# Transcendence properties of the Artin-Hasse exponential modulo $p$

Joe Kramer-Miller

###### Abstract

Let $E_{p}(x)$ denote the Artin-Hasse exponential and let
$\overline{E}_{p}(x)$ denote its reduction modulo $p$ in
$\mathbb{F}_{p}[\\![x]\\!]$. In this article we study transcendence properties
of $\overline{E}_{p}(x)$ over $\mathbb{F}_{p}[x]$. We give two proofs that
$\overline{E}_{p}(x)$ is transcendental, affirmatively answering a question of
Thakur. We also prove algebraic independence results: i) for
$f_{1},\dots,f_{r}\in x\mathbb{F}_{p}[x]$ satisfying certain linear
independence properties, we show that the
$\overline{E}_{p}(f_{1}),\dots,\overline{E}_{p}(f_{r})$ are algebraically
independent over $\mathbb{F}_{p}[x]$ and ii) we determine the algebraic
relations between $\overline{E}_{p}(cx)$, where $c\in\mathbb{F}_{p}^{\times}$.
Our proof studies the higher derivatives of $\overline{E}_{p}(x)$ and makes
use of iterative differential Galois theory.

##  1 Introduction

The Artin-Hasse exponential is defined by

$\displaystyle
E_{p}(x):=\exp\left(\sum_{n=0}^{\infty}\frac{x^{p^{n}}}{p^{n}}\right)$

and is known to have coefficients in $\mathbb{Z}_{(p)}$. This series plays an
enormous role in $p$-adic analysis, showing up in the construction of Witt
vectors (see e.g. [7]) and the $p$-adic approach to exponential sums (see e.g.
[1] and [11]). From the relation $\frac{E_{p}(x^{p})}{E_{p}(x)^{p}}=\exp(-xp)$
it is immediate that $E_{p}(x)$ is a transcendental function over
$\mathbb{C}(x)$. However, we may also consider the reduction of $E_{p}(x)$
modulo $p$, which we denote by $\overline{E}_{p}(x)$. Then
$\overline{E}_{p}(x)$ is a power series in $\mathbb{F}_{p}[\\![x]\\!]$ that in
some sense plays the role of the exponential function in characteristic $p$.
For example, one can use $\overline{E}_{p}(x)$ to translate Witt vector
addition into power series multiplication as explained in [7]. In this article
we investigate the transcendence properties of $\overline{E}_{p}(x)$. We give
two proofs that $\overline{E}_{p}(x)$ is transcendental over
$\mathbb{F}_{p}(x)$. This answers a question posed by Dinesh Thakur (see e.g.
[9] or the open questions in chapter 12 of [3]). We also prove algebraic
independence results for the evaluations of $\overline{E}_{p}(x)$ at different
values in $x\overline{\mathbb{F}}_{p}[x]$.

###  1.1 Main results and outline

Our first result is the following:

######  Theorem 1.1.

The series $\overline{E}_{p}(x)$ is transcendental over $\mathbb{F}_{p}[x]$.

We give two proofs Theorem 1.1. The first proof is given in §2. It uses
Christol’s automatic criterion for algebraicity of power series together with
the functional equation: for $c\in\mathbb{F}_{p}^{\times}$ we have
$\overline{E}_{p}(cx)=\overline{E}_{p}(x)^{[c]}$ for
$c\in\mathbb{F}_{p}^{\times}$ where $[c]$ denotes the Teichmuller lift of $c$.
This proof is relatively elementary and short, but has the disadvantage of
requiring $p$ to be at least $5$. The second proof uses iterative differential
Galois theory and iterative Frobenius equations. These are the same methods we
use for the stronger algebraic independence results, which we describe in
detail below.

To state our first algebraic independence result we need the following
definition.

######  Definition 1.2.

Let $f(x)=\sum c_{i}x^{i}\in\overline{\mathbb{F}}_{p}[x]$. We define the
_primitive_ part of $f$ to be

$\displaystyle f^{*}(x):=\sum_{\begin{subarray}{c}i\geq 0\\\ p\nmid
i\end{subarray}}c^{*}_{i}(f)x^{i},$ $\displaystyle c^{*}_{i}(f):=\sum_{k\geq
0}c_{ip^{k}}^{1/p^{k}}.$

Note that $f^{*}(x)=0$ if and only if $f(x)=g(x)^{p}-g(x)$ for some
$g(x)\in\mathbb{F}_{p}[x]$.

######  Theorem 1.3.

Let $f_{1},\dots,f_{r}\in x\overline{\mathbb{F}}_{p}[x]$ be nonzero. Assume
that one of the two conditions holds:

  1. 1.

The elements $f_{1}^{*},\dots,f_{r}^{*}$ are linearly independent over
$\mathbb{F}_{p}$.

  2. 2.

The $f_{i}^{*}$’s are not all zero and no non-trivial power product of the
$f_{i}^{\prime}s$ are contained in $\mathbb{F}_{p}(x)^{p}$. In other words, if
$f_{1}^{n_{1}}\dots f_{r}^{n_{r}}$ is a $p$-th power then
$n_{1},\dots,n_{r}\in p\mathbb{Z}$.

Then $\overline{E}_{p}(f_{1}),\dots,\overline{E}_{p}(f_{r})$ are algebraically
independent.

Our second algebraic independence result is the following.

######  Theorem 1.4.

We have

$\displaystyle\dim_{\mathbb{F}_{p}(x)}\mathbb{F}_{p}(x)(\overline{E}_{p}(cx))_{c\in\mathbb{F}_{p}^{\times}}$
$\displaystyle=\phi(p-1),$

where $\phi$ is the Euler totient function.

The proofs of Theorem 1.3 and Theorem 1.4 follow the same strategy. This
strategy can be broken into three steps, which we outline below. For
simplicity, we restrict our attention to the case where the $f_{i}$ in Theorem
1.3 have coefficients in $\mathbb{F}_{p}$.

#### Iterated Frobenius equations.

Let $\sigma$ be the lifted Frobenius endomorphism of
$\mathbb{Z}_{p}[\\![x]\\!]$ that fixes $\mathbb{Z}_{p}$ and sends $x$ to
$x^{p}$. An iterated Frobenius (abbreviated IF) equation with depth $k$ is the
system of equations:

$y_{1}^{\sigma}-y_{1}=x^{b_{1}},~{}~{}y_{2}^{\sigma}-y_{2}=x^{b_{2}}y_{1},~{}~{}\dots,~{}~{}y_{k}^{\sigma}-y_{k}=x^{b_{k}}y_{k-1}.$

where the $b_{i}\geq 1$. We say an IF-equation is primitive if each $b_{i}$ is
prime to $p$. Let $M_{k}$ denote the $\mathbb{Z}_{p}(x)$-module space
generated by all solutions to all IF-equations of depth at most $k$ and let
$M=\cup M_{k}$. In §3 we study $M$. We prove the following:

  1. 1.

The solutions to primitive IF-equations form a basis of $M$ over
$\mathbb{Z}_{p}(x)$.

  2. 2.

The $\mathbb{Z}_{p}(x)$-module $M$ is closed under multiplication and
$M_{k_{1}}M_{k_{2}}\subset M_{k_{1}+k_{2}}$.

#### Higher derivatives

Let $\partial^{(k)}$ denote the differential operator
$\frac{1}{k!}\frac{d^{k}}{d^{k}x}$ and let $\textnormal{dLog}^{(k)}$ be the
operator sending $f$ to $\frac{\partial^{(k)}f}{f}$. We regard
$\partial^{(k)}$ and $\textnormal{dLog}^{(k)}$ as operators on
$\mathbb{F}_{p}[\\![x]\\!]$ and on $\mathbb{Z}_{p}[\\![x]\\!]$. Note that if
$f$ is contained in a finite extension $L$ of $\mathbb{F}_{p}(x)$, then so is
$\textnormal{dLog}^{(k)}(f)$ for all $k\geq 0$. This gives a differential
transcendence criterion: if the $\textnormal{dLog}^{(k)}(f)$ generate an
infinite extension of $\mathbb{F}_{p}(x)$ then $f$ is transcendental over
$\mathbb{F}_{p}(x)$. In §4 we study the iterated differential equations
satisfied by $\overline{E}_{p}(x)$ and power products of the
$\overline{E}_{p}(f_{r})$. We prove that
$\textnormal{dLog}^{(k)}(\overline{E}_{p}(x))$ is contained in $M_{k}$, but
not in $M_{k-1}$. For example, when $k=1$ a quick calculation shows that
$\frac{d}{dx}\overline{E}_{p}(x)=\frac{y}{x}\overline{E}_{p}(x)$, where
$y^{\sigma}-y=x$. More precisely, we determine
$\textnormal{dLog}^{(k)}(\overline{E}_{p}(x))$ modulo $M_{k-1}$. This is
enough to show $\overline{E}_{p}(x)$ is transcendental. Next, using standard
properties of higher derivatives and results from §3, we are able to determine
$\textnormal{dLog}^{(k)}\left(\overline{E}_{p}(f_{1})^{a_{1}}\dots\overline{E}_{p}(f_{r})^{a_{r}}\right)$
modulo $M_{k-1}$, where $a_{i}\in\mathbb{Z}$ and $f_{i}\in\mathbb{F}_{p}[x]$.

