---
abstract: 'The $T$-adic exponential sum associated to a Laurent polynomial in one variable is studied. An explicit arithmetic polygon is proved to be the generic Newton polygon of the $C$-function of the T-adic exponential sum. It gives the generic Newton polygon of $L$-functions of $p$-power order exponential sums.'
address:
- 'Department of Mathematics, Shanghai Jiao Tong University, Shanghai 200240, P.R. China, E-mail: clliu@sjtu.edu.cn'
- 'School of Mathematical Sciences, Beijing Normal University, Beijing 100875, P.R. China, E-mail: wenxin8210@mail.bnu.edu.cn'
- 'School of Mathematical Sciences, Beijing Normal University, Beijing 100875, P.R. China, E-mail: czniubnu@yahoo.cn'
author:
- Chunlei Liu
- Wenxin Liu
- Chuanze Niu
title: 'Generic $T$-adic exponential sums in one variable'
---

Introduction
============

Let $W$ be the ring scheme of Witt vectors, $\mathbb{F}_q$ the field of characteristic $p$ with $q$ elements, $\mathbb{Z}_{q}=W(\mathbb{F}_q)$, and $\mathbb{Q}_q=\mathbb{Z}_q[\frac{1}{p}]$.

Let $\triangle\supsetneq\{0\}$ be an integral convex polytope in $\mathbb{R}^n$, and $I$ the set of vertices of $\triangle$ different from the origin. Let $$f(x)=\sum\limits_{u\in \triangle}(a_ux^u,0,0,\cdots)\in
W(\mathbb{F}_q[x_1^{\pm1},\cdots,x_n^{\pm1}])\text{ with }
\prod_{u\in I}a_u\neq0,$$ where $x^u=x_1^{u_1}\cdots x_n^{u_n}$ if $u=(u_1,\cdots,u_n)\in\mathbb{Z}^n$.

Let $T$ be a variable.

The sum $$S_{f}(k,T)=\sum\limits_{x\in(\mathbb{F}_{q^k}^{\times})^n}
(1+T)^{{\rm
Tr}_{\mathbb{Z}_{q^k}/\mathbb{Z}_p}(f(x))}\in\mathbb{Z}_p[[T]]$$ is call a $T$-adic exponential sum. And the function $$L_f(s,T)=\exp(\sum\limits_{k=1}^{\infty}S_f(k,T)\frac{s^k}{k})\in
1+s\mathbb{Z}_p[[T]][[s]]$$ is called an $L$-function of $T$-adic exponential sums.

We view $L_f(s, T)$ as a power series in the single variable $s$ with coefficients in the $T$-adic complete field $\mathbb{Q}_p((T))$. Let $\zeta_{p^m}$ be a primitive $p^m$-th root of unity, and $\pi_m=\zeta_{p^m}-1$. Then $L_f(s,\pi_m)$ is the $L$-function of the $p$-power order exponential sums $S_f(k,\pi_m)$ studied by Liu-Wei [@LW].

The function $$C_f(s,T) =C_f(s,T;\mathbb{F}_q) =\exp(\sum\limits_{k=1}^{\infty}-(q^k-1)^{-n}S_{f}(k,T)\frac{s^k}{k})$$ is called a $C$-function of $T$-adic exponential sums.

We have $$L_f(s,T) = \prod_{i=0}^n C_f(q^is, T)^{(-1)^{n-i+1}{n\choose i}},$$ and $$C_f(s, T)= \prod_{j=0}^{\infty} L_f(q^js,
T)^{(-1)^{n-1}{n+j-1\choose j}}.$$ So the $C$-function $C_f(s,T)$ and the $L$-function $L_f(s,T)$ determine each other. From the last identity, one sees that $$C_f(s,T)\in
1+s\mathbb{Z}_p[[T]][[s]].$$We also view $C_f(s, T)$ as a power series in the single variable $s$ with coefficients in the $T$-adic complete field $\mathbb{Q}_p((T))$. The $C$-function $C_f(s,T)$ was shown to be $T$-adic entire in $s$ by Liu-Wan [@LWn].

Let $C(\triangle)$ be the cone generated by $\triangle$, and $M(\triangle)=M(\triangle)\cap \mathbb{Z}^n$. There is a degree function $\deg$ on $C(\triangle)$ which is $\mathbb{R}_{\geq0}$-linear and takes the values $1$ on each co-dimension $1$ face not containing $0$. For $a\not\in
C(\triangle)$, we define $\deg(a)=+\infty$.

A convex function on $[0,+\infty]$ which is linear between consecutive integers with initial value $0$ is called the infinite Hodge polygon of $\triangle$ if its slopes between consecutive integers are the numbers $\deg(a)$, $a\in M(\triangle)$. We denote this polygon by $H_{\triangle}^{\infty}$.

Liu-Wan [@LWn] also proved the following.

We have $$T-\text{adic NP of }C_{f}(s,T)\geq\text{ord}_p(q)(p-1)H_{\triangle}^{\infty},$$ where NP stands for Newton polygon.

The $T^a$-adic Newton polygon of $C_f(s,T;\mathbb{F}_{p^a})$ is called the absolute Newton polygon of $C_f(s,T;\mathbb{F}_{p^a})$.

\[gent\]If $p$ is sufficiently large, then the absolute $T$-adic Newton polygon of $C_f(s,T)$ is constant for a generic $f$. We call it the generic Newton polygon of $C_f(s,T)$.

The $\pi_m^a$-adic Newton polygon of $C_f(s,\pi_m;\mathbb{F}_{p^a})$ is called the absolute Newton polygon of $C_f(s,\pi_m;\mathbb{F}_{p^a})$.

Combine results of Gelfand-Kapranov-Zelevinsky [@GKZ], Adolphson-Sperber [@AS], Liu-Wei [@LW] with Grothendieck specialization lemma [@Ka], the absolute Newton polygon of $C_f(s,\pi_m)$ is constant for a generic $f$. We call it the generic Newton polygon of $C_f(s,\pi_m)$.

\[gen\] If $p$ is sufficiently large, then the generic Newton polygon of $C_f(s,\pi_m)$ is independent of $m$, and coincides with the generic Newton polygon of $C_f(s,T)$.

