# Lines highly tangent to a hypersurface

Anand Patel  ,  Eric Riedl  ,  Geoffrey Smith  and  Dennis Tseng

###### Abstract.

We study spaces of lines that meet a smooth hypersurface $X$ in ${\bf P}^{n}$
to high order. As an application, we give a polynomial upper bound on the
number of planes contained in a smooth degree $d$ hypersurface in ${\bf
P}^{5}$ and provide a proof of a result of Landsberg without using moving
frames.

##  1\. Introduction

We work in characteristic 0. Our intention in this paper is first to
illustrate the simplifying role played by log tangent sheaves in the analysis
of lines highly tangent to a hypersurface $X\subset{\bf P}^{n}$, and then to
obtain an upper bound on the number of $2$-planes contained in a smooth
hypersurface in ${\bf P}^{5}$. The simple observation we exploit is: If
$\ell\subset{\bf P}^{n}$ is a line, then sections of the restricted log-
tangent sheaf $T_{{\bf P}^{n}}(-\log X)|_{\ell}$ parametrize those
infinitesimal deformations of $\ell$ preserving the moduli of the point-
configuration $\left(\ell,\ell\cap X\right)$.

This observation helps simplify arguments. To illustrate this, we provide a
quick proof of the following theorem of Landsberg without using moving frames:

######  Theorem 1.1.

[[6] Theorem 2] Let $X\subset{\bf P}^{n}$ be a reduced hypersurface over a
characteristic zero field, $p\in X$ a general point, and let
$\Sigma_{k}\subset\mathbf{G}(1,n)$ denote any irreducible component of the
variety of lines meeting $X$ to order at least $k$ at $p$. If
$\operatorname{dim}\Sigma_{k}$ exceeds its expected dimension $n-k$, then all
lines parametrized by $\Sigma_{k}$ are entirely contained in $X$.

A special instance of Theorem 1.1 is the fact that the flec-nodal locus $F$ of
a smooth, degree $d\geq 3$ surface $S$ in ${\bf P}^{3}$ is always one
dimensional, a fact initially proved by Salmon [9]. This fact, coupled with
the calculation of the degree of $F$, was used by Segre [10] to show that the
number of lines contained in $S$ is at most $(d-2)(11d-6)$.

Using Theorem 1.1 together with a careful geometric analysis, we can show that
for any smooth hypersurface $X$ of degree $d$ in $\mathbb{P}^{5}$, any
component of the space of lines meeting $X$ to order 5 has dimension at most 2
(see Theorem 3.8), which generalizes the $n=5$ case of the de Jong-Debarre
Conjecture [1]. Related work on lines highly tangent to hypersurfaces can also
be found in [11]. Using an intersection theory calculation, we obtain an
explicit upper bound on the space of $2$-planes in $X$, which had for $d\geq
3$ been known to be finite [2, Corollary in Appendix].

######  Theorem 1.2.

Let $X\subset\mathbb{P}^{5}$ be a smooth hypersurface of degree $d$. Then $X$
contains at most $120d^{2}-150d^{3}+35d^{4}$ 2-planes.

The Fermat hypersurface of degree $d$ in $\mathbb{P}^{5}$ is known to contain
$15d^{3}$ 2-planes, which can be compared to the $O(d^{4})$ bound that we
obtain. See [4] for some recent results about the space of planes on cubic
4-folds using lattice theory. In characteristic $p$, many of these results do
not hold. See [3] for some examples of the types of behavior that can occur.

Acknowledgements: We gratefully acknowledge conversations with Izzet Coskun,
Changho Han, Mitchell Lee, and Gwyneth Moreland. Eric Riedl was supported by
NSF CAREER grant DMS-1945944.

##  2\. Deformations of lines with fixed contact order

The goal of this section is to prove Theorem 1.1 using an analysis of the
deformation theory of the space of lines meeting $X$ to high contact order. We
begin by reviewing the deformation theory of these lines and its relation to
the log tangent sheaf. Let $X$ be defined by the homogeneous polynomial
$F(x_{0},\ldots,x_{n})$, and suppose that $X$ contains the point
$p=\left[1:0\ldots:0\right]$. We let

$T_{{\bf P}^{n}}(-\log X)\subset T_{{\bf P}^{n}}$

denote the logarithmic tangent sheaf, i.e. the sheaf of derivations
$\mathcal{O}_{{\bf P}^{n}}{\longrightarrow}\mathcal{O}_{{\bf P}^{n}}$ mapping
the ideal sheaf $\mathcal{I}_{X}$ into itself. In general, the logarithmic
tangent sheaf is reflexive if $X$ is reduced, and when $X$ is smooth (or more
generally, simple normal crossing) the logarithmic tangent sheaf is a vector
bundle. We fix homogeneous coordinates $[s:t]$ on ${\bf P}^{1}$.

Let $\iota:{\bf P}^{1}{\longrightarrow}{\bf P}^{n}$ be a linear embedding of
${\bf P}^{1}$.

######  Definition 2.1.

A first order deformation of $\iota$ is a map

$\displaystyle\iota_{\epsilon}:{\bf P}^{1}\times{\text{\rm
Spec}\,}\mathbb{K}[\epsilon]/(\epsilon^{2}){\longrightarrow}{\bf P}^{n}$ (2.1)

which restricts to $\iota$ modulo $(\epsilon)$. We say a first order
deformation $\iota_{\epsilon}$ preserves $\iota^{-1}(X)$ if the scheme
$\iota_{\epsilon}^{-1}(X)$ is the preimage of $\iota^{-1}(X)$ under the
projection to the first factor

${\bf P}^{1}\times{\text{\rm
Spec}\,}\mathbb{K}[\epsilon]/(\epsilon^{2}){\longrightarrow}{\bf P}^{1}.$

The significance of logarithmic tangent sheaves in the present setting is
captured by the following lemma.

######  Lemma 2.2 ([7, Lemma 2.1]).

Global sections of $\iota^{*}T_{{\bf P}^{n}}(-\log X)$ are in one-to-one
correspondence with first order deformations of $\iota$ preserving
$\iota^{-1}(X)$.

We take a moment to concretely describe the correspondence in 2.2 using
coordinates. Suppose $[s:t]$ are homogeneous coordinates on ${\bf P}^{1}$ and
that the map $\iota$ is defined by $[a_{0}:\ldots:a_{n}]$, where each $a_{i}$
is a linear form in $s,t$. Then, by the Euler exact sequence, the global
sections of $\iota^{*}T_{{\bf P}^{n}}$ are identified with tuples
$(b_{0},\dots,b_{n})$ of linear forms in $s,t$, considered modulo the tuple
$(a_{0},\dots,a_{n})$.

A global section of $\iota^{*}T_{{\bf P}^{n}}$ represented by such a tuple
$(b_{0},\dots,b_{n})$ corresponds to the first-order deformation
$\iota_{\epsilon}$ defined by $[s:t]\mapsto[a_{0}+\epsilon
b_{0}:\ldots:a_{n}+\epsilon b_{n}]$.

Within the space of all deformations $H^{0}({\bf P}^{1},\iota^{*}T_{{\bf
P}^{n}})$, the subspace $H^{0}({\bf P}^{1},\iota^{*}T_{{\bf P}^{n}}(-\log X))$
is represented by tuples $(b_{0},\dots,b_{n})$ obeying the condition

$\displaystyle\sum\limits_{i=0}^{n}b_{i}\cdot\frac{\partial F}{\partial
x_{i}}(a_{0},\dots,a_{n})\equiv 0\mod F(a_{0},\dots,a_{n}).$ (2.2)

The next construction will effectively allow us to replace the degree $d$
hypersurface $X$ with a degree $k<d$ hypersurface. Let $y_{j}=x_{j}/x_{0}$,
$j=1,\ldots,n$ be affine coordinates on the corresponding affine chart
$x_{0}\neq 0$ around $p$, and let $f(y_{1},\ldots,y_{n})$ denote the
dehomogenization of $F$ on this chart. We expand

$f=f_{1}+\ldots+f_{d},$

where $f_{i}$ is the degree $i$ part of $f$, and we let $F_{k}$ be the
homogenization of $f_{1}+\ldots+f_{k}$. Explicitly,

$\displaystyle F_{k}(x_{0},\ldots,x_{n})=\sum_{j=1}^{k}x_{0}^{k-j}\cdot
f_{j}(x_{1},\ldots,x_{n}).$ (2.3)

We let $X_{k}$ denote the hypersurface defined by
$F_{k}(x_{0},\ldots,x_{n})=0$, and we note that $p\in X_{k}$ is a smooth
point. Hence, $X_{k}$ is generically smooth.

######  Definition 2.3.

Suppose $\iota:{\bf P}^{1}{\longrightarrow}{\bf P}^{n}$ is a linear map such
that $\iota^{*}F\in\mathbb{K}[s,t]$ is non-zero and divisible by $s^{k}$. A
first order deformation $\iota_{\epsilon}$ of $\iota$ _preserves order $k$
contact at $[0:1]$_ if $\iota_{\epsilon}^{*}F$ is divisible by $s^{k}$ in the
ring $\mathbb{K}[\epsilon,s,t]/(\epsilon^{2})$.

######  Corollary 2.4.

Let $\iota:{\bf P}^{1}{\longrightarrow}{\bf P}^{n}$ parametrize a line such
that $\iota^{*}F$ is non-zero and divisible by $s^{k}$, and such that
$\iota([0:1])=p$. Then the global sections of $\iota^{*}T_{{\bf P}^{n}}(-\log
X_{k})$ are in one-to-one correspondence with first order deformations
$\iota_{\epsilon}$ preserving order $k$ contact at $[0:1]$.

###### Proof.

Let $[s:t]\mapsto[a_{0}:\ldots:a_{n}]$ define $\iota$, and choose coordinates
on $\mathbb{P}^{n}$ so that $a_{0}$ is not a scalar multiple of $s$, but
$a_{i}$, $i\geq 1$ are multiples of $s$. By assumption, $F(a_{0},\dots,a_{n})$
is a non-zero form divisible by $s^{k}$. This, in turn, implies that
$F_{k}(a_{0},\dots,a_{n})$ is a non-zero multiple of $s^{k}$.

Suppose now that $(b_{0},\dots,b_{n})$ is a tuple of linear forms representing
a deformation $\iota_{\epsilon}$ preserving order $k$ contact at $[0:1]$. This
condition is equivalent to the condition that the form

$\displaystyle\sum\limits_{i=0}^{n}b_{i}\cdot\frac{\partial F}{\partial
x_{i}}(a_{0},\dots,a_{n})$ (2.4)

is also divisible by $s^{k}$. Next, a direct computation shows that

$\displaystyle\frac{\partial F}{\partial x_{i}}(a_{0},\dots,a_{n})\equiv
a_{0}^{d-k}\frac{\partial F_{k}}{\partial x_{i}}\mod s^{k}$ (2.5)

Therefore, since $a_{0}$ is not divisible by $s$, (2.4) is divisible by
$s^{k}$ if and only if

$\displaystyle\sum\limits_{i=0}^{n}b_{i}\cdot\frac{\partial F_{k}}{\partial
x_{i}}(a_{0},\dots,a_{n})$ (2.6)

is also divisible by $s^{k}$. In other words

$\displaystyle\sum\limits_{i=0}^{n}b_{i}\cdot\frac{\partial F_{k}}{\partial
x_{i}}(a_{0},\dots,a_{n})\equiv 0\mod F_{k}(a_{0},\dots,a_{n})$

meaning that $(b_{0},\dots,b_{n})$ represents a global section of
$\iota^{*}T_{{\bf P}^{n}}(-\log X_{k})$, which is what we needed to prove. ∎

###  2.1. Some incidence varieties

Before recovering the result of Landsberg, we need to study the locus swept
out by a family of high-contact lines. The main goal is Theorem 2.8.

######  Definition 2.5.

Define the incidence variety

$\mathsf{Inc}\subset\mathbf{G}(1,n)\times{\bf P}^{n}$

to be the set of pairs $(\ell,p)$ such that $p\in\ell$.

Let $\lambda:\mathsf{Inc}{\longrightarrow}\mathbf{G}(1,n)$ and
$\pi:\mathsf{Inc}{\longrightarrow}{\bf P}^{n}$ denote the two projections.

The projection $\lambda$ expresses $\mathsf{Inc}$ as the universal line over
the Grassmannian $\mathbf{G}(1,n)$.

######  Definition 2.6.

Suppose $V\subset\mathsf{Inc}$ is any subset. Define

$\operatorname{Tot}(V)\subset\mathbf{G}(1,n)\times{\bf P}^{n}\times{\bf
P}^{n}$

to be the set of triples $(\ell,x,y)$ such that $(\ell,x)\in V$ and
$y\in\ell$. Define

$\overline{\operatorname{Tot}}(V)\subset{\bf P}^{n}$

to be the image of $\operatorname{Tot}(V)$ under the projection to the third
factor, i.e. $\overline{\operatorname{Tot}(V)}$ is the union of all lines in
$\lambda(V)$.

Observe that for any subset $V\subset\mathsf{Inc}$,

$\pi(V)\subset\overline{\operatorname{Tot}}(V).$

######  Definition 2.7.

Let $X\subset{\bf P}^{n}$ be a degree $d$ hypersurface. For each $k\leq d$
define the variety

$\displaystyle V_{k}(X)\subset\mathsf{Inc}$ (2.7)

to be the locus of pairs $(\ell,x)$ with $x\in X$, such that $\ell$ meets $X$
at the point $x$ to order $\geq k$.

If the hypersurface $X$ is understood in context, we will sometimes write
$V_{k}$ for $V_{k}(X)$.

######  Theorem 2.8.

Let $X\subset{\bf P}^{n}$ be a smooth hypersurface. If $W\subset V_{k}$ is an
irreducible component satisfying

$\overline{\operatorname{Tot}}(W)={\bf P}^{n},$

then $W$ has the expected dimension $2n-k-1$.

###### Proof.

That every irreducible component of $V_{k}$ has dimension at least $2n-k-1$ is
a simple dimension count which we omit. It is enough to show therefore that
the tangent space $T_{W}|_{(\ell,p)}$ of $W$ at a general point $(\ell,p)\in
W$ has dimension at most $2n-k-1$. Further, it is enough to prove the result
only for components where the general line meets $X$ at $p$ with multiplicity
exactly $k$.

To that end, suppose $(\ell,p)$ is a general point of $W$, and that

$\iota:{\bf P}^{1}{\longrightarrow}{\bf P}^{n}$

is a parametrization of $\ell$ sending $[0:1]$ to $p$, as in the setting of
2.4. Notice that we may assume that $\ell$ meets $X$ at $p$ with multiplicity
exactly equal to $k$.

We invoke the auxiliary hypersurface $X_{k}$ constructed in the last section.
Since $\ell$ meets $X_{k}$ only at $p$, and since $p$ is a smooth point of
$X_{k}$, we deduce that $\iota^{*}T_{{\bf P}^{n}}(-\log X_{k})$ is a locally
free on ${\bf P}^{1}$, and its global sections parametrize first order
deformations of the map $\iota$ preserving order $k$ contact at $[0:1]$ by
2.4.