#### Iterative differential Galois groups

Let $K$ be the algebraic closure of $\mathbb{F}_{p}(x)$ in
$\overline{\mathbb{F}}_{p}(\\!(x)\\!)$. Note that $M$ is contained in $K$. An
ID-module over $K$ is a vector space over $K$ with compatible actions of
$\partial^{(k)}$. For any $f\in x\mathbb{F}_{q}[x]$ we let $N_{f}$ denote the
ID-module associated to the differential equation satisfied by
$\overline{E}_{p}(f)$ (see §5.1 for a precise description of this ID-module).
Following Matzat and van der Put (see [8]) we may define an ID-Galois group
$\textnormal{Gal}(N_{f})$. Then $\textnormal{Gal}(N_{f})$ is a closed subgroup
of the multiplicative group $\mathbb{G}_{m}$ over $\overline{\mathbb{F}}_{p}$
and is equal to all of $\mathbb{G}_{m}$ precisely when $\overline{E}_{p}(f)$
is transcendental over $\mathbb{F}_{p}(x)$. More generally, the algebraic
independence of $\overline{E}_{p}(f_{1}),\dots,\overline{E}_{p}(f_{r})$ is
equivalent to the ID-Galois group $\textnormal{Gal}(N_{f_{1}}\oplus\dots\oplus
N_{f_{r}})$ being isomorphic to $\mathbb{G}_{m}^{r}$. Using the Tannakian
interpretation of ID-Galois groups, it is enough to prove
$\overline{E}_{p}(f_{1})^{a_{1}}\dots\overline{E}_{p}(f_{r})^{a_{r}}$ is
transcendental for any $a_{i}\in\mathbb{Z}$. We then address this
transcendentce question under the conditions of Theorem 1.4 and Theorme 1.3
using the results from §4.

###  1.2 Further questions

It would be interesting to completely determine the algebraic relations
between values $\overline{E}_{p}(f)$ for $f\in x\overline{\mathbb{F}}_{p}[x]$.
More generally, it would be interesting to determine the algebraic relations
between values of $\overline{E}_{p}(x)$ at any algebraic power series. At the
moment it is even unclear to the author what the algebraic relations are
between the $\overline{E}_{p}(\xi x)$ where $\xi$ ranges over
$\overline{\mathbb{F}}_{p}$. The reader may observe the influence of Ax’s work
on Schanuel’s conjecture in [4] on the final steps of the proof of Theorem
1.3. Thus, it is natural to ask if an analogue of Ax-Schanuel hold for the
$\overline{E}_{p}(x)$.

###  1.3 Acknowledgments

The author wishes to thank Dinesh Thakur for his enthusiasm and his comments
on some very preliminary versions.

##  2 An automatic approach to the transcendence of $\overline{E}_{p}(x)$

In this section we give a short proof of the transcendence of
$\overline{E}_{p}(x)$ using the functional equation satisfied by
$\overline{E}_{p}(x)$ and automatic methods.

###  2.1 A transcendental criterion

We recall Christol’s well-known automatic criterion for algebraicity of series
in $\mathbb{F}_{p}[\\![x]\\!]$. See [6] or [5] for proofs.

######  Theorem 2.1.

Let $f=\sum f_{n}x^{n}\in\mathbb{F}_{p}[\\![x]\\!]$. The following are
equivalent.

  1. 1.

The series $f$ is algebraic over $\mathbb{F}_{p}[x]$.

  2. 2.

There is a $p$-automaton that gives output $f_{n}$ on input $n$.

  3. 3.

There are only finitely many subsequence of $(f_{n})_{n\geq 0}$ of the form
$(f_{p^{k}n+r})_{n\geq 0}$, where $k$ is positive and $r$ is in the range
$0\leq r<p^{k}$.

This allows us to establish the following transcendence criterion:

######  Lemma 2.2.

Let $\lambda\in\mathbb{Z}_{p}$. Let $f\in 1+x\mathbb{F}_{p}[\\![x]\\!]$ be
algebraic over $\mathbb{F}_{p}[x]$. Then
$f^{\lambda}\in\mathbb{F}_{p}[\\![x]\\!]$ is algebraic over
$\mathbb{F}_{p}[x]$ if and only if $\lambda$ is rational.

###### Proof.

This is a special case of [2]. We will give a proof in the interest of having
a quick and self-contained automatic proof that $\overline{E}_{p}(x)$ is
transcendental. We easily reduce to the case where $f=1+x$. If $\lambda$ is
rational it is clear that $(1+x)^{\lambda}$ is algebraic. For the other
direction, assume $\lambda$ is irrational. Write
$\lambda=\sum_{n=0}^{\infty}a_{n}p^{n}$. For $1\leq c\leq p-1$ we define sets

$S_{c}:=\left\\{n~{}|~{}a_{n}=c\right\\}~{}~{}\text{and}~{}~{}S_{c}^{>k}:=S_{c}\cap\mathbb{Z}_{>k}.$

We then set $\lambda_{c}=\sum_{n\in S_{c}}a_{n}p^{n}.$ If $\lambda_{c}$ is
nonzero and rational, we may replace $\lambda$ with $\lambda-\lambda_{c}$. In
particular, we may assume each $\lambda_{c}$ is either zero or non-rational.
Let $c_{0}$ be the largest number such that $\lambda_{c_{0}}\neq 0$. As
$\lambda_{c_{0}}$ is irrational for each $k,m>0$ we have

$\displaystyle m+S_{c_{0}}^{k}$ $\displaystyle\neq S_{c_{0}}^{k+m},$ (1)

as otherwise the digits of $\lambda_{c_{0}}$ would eventually be periodic.

Next, write $(1+x)^{\lambda}=\sum_{n=0}b_{n}x^{n}$. By using the product
expansion $(1+x)^{\lambda}=\prod(1+x^{p^{n}})^{a_{n}}$ we see that

$\displaystyle b_{c_{0}p^{k}}$ $\displaystyle\begin{cases}1&k\in S_{c_{0}}\\\
0&\text{otherwise}\end{cases}.$

Consider the sequences $B_{k}=(b_{np^{k}})_{n\geq 1}$. By Theorem 2.1 if we
prove that the $B_{k}$ are all distinct, we will know $(1+x)^{\lambda}$ is
transcendental. Note that the $c_{0}p^{n}$-th element in $B_{k}$ is $1$ if
$n+k\in S_{c_{0}}^{k}$ and zero otherwise. In particular, if $B_{k}=B_{k+m}$,
then we must have $m+S_{c_{0}}^{k}=S_{c_{0}}^{k+m}$. However, by (1) we know
this can never happen, which proves the lemma. ∎

###  2.2 A short proof of the transcendence of $\overline{E}_{p}(x)$

The first step is to prove that $\overline{E}_{p}(x)$ satisfies a functional
equation.

######  Lemma 2.3.

For $\xi\in\mathbb{F}_{p}^{\times}$ we have $\overline{E}_{p}(\xi
x)=\overline{E}_{p}(x)^{[\xi]}$, where $[\xi]\in\mathbb{Z}_{p}$ denotes the
Teichmuller lift of $\xi$.

###### Proof.

For $f_{0}\in\mathbb{Q}_{p}[\\![x]\\!]$ and $f_{1},\dots,f_{n}\in
x\mathbb{Q}_{p}[\\![x]\\!]$ we may take the composition $f_{0}\circ\dots\circ
f_{n}\in\mathbb{Q}_{p}[\\![x]\\!]$ and this operation is associative. For
$c\in\mathbb{Z}_{p}$ we define

$g(x)=\sum_{n=1}^{\infty}\frac{x^{p^{n}}}{p^{n}},~{}~{}~{}L_{c}(x)=cx,~{}~{}\text{and}~{}~{}P_{c}(x)=(1+x)^{c}=\sum_{n=0}^{\infty}\binom{c}{n}x^{n}.$

Note that $E_{p}(x)=\exp\circ g$. Also, for $\xi\in\mathbb{F}_{p}^{\times}$ we
have the relations:

$g\circ L_{[\xi]}=L_{[\xi]}\circ g~{}~{}\text{and}~{}~{}\exp\circ
L_{[\xi]}=P_{[\xi]}\circ(\exp(x)-1).$

The first relation comes from the fact that $[\xi]^{p}=[\xi]$. Thus, we
compute

$\displaystyle E_{p}([\xi]x)$ $\displaystyle=\exp\circ g\circ L_{[\xi]}$
$\displaystyle=\exp\circ L_{[\xi]}\circ g$
$\displaystyle=P_{[\xi]}\circ(\exp(x)-1)\circ g$
$\displaystyle=P_{[\xi]}\circ(E_{p}(x)-1)=\overline{E}_{p}(x)^{[\xi]}.$

The lemma follows by reducing modulo $p$. ∎

######  Theorem 2.4.

The series $\overline{E}_{p}(x)$ is transcendental over $\mathbb{F}_{p}[x]$ if
$p\geq 5$.

###### Proof.

Take $c\in\mathbb{F}_{p}^{\times}$. Since $p\geq 5$ we can pick $c$ so that
$[c]$ is not an integer and thus not rational. Assume $\overline{E}_{p}(x)$ is
algebraic. This implies $\overline{E}_{p}(cx)$ is algebraic. Thus, by Lemma
2.3 we know that $\overline{E}_{p}(x)^{[c]}$ is algebraic. However, we know
$[c]$ is not rational, so Lemma 2.2 tells us that $\overline{E}_{p}(x)^{[c]}$
is transcendental. This is a contradiction. ∎

######  Remark 2.5.

One can get algebraic independence results using the full result of [2]. In
particular, the argument above can be modified to show
$\dim_{\mathbb{F}_{p}(x)}(\overline{E}_{p}(cx))_{c\in\mathbb{F}_{p}^{\times}}\geq\phi(p-1)-1$.
As we prove in 5 that
$\dim_{\mathbb{F}_{p}(x)}(\overline{E}_{p}(cx))_{c\in\mathbb{F}_{p}^{\times}}=\phi(p-1)$,
we omit the proof of this weaker result.

