# Differential forms on the curves associated to Appell-Lauricella
hypergeometric series and the Cartier operator on them

Ryo Ohashi and Shushi Harashita

###### Abstract

Archinard studied the curve $C$ over $\mathbb{C}$ associated to an Appell-
Lauricella hypergeometric series and differential forms on its
desingularization. In this paper, firstly as a generalization of Archinard’s
results, we describe a partial desingularization of $C$ over a perfect field
$K$ under a mild condition on its characteristic and the space of global
sections of its dualizing sheaf, especially we give an explicit basis of it.
Secondly, when the characteristic is positive, we show that the Cartier
operator on the space can be defined and describe it in terms of Appell-
Lauricella hypergeometric series.

##  Introduction

An Appell-Lauricella hypergeometric series is defined to be a period of a
family of degenerations of superelliptic curves, see Section 2 for the
details. We are interested in relations between the geometry of the associated
family of the (possibly singular) curves and the analysis of Appell-Lauricella
hypergeometric series. Among them, we shall describe Cartier-Manin matrices of
the curves in terms of Appell-Lauricella hypergeometric series (Theorem 6.5).
For this, we need to find nice explicit basis of the space of regular
differential forms on the curves.

Let us start with recalling the most classical case: a relation between
elliptic curves and Gauss’ hypergeometric series. Gauss’ hypergeometric series
is defined to be

$F(a,b,c\,;z):=\sum_{n=0}^{\infty}\frac{(a\,;n)(b\,;n)}{(c\,;n)(1\,;n)}z^{n},$
(1.1)

where $(x\,;n):=\prod_{j=1}^{n}(x+n-j)$ and
$a,b,c\in\mathbb{C},-c\notin\mathbb{N}$. It is well-known that $F(a,b,c\,;z)$
satisfies the differential equation $\mathcal{D}F(a,b,c\,;z)=0$ with

$\mathcal{D}=z(1-z)\frac{d^{2}}{dz^{2}}+\bigl{(}c-(a+b+1)z\bigr{)}\frac{d}{dz}-ab.$
(1.2)

According to [5, Theorem V.4.1], elliptic curves $E:y^{2}=x(x-1)(x-z)$ in
characteristic $p>0$ are supersingular if and only if $H_{p}(z)=0$, where

$H_{p}(z):=\sum_{i=0}^{m}\binom{m}{i}^{\\!\\!2}z^{i}$

with $m=(p-1)/2$, and $H_{p}(z)$ satisfies the differential equation
${\mathcal{D}}^{\prime}H_{p}(z)=0$ with

${\mathcal{D}}^{\prime}=4z(1-z)\frac{d^{2}}{dz^{2}}+4(1-2z)\frac{d}{dz}-1.$
(1.3)

Note that (1.3) is obtained by (1.2) for $(a,b,c)=(1/2,1/2,1)$, and $H_{p}(z)$
is obtained by truncating the series $(-1)^{(p-1)/2}F(1/2,1/2,1\,;z)$ by
degree $(p-1)/2$.

In this paper, we study the curve $C$ associated to an Appell-Lauricella
hypergeometric series (see Definition 2.3), which is a certain generalization
of Gauss’ hypergeometric series. Let $K$ be a perfect field, then the curve
$C$ is defined by the affine equation

$C:y^{N}=\prod_{i=0}^{r}(x-\lambda_{i})^{A_{i}},\quad i\neq
j\,\Rightarrow\lambda_{i}\neq\lambda_{j},\quad\lambda_{0},\ldots,\lambda_{r}\in\accentset{{\cc@style\underline{\mskip
9.0mu}}}{K}.$

Archinard [1, Section 2] gave the desingularization $X$ of $C$, especially for
$K=\mathbb{C}$ and studied the space of regular differential forms on $X$. For
later discussions, we first generalize the result to the case of a perfect
field $K$ under a mild assumption on its characteristic. We review basic
properties of the curve $C$ associated to an Appell-Lauricella hypergeometric
series. We construct the explicit desingularization of $C$ in Section 2, and
describe the space of regular differential forms on it in Section 3.