The condition

$\overline{\operatorname{Tot}}(W)={\bf P}^{n}$

implies that the map which sends a deformation $\iota_{\epsilon}$ preserving
order $k$ contact at $[0:1]$ to the deformation induced on the point
$\iota(y)\in{\bf P}^{n}$ for a general $y\in{\bf P}^{1}$ is surjective. This
means that the restriction map

$\displaystyle H^{0}({\bf P}^{1},\iota^{*}T_{{\bf P}^{n}}(-\log
X_{k})){\longrightarrow}\iota^{*}T_{{\bf P}^{n}}(-\log
X_{k})|_{y}=\iota^{*}T_{{\bf P}^{n}}|_{y}$ (2.8)

is surjective. We conclude that $\iota^{*}T_{{\bf P}^{n}}(-\log X_{k})$ is a
globally generated vector bundle. Since

$\deg\iota^{*}T_{{\bf P}^{n}}(-\log X_{k})=n+1-k,$

and

$\operatorname{rk}\iota^{*}T_{{\bf P}^{n}}(-\log X_{k})=n,$

it follows from global generation that

$\operatorname{dim}H^{0}({\bf P}^{1},\iota^{*}T_{{\bf P}^{n}}(-\log
X_{k}))=2n-k+1.$

Finally, we note that there is a $2$-dimensional space of deformations of the
map $\iota$ which induce a trivial deformation on the pair $(\ell,p)$ (There
is a $2$-dimensional group of automorphisms of ${\bf P}^{1}$ fixing a point).
Since every tangent vector of $W$ at the point $(\ell,p)$ can be lifted to a
first order deformation of $\iota$ preserving order $k$ contact at $[0:1]$, we
conclude that $\operatorname{dim}T_{W}|_{(\ell,p)}\leq(2n-k+1)-2$, as desired.
∎

We are now in position to provide the proof of Theorem 1.1:

###### Proof of Theorem 1.1.

Assume the setting in the statement of Theorem 1.1. Suppose the dimension of
$\Sigma_{k}$ exceeds $n-k$. Since $p\in X$ is a general point, this implies
the existence of an irreducible component

$W\subset V_{k}$

containing $\Sigma_{k}$ having dimension exceeding $(n-k)+(n-1)=2n-k-1$.

Theorem 2.8 now implies $\overline{\operatorname{Tot}}(W)\neq{\bf P}^{n}$.
Since $X=\pi(W)$ is a subvariety of $\overline{\operatorname{Tot}}(W)$, it
follows that $\overline{\operatorname{Tot}}(W)=X$, which is the theorem’s
conclusion. ∎

##  3\. Fifth order tangent lines

This section culminates in the proof of Theorem 3.8, which states that the
variety $V_{5}(X)$ has the expected dimension for any degree $d\geq n$ smooth
hypersurface $X\subset{\bf P}^{n}$. We note here that similar statements for
$V_{k}(X)$, $k\leq 4$ are also true – $k=1$ is trivial, $k=2$ follows because
$X$ is smooth, and $k=3$ follows from Theorem 3.1. Finally, the case $k=4$ can
be proved using a simpler version of the proof in Theorem 3.8, so we omit it.

Theorem 3.8 will then be used in § 4 to provide a bound on the number of
$m$-planes that a sufficiently large degree $2m$-dimensional hypersurface can
contain.

After recalling some classical facts about the second fundamental form and the
classification of varieties with many lines, we will prove a basic fact about
cones on scrolls, and then prove the main theorem of the section, Theorem 3.8.

###  3.1. Preliminaries on the second fundamental form

Let $x$ be a point on a smooth hypersurface $X\subset{\bf P}^{n}$ of degree
$d\geq 2$. Letting $\mathfrak{m}_{x}$ denote the maximal ideal of $x\in X$,
since $X$ is smooth, the vector space

$H^{0}(X,\mathfrak{m}_{x}^{2}\otimes\mathcal{O}_{X}(1))$

is $1$-dimensional, generated by an element $s_{x}$. Its residue in the
$\mathbb{K}$-vector space
$\left(\mathfrak{m}_{x}^{2}/\mathfrak{m}_{x}^{3}\right)(1)$ defines a
quadratic form (well-defined up to scaling, and possibly zero) denoted

$\operatorname{II}_{x}$

on the Zariski tangent space
$T_{x}X=\left(\mathfrak{m}_{x}/\mathfrak{m}_{x}^{2}\right)^{\vee}$. The
quadratic form $\operatorname{II}_{x}$ is called the second fundamental form
of $X$ at $x$.

We will need the following fundamental fact about the behavior of
$\operatorname{II}_{x}$:

######  Theorem 3.1.

([5] Theorem 7.11) For each $0\leq i\leq n-1$, the closed subset

$R_{i}\subset X$

consisting of points $x\in X$ where

$\operatorname{rk}\operatorname{II}_{x}\leq(n-1)-i$

has codimension $\geq i$.

###  3.2. Fourfolds with many lines

First, we record a well-known proposition about sweeping families of lines on
hypersurfaces. For any hypersurface $X$, we let $F_{1}(X)$ denote the Fano
scheme of lines contained in $X$. For any subscheme $A\subset F_{1}(X)$, we
let $\operatorname{Tot}(A){\longrightarrow}A$ denote the universal line over
$A$; $\operatorname{Tot}(A)$ is a ${\bf P}^{1}$-bundle over $A$. Furthermore,
we denote by $\operatorname{\overline{\operatorname{Tot}}}(A)$ the union of
all lines parametrized by $A$.

######  Proposition 3.2.

Suppose $X\subset{\bf P}^{n}$ is a smooth hypersurface, and suppose an
irreducible component $B\subset F_{1}(X)$ sweeps out $X$, i.e.
$\operatorname{\overline{\operatorname{Tot}}}(B)=X$. The normal bundle
$N_{\ell/X}$ is globally generated for a general line $\ell\in B$.

###### Proof.

Let $(\ell,p)\in\operatorname{Tot}(B)$ denote a general point. Then the
derivative of the natural projection
$\pi:\operatorname{Tot}(B){\longrightarrow}X$ is a map on Zariski tangent
spaces:

$\displaystyle\psi:T_{(\ell,p)}\operatorname{Tot}(B){\longrightarrow}T_{p}X.$
(3.1)

The sweeping hypothesis implies that (3.1) is surjective. Furthermore, (3.1)
sends the “vertical tangent subspace” space

$T_{p}\ell\subset T_{(\ell,p)}\operatorname{Tot}(B)$

isomorphically to $T_{p}\ell\subset T_{p}X$, and therefore we get an induced
surjection on quotient spaces:

$\displaystyle
T_{(\ell,p)}\operatorname{Tot}(B)/T_{p}\ell{\longrightarrow}N_{\ell/X}|_{p}.$
(3.2)

The domain vector space in (3.2) is canonically identified with the Zariski
tangent space $T_{[\ell]}B$, which by standard deformation theory is
canonically isomorphic to $H^{0}(\ell,N_{\ell/X})$. Under these
identifications, (3.2) becomes the restriction map

$\displaystyle H^{0}(\ell,N_{\ell/X}){\longrightarrow}N_{\ell/X}|_{p},$

and its surjectivity is precisely what we needed to prove. ∎

######  Corollary 3.3.

If $X\subset{\bf P}^{n}$ is a smooth hypersurface of degree $d\geq n$ then $X$
is not swept out by lines.

###### Proof.

The normal bundle of any line $\ell\subset X$ has degree $n-d-1<0$, and
therefore cannot be globally generated. The corollary now follows from 3.2. ∎

We will also need the following theorem of Segre classifying four dimensional
projective varieties containing a five dimensional family of lines:

######  Theorem 3.4 (Theorem 1 of [8]).

If $X\subset{\bf P}^{N}$ is a variety which contains a $(2\dim
X-3)$-dimensional family of lines, then $X$ is either a linear space, a
quadric, or a scroll.

###  3.3. Cones in scrolls

In this section, we record a basic result on cones contained in scrolls, which
we were unable to find a reference for.

######  Definition 3.5.

An $n$-dimensional variety $P\subset{\bf P}^{N}$ is a scroll if there exists a
smooth, irreducible, projective curve $B$, a ${\bf P}^{n-1}$-bundle
$\pi:\Lambda{\longrightarrow}B$ and a morphism
$\varphi:\Lambda{\longrightarrow}{\bf P}^{N}$ satisfying:

  1. (1)

$\varphi$ restricts to a linear embedding on each fiber of $\pi$, and

  2. (2)

$\varphi(\Lambda)=P$, and $\varphi$ is birational onto its image.

If $P$ is a scroll, and if $\pi:\Lambda{\longrightarrow}B$ is understood, we
call the linear spaces $\varphi(\Lambda_{b})\subset{\bf P}^{N}$ the rulings of
$P$. To ease notation, we will denote $\varphi(\Lambda_{b})$ by just
$\Lambda_{b}$ as long as no confusion arises.

######  Definition 3.6.

A cone in ${\bf P}^{N}$ is a pair $(C,c)$ where $C\subset{\bf P}^{N}$ is a
projective variety and $c\in C$ satisfying: For any $q\in C\setminus\\{c\\}$,
the line joining $q$ and $c$ is entirely contained in $C$. If $(C,c)$ is a
cone, we say a line $\ell\subset C$ is a cone line if $c\in\ell$.

Note that if $(C,c)$ is a cone, and if $H\subset{\bf P}^{N}$ is any hyperplane
containing $c$, then $(C\cap H,c)$ is again a cone. Furtheremore, if a cone
$(C,c)$ is irreducible and if $\dim C\geq 3$, then for a general hyperplane
$H$ containing $c$, the cone $(C\cap H,c)$ is again irreducible.

The next proposition constrains large cones in scrolls.

######  Proposition 3.7.

Suppose $P\subset{\bf P}^{N}$ is an $n$-dimensional scroll which is not a
linear space, with rulings $\Lambda_{b},b\in B$. Let $p\in P$, and suppose
$(C,p)$ is a $(n-1)$-dimensional irreducible cone contained in $P$. Assume
only finitely many of the rulings contain $p$.

Then $C$ is a linear space.

###### Proof.

Assume, for sake of contradiction, that $\deg C>1$. Then, as $\dim C=n-1$, it
follows that $C$ is not contained in any ruling $\Lambda_{b}$. Therefore, a
general cone line $\ell\subset C$ meets a general $\Lambda_{b}$ at a single
point. Since this is true for a general $\ell$, it follows that every $\ell$
meets a general $\Lambda_{b}$ – furthermore, this meeting is a single point,
because $p$ is only contained in finitely many rulings.

Now pick any hyperplane $H$ avoiding $p$, and consider projection from $p$

$\pi_{p}:P\setminus\\{p\\}{\longrightarrow}H.$

The image $W:=\pi_{p}(C\setminus\\{p\\})$ is the $n-2$-dimensional variety
$C\cap H$, which has degree $>1$ by assumption. Furthermore, the image
$\pi_{p}(\Lambda_{b})$ for $b\in B$ general, is a $(n-1)$-dimensional linear
space in $H$, and

$W\subset\pi_{p}(\Lambda_{b}).$

However, as $\deg W>1$, this implies that $\pi_{p}(\Lambda_{b})$ is
independent of $b$. Thus, the image of $\pi_{p}:P\setminus\\{p\\}$ is the
$(n-1)$-dimensional linear space $\pi_{p}(\Lambda_{b})$, forcing $P$ itself to
be a linear space, contrary to assumption. The argument by contradiction is
complete. ∎

We can now prove the main theorem of this section.

######  Theorem 3.8.

If $X\subset{\bf P}^{n}$ is a smooth hypersurface of degree $d\geq n\geq 4$,
then $V_{5}(X)$ has the expected dimension $2n-6$.

###### Proof.

We proceed by contradiction, and suppose here onward that

$W\subset V_{5}$

is an irreducible subvariety satisfying

$\dim W=2n-5.$

We let $\pi:W{\longrightarrow}X$ denote the map sending a pair $(\ell,x)$ to
the point $x\in X$. Observe that for every $x\in X$, the preimage

$\pi^{-1}(x)\subset W$

is naturally a subvariety of the $(n-2)$-dimensional projective space
parametrizing lines in ${\bf P}^{n}$ which are tangent to $X$ at $x$. Thus, if
we set

$Y:=\pi(W)\subset X,$

there are three possibilities:

  1. (1)

$Y=X$, or

  2. (2)

$Y\subset X$ has codimension $2$, or

  3. (3)

$Y\subset X$ has codimension $1$.

The proof proceeds by eliminating each of these possibilities.

######  Lemma 3.9.

$Y\neq X$.

###### Proof.

Direct application of Theorem 1.1. ∎

######  Lemma 3.10.

$Y\subset X$ cannot have codimension $2$.

###### Proof.

By contradiction. Assume $\operatorname{codim}Y=2$. In this case, for a
general (and hence every) point $y\in Y$, the fiber $\pi^{-1}(y)\subset W$ has
dimension $(n-2)$. Since, for every $(\ell,x)\in W$, the line $\ell$ meets $X$
at $x$ with contact order $>2$, it follows that the second fundamental form
$\operatorname{II}_{x}(X)$ vanishes identically (has rank $0$) for all points
$y\in Y$. This violates Theorem 3.1, providing the desired contradiction. ∎

Thus, we may and shall assume $Y\subset X$ has codimension $1$ from here
onward. First, we determine the dimension of
$\operatorname{\overline{\operatorname{Tot}}}(W)$. Theorem 2.8 implies that
$\operatorname{\overline{\operatorname{Tot}}}(W)\neq{\bf P}^{n}$. Secondly,
$\operatorname{\overline{\operatorname{Tot}}}(W)\neq Y$: Otherwise, $Y$ would
be a $(n-2)$-dimensional variety which contains at least a
$(2n-6)$-dimensional family of lines. Thus, $Y$ would be a linear space. As
$n\geq 4$, $X$ could not contain such a space.

We conclude that $\operatorname{\overline{\operatorname{Tot}}}(W)\subset{\bf
P}^{n}$ is a hypersurface. Since $d\geq n$, 3.3 implies that
$\operatorname{\overline{\operatorname{Tot}}}(W)\neq X$. Observe that, by
Segre’s classification theorem,
$\operatorname{\overline{\operatorname{Tot}}}(W)$ is either a linear space, a
quadric hypersurface, or a scroll.

Now pick a general point $y\in Y$. Let $(C,y)\subset{\bf P}^{n}$ denote the
$(n-2)$-dimensional cone swept out by the family of lines $\pi^{-1}(y)$.
Observe, by construction, that
$C\subset\operatorname{\overline{\operatorname{Tot}}}(W)$. Furthermore, if
$\mathbf{T}_{y}X$ denotes the projective tangent space (an $(n-1)$-dimensional
linear space) of $X$ at $y$, then $C\subset\mathbf{T}_{x}X$.

By Theorem 3.1, $\operatorname{II}_{y}$ defines an irreducible quadric, and
therefore for dimension reasons, $(C,y)$ must be the cone over this quadric.
This in turn implies that $C$ spans the hyperplane $\mathbf{T}_{y}X$.
Furthermore, this also eliminates the possibility that
$\operatorname{\overline{\operatorname{Tot}}}(W)$ is a scroll: the hypotheses
of 3.7 would be met, so $C$ could not be a quadric.

Thus, $\operatorname{\overline{\operatorname{Tot}}}(W)$ is either a linear
space or a quadric. Assume $\operatorname{\overline{\operatorname{Tot}}}(W)$
is a linear space. At a general point $y\in Y$, the quadric cone $(C,y)$ is
simultaneously contained in the linear space
$\operatorname{\overline{\operatorname{Tot}}}(W)$ and also spans
$\mathbf{T}_{y}X$. Thus all $\mathbf{T}_{y}X$ are equal for $y\in Y$. This
violates the finiteness of the Gauss map

$\Gamma:X{\longrightarrow}{\bf P}^{n*},$

which sends $x\in X$ to the hyperplane $\mathbf{T}_{x}X$.

Finally, assume $\operatorname{\overline{\operatorname{Tot}}}(W)$ is a quadric
hypersurface. Then, this quadric hypersurface cannot be singular at the
general point $y\in Y$, because
$\operatorname{\overline{\operatorname{Tot}}}(W)$ is irreducible. As the cone
$(C,y)$ is contained in $\operatorname{\overline{\operatorname{Tot}}}(W)$, it
follows that

$\mathbf{T}_{y}\operatorname{\overline{\operatorname{Tot}}}(W)=\mathbf{T}_{y}X$

for $y\in Y$ general. Now, pick a general complete curve $D\subset Y$. On the
one hand, the Gauss map $\Gamma$ induces a regular map of degree
$\deg(D)(d-1)$. On the other hand, this same map agrees with the (rational)
Gauss map for $\operatorname{\overline{\operatorname{Tot}}}(W)$, which is a
map defined by linear forms on ${\bf P}^{n}$. Thus, $\deg(D)(d-1)\leq\deg(D)$,
which is absurd given $d\geq n$.