##  3 Solutions to iterated Frobenius equations

Let $\mathbb{F}_{q}$ be the field with $q=p^{f}$ elements and fix
$\zeta\in\mathbb{F}_{q}$ such that $\zeta,\zeta^{p},\dots,\zeta^{p^{f-1}}$ is
a basis of $\mathbb{F}_{q}$ over $\mathbb{F}_{p}$. Let $\mathbb{Z}_{q}$ denote
the $p$-typical Witt vectors $W(\mathbb{F}_{q})$ of $\mathbb{F}_{q}$. By abuse
of notation we regard the $\zeta$ as an element of $\mathbb{Z}_{q}$ by
identifying $\zeta$ with its Teichmuller lift. It will be convenient to define
$x_{j}^{i}=\zeta^{p^{j}}x^{i}$ and the set

$\displaystyle X:=\left\\{x_{j}^{i}~{}\middle|~{}\begin{array}[]{c}1\leq j\leq
f\\\ i\geq 1,p\nmid i\end{array}\right\\}.$

###  3.1 Iterated Frobenius equations

Let $\sigma:\mathbb{Z}_{q}(\\!(x)\\!)\to\mathbb{Z}_{q}(\\!(x)\\!)$ denote the
$\mathbb{Z}_{p}$-linear endomorphism sending $x$ to $x^{p}$ and acting as the
Frobenius endomorphism on $\mathbb{Z}_{q}$. Let $f\in
x\mathbb{Z}_{q}[\\![x]\\!]$. We define $\tau(f)$ to be the unique solution to
$y^{\sigma}-y=f$ in $x\mathbb{Z}_{q}(\\!(x)\\!)$. Concretely, we have

$\tau(f)=\sum_{n=0}^{\infty}f^{\sigma^{n}}.$

We regard $\tau$ as a $\mathbb{Z}_{p}$-linear map. Note that $\tau$ satisfies
these two fundamental equations:

$\displaystyle\tau(f)\tau(g)$
$\displaystyle=\tau(f\tau(g))+\tau(g\tau(f))-\tau(fg),$ (2)
$\displaystyle\tau(f^{\sigma}\tau(g))$
$\displaystyle=\tau(f\tau(g))+\tau(f^{\sigma}g)-f\tau(g).$ (3)

######  Definition 3.1.

Let $[k]$ denote the set $\\{1,\dots,k\\}$ and let $[0]$ denote the empty set.
An _iterated Frobenius datum_ $b$ (abbreviated IF-datum) of depth $k$ is a
function

$b:[k]\to x\mathbb{Z}_{q}[\\![x]\\!].$

If the image of $b$ is in $X$ we say $b$ is a $p$-primary IF-datum. We denote
the depth of $b$ by $\text{dep}(b)$. We remark that there is exactly one IF-
datum of depth $0$. We define $D$ to be the set of all IF-data.

######  Definition 3.2.

We let $b^{tr}$ denote the restriction of $b$ to $[k-1]$ (we think of $b$ as a
list of elements and $b^{tr}$ as the truncation of the last element, hence the
$tr$ in the superscript). Also, for any $z\in x\mathbb{Z}_{q}[\\![x]\\!]$ we
let $\textnormal{cat}(b,z)$ to be the IF-datum of depth $k+1$ defined by:

$\displaystyle\textnormal{cat}(b,z)(i)$
$\displaystyle=\begin{cases}b(i)&i\in[k]\\\ z&i=k+1\end{cases}.$

######  Definition 3.3.

Let $b$ be an IF-datum of depth $k$. We define $w_{b}\in
x\mathbb{Z}_{q}[\\![x]\\!]$ by the recursive definition:

$\displaystyle w_{b}$ $\displaystyle=\begin{cases}1&\text{dep}(b)=0\\\
\tau(b(k)w_{b^{tr}})&\text{dep}(b)=k>0\end{cases}.$

We refer to $w_{b}$ as an iterative Frobenius solution (abbreviated IF-
solution).

For an IF-datum $b$ we may consider the iterative Frobenius equation:

$y_{1}^{\sigma}-y_{1}=b(1),~{}~{}y_{2}^{\sigma}-y_{2}=b(2)y_{1},~{}~{}\dots,~{}~{}y_{k}^{\sigma}-y_{k}=b(k)y_{k-1}.$

There is a unique solution vector
$(z_{1},\dots,z_{k})\in\mathbb{F}_{q}[\\![x]\\!]$ such that $z_{i}\in
x^{i}\mathbb{F}_{q}[\\![x]\\!]$. Then $w_{b}$ is $z_{k}$.

###  3.2 Spaces generated by IF-solutions

We make the following definitions:

$\displaystyle M_{k}$ $\displaystyle:=\left\\{\sum_{\begin{subarray}{c}b\in
D\\\ \text{dep}(b)\leq
k\end{subarray}}a_{b}w_{b}~{}|~{}a_{b}\in\mathbb{Z}_{q}(x)\right\\},$
$\displaystyle\widehat{M}_{k}$ $\displaystyle:=\text{$p$-adic completion of
$M_{k}$ in $\mathbb{Z}_{q}[\\![x]\\!]\langle x^{-1}\rangle$},$ $\displaystyle
A_{k}$ $\displaystyle:=\left\\{\sum_{\begin{subarray}{c}b\in D\\\
\text{dep}(b)\leq
k\end{subarray}}a_{b}w_{b}~{}|~{}a_{b}\in\mathbb{Z}_{q}[x]\right\\}.$

######  Lemma 3.4.

For $k\geq 0$ we have $\tau(xA_{k})\subset A_{k+1}$.

###### Proof.

We proceed by induction on $k$. For $k=0$ we are reduced to showing
$\tau(x_{j}^{i})\in A_{1}$. Write $i=p^{t}h$ where $h$ is coprime to $p$. Then

$\tau(x_{j}^{i})=\tau(x_{j-t}^{i-t})-\sum_{m=1}^{t}x_{j-m}^{p^{t-m}h},$

so $\tau(x_{j}^{i})\in A_{1}$. For $k>0$ it suffices to show
$\tau(x_{j}^{i}w_{b})\in A_{k+1}$ for any $w_{b}\in W$ with $\text{dep}(w)=k$
and $k\geq 1$. Again we write $i=p^{t}h$. By repeatedly using (3) we obtain

$\tau(x_{j}^{i}w_{b})=\tau(x_{j-t}^{h}w_{b})+\sum_{m=1}^{t}\tau(x_{j-m}^{p^{t-m}h}b(k)w_{b^{tr}})-\sum_{m=1}^{t}x_{j-m}^{p^{t-m}h}w_{b}.$

Note that $\tau(x_{j-t}^{h}w_{b})$ is an IF-solution of degree $k+1$, so it is
contained in $A_{k+1}$. The second summation is clearly contained in $A_{k}$.
Finally, by our inductive hypothesis $\tau(x_{j-m}^{p^{t-m}h}b(k)w_{b^{tr}})$
is contained in $A_{k}$.

∎

######  Corollary 3.5.

For any IF-datum $b$ of depth $k$ we have $w_{b}\in M_{k}$.

######  Lemma 3.6.

We have $\sigma(\widehat{M}_{k})\subset\widehat{M}_{k}$.

###### Proof.

It suffices to prove $w_{b}^{\sigma}\in M_{k}$ for any IF-datum with
$\text{dep}(b)=k$. We have $w_{b}^{\sigma}=w_{b}+b(k)w_{b^{tr}}$, which is in
$M_{k}$. ∎

######  Lemma 3.7.

We have $\frac{d}{dx}\widehat{M}_{k}\subset\widehat{M}_{k}$.

###### Proof.

We induct on $k$. When $k=0$ the statement is immediate, since
$\widehat{M}_{0}$ is the ring $\mathbb{Z}_{q}\langle x,x^{-1}\rangle$, which
is preserved by $\frac{d}{dx}$. From the definitions and the Leibnitz rule, we
reduce to showing $\frac{d}{dx}w_{b}\in\widehat{M}_{k}$ whenever
$\text{dep}(b)=k$. Applying $\frac{d}{dx}$ to the equation
$w_{b}^{\sigma}-w_{b}=b(k)w_{b^{tr}}$ and using the chain rule gives

$px^{p-1}(w_{b}^{\prime})^{\sigma}+w_{b}^{\prime}=b(k)^{\prime}w_{b^{tr}}+b(k)w_{b^{tr}}^{\prime}.$

We deduce that

$\displaystyle w_{b}^{\prime}$
$\displaystyle=\cdot\sum_{n=0}^{\infty}p^{n}x^{p^{n+1}-1}(b(k)^{\prime}w_{b^{tr}}+b(k)w_{b^{tr}}^{\prime})^{\sigma^{n}}.$
(4)

By our inductive hypothesis we know $w_{b^{tr}}^{\prime}\in\widehat{M}_{k-1}$.
Then from Lemma 3.6 we see that each summand in (4) is in $\widehat{M}_{k-1}$.
As the terms in the sum converge $p$-adically, the lemma follows. ∎

######  Lemma 3.8.

For $k_{1},k_{2}\geq 0$ we have
$\widehat{M}_{k_{1}}\widehat{M}_{k_{2}}\subset\widehat{M}_{k_{1}+k_{2}}$.

