding dessins d’enfants. In both cases
there is a totally ramified branch point (i.e. after applying a fractional-
linear transformation $w$ becomes a polynomial of degree $n$).

Consider the rational map
$w=\frac{f}{c}\colon\mathbb{P}^{1}\to\mathbb{P}^{1}$. By our assumptions it
has degree $n$, simple poles at the roots of $c$ and a pole of multiplicity
$n-m$ at infinity. Note that $\deg(f^{\prime}c-fc^{\prime})=n+m-1$ as
$m\not\equiv n\pmod{p}$.

$(i)$. Assume that $p\neq 2$, $m\equiv n\pmod{2}$ and
$g=f^{\prime}c-fc^{\prime}$ is squarefree. Assume further that
$\mathrm{Mon}(w)\neq S_{n}$, in particular $n>2$. As all zeros of $g$ are
simple and $p\neq 2$, the ramification points of $w$ are the zeros of $g$ with
ramification index 2 (Lemma 2.8) and $\infty$ with ramification index $n-m$.
In particular, $w$ is tamely ramified (as $p\neq 2$ and $p\nmid n-m$) and the
ramification type of $w$ has the form

$\left(1^{a_{1}}2^{b_{1}},\ldots,1^{a_{r}}2^{b_{r}},1^{m}(n-m)^{1}\right),$

where $r$ is the number of finite critical values of $w$ and $a_{i}+2b_{i}=n$
for all $i$. As $\sum_{i=1}^{r}b_{i}=\deg g\geq n-1>n/2$, we see that $r\geq
2$.

Note that $\mathrm{Mon}(w)$ cannot be $A_{n}$ since the last entry in the
ramification type corresponds to an odd permutation, it cannot be $C_{n}$ with
$n$ prime since then it would have only one finite critical value and it
cannot be $D_{n}$ with $n$ prime since this case occurs only if $w$ has a
totally ramified branch point (i.e. $n^{1}$ in its ramification type), but if
$n>2$ is prime then $m$ must be odd and thus $w$ is not a polynomial.
Therefore $\mathrm{Mon}(w)$ must be one of the entries in the tables.

Going through the tables we see that the only entries with a ramification type
as above (with $n-m$ even) are:

  * •

4/6.2 with $\mathrm{Mon}(w)=PGL_{2}(5)$ and ramification type
$(1^{2}2^{2},1^{2}2^{2},2^{3},1^{2}4^{1})$.

  * •

4/8.4 with $\mathrm{Mon}(w)=PGL_{2}(7)$ and ramification type
$(1^{2}2^{3},1^{2}2^{3},1^{2}2^{3},1^{2}6^{1})$.

  * •

4/9.1 with $\mathrm{Mon}(w)=AGL_{2}(3)$ and ramification type
$(1^{3}2^{3},1^{3}2^{3},1^{3}2^{3},1^{1}8^{1})$.

  * •

4/10.1 with $\mathrm{Mon}(w)=P\Gamma L_{2}(9)$ and ramification type
$(1^{4}2^{3},1^{4}2^{3},2^{5},1^{2}8^{1})$.

$(ii)$. Now assume instead that $p\notin\\{2,3\\}$, $m\not\equiv n\pmod{2}$
and $f^{\prime}c-fc^{\prime}=g^{2}$ with $g$ squarefree. As all zeros of
$g^{2}$ have multiplicity $2$ and $p\notin\\{2,3\\}$, the ramification points
of $w$ are the zeros of $g$ with ramification index 3 (Lemma 2.8) and $\infty$
with ramification index $n-m$. In particular, $w$ is tamely ramified (as
$p\neq 3$ and $p\nmid n-m$) and has a ramification type of the form

$\left(1^{a_{1}}3^{b_{1}},\ldots,1^{a_{r}}3^{b_{r}},1^{m}(n-m)^{1}\right),$

where $r$ is the number of finite critical values of $w$. Since now $n-m$ is
odd, all entries in the ramification type correspond to even permutations. So
since $\mathrm{Mon}(w)$ is generated by the inertia groups,
$\mathrm{Mon}(w)\leqslant A_{n}$, which excludes the case $S_{n}$. If $n=3$
(and $m=0$), then $r=1$ and $\mathrm{Mon}(w)=A_{3}=C_{3}$. If $n>3$, then
similarly to part (i) one can argue that $r\geq 2$ and that one can exclude
the cases $C_{n}$ and $D_{n}$ ($D_{n}$ comes with ramification index 2 at the
finite critical points).

Now once again we go over the tables and list the entries with a ramification
type of the above form (with $n-m$ odd):

  * •

8.1 with $\mathrm{Mon}(w)=A\Gamma L_{1}(8)$ and ramification type
$(1^{2}3^{2},1^{2}3^{2},1^{1}7^{1})$.

  * •

8.9 with $\mathrm{Mon}(w)=L_{2}(7)$ and ramification type
$(1^{2}3^{2},1^{2}3^{2},1^{1}7^{1})$.

  * •

9.7 with $\mathrm{Mon}(w)=P\Gamma L_{2}(8)$ and ramification type
$(1^{3}3^{2},1^{3}3^{2},9^{1})$.

  * •

12.10 with $\mathrm{Mon}(w)=M_{12}$ and ramification type
$(1^{3}3^{3},1^{3}3^{3},1^{1}11^{1})$.

  * •

24.5 with $\mathrm{Mon}(w)=M_{24}$ and ramification type
$(1^{6}3^{6},1^{6}3^{6},1^{1}23^{1})$.∎

######  Remark 4.6.

While the above proposition relies on the CFSG in general, if we assume that
$n-m$ is small, much more elementary group theory is sufficient. For example
if $n-m\leq 15$ the result can be derived from the classification by Manning
of primitive permutation groups of class $\leq 15$ (i.e. primitive groups
which have a non-identity element fixing all but $\leq 15$ points). See
[Wie64, §II.15] for the list of references.

######  Lemma 4.7.

Let $w\colon\mathbb{P}^{1}\to\mathbb{P}^{1}$ be a tamely ramified rational
function defined over a field $k$. Assume that $w$ has a multiple pole at
$\infty$ and all other poles over $\overline{k}$ are simple. Assume further
that the numerator $g\in k[X]$ of the derivative $w^{\prime}$ is irreducible
or the square of an irreducible polynomial. Then $w$ is indecomposable over
$k$.

###### Proof.

It will be convenient to introduce two new variable symbols $T,U$ and consider
three separate copies of $\mathbb{P}^{1}_{k}$ which we will denote
$\mathbb{P}^{1}_{X},\mathbb{P}^{1}_{T},\mathbb{P}^{1}_{U}$. For the sake of
contradiction, assume that $w=v\circ u$ is the composition of two rational
functions $u,v\in k(X)$ with $\deg u,\deg v>1$. Since $w(\infty)=\infty$,
assume without loss of generality that $u(\infty)=\infty$ and
$v(\infty)=\infty$. Consider the corresponding coverings

$\mathbb{P}^{1}_{X}\xrightarrow{u}\mathbb{P}^{1}_{T}\xrightarrow{v}\mathbb{P}^{1}_{U}.$

For a divisor $D=\sum_{P}n_{P}P$ on $\mathbb{P}^{1}_{X}$ we denote

$D^{0}=\sum_{w(P)\neq\infty}n_{P}P,\quad D^{\infty}=\sum_{w(P)=\infty}n_{P}P$

and similarly for a divisor $D$ on $\mathbb{P}^{1}_{T}$ we denote

$D^{0}=\sum_{v(P)\neq\infty}n_{P}P,\quad D^{\infty}=\sum_{v(P)=\infty}n_{P}P.$

Consider the differents $\mathfrak{d}_{w}$, $\mathfrak{d}_{u}$,
$\mathfrak{d}_{v}$ of $w$, $u$ and $v$ as divisors on $\mathbb{P}^{1}_{X}$,
$\mathbb{P}^{1}_{X}$ and $\mathbb{P}^{1}_{T}$, respectively. The Riemann-
Hurwitz formula implies that $\deg\mathfrak{d}_{u}=2\deg u-2$ and
$\deg\mathfrak{d}_{v}=2\deg v-2$. Since all finite poles of $w$ are simple,
the same holds for the finite poles of $u$ and $v$. Moreover, since $w$ is
tamely ramified, so are $u$ and $v$, which therefore each ramify over at least
two geometric primes. Thus, $\mathfrak{d}_{u}^{0}>0$ and
$\mathfrak{d}_{v}^{0}>0$. By the basic properties of differents we have
$\mathfrak{d}_{w}=u^{*}\mathfrak{d}_{v}+\mathfrak{d}_{u}$. Therefore, as all
divisors in this relation are effective,

(4.1) $\mathfrak{d}_{w}^{0}=u^{*}\mathfrak{d}_{v}^{0}+\mathfrak{d}_{u}^{0}.$

Note that $\mathfrak{d}_{w}^{0}$ is precisely the zero divisor of the
numerator of $w^{\prime}$, which by assumption is of the form
$\mathfrak{d}_{w}^{0}=Q$ or $\mathfrak{d}_{w}^{0}=2Q$ for some prime divisor
$Q$. The first case is already excluded by (4.1) (since
$u^{*}\mathfrak{d}_{v}^{0},\mathfrak{d}_{u}^{0}>0$, their sum cannot be
prime), so assume that $\mathfrak{d}_{w}^{0}=2Q$. Then (4.1) gives that
$u^{*}\mathfrak{d}_{v}^{0}=\mathfrak{d}_{u}^{0}=Q$. We have

$(\deg u)(\deg\mathfrak{d}_{v}^{0})=\deg u^{*}\mathfrak{d}_{v}^{0}=\deg
Q=\deg\mathfrak{d}_{u}^{0}\leq\deg\mathfrak{d}_{u}=2\deg u-2,$

which implies $(\deg u)(\deg\mathfrak{d}_{v}^{0}-2)\leq-2$, hence $\deg u=2$
and $\deg\mathfrak{d}_{v}^{0}=1$. Since
$\deg\mathfrak{d}_{v}^{0}+\deg\mathfrak{d}_{v}^{\infty}=2\deg v-2$, the latter
shows that $\mathfrak{d}_{v}^{\infty}>0$. Since
$\deg\mathfrak{d}_{u}^{0}+\deg\mathfrak{d}_{u}^{\infty}=2\deg u-2=2$ and
$\deg\mathfrak{d}_{u}^{0}=\deg Q=(\deg u)(\deg\mathfrak{d}_{v}^{0})\geq 2$ it
also follows that $\mathfrak{d}_{u}^{\infty}=0$. Thus $v$ ramifies over
$\infty$ but $u$ does not, which implies that $w=v\circ u$ has $\deg u=2$ many
multiple poles over $\overline{k}$, contradicting our assumption. ∎

######  Proposition 4.8.

Let $p$ be prime, $q$ a power of $p$, and $m\geq 0$ and $n\geq m+2$ integers
with $n\not\equiv m\pmod{p}$. Let $f,c\in\mathbb{F}_{q}[X]$ be polynomials
with ${\rm deg}(f)=n$, ${\rm deg}(c)=m$, $(f,c)=1$ and $c$ squarefree. Let
$w=\frac{f}{c}$.

  1. (i)

Assume $p>2$, $m\equiv n\pmod{2}$, $m\neq 2$ and $g=f^{\prime}c-fc^{\prime}$
is irreducible. Then $\mathrm{Mon}_{\overline{\mathbb{F}}_{q}}(w)=S_{n}$ or

     * •

$n=9,m=1$, $\mathrm{Mon}_{\overline{\mathbb{F}}_{q}}(w)=AGL_{2}(3)$ with ram.
type $(1^{3}2^{3},1^{3}2^{3},1^{3}2^{3},1^{1}8^{1})$

  2. (ii)

Assume $p>3$, $m\not\equiv n\pmod{2}$, and $f^{\prime}c-fc^{\prime}=g^{2}$
where $g\in\mathbb{F}_{q}[X]$ is irreducible. Then
$\mathrm{Mon}_{\overline{\mathbb{F}}_{q}}(w)=A_{n}$ or one of the following:

     * •

$n=8,m=1,\mathrm{Mon}_{\overline{\mathbb{F}}_{q}}(w)=A\Gamma L_{1}(8)$ or
$L_{2}(7)$ and ram. type $(1^{2}3^{2},1^{2}3^{2},1^{1}7^{1})$.

     * •

$n=9,m=0,\mathrm{Mon}_{\overline{\mathbb{F}}_{q}}(w)=P\Gamma L_{2}(8)$ and
ram. type $(1^{3}3^{2},1^{3}3^{2},9^{1})$.

     * •

$n=12,m=1,\mathrm{Mon}_{\overline{\mathbb{F}}_{q}}(w)=M_{12}$ and ram. type
$(1^{3}3^{3},1^{3}3^{3},1^{1}11^{1})$.

     * •

$n=24,m=1,\mathrm{Mon}_{\overline{\mathbb{F}}_{q}}(w)=M_{24}$ and ram. type
$(1^{6}3^{6},1^{6}3^{6},1^{1}23^{1})$.

###### Proof.

Note that
$\mathrm{Mon}_{\overline{\mathbb{F}}_{q}}(w)\unlhd\mathrm{Mon}_{\mathbb{F}_{q}}(w)$
are transitive subgroups of $S_{n}$. By Lemma 4.7, $w$ is indecomposable over
$\mathbb{F}_{q}$, hence $\mathrm{Mon}_{\mathbb{F}_{q}}(w)$ is primitive (Lemma
2.9), and Lemma 2.1(ii) applied to the infinite prime shows that
$\mathrm{Mon}_{\overline{\mathbb{F}}_{q}}(w)$ contains an $(n-m)$-cycle. We
treat parts $(i)$ and $(ii)$ simultaneously but distinguish several cases
according to the value of $m$:

Case $m=0$: In this case $w$ is a polynomial. By [FM69, Theorem 3.5] a
polynomial which is indecomposable over $\mathbb{F}_{q}$ of degree coprime
with $p$ is also indecomposable over $\overline{\mathbb{F}}_{q}$ and therefore
in our case $w$ is indecomposable over $\overline{\mathbb{F}}_{q}$. The
conditions of Proposition 4.5(i) resp. (ii) hold and therefore
$\mathrm{Mon}_{\overline{\mathbb{F}}_{q}}(w)=S_{n}$ (none of the exceptional
cases has $m=0$) resp. $\mathrm{Mon}_{\overline{\mathbb{F}}_{q}}(w)=A_{n}$ or
the exceptional case with $m=0$ appearing in Proposition 4.5(ii).