Thus, $Y\subset X$ cannot be a divisor, and we have exhausted all
possibilities. The theorem follows.

∎

##  4\. Intersection Theory computation

######  Theorem 4.1.

For $d\geq 5$, the number of 2-planes in a smooth 4-fold is at most
$120d^{2}-150d^{3}+35d^{4}$.

###### Proof.

Let $S$ be the tautological subbundle of the Grassmannian $\mathbb{G}(1,5)$
and $Z_{5,1}\subset\mathbb{P}(S)\times_{\mathbb{G}(1,5)}\mathbb{P}(S)$ consist
of the locus $(\ell,p_{1},p_{2})$ where $\ell$ is a line in $\mathbb{P}^{5}$
meeting $X$ at points $p_{1}$ and $p_{2}$ with orders 5 and 1, respectively.

By Theorem 3.8, $Z_{5,1}$ is 4-dimensional. Let $H$ be the $\mathscr{O}(1)$
class of $\mathbb{P}(S)$, which is also the hyperplane class of
$\mathbb{P}^{5}$ pulled back under
$\mathbb{P}(S){\longrightarrow}\mathbb{P}^{5}$. Let $H_{1}$ and $H_{2}$ be $H$
pulled back to the two factors of
$\mathbb{P}(S)\times_{\mathbb{G}(1,5)}\mathbb{P}(S)$ respectively under the
projections
$\pi_{1},\pi_{2}:\mathbb{P}(S)\times_{\mathbb{G}(1,5)}\mathbb{P}(S){\longrightarrow}\mathbb{P}(S)$.

To compute $Z_{5,1}\subset\mathbb{P}(S)\times_{\mathbb{G}(1,5)}\mathbb{P}(S)$,
we note that it is the zero locus of a section of the sheaf of relative
principal parts
$\pi_{1}^{*}P_{\mathbb{P}(S)/\mathbb{G}(1,5)}^{4}(\mathscr{O}(d))\times\pi_{2}^{*}\mathscr{O}(d)$
as defined in [5, Section 11.1.1]. The relative canonical divisor of
$\mathbb{P}(S){\longrightarrow}\mathbb{G}(1,5)$ is $-2H+\sigma_{1}$.
Therefore, the class of $Z_{5,1}$ is

$\displaystyle
dH_{1}((d-2)H_{1}+\sigma_{1})((d-4)H_{1}+2\sigma_{1})((d-6)H_{1}+3\sigma_{1})((d-8)H_{1}+4\sigma_{1})dH_{2}$

inside $A^{\bullet}(\mathbb{P}(S)\times_{\mathbb{G}(1,5)}\mathbb{P}(S))$ by
[5, Theorem 11.2].

We will intersect $[Z_{5,1}]\in
A^{\bullet}(\mathbb{P}(S)\times_{\mathbb{G}(1,5)}\mathbb{P}(S))$ with the
pullback of $\mathbb{G}(1,4)\cong\sigma_{1,1}\subset\mathbb{G}(1,5)$ and the
product $H_{1}H_{2}$. This must be nonnegative on each component of $Z_{5,1}$
by Kleiman’s transversality [5, Theorem 1.7] and yields

$\displaystyle\int_{\mathbb{P}(S)\times_{\mathbb{G}(1,5)}\mathbb{P}(S)}\sigma_{1,1}H_{1}H_{2}(dH_{1}((d-2)H_{1}+\sigma_{1})\cdots((d-8)H_{1}+4\sigma_{1})dH_{2}=$
$\displaystyle\int_{(\mathbb{P}(S)\times_{\mathbb{G}(1,5)}\mathbb{P}(S))|_{\mathbb{G}(1,4)}}H_{1}H_{2}(dH_{1}((d-2)H_{1}+\sigma_{1})\cdots((d-8)H_{1}+4\sigma_{1})dH_{2}.$
(4.1)

In the final integral, the image of $H_{1}$ in
$A^{\bullet}((\mathbb{P}(S)\times_{\mathbb{G}(1,5)}\mathbb{P}(S))|_{\mathbb{G}(1,4)})$
can be identified as the image of $H_{1}$ under the map
$A^{\bullet}(\mathbb{P}^{5}){\longrightarrow}A^{\bullet}((\mathbb{P}(S)\times_{\mathbb{G}(1,5)}\mathbb{P}(S))|_{\mathbb{G}(1,4)})$
in the commutative diagram of pullback maps of Chow rings

${A^{\bullet}(\mathbb{P}^{5})}$${A^{\bullet}(\mathbb{P}(S)\times_{\mathbb{G}(1,5)}\mathbb{P}(S))}$${A^{\bullet}(\mathbb{P}^{4})}$${A^{\bullet}((\mathbb{P}(S)\times_{\mathbb{G}(1,5)}\mathbb{P}(S))|_{\mathbb{G}(1,4)}).}$

Since $H_{1}^{5}=0$ in $A^{\bullet}(\mathbb{P}^{4})$, the same is true in
$A^{\bullet}((\mathbb{P}(S)\times_{\mathbb{G}(1,5)}\mathbb{P}(S))|_{\mathbb{G}(1,4)})$.
Therefore, the numerical answer in (4.1) is $O(d^{4})$. To get the exact
numerical value for (4.1), we expand and evaluate with the help of Mathematica
and the presentation of the Chow ring of a projective bundle [5, Theorem 9.6]
to obtain

$\displaystyle 120d^{2}-150d^{3}+35d^{4}.$

To finish, we note that every 2-plane $\Lambda$ in $X$ gives rise to a
component of $Z_{5,1}$ whose class is the pullback of $\sigma_{2,2}$, the
lines in $\Lambda$. Since $\sigma_{2,2}\sigma_{1,1}$ is a point on
$\mathbb{G}(1,5)$, $\sigma_{2,2}\sigma_{1,1}H_{1}H_{2}$ evaluates to 1 in
$A^{\bullet}(\mathbb{P}(S)\times_{\mathbb{G}(1,5)}\mathbb{P}(S))$. ∎

######  Remark 4.2.

If one could show that the space $Z_{6}$ consisting of pairs $(p,\ell)$ where
$\ell$ meets $X$ to order 6 at $p$ is of expected dimension 3, then the
argument in the proof of Theorem 4.1 would imply the number of 2-planes in $X$
is at most

$\displaystyle 1800d-1370d^{2}+225d^{3},$

which one can compare with the $15d^{3}$ 2-planes in $X$ in the case $X$ is a
Fermat hypersurface.

## References

  * [1] Roya Beheshti.  Lines on projective hypersurfaces.  J. Reine Angew. Math., 592:1–21, 2006.
  * [2] T. D. Browning, D. R. Heath-Brown, and J. M. Starr.  The density of rational points on non-singular hypersurfaces, II.  Proceedings of the London Mathematical Society, 93(2):273–303, 2016\.
  * [3] Raymond Cheng.  Geometry of $q$-bic hypersurfaces.  arXiv:2205.05273v1.
  * [4] Alex Degtyarev, Ilia Itenberg, and John Christian Ottem.  Planes in cubic fourfolds, arxiv:2105.13951.
  * [5] David Eisenbud and Joe Harris.  3264 and all that—a second course in algebraic geometry.  Cambridge University Press, Cambridge, 2016.
  * [6] J. M. Landsberg.  Lines on projective varieties.  J. Reine Angew. Math., 562:1–3, 2003.
  * [7] Anand Patel, Eric Riedl, and Dennis Tseng.  Moduli of linear slices of high degree hypersurfaces.  arXiv preprint arXiv:2005.03689, 2020.
  * [8] Enrico Rogora.  Varieties with many lines.  Manuscripta Math., 82(2):207–226, 1994.
  * [9] George Salmon.  A treatise on the analytic geometry of three dimensions.  Chelsea Publishing Co., New York, 1958.  Revised by R. A. P. Rogers, 7th ed. Vol. 1, Edited by C. H. Rowe.
  * [10] B. Segre.  The maximum number of lines lying on a quartic surface.  Quart. J. Math. Oxford Ser., 14:86–96, 1943.
  * [11] Lloyd Smith.  The kontsevich space of rational curves on cyclic covers of projective space.  Ph.D. thesis.


# The Least Common Multiple of Polynomial Values over Function Fields

Alexei Entin  Raymond and Beverly Sackler School of Mathematical Sciences, Tel
Aviv University, Tel Aviv 69978, Israel<EMAIL_ADDRESS>and  Sean
Landsberg<EMAIL_ADDRESS>
###### Abstract.

Cilleruelo conjectured that for an irreducible polynomial $f\in\mathbb{Z}[X]$
of degree $d\geq 2$ one has

$\log\left(\mathrm{lcm}(f(1),f(2),\ldots f(N))\sim(d-1)\right)N\log N$

as $N\to\infty$. He proved it in the case $d=2$ but it remains open for every
polynomial with $d>2$.

We investigate the function field analogue of the problem by considering
polynomials over the ring ${\mathbb{F}}_{q}[T]$. We state an analog of
Cilleruelo’s conjecture in this setting: denoting by

$\deg L_{f}(n):=\mathrm{lcm}\left(f\left(Q\right)\ :\
Q\in{\mathbb{F}}_{q}[T]\mbox{ monic},\,\deg Q=n\right)$

we conjecture that

(1) $\deg L_{f}(n)\sim c_{f}\left(d-1\right)nq^{n},\ n\to\infty$

($c_{f}$ is an explicit constant dependent only on $f$, typically $c_{f}=1$).
We give both upper and lower bounds for $L_{f}(n)$ and show that the
asymptotic (1) holds for a class of “special” polynomials, initially
considered by Leumi in this context, which includes all quadratic polynomials
and many other examples as well. We fully classify these special polynomials.
We also show that $\deg L_{f}(n)\sim\deg\mathrm{rad}\left(L_{f}(n)\right)$ (in
other words the corresponding LCM is close to being squarefree), which is not
known over ${\mathbb{Z}}$

##  1\. Introduction

While studying the distribution of prime numbers, Chebychev estimated the
least common multiple of the first $N$ integers. This was an important step
towards the prime number theorem. In fact the condition $\textrm{log
lcm}\left(1,\ldots,N\right)\sim N$ is equivalent to the prime number theorem.
This problem later inspired a more general problem of studying the least
common multiple of polynomial sequences. For a linear polynomial $f(X)=kX+h,\
k,h\in{\mathbb{Z}},\ k>0,h+k>0$ it was observed by Bateman [bateman2002limit],
that $\textrm{log lcm}\left(f\left(1\right),\ldots,f\left(N\right)\right)\sim
c_{f}N$ as $N\to\infty$, where $c_{f}=\frac{k}{\varphi(k)}\sum_{1\leq m\leq
k,(m,k)=1}\frac{1}{m}$, which is a consequence of the Prime Number Theorem for
arithmetic progressions.

Cilleruelo conjectured [cilleruelo2011least] that for any irreducible
polynomial $f\in\mathbb{Z}[X]$ of degree $d\geq 2$ the following estimate
holds:

(1.1) $\textrm{log
lcm}\left(f\left(1\right),...,f\left(N\right)\right)\sim\left(d-1\right)N\log
N$

as $f$ is fixed and $N\to\infty$.

Convention. Throughout the rest of the paper in all asymptotic notation the
polynomial $f$, and the parameter $q$ to appear later, are assumed fixed,
while the parameter $N$ (or $n$) is taken to infinity. If any parameters other
than $f,q,N,n$ appear in the notation, the implied constant or rate of
convergence are uniform in these parameters.

Cilleruelo proved (1.1) for quadratic polynomials and there are no known
examples of polynomials of degree $d>2$ for which the conjecture is known to
hold. Cilleruelo’s argument also shows the predicted upper bound, i.e. if
$f\in\mathbb{Z}[X]$ is an irreducible polynomial of degree $d\geq 2$, then

$\log\mathrm{lcm}\,\left(f(1),\ldots,f(N)\right)\lesssim(d-1)N\log N.$

Maynard and Rudnick [maynard2019lower] provided a lower bound of the correct
order of magnitude:

(1.2) $\textrm{log lcm}\left(f(1),\ldots,f(N)\right)\gtrsim\frac{1}{d}N\log
N.$

Sah [sah2020improved] improved upon this lower bound while also providing a
lower bound for the radical of the least common multiple:

$\textrm{log lcm}\left(f\left(1\right),\ldots,f\left(N\right)\right)\gtrsim
N\log N,$

(1.3) $\textrm{log
rad}\left[\textrm{lcm}\left(f\left(1\right),...,f\left(N\right)\right)\right]\gtrsim\frac{2}{d}N\log
N.$

Rudnick and Zehavi [rudnick2020cilleruelo] established an averaged form of
(1.1) with $f$ varying in a suitable range. Leumi [leumi2021lcm] studied a
function field analogue of the problem. In the present work we expand upon and
generalize the results and conjectures in [leumi2021lcm], as well as correct
some erroneous statements and conjectures from the latter work. Despite some
overlap, we have kept our exposition self-contained and independent of
[leumi2021lcm].

###  1.1. The function field analogue.

Let $q=p^{k}$ be a prime power. For a polynomial $f\in\mathbb{F}_{q}[T][X]$ of
degree $d\geq 1$ of the form

$f(X)=f_{d}X^{d}+f_{d-1}X^{d-1}+...+f_{0},\ f_{i}\in{\mathbb{F}}_{q}[T]$

set

(1.4) $L_{f}(n):=\mathrm{lcm}\left(f\left(Q\right):\ Q\in M_{n}\right),$

where

$M_{n}:=\\{Q\in\mathbb{F}_{q}[T]\text{ monic},\ \deg Q=n\\}.$

Also denote

$V_{f}:=\left\\{g\in\mathbb{F}_{q}[T]:\
f\left(X+g\right)=f\left(X\right)\right\\}.$

The set $V_{f}$ is a finite-dimensional ${\mathbb{F}}_{p}$-linear subspace of
$\mathbb{F}_{q}[T]$ (see Lemma 2.1 below). Denote

$c_{f}:=\frac{1}{|V_{f}|}.$

We now state a function field analog of Cilleruelo’s conjecture.

######  Conjecture 1.1.

Let $f\in\mathbb{F}_{q}[T][X]$ be a fixed irreducible polynomial with
$\deg_{X}f=d\geq 2$. Then

$\deg L_{f}(n)\sim c_{f}\left(d-1\right)nq^{n},\ n\to\infty.$

######  Remark 1.2.

The expression $(d-1)nq^{n}$ is directly analogous to $(d-1)N\log N$ appearing
in (1.1). Over the integers (or generally in characteristic 0) if $f$ is not
constant then $V_{f}$ is always trivial and no constant $c_{f}$ appears in
(1.1). This is because if $0\neq g\in V_{f}$ then $2g,...,dg$ are also in
$V_{f}$ implying $f\left(0\right)=f\left(g\right)=...=f\left(dg\right)$. Since
$f\left(x\right)-f\left(0\right)$ has at most $d$ roots, we reach a
contradiction. Even over ${\mathbb{F}}_{q}[T]$, the typical case that occurs
for “most” polynomials is $c_{f}=1$ (i.e. $V_{f}$ is trivial). A heuristic
justification of 1.1 will be given in Section 4.