###### Proof.

We will show $A_{k_{1}}A_{k_{2}}\subset A_{k_{1}+k_{2}}$. The lemma will
follow by inverting $x$ and taking the $p$-adic completion. We proceed by
induction on $k=k_{1}+k_{2}$. The case $k=0$ is immediate. Let $k>0$ and
assume the result holds for $k-1$. Let $b_{i}$ be an IF-datum of degree
$k_{i}$. Then using (2) we have

$\displaystyle w_{b_{1}}w_{b_{2}}$
$\displaystyle=\tau(b_{1}(k_{1})w_{b_{1}^{tr}}w_{b_{2}})+\tau(b_{2}(k_{2})w_{b_{2}^{tr}}w_{b_{1}})-\tau(b_{1}(k_{1})w_{b_{1}^{tr}}\tau(b_{2}(k_{2})w_{b_{2}^{tr}}).$

By lemma follows from our inductive hypothesis and Lemma 3.4. ∎

###  3.3 IF-solutions modulo $p$.

We now study the space of IF-solutions modulo $p$.

######  Definition 3.9.

Let $b$ be an IF-datum. We define $\overline{w}_{b}$ to be the reduction of
$w_{b}$ modulo $p$ in $\mathbb{F}_{q}[\\![x]\\!]$.

We define the spaces

$\displaystyle\overline{M}_{k}$
$\displaystyle:=\left\\{\sum_{\begin{subarray}{c}b\in D\\\ \text{dep}(b)\leq
k\end{subarray}}a_{b}\overline{w}_{b}~{}|~{}a_{b}\in\mathbb{F}_{q}(x)\right\\}$
$\displaystyle\overline{M}$ $\displaystyle=\bigcup_{n\geq 1}\overline{M}_{k}.$

Our main result of this section is the following proposition.

######  Proposition 3.10.

The elements $\overline{w}_{b}$, where $b$ ranges over $b\in D$, are linearly
independent over $\mathbb{F}_{q}(x)$. In particular, the set
$\\{\overline{w}_{b}\\}_{b\in D}$ is a basis of $\overline{M}$ over
$\mathbb{F}_{q}(x)$.

###### Proof.

It will be useful for us to introduce an ordering on $D$. We set $b_{1}>b_{2}$
if $\text{dep}(b_{1})>\text{dep}(b_{2})$ and for IF-data with the same depth
$k$ we use the lexicographical ordering on $(b(k),\dots,b(1))$. Consider
$c=\\{c_{b}\\}_{b\in B}$ with $c_{b}\in\mathbb{F}_{q}(x)$, where almost all
the $c_{b}=0$. We define $\max(c)$ to be the largest $b$ such that $c_{b}\neq
0$. Assume there exists such a $c$ such that

$\displaystyle\sum_{b\in B}c_{b}\overline{w}_{b}$ $\displaystyle=0.$ (5)

We may further assume that $\max(c)$ is minimal in the sense that for any
other $c^{\prime}=\\{c_{b}^{\prime}\\}_{b\in B}$ where
$\max(c^{\prime})<\max(c)$ the corresponding sum $\sum
c_{b}^{\prime}\overline{w}_{b}$ is nonzero. Set $b_{max}=\max(c)$ and
$k_{max}=\text{dep}(b_{max})$. By multiplying (5) with a constant we may
assume $c_{b_{max}}=1$.

Let $F$ be the $p$-th power Frobenius. For $b\in D$ with $\text{dep}(b)=k$ we
have

$(F-1)c_{b}\overline{w}_{b}=c_{b}^{p}\overline{w}_{b}^{p}-c_{b}\overline{w}_{b}=(c_{b}^{p}-c_{b})\overline{w}_{b}+c_{b}^{p}b(k)\overline{w}_{b^{tr}}.$

Thus, if we $F-1$ to (5) we get

$\displaystyle\sum_{b\in D}d_{b}\overline{w}_{b}$ $\displaystyle=0,$

where for $b$ of depth $k$ we have

$\displaystyle d_{b}$
$\displaystyle:=(c_{b}^{p}-c_{b})+\sum_{\begin{subarray}{c}b_{0}\in D\\\
b_{0}^{tr}=b\end{subarray}}c_{b_{0}}b_{0}(k+1).$

It is clear that $d_{b_{max}}=0$ and $\max(d)<\max(c)$. Thus, we are reduced
to show that not all the $d_{b}$’s are zero, as this will contradict our
minimality condition on $c$.

First assume each $c_{b}\in\mathbb{F}_{p}$. Then we have

$\displaystyle d_{b_{max}^{tr}}$ $\displaystyle=\sum_{\begin{subarray}{c}b\in
D\\\ b^{tr}=b_{max}^{tr}\end{subarray}}c_{b}b(k_{max}),$

where the sum on the right is over all $b\in D$ that have the same truncation
as $b^{tr}$. The terms $b_{0}(k_{max})$ are all distinct and of the form
$x_{j}^{i}$. In particular, they are linearly independent over
$\mathbb{F}_{p}$. Not all of the $c_{b}$’s are equal to zero, since we know
$c_{b_{max}}\neq 0$, so we see that $d_{b_{max}^{tr}}\neq 0$. Next, assume
that not all the $c_{b}$’s are in $\mathbb{F}_{p}$. Let $b_{0}$ be the largest
IF-datum such that $c_{b_{0}}\not\in\mathbb{F}_{p}$ and let
$k_{0}=\text{dep}(b_{0})$. If $d_{b_{0}}=0$ we have

$\displaystyle c_{b_{0}}^{p}-c_{b_{0}}=-\sum_{\begin{subarray}{c}b\in D\\\
b^{tr}=b_{0}\end{subarray}}c_{b}b(k_{0}+1).$

We know that the left side is nonzero. In particular, the right side cannot be
zero. However, the right side is an $\mathbb{F}_{p}$-linear combination of
terms of the form $x_{j}^{i}=\zeta^{p^{j}}x^{i}$, where $p\nmid i$. In
particular, the right side is a polynomial $f(x)\in\mathbb{F}_{q}[x]$ whose
degree is coprime to $p$. However, for such an $f$, the equation
$y^{p}-y=f(x)$ cannot have a solution in $\mathbb{F}_{q}(x)$, which gives a
contradiction. ∎

######  Corollary 3.11.

For all $k,m\geq 0$ we have
$p^{m}\mathbb{Z}_{q}(\\!(x)\\!)\cap\widehat{M}_{k}=p^{m}\widehat{M}_{k}$. In
particular, let $x,y\in\widehat{M}_{k+1}$. If $p^{m}x\equiv
p^{m}y\mod\widehat{M}_{k}$ then $x\equiv y\mod\widehat{M}_{k}$.

In light of Proposition 3.10 we make the following definitions.

######  Definition 3.12.

Let $f\in\overline{M}$ and write $f=\sum a_{b}\overline{b}$, where almost all
$a_{b}=0$. We define the depth of $f$, written as $\text{dep}(f)$, to be the
largest $k$ such that for some $b$ with $\text{dep}(b)=k$ we have $a_{b}\neq
0$. In particular the depth of $f\in\overline{M}$ is the unique $k\geq 0$ such
that $f\in\overline{M}_{k}$ and $f\not\in\overline{M}_{k-1}$.

######  Corollary 3.13.

Let $\alpha_{1},\alpha_{2},\dots$ be a sequence of elements in $\overline{M}$
with $\text{dep}(\alpha_{k})=k$ for each $k$. Then the $\alpha_{k}$’s are
linearly independent over $\mathbb{F}_{p}(x)$.

######  Definition 3.14.

Let $\\{\dot{w}_{b}\\}_{b\in D}$ be the basis of
$\text{Hom}_{\mathbb{F}_{p}(\\!(x)\\!)}(\overline{M},\mathbb{F}_{p}(\\!(x)\\!))$
that is dual to $\\{\overline{w}_{b}\\}_{b\in D}$. That is, for
$b,b^{\prime}\in D$ we have $\dot{w}_{b}(\overline{w}_{b^{\prime}})$ is one if
$w=w^{\prime}$ and zero otherwise.

###  3.4 Products of IF-solutions

for the the product $w_{b_{1}}\dots w_{b_{r}}$ in terms of the IF-data
$b_{1},\dots,b_{r}$.

######  Definition 3.15.

Let $k_{1},\dots,k_{r}\geq 1$ and set $k=k_{1}+\dots+k_{r}$. We define

$\displaystyle[k_{1},\dots,k_{r}]:=\left\\{(i,j)\middle|\begin{array}[]{c}1\leq
i\leq r\\\ j\in[k_{i}]\end{array}\right\\}.$

A splicing function is a bijection $\eta:[k]\to[k_{1},\dots,k_{r}]$ such that
for any $1\leq j_{1}<j_{2}\leq k_{i}$ we have
$\eta^{-1}(i,j_{1})<\eta^{-1}(i,j_{2})$. We let $C(k_{1},\dots,k_{r})$ denote
the set of all splicing functions. Note that $C(k_{1},\dots,k_{r})$ has
cardinality $\binom{k}{k_{1},\dots,k_{r}}$.