The generic Newton polygon of $C_f(s,\pi_m)$ for $m=1$ was studied by Wan [@Wa1; @Wa2].

In the rest of this section we assume that $\triangle\subset\mathbb{Z}$.

Let $0\neq a\in M(\triangle)$. We define $$\delta_{\in}(a)=\left\{
                           \begin{array}{ll}
                            1, & \hbox{ }\{\deg(a)\}=\{\deg(pi)\}\text{ for some }i\text{ with }ia>0,\deg(i)<\{\deg(a)\}, \\
                            0, & \hbox{ }{\rm otherwise},
                            \end{array}
                            \right.$$ where $\{\cdot\}$ is the fractional part of a real number. We also define $\delta_{\in}(0)=0$.

A convex function on $[0,+\infty]$ which is linear between consecutive integers with initial value $0$ is called the arithmetic polygon of $\triangle$ if its slopes between consecutive integers are the numbers $$\varpi_{\triangle}(a)=\lceil(p-1)\deg(a)\rceil-\delta_{\in}(a), a\in M(\triangle),$$ where $\lceil\cdot\rceil$ is the least integer equal or greater than a real number. We denote this polygon by $p_{\triangle}$.

We can prove the following.

\[arithhodge\]We have$$p_{\triangle}\geq (p-1)H_{\triangle}^{\infty}.$$ Moreover, they coincide at the point ${\rm Vol}(\triangle)$.

Let $D$ be the least common multiple of the nonzero endpoint(s) of $\triangle$. The main results of this paper are the following theorems.

\[main1\]If $p>3D$, then $$T-\text{adic NP of }C_{f}(s,T)\geq {\rm ord}_p(q)p_{\triangle}.$$

\[main2\]Let $f(x)=\sum\limits_{u\in\triangle}(a_ux^u,0,0,\cdots)$, and $p>3D$. Then there is a non-zero polynomial $H(y)\in\mathbb{F}_q[y_u\mid
u\in\triangle]$ such that $$T-\text{adic NP of }C_{f}(s,T)={\rm
ord}_p(q)p_{\triangle}$$ if and only if $H((a_u)_{u\in\triangle})\neq0$.

The above theorem implies Conjecture \[gent\] for $\triangle\subset\mathbb{Z}$.

\[main3\]Let $f(x)=\sum\limits_{u\in\triangle}(a_ux^u,0,0,\cdots)$, $p>3D$, and $m\geq1$. Then $$\pi_m-\text{adic NP of }C_{f}(s,\pi_m)={\rm
ord}_p(q)p_{\triangle}$$ if and only if $H((a_u)_{u\in\triangle})\neq0$.

The above theorem implies Conjecture \[gen\] for $\triangle\subset\mathbb{Z}$.

Note that, if $p\nmid D$, then $L(s,\pi_m)$ is a polynomial of degree $p^{m-1}{\rm Vol}(\triangle)$ for all $m\geq1$ by a result of Adolphson-Sperber [@AS] and a result of Liu-Wei [@LW]. We shall prove the following.

\[expsum\]Let $f(x)=\sum\limits_{u\in\triangle}(a_ux^u,0,0,\cdots)$, and $p>3D$. Then $$\pi_m-\text{adic NP of }L_{f}(s,\pi_m)\geq{\rm
ord}_p(q)p_{\triangle}\text{ on }[0,p^{m-1}{\rm Vol}(\triangle)]$$ with equality holding if and only if $H((a_u)_{u\in\triangle})\neq0$.

Zhu [@Zh1; @Zh2] and Blache-Férard [@BF] studied the Newton polygon of the $L$-function $L_{f}(s,\pi_m)$ for $m=1$.

Arithmetic estimate
===================

In this section $\triangle\subset\mathbb{Z}$, $A$ is a finite subset of $M(\triangle)\times\mathbb{Z}/(b),$ and $\tau$ is a permutation of $A$. We shall estimate $$\sum\limits_{a\in A}\lceil
\deg(pa-\tau(a)\rceil,$$ where $\deg(i,u)=\deg(i)$.

However, except in the ending paragraph, we assume that $\triangle=[0,d]$, $A$ is a finite subset of $\{1,2,\cdots\}\times\mathbb{Z}/(b)$.

Write $$\{x\}'=1+x-\lceil x\rceil=\left\{
                                                   \begin{array}{ll}
                                                     \{x\}, & \hbox{ if } \{x\}\neq0,\\
                                                     1, & \hbox{ if
}\{x\}=0.
                                                   \end{array}
                                                 \right.$$

We have$$\sum\limits_{a=0}^{m}(\delta_{<}-\delta_{\in})(a)=\#\{1\leq a\leq
d\{\frac{m}{d}\}'\mid \{\frac{m}{d}\}'<\{\frac{pa}{d}\}'\},$$ where $$\delta_{<}(a)=\left\{
                  \begin{array}{ll}
                    1, & \hbox{ } \{\frac{a}{d}\}'<\{\frac{pa}{d}\}'\\
                    0, & \hbox{ }{\rm otherwise}.
                  \end{array}
                \right.$$

Note that both $\delta_{\in}$ and $\delta_{<}$ have a period $d$ and have initial value $0$. So we may assume that $m<d$. We have $$\sum\limits_{a=0}^m\delta_{\in}(a)=\sum\limits_{a=1}^m
\sum\limits_{\stackrel{i=1}{pi\equiv a(
d)}}^{a-1}1$$$$=\sum\limits_{i=1}^{m-1}
\sum\limits_{\stackrel{a=i+1}{pi\equiv a( d)}}^{m}1=\#\{1\leq
i<m\mid i<d\{\frac{pi}{d}\}\leq m\}.$$And, by definition, $$\sum\limits_{a=0}^m\delta_{<}(a)=\#\{1\leq a\leq m\mid
a<d\{\frac{pa}{d}\}\}.$$The lemma now follows.