In the present paper we prove that Conjecture 1.1 gives the correct upper
bound:

######  Theorem 1.3.

$\deg L_{f}(n)\lesssim c_{f}(d-1)nq^{n}.$

The proof of 1.3 will be given in Section 3. We also give a lower bound of the
correct order of magnitude (this bound is comparable to the bound in
[sah2020improved, Theorem 1.3] under a mild assumption on $f$):

######  Theorem 1.4.

  1. (i)

$\deg L_{f}(n)\gtrsim\frac{d-1}{d}nq^{n}.$

  2. (ii)

If $p\nmid d$ or $f_{d}\nmid f_{d-1}$ then

$\deg L_{f}(n)\gtrsim nq^{n}.$

The proof of 1.4 will be given in Section 5. We note that by Lemma 2.1 below
we always have $c_{f}\geq 1/d$ and $c_{f}\geq 1/(d-1)$ if $p\nmid d$ or
$f_{d}\nmid f_{d-1}$, so this is consistent with Conjecture 1.1.

Regarding the radical of the LCM

$\ell_{f}(n):=\mathrm{rad}\ \mathrm{lcm}(f(Q):\ Q\in M_{n})$

we prove the following

######  Theorem 1.5.

$\deg\ell_{f}(n)\sim\deg L_{f}(n).$

The proof of 1.5 will be given in Section 6. As a consequence, our lower
bounds for $L_{f}(n)$ apply also to $\ell_{f}(n)$. The analogous statement
over ${\mathbb{Z}}$ is not known and the best lower bound over ${\mathbb{Z}}$,
namely (1.3), has a smaller constant (if $d>2$) than the best known lower
bound for $L_{f}(N)$ given by (1.2). The key ingredient in the proof which is
unavailable over ${\mathbb{Z}}$ is the work of Poonen [Poo03] on squarefree
values of polynomials over ${\mathbb{F}}_{q}[T]$ (later generalized by Lando
[Lan15] and Carmon [carmon2021square]).

For a class of polynomials $f\in{\mathbb{F}}_{q}[T][X]$ we call _special_
(first introduced by Leumi in [leumi2021lcm]) it is possible to establish
Conjecture 1.1 in full. This class includes all quadratic polynomials, but
also many polynomials of higher degree. We now define special polynomials over
an arbitrary unique factorization domain (UFD). This definition was introduced
in [leumi2021lcm].

######  Definition 1.6.

A polynomial $f\in R[X]$ of degree $d=\deg f\geq 2$ is called _special_ in if
the bivariate polynomial $f(X)-f(Y)$ factors into a product of linear terms in
$R[X,Y]$:

$f(X)-f(Y)=\prod_{i=1}^{d}(a_{i}X+b_{i}Y+c_{i}),\quad a_{i},b_{i},c_{i}\in R.$

######  Example 1.7.

  1. (i)

A quadratic polynomial is always special because

$AX^{2}+BX+C-(AY^{2}+BY+C)=(X-Y)(AX+AY+B).$

  2. (ii)

If $R={\mathbb{F}}_{p}$ then $f=X^{p}$ is special because

$X^{p}-Y^{p}=\prod_{a\in{\mathbb{F}}_{p}}(X-aY).$

For a special polynomial $f$ 1.1 can be established in full.

######  Theorem 1.8.

If $f\in{\mathbb{F}}_{q}[T][X]$ is irreducible and special then Conjecture 1.1
holds for $f$.

The proof of 1.8 will be given in section Section 7.

######  Example 1.9.

The polynomial $f=X^{p}-T\in{\mathbb{F}}_{p}[T][X]$ is irreducible (since it
is linear in $T$) and special (similarly to 1.7(ii)). Hence 1.1 holds for it.

We fully classify the set of special polynomials over an arbitrary UFD $R$.

######  Theorem 1.10.

Let $R$ be a UFD, $K$ its field of fractions and $p=\mathrm{char}(K)$.

  1. (i)

Assume $p=0$. Then $f\in R[X]$ is special iff it is of the form

$f(X)=f_{d}(X+A)^{d}+C,\quad 0\neq f_{d}\in R,\quad A,C\in K,$

where $d\geq 2$ is such that there exists a primitive $d$-th root of unity in
$K$.

  2. (ii)

Assume $p>0$. Then $f\in R[X]$ of degree $\deg f=d=p^{l}m\geq 2,(m,p)=1$ is
special iff it is of the form

$f(X)=f_{d}\prod_{i=1}^{p^{v}}(X-b_{i}+A)^{mp^{l-v}}+C,$ $0\neq f_{d}\in
R,\quad A,C\in K,\quad 0\leq v\leq l,\quad\zeta\in K,\quad
V=\\{b_{1},...,b_{p^{v}}\\}\subset K,$

where $\zeta$ is a primitive $m$-th root of unity and $V$ is an
$\mathbb{F}_{p}(\zeta)$-linear subspace of $K$ with $|V|=p^{v}$.

The proof of 1.10 will be given in Section 8.

We now briefly discuss how our main conjecture and results compare with the
work of Leumi [leumi2021lcm], which also studied the function field analog of
Cilleruelo’s conjecture and influenced the present work. First, [leumi2021lcm]
states Conjecture 1.1 without the constant $c_{f}$. This is certainly false by
Theorem 1.3 because it can happen that $c_{f}<1$ (see Example 2.2 below). It
seems to have been overlooked that when $g\in V_{f}$ and $Q\in M_{n},\,n>\deg
g$ then since $f(Q+g)=f(Q)$, the value $f(Q+g)$ contributes nothing new to the
LCM on the RHS of (1.4) over the contribution of $f(Q)$. Once one accounts for
this redundancy with the constant $c_{f}=1/|V_{f}|$, Cilleruelo’s original
heuristic carries over to the function field case giving rise to Conjecture
1.1. Second, all results and conjectures in [leumi2021lcm] are stated only for
an absolutely irreducible and separable $f\in{\mathbb{F}}_{q}[T][X]$. We do
not make these restrictions here and it takes additional technical work to
treat the general case. Third, the lower bound on $L_{f}(n)$ given in
[leumi2021lcm] has a smaller constant than ours (thus the bound is weaker),
comparable to the RHS of (1.3), and a lower bound for the radical of
$L_{f}(n)$ is not stated explicitly. Fourth, in [leumi2021lcm] our Theorem 1.8
is stated without the constant $c_{f}$, which is incorrect in general for the
same reasons explained above, although the arguments therein are essentially
correct in the case $c_{f}=1$. Finally, [leumi2021lcm] gives a classification
of special polynomials only in the case $p\nmid d$ (and it is stated only for
the ring ${\mathbb{F}}_{q}[T]$), whereas we treat the general case.

Acknowledgments. The authors would like to thank Zeév Rudnick for spotting a
few small errors in a previous draft of the paper. Both authors were partially
supported by Israel Science Foundation grant no. 2507/19.

##  2\. Preliminaries

For background on the arithmetic of function fields see [rosen2002number]. For
background on resultants, which we will use below, see [Lang2002, §IV.8].

###  2.1. Notation

We now introduce some notation which will be used throughout Sections 2-7. Let
$p$ be a prime, $q=p^{k}$. For $Q\in{\mathbb{F}}_{q}[T]$ we denote by
$|Q|=q^{\deg Q}$ the standard size of $Q$. For $P\in{\mathbb{F}}_{q}[T]$ prime
and $Q\in{\mathbb{F}}_{q}[T]$ we denote by $v_{P}(Q)$ the exponent of $P$ in
the prime factorization of $Q$.

We will always fix a polynomial

$f(X)=\sum_{i=0}^{d}f_{i}X^{i}\in{\mathbb{F}}_{q}[T][X],\,f_{d}\neq 0$

of degree $d$.

We also adopt the following conventions about notation.

  * •

For a polynomial $Q\in{\mathbb{F}}_{q}[T]$ we denote by $\deg Q$ its degree in
$T$. For a polynomial $g\in{\mathbb{F}}_{q}[T][X]$ we denote by $\deg
g=\deg_{X}g$ its degree in the variable $X$.

  * •

For a polynomial $g\in{\mathbb{F}}_{q}[T][X]$ we denote by
$g^{\prime}=\frac{\partial g}{\partial X}$ its derivative in the variable $X$.
The derivative in the variable $T$ will be written explicitly $\frac{\partial
g}{\partial T}$.

  * •

$g\in{\mathbb{F}}_{q}[T][X]$ is called separable if it is separable as a
polynomial in the variable $X$, equivalently
$g\not\in{\mathbb{F}}_{q}[T][X^{p}]$.

  * •

For two polynomials $g,h\in{\mathbb{F}}_{q}[T][X]$ we denote by
$\mathrm{Res}(g,h)=\mathrm{Res}_{X}(g,h)$ their resultant in the variable $X$.

###  2.2. The space $V_{f}$

Recall that $V_{f}=\left\\{g\in\mathbb{F}_{q}[T]:\
f\left(X+g\right)=f\left(X\right)\right\\}$.

######  Lemma 2.1.

Assume $d\geq 1$.

  1. (i)

$|V_{f}|\leq d$.

  2. (ii)

$V_{f}$ an ${\mathbb{F}}_{p}$-linear subspace of ${\mathbb{F}}_{q}[T]$

  3. (iii)

$|V_{f}|$ is a power of p.

  4. (iv)

If $p\nmid d$ then $V_{f}$ is trivial.

  5. (v)

If $f_{d}\nmid f_{i}$ for some $1\leq i\leq d-1$ then $|V_{f}|\leq d-1$.

###### Proof.

  1. (i)

Assume by way of contradiction that $|V_{f}|\geq d+1$. Then $f(g)=f(0)$ for
every $g\in V_{f}$. Since $f(x)-f(0)$ has at most $d$ roots,
$f\left(x\right)=f\left(0\right)$ and $f$ is constant, a contradiction.

  2. (ii)

It is obvious that $0\in V_{f}$. Now let $a,b\in V_{f}$ and let us prove
$\alpha a+\beta b\in V_{f}$ where $\alpha,\beta\in\\{0,1,...,p-1\\}$.
Recursively applying $f(X+a)=f(X+b)=f(X)$ we obtain

$f\left(X+\alpha a+\beta
b\right)=f\left(X+\sum_{i=1}^{\alpha}a+\sum_{i=1}^{\beta}b\right)=f\left(X\right).$

  3. (iii)

Obvious from (i) and (ii).

  4. (iv)

Let $g\in V_{f}$, so $f(X+g)=f(X)$. Comparing coefficients at $X^{d-1}$ we
find $dg=0$. If $p\nmid d$ then $g=0$, so in this case $V_{f}=\\{0\\}$ and the
claim follows.

  5. (v)

If $|V_{f}|=d$ then since all elements $g\in V_{f}$ are roots of $f(X)-f(0)$,
we have $f=f_{d}\prod_{g\in V_{f}}(X-g)+f(0)$ and $f_{d}|f_{i}$ for all $1\leq
i\leq d-1$, contradicting the assumption.

∎

######  Example 2.2.

(Example of a polynomial with $|V_{f}|=d=\deg f$) Let
$V=\\{b_{1},...,b_{p^{v}}\\}\subset{\mathbb{F}}_{q}[T]$ be an
${\mathbb{F}}_{p}$-linear subspace and $C\in{\mathbb{F}}_{q}[T]$. Then the
polynomial

$f(X)=\prod_{i=1}^{p^{v}}(X-b_{i})+C$

has $V_{f}=V$. Thus $|V_{f}|=|V|=p^{v}=d$.

###  2.3. Roots of $f$ modulo prime powers

In this subsection we study the quantity

$\rho_{f}(P^{k}):=\left|\\{Q\bmod P^{k}:f(Q)\equiv 0\bmod P^{k}\\}\right|,$

i.e. the number of roots of $f$ modulo a prime power $P^{k}$.

######  Lemma 2.3.

Let $g,h\in\mathbb{F}_{q}[T][X]$ be polynomials and let
$P\in{\mathbb{F}}_{q}[T]$ prime. If $g,h$ have a common root modulo $P^{m}$
then $P^{m}\mid R:=\mathrm{Res}(g,h)$.

###### Proof.

We can express $R$ as $R=a(X)f(X)+b(X)g(X)$ for some
$a(X),b(X)\in\mathbb{F}_{q}[T][X]$ (see [Lang2002, §IV.8]). Therefore, if
there exists a $Q\in\mathbb{F}_{q}[T]$ such that $P^{m}\mid f(Q)$ and
$P^{m}\mid g(Q)$, then $P^{m}$ must also divide $a(Q)f(Q)+b(Q)g(Q)=R$. ∎

The proof of the next two lemmas is similar to the analogous proof for the
integer case in [TrygveNagel1921, proof of Theorem II].

######  Lemma 2.4.

Assume $f$ is separable and denote $R:=\mathrm{Res}_{X}(f,f^{\prime})\neq 0$.
Denote $\mu=v_{P}(R)$. Let $x_{0},x_{1}\in{\mathbb{F}}_{q}[T]$ be such that
the following conditions hold:

  1. (a)

$f(x_{0})\equiv 0\pmod{P^{\mu+1}}$.

  2. (b)

$f(x_{1})\equiv 0\pmod{P^{\beta+1}}$, where $\beta:=v_{P}(f^{\prime}(x_{0}))$.

  3. (c)

$x_{1}\equiv x_{0}\mod P^{\mu+1}$.

Then $\beta\leq\mu$ and $v_{P}(f^{\prime}(x_{1}))=\beta$.

###### Proof.

Since $f(x_{0})\equiv 0\mod P^{\mu+1}$ and $P^{\mu+1}\nmid R$, by Lemma 2.3 we
must have $P^{\mu+1}\nmid f^{\prime}(x_{0})$. Hence $\beta\leq\mu$. Now
writing $x_{1}=x_{0}+tP^{\mu+1},\,t\in{\mathbb{F}}_{q}[T]$ and using
$\beta\leq\mu$ and conditions (a),(b) we have

$f^{\prime}(x_{1})=f^{\prime}(x_{0}+tP^{\mu+1})\equiv f^{\prime}(x_{0})\equiv
0\pmod{P^{\beta}},$ $f^{\prime}(x_{1})=f^{\prime}(x_{0}+tP^{\mu+1})\equiv
f^{\prime}(x_{0})\not\equiv 0\pmod{P^{\beta+1}},$

so $v_{P}(f^{\prime}(x_{1}))=\beta$. ∎

######  Lemma 2.5.

In the setup of Lemma 2.4 let $\alpha>\mu$ be an integer and assume that in
fact $f(x_{1})\equiv 0\pmod{P^{\alpha+\beta}}$. Consider the set

$S_{1}:=\left\\{x_{1}+uP^{\alpha}\mid
u\in\mathbb{F}_{q}[T]/P^{\beta}\right\\}\subset{\mathbb{F}}_{q}[T]/P^{\alpha+\beta}.$

Then

  1. (i)

The elements of $S_{1}$ are roots of $f$ modulo $P^{\alpha+\beta}$.

  2. (ii)

The number of roots of $f$ modulo $P^{\alpha+\beta+1}$ that reduce modulo
$P^{\alpha+\beta}$ to an element of $S_{1}$ is equal to $|S_{1}|=q^{\beta\deg
P}$.

###### Proof.

To prove (i) we note that

$f(x_{1}+uP^{\alpha})\equiv
f(x_{1})+uP^{\alpha}f^{\prime}(x_{1})\pmod{P^{2\alpha}}.$

Thus $P^{\alpha+\beta}\mid f(x_{1}+uP^{\alpha})$ as $P^{\alpha+\beta}\mid
f(x_{1})$ and by Lemma 2.4 we have $P^{\beta}\mid f^{\prime}(x_{1})$ and
$\alpha+\beta\leq\alpha+\mu<2\alpha$.

To show (ii), consider the set of possible lifts from $S_{1}$ to
${\mathbb{F}}_{q}[T]/P^{\alpha+\beta+1}$, i.e.