Case $m=1$: $\mathrm{Mon}_{\overline{\mathbb{F}}_{q}}(w)$ is transitive and
contains an $(n-1)$-cycle and is therefore primitive, hence $w$ is
indecomposable over $\overline{\mathbb{F}}_{q}$ (Lemma 2.9). The conditions of
Proposition 4.5(i) resp. (ii) hold and therefore
$\mathrm{Mon}_{\overline{\mathbb{F}}_{q}}(w)=S_{n}$ or the exceptional case
with $m=1$ in Proposition 4.5(i) resp.
$\mathrm{Mon}_{\overline{\mathbb{F}}_{q}}(w)=A_{n}$ or one of the exceptional
cases with $m=1$ listed in Proposition 4.5(ii).

Case $m=2$: Note that this case is allowed only in part $(ii)$. Since $n\geq
5$ is odd by assumption, $(n-2)$ is coprime with $n$ and so
$\mathrm{Mon}_{\overline{\mathbb{F}}_{q}}(w)$ must be primitive since it is
transitive and contains an $(n-2)$-cycle. Hence $w$ is indecomposable over
$\overline{\mathbb{F}}_{q}$ (Lemma 2.9). The conditions of Proposition 4.5(ii)
hold and therefore $\mathrm{Mon}_{\overline{\mathbb{F}}_{q}}(w)=A_{n}$.

Case $m>2$: Here $\mathrm{Mon}_{\mathbb{F}_{q}}(w)$ is primitive and contains
an $(n-m)$-cycle, so by Theorem 2.3(i) it is $S_{n}$ or $A_{n}$. Note that
$n\geq m+2\geq 5$, so $A_{n}$ is simple and $S_{n}$ has a unique nontrivial
cyclic quotient. Now
$\mathrm{Mon}_{\overline{\mathbb{F}}_{q}}(w)\unlhd\mathrm{Mon}_{\mathbb{F}_{q}}(w)$
and
$\mathrm{Mon}_{\mathbb{F}_{q}}(w)/\mathrm{Mon}_{\overline{\mathbb{F}}_{q}}(w)$
is cyclic, hence also $\mathrm{Mon}_{\overline{\mathbb{F}}_{q}}(w)$ is $S_{n}$
or $A_{n}$. Under the assumptions of part $(i)$, the $(n-m)$-cycle is odd, and
thus $\mathrm{Mon}_{\overline{\mathbb{F}}_{q}}(w)=S_{n}$. Under the
assumptions of part $(ii)$, the $(n-m)$-cycle is even, as are all the other
inertia subgroups (the ramification indices at the finite ramified primes all
equal $3$), so since $\mathrm{Mon}_{\overline{\mathbb{F}}_{q}}(w)$ is
generated by the inertia subgroups,
$\mathrm{Mon}_{\overline{\mathbb{F}}_{q}}(w)=A_{n}$. ∎

For a squarefree polynomial $c\in\mathbb{F}_{q}[X]$ we denote

$\mathcal{H}_{c}=\left\\{R^{p}:R\in\left(\mathbb{F}_{q}[X]/c^{2}\mathbb{F}_{q}[X]\right)^{\times}\right\\}\leqslant\left(\mathbb{F}_{q}[X]/c^{2}\mathbb{F}_{q}[X]\right)^{\times}.$

Denote also

(4.2)
$\pi_{c}=\prod_{i=1}^{m}(X-\alpha_{i})^{2}\cdot\sum_{i=1}^{m}\frac{1}{(X-\alpha_{i})^{2}}=\sum_{i=1}^{m}\prod_{j\neq
i}(X-\alpha_{j})^{2}\in\mathbb{F}_{q}[X],$

where $\alpha_{1},\dots,\alpha_{m}\in\overline{\mathbb{F}}_{q}$ are the roots
of $c$. Note that $(\pi_{c},c)=1$. In short we write $\pi_{c}\mathcal{H}_{c}$
for the coset $(\pi_{c}+c^{2}\mathbb{F}_{q}[X])\mathcal{H}_{c}$ in
$(\mathbb{F}_{q}[X]/c^{2})^{\times}$.

######  Lemma 4.9.

Let $c\in\mathbb{F}_{q}[X],\deg c=m$ be a squarefree polynomial and
$g\in\mathbb{F}_{q}[X]$ a polynomial such that $(g,c)=1$ and $\deg g\geq 2m$.
Denote $n=\deg(g)-m+1$ and assume that $p=\mathrm{char}(\mathbb{F}_{q})>n-m$.
Assume further that $g\bmod c^{2}\in\pi_{c}\mathcal{H}_{c}$. Then there exists
$f\in\mathbb{F}_{q}[X],\deg f=n,(f,c)=1$ such that
$g=f^{\prime}c-fc^{\prime}$.

###### Proof.

Without loss of generality, $c$ is monic. Write
$g=uc^{2}+v,u,v\in\mathbb{F}_{q}[X],\deg v<\deg c^{2}=2m$. Let
$\alpha_{1},\ldots,\alpha_{m}\in\overline{\mathbb{F}}_{q}$ be the roots of $c$
(they are pairwise distinct since $c$ is squarefree). By assumption
$v\equiv\pi_{c}w\pmod{c^{2}}$, where $w$ is a $p$-th power modulo each
$(X-\alpha_{i})^{2}$, the latter condition implying $w\equiv
a_{i}\pmod{(X-\alpha_{i})^{2}}$ for a (uniquely determined) $0\neq
a_{i}\in\overline{\mathbb{F}}_{q}$. It follows using (4.2) that $v\equiv
a_{i}\prod_{j\neq i}(X-\alpha_{j})^{2}\pmod{(X-\alpha_{i})^{2}}$ for each
$1\leq i\leq m$, hence $v\equiv\sum_{i=1}^{m}a_{i}\prod_{j\neq
i}(X-\alpha_{j})^{2}\pmod{c^{2}}$. Since both sides of this congruence are of
degree less than $\deg c^{2}$, we get the following partial fraction
decomposition over $\overline{\mathbb{F}}_{q}$:

$\frac{v}{c^{2}}=\sum_{i=1}^{m}\frac{a_{i}}{(X-\alpha_{i})^{2}}.$

We therefore can write

$-\sum_{i=1}^{m}\frac{a_{i}}{X-\alpha_{i}}=\frac{\psi}{c}\quad\mbox{ with
}\quad\psi\in\overline{\mathbb{F}}_{q}[X],\deg\psi<m,$

and since the map
$\\{\alpha_{1},\dots,\alpha_{m}\\}\rightarrow\\{a_{1},\dots,a_{m}\\}$,
$\alpha_{i}\mapsto a_{i}$, is a morphism of ${\rm
Gal}(\overline{\mathbb{F}}_{q}/\mathbb{F}_{q})$-sets, in fact
$\psi\in\mathbb{F}_{q}[X]$. We have $(\psi/c)^{\prime}=v/c^{2}$. Now $\deg
u=n-m-1<p-1$ and therefore there exists
$\phi\in\mathbb{F}_{q}[X],\deg\phi=n-m$ such that $\phi^{\prime}=u$. Denote
$f=\phi c+\psi$. We have $\deg f=n$ and

$\left(\frac{f}{c}\right)^{\prime}=\phi^{\prime}+\left(\frac{\psi}{c}\right)^{\prime}=u+\frac{v}{c^{2}}=\frac{g}{c^{2}},$

from which it follows that $g=f^{\prime}c-fc^{\prime}$. ∎

######  Proposition 4.10.

Let $p$ be prime, $q$ a power of $p$, $m\geq 0,n\geq m+2$ integers such that
$p>n-m$. Let $c\in\mathbb{F}_{q}[X],\deg c=m$ be squarefree.

  1. (i)

Assume that $q^{\frac{n-m-1}{2}}>2m+1$. Then there exists
$f\in\mathbb{F}_{q}[X]$, $\deg f=n$, $(f,c)=1$, such that
$g=f^{\prime}c-fc^{\prime}$ is irreducible in $\mathbb{F}_{q}[X]$.

  2. (ii)

Assume $n\not\equiv m\pmod{2}$ and $q^{\frac{n-3m-1}{4}}>2m+1$. Then there
exists $f\in\mathbb{F}_{q}[X]$, $\deg f=n$, $(f,c)=1$, such that
$f^{\prime}c-fc^{\prime}=g^{2}$ with $g\in\mathbb{F}_{q}[X]$ irreducible.

###### Proof.

In case (i) let $N=n+m-1$ and $\psi_{c}=\pi_{c}$; in case (ii) let
$N=\frac{n+m-1}{2}$ and choose $\psi_{c}\in\mathbb{F}_{q}[X]$ with
$\psi_{c}^{2}\equiv\pi_{c}\pmod{c^{2}}$. The latter exists since $\pi_{c}$ is
a square modulo $c$ (e.g. $\pi_{c}\equiv\delta^{2}\pmod{c}$ with
$\delta=\prod_{i=1}^{m}(X-\alpha_{i})\cdot\sum_{i=1}^{m}\frac{1}{(X-\alpha_{i})}$)
and the kernel of
$(\mathbb{F}_{q}[X]/c^{2})^{\times}\rightarrow(\mathbb{F}_{q}[X]/c)^{\times}$
has order $q^{m}$, which is odd.

By Lemma 4.9 it suffices to show that there is a monic irreducible polynomial
$g\in\mathbb{F}_{q}[X]$ with $\deg g=N$ such that $g\bmod
c^{2}\in\mathcal{\psi}_{c}\mathcal{H}_{c}$ (for case $(ii)$ note that $g\bmod
c^{2}\in\psi_{c}\mathcal{H}_{c}$ implies $g^{2}\bmod
c^{2}\in\pi_{c}\mathcal{H}_{c}$). Denote by $\mathcal{H}_{c}^{\perp}$ the
group of Dirichlet characters modulo $c^{2}$ that are trivial on
$\mathcal{H}_{c}$ (this is the orthogonal group of $\mathcal{H}_{c}$). We have
$|\mathcal{H}_{c}^{\perp}|=[(\mathbb{F}_{q}[X]/c^{2})^{\times}:\mathcal{H}_{c}]=q^{m}$.
Below when we sum over $g$ we will always restrict $g$ to be monic, and
$\Lambda$ denotes the polynomial von Mangoldt function, i.e. $\Lambda(g)=\deg
P$ if $g=P^{\nu}$ for a monic irreducible $P\in\mathbb{F}_{q}[X]$ and $\nu\geq
1$, and $\Lambda(g)=0$ otherwise. By the second orthogonality relation [Ser73,
Corollary on p. 63] applied to the group
$\left(\mathbb{F}_{q}[X]/c^{2}\right)^{\times}/\mathcal{H}_{c}$ (the dual of
this group can be identified with $\mathcal{H}_{c}^{\perp}$) and the element
$g/\psi_{c}$, we have

$\sum_{\chi\in\mathcal{H}_{c}^{\perp}}\overline{\chi(\psi_{c})}\chi(g)=\left[\begin{array}[]{ll}q^{m},&\mbox{if
}g\bmod c^{2}\in\psi_{c}\mathcal{H}_{c},\\\
0,&{\mathrm{otherwise}},\end{array}\right.$

and hence

$\sum_{\deg g=N\atop{g\bmod
c^{2}\in\psi_{c}\mathcal{H}_{c}}}\Lambda(g)=\frac{1}{q^{m}}\sum_{\chi\in\mathcal{H}_{c}^{\perp}}\overline{\chi(\psi_{c})}\sum_{\deg
g=N}\chi(g)\Lambda(g).$

Let $\chi_{1}$ denote the trivial character modulo $c^{2}$. For all
$\chi\neq\chi_{1}$, Weil’s Riemann Hypothesis for function fields gives
that222For example using the notation of [Ros02, p. 41-42]: $\sum_{\deg
g=N}\chi(g)\Lambda(g)=c_{N}(\chi)=-\sum_{k=1}^{M-1}\alpha_{k}(\chi)^{N}$ where
$M=\deg c^{2}$ and $|\alpha_{k}(\chi)|\leq\sqrt{q}$.

$\left|\sum_{\deg g=N}\chi(g)\Lambda(g)\right|\leq(2m-1)q^{\frac{N}{2}}.$

Thus together with the elementary relation $\sum_{\deg g=N}\Lambda(g)=q^{N}$
(see [Ros02, Proposition 2.1]) we obtain

$\sum_{\deg g=N\atop{g\bmod
c^{2}\in\psi_{c}\mathcal{H}_{c}}}\Lambda(g)\geq\frac{1}{q^{m}}\sum_{\deg
g=N\atop{(g,c)=1}}\Lambda(g)-\frac{q^{m}-1}{q^{m}}(2m-1)q^{\frac{N}{2}}\geq
q^{N-m}-1-(2m-1)q^{\frac{N}{2}}.$

By the same elementary relation,

$\sum_{\deg g=N\atop{g\bmod
c^{2}\in\mathcal{\psi}_{c}H_{c}\atop{g\,\mathrm{not\,irreducible}}}}\Lambda(g)\leq\sum_{d|N\atop{d\neq
N}}q^{d}\leq 2q^{\frac{N}{2}}-1,$

and therefore as long as

$q^{N-m}>(2m+1)q^{\frac{N}{2}}$

there must be at least one irreducible $g$ with $\deg g=N$ and $g\bmod
c^{2}\in\mathcal{\psi}_{c}\mathcal{H}_{c}$. To conclude the proof note that
$\frac{N}{2}-m$ equals $\frac{n-m-1}{2}$ in case $(i)$ and $\frac{n-3m-1}{4}$
in case $(ii)$. ∎

######  Theorem 4.11.

Let $p$ be a prime, $q$ a power of $p$, $0\leq m\leq n-2$ integers such that
$n-m<p$. Assume that either

  1. (i)

$G=S_{n}$ and the following conditions hold

    1. (a)

$p>2$,

    2. (b)

$m\equiv n\pmod{2}$ and $m\neq 2$,

    3. (c)

$(n,m)\neq(9,1)$, and

    4. (d)

$q^{\frac{n-m-1}{2}}>2m+1$,

or

  2. (ii)

$G=A_{n}$ and the following conditions hold

    1. (a)

$p>3$,

    2. (b)

$m\not\equiv n\pmod{2}$ and $(n,p)=1$,

    3. (c)

$(n,m)\neq(8,1),(9,0),(12,1),(24,1)$,

    4. (d)

$q^{\frac{n-3m-1}{4}}>2m+1$, and

    5. (e)

$m\geq 2$ or $\left(\frac{q}{n-m}\right)=1$.

Then there exist $f,c\in\mathbb{F}_{q}[X]$ with ${\rm deg}(f)=n$, ${\rm
deg}(c)=m$, and $c$ squarefree such that the splitting field of $f(X)-Tc(X)$
over $\mathbb{F}_{q}(T)$ is geometric with Galois group $G$ and is ramified
over only one finite prime $\mathcal{F}$, with $\Delta_{X}(f-Tc)$ a power of
$\mathcal{F}$ and ${\rm deg}(\Delta_{X}(f-Tc))=n+m-1$. In particular,
$r_{\mathbb{F}_{q}(T)}(G)\leq 2$.