Let $A,B,C$ be sets with $A$ finite, and $\tau$ a permutation of $A$. Then $$\#\{a\in A\mid \tau(a)\in B,a\in C\}$$$$\geq\#\{a\in A\mid a\in B,a\in C\}-\#\{a\in A\mid a\not\in
B,a\not\in C\}.$$

We have$$\#\{a\in A \mid a\in B,\tau(a)\in B,a\in C\}$$$$\geq\#\{a\in A \mid a\in B,a\in C\}-\#\{a\in A\mid a\in
B,\tau(a)\not\in B\},$$ and $$\#\{a\in A\mid a\not\in B,\tau(a)\in B,a\in C\}$$$$\geq\#\{a\in
A\mid a\not\in B,\tau(a)\in B\}-\#\{a\in A\mid a\not\in B,a\not\in
C\}.$$ Note that $$\#\{a\in A\cap B\mid \tau(a)\not\in B\}=\#\{a\in A\setminus B\mid \tau(a)\in B\}.$$ So $$\#\{a\in A \mid \tau(a)\in B,a\in C\}$$$$\geq\#\{a\in A\mid a\in B,a\in C\}-\#\{a\in A\mid a\not\in
B,a\not\in C\}.$$The lemma is proved.For $a\in A$, we define $$\delta_{<}^{\tau}(a)=\left\{
                         \begin{array}{ll}
                           1, & \hbox{ } \{\deg(\tau(a))\}'<\{p\deg(a)\}',\\
                           0, & \hbox{otherwise.}
                         \end{array}
                       \right.$$

If $p>3d$, and $$A=(\{1,\cdots,m-1\}\times\mathbb{Z}/(b))\cup\{(m,i_0),\cdots,(m,i_{l-1})\},$$ then$$\sum\limits_{a\in A}\delta_{<}^{\tau}(a)\geq
b\sum\limits_{a=0}^{m-1}(\delta_{<}-\delta_{\in})(a)+l(\delta_{<}-\delta_{\in})(m).$$

First we assume that $l=0$. We have $$\sum\limits_{a\in
A}\delta_{<}^{\tau}(a)\geq\#\{a\in
A,\{\deg(\tau(a))\}'\leq\{\frac{m-1}{d}\}'<\{p\deg(a)\}'\}.$$ Applying the last lemma with $$B=\{a\in
A,\{\deg(a)\}'\leq\{\frac{m-1}{d}\}'\},$$ and $$C=\{a\in
A,\{\frac{m-1}{d}\}'<\{p\deg(a)\}'\},$$ we get $$\sum\limits_{a\in
A}\delta_{<}^{\tau}(a)\geq\#\{a\in
A,\{\deg(a)\}'\leq\{\frac{m-1}{d}\}'<\{p\deg(a)\}'\}$$$$-\#\{a\in
A,\{\deg(a)\}'>\{\frac{m-1}{d}\}'\geq\{p\deg(a)\}'\}.$$ We have $$\#\{a\in
A,\{\deg(a)\}'\leq\{\frac{m-1}{d}\}'<\{p\deg(a)\}'\}$$ $$=b([\frac{m}{d}]+1)\#\{1\leq a\leq d\{\frac{m-1}{d}\}'\mid \{\frac{m-1}{d}\}'<\{\frac{pa}{d}\}'\}.$$ We also have $$\#\{a\in
A,\{\deg(a)\}'>\{\frac{m-1}{d}\}'\geq\{p\deg(a)\}'\}$$ $$=b[\frac{m}{d}]\#\{d\geq a>d\{\frac{m-1}{d}\}'\mid \{\frac{m-1}{d}\}'\geq\{\frac{pa}{d}\}'\}.$$ $$=b[\frac{m}{d}]\#\{1\leq a\leq d\{\frac{m-1}{d}\}'\mid \{\frac{m-1}{d}\}'<\{\frac{pa}{d}\}'\}.$$

It follows that $$\sum\limits_{a\in
A}\delta_{<}^{\tau}(a)\geq b\#\{1\leq a\leq d\{\frac{m-1}{d}\}'\mid
\{\frac{m-1}{d}\}'<\{\frac{pa}{d}\}'\}$$$$\geq
b\sum\limits_{a=0}^{m-1}(\delta_{<}-\delta_{\in})(a).$$

Secondly we assume that $\delta_{\in}(m)=0$. Extend the action of $\tau$ trivially from $A$ to $\{1,\cdots,m\}\times\mathbb{Z}/(b)$. By what we just proved, $$\sum\limits_{a\in
\{1,\cdots,m\}\times\mathbb{Z}/(b)}\delta_{<}^{\tau}(a)\geq
b\sum\limits_{a=0}^{m}(\delta_{<}-\delta_{\in})(a).$$ It follows that$$\sum\limits_{a\in A}\delta_{<}^{\tau}(a)\geq
b\sum\limits_{a=0}^{m}(\delta_{<}-\delta_{\in})(a)-(b-l)\delta_{<}(m)$$$$\geq
b\sum\limits_{a=0}^{m-1}(\delta_{<}-\delta_{\in})(a)+l(\delta_{<}-\delta_{\in})(m).$$