$S_{2}:=\left\\{x_{1}+uP^{\alpha}+vP^{\alpha+\beta}\mid
u\in\mathbb{F}_{q}[T]/P^{\beta},v\in\mathbb{F}_{q}[T]/P\right\\}.$

We will now determine for which $u,v$ the element
$x_{1}+uP^{\alpha}+vP^{\alpha+\beta}$ is a root of $f$ modulo
$P^{\alpha+\beta+1}$. Using $2\alpha>\alpha+\beta$ we have

$f(x_{1}+uP^{\alpha}+vP^{\alpha+\beta})\equiv
f(x_{1})+uP^{\alpha}f^{\prime}(x_{1})+vP^{\alpha+\beta}f^{\prime}(x_{1})\pmod{P^{\alpha+\beta+1}}.$

Hence $x_{1}+uP^{\alpha}+vP^{\alpha+\beta}$ is a root of $f\bmod
P^{\alpha+\beta+1}$ iff

(2.1)
$f(x_{1})+uP^{\alpha}f^{\prime}(x_{1})+vP^{\alpha+\beta}f^{\prime}(x_{1})\equiv
0\pmod{P^{\alpha+\beta+1}}.$

As $P^{\alpha+\beta}\mid f(x_{1})$ and $v_{P}(f^{\prime}(x_{1}))=\beta$ (Lemma
2.4), we have (2.1) iff

$u\equiv-\left(\frac{f^{\prime}(x_{1})}{P^{\beta}}\right)^{-1}\left(\frac{f(x_{1})}{P^{\alpha+\beta}}+vf^{\prime}(x_{1})\right)\pmod{P}$

Thus we have $|P|$ possible values of $v$ and for each of these
$|P^{\beta}|/|P|$ possible values of $u$. Overall we have
$|P^{\beta}|=q^{\beta\deg P}=|S_{1}|$ possible values of $(u,v)$ and the
assertion follows. ∎

######  Lemma 2.6.

Assume that $f$ is separable, $R=\mathrm{Res}(f,f^{\prime})\neq 0$. Let $P$ be
a prime in $\mathbb{F}_{q}[T]$ and $\mu=v_{P}(R)$. Then
$\rho_{f}(P^{2\mu+k})=\rho_{f}(P^{2\mu+1})$ for all $k\geq 1$.

###### Proof.

For each root $x_{1}$ of $f$ modulo $P^{2\mu+k}$ we can apply Lemma 2.5 with
$\alpha=2\mu+k-\beta$, where $\beta=v_{P}(f^{\prime}(x_{1}))$ (the condition
$\alpha>\mu$ holds because $\beta\leq\mu$ by Lemma 2.4 applied with
$x_{0}=x_{1}$) and obtain that the number of roots of $f$ modulo
$P^{2\mu+k+1}$ equals the number of roots of $f$ modulo $P^{2\mu+k}$, i.e.
$\rho_{f}(P^{2\mu+k+1})=\rho_{f}(P^{2\mu+k})$. The assertion now follows by
induction on $k$. ∎

######  Lemma 2.7.

Assume $f\in{\mathbb{F}}_{q}[T][X^{p}]$ is inseparable and irreducible. Then
$U:=\mathrm{Res}(f,\frac{\partial f}{\partial T})\neq 0$.

###### Proof.

Assume by way of contradiction that $U=0$. Then $f,\frac{\partial f}{\partial
T}$ have a common factor and since $f$ is irreducible we have
$f\mid\frac{\partial f}{\partial T}$. Comparing degrees in $T$ we must have
$\frac{\partial f}{\partial T}=0$. This means that
$f\in{\mathbb{F}}_{q}[T^{p},X^{p}]$ is a $p$-th power, contradicting its
irreducibility. ∎

######  Lemma 2.8.

Assume that $f$ is inseparable and irreducible, and $P^{m}\nmid
U:=\mathrm{Res}(f,\frac{\partial f}{\partial T})\in\mathbb{F}_{q}[T]$ for some
$P\in{\mathbb{F}}_{q}[T]$ prime and $m\geq 1$. Then $\rho_{f}(P^{k})=0$ for
every $k\geq m+1$.

###### Proof.

Assuming the existence of $Q\in\mathbb{F}_{q}[T]$ such that $P^{m}\mid f(Q)$
(if $f$ has no roots modulo $P^{m}$ we are done), we will now prove that
$P^{m+1}\nmid f(Q)$. Since $P^{m}\mid f(Q)$, we know that
$P^{m}\nmid\frac{\partial f}{\partial T}(Q)$; otherwise, by Lemma 2.3 we would
have $P^{m}|U$, contradicting our assumption.

Since $f$ is inseparable and irreducible we have $f^{\prime}=0$ and therefore

$\frac{\partial f(Q)}{\partial T}=\frac{\partial f}{\partial
T}(Q)+\frac{\partial f}{\partial X}(Q)\frac{dQ}{dT}=\frac{\partial f}{\partial
T}(Q).$

If $P^{m+1}\mid f(Q)$, write $f(Q)=P^{m+1}C$ and then

$\begin{split}\frac{\partial f(T,Q(T))}{\partial
T}&=\frac{\partial(P^{m+1}C)}{\partial T}=P^{m+1}\frac{\partial C}{\partial
T}+(m+1)P^{m}\frac{\partial P}{\partial T}C.\\\ \end{split}$

Thus $P^{m}\mid\frac{\partial f(T,Q(T))}{\partial T}=\frac{\partial
f}{\partial T}(Q)$, contradicting the above observation
$P^{m}\nmid\frac{\partial f}{\partial T}(Q)$. ∎

######  Proposition 2.9.

Assume that $f$ is irreducible. Then $\rho_{f}(P^{m})\ll 1$ for all $P$ prime
and $m\geq 1$.

###### Proof.

Since $\mathbb{F}_{q}[T]/P$ is a field we have $\rho_{f}(P)\leq d$. Thus it
remains to handle the case $m\geq 2$.

If $f$ is inseparable then using Lemmas 2.8 and 2.7 we see that there are only
finitely many pairs $P,m$ such that $m\geq 2$ and $\rho_{f}(P^{m})\neq 0$ (as
they must satisfy $P^{m}\,|\,\mathrm{Res}(f,\frac{\partial f}{\partial T})\neq
0$).

Now if $f$ is separable then by Lemma 2.6 there are only finitely many pairs
of $(P,m)$ such that $m\geq 2$ and $\rho_{f}(P^{m})\neq\rho_{f}(P^{m-1})$,
Denote these pairs by $(P_{i},m_{i})_{i=1}^{s}$. Then for all primes $P$ and
$m\geq 1$

$\rho_{f}(P^{m})\leq\max\\{d,\rho_{f}(P_{i}^{m_{i}})_{i=1}^{s}\\}\ll 1.$

∎

######  Lemma 2.10.

Let $f\in{\mathbb{F}}_{q}[T][X^{p}]$ be inseparable and irreducible. Then
there exist a separable and irreducible $h\in\mathbb{F}_{q}[T][X]$ such that
$\rho_{f}(P)=\rho_{h}(P)$ for all primes $P\in\mathbb{F}_{q}[T]$.

###### Proof.

Write $f(X)=h(X^{p^{m}})$ for some $m\geq 1$ such that
$h\in{\mathbb{F}}_{q}[T][X]\setminus{\mathbb{F}}_{q}[T][X^{p}]$ is separable.
For any prime $P$, he $m$-fold Frobenius isomorphism of
${\mathbb{F}}_{q}[T]/P$ given by $x\mapsto x^{p^{m}}$ gives a one-to-one
correspondence between the roots of $f$ modulo $P$ and the roots of $h$ modulo
$P$. Hence $\rho_{f}(P)=\rho_{h}(P)$. ∎

##  3\. The upper bound

Throughout this section the polynomial $f$ is assumed _irreducible_. To obtain
the upper and lower bounds on $L_{f}(n)$ (recall that this quantity was
defined in (1.4)) we set

$P_{f}(n):=\prod_{Q\in M_{n}}f(Q)$

and write

(3.1) $\deg L_{f}(n)=c_{f}\deg P_{f}(n)-(c_{f}\deg P_{f}(n)-\deg L_{f}(n)),$

where $c_{f}=1/|V_{f}|$.

######  Lemma 3.1.

$\deg P_{f}(n)=dnq^{n}+O(q^{n})$.

###### Proof.

For sufficiently large $n$ and $Q\in M_{n}$ we have $\deg f(Q)=dn+\deg f_{d}$,
hence

$\deg P_{f}(n)=\deg\prod_{Q\in M_{n}}f(Q)=\sum_{Q\in M_{n}}\deg
f(Q)=\sum_{Q\in M_{n}}(dn+\deg f_{d})=dnq^{n}+O(q^{n}).$

∎

Convention. Throughout the rest of the paper $P$ will always denote a prime of
${\mathbb{F}}_{q}[T]$ and $\sum_{P}$ (resp. $\prod_{P}$) will denote a sum
(resp. product) over all monic primes of ${\mathbb{F}}_{q}[T]$. A sum of the
form $\sum_{a\leq\deg P\leq b}$ is over all monic primes in the corresponding
degree range (and the same for products).

To estimate $c_{f}\deg P_{f}(n)-\deg L_{f}(n)$ we write the prime
decomposition of $L_{f}(n)$ and $P_{f}(n)$ as

$L_{f}(n)=\prod_{P}P^{\beta_{P}(n)},\quad
P_{f}(n)=\prod_{P}P^{\alpha_{P}(n)},$

where the products are over all (monic) primes in ${\mathbb{F}}_{q}[T]$.

Convention. Throughout Sections 3-7 we will always assume that $n$ is large
enough so that $f(Q)\neq 0$ for all $Q\in M_{n}$. Thus $L_{f}(n),P_{f}(n)\neq
0$ and $\alpha_{P}(n),\beta_{P}(n)$ are always finite.

We have

(3.2) $\alpha_{P}(n)=\sum_{Q\in M_{n}}v_{P}(f(Q)),\
\beta_{P}(n)=\max\\{v_{P}(f(Q)):Q\in M_{n}\\},$

(recall that $v_{P}(Q)$ is the exponent of $P$ in the prime factorization of
$Q$). Combining (3.1) with 3.1 we have

$\deg L_{f}(n)=c_{f}dnq^{n}-\sum_{P}(c_{f}\alpha_{P}(n)-\beta_{P}(n))\deg
P+O(q^{n}).$

######  Lemma 3.2.

For sufficiently large n

$\beta_{P}(n)\leq c_{f}\alpha_{P}(n).$

###### Proof.

Denote by $Q_{max}\in M_{n}$ an element such that
$\beta_{P}(n)=v_{P}(f(Q_{max}))$. Then for each element $g\in V_{f}$ we have
$f(Q_{max}+g)=f(Q_{max})$ and hence
$\beta_{P}(n)=v_{P}(f(Q_{max}))=v_{P}(f(Q_{max}+g)).$ For $n$ sufficiently
large $Q_{max}+g\in M_{n}$, so

$|V_{f}|\beta_{P}(n)=\sum_{g\in V_{f}}v_{P}(f(Q_{max}+g))\leq\sum_{Q\in
M_{n}}v_{P}(f(Q))=\alpha_{P}(n).$

The assertion follows since $c_{f}=1/|V_{f}|$. ∎

The next lemma is the main tool for estimating $\alpha_{P}(n),\beta_{P}(n)$.
For its proof we introduce the following notation, which will be used in the
sequel as well:

(3.3) $s_{f}(P^{k},n)=\left|\\{Q\in M_{n}:f(Q)\equiv 0\bmod P^{k}\\}\right|,$

where $P$ is prime and $k\geq 1$.

######  Lemma 3.3.

Let $P\in\mathbb{F}_{q}[T]$ be a prime.

  1. (i)

$\beta_{P}(n)=O\left(\frac{n}{\deg P}\right)$.

  2. (ii)

If $f$ is separable and $P\nmid\mathrm{Res}(f,f^{\prime})$, then

$\alpha_{P}(n)=q^{n}\frac{\rho_{f}(P)}{|P|-1}+O\left(\frac{n}{\deg P}\right).$

  3. (iii)

If $f$ is inseparable and $P\nmid\mathrm{Res}(f,\frac{\partial f}{\partial
T})$, then

$\alpha_{P}(n)=q^{n}\frac{\rho_{f}(P)}{|P|}+O\left(\frac{n}{\deg P}\right).$

  4. (iv)

For the finitely many “bad” primes where neither (ii) nor (iii) hold we have

$\alpha_{P}(n)=O(q^{n}).$

###### Proof.

To prove (i) we note that since there exists $Q\in M_{n}$ such that
$P^{\beta_{P}(n)}\mid f(Q)$ and $f(Q)\neq 0$ we get:

$\beta_{P}(n)\deg P\leq\deg f(Q).$

For sufficiently large $n$ we have $\deg f(Q)=dn+\deg f_{d}$, thus

$\beta_{P}(n)\ll\frac{n}{\deg P},$

establishing (i).

Using the notation (3.3) we have

$\alpha_{P}(n)=\sum_{f(Q)\in M_{n}}v_{P}(Q)=\sum_{Q\in
M_{n}}\sum_{\begin{subarray}{c}P,\,k\geq 1\atop{P^{k}\mid
f(Q)}\end{subarray}}1=\sum_{k\geq 1{\deg P}}\sum_{\begin{subarray}{c}Q\in
M_{n}\atop{P^{k}\mid f(Q)}\end{subarray}}1=\sum_{k\geq 1}s_{f}(P^{k},n).$

Note that

$s_{f}(P^{k},n)=q^{n}\frac{\rho_{f}(P^{k})}{|P^{k}|}+O(1),$

since if $Q$ is a root of $f$ modulo $P^{k}$ and $\deg P^{k}\leq n$ then there
are exactly $\frac{q^{n}}{|P|^{k}}$ element in $M_{n}$ that reduce to $Q$
modulo $P^{k}$. And if $\deg P^{k}>n$ then
$s_{f}(P^{k},n)\leq\rho_{f}(P^{k})\ll 1$ by Proposition 2.9. Hence

$\alpha_{P}(n)=\sum_{k\geq 1}s_{f}(P^{k},n)=\sum_{k\geq
1}\left[q^{n}\frac{\rho_{f}(P^{k})}{|P|^{k}}+O(1)\right]=q^{n}\sum_{k\geq
1}\frac{\rho_{f}(P^{k})}{|P|^{k}}+O\left(\frac{n}{\deg P}\right).$

Let us look at the different cases:

  * •

If $f$ is separable and $P\nmid\mathrm{Res}(f,f^{\prime})$ then we have by 2.6
that $\rho_{f}(P^{k})=\rho_{f}(P)$, thus

$\begin{split}\alpha_{P}(n)&=q^{n}\sum_{k\geq
1}\frac{\rho_{f}(P)}{|P|^{k}}+O\left(\frac{n}{\deg P}\right)\\\
&=q^{n}\rho_{f}(P)\sum_{k\geq
1}\left(\frac{1}{|P|}\right)^{k}+O\left(\frac{n}{\deg P}\right)\\\
&=q^{n}\frac{\rho_{f}(P)}{|P|-1}+O\left(\frac{n}{\deg P}\right).\end{split}$

  * •

If $f$ is inseparable and $P\nmid\mathrm{Res}_{X}(f,\frac{\partial f}{\partial
T})$ then using Lemma 2.8 we get

$\alpha_{P}(n)=q^{n}\frac{\rho_{f}(P)}{|P|}+O\left(\frac{n}{\deg P}\right).$

  * •

For a general irreducible $f$, using the fact that $\text{for every }k\geq 1,\
\rho_{f}(P^{k})\ll 1$ (2.9) we get

$\begin{split}\alpha_{P}(n)=q^{n}\sum_{k\geq
1}\frac{\rho_{f}(P^{k})}{|P|^{k}}+O\left(\frac{n}{\deg P}\right)\ll
q^{n}\sum_{k\geq 1}\frac{1}{|P|^{k}}+O\left(\frac{n}{\deg P}\right)\ll
O(q^{n}).\end{split}$

∎

######  Lemma 3.4.

$\deg\left(\prod_{n<\deg P\leq n+\deg
f_{d}}P^{\alpha_{P}(n)}\right)=O(q^{n}).$

###### Proof.

By Lemma 3.3 for sufficiently large $n$ and $P$ a prime such that $\deg P>n$,
we have $\alpha_{P}(n)\leq q^{n}\frac{\rho_{f}(P)}{|P|-1}+O\left(\frac{n}{\deg
P}\right)\ll\rho_{f}(P)+O(1)\ll 1$. Using the Prime Polynomial Theorem, we
have