###### Proof.

$(i)$. First assume that $G=S_{n}$, $p>2$, $m\neq 2$, $n\equiv m\pmod{2}$,
$(n,m)\neq(9,1)$ and $q^{\frac{n-m-1}{2}}>2m+1.$ Take any squarefree
$c\in\mathbb{F}_{q}[X],\deg c=m$. By Proposition 4.10(i) one can find an
$f\in\mathbb{F}_{q}[X],\deg f=n$ such that $g=f^{\prime}c-fc^{\prime}$ is
irreducible of degree $n+m-1$. By Proposition 4.8(i) the splitting field $K$
of $f(X)-Uc(X)$ over $\mathbb{F}_{q}(U)$ is geometric with Galois group
$S_{n}$. Let $w=\frac{f}{c}\in\mathbb{F}_{q}(X)$.

Consider the discriminant $D(U)=\Delta_{X}(f(X)-Uc(X))$. Let

$\alpha_{1},\alpha_{2}=\alpha_{1}^{q},\ldots,\alpha_{n+m-1}=\alpha_{1}^{q^{n+m-2}}\in\mathbb{F}_{q^{n+m-1}}$

be the roots of $g$. Then by Lemma 2.8 the roots of $D$ are
${w(\alpha_{i})},1\leq i\leq n+m-1$ (including multiplicity). The Frobenius
map ${\rm Fr}_{q}$ acts cyclically on ${w(\alpha_{i})}$ and hence
$D=\mathcal{F}^{r}$ is a power of some prime
$\mathcal{F}\in\mathbb{F}_{q}[U]$. By Lemma 2.8, $w$ is ramified only over
$\mathcal{F}$ and (possibly) $\infty$, which concludes the proof in the case
$G=S_{n}$.

$(ii)$. Now assume that $G=A_{n}$, $p>3$, $n\not\equiv m\pmod{2}$,
$(n,m)\neq(8,1)$, $(9,0)$, $(12,1)$, $(24,1)$ and $q^{\frac{n-3m-1}{4}}>2m+1$.
Let $c\in\mathbb{F}_{q}[X],\deg c=m$ be a monic squarefree polynomial. By
Proposition 4.10(ii) there exists a monic irreducible $g\in\mathbb{F}_{q}[X]$
and a monic $f\in\mathbb{F}_{q}[X],\deg f=n$ such that
$f^{\prime}c-fc^{\prime}=(n-m)g^{2}$ (we can always adjust the leading
coefficients this way). Note that $\deg f^{\prime}=n-1$ since $(n,p)=1$. Let
$w=\frac{f}{c}\in\mathbb{F}_{q}(X)$ and let $K$ be the splitting field of
$f(X)-Uc(X)$ over $\mathbb{F}_{q}(U)$. By Proposition 4.8(ii),
$\mathrm{Mon}_{\overline{\mathbb{F}}_{q}}(w)=A_{n}$, hence ${\rm
Gal}(K/\mathbb{F}_{q}(U))=\mathrm{Mon}_{\mathbb{F}_{q}}(w)$ is either $A_{n}$
or $S_{n}$ and $K/\mathbb{F}_{q}(U)$ is geometric if it is $A_{n}$, which
happens precisely when $D(U)=\Delta_{X}(f(X)-Uc(X))$ is a square in
$\mathbb{F}_{q}(U)$ (Lemma 2.6).

To shorten notation, we omit the variable $X$ in the following calculation.
All discriminants and resultants are with respect to the variable $X$. Terms
which are squares in $\mathbb{F}_{q}(U)$ are of no consequence and are simply
denoted by $\square$. If $\deg c\geq 1$ then using the properties of
resultants and discriminants we calculate:

$\displaystyle D(U)$
$\displaystyle\stackrel{{\scriptstyle(\ref{eq:disc})}}{{=}}$
$\displaystyle(-1)^{\frac{n(n-1)}{2}}\mathrm{Res}(f^{\prime}-Uc^{\prime},f-Uc)$
$\displaystyle\stackrel{{\scriptstyle(\ref{eq:resbimult})}}{{=}}$
$\displaystyle(-1)^{\frac{n(n-1)}{2}}\mathrm{Res}(c,f-Uc)^{-1}\mathrm{Res}(f^{\prime}c-Uc^{\prime}c,f-Uc)$
$\displaystyle\stackrel{{\scriptstyle(\ref{eq:resalt})}}{{=}}$
$\displaystyle\square\cdot(-1)^{\frac{n(n-1)}{2}}\mathrm{Res}(c,f)\mathrm{Res}(f^{\prime}c-fc^{\prime}+c^{\prime}(f-Uc),f-Uc)$
$\displaystyle\stackrel{{\scriptstyle(\ref{eq:resalt}),(\ref{eq:ressym})}}{{=}}$
$\displaystyle\square\cdot(-1)^{\frac{n(n-1)}{2}}\mathrm{Res}(c,f)\mathrm{Res}(f^{\prime}c-fc^{\prime},f-Uc)$
$\displaystyle\stackrel{{\scriptstyle(\ref{eq:resbimult}),(\ref{eq:resdefinition})}}{{=}}$
$\displaystyle\square\cdot(-1)^{\frac{n(n-1)}{2}}\mathrm{Res}(c,c^{\prime})\mathrm{Res}(c,c^{\prime}f)(n-m)^{n}\prod_{g(\alpha)=0}(f(\alpha)-Uc(\alpha))^{2}$
$\displaystyle\stackrel{{\scriptstyle(\ref{eq:resalt})}}{{=}}$
$\displaystyle\square\cdot(-1)^{\frac{n(n-1)}{2}}\mathrm{Res}(c,c^{\prime})\mathrm{Res}(c,c^{\prime}f-cf^{\prime})(n-m)^{n}$
$\displaystyle\stackrel{{\scriptstyle(\ref{eq:resbimult})}}{{=}}$
$\displaystyle\square\cdot(-1)^{\frac{n(n-1)}{2}}\mathrm{Res}(c,c^{\prime})\mathrm{Res}(c,m-n)\mathrm{Res}(c,g)^{2}(n-m)^{n}$
$\displaystyle\stackrel{{\scriptstyle(\ref{eq:resdefinition})}}{{=}}$
$\displaystyle\square\cdot(-1)^{\frac{n(n-1)}{2}+m}\mathrm{Res}(c,c^{\prime})(n-m)^{n+m}$
$\displaystyle\stackrel{{\scriptstyle(\ref{eq:disc}),(\ref{eq:ressym})}}{{=}}$
$\displaystyle\square\cdot(-1)^{\frac{n(n-1)}{2}+\frac{m(m+1)}{2}}(n-m)\Delta(c).$

If $c=1$ a similar calculation gives the same final expression for $D(U)$.

Therefore, $D(U)$ is a square in $\mathbb{F}_{q}(U)$ if and only if

$\delta:=(-1)^{\frac{n(n-1)}{2}+\frac{m(m+1)}{2}}(n-m)\Delta(c)$

is a square in $\mathbb{F}_{q}$. If $m<2$, then $\Delta(c)=1$, and by
quadratic reciprocity,

$\left(\frac{(-1)^{\frac{n(n-1)}{2}+m}(n-m)}{p}\right)=(-1)^{\frac{p-1}{4}(n^{2}+m-1)}\left(\frac{p}{n-m}\right)=\left(\frac{p}{n-m}\right)$

as $n^{2}+m-1\equiv 0\pmod{4}$ if either $m=0$ and $n\equiv 1\pmod{2}$, or
$m=1$ and $n\equiv 0\pmod{2}$, thus $\delta$ is a square in $\mathbb{F}_{q}$
if and only if $\left(\frac{q}{n-m}\right)=1$, which is precisely our
assumption (e) for this case. If $m\geq 2$, then $\Delta(c)$ is a square in
$\mathbb{F}_{q}$ if and only if $(-1)^{m}\mu(c)=1$ where $\mu(c)$ is the
polynomial Möbius function, see [Con05, Lemma 4.1]. Therefore we can choose
$c$ accordingly with an odd or an even number of irreducible factors to get
the value we need so that $\delta$ is a square in $\mathbb{F}_{q}$.

This shows that $K/\mathbb{F}_{q}(U)$ is geometric with Galois group $A_{n}$.
Arguing as in $(i)$ one sees that $D=\mathcal{F}^{r}$ with $\mathcal{F}$ prime
and that $w$ is ramified only over $\mathcal{F}$ and possibly $\infty$. ∎

###  4.3. Eliminating tame ramification at infinity

We will now explain how to eliminate the tame ramification at infinity of the
extensions obtained in the previous subsections, conditional on a weak
consequence of the function field analogue of Schinzel’s hypothesis H that can
actually be proven in many cases. The classical Hypothesis H states the
following:

######  Conjecture 4.12 (Schinzel’s hypothesis H).

Let $f_{1},\dots,f_{r}\in\mathbb{Z}[X]$ irreducible. If $f_{1}\cdots f_{r}$ is
not the zero function modulo any prime number $p$, then there exist infinitely
many $n\in\mathbb{Z}$ with $f_{1}(n),\dots,f_{r}(n)$ simultaneously prime.

In the function field setting, the naive analogue fails due to inseparability
issues, see [CCG08]. Taking this into account, one conjectures:

######  Conjecture 4.13.

Let $\mathcal{F}_{1},\dots,\mathcal{F}_{r}\in\mathbb{F}_{q}[T][X]$ irreducible
and separable in $X$. If $\mathcal{F}_{1}\cdots\mathcal{F}_{r}$ is not the
zero function modulo any prime polynomial $P\in\mathbb{F}_{q}[T]$, then there
exist infinitely many $h\in\mathbb{F}_{q}[T]$ with
$\mathcal{F}_{1}(h),\dots,\mathcal{F}_{r}(h)$ simultaneously irreducible in
$\mathbb{F}_{q}[T]$.

For partial results on Conjecture 4.13 see for example [Pol08, Bar12, Ent16,
Ent19, SS20]. We will work with the following weaker assumption:

######  Definition 4.14.

We say that property $H(q,d,e)$ holds if for any irreducible
$\mathcal{F}\in\mathbb{F}_{q}[X]$ with $\deg(\mathcal{F})|d$ there exists
$h\in\mathbb{F}_{q}[T]$ with $e|\deg(h)$ such that $\mathcal{F}(h)$ is
irreducible in $\mathbb{F}_{q}[T]$.

We write ${\rm rad}(e)$ for the radical of $e$ and $\omega(e)$ for the number
of distinct prime divisors of $e$. We define ${\rm
rad}^{\prime}(e)=\frac{(4,e)}{(2,e)}{\rm rad}(e)$, i.e. ${\rm
rad}^{\prime}(e)$ equals $2{\rm rad}(e)$ if $4|e$ and ${\rm rad}(e)$
otherwise. We denote by $v_{\ell}$ the $\ell$-adic valuation.

######  Proposition 4.15.

$H(q,d,e)$ holds in each of the following cases:

  1. (1)

Conjecture 4.13 holds for $\mathbb{F}_{q}$.

  2. (2)

$q$ is sufficiently large with respect to $d$ and $e$.

  3. (3)

${\rm rad}^{\prime}(e)|q-1$ and $(d,e)=1$.

  4. (4)

${\rm rad}^{\prime}(e)|q^{\ell}-1$ for every prime divisor $\ell$ of $d$, and
$q\geq(d-1)^{2}(2^{\omega(e)}-1)^{2}$.

  5. (5)

${\rm rad}(e)|q-1$ and
$q\geq(2^{\max\\{v_{2}(e)-1,0\\}}d-1)^{2}(2^{\omega(e)}-1)^{2}$.

More precisely, for any irreducible $\mathcal{F}\in\mathbb{F}_{q}[X]$ with
$\deg(\mathcal{F})|d$ there exists $h\in\mathbb{F}_{q}[T]$ with $e|\deg(h)$
such that $\mathcal{F}(h)$ is irreducible in $\mathbb{F}_{q}[T]$, where

  1. (a)

there exists such $h$ with ${\rm deg}(h)=e$ in cases (2), (3), (4) and (5),

  2. (b)

there exists such $h$ that is monic in cases (1), (2), (4) and (5), and

  3. (c)

there exists such $h$ that is monic and satisfies ${\rm deg}(h)=e$ in cases
(4) and (5), as well as in case (2) if $q$ is odd or $e\geq 4$.

###### Proof.

Let $e,d\geq 1$ and $\mathcal{F}\in\mathbb{F}_{q}[X]$ irreducible with ${\rm
deg}(\mathcal{F})=n|d$ be given.

(1): Without loss of generality assume that $q-1|e$. The polynomial

$g:=\mathcal{F}(X^{e}+A_{e-1}X^{e-1}+\dots+A_{0})\in\mathbb{F}_{q}[A_{0},\dots,A_{e-1},X]$

is separable in $X$ and irreducible: Indeed, $g$ is obtained from
$\mathcal{F}(A_{0})$ by the invertible change of variables $A_{0}\mapsto
X^{e}+A_{e-1}X^{e-1}+\dots+A_{0}$. Let $S$ be the set of primes
$P\in\mathbb{F}_{q}[T]$ with ${\rm deg}(P)\leq\log_{q}{\rm deg}_{X}g$ and let
$M=\prod_{P\in S}P$. By [BE21, Theorem 1.1] there exist
$a_{0}\in\mathbb{F}_{q}+M\mathbb{F}_{q}[T]$ and
$a_{1},\dots,a_{e-1}\in\mathbb{F}_{q}[T]$ with
$\tilde{g}(X):=g(a_{0},\dots,a_{e-1},X)\in\mathbb{F}_{q}[T][X]$ separable in
$X$ and irreducible in $\mathbb{F}_{q}(T)[X]$, hence also irreducible in
$\mathbb{F}_{q}[T,X]$ (since it is primitive in $X$). In particular,
$\tilde{g}(0)=\mathcal{F}(a_{0})\not\equiv 0\mbox{ mod }P$ for every $P\in S$,
and modulo a prime $P\in\mathbb{F}_{q}[T]$ not in $S$, $\tilde{g}$ has at most
${\rm deg}_{X}\tilde{g}<q^{{\rm deg}(P)}$ many zeros. Therefore, Conjecture
4.13 predicts infinitely many $h\in\mathbb{F}_{q}[T]$ such that, with
$\tilde{h}:=h^{e}+a_{e-1}h^{e-1}+\dots+a_{0}$, the polynomial
$\tilde{g}(h)=\mathcal{F}(\tilde{h})$ is irreducible. For ${\rm deg}(h)$
sufficiently large, ${\rm deg}(\tilde{h})={\rm deg}(h^{e})$ is divisible by
$e$, and ${\rm lc}(\tilde{h})={\rm lc}(h^{e})$, so $\tilde{h}$ is monic as
$q-1|e$.