Finally we assume that $\delta_{\in}(m)=1$. We have $$\sum\limits_{a\in
A}\delta_{<}^{\tau}(a)\geq\#\{a\in
A,\{\deg(\tau(a))\}'\leq\{\frac{m-1}{d}\}'<\{p\deg(a)\}'\}.$$ Applying the last lemma with $$B=\{a\in
A,\{\deg(a)\}'\leq\{\frac{m-1}{d}\}'\},$$ and $$C=\{a\in
A,\{\frac{m-1}{d}\}'<\{p\deg(a)\}'\},$$ we get $$\sum\limits_{a\in
A}\delta_{<}^{\tau}(a)\geq\#\{a\in
A,\{\deg(a)\}'\leq\{\frac{m-1}{d}\}'<\{p\deg(a)\}'\}$$$$-\#\{a\in
A,\{\deg(a)\}'>\{\frac{m-1}{d}\}'\geq\{p\deg(a)\}'\}.$$ We have $$\#\{a\in
A,\{\deg(a)\}'\leq\{\frac{m-1}{d}\}'<\{p\deg(a)\}'\}$$ $$=b([\frac{m}{d}]+1)\#\{1\leq a\leq d\{\frac{m-1}{d}\}'\mid \{\frac{m-1}{d}\}'<\{\frac{pa}{d}\}'\}.$$ We also have $$\#\{a\in
A,\{\deg(a)\}'>\{\frac{m-1}{d}\}'\geq\{p\deg(a)\}'\}$$ $$=b[\frac{m}{d}]\#\{d\geq a>d\{\frac{m-1}{d}\}'\mid \{\frac{m-1}{d}\}'\geq\{\frac{pa}{d}\}'\}
+l1_{\{\frac{m-1}{d}\}'\geq\{\frac{pm}{d}\}'}.$$ $$=b[\frac{m}{d}]\#\{1\leq a\leq d\{\frac{m-1}{d}\}'\mid \{\frac{m-1}{d}\}'<\{\frac{pa}{d}\}'\}
+l1_{\{\frac{m-1}{d}\}'\geq\{\frac{pm}{d}\}'}.$$

It follows that $$\sum\limits_{a\in
A}\delta_{<}^{\tau}(a)\geq b\#\{1\leq a\leq d\{\frac{m-1}{d}\}'\mid
\{\frac{m-1}{d}\}'<\{\frac{pa}{d}\}'\}$$ $$-l1_{\{\frac{m-1}{d}\}'\geq\{\frac{pm}{d}\}'}\geq
b\sum\limits_{a=0}^{m-1}(\delta_{<}-\delta_{\in})(a)+l(\delta_{<}-\delta_{\in})(m).$$ The proof of the theorem is completed.

If $p>3d$, $A$ is of cardinality $bm+l$ with $0\leq l<b$, then $$\sum\limits_{a\in A}(\lceil
p\deg(a)\rceil-\lceil\deg(a)\rceil+\delta_{<}^{\tau}(a))\geq
bp_{\triangle}(m)+l\varpi(m).$$ Moreover, the strict inequality holds if $A$ is not of the form $$(\{1,\cdots,m-1\}\times\mathbb{Z}/(b))\cup\{(m,i_0),\cdots,(m,i_{l-1})\}.$$

We may assume that $A$ is not of the form $$(\{1,\cdots,m-1\}\times\mathbb{Z}/(b))\cup\{(m,i_0),\cdots,(m,i_{l-1})\}.$$ There is an element $a_0\in A$ with $\deg(a_0)>\deg(m)$. Set $A'=A\setminus\{a_0\}$ and $$\tau'(a)=\left\{
             \begin{array}{ll}
               \tau(a), & \hbox{ } a\neq\tau^{-1}(a_0),\\
               \tau(a_0), & \hbox{}a=\tau^{-1}(a_0).
             \end{array}
           \right.$$ Then $$\sum\limits_{a\in A}(\lceil
p\deg(a)\rceil-\lceil\deg(a)\rceil+\delta_{<}^{\tau}(a))$$$$>\sum\limits_{a\in
A'}(\lceil
p\deg(a)\rceil-\lceil\deg(a)\rceil+\delta_{<}^{\tau'}(a))+\varpi(m).$$The theorem now follows by induction.

We now assume that $\triangle\subset\mathbb{Z}$, and $A$ is a finite subset of $M(\triangle)\times\mathbb{Z}/(b).$ For an integer $m\geq1$, we define $$A_m=\{a\in M(\triangle)\mid \varpi(a)\leq p_{\triangle}(m)-p_{\triangle}(m-1)\}.$$

\[order-estimate\]If $p>3D$, then $$\sum\limits_{a\in A}\lceil\deg(pa-\tau(a))\rceil\geq bp_{\triangle}(m).$$ Moreover, the strict inequality holds if $m$ is a turning point of $p_{\triangle}$, and $A\neq A_m\times\mathbb{Z}/(b)$.

We define ${\rm sgn}((i,u))={\rm sgn}(i)$. Write $$A_i=\{a\in A\mid {\rm sgn}(a)=(-1)^i\},$$ and $$A_{ij}=\{a\in A\mid {\rm sgn}(a)=(-1)^i,{\rm
sgn}(\tau(a))=(-1)^j\}.$$ Define a new permutation $\tau_0$ as follows:

-   $\tau_0$ is identity on $A_0$.

-   $\tau_0=\tau$ on $A_{11}$ and $A_{22}$.

-   $\tau_0$ maps $A_1\setminus A_{11}$ to $A_1\setminus\tau(A_1)$.

-   $\tau_0$ maps $A_2\setminus A_{22}$ to $A_2\setminus\tau(A_2)$.

We have $$\sum\limits_{a\in A}\lceil\deg(pa-\tau(a))\rceil\geq
\sum\limits_{a\in A}(\lceil
p\deg(a)\rceil-\lceil\deg(\tau_0(a))\rceil+1_{\{\deg(\tau_0(a))\}'<\{p\deg(a)\}'}).$$The theorem now follows the last one.

The $T$-adic Dwork Theory
=========================

In this section we review the $T$-adic analogue of Dwork theory on exponential sums.