As well as nonsingular curves, we often need to study singular curves. We
shall deal with $C$ itself in Section 4 and the partial desingularization only
at $\infty$ in Section 5. In particular, the latter objects (written as
$\widetilde{C}$) appear when we consider degenerations of hyperelliptic curves
or superelliptic curves. The second aim of this paper is to give an explicit
basis of the regular differential module on $C$ and $\widetilde{C}$. Now the
notion of the regular differential forms of singular curves is defined by [4,
Section IV.3], see Section 4 for the details.

As an application, in Section 6 we introduce the modified Cartier operator on
the regular differential modules on $X$, $C$ and $\widetilde{C}$ (Theorem
6.2). The last aim of this paper is to describe an relation between Appell-
Lauricella hypergeometric series and the modified Cartier operator (Theorem
6.5). This result is a generalization of the fact that the coefficient of
$x^{p-1}$ in the polynomial $\\{x(x-1)(x-z)\\}^{(p-1)/2}$) is equal to a
truncation of $F(1/2,1/2,1\,;z)$.

This paper is organized as follows. In Section 2, we study the fundamentals of
curves associated to Appell-Lauricella hypergeometric series and give their
explicit desingularizations. In Section 3, we describe the space of regular
differential forms on them. In Section 4, we describe the space of regular
differential forms on $C$. In Section 5, we describe the space of regular
differential forms on $\widetilde{C}$. Finally in Section 6, we show the
modified Cartier-Manin operator stabilizes the space of regular differential
forms on $\widetilde{C}$ and so on, and describe the relation between the
Cartier-Manin operator and Appell-Lauricella hypergeometric series.

### Acknowledgements

This work was supported by JSPS Grant-in-Aid for Scientific Research (C)
17K05196 and 21K03159.

##  The curves associated to Appell-Lauricella hypergeometric series

In this section, we recall the definition of the curves associated to Appell-
Lauricella hypergeometric series and properties of them. Let $K$ be a perfect
field.

######  Definition 2.1.

Let $N\geq 1$ be a positive integer which is coprime to the characteristic of
$K$. Then a curve associated to Appell-Lauricella hypergeometric series is the
1-dimensional algebraic set

$C:y^{N}=f(x)$

such that $f(x)\in K[x]$ which can be factorized into

$f(x)=\prod_{i=0}^{r}(x-\lambda_{i})^{A_{i}},\quad\lambda_{0},\ldots,\lambda_{r}\in\accentset{{\cc@style\underline{\mskip
12.0mu}}}{K}$

with $A_{i}\geq 1$ and we give the conditions $\lambda_{i}\neq\lambda_{j}$ for
$i\neq j$.

######  Remark 2.2.

The curve $C$ above (or more precisely its desingularization) is called
superelliptic if $A_{i}=1$ for all $i\in\\{0,\ldots,r\\}$. If $N=2$ and $r>3$
in addition, the curve $C$ is called hyperelliptic. A curve associated to
Appell-Lauricella hypergeometric series is a certain generalization of these
curves.

######  Definition 2.3.

Appell-Lauricella hypergeometric series is defined to be

${\mathcal{F}}(a,b_{1},\ldots,b_{d},c\,;z_{1},\ldots,z_{d}):=\sum_{n_{1}=0}^{\infty}\cdots\sum_{n_{d}=0}^{\infty}\frac{(a\,;\sum{n_{j}})\prod(b_{j}\,;n_{j})}{(c\,;\sum{n_{j}})\prod(1\,;n_{j})}\prod_{j=1}^{d}{z_{j}}^{n_{j}}$
(2.1)

where $a,b_{1},\ldots,b_{d},c\in\mathbb{C},-c\notin\mathbb{N}$.

It is obvious that ${\mathcal{F}}(a,b,c\,;z)=F(a,b,c\,;z)$ when $d=1$, hence
Appell-Lauricella hypergeometric series
${\mathcal{F}}(a,b_{1},\ldots,b_{d},c\,;z_{1},\ldots,z_{d})$ can be regarded
as a certain generalization of Gauss’ hypergeometric series $F(a,b,c\,;z)$. If
$0<\mathfrak{R}a<\mathfrak{R}c$, then it is known that
${\mathcal{F}}(a,b_{2},\ldots,b_{r},c\,;\lambda_{2},\ldots,\lambda_{r})$ has
the integral representation as below:

${\mathcal{F}}(a,b_{2},\ldots,b_{r},c\,;\lambda_{2},\ldots,\lambda_{r})=\frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}\int_{1}^{\infty}\prod_{i=0}^{r}(x-\lambda_{i})^{-\mu_{i}}dx$
(2.2)

where $\lambda_{0}=0,\,\lambda_{1}=1$ and
$\mu_{0}=c-\sum_{j=2}^{r}{b_{j}},\,\mu_{1}=1+a-c,\,\mu_{j}=b_{j}~{}(j=2,\ldots,r)$.