(2): For large $q$, $H(q,d,e)$ follows immediately from [Ent16, Theorem 1.1].
That one can choose $h$ monic of degree $e$ follows in the case that $q$ is
large and odd from [Bar12, Theorem 1.4]. If $q$ is large and $e\geq 4$, the
existence of a suitable monic $h$ of degree $e$ follows from [Ent19,
Proposition 6.1]: If $\mathcal{F}(X)$ is irreducible, then so is
$\mathcal{F}(T^{e}+X)$, and [Ent19, Proposition 6.1] gives a $h_{0}$ of degree
$e-1\geq 3$ with $\mathcal{F}(T^{e}+h_{0})$ irreducible. This also implies
that if $q$ is large one can always obtain a suitable $h$ that is monic, by
replacing $e$ by $4e$ if necessary.

(3): Let $\alpha\in\mathbb{F}_{q^{n}}$ be a root of $\mathcal{F}$. We will
show that there exists $c\in\mathbb{F}_{q}^{\times}$ such that $c\alpha$ is
not an $\ell$-th power in $\mathbb{F}_{q^{n}}^{\times}$ for any prime divisor
$\ell$ of $e$. Then $T^{e}-c\alpha\in\mathbb{F}_{q^{n}}[T]$ is irreducible,
cf. [Lan02, Chapter VI, Theorem 9.1], where in the case $4|e$ one has to
observe that the assumption $2{\rm rad}(e)|q-1$ implies that $-4$ is a square
in $\mathbb{F}_{q}^{\times}$, hence $-4c\alpha$ is in particular not a $4$-th
power in $\mathbb{F}_{q^{n}}^{\times}$. This will show that $\mathcal{F}(h)$
is irreducible for $h=c^{-1}T^{e}$.

Let $y=N_{\mathbb{F}_{q^{n}}/\mathbb{F}_{q}}(\alpha)$. By the assumptions that
${\rm rad}(e)|q-1$ and that $n$ and $e$ are coprime we can apply the
isomorphism
$\mathbb{F}_{q}^{\times}\cong\mathbb{Z}/(q-1)\mathbb{Z}\cong\prod_{\ell|q-1}\mathbb{Z}/\ell^{v_{\ell}(q-1)}\mathbb{Z}$
to obtain $c$ such that $c^{n}y$ is not an $\ell$-th power in
$\mathbb{F}_{q}^{\times}$ for any prime divisor $\ell$ of $e$. As the norm is
multiplicative and $N_{\mathbb{F}_{q^{n}}/\mathbb{F}_{q}}(c\alpha)=c^{n}y$,
the element $c\alpha$ is then not an $\ell$-th power in
$\mathbb{F}_{q^{n}}^{\times}$ as well.

(4): The case ${\rm deg}(\mathcal{F})=1$ is trivial, so assume that $n>1$ and
note that the assumption implies that ${\rm rad}^{\prime}(e)|q^{n}-1$. Let
$\alpha\in\mathbb{F}_{q^{n}}$ be a root of $\mathcal{F}$. Cohen’s result
[Coh10, Corollary 2.3] gives that the number of $c\in\mathbb{F}_{q}$ for which
$\alpha+c$ is not an $\ell$-th power in $\mathbb{F}_{q^{n}}$ for any prime
divisor $\ell$ of $e$ is bigger than $\frac{\phi({\rm rad}(e))}{{\rm
rad}(e)}(q-(n-1)(2^{\omega(e)}-1)\sqrt{q})$, in particular this number is
positive as soon as $q\geq(n-1)^{2}(2^{\omega(e)}-1)^{2}$, which is satisfied
by our assumption. As in (3) we conclude that
$T^{e}-(\alpha+c)\in\mathbb{F}_{q^{n}}[T]$ is irreducible, hence
$\mathcal{F}(h)$ is irreducible for $h=T^{e}-c$.

(5): Let $s=v_{2}(e)$. If $s\leq 1$, then ${\rm rad}^{\prime}(e)={\rm
rad}(e)$, so the claim follows from (4). If $s\geq 2$, then $q$ is odd and we
first apply (4) with $e=2$ to obtain a monic $h_{1}\in\mathbb{F}_{q}[T]$ of
degree $2$ with $\mathcal{F}_{1}(T):=\mathcal{F}(h_{1}(T))$ irreducible of
degree $2n$. We iterate this until we obtain
$\mathcal{F}_{s-1}=\mathcal{F}(h_{1}(\dots h_{s-1}(T)))$ irreducible of degree
$2^{s-1}n$. Then ${\rm rad}^{\prime}(e/2^{s-1})={\rm rad}(e/2^{s-1})$, so we
can apply (4) to obtain a monic $h_{s}$ of degree $e/2^{s-1}$ with
$\mathcal{F}(h_{1}(\dots h_{s}(T)))$ irreducible. The polynomial
$h=h_{1}\circ\dots\circ h_{s}$ has degree $e$ and satisfies the claim. ∎

######  Lemma 4.16.

Let $k$ be a perfect field and let $h\in k[T]$ be a non-constant polynomial.
Denote $U=h(T)$ and let $K$ be a Galois extension of $k(U)$ ramified only over
$U=\infty$ and the finite primes $\mathcal{F}_{1},\dots,\mathcal{F}_{r}\in
k[U]$, with tame ramification over $\infty$. Assume that each
$\mathcal{F}_{i}(h(T))\in k[T]$ is separable. Then the extensions $K,k(T)$ of
$k(U)$ are linearly disjoint.

###### Proof.

Denote $D(U)=\Delta_{T}(h(T)-U)$. By Lemma 2.8 the finite branching primes of
the extension $k(T)/k(U)$ are exactly the primes dividing $D(U)$. If
$\mathcal{F}_{i}|D$ for some $i$, then 0 would be a critical value of
$\mathcal{F}_{i}\circ h\colon\mathbb{P}^{1}\to\mathbb{P}^{1}$, which means
that $\mathcal{F}_{i}(h)$ is not separable, contradicting our assumption. Thus
$\mathcal{F}_{i}\nmid D$ for each $i$, so the finite branching loci of
$k(T)/k(U)$ and $K/k(U)$ are disjoint, hence the extension $K\cap k(T)/k(U)$
is tamely ramified and ramified only over $\infty$, and is therefore trivial.
Since $K/k(U)$ is Galois, this already implies that $K$ and $k(T)$ are
linearly disjoint over $k(U)$. ∎

######  Lemma 4.17.

Let $k$ be a perfect field of characteristic $p\geq 0$, and let $f\in k[U,X]$.
Assume that the splitting field of $f$ over $k(U)$ is ramified only over the
primes $\mathcal{F}_{1},\dots,\mathcal{F}_{r}\in k[U]$ and over the infinite
prime with ramification index $e$ not divisible by $p$. If $h\in k[T]$ is such
that $e|{\rm deg}(h)$, then the splitting field of $f(h(T),X)$ over $k(T)$ is
ramified at most over the prime factors of
$\mathcal{F}_{1}(h(T)),\dots,\mathcal{F}_{r}(h(T))$.

###### Proof.

We identify $U$ with $h(T)$ and $k(U)=k(h(T))\subseteq k(T)$. Let $K$ be the
splitting field of $f(U,X)$ over $k(U)$, and consider the splitting field $L$
of $f(h(T),X)$ over $k(T)$, which is the compositum of $K$ and $k(T)$. Since
$k(T)/k(U)$ is totally ramified over $\infty$ of degree ${\rm deg}(h)$
divisible by $e$, by Abhyankar’s Lemma (Lemma 2.2) the base change from $k(U)$
to $k(T)$ eliminates the ramification over infinity. Every finite prime of
$k(T)$ that ramifies in $L$ lies over a finite prime of $k(U)$ that ramifies
in $K$, i.e. is a prime factor of some $\mathcal{F}_{i}(h)$. ∎

######  Theorem 4.18.

Let $p$ be a prime, $q$ a power of $p$, $0\leq m\leq n-2$ integers such that
$n-m<p$. Assume that either

  1. (i)

$G=S_{n}$ and the following conditions hold

    1. (a)

$p>2$,

    2. (b)

$m\equiv n\pmod{2}$ and $m\neq 2$,

    3. (c)

$(n,m)\neq(9,1)$,

    4. (d)

$q^{\frac{n-m-1}{2}}>2m+1$, and

    5. (e)

$H(q,n+m-1,n-m)$,

or

  2. (ii)

$G=A_{n}$ and the following conditions hold

    1. (a)

$p>3$,

    2. (b)

$m\not\equiv n\pmod{2}$ and $(n,p)=1$,

    3. (c)

$(n,m)\neq(8,1),(9,0),(12,1),(24,1)$,

    4. (d)

$q^{\frac{n-3m-1}{4}}>2m+1$,

    5. (e)

$m\geq 2$ or $\left(\frac{q}{n-m}\right)=1$, and

    6. (f)

$H\left(q,\frac{n+m-1}{2},n-m\right)$.

Then there exist $f,c\in\mathbb{F}_{q}[X]$ and $h\in\mathbb{F}_{q}[T]$ with
${\rm deg}(f)=n$, ${\rm deg}(c)=m$ and $(n-m)|{\rm deg}(h)$ such that the
splitting field of $f-hc$ over $\mathbb{F}_{q}(T)$ is geometric with Galois
group $G$ and is ramified over only one prime of $\mathbb{F}_{q}(T)$. In
particular, $r_{\mathbb{F}_{q}(T)}(G)=1$.

###### Proof.

In $(i)$, Theorem 4.11$(i)$ provides $f,c$ of the right degree such that the
splitting field $K$ of $f-Uc$ over $\mathbb{F}_{q}(U)$ is geometric with
Galois group $S_{n}$ and is ramified over only one finite prime $\mathcal{F}$
of degree dividing $n+m-1$ and over the infinite prime, with ramification
index $n-m$. Thus $H(q,n+m-1,n-m)$ provides a suitable $h$ with
$\mathcal{F}(h)$ irreducible, hence Lemma 4.16 gives that the splitting field
$K\mathbb{F}_{q}(T)$ of $f-hc$ over $\mathbb{F}_{q}(T)$ is geometric with
Galois group $S_{n}$, and Lemma 4.17 gives that it is ramified only over
$\mathcal{F}(h)$.

In $(ii)$, Theorem 4.11$(ii)$ similarly provides $f,c$ of the right degree
such that the splitting field of $f-Uc$ over $\mathbb{F}_{q}(U)$ is geometric
with Galois group $A_{n}$ and is ramified over only one finite prime
$\mathcal{F}$, with $\Delta_{X}(f-Uc)=\mathcal{F}^{r}$ of degree $n+m-1$, and
over the infinite prime, with ramification index $n-m$. As $G=A_{n}$ implies
that $\Delta_{X}(f-Uc)$ is a square, $r$ is even, and so in particular ${\rm
deg}(\mathcal{F})$ divides $\frac{n+m-1}{2}$. Thus $H(q,\frac{n+m-1}{2},n-m)$
provides a suitable $h$ with $\mathcal{F}(h)$ irreducible, and we conclude
again with Lemma 4.16 and Lemma 4.17. ∎

######  Remark 4.19.

Note that in the cases (2)-(5) of Proposition 4.15, one can obtain $h$ in
Theorem 4.18 to be of degree equal to $n-m$.

###  4.4. Two ramified primes via small prime gaps

In this final subsection we present a different approach to produce extensions
of group $S_{n}$ or $A_{n}$ with two ramified primes. For this we will need
the following result on primes with small gaps in $\mathbb{F}_{q}[T]$, which
follows directly from [CHLPT15, Theorem 1.3], which is a function field
adaptation of the method of Maynard [May15].

######  Proposition 4.20.

Assume $q\geq 107$ and fix $b\in\mathbb{F}_{q}^{\times}$. For all sufficiently
large $d$ (in terms of $q$) there exist polynomials $h\in\mathbb{F}_{q}[T]$ of
degree $d$ (not necessarily monic) such that $h$ and $h-b$ are both
irreducible.

###### Proof.

By [CHLPT15, Theorem 1.3] (applied with $m=2$ and $h_{1},\dots,h_{k}$ an
enumeration of $\mathbb{F}_{q}$) and the remark following it, as long as
$q>k_{0}(2)=105$ (using the notation of the cited theorem) and $d\geq
d_{0}(q)$ is sufficiently large, one can find $f\in\mathbb{F}_{q}[T],\deg f=d$
such that $f,f-a$ are irreducible for some $a\in\mathbb{F}_{q}^{\times}$ (the
statement of the cited theorem only asserts the existence of infinitely many
such $f$, but the proof shows that such an $f$ exists with any sufficiently
large degree). Then $h=\frac{b}{a}f$ satisfies the assertion of the
proposition. ∎

######  Theorem 4.21.

Let $p$ be a prime number, $q$ a power of $p$, and $3\leq n<p$. Then
$r_{\mathbb{F}_{q}(T)}(S_{n})\leq 2$ and $r_{\mathbb{F}_{q}(T)}(A_{n})\leq 2$.

###### Proof.

Consider the polynomial

$f=X^{n}-X^{n-1}-U\in\mathbb{F}_{q}[U][X].$

Using (2.4) and $f^{\prime}=nX^{n-2}(X-1+n^{-1})$ it is easy to show that
$D(U):=\Delta_{X}(f)=aU^{n-2}(U-b)$ with $a\in\mathbb{F}_{p}^{\times}$ and
$b=(1-n^{-1})^{n}-(1-n^{-1})^{n-1}\in\mathbb{F}_{p}^{\times}$. Hence the
splitting field of $f$ over $\overline{\mathbb{F}}_{q}(U)$ is ramified over
(at most) $0,b,\infty$.