Let $$E(t)=\exp(\sum_{i=0}^{\infty}\frac{t^{p^i}}{p^i})=\sum\limits_{i=0}^{+\infty}\lambda_it^i
\in 1+t{\mathbb Z}_p[[t]]$$ be the $p$-adic Artin-Hasse exponential series. Define a new $T$-adic uniformizer $\pi$ of ${\mathbb
Q}_p((T))$ by the formula $E(\pi)=1+T$. Let $\pi^{1/D}$ be a fixed $D$-th root of $\pi$. Let $$L=\{\sum_{i\in M(\triangle)}c_i\pi^{\deg(i)}x^i:\
 c_i\in\mathbb{Z}_q[[\pi^{1/D}]] \}.$$

Let $a\mapsto\hat{a}$ be the Teichmüller lifting. One can show that the series $$E_f(x) :=\prod\limits_{a_i\neq0}E(\pi \hat{a}_ix^i)\in L.$$ Note that the Galois group of $\mathbb{Q}_q$ over $\mathbb{Q}_p$ can act on $L$ but keeping $\pi^{1/D}$ as well as the variable $x$ fixed. Let $\sigma$ be the Frobenius element in the Galois group such that $\sigma(\zeta)=\zeta^p$ if $\zeta$ is a $(q-1)$-th root of unity. Let $\Psi_p$ be the operator on $L$ defined by the formula $$\Psi_p(
 \sum\limits_{i\in
M(\triangle)} c_ix^i)=\sum\limits_{i\in M(\triangle)} c_{pi}x^i.$$ Then $\Psi:=\sigma^{-1}\circ\Psi_p\circ E_f$ acts on the $T$-adic Banach module $$B=\{\sum\limits_{i\in M(\triangle)}c_i\pi^{\deg(i)}x^i \in L,\
 \text{\rm ord}_T(c_i)\rightarrow+\infty
 \text{ if }\deg(i)\rightarrow+\infty\}.$$ We call it Dwork’s $T$-adic semi-linear operator because it is semi-linear over $\mathbb{Z}_q[[\pi^{\frac{1}{D}}]]$ .

Let $b=\log_pq$. Then the $b$-iterate $\Psi^b$ is linear over $\mathbb{Z}_q[[\pi^{1/D}]]$, since $$\Psi^{b}=\Psi_p^{b}\circ
\prod\limits_{i=0}^{b-1}E_{f}^{\sigma^i}(x^{p^i}).$$ One can show that $\Psi$ is completely continuous in the sense of Serre [@Se]. So $\det(1-\Psi^bs\mid
B/\mathbb{Z}_{q}[[\pi^{\frac{1}{D}}]])$ and $\det(1-\Psi s\mid
B/\mathbb{Z}_p[[\pi^{\frac{1}{D}}]])$ are well-defined.

We now state the $T$-adic Dwork trace formula[@LWn].

We have $$C_f(s,T)=\det(1-\Psi^bs\mid
B/\mathbb{Z}_{q}[[\pi^{\frac{1}{D}}]]).$$

The Dwork semi-linear operator
==============================

Write $$E_f(x)= \sum\limits_{i\in M(\triangle)}\gamma_ix^i,$$ and $$\det(1-\Psi s\mid B/\mathbb{Z}_p[[\pi^{\frac{1}{D}}]])=\sum\limits_{i=0}^{+\infty}(-1)^ic_is^i.$$ Let $O(\pi^{\alpha})$ denotes any element of $\pi$-adic order $\geq
\alpha$. In this section we prove the following.

\[general-points\]Let $p>3D$. Then $$\text{ord}_{\pi}(c_{bm})\geq
bp_{\triangle}(m).$$ Moreover, if $m<{\rm Vol}(\triangle)$ is a turning point of $p_{\triangle}$, then $$c_{bm}=\pm{\rm Norm}(\det(\gamma_{pi-j})_{i,j\in A_m})+O(\pi^{bp_{\triangle}(m)+1/D}),$$ where Norm is the norm map from $\mathbb{Q}_q(\pi^{1/D})$ to $\mathbb{Q}_p(\pi^{1/D})$.

Fix a normal basis $\bar{\xi}_u$, $u\in\mathbb{Z}/(b)$ of $\mathbb{F}_q$ over $\mathbb{F}_p$. Let $\xi_u$ be their Terchmüller lift of $\bar{\xi}_u$. Then $\xi_u$, $u\in\mathbb{Z}/(b)$ is a normal basis of $\mathbb{Q}_q$ over $\mathbb{Q}_p$, and $\sigma$ acts on the basis $\xi_u$ , $u\in\mathbb{Z}/(b)$ as a permutation. Let $(\gamma_{(i,u),(j,\omega)})_{i,j\in M(\triangle),1\leq u,\omega\leq
b}$ be the matrix of $\Psi$ on $B\otimes_{\mathbb{Z}_p}\mathbb{Q}_p(\pi^{1/D})$ with respect to the basis $\{\xi_ux^i\}_{i\in M(\triangle),1\leq u\leq b}$. Then $$c_{bm}=\sum\limits_{A}\text{det}((\gamma_{(i,u),(j,\omega)})_{(i,u),(j,\omega)\in
A}),$$ where $A$ runs over all subsets of $
M(\triangle)\times\mathbb{Z}/(b)$ with cardinality $bm$. One can show that $$\det(\gamma_{i,j})_{i,j\in
A_m\times\mathbb{Z}/(b)}=\pm{\rm Norm}(\det((\gamma_{pi-j})_{i,j\in
A_m})).$$Therefore the theorem follows from the following.

\[det\] Let $A\subset M(\triangle)\times\mathbb{Z}/(b)$ be a subset of cardinality $bm$. If $p>3D$, then $${\rm ord}_T(\det(\gamma_{(i,u),(j,\omega)})_{(i,u),(j,\omega)\in
A})\geq bp_{\triangle}(m).$$ Moreover, if $m<{\rm Vol}(\triangle)$ is a turning point of $p_{\triangle}$, and $A\neq
A_m\times\mathbb{Z}/(b)$, then the strict inequality holds.

one can show that $\gamma_i=O(\pi^{\lceil\deg(i)\rceil})$. And, from the equality $$(\xi_{u}\gamma_{pr-l})^{\sigma^{-1}}=\sum\limits_{\omega=1}^{b}\gamma_{(r,w),(l,u)}\xi_{\omega},$$ we infer that $$\gamma_{i,j}=O(\pi^{\lceil\deg(pi-j)\rceil}).$$ So we have $$\sum\limits_{a\in A}{\rm ord}_{\pi}(\gamma_{a,\tau(a)})\geq\sum\limits_{a\in A}\lceil\deg(pa-\tau(a))\rceil\geq
bp_{\triangle}(m).$$ Moreover, if $m$ is a turning point of $p_{\triangle}$, and $A\neq A_m\times\mathbb{Z}/(b)$, then the strict inequality holds.