In the case that all $\mu_{i}$ are positive rational numbers, by setting the
integrand of (2.2) as $1/y$, we see that the hypergeometric function is
associated to the curve

$y^{N}=\prod_{i=0}^{r}(x-\lambda_{i})^{A_{i}}$

where we set $N$ to be the least common multiple of the denominators of
$\mu_{0},\ldots,\mu_{r}$ and set $A_{i}=N\mu_{i}$ so that
$(N,A_{0},\ldots,A_{r})=1$ holds.

First, we discuss the condition for $C$ to be a curve (i.e. irreducible).

######  Proposition 2.4.

An algebraic set $C$ is irreducible over $K$ (and over $\overline{K}$) if and
only if $(N,A_{0},\ldots,A_{r})=1$.

###### Proof.

Firstly, we show the “only if”-part. Suppose that
$d=(N,A_{0},\ldots,A_{r})>1$, there is
$g(x)\in\accentset{{\cc@style\underline{\mskip 12.0mu}}}{K}[x]$ which
satisfies $f(x)=g(x)^{d}$. Then the polynomial $y^{N}-f(x)$ can be factorized
with $\accentset{{\cc@style\underline{\mskip 12.0mu}}}{K}$-coefficients, and
it suffices to claim that $g(x)\in K[x]$. Now we may divide both $f(x)$ and
$g(x)$ by $x$ in order that $f(x)=\sum a_{i}x^{i},g(x)=\sum b_{i}x^{i}$ such
that $a_{0},b_{0}\neq 0$ if necessary. Since $x^{d}-a_{0}\in K[x]$ is a
separable polynomial, the minimal polynomial of $b_{0}$ does not have any
multiple roots. This means $b_{0}\in K_{\rm sep}$, and $b_{i}$ can be written
as a fractional expression of $b_{0},\ldots,b_{i-1},a_{i}$ by comparing
coefficients of $x^{i}$. Therefore we obtain that $b_{i}\in K_{\rm sep}$ for
all $i$ by inductive steps. For any $\sigma\in{\rm Gal}(K_{\rm sep}/K)$, let
$f^{\sigma}(x)$ and $g^{\sigma}(x)$ be polynomials obtained from $f(x)$ and
$g(x)$ by actions of $\sigma$ on its respective coefficients, namely set
$f^{\sigma}(x):=\sum\sigma(a_{i})x^{i}$ and
$g^{\sigma}(x):=\sum\sigma(b_{i})x^{i}$. Then we have
$f^{\sigma}(x)={g^{\sigma}(x)}^{d}$ clearly, and $f(x)=f^{\sigma}(x)$ since
$a_{i}\in K$ for all $i$. Moreover the leading coefficients of $g(x)$ and
$g^{\sigma}(x)$ are both 1, so it follows that $g(x)=g^{\sigma}(x)$. This
implies that $b_{i}\in K$ for all $i$, then we get $g(x)\in K[x]$.

Secondly, we show the “if”-part. Let $y^{N}=f(x)$ be satisfied. Let $e$ be the
smallest integer $>0$ which satisfies $y^{e}\in K[x]$. Then $e$ is a divisor
of $N$ according to Proposition A.2. Hence there exists a positive integer
$d>0$ such that $N=de$, and we get $g(x)^{d}=f(x)$ with $g(x):=y^{e}$. It
follows from uniqueness of factorization that $(A_{0},\ldots,A_{r})$ is a
multiple of $d$, since $g(x)^{d}=\prod(x-\lambda_{i})^{A_{i}}$. This shows
$d=1$, so $y^{N}-f(x)$ is an irreducible polynomial with
$\accentset{{\cc@style\underline{\mskip 12.0mu}}}{K}$-coefficients. The proof
is completed. ∎