$\begin{split}\deg\left(\prod_{n<\deg P\leq n+\deg
f_{d}}P^{\alpha_{P}(n)}\right)&=\sum_{\deg n<\deg P\leq n+\deg
f_{d}}\alpha_{P}(n)\deg P\\\ &\ll\sum_{\deg n<\deg P\leq n+\deg f_{d}}\deg
P=\sum_{k=n+1}^{n+\deg f_{d}}\sum_{\deg P=k}k\\\ &\ll\sum_{k=n+1}^{n+\deg
f_{d}}k\frac{q^{k}}{k}=\frac{q^{n+\deg f_{d}+1}-q^{n+1}}{q-1}\\\ &\ll
q^{n}.\end{split}$

∎

######  Proposition 3.5.

Denote $R_{f}(n)=\prod_{\deg P\leq n+\deg f_{d}}P^{\alpha_{P}(n)}$. Then

$\deg R_{f}(n)=nq^{n}+O(q^{n}).$

###### Proof.

Let us assume first that $f$ is separable. We will handle the inseparable case
at the end of the proof. We note that we may ignore the $O(1)$ “bad” primes
(as defined in Lemma 3.3) with an error term of $O(q^{n})$ and by 3.4 we can
ignore the primes with $n<\deg P\leq n+\deg f_{d}$ with the same error term.
Thus by 3.3 we obtain

(3.4) $\begin{split}\deg R_{f}(n)=\sum_{\deg P\leq n}\alpha_{P}(n)\deg
P+O(q^{n})=\sum_{\deg P\leq n}q^{n}\frac{\rho_{f}(P)}{|P|-1}\deg P+\sum_{\deg
P\leq n}O(n)+O(q^{n}).\end{split}$

We bound the error term using the Prime Polynomial Theorem:

(3.5) $\begin{split}\sum_{\deg P\leq n}n=n\sum_{k=1}^{n}\sum_{\deg P=k}1\ll
n\sum_{k=1}^{n}\frac{q^{k}}{k}\ll q^{n}.\end{split}$

Now to estimate $\sum_{\deg P\leq n}q^{n}\frac{\rho_{f}(P)}{|P|-1}\deg P$ we
will use the Chebotarev Density Theorem in function fields [Fried2023,
Proposition 7.4.8]. First we introduce some notation and recall some
terminology. Let $E/{\mathbb{F}}$ be the splitting field of $f$ and denote by
$G$ the Galois group of $f$. For each prime $P$ of $\mathbb{F}_{q}[T]$
unramified in $E/{\mathbb{F}}_{q}(T)$ the Frobenius class of $P$ is defined to
be:

$\begin{split}\left(\frac{E/{\mathbb{F}}_{q}(T)}{P}\right)=\left\\{\sigma\in
G\,:\,\exists\ \mathfrak{P}/P\text{ prime of }E\text{ such that }x^{|P|}\equiv
x\pmod{\mathfrak{P}}\text{ for all }x\text{ with }v_{\mathfrak{P}}(x)\geq
0\right\\}.\end{split}$

The Frobenius class $\left(\frac{E/{\mathbb{F}}_{q}(T)}{P}\right)$ is a
conjugacy class in $G$. Denote by $S$ the set of all conjugacy classes in $G$.
Given a conjugacy class $C\in S$, we set

$\pi_{C}(n)=\left|\left\\{P\ \mathrm{prime}:\deg
P=n,\left(\frac{E/{\mathbb{F}}_{q}(T)}{P}\right)=C\right\\}\right|.$

Let $K=\mathbb{F}_{q^{v}}$ be the algebraic closure of $\mathbb{F}_{q}$ in
$E$. We have $G_{0}:=\mathrm{Gal}(K/\mathbb{F}_{q})=\langle\phi\rangle$, where
$\phi(x)=x^{q}$ is the $q$-Frobenius. Denote the restriction of automorphisms
map by $\varphi:G\to G_{0}$. Since $G_{0}=\langle\phi\rangle$ is cyclic and in
particular abelian, we have for all $C\in S$ that
$\varphi(C)=\\{\phi^{n_{C}}\\}$ for some $n_{C}\in{\mathbb{Z}}$. Define

$S_{k}:=\\{C\in S\mid\varphi(C)=\\{\phi^{k}\\}\\}.$

Now the Chebotarev Density Theorem in function fields says that if $C\in
S_{k}$ then

$\pi_{C}(k)=\left\\{\begin{array}[]{ll}v\frac{|C|}{|G|}\frac{q^{k}}{k}+O\left(\frac{q^{k/2}}{k}\right),&C\in
S_{k},\\\ 0,&C\notin S_{k}.\end{array}\right.$

Since the Galois group acts on the roots of $f$, we can define
$\mathrm{Fix}(C)$ to be the number of roots fixed by any element in the
conjugacy class $C$ (this number is the same for all $\sigma\in C$). Assuming
$P$ is unramified in $E/{\mathbb{F}}_{q}(T)$, we have
$\rho_{f}(P)=\mathrm{Fix}\left(\left(\frac{E/{\mathbb{F}}_{q}(T)}{P}\right)\right)$.
In the calculations throughout the rest of the proof summation over $P$
denotes summation over primes $P\nmid\mathrm{Res}_{X}(f,f^{\prime})$ if $f$ is
separable and $P\nmid\mathrm{Res}(f,\frac{\partial f}{\partial T})$ otherwise.
In the case when $f$ is separable this conditions insures that $P$ is
unramified in $E/{\mathbb{F}}_{q}(T)$. As we observed above the excluded
primes contribute $O(q^{n})$.

We have:

$\sum_{\deg P=k}\frac{\rho_{f}(P)}{q^{k}-1}k=\sum_{\deg
P=k}\frac{\mathrm{Fix}\left(\left(\frac{E/{\mathbb{F}}_{q}(T)}{P}\right)\right)}{q^{k}-1}k=\sum_{C\in
S}\frac{\mathrm{Fix}(C)\pi_{C}(k)}{q^{k}-1}k=\sum_{C\in
S_{k}}\left[\frac{\mathrm{Fix}(C)q^{k}}{(q^{k}-1)k}\frac{v|C|}{|G|}k+O(q^{-k/2})\right]\\\
=\sum_{C\in S_{k}}v\frac{\mathrm{Fix}(C)|C|}{|G|}+O(q^{-k/2})$

and therefore

(3.6) $\sum_{\deg P\leq
n}q^{n}\frac{\rho_{f}(P)}{q^{k}-1}k=q^{n}\sum_{k=1}^{n}\sum_{\deg
P=k}\frac{\rho_{f}(P)}{q^{k}-1}k=q^{n}\sum_{k=1}^{n}\left[\sum_{C\in
S_{k}}v\frac{\mathrm{Fix}(C)|C|}{|G|}+O(q^{-k/2})\right]\\\
=q^{n}\sum_{k=1}^{n}\sum_{C\in
S_{k}}v\frac{\mathrm{Fix}(C)|C|}{|G|}+O(q^{n})=q^{n}\sum_{l=1}^{\lfloor\frac{n}{v}\rfloor}v\sum_{C\in
S}\frac{\mathrm{Fix}(C)|C|}{|G|}+O(q^{n}).$

Using Burnside’s lemma [rotman2012introduction, Theorem 3.22] and the
transitivity of $G$ (which is a consequence of $f$ being irreducible) we get

$1=\sum_{\sigma\in G}\frac{\mathrm{Fix}(\sigma)}{|G|}=\sum_{C\in
S}\frac{\mathrm{Fix}(C)|C|}{|G|}.$

Pluggin this into (3.6) and recalling (3.4),(3.5) we obtain

$\deg R_{f}(n)=\sum_{\deg P\leq n}q^{n}\frac{\rho_{f}(P)}{|P|-1}\deg
P+O(q^{n})=q^{n}\sum_{l=1}^{\lfloor\frac{n}{v}\rfloor}v+O(q^{n})=nq^{n}+O(q^{n}).$

This completes the proof in the separable case. To handle the case when $f$ is
inseparable we again do the same calculations (using Lemma 3.3(iii) this
time):

$\begin{split}\deg R_{f}(n)&=\sum_{\deg P\leq n}\alpha_{P}(n)\deg
P+O\left(q^{n}\right)\\\ &=\sum_{\deg P\leq n}q^{n}\frac{\rho_{f}(P)}{|P|}\deg
P+\sum_{\deg P\leq n}O(n)+O(q^{n}).\end{split}$

We can replace $\rho_{f}(P)$ with $\rho_{h}(P)$, where $h$ is the polynomial
given by Lemma 2.10. Since $h$ is separable, the argument above (with $f$
replaced by $h$) yields:

$\deg R_{f}(n)=\sum_{\deg P\leq
n}q^{n}\frac{\rho_{h}(P)}{|P|}+O(q^{n})=nq^{n}+O(q^{n}),$

completing the proof. ∎

We are now ready to prove the upper bound on $\deg L_{f}(n)$.

###### Proof of Theorem 1.3.

Using Lemma 3.2 we have $c_{f}\alpha_{P}(n)\leq\beta_{P}(n)$, for all $P$
prime. We have

$\displaystyle\deg L_{f}(n)$ $\displaystyle=c_{f}\deg P_{f}(n)-(c_{f}\deg
P_{f}(n)-\deg L_{f}(n))=$
$\displaystyle=c_{f}dnq^{n}-\sum_{P}(c_{f}\alpha_{P}(n)-\beta_{P}(n))\deg P$
(3.7) $\displaystyle\leq c_{f}dnq^{n}-\sum_{\deg P\leq n+\deg
f_{d}}(c_{f}\alpha_{P}(n)-\beta_{P}(n))\deg P$
$\displaystyle=c_{f}dnq^{n}-c_{f}\sum_{\deg P\leq n+\deg
f_{d}}\alpha_{P}(n)\deg P+\sum_{\deg P\leq n+\deg f_{d}}\beta_{P}(n)\deg P$
(3.8) $\displaystyle=c_{f}(d-1)nq^{n}+\sum_{\deg P\leq n+\deg
f_{d}}\beta_{P}(n)\deg P+O(q^{n})$

where (3.7) comes from removing negative terms and (3.8) from Proposition 3.5
and Lemma 3.1.

It remains to prove that $\sum_{\deg P\leq n+\deg f_{d}}\beta_{P}(n)\deg
P=O(q^{n})$. From Lemma 3.3(i) and the Prime Polynomial Theorem we get

(3.9) $\sum_{\deg P\leq n+\deg f_{d}}\beta_{P}(n)\deg P\ll\sum_{\deg P\leq
n+\deg f_{d}}\frac{n}{\deg P}\deg P=\sum_{\deg P\leq n+\deg f_{d}}n\ll q^{n}.$

∎

##  4\. Heuristic justification for 1.1

Throughout the present section we maintain the assumption that
$f=\sum_{i=0}^{d}f_{i}X^{i}\in{\mathbb{F}}_{q}[T][X]$ is irreducible. Next we
want to study the difference $\deg L_{f}(n)-c_{f}(d-1)nq^{n}$ and we will do
so by relating it to a slightly different quantity defined by (4.1) below. To
this end let us define an equivalence relation on $M_{n}$ by

$\ Q_{1}\sim Q_{2}\iff f(X+Q_{1}-Q_{2})=f(X)\iff Q_{1}-Q_{2}\in V_{f}.$

We note that for sufficiently large $n$ the size of each equivalence class is
$|V_{f}|$ and the number of such classes is $c_{f}q^{n}$. Consider

(4.1) $S_{f}(n)=\left|\left\\{P\in\mathbb{F}_{q}[T]\mbox{
prime}:\begin{aligned} \deg P>&n+\deg f_{d}.\\\ \exists\,Q_{1}\nsim Q_{2}\in
M_{n}&\text{ such that }P\mid f(Q_{1}),f(Q_{2})\end{aligned}\right\\}\right|.$

Note that the condition in the bottom line on the RHS of (4.1) is equivalent
to $\beta_{P}(n)\neq c_{f}\alpha_{P}(n)$ (the negation of this condition is
that $P$ occurs precisely in all $f(Q)$ where $Q$ runs over a single
equivalence class of $\sim$, equivalently
$|V_{f}|\beta_{P}(n)=\alpha_{P}(n)$), hence

(4.2) $S_{f}(n)=\sum_{\deg P>n+\deg
f_{d}\atop{c_{f}\alpha_{P}(n)\neq\beta_{P}(n)}}1.$

######  Proposition 4.1.

Let $f\in\mathbb{F}_{q}[T][X]$ be irreducible with $\deg_{X}f=d\geq 2$.

  1. (i)

$S_{f}(n)\ll q^{n}.$

  2. (ii)

1.1 is equivalent to $S_{f}(n)=o(q^{n}).$

###### Proof.

The first part follows immediately from the definition of $S_{f}(n)$, since
there are $q^{n}$ possible values of $f(Q)$ and each has $O(1)$ prime factors
of degree $\deg P>n+\deg f_{d}$. It remains to prove the second part.

In the main calculation in the proof of Theorem 1.3 there is only one
inequality, namely

$c_{f}dnq^{n}-\sum_{P}(c_{f}\alpha_{P}(n)-\beta_{P}(n))\deg P\leq
c_{f}dnq^{n}-\sum_{\deg P\leq n+\deg
f_{d}}(c_{f}\alpha_{P}(n)-\beta_{P}(n))\deg P.$

equivalently

$-\sum_{\deg>n+\deg f_{d}}(c_{f}\alpha_{P}(n)-\beta_{P}(n))\deg P\leq 0.$

Since the RHS in Conjecture 1.1 is $\gg nq^{n}$, the conjecture is now seen to
be equivalent to

$\sum_{\deg>n+\deg f_{d}}(c_{f}\alpha_{P}(n)-\beta_{P}(n))=o(nq^{n}).$

Hence is suffices to prove that

(4.3) $S_{f}(n)=o(q^{n})\iff\sum_{\deg P>n+\deg
f_{d}}(c_{f}\alpha_{P}(n)-\beta_{P}(n))\deg P=o(nq^{n}).$

For $P\in\mathbb{F}_{q}[T]$ prime set

$\delta_{P}(n)=\begin{cases}1,&c_{f}\alpha_{P}(n)\neq\beta_{P}(n),\\\
0,&\text{otherwise}.\end{cases}$

Note that for $n$ sufficiently large if $\deg P>dn+\deg f_{d}$ then
$c_{f}\alpha_{P}(n)=\beta_{P}(n)=0$. Hence, for $n$ sufficiently large using
(4.2) we have

$S_{f}(n)=\sum_{n+\deg f_{d}<\deg P\leq dn+\deg f_{d}}\delta_{P}(n)$

From 3.3(ii-iii) and 2.9, for sufficiently large $n$ and $P$ not one of $O(1)$
bad primes, if $\deg P>n$ then $\alpha_{P}(n)\ll 1$, so

(4.4) $\sum_{n+\deg f_{d}<\deg P}(c_{f}\alpha_{P}(n)-\beta_{P}(n))\deg
P\ll\sum_{n+\deg f_{d}<\deg P\leq dn+\deg f_{d}}\delta_{P}(n)\deg P\\\
\ll\sum_{n+\deg f_{d}<\deg P\leq dn+\deg f_{d}}\delta_{P}(n)n=S_{f}(n)n.$

On the other hand, once again using the fact that
$c_{f}\alpha_{P}(n)=\beta_{P}(n)$ if $\deg P>dn+\deg f_{d}$,

(4.5) $\sum_{n+\deg f_{d}<\deg P}(c_{f}\alpha_{P}(n)-\beta_{P}(n))\deg P\\\
\geq\sum_{n+\deg f_{d}<\deg P\leq dn+\deg f_{d}}\delta_{P}(n)\deg
P\geq\sum_{n+\deg f_{d}<\deg P\leq dn+\deg f_{d}}\delta_{P}(n)n=S_{f}(n)n,$

From (4.4) and (4.5) the equivalence (4.3) follows, which completes the proof.
∎

Heuristically one can estimate $S_{f}(n)$ by arguing that the “probability”
that for $Q_{1}\not\sim Q_{2}\in M_{n}$ the values $f(Q_{1}),f(Q_{2})$ share a
prime factor of degree $\deg P>n+\deg f_{d}$ is (using the Prime Polynomial
Theorem)

$\sum_{m=n+\deg f_{d}+1}^{\infty}\sum_{\deg
P=m}\frac{1}{|P|^{2}}=\sum_{m=n+\deg
f_{d}+1}^{\infty}\frac{1}{mq^{m}}=O\left(\frac{1}{nq^{n}}\right).$

Since there are $O(q^{2n})$ pairs $Q_{1},Q_{2}$ we expect

$S_{f}(n)\ll\frac{1}{nq^{n}}\cdot q^{2n}\ll\frac{q^{n}}{n}=o(q^{n}).$

By 4.1 this is equivalent to 1.1.