The Hasse polynomial
====================

In this section we study $\det(\gamma_{pi-j})_{i,j\in A_m}$.

For each positive integer $m$, we define $S_m^0$ to be the set of permutations $\tau$ of $A_m$ satisfying $\tau(0)=0$, and $$\frac{\tau(a)}{d({\rm sgn}(a))}\geq \deg(pa)-\lceil\deg(pa)-\deg(n)\rceil,\ a\neq0,$$ where $n$ is the element of maximal degree in $A_m\cap({\rm
sgn}(a)\mathbb{N})$.

Let $p>3D$, $m<{\rm Vol}(\triangle)$ a turning point of $p_{\triangle}$, and $\tau$ a permutation of $A_m$. Then $$\sum\limits_{a\in A_m}\lceil\deg(pa-\tau(a))\rceil\geq
 p_{\triangle}(m),$$ with equality holding if and only if $\tau\in S_m^0$.

We assume that $M(\triangle) =\mathbb{N}$. The other cases can be proved similarly. In this case, ${\rm sgn}(a)=+1$ if $a\neq0$, and the element $n$ of maximal degree in $A_m\cap{\rm
sgn}(a)\mathbb{N}$ is $m-1$. Let $d$ be the nonzero endpoint of $\triangle$. For $a=0$, we have $$\lceil\deg(pa-\tau(a))\rceil\geq0$$ with equality holding if and only if $\tau(a)=0$. For $a\neq0$, we have $$\lceil\frac{pa-\tau(a)}{d}\rceil\geq\lceil\frac{pa-n}{d}\rceil$$ with equality holding if and only if $$\frac{\tau(a)}{d}\geq \frac{pa}{d}-\lceil\frac{pa-n}{d}\rceil.$$ It follows that$$\sum\limits_{a\in
A_m}\lceil\deg(pa-\tau(a))\rceil\geq\sum\limits_{a=1}^n\lceil\deg(pa-n)\rceil=
 p_{\triangle}(m)$$ with equality holding if and only if $\tau\in S_m^0$. The theorem is proved.

We have $$\gamma_i=\pi^{\lceil\deg(i)\rceil}\sum\limits_{\stackrel{\sum\limits_{j\in\triangle}jn_j=i}
{\sum\limits_{j\in\triangle}n_j =\lceil\deg(i)\rceil}}
\prod_{j\in\triangle}\lambda_{n_j}\hat{a}_j^{n_j}
+O(\pi^{\lceil\deg(i)\rceil+1}).$$

We have $$\gamma_i=\sum\limits_{\sum\limits_{j\in\triangle}jn_j=i,n_j\geq0}\pi^{\sum\limits_{j\in\triangle}n_j}
\prod_{j\in\triangle}\lambda_{n_j}\hat{a}_j^{n_j}.$$ We also have that$$\sum\limits_{j\in\triangle}n_j \geq\lceil\deg(i)\rceil\text{
if }\sum\limits_{j\in\triangle}jn_j=i.$$ The lemma now follows.

For each positive integer $m$, we define $$H_m(y)=\sum\limits_{\tau\in
S_m^0}\text{sgn}(\tau)\prod_{i\in A_m}
\sum\limits_{\stackrel{\sum\limits_{j\in\triangle}jn_j=pi-\tau(i)}{\sum\limits_{j\in\triangle}n_j
=\lceil\deg(pi-\tau(i))\rceil}}
\prod_{j\in\triangle}\lambda_{n_j}y_j^{n_j}\in\mathbb{Z}_p[y_j\mid
j\in\triangle].$$

\[special-points\]Let $p>3D$, and $m<{\rm Vol}(\triangle)$ a turning point of $p_{\triangle}$. Then $$\det(\gamma_{pi-j})_{i,j\in A_m}=H_m((\hat{a}_j)_{j\in\triangle})\pi^{p_{\triangle}(m)}+O(\pi^{p_{\triangle}(m)+1/D}).$$

Let $S_m$ be the set of permutations of $A_m$. We have $$\det(\gamma_{pi-j})_{i,j\in A_m}=\sum\limits_{\tau\in S_m}\prod_{i\in A_m}\gamma_{pi-\tau(i)}.$$ The theorem now follows from the last two lemmas.

The reduction of $H_m$ modulo $p$ is denoted as $\overline{H}_m$, and is called the Hasse polynomial of $\triangle$ at $m$.

If $p>3D$, and $m<\text{Vol}(\triangle)$ is a turning point of $p_{\triangle}$, then $\overline{H}_m$ is non-zero.

Define $\deg(y_j)=|j|$. Then $$\prod_{i\in A_m}
\sum\limits_{\stackrel{\sum\limits_{j\in\triangle}jn_j=pi-\tau(i)}{\sum\limits_{j\in\triangle}n_j
=\lceil\deg(pi-\tau(i))\rceil}}
\prod_{j\in\triangle}\lambda_{n_j}y_j^{n_j}$$ has degree $$\sum\limits_{i\in A_m}|pi-\tau(i)|=\sum\limits_{i\in A_m}{\rm sgn}(i)(pi-\tau(i))$$ $$=p\sum\limits_{i\in A_m}{\rm sgn}(i)i-\sum\limits_{i\in A_m}{\rm sgn}(i)\tau(i)$$ $$\geq p\sum\limits_{i\in A_m}{\rm sgn}(i)i-\sum\limits_{i\in A_m}{\rm sgn}(\tau(i))\tau(i)$$ $$\geq (p-1)\sum\limits_{i\in A_m}{\rm sgn}(i)i,$$ with equality holding if and only if $\tau$ preserves the sign. Therefore it suffices to show that the reduction of $$\sum\limits_{\tau\in
S_m^1}\text{sgn}(\tau)\prod_{i\in A_m}
\sum\limits_{\stackrel{\sum\limits_{j\in\triangle}jn_j=pi-\tau(i)}{\sum\limits_{j\in\triangle}n_j
=\lceil\deg(pi-\tau(i))\rceil}}
\prod_{j\in\triangle}\lambda_{n_j}y_j^{n_j},$$ where $S_m^1$ consists of the sign-preserving permutations of $S_m^0$, is nonzero. One can can prove this by the maximal-monomial-locating technique of Zhu [@Zh1], as was used by Blache-Férard [@BF].