From now on, we suppose that $(N,A_{0},\ldots,A_{r})=1$, since we are
interested in the case that $C$ is a curve (i.e. irreducible). Now we regard
$C$ as a projective variety in
${\mathbb{P}}^{2}(K)=\operatorname{Proj}K[x_{0},x_{1},x_{2}]$. Set
$A_{\infty}:=\bigl{|}N-\sum_{k=0}^{r}{A_{k}}\bigr{|}$. The projective equation
of $C$ reads

  * •

Case 1: $N-\sum_{k=0}^{r}{A_{k}}>0$.
${x_{2}}^{N}={x_{0}}^{A_{\scalebox{0.6}[0.65]{$\infty$}}}\prod(x_{1}-\lambda_{i}x_{0})^{A_{i}}$;

  * •

Case 2: $N-\sum_{k=0}^{r}{A_{k}}<0$.
${x_{2}}^{N}{x_{0}}^{A_{\scalebox{0.6}[0.65]{$\infty$}}}=\prod(x_{1}-\lambda_{i}x_{0})^{A_{i}}$;

  * •

Case 3: $N-\sum_{k=0}^{r}{A_{k}}=0$.
${x_{2}}^{N}=\prod(x_{1}-\lambda_{i}x_{0})^{A_{i}}$.

Put $P_{j}=(1:\lambda_{j}:0)$ for $j\in\\{0,\ldots,r\\}$ and set
${\mathcal{P}}_{\rm fin}:=\\{P_{0},\ldots,P_{r}\\}$. Put $P_{\infty}=(0:1:0)$
in Case 1 and $P_{\infty}=(0:0:1)$ in Case 2. We set
${{\mathcal{P}}}_{\infty}:=\\{P_{\infty}\\}$ for Cases 1 and 2 and
${{\mathcal{P}}}_{\infty}:=\emptyset$ for Case 3. Set

$\mathcal{P}={\mathcal{P}}_{\rm fin}\cup{\mathcal{P}}_{\infty}.$

By the Jacobian criterion and the assumption that $N$ and $p$ are coprime, it
is straightforward to see where $C$ has singularities.

######  Lemma 2.5.

Any singular point of $C$ belongs to ${\mathcal{P}}$. Moreover $C$ is singular
at $P_{i}$ for each $i\in\\{0,\ldots,r,\infty\\}$ if and only if $A_{i}>1$.

Next we review the explicit description of the desingularization $X$ of $C$
obtained by Archinard [1, 3.1], which works also in positive characteristic.
Let $g_{i}:=(N,A_{i})$ for each $i\in\\{0,\ldots,r,\infty\\}$, hence
${N}_{i}=N/{g_{i}}$ and ${A^{\prime}}_{\\!i}={A_{i}}/{g_{i}}$ are coprime non-
negative integers. Thus, there exist $m_{i},n_{i}\in\mathbb{Z}$ such that
${m_{i}}{A^{\prime}}_{\\!i}+{n_{i}}{N}_{i}=1$.

######  Proposition 2.6 (Archinard [1, 3.1.1]).

Let $j\in\\{0,\ldots,r\\}$ and put $f_{j}(x):=\prod_{i\neq
j}(x-\lambda_{i})^{A_{i}}$. Let $D(f_{j})$ be the open subscheme obtained by
excluding the part of $f_{j}(x)=0$ from
$\mathbb{A}^{3}(K)=\operatorname{Spec}K[x,u,z]$ and set $X_{j}$ to be the
closed subscheme of $D(f_{j})$ defined by

$X_{j}:z^{{N}_{j}}=(x-\lambda_{j})u^{m_{j}},\ u^{g_{j}}=f_{j}(x).$

Then $X_{j}$ is nonsingular with (birational) morphism

$\pi_{j}:X_{j}\rightarrow
C\\!\smallsetminus\\!({\mathcal{P}}\\!\smallsetminus\\!\\{P_{j}\\})\,;\,(x,u,z)\mapsto(1:x:u^{n_{j}}z^{{A^{\prime}}_{\\!j}}).$

Moreover $\pi_{j}$ induces an isomorphism
$X_{j}\\!\smallsetminus\\!\pi^{-1}(\\{P_{j}\\})\overset{\\!\cong}{\longrightarrow}C\\!\smallsetminus\\!{\mathcal{P}}$,
whose inverse is