The polynomial $f$ is irreducible over $\overline{\mathbb{F}}_{q}(U)$ since it
is primitive and linear in the variable $U$, in particular ${\rm
Gal}(f/\overline{\mathbb{F}}_{q}(U))\leqslant S_{n}$ is transitive. Let
$\alpha$ be a root of $f$ in an algebraic closure of $\mathbb{F}_{q}(U)$. Then
$U=(\alpha-1)\alpha^{n-1}$ and we see that in the extension
$\overline{\mathbb{F}}_{q}(\alpha)/\overline{\mathbb{F}}_{q}(U)$ the point
$U=0$ splits into an unramified point $\alpha=1$ and a point $\alpha=0$ with
ramification index $n-1$, while $\infty$ is totally ramified. Hence, as
$(n-1,p)=1$, by Lemma 2.1(ii) the group ${\rm
Gal}(f/\overline{\mathbb{F}}_{q}(U))$ contains an $(n-1)$-cycle, in particular
it is primitive. Since $U-b$ divides $D(U)$ exactly once, by Lemma 2.7 the
prime $U-b$ splits in the extension
$\overline{\mathbb{F}}_{q}(\alpha)/\overline{\mathbb{F}}_{q}(U)$ into $n-2$
unramified primes and one prime with ramification index 2, hence ${\rm
Gal}(f/\overline{\mathbb{F}}_{q}(U))$ contains a transposition by Lemma
2.1(ii). Thus by Lemma 2.5, ${\rm Gal}(f/\overline{\mathbb{F}}_{q}(U))=S_{n}$.
Since ${\rm Gal}(f/\overline{\mathbb{F}}_{q}(U))\leqslant{\rm
Gal}(f/\mathbb{F}_{q}(U))\leqslant S_{n}$ we must have ${\rm
Gal}(f/\mathbb{F}_{q}(U))={\rm Gal}(f/\overline{\mathbb{F}}_{q}(U))=S_{n}$.
Therefore the splitting field $K$ of $f$ over $\mathbb{F}_{q}(U)$ is geometric
with Galois group $S_{n}$ and ramified over infinity and at most the two
finite primes $U$ and $U-b$. We now fix some $h\in\mathbb{F}_{q}[T]$ with
$n|\deg h$ such that $h$ and $h-b$ are both irreducible – for $q\geq 107$ such
an $h$ is provided by Proposition 4.20, and for $q<107$, a quick computer
search (that only needs to go through the finitely many pairs $(n,q)$ with
$3\leq n<p\leq q<107$) verifies that there always exists such an $h$ even with
$\deg h=n$. Identify $U=h(T)$ and let $L=K\mathbb{F}_{q}(T)$ be the splitting
field of $f(h(T),X)$ over $\mathbb{F}_{q}(T)$. Lemma 4.16 gives that
$L/\mathbb{F}_{q}(T)$ is geometric with Galois group $S_{n}$, and Lemma 4.17
gives that it is ramified only over the primes $h$ and $h-b$. In particular,
$r_{\mathbb{F}_{q}(T)}(S_{n})\leq 2$.

Now viewing $A_{n}$ as a subgroup of ${\rm Gal}(K/\mathbb{F}_{q}(U))$ consider
the fixed field $F=K^{A_{n}}$. The extension $F/\mathbb{F}_{q}(U)$ is
quadratic and ramified over at most three points $0,b,\infty$. Explicitly
$F=\mathbb{F}_{q}(U,\sqrt{D(U)})$ (cf. Lemma 2.6) and since $U-b$ divides
$D(U)$ exactly once, by Lemma 2.7 we see that $U-b$ is ramified in
$F/\mathbb{F}_{q}(U)$. On the other hand the unique prime of $F$ lying over
$U-b$ is unramified in the extension $K/F$ since the ramification index of
$U-b$ in $K/\mathbb{F}_{q}(U)$ is 2. By the Riemann-Hurwitz formula we see
that in fact $F/\mathbb{F}_{q}(U)$ is ramified over exactly one of $0,\infty$
(say $\infty$, the other case is similar) and $F$ has genus 0, hence
$F=\mathbb{F}_{q}(U^{\prime})$ for some $U^{\prime}\in K$ [Sti09, Corollary
5.1.12]. We have ${\rm Gal}(K/F)=A_{n}$ and the only primes of $F$ that ramify
in $K/F$ are the primes lying over $U=0,\infty$. If 0 is inert in
$F/\mathbb{F}_{q}(U)$ then only two primes of $F$ ramify in $K$, and therefore
$r_{\mathbb{F}_{q}(U)}(A_{n})=2$. Otherwise three primes of $F$ ramify in $K$,
all of degree one, and applying a linear fractional transformation we may
assume that these are $U^{\prime}=0,1,\infty$. In this case we repeat the
argument we used above for $S_{n}$ (taking $b=1$) to conclude that
$r_{\mathbb{F}_{q}(T)}(A_{n})\leq 2$. ∎

##  5\. $S_{n}$ and $A_{n}$ when $p\leq n$

Let $p$ be a prime number and $q$ a power of $p$. The main goal of this
section is to prove Theorem 1.8, which contains Theorems 1.6(2a), 1.7(2a) and
1.7(2b) as special cases. Theorem 1.8 will follow from Theorem 5.7 and
Proposition 5.9 which will appear below. The proof will be given in Section
5.2.

######  Remark 5.1.

The proof of Theorem 1.8 (and Theorem 5.7 from which it follows) uses the
Classification of Finite Simple Groups (via Theorem 2.3) to handle the case
$n=p+2,p>2$ (for both $A_{n}$ and $S_{n}$). In all other cases only elementary
group theory (mainly Jordan’s theorem on primitive permutation groups
containing a prime cycle) is used.

###  5.1. $L_{1}^{\mathrm{gt}}(q)$-realizable groups

######  Definition 5.2.

A finite group $G$ is called _quasi_ -$p$ if it is generated by its $p$-Sylow
subgroups, i.e. $G=p(G)$. A finite group $G$ is called _cyclic-by-quasi- $p$_
if it is an extension of a cyclic group by a quasi-$p$ group, i.e.
$d(G/p(G))\leq 1$.

######  Example 5.3.

The groups $S_{n}$ and $A_{n}$ are cyclic-by-quasi-$p$ if $n\geq p$.

######  Definition 5.4.

We say that a finite group $G$ is _$L_{1}^{\mathrm{gt}}(q)$ -realizable_ if
there exists a geometric Galois extension $L/\mathbb{F}_{q}(T)$ with ${\rm
Gal}(L/\mathbb{F}_{q}(T))=G$ that is ramified over at most two primes of
$\mathbb{F}_{q}(T)$, both of degree one, and with wild ramification over at
most one of them. An extension $L$ with this property is called an
$L_{1}^{\mathrm{gt}}(q)$-_realization_ of $G$.333Abhyankar introduced the
notation $L_{1}$ for the once-punctured affine line. The letters $\mathrm{gt}$
remind us that (up to Möbius transformations) we only consider geometric
Galois $\mathbb{F}_{q}$-covers of $L_{1}$ which are tamely ramified at
$\infty$.

Note that an $L_{1}^{\mathrm{gt}}(q)$-realizable group is automatically
$L_{1}^{\mathrm{gt}}(q^{r})$-realizable for any $r\geq 1$.

######  Lemma 5.5.

Let $G$ be an $L_{1}^{\mathrm{gt}}(q)$-realizable group. Then $G$ is cyclic-
by-quasi-$p$ and satisfies $[G:p(G)]\mid q-1$, and there exists an
$L_{1}^{\mathrm{gt}}(q)$-realization of $p(G)$ that is ramified over a single
prime of degree one.

###### Proof.

Let $L/\mathbb{F}_{q}(T)$ be an $L_{1}^{\mathrm{gt}}(q)$-realization of $G$.
We may assume that it is ramified only over $T=0,\infty$, tamely ramified over
$\infty$. Let $H=p(G)\unlhd G$ and denote $K=L^{H}$. We have ${\rm
Gal}(K/\mathbb{F}_{q}(T))=C:=G/H$. Since $(|C|,p)=1$ the extension
$K/\mathbb{F}_{q}(T)$ is tamely ramified and by assumption it is ramified at
most over $T=0,\infty$, so it must be of the form $K=\mathbb{F}_{q}(T^{1/n})$
with $n=|C|$ and $n|q-1$ (e.g. since $K(T^{1/n})/\mathbb{F}_{q}(T^{1/n})$ is
geometric, and unramified by Lemma 2.2, hence trivial). In particular, $C={\rm
Gal}\left(\mathbb{F}_{q}(T^{1/n})/\mathbb{F}_{q}(T)\right)$ is cyclic, so $G$
is cyclic-by-quasi-$p$ and $[G:p(G)]=n\mid q-1$.

Let $U=T^{1/n}$ and let $e$ be the ramification index of the prime $U=\infty$
in $L/\mathbb{F}_{q}(U)$. Note that $(e,p)=1$ and denote
$K^{\prime}=\mathbb{F}_{q}(S)$, where $S^{e}=U$. The extension
$K^{\prime}\overline{\mathbb{F}}_{q}/\overline{\mathbb{F}}_{q}(U)$ is Galois
with ${\rm
Gal}(K^{\prime}\overline{\mathbb{F}}_{q}/\overline{\mathbb{F}}_{q}(U))=\mathbb{Z}/e\mathbb{Z}$
and since $L/\mathbb{F}_{q}(U)$ is geometric we have that ${\rm
Gal}(L\overline{\mathbb{F}}_{q}/\overline{\mathbb{F}}_{q}(U))=p(G)$. Since
$p(G)$ is quasi-$p$, the groups $p(G)$ and $\mathbb{Z}/e\mathbb{Z}$ have no
nontrivial common quotients and therefore the extensions
$K^{\prime}\overline{\mathbb{F}}_{q},L\overline{\mathbb{F}}_{q}$ of
$\overline{\mathbb{F}}_{q}(U)$ are linearly disjoint. Consequently the
geometric extensions $K^{\prime},L$ of $\mathbb{F}_{q}(U)$ are also linearly
disjoint and we have ${\rm Gal}(L\mathbb{F}_{q}(S)/\mathbb{F}_{q}(S))={\rm
Gal}(L/\mathbb{F}_{q}(U))=p(G)$. By Lemma 4.17,
$L\mathbb{F}_{q}(S)/\mathbb{F}_{q}(S)$ is ramified only over the prime $S$. ∎

######  Question 5.6.

Is a cyclic-by-quasi-$p$ group $G$ with $[G:p(G)]\mid q-1$ always
$L_{1}^{\mathrm{gt}}(q)$-realizable?

This question is a variant of the arithmetic Abhyankar conjecture (Conjecture
1.9) and is suggested by Abhyankar’s general philosophy regarding ramified
covers in characteristic $p$. We remark that Harbater and van der Put [HP02,
Theorem 5.5] have given a non-obvious necessary condition for a group to be
the Galois group of an extension of $\mathbb{F}_{q}(T)$ ramified over two
geometric points, but this condition is automatically satisfied for cyclic-by-
quasi-$p$ groups, as can easily be shown using Frattini’s argument. The next
theorem answers Question 5.6 affirmatively for the symmetric and alternating
groups (with some mild restrictions on $n,p$).

######  Theorem 5.7.

Let $p$ be a prime, $q$ a power of $p$, and $n\geq p$. The following groups
are $L_{1}^{\mathrm{gt}}(q)$-realizable:

  1. (1)

$S_{n}$ if $n\neq p+1$ or $p=2$.

  2. (2)

$A_{n}$ if $p>2$ and either $n\neq p+1$ or
$\mathbb{F}_{q}\supseteq\mathbb{F}_{p^{2}}$ or $p=3$.

  3. (3)

$A_{n}$ if $p=2$ and either $\mathbb{F}_{q}\supseteq\mathbb{F}_{4}$ or $10\neq
n\geq 8$ and $n\equiv 0,1,2,6,7\pmod{8}$.

Theorem 5.7 will be proved in Section 5.3. Note that in the case of
$A_{n},n\geq p>2$ Theorem 5.7 combined with Lemma 5.5 and Example 5.3
immediately imply Theorem 1.10.

###  5.2. Application to the minimal ramification problem

We note that $L_{1}^{\mathrm{gt}}(q)$-realizability implies a positive answer
to the minimal ramification problem:

######  Lemma 5.8.

Let $G$ be an $L_{1}^{\mathrm{gt}}(q)$-realizable nontrivial finite group.
Then for infinitely many primes $h\in\mathbb{F}_{q}[T]$ there exists a
geometric Galois extension $L/\mathbb{F}_{q}(T)$ with ${\rm
Gal}(L/\mathbb{F}_{q}(T))=G$ which is ramified only over $h$. In particular,
$r_{\mathbb{F}_{q}(T)}(G)=1$ and $G$ satisfies Conjecture 1.4.

###### Proof.

Let $M/\mathbb{F}_{q}(U)$ be a geometric realization of $G$ ramified over
$U=0$ and (at most) tamely ramified over $U=\infty$. Denote by $e$ the
ramification index over $\infty$, which by assumption is not divisible by $p$.
Let $h\in\mathbb{F}_{q}[T]$ irreducible of degree divisible by $e$ and
identify $U=h(T)$. By Lemma 4.16, $M$ and $\mathbb{F}_{q}(T)$ are linearly
disjoint over $\mathbb{F}_{q}(U)$, and $M\mathbb{F}_{q}(T)/\mathbb{F}_{q}(T)$
is ramified only over $h$ by Lemma 4.17. Therefore,
$M\mathbb{F}_{q}(T)/\mathbb{F}_{q}(T)$ is geometric Galois of Galois group $G$
and ramified only over the single prime $h$, hence
$r_{\mathbb{F}_{q}(T)}(G)=1$. ∎

In this subsection we extend this to (certain) products of
$L_{1}^{\mathrm{gt}}(q)$-realizable groups, which we then combine with Theorem
5.7 to prove Theorem 1.8:

######  Proposition 5.9.

Let $G=G_{1}\times\ldots\times G_{m}$ be a product with each $G_{i}$ being
$L_{1}^{\mathrm{gt}}(q)$-realizable. Assume that there is a prime number
$\ell$ such that for each $1\leq i\leq m$ either $G_{i}$ is quasi-$p$ or
$\ell\mid[G_{i}:p(G_{i})]$. Then Conjecture 1.4 holds for $G$ over
$\mathbb{F}_{q}(T)$.

By a $k$-variety (for a field $k$) we mean a reduced $k$-scheme of finite
type. In what follows all maps are morphisms of $\mathbb{F}_{q}$-varieties,
and $\mathbb{A}^{n}$ (resp. $\mathbb{P}^{n}$) always denotes the affine (resp.
projective) $n$-space over $\mathbb{F}_{q}$ considered as an
$\mathbb{F}_{q}$-variety. The function field of an irreducible
$\mathbb{F}_{q}$-variety $X$ is denoted by $\mathbb{F}_{q}(X)$. We will also
use the notion of a Galois cover of $\mathbb{F}_{q}$-varieties and its Galois
group, see [Mil80, §I.5] for the definition and basic properties. In
particular, if $w\colon Y\rightarrow X$ is a Galois cover of
$\mathbb{F}_{q}$-varieties, then $Y$ and $X$ are connected and $w$ is étale,
and we denote its Galois group by ${\rm Gal}(Y/X)$. We call $w$ a _geometric_
Galois cover if $Y\times\mathrm{Spec}\,\overline{\mathbb{F}}_{q}\to
X\times\mathrm{Spec}\,\overline{\mathbb{F}}_{q}$ is also a Galois cover (if
$X$ is geometrically irreducible this is equivalent to $Y$ being geometrically
irreducible).