We define $H=\prod_m\overline{H}_m$, where the product is over all turning points $m<{\rm Vol}(\triangle)$ of $p_{\triangle}$.

\[non-zero\]If $p>3D$, then $H$ is non-zero.

This follows from the last theorem.

Proof of the main theorem
=========================

In this section we prove the main theorems of this paper.

The Newton polygon of $\det(1-\Psi^bs^b\mid
B/\mathbb{Z}_{q}[[\pi^{\frac{1}{D}}]])$ coincides with that of $\det(1-\Psi s\mid B/\mathbb{Z}_p[[\pi^{\frac{1}{D}}]])$.

Note that $$\det(1-\Psi^b s\mid B/\mathbb{Z}_p[[\pi^{\frac{1}{D}}]])
={\rm Norm}(\det(1-\Psi^b s\mid
B/\mathbb{Z}_q[[\pi^{\frac{1}{D}}]])),$$ where Norm is the norm map from $\mathbb{Z}_q[[\pi^{\frac{1}{D}}]]$ to $\mathbb{Z}_p[[\pi^{\frac{1}{D}}]]$. The lemma now follows from the equality $$\prod\limits_{\zeta^b=1}\det(1-\Psi\zeta s\mid B/\mathbb{Z}_p[[\pi^{\frac{1}{D}}]])
=\det(1-\Psi^b s^b\mid B/\mathbb{Z}_p[[\pi^{\frac{1}{D}}]]).$$

The $T$-adic Newton polygon of $\det(1-\Psi^bs\mid
B/\mathbb{Z}_{q}[[\pi^{\frac{1}{D}}]])$ is the lower convex closure of the points $$(m,\text{ord}_{T}(c_{bm})),\ m=0,1,\cdots.$$

By the last lemma, the $T$-adic Newton polygon of the power series $\det(1-\Psi^bs^b\mid B/\mathbb{Z}_{q}[[\pi^{\frac{1}{D}}]])$ is the lower convex closure of the points $$(i,\text{ord}_{T}(c_i)),\ i=0,1,\cdots.$$ It is clear that $(i,\text{ord}_{T}(c_i))$ is not a vertex of that polygon if $b\nmid i$. So that Newton polygon is the lower convex closure of the points $$(bm,\text{ord}_{T}(c_{bm})),\ m=0,1,\cdots.$$ It follows that the $T$-adic Newton polygon of $\det(1-\Psi^bs\mid
B/\mathbb{Z}_{q}[[\pi^{\frac{1}{D}}]])$ is the lower convex closure of the points $$(m,\text{ord}_{T}(c_{bm})),\ m=0,1,\cdots.$$ We now prove Theorem \[main1\], which says that $$T-\text{adic NP of }C_{f}(s,T)\geq
{\rm ord}_p(q)p_{\triangle}\text{ if }p>3D.$$

[*Proof of Theorem \[main1\].* ]{}Combine the last theorem with the $T$-adic Dwork’s trace formula, we see that the $T$-adic Newton polygon of $C_{f}(s,T)$ is the lower convex closure of the points $$(m,\text{ord}_{T}(c_{bm})),\ m=0,1,\cdots.$$ The theorem now follows from the estimate $$\text{ord}_{\pi}(c_{bm})\geq bp_{\triangle}(m).$$

We now prove Theorem \[arithhodge\], which says that $$p_{\triangle}\geq(p-1)H_{\triangle}^{\infty}$$ with equality holding at the point ${\rm Vol}(\triangle)$.

[*Proof of Theorem \[arithhodge\].* ]{}We assume that $\triangle=[0,d]$. The other cases can be proved similarly. It suffices to show that $$p_{\triangle}\geq(p-1)H_{\triangle}^{\infty}\text{ on }[0,d]$$ with equality holding at the point $d$. Let $0<m\leq d$. We have $$p_{\triangle}(m)=\sum\limits_{0\leq a<m}(\lceil\frac{(p-1)a}{d}\rceil-\delta_{\in}(a))$$ $$=\sum\limits_{0\leq a<m}(\lceil \frac{pa}{d}\rceil-\lceil \frac{a}{d}\rceil+\delta_{<}(a)-\delta_{\in}(a))$$ $$=\sum\limits_{1\leq a<m}(\lceil \frac{pa}{d}\rceil-1)+\sum\limits_{1\leq a<m}(\delta_{<}-\delta_{\in})(a)$$ $$=\sum\limits_{1\leq a<m}[\frac{pa}{d}]+\sum\limits_{1\leq a<m}(\delta_{<}-\delta_{\in})(a)$$ $$=p\sum\limits_{1\leq a<m}\frac{a}{d}-\sum\limits_{1\leq a<m}\{\frac{pa}{d}\}
+\sum\limits_{\stackrel{1\leq a<m}{d\{\frac{pa}{d}\}\geq m}}1$$ $$=(p-1)H_{\triangle}^{\infty}(m)+\sum\limits_{1\leq a<m}\frac{a}{d}-\sum\limits_{1\leq a<m}\{\frac{pa}{d}\}+\#\{1\leq
a<m\mid d\{\frac{pa}{d}\}\geq m\}.$$ In particular, we have $$p_{\triangle}(d)=(p-1)H_{\triangle}^{\infty}(d).$$ Note that $$\sum\limits_{1\leq a<m}\frac{a}{d}-\sum\limits_{1\leq a<m}\{\frac{pa}{d}\}+\#\{1\leq
a<m\mid d\{\frac{pa}{d}\}\geq m\}$$$$\geq\sum\limits_{1\leq
a<m}\frac{a}{d}-\sum\limits_{\stackrel{1\leq
a<m}{d\{\frac{pa}{d}\}<m}}\{\frac{pa}{d}\}\geq0.$$ It follows that $$p_{\triangle}(m)\geq(p-1)H_{\triangle}^{\infty}(m).$$

\[spe\] Let $A(s,T)$ be a $T$-adic entrie series in $s$ with unitary constant term. If $0\neq|t|_p<1$, then $$t-adic\text{ NP of
}A(s,t)\geq T-adic\text{ NP of }A(s,T),$$where NP is the short for Newton polygon. Moreover, the equality holds for one $t$ if and only if it holds for all $t$.