$\rho_{j}:C\\!\smallsetminus\\!{\mathcal{P}}\to
X_{j}\\!\smallsetminus\\!\pi^{-1}(\\{P_{j}\\})\,;\,(1:x:y)\mapsto(x,y^{{N}_{j}}(x-\lambda_{j})^{-{A^{\prime}}_{\\!j}},y^{m_{j}}(x-\lambda_{j})^{n_{j}}).$

######  Proposition 2.7 (Archinard [1, 3.1.2]).

Let $f_{\infty}(x):=\prod_{i=0}^{r}(1-\lambda_{i}x)^{A_{i}}$. We define
$X_{\infty}$ and $\pi_{\infty}$ in each case as follows. Then $X_{\infty}$ is
nonsingular, and birationally equivalent to $C$ under a rational map
$\pi_{\infty}$.

  * •

Case 1: $N-\sum{A_{k}}>0$. In this case, $x={x_{0}}/{x_{1}}$ and
$y={x_{2}}/{x_{1}}$ are regular on $\infty$. Then we write $D(f_{\infty})$ be
the open subscheme obtained by excluding the part of $f_{\infty}(x)=0$ from
$\mathbb{A}^{3}(K)=\operatorname{Spec}K[x,u,z]$ and set $X_{\infty}$ to be the
closed subscheme of $D(f_{\infty})$ defined by

$X_{\infty}:z^{{N}_{\scalebox{0.6}[0.65]{$\infty$}}}=xu^{m_{\scalebox{0.6}[0.65]{$\infty$}}},\
u^{g_{\scalebox{0.6}[0.65]{$\infty$}}}=f_{\infty}(x).$

Then $X_{\infty}$ is nonsingular with (birational) morphism

$\pi_{\infty}:X_{\infty}\rightarrow C\\!\smallsetminus\\!{\mathcal{P}}_{\rm
fin}\,;\,(x,u,z)\mapsto(x:1:u^{n_{\scalebox{0.6}[0.65]{$\infty$}}}z^{{A^{\prime}}_{\\!\scalebox{0.6}[0.65]{$\infty$}}}).$

which induces an isomorphism
$X_{\infty}\\!\smallsetminus\\!\pi^{-1}({{\mathcal{P}}}_{\infty})\overset{\cong}{\longrightarrow}C\\!\smallsetminus\\!{\mathcal{P}}$,
whose inverse is

$\rho_{\infty}:C\\!\smallsetminus\\!{\mathcal{P}}\rightarrow
X_{\infty}\\!\smallsetminus\\!\pi^{-1}({{\mathcal{P}}}_{\infty})\,;\,(x:1:y)\mapsto(x,x^{-{A^{\prime}}_{\\!\scalebox{0.6}[0.65]{$\infty$}}}y^{{N}_{\scalebox{0.6}[0.65]{$\infty$}}},x^{n_{\scalebox{0.6}[0.65]{$\infty$}}}y^{m_{\scalebox{0.6}[0.65]{$\infty$}}}).$

  * •

Case 2: $N-\sum{A_{k}}<0$. In this case, $x={x_{0}}/{x_{2}}$ and
$y={x_{1}}/{x_{2}}$ are regular on $\infty$. Then we write $D(f_{\infty})$ be
the open subscheme obtained by excluding the part of $f_{\infty}(u)=0$ from
$\mathbb{A}^{4}(K)=\operatorname{Spec}K[u,v,w,z]$ and set $X_{\infty}$ to be
the closed subscheme of $D(f_{\infty})$ defined by

$X_{\infty}:u^{{A^{\prime}}_{\\!\scalebox{0.6}[0.65]{$\infty$}}}=v^{{N}_{\scalebox{0.6}[0.65]{$\infty$}}}w,\
z^{{A^{\prime}}_{\\!\scalebox{0.6}[0.65]{$\infty$}}}=vw^{n_{\scalebox{0.6}[0.65]{$\infty$}}},\
w^{g_{\scalebox{0.6}[0.65]{$\infty$}}}=f_{\infty}(u).$