######  Proposition 5.10.

Let $U\subseteq\mathbb{A}^{m}$ be an open $\mathbb{F}_{q}$-subvariety and
$Y\to U$ a geometric Galois cover of $\mathbb{F}_{q}$-varieties with ${\rm
Gal}(Y/U)=G$. Then there exists a morphism
$\psi\colon\mathbb{A}^{1}\to\mathbb{A}^{m}$ such that
$\psi^{-1}(U)\neq\emptyset$ and $Y\times_{U}\psi^{-1}(U)\to\psi^{-1}(U)$ is a
geometric Galois cover with Galois group $G$.

###### Proof.

Denote $\nu=|G|$ and
$Y^{\prime}=Y\times_{\mathbb{F}_{q}}\mathrm{Spec}\,\mathbb{F}_{q^{\nu}},U^{\prime}=U\times_{\mathbb{F}_{q}}\mathrm{Spec}\,\mathbb{F}_{q^{\nu}}$.
Since $Y\rightarrow U$ is geometric, $Y^{\prime}$ is irreducible and the cover
$Y^{\prime}=Y\times_{U}U^{\prime}\to U$ is Galois with ${\rm
Gal}(Y^{\prime}/U)=G\times{\rm Gal}(\mathbb{F}_{q^{\nu}}/\mathbb{F}_{q})$.
(Generally, if $Z\rightarrow X$ and $W\rightarrow X$ are Galois covers then
$Z\times_{X}W$ is a union of isomorphic Galois covers with Galois group a
subgroup of ${\rm Gal}(Z/X)\times{\rm Gal}(W/X)$, and if $Z\times_{X}W$ is
irreducible we have ${\rm Gal}(Z\times_{X}W/X)\cong{\rm Gal}(Z/X)\times{\rm
Gal}(W/X)$. This fact can be deduced from [Mil80, Theorem 5.3] or by comparing
with the Galois groups of the corresponding function field extensions.) We
will show that one can choose $h_{1},\dots,h_{m}\in\mathbb{F}_{q}[T]$ such
that the corresponding morphism
$\psi\colon\mathbb{A}^{1}\rightarrow\mathbb{A}^{m}$ satisfies
$\psi^{-1}(U)\neq\emptyset$ and
$Y^{\prime}\times_{U}\psi^{-1}(U)\to\psi^{-1}(U)$ is a Galois cover with
Galois group ${\rm Gal}(Y^{\prime}/U)=G\times{\rm
Gal}(\mathbb{F}_{q^{\nu}}/\mathbb{F}_{q})$, which will imply that
$Y\times_{U}\psi^{-1}(U)\to\psi^{-1}(U)$ is a geometric Galois cover with
Galois group $G$, since any extension of the field of constants occurring in
this cover must be of degree dividing $\nu=|G|$.

First note that $Y^{\prime}\times_{U}\psi^{-1}(U)\to\psi^{-1}(U)$ is again
étale for any $\psi$, so it suffices to show that $h_{1},\dots,h_{m}$ can be
chosen so that it is a Galois cover with Galois group ${\rm
Gal}(Y^{\prime}/U)$, for which we are allowed to replace $U$ and $Y^{\prime}$
by dense open subvarieties. So since an étale morphism is locally standard
étale (see e.g. [Mil80, Proposition I.3.19]), we can assume that
$U=\\{g(T_{1},\ldots,T_{m})\neq 0\\}$ with $0\neq
g\in\mathbb{F}_{q}[T_{1},\ldots,T_{m}]$ and

$Y^{\prime}=\\{f(T_{1},\ldots,T_{m},X)=0,g(T_{1},\ldots,T_{m})\neq
0\\}\subseteq\mathbb{A}^{m+1}$

with $f\in\mathbb{F}_{q}[T_{1},\ldots,T_{m},X]$, and $Y^{\prime}\to U$ is the
projection to the first $m$ coordinates. By [FJ08, Theorem 13.3.5 and
Proposition 16.1.5] there exist $h_{1},\ldots,h_{m}\in\mathbb{F}_{q}[T]$ with
$g(h_{1},\ldots,h_{m})\neq 0$, $f(h_{1},\dots,h_{m},X)\in\mathbb{F}_{q}(T)[X]$
is irreducible and

${\rm Gal}(f(h_{1},\ldots,h_{m},X)/\mathbb{F}_{q}(T))={\rm
Gal}(f(T_{1},\ldots,T_{m},X)/\mathbb{F}_{q}(T_{1},\ldots,T_{m}))={\rm
Gal}(Y^{\prime}/U).$

The corresponding $\psi$ satisfies $\psi^{-1}(U)\neq\emptyset$,
$Y^{\prime}\times_{U}\psi^{-1}(U)$ is irreducible, and

${\rm Gal}(Y^{\prime}\times_{U}\psi^{-1}(U)/\psi^{-1}(U))={\rm
Gal}(f(h_{1},\ldots,h_{m},X)/\mathbb{F}_{q}(T))={\rm Gal}(Y^{\prime}/U),$

concluding the proof. ∎

The next proposition combined with Lemma 5.8 implies Proposition 5.9 in the
special case when all $G_{i}$ are quasi-$p$.

######  Proposition 5.11.

Let $G_{1},G_{2}$ be $L_{1}^{\mathrm{gt}}(q)$-realizable groups with $G_{2}$
quasi-$p$. Then $G_{1}\times G_{2}$ is $L_{1}^{\mathrm{gt}}(q)$-realizable.

###### Proof.

First assume that both $G_{1},G_{2}$ are quasi-$p$. Let $T_{1},T_{2}$ be
independent variables and let $L_{i}/\mathbb{F}_{q}(T_{i})$ be an
$L_{1}^{\mathrm{gt}}(q)$-realization of $G_{i}$ ramified only over $\infty$
(we may assume this by Lemma 5.5). It corresponds to a geometric Galois cover
$Y_{i}\to\mathbb{A}^{1}=\mathrm{Spec}(\mathbb{F}_{q}[T_{i}])$ with ${\rm
Gal}(Y_{i}/\mathbb{A}^{1})=G_{i}$. Taking the product of these covers we
obtain a geometric Galois cover $Y=Y_{1}\times Y_{2}\to\mathbb{A}^{2}$ with
${\rm Gal}(Y/\mathbb{A}^{2})=G_{1}\times G_{2}$. By Proposition 5.10 there
exists a morphism $h\colon\mathbb{A}^{1}\to\mathbb{A}^{2}$ such that
$Z=Y\times_{\mathbb{A}^{2}}\mathbb{A}^{1}\to\mathbb{A}^{1}$ is a geometric
Galois cover with ${\rm Gal}(Z/\mathbb{A}^{1})={\rm Gal}(Y/\mathbb{A}^{2})=G$.
The extension $L=\mathbb{F}_{q}(Z)$ of $\mathbb{F}_{q}(\mathbb{A}^{1})$ now
gives a geometric realization of $G$ ramified only over $\infty$.

Now let $G_{1},G_{2}$ be $L_{1}^{\mathrm{gt}}(q)$-realizable with only $G_{2}$
assumed quasi-$p$. Iterating the above claim we see that an arbitrary product
of $L_{1}^{\mathrm{gt}}(q)$-realizable quasi-$p$ groups is again
$L_{1}^{\mathrm{gt}}(q)$-realizable (and of course quasi-$p$ as well). In
particular since $G_{2}$ is quasi-$p$ and $L_{1}^{\mathrm{gt}}(q)$-realizable,
for every $m$ we can find a geometric Galois extension $L/\mathbb{F}_{q}(T)$
with Galois group $(G_{2})^{m}$ ramified only over $\infty$, and it contains
linearly disjoint subextensions $L_{1},\ldots,L_{m}$ such that ${\rm
Gal}(L_{i}/\mathbb{F}_{q}(T))=G_{2}$. Let $L/\mathbb{F}_{q}(T)$ be a geometric
realization of $G_{1}$ ramified over $\infty$, tamely ramified over 0 and
unramified elsewhere. For $m$ sufficiently large, one of the $L_{i}$ is
linearly disjoint from $L$ over $\mathbb{F}_{q}(T)$, and then $K=LL_{i}$ is a
geometric Galois extension of $\mathbb{F}_{q}(T)$ with ${\rm
Gal}(K/\mathbb{F}_{q}(T))=G_{1}\times G_{2}$, ramified only over $0,\infty$,
tamely over $0$. ∎

###### Proof of Proposition 5.9.

Let $G=G_{1}\times\dots\times G_{m}$ be a product of
$L_{1}^{\mathrm{gt}}(q)$-realizable groups. If $G$ is quasi-$p$, then by
Proposition 5.11 the group $G$ is $L_{1}^{\mathrm{gt}}(q)$-realizable and by
Lemma 5.8 it satisfies Conjecture 1.4.

Otherwise we can use Proposition 5.11 to absorb all the quasi-$p$ factors into
one of the non-quasi-$p$ factors and reduce to the case when none of the
$G_{i}$ are quasi-$p$. Assuming none of the $G_{i}$ are quasi-$p$ we have
$d\left((G/p(G))^{\mathrm{ab}}\right)=m$, since by Lemma 5.5 the $G_{i}$ are
cyclic-by-quasi-$p$ and the assumptions of the proposition imply that each
$G_{i}$ has a quotient isomorphic to $\mathbb{Z}/\ell\mathbb{Z}$. By Lemma
5.8, for each $G_{i}$ there exists a geometric Galois extension
$L_{i}/\mathbb{F}_{q}(T)$ with Galois group $G_{i}$ branched only over a
finite prime $h_{i}\in\mathbb{F}_{q}[T]$ and we may take the
$h_{i},i=1,\ldots,m$ to be pairwise distinct. Then the extensions
$L_{i}/\mathbb{F}_{q}(T)$ are pairwise linearly disjoint, since their branch
loci are disjoint and $\mathbb{F}_{q}(T)$ has no unramified geometric
extensions. Taking $L=L_{1}\cdots L_{m}$ we see that ${\rm
Gal}(L/\mathbb{F}_{q}(T))=G$ and $L/\mathbb{F}_{q}(T)$ is ramified exactly
over $h_{1},\ldots,h_{m}$, so $r_{\mathbb{F}_{q}(T)}(G)\leq m$. Since by
Corollary 3.4 we also have $r_{\mathbb{F}_{q}(T)}(G)\geq m$, we conclude that
$r_{\mathbb{F}_{q}(T)}(G)=m$, which is what Conjecture 1.4 predicts. ∎

###### Proof of Theorem 1.8.

Let $G=G_{1}\times\ldots\times G_{m}$ with each $G_{i}$ a symmetric or
alternating group satisfying the assumptions of Theorem 1.8. Since Theorem 5.7
has the same assumptions, each $G_{i}$ is $L_{1}^{\mathrm{gt}}(q)$-realizable.
If $p>2$ we may take $\ell=2$ and then each $G_{i}$ is either an alternating
group and thus quasi-$p$ or is a symmetric group and then $\ell\mid
2=[G_{i}:p(G_{i})]$. Thus the conditions of Proposition 5.9 are satisfied and
the conclusion follows. If $p=2$ then we take $\ell=3$ and $G_{i}$ is quasi-2
unless $G_{i}=A_{3},A_{4}$ in which case $\ell\mid 3=[G_{i}:p(G_{i})]$ and
again Proposition 5.9 applies and the conclusion follows. ∎

###  5.3. Proof of Theorem 5.7

Let $n\geq p$. In the present subsection we prove for $G\in\\{S_{n},A_{n}\\}$
and each of the pairs $(n,q)$ from Theorem 5.7 that the cyclic-by-quasi-$p$
group $G$ is $L_{1}^{\mathrm{gt}}(q)$-realizable. We construct realizations
for these groups by writing down explicit equations. Our constructions are
inspired by the work of Abhyankar [Abh92], and in the case $p=2$ we directly
use the results of Abhyankar, Ou, Sathaye and Yie [AOS94, AY94]. Some of the
Galois group calculations rely on the classification of multiply transitive
groups, which in turn relies on the Classification of Finite Simple Groups
(CFSG). We will indicate when CFSG and its applications are used. In all cases
we will write an explicit polynomial $f\in\mathbb{F}_{q}(T)[X]$ of degree $n$
and show that ${\rm Gal}(f/\mathbb{F}_{q}(T))=G$ and that the splitting field
of $f$ over $\mathbb{F}_{q}(T)$ is geometric and is ramified over a single
prime of degree one if $G$ is quasi-$p$ and over two primes of degree one with
tame ramification over at least one of them if $G$ is cyclic-by-quasi-$p$ but
not quasi-$p$. Note that we always have the inclusions ${\rm
Gal}(f/\overline{\mathbb{F}}_{q}(T))\leqslant{\rm
Gal}(f/\mathbb{F}_{q}(T))\leqslant S_{n}$, so for $G=S_{n}$, proving that
${\rm Gal}(f/\overline{\mathbb{F}}_{q}(T))=S_{n}$ shows both that ${\rm
Gal}(f/\mathbb{F}_{q}(T))=G$ and that the splitting field of $f$ is geometric,
and for $G=A_{n}$, proving that ${\rm
Gal}(f/\overline{\mathbb{F}}_{q}(T))\geqslant A_{n}$ and ${\rm
Gal}(f/\mathbb{F}_{q}(T))\leqslant A_{n}$ shows that ${\rm
Gal}(f/\mathbb{F}_{q}(T))=G$ and that the splitting field of $f$ is geometric.
Recall from Lemma 2.6 that if $p>2$ then ${\rm
Gal}(f/\mathbb{F}_{q}(T))\leqslant A_{n}$ if and only if $\Delta(f)$ is a
square in $\mathbb{F}_{q}(T)$ (where we always take the discriminant with
respect to $X$). We distinguish four main cases according to $G=S_{n}$ or
$A_{n}$ and $p=2$ or $\neq 2$, and several subcases according to $n$ and $q$.

$G=S_{n},p>2$. In this case $G$ is cyclic-by-quasi-$p$ but not quasi-$p$, so
we are looking for a polynomial $f\in\mathbb{F}_{q}(T)[X],\deg f=n$ with ${\rm
Gal}(f/\overline{\mathbb{F}}_{q}(T))=S_{n}$ with splitting field ramified
exactly over $T=0,\infty$, tamely over one of these points.