######  Definition 3.16.

Let $b_{1},\dots,b_{r}$ be IF-data with $\text{dep}(b_{i})=k_{i}$. For
$\eta\in S(k_{1},\dots,k_{r})$ we define the splicing of $b_{1},\dots,b_{r}$
according to $\eta$ to be the depth $k$ IF-datum defined by the composition

$b_{\eta}:[k]\xrightarrow{\eta}[k_{1},\dots,k_{r}]\xrightarrow{(i,j)\mapsto
b_{i}(j)}x\mathbb{F}_{q}[\\![x]\\!].$

We let $S(b_{1},\dots,b_{r})$ denote the multiset of all possible splicings of
$b_{1},\dots,b_{r}$. That is,

$\displaystyle S(b_{1},\dots,b_{r}):=\left\\{b_{\eta}~{}\middle|~{}\eta\in
C(k_{1},\dots,k_{r})\right\\}.$

The following lemma follows immediately from these definitions.

######  Lemma 3.17.

Let $b_{i}$ and $k_{i}$ be as in Definition 3.16. Then

$\displaystyle S(b_{1},\dots,b_{r})$
$\displaystyle=\bigsqcup_{i=1}^{r}\left\\{\textnormal{cat}(b_{\eta},b_{i}(k_{i}))~{}\middle|~{}\eta\in
S(b_{1},\dots,b_{i-1},b_{i}^{tr},b_{i+1},\dots,b_{r})\right\\}.$

######  Lemma 3.18.

Continuing with the same notation, we define

$\mathcal{B}_{r}:=\prod_{i=1}^{r}w_{b_{i}}~{}~{}\text{ and
}~{}~{}\mathcal{B}_{r}^{(j)}:=w_{b_{j}^{tr}}\prod_{\begin{subarray}{c}1\leq
i\leq r\\\ i\neq j\end{subarray}}w_{b_{i}}.$

Then we have

$\displaystyle\mathcal{B}_{r}$
$\displaystyle=\sum_{i=1}^{r}\tau\left(b_{i}(k_{i})\mathcal{B}_{r}^{(i)}\right)\mod\widehat{M}_{k-1}.$

###### Proof.

We proceed by induction on $r$ with the base case $r=1$ being immedaite. Let
$r>1$ and assume the lemma holds for $r-1$. In particular, we have

$\displaystyle\mathcal{B}_{r-1}$
$\displaystyle\equiv\sum_{i=1}^{r-1}\tau\left(b_{i}(k_{i})\mathcal{B}_{r-1}^{(i)}\right)\mod\widehat{M}_{k-k_{r}-1}.$
(6)

We know $\widehat{M}_{k-k_{r}-1}\widehat{M}_{k_{r}}\subset\widehat{M}_{k-1}$
by Lemma 3.8. Thus, multiplying (6) by $w_{b_{r}}$:

$\displaystyle\mathcal{B}_{r}$ $\displaystyle\equiv
w_{b_{r}}\sum_{i=1}^{r-1}\tau\left(b_{i}(k_{i})\mathcal{B}_{r-1}^{(i)}\right)\mod\widehat{M}_{k-1}.$
(7)

For $1\leq i\leq r-1$ we know by (2) that

$\displaystyle w_{b_{r}}\tau(b_{i}(k_{i})\mathcal{B}_{r-1}^{(i)})$
$\displaystyle=\tau\left(b_{r}(k_{r})w_{b_{r}^{tr}}\tau\left(b_{i}(k_{i})\mathcal{B}_{r-1}^{(i)}\right)\right)+\tau\left(b_{i}(k_{i})\mathcal{B}_{r-1}^{(i)}w_{b_{r}}\right)-\tau\left(b_{r}(k_{r})w_{b_{r}^{tr}}b_{i}(k_{i})\mathcal{B}_{r-1}^{(i)}\right).$
(8)

We know $\mathcal{B}_{r-1}^{(i)}\in\widehat{M}_{k-k_{r}-1}$ and
$w_{b_{r}^{tr}}\in\widehat{M}_{k_{r}-1}$, so that the last term in the right
side of (8) is in $\widehat{M}_{k-1}$. As
$\mathcal{B}_{r-1}^{(i)}w_{b_{r}}=\mathcal{B}_{r}^{(i)}$ we have

$\displaystyle w_{b_{r}}\tau(b_{i}(k_{i})\mathcal{B}_{r-1}^{(i)})$
$\displaystyle\equiv\tau\left(b_{r}(k_{r})w_{b_{r}^{tr}}\tau\left(b_{i}(k_{i})\mathcal{B}_{r-1}^{(i)}\right)\right)+\tau\left(b_{i}(k_{i})\mathcal{B}_{r}^{(i)}\right)\mod\widehat{M}_{k-1}.$
(9)

Next, by our inductive hypothesis we know

$\displaystyle\mathcal{B}_{r-1}$
$\displaystyle\equiv\sum_{i=1}^{r-1}\tau\left(b_{i}(k_{i})\mathcal{B}_{r-1}^{(i)}\right)\mod\widehat{M}_{k-k_{r}-1}.$

We can combine this with (9) and (7) to get

$\displaystyle\mathcal{B}_{r}$
$\displaystyle\equiv\tau\left(b_{r}(k_{r})w_{b_{r}^{tr}}\mathcal{B}_{r-1}\right)+\sum_{i=1}^{r-1}\tau\left(b_{i}(k_{i})\mathcal{B}_{r}^{(i)}\right)\mod\widehat{M}_{k-k_{r}-1}.$

The result follows by observing that
$\mathcal{B}_{r}^{(r)}=w_{b_{r}^{tr}}\mathcal{B}_{r-1}$. ∎

######  Proposition 3.19.

Continue with the notation from Lemma 3.18. We have

$\displaystyle\mathcal{B}_{r}$ $\displaystyle\equiv\sum_{c\in
S(b_{1},\dots,b_{r})}w_{c}\mod\widehat{M}_{k-1}.$

###### Proof.

We induct on $r$. When $r=1$ the result is immediate. Let $r>1$ and assume the
result holds for $r-1$. From Lemma 3.18 we know

$\displaystyle\mathcal{B}_{r}$
$\displaystyle=\sum_{i=1}^{r}\tau\left(b_{i}(k_{i})\mathcal{B}_{r}^{(i)}\right)\mod\widehat{M}_{k-1}.$

By our inductive hypothesis we have

$\displaystyle\mathcal{B}_{r}^{(i)}$ $\displaystyle\equiv\sum_{c\in
S(b_{1},\dots,b_{i-1},b_{i}^{tr},b_{i+1},\dots,b_{r})}w_{c}\mod\widehat{M}_{k-2}.$

The proposition then follows from Lemma 3.17 and Corollary 3.4 ∎

We will now give some corollaries of Proposition 3.19. First we need a
definition.

######  Definition 3.20.

Let $k>0$ and $z\in x\mathbb{F}_{q}[\\![x]\\!]$. Let $\beta_{z}^{k}$ to be the
IF-datum of depth $k$ such that $\beta_{z}^{k}(i)=z$ for each $1\leq i\leq k$.
We define $\omega_{z}^{k}$ to be $w_{\beta_{z}^{k}}$. We let
$\overline{\omega}_{z}^{k}$ be the reduction of $\omega_{z}^{k}$ modulo $p$
and we let $\dot{\omega}_{z}^{k}$ be $\dot{w_{\beta_{z}^{k}}}$ as defined in
Definition 3.14.

We readily compute that

$\displaystyle\omega^{k}_{z}$ $\displaystyle=\sum_{i_{k}\geq\dots\geq
i_{1}\geq 0}z^{bp^{i_{1}+\dots+i_{k}}}.$

The elements $\omega_{z}^{k}$ and $\overline{\omega}^{k}_{z}$ will appear when
studying the higher derivatives of the Artin-Hasse exponential. We have the
following corollaries of Proposition 3.19.

######  Corollary 3.21.

Fix $z\in\mathbb{F}_{q}[\\![x]\\!]$. Let $k_{1},\dots,k_{r}\geq 1$ and set
$k=k_{1}+\dots+k_{r}$. Then

$\displaystyle\prod_{i=1}^{r}{\omega}^{k_{1}}_{z}$
$\displaystyle\equiv\binom{k}{k_{1},\dots,k_{r}}{\omega}^{k}_{z}\mod\widehat{M}_{k-1}.$

###### Proof.

This follows from Proposition 3.19 and the fact that there are
$\binom{k}{k_{1},\dots,k_{r}}$ splicing functions. ∎

######  Corollary 3.22.

We have $\tau(z)^{k}\equiv k!{\omega}^{k}_{z}\mod\widehat{M}_{k-1}$.

######  Corollary 3.23.

Let $b_{1},\dots,b_{r}\in D$, let $k_{i}=\text{dep}(b_{i})$, and set
$k=k_{1}+\dots+k_{r}$. Then for any $z\in X$ we have

$\displaystyle\dot{\omega}_{z}^{k}(\overline{w}_{b_{1}}\dots\overline{w}_{b_{r}})$
$\displaystyle=\begin{cases}\binom{k}{k_{1},\dots,k_{r}}&\text{ each
}\overline{w}_{i}=\overline{\omega}_{z}^{k_{i}}\\\ 0&\text{
otherwise}\end{cases}.$

##  4 Higher derivatives and the Artin-Hasse exponential

###  4.1 Higher derivatives and a transcendental criterion

We define the _higher derivatives_ on $\mathbb{F}_{q}(\\!(x)\\!)$ and
$\mathbb{Z}_{q}(\\!(x)\\!)$:

$\partial^{(k)}=\frac{1}{k!}\frac{d^{k}}{d^{k}x}.$

We define the _higher logarithmic derivatives_ of
$f\in\mathbb{F}_{q}(\\!(x)\\!)$ or $f\in\mathbb{Z}_{q}(\\!(x)\\!)$ by

$\textnormal{dLog}^{(k)}(f)=\frac{\partial^{(k)}(f)}{f}.$

Note that $\partial^{(k)}$ and $\textnormal{dLog}^{(k)}$ restrict to maps on
$\mathbb{F}_{p}(x)$. Furthermore, if $K$ is a finite separable extension of
$\mathbb{F}_{p}(x)$, then each $\partial^{(k)}$ and $\textnormal{dLog}^{(k)}$
extends uniquely to $K$ by a theorem of Schmidt (see [8, §2]). We begin with
the following general transcendence result.