Write $$A(s,T)=\sum\limits_{i=0}^{\infty}a_i(T)s^i.$$ The inequality follows from the fact that $a_i(T)\in\mathbb{Z}_q[[T]]$. Moreover, $$t-adic\text{ NP of
}A(s,t)=T-adic\text{ NP of }A(s,T)$$if and only if $$a_i(T)\in T^e\mathbb{Z}_q[[T]]^{\times}$$ for every turning point $(i,e)$ of the $T$-adic Newton polygon of $A(s,T)$. It follows that the equality holds for one $t$ iff it holds for all $t$.

Let $f(x)=\sum\limits_{u\in\triangle}(a_ux^u,0,0,\cdots)$, and $p>3D$. If the equality $$\pi_m-\text{adic NP of }C_{f}(s,\pi_m)={\rm
ord}_p(q)p_{\triangle}$$ for one $m\geq1$, then it holds for all $m\geq1$, and we have $$T-\text{adic NP of }C_{f}(s,T)={\rm
ord}_p(q)p_{\triangle}.$$

This follows from Theorems \[general-points\] and \[spe\].

Let $f(x)=\sum\limits_{u\in\triangle}(a_ux^u,0,0,\cdots)$. Then $$\pi_m-\text{adic NP of }C_{f}(s,\pi_m)={\rm
ord}_p(q)p_{\triangle}$$ if and only if $$\pi_m-\text{adic NP of }L_{f}(s,\pi_m)={\rm
ord}_p(q)p_{\triangle}\text{ on }[0,p^{m-1}{\rm Vol}(\triangle)].$$

Assume that $ L_f(s,
\pi_m)=\prod\limits_{i=1}^{p^{m-1}d}(1-\beta_is) $. Then $$C_f(s,\pi_m )= \prod\limits_{j=0}^{\infty} L_f(q^js, \pi_m)=
\prod\limits_{j=0}^{\infty}\prod\limits_{i=1}^{p^{m-1}d}(1-\beta_iq^js).$$ Therefore the slopes of the $q$-adic Newton polygon of $C_{f}(s,\pi_m)$ are the numbers $$j+{\rm ord}_q(\beta_i),\ 1\leq
i\leq p^{m-1}{\rm Vol}(\triangle), j=0,1,\cdots.$$ One can show that the slopes of $p_{\triangle}$ are the numbers $$j(p-1)+p_{\triangle}(i)-p_{\triangle}(i-1),\ 1\leq i\leq
p^{m-1}{\rm Vol}(\triangle), j=0,1,\cdots.$$ It follows that $$\pi_m-\text{adic NP of }C_{f}(s,\pi_m)={\rm
ord}_p(q)p_{\triangle}$$ if and only if $$\pi_m-\text{adic NP of }L_{f}(s,\pi_m)={\rm
ord}_p(q)p_{\triangle}\text{ on }[0,p^{m-1}{\rm Vol}(\triangle)].$$

We now prove Theorems \[main2\], \[main3\] and \[expsum\]. By the above theorems, it suffices to prove the following.

Let $f(x)=\sum\limits_{u\in\triangle}(a_ux^u,0,0,\cdots)$, and $p>3D$. Then $$\pi_1-\text{adic NP of }L_{f}(s,\pi_1)={\rm
ord}_p(q)p_{\triangle}\text{ on }[0,{\rm Vol}(\triangle)]$$ if and only if $H((a_u)_{u\in\triangle})\neq0$.

It is known that the $q$-adic Newton polygon of $L_{f}(s,\pi_1)$ coincides with $H_{\triangle}^{\infty}$ at the point ${\rm Vol}(\triangle)$. By Theorem \[arithhodge\], $p_{\triangle}$ coincide with $(p-1)H_{\triangle}^{\infty}$ at the point ${\rm Vol}(\triangle)$, It follows that the $\pi_1$-adic Newton polygon of $L_{f}(s,\pi_1)$ coincides with ${\rm ord}_p(q)p_{\triangle}$ at the point ${\rm
Vol}(\triangle)$. Therefore it suffices to show that $$\pi_1-\text{adic NP of }L_{f}(s,\pi_1)={\rm
ord}_p(q)p_{\triangle}\text{ on }[0,{\rm Vol}(\triangle)-1]$$ if and only if $H((a_u)_{u\in\triangle})\neq0$.

From the identity $$C_f(s,\pi_1 )= \prod\limits_{j=0}^{\infty}
L_f(q^js, \pi_1),$$ and the fact the $q$-adic orders of the reciprocal roots of $L_{f}(s,\pi_1)$ are no greater than $1$, we infer that $$\pi_1-\text{adic NP of }L_{f}(s,\pi_1) =\pi_1-\text{adic NP of
}C_{f}(s,\pi_1)\text{ on }[0,{\rm Vol}(\triangle)-1].$$Therefore it suffices to show that $$\pi_1-\text{adic NP of }C_{f}(s,\pi_1)={\rm
ord}_p(q)p_{\triangle}\text{ on }[0,{\rm Vol}(\triangle)-1]$$ if and only if $H((a_u)_{u\in\triangle})\neq0$. The theorem now follows from the $T$-adic Dwork trace formula and Theorems \[general-points\] and \[special-points\].

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