$n$ odd, $n>p,p\nmid n+1$. Let $g=X^{n+1}-(1+T)X^{p}+T\in\mathbb{F}_{p}[T,X]$
and

(5.1)
$f=\frac{g}{X-1}=X^{p}\frac{X^{n+1-p}-1}{X-1}-(X-1)^{p-1}T\in\mathbb{F}_{p}[T,X].$

Note that $f$ is a monic polynomial of degree $n$ in $X$. It is irreducible
over $\overline{\mathbb{F}}_{q}(T)$ because it is linear in the variable $T$
with coprime coefficients $X^{p}\frac{X^{n+1-p}-1}{X-1}$ and $-(X-1)^{p-1}$
(here we use the assumption $p\nmid n+1$). Since $g^{\prime}=(n+1)X^{n}$,
using (2.5),(2.4) and (2.2) we have

$\Delta(f)=\Delta(g)/\mathrm{Res}(X-1,f)^{2}=\pm(n+1)^{n+1}T^{n}/f(1)^{2}=\pm(n+1)^{n-1}T^{n}.$

Let $\alpha$ be a root of $f$ in an algebraic closure of $\mathbb{F}_{q}(T)$.
Since $\Delta(f)=\pm(n+1)^{n-1}T^{n}$, the extension
$\overline{\mathbb{F}}_{q}(T,\alpha)/\overline{\mathbb{F}}_{q}(T)$ (and
therefore its normal closure) is ramified at most over $0,\infty$. We now
compute the ramification indices of this extension over 0. To this end observe
that by (5.1) we have

(5.2)
$T=\frac{\alpha^{p}(\frac{\alpha^{n+1-p}-1}{\alpha-1})}{(\alpha-1)^{p-1}}$

and we see that the divisor of zeros of $T$ over
$\overline{\mathbb{F}}_{q}(T,\alpha)=\overline{\mathbb{F}}_{q}(\alpha)$ is
composed of one prime of multiplicity $p$ and other primes of multiplicity 1.
Therefore the ramification indices over $T=0$ are $p,1,\ldots,1$, so by Lemma
2.1(ii) $G={\rm Gal}(f/\overline{\mathbb{F}}_{q}(T))$ contains a cycle of
length $p$. Note also that by (5.2), the ramification indices over $T=\infty$
are $p-1,n+1-p$, in particular the ramification over $T=\infty$ is tame.
Therefore the action of ${\rm Gal}(f/\overline{\mathbb{F}}_{q}(T))$ on the
roots of $f$ is primitive by Lemma 2.12, so since it contains a cycle of
length $p$ we can apply Theorem 2.3 with $\ell=p$.

If $n>p+2$ then Jordan’s Theorem (see Remark 2.4) implies that ${\rm
Gal}(f/\overline{\mathbb{F}}_{q}(T))\geqslant A_{n}$ and since $\Delta(f)$ is
not a square we have ${\rm Gal}(f/\mathbb{F}_{q}(T))={\rm
Gal}(f/\overline{\mathbb{F}}_{q}(T))=S_{n}$ and the polynomial $f$ satisfies
all the required properties.

If $n=p+2$ then we use Theorem 2.3(ii) to conclude that either ${\rm
Gal}(f/\overline{\mathbb{F}}_{q}(T))\geqslant A_{n}$ (and then we argue as
before) or $p=2^{k}-1$ is a Mersenne prime and ${PGL}_{2}(2^{k})\leqslant{\rm
Gal}(f/\overline{\mathbb{F}}_{q}(T))\leqslant{P\Gamma L}_{2}(2^{k})$ with its
action on the roots being the standard action of such a group on
$\mathbb{P}^{1}(\mathbb{F}_{2^{k}})$. However, the ramification indices over
$\infty$ are $3,p-1$ and since $p>3$ (as $p\nmid n+1=p+3$), by Lemma 2.1(i),
${\rm Gal}(f/\overline{\mathbb{F}}_{q}(T))$ contains a permutation with cycle
structure $3,p-1$. The group $P\Gamma L_{2}(2^{k})$ with its standard action
on $\mathbb{P}^{1}(\mathbb{F}_{2^{k}})$ has no such element (see [GMPS16,
Corollary 4.6]), a contradiction.

######  Remark 5.12.

Note that in treating the case $n=p+2$ we made use of Theorem 2.3(ii), which
makes use of the CFSG.

$n=p$. Let

$f=X^{p}+X^{2}-T\in\mathbb{F}_{p}[T,X].$

We have $f^{\prime}=2X$ and therefore by (2.4) we have
$\Delta(f)=aT,a\in\mathbb{F}_{p}^{\times}$, hence the splitting field of $f$
over $\overline{\mathbb{F}}_{q}(T)$ is ramified at most over 0 and $\infty$.
The polynomial $f$ is irreducible over $\overline{\mathbb{F}}_{q}(T)$ since it
is monic (up to sign) and linear in the variable $T$.

Let $\alpha$ be a root of $f$ in an algebraic closure of $\mathbb{F}_{q}(T)$.
Then $T=\alpha^{2}(\alpha^{p-2}+1)$ and we get (reasoning as in the previous
case) that in the extension
$\overline{\mathbb{F}}_{q}(\alpha)=\overline{\mathbb{F}}_{q}(T,\alpha)/\overline{\mathbb{F}}_{q}(T)$
the prime $T=0$ splits into $p-2$ unramified primes and the prime $\alpha=0$
with ramification index 2. In particular, the splitting field of $f$ is tamely
ramified over $T=0$ and ${\rm Gal}(f/\overline{\mathbb{F}}_{q}(T))\leqslant
S_{n}$ contains a transposition by Lemma 2.1. On the other hand $T=\infty$ is
totally ramified with ramification index $p$. By Lemma 2.12, ${\rm
Gal}(f/\overline{\mathbb{F}}_{q}(T))$ is primitive, so by Lemma 2.5 we have
${\rm Gal}(f/\overline{\mathbb{F}}_{q}(T))=S_{n}$ and $f$ is as desired.

$n$ odd, $p|n+1$, $n\neq 2p-1$. Note that these conditions imply $n\geq 4p-1$.
Choose $h\in\mathbb{F}_{p}[X],\deg h=3$ monic irreducible. There exists
$u\in\mathbb{F}_{p}[X],\deg u=3$ with $X^{n+3-p}\equiv u^{p}\pmod{h}$, since
$a\mapsto a^{p}$ is an automorphism of $\mathbb{F}_{p}[X]/h$ and we can assume
without loss of generality that $X\nmid u$ by adding $\pm h$ if necessary. Let
$g=X^{n+3}-u(X)^{p}X^{p}-h(X)^{p}T$ and

(5.3)
$f=\frac{g}{h}=\frac{X^{n+3-p}-u(X)^{p}}{h(X)}X^{p}-h(X)^{p-1}T\in\mathbb{F}_{p}[T,X].$

The polynomial $f$ is monic in $X$ with $\deg_{X}f=n$. Let
$v=X^{n+3-p}-u(X)^{p}$ and note that $v^{\prime}=(n+3)X^{n+2-p}\neq 0$, in
particular $v$ is separable (as $X\nmid u$), which also implies that
$h^{2}\nmid v$ and so $f$ is linear in $T$ with coprime coefficients, hence
irreducible. Using (2.5),(2.4) and (2.2) we compute

$\Delta(f)=\Delta(g)/(\mathrm{Res}(h,f)^{2}\Delta(h))=aT^{n+2}$

for some $a\in\mathbb{F}_{p}^{\times}$, as $\mathrm{Res}(h,f)$ is independent
of $T$ and non-zero since $h^{2}\nmid v$. Consequently the splitting field of
$f$ is ramified only over $T=0,\infty$. Since $n$ is odd, $\Delta(f)$ is not a
square in $\overline{\mathbb{F}}_{q}(T)$ and so ${\rm
Gal}(f/\overline{\mathbb{F}}_{q}(T))\not\leqslant A_{n}$.

Let $\alpha$ be a root of $f$ in an algebraic closure of $\mathbb{F}_{q}(T)$.
By (5.3) we have

(5.4)
$T=\frac{\left(\frac{\alpha^{n+3-p}-u(\alpha)^{p}}{h(\alpha)}\right)\alpha^{p}}{h(\alpha)^{p-1}}$

and therefore
$\overline{\mathbb{F}}_{q}(T,\alpha)=\overline{\mathbb{F}}_{q}(\alpha)$. Now
since $n\geq 4p-1$ we see from (5.4) that the primes of
$\overline{\mathbb{F}}_{q}(\alpha)$ over $T=0$ are $\alpha=0$ with
ramification index $p$ and the roots of $v/h$ with ramification index 1, so by
Lemma 2.1 the group ${\rm Gal}(f/\overline{\mathbb{F}}_{q}(T))$ contains a
cycle of length $p$. Similarly, the primes of
$\overline{\mathbb{F}}_{q}(\alpha)$ over $T=\infty$ are the 3 roots of $h$
with multiplicity $p-1$ each, and $\alpha=\infty$ with multiplicity
$n-3(p-1)\equiv 2\pmod{p}$, in particular the splitting field of $f$ is tamely
ramified over $T=\infty$. By Lemma 2.12, ${\rm
Gal}(f/\overline{\mathbb{F}}_{q}(T))$ acts primitively on the roots of $f$. By
Jordan’s Theorem (Theorem 2.3(i) in the elementary case $\ell=p\leq n-3$
prime) we have ${\rm Gal}(f/\overline{\mathbb{F}}_{q}(T))\geqslant A_{n}$ and
therefore ${\rm Gal}(f/\overline{\mathbb{F}}_{q}(T))=S_{n}$.

$n=2p-1$. Let

$f=\frac{X^{2p}-TX^{p}-X^{2}+T}{X-1}=\frac{X^{2p-2}-1}{X-1}X^{2}-T(X-1)^{p-1}\in\mathbb{F}_{p}[T,X].$

The polynomial $f$ is monic in $X$ and irreducible over
$\overline{\mathbb{F}}_{q}(T)$, since it is linear in $T$ with coprime
coefficients. Using (2.5),(2.4) and (2.2) we compute

$\Delta(f)=\Delta(X^{2p}-TX^{p}-X^{2}+T)/\mathrm{Res}(X-1,f)^{2}=aT$

for some $a\in\mathbb{F}_{p}^{\times}$. We see that the splitting field of $f$
over $\overline{\mathbb{F}}_{q}(T)$ is ramified only over $0,\infty$. Let
$\alpha$ be a root of $f$ in an algebraic closure of $\mathbb{F}_{q}(T)$. We
have

$T=\frac{\left(\frac{\alpha^{2p-2}-1}{\alpha-1}\right)\alpha^{2}}{(\alpha-1)^{p-1}},$

from which we see that in the extension
$\overline{\mathbb{F}}_{q}(\alpha)=\overline{\mathbb{F}}_{q}(T,\alpha)$ of
$\overline{\mathbb{F}}_{q}(T)$ there are $2p-2$ primes lying over $T=0$ with
one of them having ramification index $2$ and the other unramified, while over
$T=\infty$ we have the primes $\alpha=\infty$ with ramification index $p$, and
$\alpha=1$ with ramification index $p-1$. Therefore, by Lemma 2.1, ${\rm
Gal}(f/\overline{\mathbb{F}}_{q}(T))\leqslant S_{n}$ contains both a
transposition and cycle of length $p$. The latter implies that it is
primitive, and so ${\rm Gal}(f/\overline{\mathbb{F}}_{q}(T))=S_{n}$ by Lemma
2.5.

$n$ even, $p\nmid n$, $n>p+1$. Let

$f=X^{n}+X^{p}-T\in\mathbb{F}_{p}[T,X].$

Since $f$ is monic (up to sign) and linear in $T$, it is irreducible over
$\overline{\mathbb{F}}_{q}(T)$. Since $f^{\prime}=nX^{n-1}$, by (2.4) we have
$\Delta(f)=aT^{n-1},a\in\mathbb{F}_{p}^{\times}$. Therefore the splitting
field of $f$ is only ramified over $T=0$ and $T=\infty$, and $\Delta(f)$ is
not a square in $\overline{\mathbb{F}}_{q}(T)$ (since $n$ is even) and so
${\rm Gal}(f/\overline{\mathbb{F}}_{q}(T))\not\leqslant A_{n}$. Denoting by
$\alpha$ a root of $f$ in an algebraic closure of $\mathbb{F}_{q}(T)$, we see
that $T=\alpha^{p}(\alpha^{n-p}+1)$. The polynomial $X^{n-p}+1$ is separable
by the assumption $p\nmid n$, and so the ramification indices of
$\overline{\mathbb{F}}_{q}(T,\alpha)=\overline{\mathbb{F}}(\alpha)$ over $T=0$
are $p,1,\ldots,1$, and $T=\infty$ is totally ramified with ramification index
$n$ (in particular tame). Therefore, ${\rm
Gal}(f/\overline{\mathbb{F}}_{q}(T))\leqslant S_{n}$ is primitive by Lemma
2.12 and contains a cycle of length $p$ by Lemma 2.1, so by Jordan’s Theorem
(i.e. Theorem 2.3(i) in the case $\ell=p\leq n-3$ prime) we get that ${\rm
Gal}(f/\overline{\mathbb{F}}_{q}(T))=S_{n}$.

$n$ even, $p|n$. By our assumptions, $n\geq 2p$. Choose
$h(X)\in\mathbb{F}_{p}[X]$ monic irreducible with $\deg h=2$ and
$h^{\prime}(0)\neq 0$ (which always exists). Since $a\mapsto a^{p}$ is an
automorphism of $\mathbb{F}_{p}[X]/h$, there exists $u(X)\in\mathbb{F}_{p}[X]$
with $\deg u<n/p$ and $u^{p}\equiv X^{n+2-p}\pmod{h}$. We can assume without
loss of generality that $u(0)\neq 0$, since if $n>2p$ we can add $\pm h$ to
$u$ if necessary, while if $n=2p$ and we take the unique $u$ with $\deg u\leq
1$ such that $u^{p}\equiv X^{p+2}\pmod{h}$ then automatically $u(0)\neq 0$,
since otherwise $u=cX,c\in\mathbb{F}_{p}^{\times}$ and then $h=X^{2}-c$,
contradicting our assumption $h^{\prime}(0)\neq 0$. Let
$g=X^{n+2}-u(X)^{p}X^{p}-Th(X)^{p}\in\mathbb{F}_{p}[T,X]$ and

(5.5)
$f=\frac{g}{h}=\frac{X^{n+2-p}-u(X)^{p}}{h(X)}X^{p}-Th(X)^{p-1}\in\mathbb{F}_{p}[T,X].$

The polynomial $f$ is monic in $X$ of degree $n$. Let $v=X^{n+2-p}-u(X)^{p}$.
As $v^{\prime}=2X^{n+1-p}$ and $u(0)\neq 0$, we have that $v$ is separable, in
particular $h^{2}\nmid v$. So $f$ is linear in $T$ with coprime coefficients,
hence irreducible over $\overline{\mathbb{F}}_{q}(T)$. Using (2.5),(2.4) and
(2.2), we compute

$\Delta(f)=\Delta(g)/(\mathrm{Res}(h(X),f)^{2}\Delta(h))=aT^{n+1},a\in\mathbb{F}_{p}^{\times}$

and conclude that the splitting field of $f$ over
$\overline{\mathbb{F}}_{q}(T)$ is ramified at most over $T=0,\infty$ and that
${\rm Gal}(f/\overline{\mathbb{F}}_{q}(T))\not\leqslant A_{n}$ (since $n$ is
even).