At this point one may ask whether the equivalence relation $\sim$ is really
the dominant source of pairs $Q_{1},Q_{2}\in M_{n}$ with $f(Q_{1})=f(Q_{2})$,
otherwise we would definitely not have $S_{f}(n)=o(q^{n})$. Over the integers
it cannot happen that $f(n_{1})=f(n_{2})$ for $n_{1}\neq n_{2}\in{\mathbb{Z}}$
sufficiently large (since a polynomial is a monotone function at sufficiently
large arguments), but in finite characteristic this is less obvious.
Nevertheless this is guaranteed by the following proposition, which will be
needed for the proof of 1.8 as well.

######  Proposition 4.2.

$\biggl{|}\biggl{\\{}(Q_{1},Q_{2})\,:\,Q_{1}\not\sim Q_{2}\in
M_{n},\,f(Q_{1})=f(Q_{2})\biggr{\\}}\biggr{|}=O(q^{n/2}).$

Before we prove 4.2 we need a couple of auxiliary results.

######  Lemma 4.3.

Let $g(X,Y)\in{\mathbb{F}}_{q}[T][X,Y]$ be a fixed irreducible polynomial with
$\deg_{X,Y}g\geq 2$. The number of pairs $Q_{1},Q_{2}\in M_{n}$ such that
$g(Q_{1},Q_{2})=0$ is $O(q^{n/2})$.

###### Proof.

If $\deg_{X}g=\deg_{Y}g=1$ then $g=AXY+BX+CY+D,\,A\neq
0,B,C,D\in{\mathbb{F}}_{q}[T]$ and one cannot have $g(Q_{1},Q_{2})=0$ for
$Q_{1},Q_{2}\in M_{n},n>\max(\deg B,\deg C,\deg D)$ since the term
$AQ_{1}Q_{2}$ has higher degree than the other terms. Hence assume WLOG that
$\deg_{Y}g\geq 2$, otherwise switch $X$ and $Y$. It follows from the
quantitative Hilbert Irreducibility Theorem in function fields [BaEn21,
Theorem 1.1] that for all but $O(q^{n/2})$ polynomials $Q_{1}\in M_{n}$, the
polynomial $g(Q_{1},Y)\in{\mathbb{F}}_{q}[T][Y]$ is irreducible of degree
$\geq 2$ and therefore has no root $Q_{2}\in M_{n}$. For the remaining
$O(q^{n/2})$ values of $Q_{1}$ there are at most $\deg_{Y}g=O(1)$ roots for
each $Q_{1}$, so overall there are $O(q^{n/2})$ pairs $Q_{1},Q_{2}\in M_{n}$
with $g(Q_{1},Q_{2})=0$. ∎

######  Lemma 4.4.

Assume that the polynomial $f(X)-f(Y)\in{\mathbb{F}}_{q}[T][X,Y]$ is divisible
by $aX+bY+c$ for some $a,b,c\in{\mathbb{F}}_{q}[T]$ with $b\neq 0$. Then

  1. (i)

$\zeta:=-a/b$ satisfies $\zeta^{d}=1$ and $\zeta\in{\mathbb{F}}_{q}$.

  2. (ii)

If $\zeta=1$ then $c/b\in V_{f}$.

###### Proof.

The divisibility assumption implies
$f(X)=f\left(-\frac{a}{b}X-\frac{c}{b}\right)$. Comparing coefficients at
$X^{d}$ gives $\zeta^{d}=1$. Since $\zeta\in{\mathbb{F}}_{q}(T)$ and
${\mathbb{F}}_{q}$ is algebraically closed in ${\mathbb{F}}_{q}(T)$ we have
$\zeta\in{\mathbb{F}}_{q}$, establishing (i). If $\zeta=-1$ the above identity
becomes $f(X)=f\left(X-\frac{c}{b}\right)$, i.e. $-c/b\in V_{f}$. Since
$V_{f}$ is a vector space we obtain (ii). ∎

We are ready to prove 4.2.

###### Proof of 4.2.

Write $f(X)-f(Y)=c\prod_{i=1}^{m}g_{i}(X,Y)$ with $c\in{\mathbb{F}}_{q}[T]$
and each $g_{i}\in{\mathbb{F}}_{q}[T][X,Y]$ irreducible with
$\deg_{X,Y}g_{i}\geq 1$. It is enough to show that for each factor
$g_{i}(X,Y)$ there are $\ll q^{n/2}$ pairs $Q_{1}\not\sim Q_{2}\in M_{n}$ with
$g_{i}(Q_{1},Q_{2})=0$. For a given $i$, if we have $\deg_{X,Y}g_{i}\geq 2$
this follows from 4.3. If $\deg_{X,Y}g_{i}=1$, write $g_{i}=aX+bY+c$, assume
WLOG that $b\neq 0$ and denote $\zeta=-a/b$. By 4.4 we have $\zeta^{d}=1$.

We distinguish two cases: if $\zeta\neq 1$ then $a\neq-b$ and the leading
coefficient of $g_{i}(Q_{1},Q_{2})$ is $a+b\neq 0$ for any $Q_{1},Q_{2}\in
M_{n}$ with $n>\deg c$, and therefore there are no such pairs with
$g_{i}(Q_{1},Q_{2})=0$. If on the other hand $\zeta=1$ (i.e. $a=-b$) then by
4.4 we have $c/b\in V_{f}$ and $g_{i}(Q_{1},Q_{2})=aQ_{1}+bQ_{2}+c=0$ is
equivalent to $Q_{1}+c/b=Q_{2}$, so $Q_{1}\sim Q_{2}$ and there are no pairs
with $g_{i}(Q_{1},Q_{2})=0$, $Q_{1}\not\sim Q_{2}\in M_{n}$. This completes
the proof. ∎

######  Remark 4.5.

It is readily seen from the proof that if $f$ is special in the sense of 1.6
then we may put 0 on the RHS of 4.2 for $n$ sufficiently large, since in this
case $\deg_{X,Y}g_{i}=1$ for all $i$.

##  5\. The lower bound

######  Lemma 5.1.

Let $P\in\mathbb{F}_{q}[T]$ be a prime such that

  1. (a)

$\deg P>\deg f_{d}+n$.

  2. (b)

$P\nmid\mathrm{Res}(f,f^{\prime})$ if $f$ is separable.

  3. (c)

$P\nmid\mathrm{Res}(f,\frac{\partial f}{\partial T})$ if $f$ is inseparable.

Denote

$B_{i}=B_{i}(P):=\left|\\{Q\in M_{n}:P^{i}\mid f(Q)\neq 0\\}\right|.$

Then

  1. (i)

$B_{1}\leq d$.

  2. (ii)

$B_{i}\leq B_{1},\,i\geq 1$.

  3. (iii)

$B_{i}=0$ if $i>d$.

###### Proof.

Observe that since $n+\deg f_{d}<\deg P^{i}$, $M_{n}$ is mapped injectivly
into $\mathbb{F}_{q}[T]/P^{i}$ by $Q\mapsto f(Q)\bmod P^{i}$.

(i). By the observation above $B_{1}\leq\rho_{f}(P)\leq\deg(f\bmod P)=d$, so
$B_{1}\leq d$.

(ii). If $f$ is separable then 2.4 combined with assumption (b) applied
iteratively with $\alpha=1,2,\ldots,i-1$ implies that each root of $f$ modulo
$P$ has a unique lift to a root modulo $P^{i}$. Combined with the observation
above this implies $B_{i}\leq B_{1}$. If $f$ is inseparable then by 2.8 and
assumption (c) we have $B_{i}\leq\rho_{f}(P^{i})=0\leq B_{1}$ for $i>1$.

(iii). Finally suppose that $i>d$. If there exists $Q\in\mathbb{F}_{q}[T]$
such that $P^{i}\mid f(Q)\neq 0$, then $\deg f(Q)\leq dn+\deg f_{d}$, while
$dn+\deg f_{d}<i\deg P=\deg P^{i}$, a contradiction. Hence $B_{i}=0$ in this
case. ∎

We are ready to prove the first part of 1.4.

###### Proof of 1.4(i).

From 5.1 we get that for all but $O(1)$ primes $P\in\mathbb{F}_{q}[T]$ (note
that by 2.7 conditions (b-c) of 5.1 are satisfied for all but $O(1)$ primes)
such that $\deg P>n+\deg f_{d}$ we have

$\alpha_{P}(n)=\sum_{i\geq 1}\sum_{Q\in M_{n}\atop{P^{i}|f(Q)}}1=\sum_{i\geq
1\atop{B_{i}>0}}B_{i}\leq d\sum_{i\geq 1\atop{B_{i}>0}}1\leq d\beta_{P}(n).$

Now using 3.1 and 3.5 we obtain

$\begin{split}(d-1)nq^{n}\lesssim\deg\frac{P_{f}(n)}{R_{f}(n)}=\sum_{\deg
P>n+\deg f_{d}}\alpha_{P}(n)\deg P\leq d\sum_{\deg P>n+\deg
f_{d}}\beta_{P}(n)\deg P\leq d\deg L_{f}(n),\end{split}$

hence $\deg L_{f}(n)\gtrsim\frac{d-1}{d}nq^{n}$ as required. ∎

######  Remark 5.2.

There exist polynomials $f$ for which the lower bound given by in 1.4(i) and
the upper bound given by 1.3 are equal. Indeed, if $|V_{f}|=d$ (as in 2.2),
then combining 1.3 and 1.4(ii) gives $\deg L_{f}(n)\sim\frac{d-1}{d}nq^{n}$ as
$n\to\infty$. Hence the lower bound is tight in this case.

To prove the second part of 1.4 we first need the following

######  Lemma 5.3.

Assume $p\nmid d$ or $f_{d}\nmid f_{d-1}$. Let $n$ be sufficiently large and
$P\in\mathbb{F}_{q}[T]$ a prime such that $\deg P>\deg f_{d}+n$. Then in the
notation of 5.1 we have $B_{1}(P)\leq d-1$.

###### Proof.

Assume by way of contradiction that there exist distinct $Q_{1},...,Q_{d}$
such that $P\mid f(Q_{1}),...,f(Q_{d})$. The assumption $\deg P>\deg f_{d}+n$
implies that $Q_{i}\bmod P$ are also distinct and therefore

$f(X)\equiv f_{d}\prod_{j=1}^{d}(x-Q_{j})\pmod{P}$

Comparing coefficients at $X^{d-1}$ we obtain

$f_{d-1}\equiv-f_{d}\sum_{j=1}^{d}Q_{j}\pmod{P}$

or

$f_{d-1}+f_{d}\sum_{j=1}^{d}Q_{j}\equiv 0\pmod{P}.$

Thus $P\mid f_{d-1}+f_{d}\sum_{j=1}^{d}Q_{j}$ and if
$f_{d-1}+f_{d}\sum_{j=1}^{d}Q_{j}\neq 0$ we would have $\deg P\leq n+\deg
f_{d}$ (if $n\geq f_{d-1}$), a contradiction.

Now we observe that as $Q_{1},...,Q_{d}$ are monic, if
$\deg\sum_{j=1}^{d}Q_{j}<n$ then $p\mid d$. Therefore for a sufficiently large
$n$ if $f_{d-1}+f_{d}\sum_{j=1}^{d}Q_{j}=0$ then $p\mid d$ and additionally
$f_{d}\sum_{j=1}^{d}Q_{j}=-f_{d-1}$, hence $f_{d}\mid f_{d-1}$. This
contradicts our initial assumption and we have reached a contradiction in all
cases. ∎

We are ready to prove the second part of 1.4.

###### Proof of 1.4(ii).

From 5.3 we get that for $P\in\mathbb{F}_{q}[T]$ prime such that $\deg
P>n+\deg f_{d}$ we have $B_{1}(P)\leq d-1$. The rest of the proof is now the
same as the proof of 1.4(i), with the inequality $\alpha_{P}(n)\leq
d\beta_{P}(n)$ replaced by $\alpha_{P}(n)\leq(d-1)\beta_{P}(n)$, which has the
effect of replacing the constant $\frac{d-1}{d}$ by 1 in the conclusion. ∎

##  6\. The radical of $L_{f}(n)$

We assume for simplicity of exposition that $f$ is irreducible, although this
assumption can be dispensed with in Theorem 1.5 and its proof, which we give
in the present section. Recall that we denote
$\ell_{f}(n)=\mathrm{rad}\,L_{f}(n)$ and observe that

$\deg\ell_{f}(n)=\sum_{P\atop{\beta_{P}(n)>0}}\deg P.$

By 1.3 and 1.4(i) we have $nq^{n}\ll\deg L_{f}(n)\ll nq^{n}$ and the
asymptotic $\deg\ell_{f}(n)\sim\deg L_{f}(n)$ (1.5) is equivalent to the
following

######  Proposition 6.1.

$\deg L_{f}(n)-\deg\ell_{f}(n)=o(nq^{n}).$

###### Proof.

Let us write

$\ell_{f}(n)=\sum_{\deg P>n\atop{\beta_{P}(n)>0}}\deg P+\sum_{\deg P\leq
n\atop{\beta_{P}(n)>0}}\deg P,$ $L_{f}(n)=\sum_{\deg P>n}\beta_{P}(n)\cdot\deg
P+\sum_{\deg P\leq n}\beta_{P}(n)\cdot\deg P.$

Now using (3.9) we obtain

$\sum_{\deg P\leq n\atop{\beta_{P}(n)>0}}\deg P\leq\sum_{P\leq n+\deg
f_{d}}\beta_{P}(n)\cdot\deg P=O(q^{n})$

and therefore

$\ell_{f}(n)=\sum_{\deg P>n\atop{\beta_{P}(n)>0}}\deg P+O(q^{n}),$
$L_{f}(n)=\sum_{\deg P>n\atop{\beta_{P}(n)>0}}\beta_{P}(n)\deg P+O(q^{n}).$

Denote

$S:=\\{Q\in M_{n}:\exists P\text{ prime, }\deg P>n,\,P^{2}\mid f(Q)\\}.$

We have

$\begin{split}0\leq\deg L_{f}(n)-\deg\ell_{f}(n)&=\sum_{\deg
P>n}\beta_{P}(n)\deg P-\sum_{\deg P>n\atop{\beta_{P}(n)>0}}\deg P+O(q^{n})\\\
&\leq\sum_{\deg P>n\atop{\beta_{P}(n)\geq 2}}\beta_{P}(n)\deg P+O(q^{n})\\\
&\leq\sum_{Q\in S}\deg f(Q)+O(q^{n})\\\ &\ll|S|n+q^{n}.\end{split}$

It remains to prove that $|S|=o(q^{n})$ but by [Poo03, Lemma 7.1], the set

$W=\\{\deg Q\leq n:\exists P\text{ prime, }\deg P>n/2,\,P^{2}\mid f(Q)\\}$

is of size $o(q^{n})$ (this is the key ingredient of the proof; see also
[Lan15, Proposition 3.3] and [carmon2021square, (4.7)] for refinements of this
statement). Since $S\subset W$ we have $|S|=o(q^{n})$ and we are done. ∎

As noted above, this concludes the proof of 1.5.