######  Theorem 4.1.

Let $\alpha_{1},\alpha_{2},\dots$ be a sequence of power series in
$\mathbb{F}_{p}\llbracket x\rrbracket$. Let $f(x)\in\mathbb{F}_{p}\llbracket
x\rrbracket$ be a power series satisfying the iterative differential equation

$\textnormal{dLog}^{(k)}(f)=\alpha_{k}.$

If $\mathbb{F}_{p}(x)[\alpha_{i}]_{i\geq 1}$ is an infinite extension of
$\mathbb{F}_{p}(x)$, then $f(x)$ is transcendental over $\mathbb{F}_{p}[x]$.

###### Proof.

Assume $f(x)$ is algebraic (necessarily separable) and contained in a field
$K$. As $\textnormal{dLog}^{(k)}$ extends to $K$, we see that $\alpha_{n}$ is
contained in $K$ as well. This contradicts our assumption that
$\mathbb{F}_{p}(x)(\alpha_{1},\alpha_{2},\dots)$ is an infinite extension of
$\mathbb{F}_{p}(x)$. ∎

######  Corollary 4.2.

Let $\alpha_{1},\alpha_{2},\dots$ be a sequence of power series contained in
$\overline{M}$ with $\text{dep}(\alpha_{k})=k$ for each $k$. Let
$f(x)\in\mathbb{F}_{p}\llbracket x\rrbracket$ be a power series satisfying the
iterative differential equation

$\textnormal{dLog}^{(k)}(f)=\alpha_{k}.$

Then $f(x)$ is transcendental.

###### Proof.

From Corollary 3.13 we know that the $\alpha_{k}$’s are linearly independent
over $\mathbb{F}_{p}(x)$, so they cannot be contained in a finite extension.
Then use Theorem 4.1. ∎

###  4.2 Estimating the higher derivatives of $E_{p}(x)$.

Let $L$ denote the differential operator $\frac{d}{dx}-\frac{\tau(x)}{x}$.
Using the description
$E_{p}(x)=\exp\left(\sum_{n=0}^{\infty}\frac{x^{p^{n}}}{p^{n}}\right)$ we see
that for $k\geq 1$ we have

$\displaystyle\textnormal{dLog}^{(k)}(E_{p}(x))$
$\displaystyle=L^{k-1}\Big{(}\frac{\tau(x)}{x}\Big{)}.$ (10)

We have the following proposition.

######  Proposition 4.3.

For $k\geq 1$ we have

$\displaystyle\textnormal{dLog}^{(k)}(E_{p}(x))\equiv\frac{\tau(x)^{k}}{x^{k}}\mod\widehat{M}_{k-1}.$

###### Proof.

We proceed by induction on $k$. When $k=1$ the result is immediate. Assume the
result holds for $k$. Then we have

$\displaystyle
L^{k+1}\Big{(}\frac{\tau(x)}{x}\Big{)}=L\Big{(}\frac{\tau(x)^{k}}{x^{k}}+c\Big{)},$

where $c\in\widehat{M}_{k-1}$. We know
$\frac{d}{dx}(\frac{\tau(x)^{k}}{x^{k}}+c)\in\widehat{M}_{k}$ from Lemma 3.7.
We also know $\frac{\tau(x)}{x}c\in\widehat{M}_{k}$ by Lemma 3.8. Thus,

$\displaystyle L\Big{(}\frac{\tau(x)^{k}}{x^{k}}+c\Big{)}$
$\displaystyle\equiv\frac{\tau(x)^{k+1}}{x^{k+1}}\mod\widehat{M}_{k}.$

The Proposition follows from (10). ∎

######  Corollary 4.4.

For $k\geq 1$ we have

$\displaystyle\textnormal{dLog}^{(k)}(E_{p}(x))\equiv\frac{\omega_{x}^{k}}{x^{k}}\mod\widehat{M}_{k-1},$
$\displaystyle\textnormal{dLog}^{(k)}(\overline{E}_{p}(x))\equiv\frac{\overline{\omega}^{k}_{x}}{x^{k}}\mod\overline{M}_{k-1}.$

###### Proof.

From Corollary 3.22 and Propostion 4.3 we know
$\textnormal{dLog}^{(k)}(E_{p}(x))\equiv\frac{k!\omega_{x}^{k}}{x^{k}}\mod\widehat{M}_{k-1}$.
The equation about $\textnormal{dLog}^{(k)}(E_{p}(x))$ then follows from
Corollary 3.11. The equation about
$\textnormal{dLog}^{(k)}(\overline{E}_{p}(x))$ comes by reducing modulo $p$. ∎

At this point we may establish the transcendence of $\overline{E}_{p}(x)$.

######  Theorem 4.5.

The series $\overline{E}_{p}(x)$ is transcendental over $\mathbb{F}_{p}(x)$.

###### Proof.

Set $\alpha_{k}=\textnormal{dLog}^{(k)}(\overline{E}_{p}(x))$. From Corollary
4.4 we see that $\alpha_{k}\in\overline{M}$ and $\text{dep}(\alpha_{k})=k$.
The theorem follows from Corollary 4.2. ∎

###  4.3 Estimating
$\textnormal{dLog}^{(k)}\left[\overline{E}_{p}(f_{1})^{n_{1}}\dots\overline{E}_{p}(f_{r})^{n_{r}}\right]$

Recall from the introduction that for $f=\sum c_{i}x^{i}\in
x\mathbb{F}_{q}[x]$ we define

$f^{*}=\sum_{\begin{subarray}{c}i\geq 0\\\ p\nmid
i\end{subarray}}c_{i}^{*}(f)x^{i},~{}~{}~{}\text{
with}~{}~{}~{}c_{i}^{*}(f)=\sum_{k\geq 0}c_{ip^{k}}^{1/p^{k}}.$

We can express $c_{i}^{*}(f)$ uniquely as
$\sum_{j=0}^{f-1}c_{x_{j}^{i}}^{*}\zeta^{p^{j}}$ where
$c_{x_{j}^{i}}^{*}\in\mathbb{F}_{p}$. In particular, we have

$\displaystyle f^{*}$ $\displaystyle=\sum_{z\in X}c_{z}^{*}z.$

In this subsection we prove the following proposition.

######  Proposition 4.6.

Let $k\geq 0$ and let $n_{1},\dots,n_{r}\geq 0$. Let $f_{1},\dots,f_{r}\in
x\mathbb{F}_{q}[x]$. Then
$\textnormal{dLog}^{(k)}\left[\overline{E}_{p}(f_{1})^{n_{1}}\dots\overline{E}_{p}(f_{r})^{n_{r}}\right]$
is in $\overline{M}_{k}$ and for any $z\in X$ we have

$\displaystyle\dot{\omega}^{k}_{z}\left(\textnormal{dLog}^{(k)}\left[\overline{E}_{p}(f_{1})^{n_{1}}\dots\overline{E}_{p}(f_{r})^{n_{r}}\right]\right)$
$\displaystyle=\left(\sum_{i=1}^{r}n_{i}c_{z}^{*}(f_{i})\textnormal{dLog}^{(1)}(f_{i})\right)^{k}.$
(11)

In particular,
$\textnormal{dLog}^{(k)}\left[\overline{E}_{p}(f_{1})^{n_{1}}\dots\overline{E}_{p}(f_{r})^{n_{r}}\right]$
has depth exactly $k$ if there exists $z\in X$ such that
$\displaystyle\sum_{i=1}^{r}n_{i}c_{z}^{*}(f_{i})\textnormal{dLog}^{(1)}(f_{i})$
is nonzero.

The proof of Proposition 4.6 is broken into several steps. The main
ingredients are Corollary 4.4 and the following well known identities on
higher derivatives.