Denoting by $\alpha$ a root of $f$ in an algebraic closure of
$\mathbb{F}_{q}(T)$ we have (by (5.5))

$T=\frac{\alpha^{p}\left(\frac{\alpha^{n+2-p}-u(\alpha)^{p}}{h(\alpha)}\right)}{h(\alpha)^{p-1}}.$

The polynomial $v$ is separable, hence the primes of
$\overline{\mathbb{F}}_{q}(T,\alpha)=\overline{\mathbb{F}}_{q}(\alpha)$ over
$T=0$ are $\alpha=0$ with ramification index $p$ and $n-p$ primes with
ramification index 1. The primes over $T=\infty$ are the two roots of $h$ with
ramification index $p-1$ and $\alpha=\infty$ with ramification index
$n-2(p-1)\equiv 2\pmod{p}$, in particular $T=\infty$ is tamely ramified.
Therefore again, ${\rm Gal}(f/\overline{\mathbb{F}}_{q}(T))\leqslant S_{n}$ is
primitive by Lemma 2.12 and contains a cycle of length $p$ by Lemma 2.1, so by
Jordan’s Theorem (i.e. Theorem 2.3(i) in the case $\ell=p\leq n-p\leq n-3$
prime) we get that ${\rm Gal}(f/\overline{\mathbb{F}}_{q}(T))=S_{n}$.

$G=S_{n},p=2$. In this case $G$ is quasi-$p$.

$n$ odd. We have $n\geq 3$. Let

$f(X)=X^{n}+TX^{n-2}+1\in\mathbb{F}_{2}[T,X].$

By [Abh92, §11.I.5] (with $t=n-2$, $a=1$, $s=1$), we have ${\rm
Gal}(f/\overline{\mathbb{F}}_{2}(T))=S_{n}$ and the splitting field of $f$ is
ramified only over $T=\infty$.

$n$ even. Let

$f(X)=\left((X+1)^{n-1}+X^{n-1}\right)(X+1)^{2}+T^{n-1}X^{n-1}\in\mathbb{F}_{2}[T,X].$

By [Abh92, §12.IV.4] (with $t=n-1$, $s=n-1$, $a=1$, $b=1$) we have ${\rm
Gal}(f/\overline{\mathbb{F}}_{2}(T))=S_{n}$ and the splitting field of $f$ is
ramified only over $T=0$.

$G=A_{n},p>2$.

$n\neq p+1$. In this case we have shown that $S_{n}$ is
$L_{1}^{\mathrm{gt}}(q)$-realizable, and since $p>2$ we have $p(S_{n})=A_{n}$.
Therefore by Lemma 5.5, the group $A_{n}=p(S_{n})$ is also
$L_{1}^{\mathrm{gt}}(q)$-realizable.

$p=3,n=4$. Let

$f=X^{4}-TX^{3}+1\in\mathbb{F}_{3}[T,X].$

As $f^{\prime}=X^{3}$ and $f(0)=1$ we compute that $\Delta(f)=1$, and thus the
splitting field of $f$ is ramified only over $T=\infty$ and ${\rm
Gal}(f/\mathbb{F}_{q}(T))\leqslant A_{4}$. The group ${\rm
Gal}(f/\overline{\mathbb{F}}_{q}(T))\leqslant A_{4}$ is transitive on the
roots (since $f$ is irreducible, being linear in the variable $T$) and has
order divisible by $3$ (otherwise the splitting field of $f$ over
$\overline{\mathbb{F}}_{q}(T)$ would be tamely ramified and ramified only over
infinity, which implies it is trivial). Thus ${\rm
Gal}(f/\mathbb{F}_{q}(T))={\rm Gal}(f/\overline{\mathbb{F}}_{q}(T))=A_{4}$.

$n=p+1,p>3,\mathbb{F}_{q}\supseteq\mathbb{F}_{p^{2}}$. Let $s\geq 1$ and
$2\leq a\leq p-1$ be integers and let

$f=(X+1)\left(X+\frac{a}{a-1}\right)^{p}-T^{s}X^{a}\in\mathbb{F}_{p}[T,X].$

By [Abh92, §22] (with $\tau=a$, $Y=T^{s}$, $b=\frac{a}{a-1}$, but note the
nonstandard sign convention in the definition of the discriminant),

(5.6)
$\Delta(f)=(-1)^{(p+1)/2}\frac{a^{2a-1}}{(a-1)^{2a-3}}T^{s(p+1)}\in\mathbb{F}_{p}[T].$

It is shown in [Abh92, §12.IV.3] (with $t=a$, $b=\frac{a}{a-1}$) that if
$p>5$, $2\leq a\leq\frac{p-1}{2},(a,p+1)=1$ and $a(p+1-a)|s$, then ${\rm
Gal}(f/\overline{\mathbb{F}}_{q}(T))=A_{n}$ and the splitting field of $f$ is
ramified only over $T=0$. If additionally
$\mathbb{F}_{q}\supseteq\mathbb{F}_{p^{2}}$, then $\Delta(f)$ is a square in
$\mathbb{F}_{q}[T]$ and we have ${\rm Gal}(f/\mathbb{F}_{q}(T))=A_{n}$. Note
that for all $p>5$, an $a$ with $(a,p+1)=1$ and $a\not\equiv\pm 1\pmod{p+1}$
can be found (as $\phi(x)>2$ for $x>6$), and after replacing $a$ with $p+1-a$
if necessary to assume that $2\leq a\leq\frac{p-1}{2}$, we can set
$s=a(p+1-a)$.

If $p=5,n=6,\mathbb{F}_{q}\supseteq\mathbb{F}_{25}$ we take the polynomial

$f=(X+1)(X+2)^{5}-T^{4}X^{2}\in\mathbb{F}_{p}[T,X].$

The discriminant of $f$ is computed by (5.6) with $a=2,s=4$ and equals
$\Delta(f)=2T^{24}$, which is a square in $\mathbb{F}_{25}(T)$ and hence ${\rm
Gal}(f/\mathbb{F}_{q}(T))\leqslant A_{5}$. On the other hand, by [Abh92,
§12.IV.2] (with $t=2,b=2$), the splitting field of $f$ over
$\overline{\mathbb{F}}_{5}(T)$ is ramified only over $T=0$, and its Galois
group is $A_{5}$, hence ${\rm Gal}(f/\mathbb{F}_{q}(T))={\rm
Gal}(f/\overline{\mathbb{F}}_{q}(T))=A_{5}$.

######  Remark 5.13.

If we want to drop the assumption $\mathbb{F}_{q}\supseteq\mathbb{F}_{p^{2}}$
in the case $p>5$ we would need to find an $a$ with $2\leq
a\leq\frac{p-1}{2},(a,p+1)=1$ such that $(-1)^{(p+1)/2}a(a-1)$ is a square
modulo $p$. It is not hard to show (using a Pólya-Vinogradov-type inequality
and an elementary sieve argument) that this is possible for all $p$
sufficiently large, and we conjecture that such an $a$ exists for all $p>13$.
While proving this should be doable by means of a careful analysis and
sufficiently large computer search, we did not pursue this.

$G=A_{n},p=2$. The group $A_{n}$ is always cyclic-by-quasi-2 and it is quasi-2
iff $n\neq 3,4$. Most of the required realizations were constructed by
Abhyankar, Ou, Sathaye and Yie [Abh92, Abh93, AOS94, AY94].

$n=3,4$, $\mathbb{F}_{q}\supseteq\mathbb{F}_{4}$. For
$A_{3}\cong\mathbb{Z}/3\mathbb{Z}$ we can take the extension
$K=\mathbb{F}_{q}(s)/F_{q}(T)$ with $s^{3}=T$ which is Galois with group
$A_{3}$ if $\mathbb{F}_{q}\supseteq\mathbb{F}_{4}$ and ramified only over
$T=0,\infty$. Denote by $\zeta$ any element of
$\mathbb{F}_{4}\setminus\mathbb{F}_{2}$, and consider the splitting field $L$
of the polynomial

$f=(X^{2}+X+s)(X^{2}+X+\zeta s)(X^{2}+X+\zeta^{2}s)\in\mathbb{F}_{2}(s)[X]$

over $K$. The extension $L/\mathbb{F}_{q}(T)$ is Galois (since $s,\zeta
s,\zeta^{2}s$ are a Galois orbit over $\mathbb{F}_{q}(T)$) and by Artin-
Schreier theory ${\rm
Gal}(L/K)=\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$ with ${\rm
Gal}(L/\mathbb{F}_{q}(T))$ acting nontrivially on the order 2 subgroups of
${\rm Gal}(L/K)$ by conjugation, in particular ${\rm
Gal}(L/\mathbb{F}_{q}(T))$ is non-abelian. The only non-abelian extension of
$\mathbb{Z}/3\mathbb{Z}$ by
$\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$ is $A_{4}$. Finally
observe that $L/\mathbb{F}_{q}(T)$ is ramified only over $0,\infty$ and is
therefore an $L_{1}^{\mathrm{gt}}(q)$-realization of $A_{4}$.

######  Remark 5.14.

Note that by Lemma 5.5, both for $n=3$ and for $n=4$ the condition
$\mathbb{F}_{q}\supseteq\mathbb{F}_{4}$ is necessary for $A_{n}$ to be
$L_{1}^{\mathrm{gt}}(q)$-realizable since $A_{3},A_{4}$ have cyclic quotients
of order 3.

$n=5$, $\mathbb{F}_{q}\supseteq\mathbb{F}_{4}$. Let

$f=X^{5}+TX+1\in\mathbb{F}_{2}[T,X].$

It follows from [Abh92, §11.III.1] (with $q=4$, $t=s=1$, $a=-1$, in the
notation used there) that ${\rm Gal}(f/\overline{\mathbb{F}}_{2}(T))\cong
PSL_{2}(4)\cong A_{5}$ and the splitting field of $f$ is ramified only over
$T=\infty$. By [AOS94, 2.23] (with $K=\mathbb{F}_{q}(T)$, $d=4$, $e=1$,
$\bar{b}_{d}=T$, $\bar{b}_{n}=1$), $\mathbb{F}_{q}\supseteq\mathbb{F}_{4}$
implies that ${\rm Gal}(f/\mathbb{F}_{q}(T))\leqslant A_{5}$. Thus ${\rm
Gal}(f/\mathbb{F}_{q}(T))={\rm Gal}(f/\overline{\mathbb{F}}_{q}(T))=A_{5}$.

$n=6,7$, $\mathbb{F}_{q}\supseteq\mathbb{F}_{4}$. Consider the polynomials

$\displaystyle f_{6}$ $\displaystyle=$ $\displaystyle
X^{6}+T^{27}X^{5}+T^{54}X^{4}+(T^{18}+T^{36})X^{3}+T^{108}X^{2}+(T^{90}+T^{135})X+T^{162},$
$\displaystyle f_{7}$ $\displaystyle=$ $\displaystyle X^{7}+TX^{4}+X^{2}+1.$

The polynomials $f_{6},f_{7}$ were found by Abhyankar and Yie [AY94, Theorems
2.10 and 2.11], who showed that the splitting fields of $f_{n},n=6,7$ are
ramified only over $\infty$ and, under the assumption
$\mathbb{F}_{q}\supseteq\mathbb{F}_{4}$, that ${\rm
Gal}(f_{n}/\mathbb{F}_{q}(T))={\rm
Gal}(f_{n}/\overline{\mathbb{F}}_{q}(T))=A_{n}$.

$n\geq 9$ odd, $n\equiv 1,7\pmod{8}$ or
$\mathbb{F}_{q}\supseteq\mathbb{F}_{4}$. Let

$f=X^{n}+TX^{n-4}+1\in\mathbb{F}_{2}[T,X].$

By [Abh93, Theorem 2] (with $t=n-4$, $q=4$ in the notation of the cited
paper), the splitting field of $f$ is ramified only over $\infty$ and we have
${\rm Gal}(f/\mathbb{F}_{q}(T))\geqslant{\rm
Gal}(f/\overline{\mathbb{F}}_{q}(T))\geqslant A_{n}$. Conversely, by [AOS94,
(2.27)] (with $t=n-4$, $b_{n-t}^{*}=T$, $b_{n}^{*}=1$) we have ${\rm
Gal}(f/\mathbb{F}_{q}(T))\leqslant A_{n}$ if either
$\mathbb{F}_{q}\supseteq\mathbb{F}_{4}$ or $n\equiv 1,7\pmod{8}$. In these
cases we conclude that ${\rm Gal}(f/\mathbb{F}_{q}(T))={\rm
Gal}(f/\overline{\mathbb{F}}_{2}(T))=A_{n}$.

$n\geq 8$ even, $10\neq n\equiv 0,2,6\pmod{8}$ or
$\mathbb{F}_{q}\supseteq\mathbb{F}_{4}$. Let $1\leq t\leq n$ with $(t,n)=1$,
and

$f=X^{n}+X^{t}+T^{t}\in\mathbb{F}_{2}[T,X].$

By [Abh92, §11.II.5] (with $s=t$, $a=1$), if $2\leq t\leq n-4$ then ${\rm
Gal}(f/\mathbb{F}_{q}(T))\geqslant{\rm
Gal}(f/\overline{\mathbb{F}}_{2}(T))\geqslant A_{n}$ and the splitting field
of $f$ is ramified only over $\infty$. Conversely, by [AOS94, Theorem 2.27]
(with $b_{n-t}^{*}=1$, $b_{n}^{*}=T^{t}$) we have ${\rm
Gal}(f/\mathbb{F}_{q}(T))\leqslant A_{n}$ if either
$\mathbb{F}_{q}\supseteq\mathbb{F}_{4}$ or $n\equiv 0\pmod{8}$ or $n\equiv
2,6\pmod{8},2t\equiv n\pmod{8}$. In each of these cases we can choose a
suitable $t$: For $n\geq 8$ there exists $2\leq t\leq n-4$ with $(t,n)=1$
since $\phi(n)>2$, and for $n\equiv 2,6\pmod{8},n>10$, the choice
$t=\frac{n}{2}-4$ satisfies $2\leq t\leq n-4$, $(t,n)=1$ and $2t\equiv
n\pmod{8}$.

##  6\. Summary and application to the minimal ramification problem over
$\mathbb{Q}$

###  6.1. Summary

We first summarize some of the results for the groups $G=S_{n}$ and $G=A_{n}$.
These results are obtained by combining Theorem 1.8 (proved in Section 5.2