Then $X_{\infty}$ is nonsingular with birational morphism

$\pi_{\infty}:X_{\infty}\rightarrow C\\!\smallsetminus\\!{{\mathcal{P}}}_{\rm
fin}\,;\,(u,v,w,z)\mapsto(vw^{m_{\scalebox{0.6}[0.65]{$\infty$}}}z^{{N}_{\scalebox{0.6}[0.65]{$\infty$}}}:v:1),$

which induces an isomorphism
$X_{\infty}\\!\smallsetminus\\!\pi^{-1}({{\mathcal{P}}}_{\infty})\overset{\cong}{\longrightarrow}C\\!\smallsetminus\\!{\mathcal{P}}$,
whose inverse is

$\rho_{\infty}:C\\!\smallsetminus\\!{\mathcal{P}}\rightarrow
X_{\infty}\\!\smallsetminus\\!\pi^{-1}({{\mathcal{P}}}_{\infty})\,;\,(x:y:1)\mapsto(xy^{-1},y,x^{{A^{\prime}}_{\\!\scalebox{0.6}[0.65]{$\infty$}}}y^{{N}_{\scalebox{0.6}[0.65]{$\infty$}}-{A^{\prime}}_{\\!\scalebox{0.6}[0.65]{$\infty$}}},x^{n_{\scalebox{0.6}[0.65]{$\infty$}}}y^{m_{\scalebox{0.6}[0.65]{$\infty$}}-n_{\scalebox{0.6}[0.65]{$\infty$}}}).$

Note that in Case 3: $N-\sum A_{i}=0$, the points at infinity $(0:1:\zeta)$
with $\zeta^{N}=1$ are nonsingular by Lemma 2.5. Hence we do not need to
consider $X_{\infty}$ as Archinard excluded Case 3.

######  Remark 2.8.

Note that $X_{i}$ does not depend on choice of $m_{i},n_{i}$ up to
isomorphism. More precisely, suppose that
$({m^{\prime}}_{\\!i},{n^{\prime}}_{\\!i})$ is another pair of integers
satisfying
${m^{\prime}}_{\\!i}{A^{\prime}}_{\\!i}+{n^{\prime}}_{\\!i}{N}_{i}=1$ and let
${X^{\prime}}_{\\!i}$ be the set obtained from
$({m^{\prime}}_{\\!i},{n^{\prime}}_{\\!i})$ in the same way, then
${X^{\prime}}_{\\!i}$ is isomorphic to $X_{i}$. We give a proof of the case
$i\in\\{0,\ldots,r\\}$. Now there is a relation
$({n_{i}}-{n^{\prime}}_{\\!i}){N}_{i}=-({m_{i}}-{m^{\prime}}_{\\!i}){A^{\prime}}_{\\!i}$,
then we have $e:=(n_{i}-{n^{\prime}}_{\\!i})/{A^{\prime}}_{\\!i}\in\mathbb{Z}$
since ${N}_{i}$ and ${A^{\prime}}_{\\!i}$ are coprime non-negative integers.
Hence consider the morphism
$X_{i}\rightarrow{X^{\prime}}_{\\!i}\,;\,(x,u,z)\mapsto(x,u,u^{e}z)$ which has
the obvious inverse, thus we conclude that $X_{i}\cong{X^{\prime}}_{\\!i}$.
Also the uniqueness of $X_{\infty}$ can be proved similarly.

######  Remark 2.9.

The action on $C$ of the group $\mu_{N}$ of $N$-th roots defined by
$(x,y)\to(x,\zeta y)$ for $\zeta\in\mu_{N}$ is extended to that on $X_{j}$ by
$(x,u,z)\to(x,\zeta^{{N}_{j}}u,\zeta^{m_{j}}z)$. Also a similar thing holds
for $P_{\infty}$.

We define the desingularization $X$ obtained by gluing $X_{0},\ldots,X_{r}$
and $X_{\infty}$ along $X_{i}\\!\smallsetminus\\!\pi^{-1}(\\{P_{i}\\})$ and
$X_{j}\\!\smallsetminus\\!\pi^{-1}(\\{P_{j}\\})$ via the isomorphisms

$X_{i}\\!\smallsetminus\\!\pi^{-1}(\\{P_{i}\\})\overset{\pi_{i}}{\longrightarrow}C\\!\smallsetminus\\!{\mathcal{P}}\overset{\pi_{j}}{\longleftarrow}X_{j}\\!\smallsetminus\\!\pi^{-1}(\\{P_{j}\\}).$

Then by gluing $\pi_{i}:X_{i}\rightarrow C$, we also obtain a morphism
$\pi:X\rightarrow C$ such that $\pi|_{X_{i}}=\pi_{i}$ for all
$i\in\\{0,\ldots,r,\infty\\}$. As can be found in [1, Section 3.2], $X$ is the
desingularization of $C$ under $\pi$.