$\displaystyle\partial^{(k)}(f_{1}\dots f_{n})$
$\displaystyle=\sum_{\begin{subarray}{c}k_{1},\dots,k_{n}\geq 0\\\
k_{1}+\dots+k_{n}=k\end{subarray}}\partial^{(k_{1})}(f_{1})\cdot\dots\cdot\partial^{(k_{n})}(f_{n}),$
(12) $\displaystyle\partial^{(k)}(f(g))$
$\displaystyle=\sum_{j=1}^{k}\partial^{(j)}(f)(g)\cdot\sum_{\begin{subarray}{c}k_{1},\dots,k_{j}\geq
1\\\
k_{1}+\dots+k_{j}=k\end{subarray}}\partial^{(k_{1})}(g)\cdot\dots\cdot\partial^{(k_{j})}(g).$
(13)

######  Lemma 4.7.

Let $n$ be any $p$-adic integer. We have

$\displaystyle\textnormal{dLog}^{(k)}(\overline{E}_{p}(x)^{n})$
$\displaystyle\equiv
n^{k}\frac{\overline{\omega}_{x}^{k}}{x^{k}}\mod\overline{M}_{k-1}$

###### Proof.

First assume $n$ is a positive integer. By Corollary 4.4, Corollary 3.21, and
(12) we have

$\displaystyle\textnormal{dLog}^{(k)}(\overline{E}_{p}(x)^{n})$
$\displaystyle\equiv\sum_{\begin{subarray}{c}k_{1},\dots,k_{n}\geq 0\\\
k_{1}+\dots+k_{n}=k\end{subarray}}\frac{1}{x^{k}}\overline{\omega}^{k_{1}}_{x}\dots\overline{\omega}^{k_{n}}_{x}\mod\overline{M}_{k-1}$
$\displaystyle\equiv\sum_{\begin{subarray}{c}k_{1},\dots,k_{n}\geq 0\\\
k_{1}+\dots+k_{n}=k\end{subarray}}\binom{k}{k_{1},\dots,k_{n}}\frac{\overline{\omega}^{k}_{x}}{x^{k}}\mod\overline{M}_{k-1}$
$\displaystyle\equiv
n^{k}\frac{\overline{\omega}^{k}_{x}}{x^{k}}\mod\overline{M}_{k-1}.$

For the general result, write $n=a+p^{m}b$. If $m>v_{p}(k)$ we have
$\textnormal{dLog}^{(k)}(\overline{E}_{p}(x)^{n})=\textnormal{dLog}^{(k)}(\overline{E}_{p}(x)^{a})$,
since $\partial^{(k)}(f^{p^{m}})=0$. The general result follows. ∎

######  Corollary 4.8.

For $f\in x\mathbb{F}_{p}[x]$ we have

$\displaystyle\textnormal{dLog}^{(k)}(\overline{E}_{p}(f)^{n})$
$\displaystyle\equiv
n^{k}\frac{\overline{\omega}_{x}^{k}(f)}{x^{k}}(\textnormal{dLog}^{(1)}f)^{k}\mod\overline{M}_{k-1}$

###### Proof.

From Lemma 4.7 we see that
$\textnormal{dLog}^{(i)}(\overline{E}_{p}(x)^{n})\in\overline{M}_{k-1}$ for
$1\leq i<k$. Note that for $g\in\overline{M}_{k-1}$ and $h\in
x\mathbb{F}_{p}[x]$ we have $g\circ h\in\overline{M}_{k-1}$, so that
$\textnormal{dLog}^{(i)}(\overline{E}_{p}(x)^{n})\circ f\in\overline{M}_{k-1}$
for $1\leq i<k$. The Corollary follows from (13) and Lemma 4.7. ∎

######  Lemma 4.9.

Let $f\in x\mathbb{F}_{p}[x]$. Then

$\displaystyle\overline{\omega}_{x}^{k}(f)$
$\displaystyle\equiv\sum_{\begin{subarray}{c}b\in D\\\
\text{dep}(b)=k\end{subarray}}c_{b}^{*}(f)\overline{w}_{b}\mod\overline{M}_{k-1},\text{
where}$ $\displaystyle c_{b}^{*}(f)$
$\displaystyle=\prod_{i=1}^{k}c^{*}_{b(i)}(f).$

###### Proof.

First note that
$\overline{\omega}_{x}^{k}(f)\equiv\overline{\omega}_{x}^{k}(f^{*})\mod\overline{M}_{k-1}$.
This can be proven by inducting on $k$ and using (3). We are therefore reduced
to the case where $f=f^{*}$. The result follows from the
$\mathbb{F}_{p}$-linearity of $\tau$ modulo $p$ and unraveling the nested
$\tau$’s. ∎

######  Corollary 4.10.

We have

$\displaystyle\dot{\omega}_{z}^{k}\left(\overline{\omega}_{x}^{k}(f)\right)$
$\displaystyle=c_{z}^{*}(f)^{k}.$

######  Lemma 4.11.

Let $k_{1},\dots,k_{r}\geq 1$ with $k_{1}+\dots+k_{r}=k$. Then

$\displaystyle\dot{\omega}_{z}^{k}\left(\overline{\omega}_{x}^{k_{1}}(f_{1}(x))\dots\overline{\omega}_{x}^{k_{r}}(f_{r}(x))\right)=\binom{k}{k_{1},\dots,k_{r}}\prod_{i=1}^{r}c_{z}^{*}(f_{i})^{k_{i}}$

###### Proof.

This follows from Corollary 4.10 and Corollary 3.23. ∎

###### Of Proposition 4.6.

From (12) and Corollary 4.8 we have

$\displaystyle\textnormal{dLog}^{(k)}\left[\overline{E}_{p}(f_{1})^{n_{1}}\dots\overline{E}_{p}(f_{r})^{n_{r}}\right]$
$\displaystyle=\sum_{\begin{subarray}{c}k_{1},\dots,k_{r}\geq 0\\\
k_{1}+\dots+k_{r}=k\end{subarray}}\prod_{i=1}^{r}\textnormal{dLog}^{(k_{i})}(\overline{E}_{p}(f_{i})^{n_{i}})$
$\displaystyle\equiv\sum_{\begin{subarray}{c}k_{1},\dots,k_{r}\geq 0\\\
k_{1}+\dots+k_{r}=k\end{subarray}}\prod_{i=1}^{r}n_{i}^{k_{i}}(\textnormal{dLog}^{(1)}(f_{i}))^{k_{i}}\prod_{i=1}^{r}\overline{\omega}_{x}^{k_{i}}(f)\mod\overline{M}_{k-1}.$

Applying Lemma 4.11 to each
$\displaystyle\prod_{i=1}^{r}\overline{\omega}_{x}^{k_{i}}(f)$ gives

$\displaystyle\dot{\omega}_{z}^{k}\left(\textnormal{dLog}^{(k)}\left[\overline{E}_{p}(f_{1})^{n_{1}}\dots\overline{E}_{p}(f_{r})^{n_{r}}\right]\right)$
$\displaystyle=\sum_{\begin{subarray}{c}k_{1},\dots,k_{r}\geq 0\\\
k_{1}+\dots+k_{r}=k\end{subarray}}\binom{k}{k_{1},\dots,k_{r}}\prod_{i=1}^{s}\left[n_{i}\textnormal{dLog}^{(1)}(f_{i})c_{z}^{*}(f_{i})\right]^{k_{i}}$
$\displaystyle=\left(\sum_{i=1}^{r}n_{i}\textnormal{dLog}^{(1)}(f_{i})c_{z}^{*}(f_{i})\right)^{k}.\qed$

##  5 Algebraic independence results

In this section we prove Theorem 1.3 and Theorem 1.4.

###  5.1 Iterative differential Galois theory and transcendence

We will make use of iterative differential (ID-for short) Galois theory, as
expounded on by Matzat and van der Put in [8] (in particular §3 and §4 in this
article.) Let $K$ be the algebraic closure of $\mathbb{F}_{p}(x)$ in
$\overline{\mathbb{F}}_{p}(\\!(x)\\!)$. The higher derivatives
$\partial^{(k)}$ make $K$ an ID-field. For any $f\in\mathbb{F}_{p}[x]$ we
recursively define a sequence $A_{f,0},A_{f,1},\dots\in K$ by

$A_{f,0}=1~{}\text{ and
}A_{f,k}=-\sum_{i=0}^{k-1}A_{f,i}\textnormal{dLog}^{(k-i)}(\overline{E}_{p}(f)).$

We then define an ID-module $N_{f}=Ke_{f}$ over $K$ of rank one by the rule:

$\partial^{(k)}(e_{f})=A_{f,k}e_{f}.$

Note that $N_{f}$ trivializes over the ring
$K\left(\overline{E}_{p}(f),\frac{1}{\overline{E}_{p}(f)}\right)$. For
$f_{1},\dots,f_{r}\in x\overline{F}_{p}(\\!(x)\\!)$ we define the rank $r$ ID-
module:

$\displaystyle N:=\bigoplus_{i=1}^{r}N_{f_{1}}.$

Then $N$ has a full system of solutions over the ring

$\displaystyle
R_{0}:=K\left(\overline{E}_{p}(f_{1}),\dots,\overline{E}_{p}(f_{r}),\frac{1}{\overline{E}_{p}(f_{1})\dots\overline{E}_{p}(f_{r})}\right).$

The Picard-Vessiot ring of $N$ is $R=R_{0}/I$, where $I$ is a maximal ID-ideal
(i.e. $I$ is a maximal element in the ordered set of ideals of $R_{0}$
satisfying $\partial^{(k)}(I)\subset I$.) The ID-Galois group $\mathscr{G}(N)$
is the group of ID-automorphisms of $R$ that fix $K$. A key fact is that
$\mathscr{G}(N)$ is reduced algebraic group over the field of constants
$\overline{\mathbb{F}}_{p}$ and that $\text{Spec}(R)$ is an
$\mathscr{G}(N)$-torsor over $K$. In particular we have

$\dim_{K}(\textnormal{Frac}(R_{0}))\geq\dim_{K}(\textnormal{Frac}(R))=\dim_{\overline{\mathbb{F}}_{p}}(\mathscr{G}(N)).$

Thus, it suffices to show $\mathscr{G}(N)$ has dimension $r$.