##  7\. Asymptotics of $L_{f}(n)$ for special polynomials

In the present section we assume that
$f=\sum_{i=0}^{d}f_{i}X^{i}\in{\mathbb{F}}_{q}[T][X],\,\deg f=d\geq 2$ is
irreducible and special in the sense of 1.6. We will show that $\deg
L_{f}(n)\sim c_{f}nq^{n}$ as $n\to\infty$, thus proving 1.8. As in Section 4
we consider the equivalence relation on $M_{n}$ given by $Q_{1}\sim Q_{2}\iff
Q_{1}-Q_{2}\in V_{f}$ and the quantity $S_{f}(n)$ defined by (4.1).

By the definition of a special polynomial we may write

(7.1) $f(X)-f(Y)=\prod_{i=1}^{d}(a_{i}X+b_{i}Y+c_{i}),\quad
a_{i},b_{i},c_{i}\in{\mathbb{F}}_{q}[T].$

Comparing degrees in $X$ and $Y$ shows that $a_{i},b_{i}\neq 0$. Comparing
coefficients at $X^{d}$ and $Y^{d}$ in (7.1) we see that $\deg a_{i},\deg
b_{i}\leq\deg f_{d}$. Therefore if $Q_{1},Q_{2}\in M_{n}$ with $n>\deg c_{i}$,
we have

(7.2) $\deg(a_{i}Q_{1}+b_{i}Q_{2}+c_{i})\leq n+\deg f_{d}.$

Let $Q_{1},Q_{2}\in M_{n}$ with $Q_{1}\not\sim Q_{2}$ and $P$ a prime with
$\deg P>n+\deg f_{d}$ such that $P\,|\,f(Q_{1}),f(Q_{2})$. Then in particular
$P\,|\,f(Q_{1})-f(Q_{2})$ and by (7.1) we have
$P\,|\,a_{i}Q_{1}+b_{i}Q_{2}+c_{i}\neq 0$ for some $1\leq i\leq d$. By (7.2)
and the condition $\deg P>n+\deg f_{d}$ we must have
$a_{i}Q_{1}+b_{i}Q_{2}+c_{i}=0$ and therefore by (7.1) we have
$f(Q_{1})=f(Q_{2})$. By 4.2 combined with 4.5 this is impossible for $n$
sufficiently large. Hence we have $S_{f}(n)=0$ for $n$ sufficiently large and
by 4.1 it follows that 1.1 holds for $f$, concluding the proof of 1.8.

##  8\. Classification of special polynomials

In the present section we classify the special polynomials (1.6) over an
arbitrary unique factorization domain (UFD) $R$, establishing 1.10. We denote
by $p$ the characteristic of $R$ (as $R$ is a domain, $p$ is zero or a prime)
and by $K$ its field of fractions. Also if $p>0$ we will denote by
$\mathbb{F}_{p}$ its prime subfield. Note that as $R$ is a UFD, so are the
polynomial rings $R[X],R[X,Y]$. For a polynomial $g\in R[X,Y]$ we denote by
$\deg g$ its total degree. Special polynomials $f\in R[X]$ were defined in
1.6.

######  Lemma 8.1.

Let $f\in R[X]$ be a special polynomial of degree $d$. Then we may write

(8.1) $f(X)-f(Y)=\prod_{i=1}^{d}(a_{i}X+b_{i}Y+c_{i}),\quad a_{i}\neq
0,b_{i}\neq 0,c_{i}\in R.$

###### Proof.

This is immediate from the definition, except for the condition
$a_{i},b_{i}\neq 0$ which follows by comparing degrees in $X$ and $Y$ in
(8.1). ∎

######  Lemma 8.2.

Let $f=\sum_{i=0}^{d}f_{i}X^{i}\in R[X]$ be a special polynomial of degree
$d$.

  1. (i)

If $p=0$ then

$f(X)-f(Y)=f_{d}\prod_{j=1}^{d}\left(X-\zeta^{j}Y-b^{(j)}\right),$

where $\zeta\in R$ is a primitive $d$-th root of unity and $b^{(j)}\in K$.

  2. (ii)

If $p>0$ write $d=p^{l}m$ with $(m,p)=1$. Then

$f(X)-f(Y)=f_{d}\prod_{i=1}^{p^{l}}\prod_{j=1}^{m}\left(X-\zeta^{j}Y-b_{i}^{(j)}\right),$

where $\zeta\in R$ is a primitive $m$-th root of unity and $b_{i}^{(j)}\in K$.

###### Proof.

For brevity we only treat the case $p>0$, the case $p=0$ being similar to the
case $p>0,\,l=0$. Write $f(X)-f(Y)=\prod_{i=1}^{d}(a_{i}X+b_{i}Y+c_{i})$ as in
8.1. Note that

$\prod_{i=1}^{d}(a_{i}X+b_{i}Y+c_{i})=\prod_{i=1}^{d}(a_{i}X+b_{i}Y)+A[X,Y],$

where $A[X,Y]\in R[X,Y],\deg A<d$. Since $\prod_{i=1}^{d}(a_{i}X+b_{i}Y)$ is
homogeneous of degree $d$, we have

$f_{d}(X^{d}-Y^{d})=\prod_{i=1}^{d}(a_{i}X+b_{i}Y)=\prod_{i=1}^{d}a_{i}\prod_{i=1}^{d}(X-\mu_{i}Y),$

where $\mu_{i}=-b_{i}/a_{i}\in K$. Plugging in $Y=1$ and noting that
$\prod_{i=1}^{d}a_{i}=f_{d}$ (by comparing coefficients at $X^{d}$) we obtain

$X^{d}-1=\prod_{i=1}^{d}(X-\mu_{i}).$

In particular we see that all the $d$-th roots of unity in the algebraic
closure of $K$ lie in $K$, and since $R$ is integrally closed (because it is a
UFD) they lie in $R$. Since $p=\mathrm{char}(K)>0$ these are actually the
$m$-th roots of unity, where $d=mp^{l},\,(m,p)=1$. ∎

We are ready to prove the forward direction of 1.10.

######  Proposition 8.3.

Let $f\in R[X]$ be special. Then it has the form stated in 1.10.

###### Proof.

For brevity we only treat the case $p>0$, the case $p=0$ being similar to the
case $p>0,\,l=0$ we treat below (using part (i) of Lemma 8.2 instead of part
(ii) and the fact that $V_{f}=\\{0\\}$ (see 1.2). Write $d=\deg
f=p^{l}m,(m,p)=1$. Denote by $\zeta$ a primitive $m$-th root of unity. From
8.2(ii) we have $\zeta\in R$ and

(8.2)
$f(X)-f(Y)=f_{d}\prod_{i=1}^{p^{l}}\prod_{j=1}^{m}\left(X-\zeta^{j}Y-b_{i}^{(j)}\right),\quad
b_{i}^{(j)}\in K.$

Next consider the shifted polynomial $g(X)=f(X+w)\in K[X]$ with

$w=\left\\{\begin{array}[]{ll}\frac{b_{1}^{(m-1)}}{1-\zeta^{-1}},&m>1,\\\
0,&m=1.\end{array}\right.$

Note that $g$ is also special but over the ring $K$ instead of $R$ (this is
immediate from the definition) and also by the choice of $w$ we have

$g(\zeta X)=g(X),$

because

$g(X)-g(\zeta X)=f(X+w)-f(\zeta
X+w)=f_{d}\prod_{i=1}^{p^{l}}\prod_{j=1}^{m}\left(X+w-\zeta^{j}(\zeta
X+w)-b_{i}^{(j)}\right)$

and if $m>1$ then $X+w-\zeta^{m-1}(\zeta X+w)-b_{1}^{(m-1)}=0$ (using
$\zeta^{m}=1$). If we can show that $g$ has the form asserted by 1.10 except
the coefficients are in $K$ instead of $R$, then $f(X)=g(X-w)$ would have the
form asserted by the theorem. Replacing $f$ with $g$, we see that it is enough
to prove the assertion under the additional assumptions that $R=K$ is a field
and $f(X)=f(\zeta X)$.

Thus from now on we assume that $R=K$ is a field and $f(X)=f(\zeta
X)=f(\zeta^{2}X)=...=f(\zeta^{m-1}X)$. Next let us prove that for all $j$ the
multisets

$S_{j}:=\left\\{b_{i}^{(j)}:1\leq i\leq p^{l}\right\\}$

are the same for all $1\leq j\leq m$. To this end let us identify the indices
$1\leq j\leq m$ with their corresponding residues in ${\mathbb{Z}}/m$ and
observe that by (8.2) we have

(8.3) $\prod_{j\bmod m}\prod_{b\in
S_{j}}\left(X-\zeta^{j}Y-b\right)=f(X)-f(Y)=f(X)-f(\zeta^{k}Y)=\prod_{j\bmod
m}\prod_{b\in S_{j}}\left(X-\zeta^{j+k}Y-b\right)\\\ =\prod_{j\bmod
m}\prod_{b\in S_{j-k}}\left(X-\zeta^{j}Y-b\right)$

for each $k\in{\mathbb{Z}}/m$. From the first equality in (8.3) we see that
the number of times an element $b$ appears in the multiset $S_{j}$ equals the
multiplicity of the factor $X-\zeta^{j}Y-b$ in the factorization of
$f(X)-f(Y)$ (recall that $R[X,Y]$ is a UFD) and from the last equality it also
equals the number of times $b$ appears in $S_{j-k}$. Hence the multisets
$S_{1},\ldots,S_{m}$ are all equal and we may rewrite (8.2) as

(8.4) $\begin{split}f(X)-f(Y)&=f_{d}\prod_{i=1}^{p^{l}}\prod_{j\bmod
m}(X-\zeta^{j}Y-b_{i})=\prod_{b\in S_{1}}\prod_{j\bmod
m}(X-\zeta^{j}Y-b),\end{split}$

where $b_{i}=b_{i}^{(1)}$.

Let us prove the following properties of the multiset
$S_{1}=\\{b_{1},\ldots,b_{m}\\}$:

  1. (1)

The underlying set of $S_{1}$ is $V_{f}:=\\{b\in R:f(X+b)=f(X)\\}$. It is an
${\mathbb{F}}_{p}$-linear subspace of $R$, with $|V_{f}|=p^{v}$ for some
$0\leq v\leq l$.

  2. (2)

Multiplication by $\zeta$ permutes $S_{1}$ as a multiset.

  3. (3)

$V_{f}$ is furthermore an ${\mathbb{F}}_{p}(\zeta)$-linear subspace of $R$.

  4. (4)

All elements of $S_{1}$ have the same multiplicity $p^{l-v}$.

For (1) observe that by (8.4) we have

$b\in S_{1}\iff X-Y-b\,\mid\,f(X)-f(Y)\iff f(X+b)=f(X),$

hence $V_{f}$ is the underlying set of $S_{1}$. Arguing as in the proof of 2.1
one sees that $V_{f}$ is an ${\mathbb{F}}_{p}$-linear subspace of $R$. Since
$|V_{f}|\leq|S_{1}|=p^{l}$ we see that $|V_{f}|=p^{v}$ with $0\leq v\leq l$.
Replacing $Y$ with $\zeta Y$ in (8.4) and using $f(Y)=f(\zeta Y)$ gives (2).
The properties (1),(2) imply (3).

To prove (4) observe that by (8.4) the multiplicity of $b\in V_{f}$ in $S_{1}$
equals the exponent of the factor $X-Y-b$ in the factorization of $f(X)-f(Y)$,
but since $f(Y+b)=f(Y)$ it also equals the exponent of $X-Y$ in the
factorization of $f(X)-f(Y)$ and is therefore independent of $b$. We have
established properties (1-4).

Now assuming WLOG that $V_{f}=\\{b_{1},\ldots,b_{p^{v}}\\}$ we can rewrite
(8.4) as

$f(X)-f(Y)=f_{d}\prod_{i=1}^{p^{v}}\prod_{j=1}^{m}(X-\zeta^{j}Y-b_{i}))^{p^{l-v}}=f_{d}\prod_{i=1}^{p^{v}}\left((X-b_{i})^{mp^{l-v}}-b_{i}^{mp^{l-v}}Y^{mp^{l-v}}\right).$

Setting $Y=0$ we see that $f$ has the form asserted in 1.10 with $A=0,C=f(0)$.
This completes the proof. ∎

Finally we prove the converse direction in 8.3.

######  Proposition 8.4.

  1. (i)

Assume that $p=0$, that $K$ contains a primitive $d$-th root of unity $\zeta$
and

$f=f_{d}(X+A)^{d}+C\in R[X]$

with $A,C\in K$. Then $f$ is special.

  2. (ii)

Assume that $p>0$, $d=p^{l}m,\,(m,p)=1$ and $K$ contains a primitive $m$-th
root of unity. Suppose

$f(X)=f_{d}\prod_{i=1}^{p^{v}}(X-b_{i}+A)^{mp^{l-v}}+C\in R[X],$

where $A,C\in K$ and $V=\\{b_{1},\ldots,b_{p^{v}}\\},\,0\leq v\leq l$ ($b_{i}$
distinct) is an ${\mathbb{F}}_{p}(\zeta)$-linear subspace of $K$. Then $f$ is
special.

###### Proof.

(i). Denote by $\zeta$ a primitive $d$-th root of unity. Then we have the
factorization

$f(X)-f(Y)=f_{d}\prod_{i=0}^{m-1}\left(X+A-\zeta^{i}(Y+A)\right)$

and $f$ is special.

(ii). Since $R$ is a UFD, by Gauss’s lemma $f(X)-f(Y)$ factors into linear
polynomials over $R$ iff it does over $K$. Hence we assume WLOG that $R=K$ is
a field and $A,C,b_{i}\in R$. It is then immediate from the definition that
$f(X)$ is special iff $\frac{1}{f_{d}}(f(X-A)-C)$ is, hence we assume WLOG
that $A=C=0,\,f_{d}=1$, i.e. $f=g^{mp^{l-v}}$ where

$g=\prod_{i=1}^{p^{v}}(X-b_{i}).$

We now show that the following factorization holds:

(8.5)
$f(X)-f(Y)=\prod_{i=1}^{p^{v}}\prod_{j=1}^{m}(X-\zeta^{j}Y+b_{i})^{p^{l-v}}.$

Since the Frobenius map is a ring homomorphism this is equivalent to

(8.6)
$g(X)^{m}-g(Y)^{m}=\prod_{i=1}^{p^{v}}\prod_{j=1}^{m}(X-\zeta^{j}Y+b_{i}).$

Note that since $V$ is an ${\mathbb{F}}_{p}(\zeta)$-linear subspace, the map
$x\mapsto\zeta^{-j}(x+b_{i})$ permutes $V$ and we have

$g(\zeta^{-j}(X+b_{i}))=\zeta^{-jp^{v}}g(X)$

and therefore $g(\zeta^{-j}(X+b_{i}))^{m}=g(X)^{m}$. It follows that
$X-\zeta^{j}Y+b_{i}\,\mid\,g(X)^{m}-g(Y)^{m}$ and therefore the RHS of (8.6)
divides the LHS. Since both sides of (8.6) have the same degree $mp^{v}$ and
are monic in $X$, we must have an equality. This establishes (8.6) and
therefore (8.5), which completes the proof.

∎

######  Remark 8.5.

Proposition 8.3 and its proof work under the weaker assumption that $R$ is an
integrally closed domain (not necessarily a UFD). Proposition 8.4 does require
the UFD assumption, but it can be replaced with $R$ being only integrally
closed if one assumes that $f_{d}=1$, because Gauss’s lemma works for monic
polynomials over any integrally closed domain.

## References