On the other hand, there is a Tannakian interpretation of the ID-Galois group.
Let $\textnormal{ID}_{K}$ denote the category of ID-modules over $K$. Then
$\textnormal{ID}_{K}$ is a $\overline{\mathbb{F}}_{p}$-linear tensor category.
For any object $N$ in $\textnormal{ID}_{K}$ we let $[N]$ be the full
subcategory of $\textnormal{ID}_{K}$ generated by tensor powers of $N$ and its
dual. Then $[N]$ is again a $\overline{\mathbb{F}}_{p}$-linear tensor category
and thus isomorphic to the category of representations for an affine algebraic
group $\textnormal{Gal}(N)$ defined over $\overline{\mathbb{F}}_{p}$. Then by
standard arguments (see [10] for the characteristic $0$ case and see [8] for
our precise situation) we have

$\displaystyle\mathscr{G}(N)\cong\textnormal{Gal}(N).$

Note that since $N_{f_{i}}$ has rank one we know $\textnormal{Gal}(N_{f_{i}})$
is a closed subgroup of $\mathbb{G}_{m}$. In particular, we see that
$\textnormal{Gal}(N)$ is a closed subgroup of $\mathbb{G}_{m}^{r}$. To prove
that $\textnormal{Gal}(N)\cong\mathbb{G}_{m}^{r}$ it is enough to show that
for any $n_{1},\dots,n_{r}\in\mathbb{Z}$ that are not all zero, the ID-module
$N_{f_{1}}^{\otimes n_{1}}\otimes\dots\otimes N_{f_{r}}^{\otimes n_{r}}$ is
nontrivial. This amounts to proving
$\overline{E}_{p}(f_{1})^{n_{1}}\dots\overline{E}_{p}(f_{r})^{n_{r}}$ is
transcendental over $K$. Furthermore, we are easily reduced to the case where
not all the $n_{i}$’s are divisible by $p$.

###  5.2 Proof of Theorem 1.3

Let $f_{1},\dots,f_{r}$ satisfy one of the two conditions from Theorem 1.3.
From the discussion in §5.1 we must show
$\overline{E}_{p}(f_{1})^{n_{1}}\dots\overline{E}_{p}(f_{r})^{n_{r}}$ is
transcendental, where we may assume that not all the $n_{i}$’s are divisible
by $p$. From Corollary 4.2 and Proposition 4.6 we are reduced to showing that
for some $z\in X$ we have

$\displaystyle\sum_{i=1}^{r}n_{i}c_{z}^{*}(f_{i})\textnormal{dLog}^{(1)}(f_{i})$
$\displaystyle\neq 0.$ (14)

#### The case where the $f_{i}^{*}$’s are not all zero and no non-trivial
power product of the $f_{i}^{\prime}s$ are contained in
$\mathbb{F}_{p}(x)^{p}$:

We have

$\displaystyle\sum_{i=1}^{r}n_{i}c_{z}^{*}(f_{i})\textnormal{dLog}^{(1)}(f_{i})$
$\displaystyle=\textnormal{dLog}^{(1)}\left[\prod_{i=1}^{r}f_{i}^{n_{i}c_{z}^{*}(f_{i})}\right].$

By our assumption we know the product in the $\textnormal{dLog}^{(1)}$ is not
in $\mathbb{F}_{p}(x)^{p}$. The theorem follows by observing that
$\ker(\textnormal{dLog}^{(1)})=(\mathbb{F}_{p}(x)^{p})^{\times}$.

#### The case where $f_{1}^{*},\dots,f_{r}^{*}$ are linearly independent:

Without loss of generality assume $p\nmid n_{1}$. We may also assume that none
of the $f_{i}$’s are $p$-th powers. Let $g\in\mathbb{F}_{p}[x]$ be an
irreducible polynomial dividing $f_{1}$ and let $v_{g}$ denote the valuation
associated to $g$. Write $f_{i}=g^{b_{i}}h_{i}$, where $g\nmid h_{i}$. We can
find $g$ so that $p\nmid b_{1}$, as otherwise $f_{1}$ would be a $p$-th power.
Since $f_{1}^{*},\dots,f_{r}^{*}$ are linearly dependent over
$\mathbb{F}_{p}$, we know

$\displaystyle n_{1}b_{1}f_{1}^{*}+\dots+n_{r}b_{r}f_{r}^{*}$
$\displaystyle\neq 0.$

In particular, there exists $z$ coprime to $p$ such that

$\displaystyle
D_{z}:=n_{1}b_{1}c_{z}^{*}(f_{1})+\dots+n_{r}b_{r}c_{z}^{*}(f_{r})$
$\displaystyle\neq 0.$

The we have

$\displaystyle\sum_{i=1}^{r}n_{i}c_{z}^{*}(f_{i})\textnormal{dLog}^{(1)}(f_{i})$
$\displaystyle=D_{z}\textnormal{dLog}^{(1)}(g)+\sum_{i=1}^{r}n_{i}c_{z}^{*}(f_{i})\textnormal{dLog}^{(1)}(h_{i}).$

If $z$ is a root of $g$, we see that $D_{z}\textnormal{dLog}^{(1)}(g)$ has a
simple pole at $z$, while $\textnormal{dLog}^{(1)}(h_{i})$ is regular at $z$.
In particular, the right side of this sum has a simple pole at $z$ and thus is
nonzero.

###  5.3 Proof of Theorem 1.4

We now prove Theorem 1.4. Let $\xi$ be a primitive $p-1$-th root of unity. Set
$r=\phi(p-1)$. First, we show that
$\overline{E}_{p}(x),\dots,\overline{E}_{p}(\xi^{r-1}x)$ are algebraically
independent. From the ID-Galois group argument from §5.1 it suffices to show

$A(x):=\prod_{i=0}^{r-1}\overline{E}_{p}(\xi^{i}x)^{n_{i}}$

is transcendental, where $n_{0},\dots,n_{r-1}$ are integers not all divisible
by $p$. Indeed, this implies that the ID-Galois group of the ID-module
$N_{x}\oplus N_{\xi x}\oplus\dots\oplus N_{\xi^{r-1}x}$ is all of
$\mathbb{G}_{m}^{r}$. From Lemma 2.3 we have

$\displaystyle A(x)$
$\displaystyle=\overline{E}_{p}(x)^{n_{0}+n_{1}\xi+\dots+n_{r-1}\xi^{r-1}}.$

As the $1,\dots,\xi^{r-1}$ are independent over $\mathbb{Z}$, we have
$A(x)=\overline{E}_{p}(x)^{n}$, where $n$ is a nonzero $p$-adic number. Thus,
it suffices to show $\overline{E}_{p}(x)^{n}$ is transcendental for any
nonzero $p$-adic number with $p\nmid n$. Then from Lemma 4.7 we see that
$\textnormal{dLog}^{(k)}(\overline{E}_{p}(x)^{n})$ has depth $k$. It follows
from Corollary 4.2 that $\overline{E}_{p}(x)^{n}$ is transcendental over
$\mathbb{F}_{p}(x)$. Thus, we have proven

$\displaystyle\dim_{\mathbb{F}_{p}(x)}\mathbb{F}_{p}(x)(\overline{E}_{p}(\xi^{i}x))_{0\leq
i\leq r-1}$ $\displaystyle=r.$

For $j>r-1$, we can write $\xi^{j}=c_{0}+c_{1}\xi+\dots+c_{r-1}\xi^{r-1}$.
Then Lemma 2.3 gives

$\displaystyle\overline{E}_{p}(\xi^{j}x)$
$\displaystyle=\prod_{i=0}^{r-1}\overline{E}_{p}(\xi^{i}x)^{c_{i}},$

so that

$\displaystyle\dim_{\mathbb{F}_{p}(x)}\mathbb{F}_{p}(x)(\overline{E}_{p}(\xi^{i}x))_{0\leq
i\leq p-2}$ $\displaystyle=r.$

## References

  * [1] Alan Adolphson and Steven Sperber.  Exponential sums and Newton polyhedra: cohomology and estimates.  Ann. of Math. (2), 130(2):367–406, 1989.
  * [2] Jean-Paul Allouche, M. Mendès France, and A. J. van der Poorten.  Indépendance algébrique de certaines séries formelles.  Bull. Soc. Math. France, 116(4):449–454, 1988.
  * [3] Jean-Paul Allouche and Jeffrey Shallit.  Automatic sequences.  Cambridge University Press, Cambridge, 2003.  Theory, applications, generalizations.
  * [4] James Ax.  On Schanuel’s conjectures.  Ann. of Math. (2), 93:252–268, 1971.
  * [5] G. Christol, T. Kamae, M. Mendès France, and G. Rauzy.  Suites algébriques, automates et substitutions.  Bull. Soc. Math. France, 108(4):401–419, 1980.
  * [6] Gilles Christol.  Ensembles presque periodiques $k$-reconnaissables.  Theoret. Comput. Sci., 9(1):141–145, 1979.
  * [7] Jean Dieudonné.  On the Artin-Hasse exponential series.  Proc. Amer. Math. Soc., 8:210–214, 1957.
  * [8] B. Heinrich Matzat and Marius van der Put.  Iterative differential equations and the Abhyankar conjecture.  J. Reine Angew. Math., 557:1–52, 2003.
  * [9] Dinesh S. Thakur.  Automata methods in transcendence.  In $t$-motives: Hodge structures, transcendence and other motivic aspects, EMS Ser. Congr. Rep., pages 351–372. EMS Publ. House, Berlin, [2020] ©2020.
  * [10] Marius van der Put and Michael F. Singer.  Galois theory of linear differential equations, volume 328 of Grundlehren der mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences].  Springer-Verlag, Berlin, 2003.
  * [11] Daqing Wan.  Newton polygons of zeta functions and $L$ functions.  Ann. of Math. (2), 137(2):249–293, 1993.