######  Theorem 2.10.

The genus of $X$ is given as below:

$g(X)=1+\frac{1}{2}\biggl{(}rN-\sum_{j=0}^{r}(N,A_{j})-\Bigl{(}N,N-\sum_{k=0}^{r}A_{k}\Bigr{)}\biggr{)}.$

###### Proof.

Let the projection
$C\rightarrow\mathbb{P}^{1}(K)\,;\,(x_{0}:x_{1}:x_{2})\mapsto(x_{0}:x_{1})$
except for $(0:0:1)\mapsto(0:1)$ and compose with $\pi:X\rightarrow C$, then
we obtain a finite separable morphism $X\rightarrow\mathbb{P}^{1}(K)$.

point $P$ of $C$ | # of $\pi$-preimages $Q$ of $P$ | ramification index at $Q$
---|---|---
$(1:\lambda_{j}:0)$ | $(N,A_{j})$ | $N/(N,A_{j})$
$\infty$ | $(N,N-\sum{A_{k}})$ | $N/(N,N-\sum{A_{k}})$
other points | 1 | 1

The genus of a projective line $\mathbb{P}^{1}(K)$ is 0, so directly we see
that

$\displaystyle 2g(X)-2$
$\displaystyle=-2N+\sum_{j=0}^{r}(N,A_{j})\\!\biggl{(}\frac{N}{(N,A_{j})}-1\biggr{)}+\bigl{(}N,N-\sum
A_{k}\bigr{)}\\!\biggl{(}\frac{N}{(N,N-\sum{A_{k}})}-1\biggr{)}$
$\displaystyle=rN-\bigl{(}N,N-\sum A_{k}\bigr{)}-\sum_{j=0}^{r}(N,A_{j}).$

This is the desired conclusion. ∎

##  The space of regular differential forms on $X$

In this section, for the desingularization map $\pi:X\to C$ constructed as in
Section 2, we describe the regularity condition of rational differential forms
on $X$. This enables us to give an explicit basis of the space $\varOmega[X]$
of regular differential forms on $X$, where “regular” is often called “of
first kind”. Note that $\varOmega[X]$ is realized as a subspace of the space
$\varOmega(C)$ of rational differential forms on $C$.

In [2, Theorem 12], Gorenstein gives a description of regular differential
forms on the projective nonsingular model of a plane curve $\varGamma$, but
our case does not satisfy his assumption: $y$ is an integral element over
$K(x)$. This would mean that the Zariski closure of $\varGamma$ in
${\mathbb{P}}^{2}(K)$ is regular at the infinity. So let us formulate a lemma,
which works in our case. The proof was done in the proof of [2, Theorem 12].

######  Theorem 3.1 (Gorenstein).

Let $\varGamma$ be a plane curve ${\rm Spec}\,R$ with $R=K[x,y]/(F)$ for an
irreducible element $F$ of $K[x,y]$. Let $L$ be the function field of
$\varGamma$ and $X$ the nonsingular projective curve having the same function
field $L$. Assume that $x$ considered as an element of $L$ is transendental
over $K$ and $y$ considered as an element of $L$ is separable over $K(x)$. A
rational differential form $\omega$ is regular on $X$ if and only if it can be
written in the form

$\frac{\phi(x,y)}{(\partial F/\partial y)(x,y)}dx$

such that $\phi(x,y)$ is an adjoint element, whose precise meaning is

  1. (i)

$\phi(x,y)\in{\mathfrak{C}}$, where $\mathfrak{C}$ is the conductor of
$R=K[x,y]/(F)$ in the integral closure $\accentset{{\cc@style\underline{\mskip
12.0mu}}}{R}$ of $R$ in $L$, i.e. $\mathfrak{C}:=\\{z\in
R\,;\,z{\accentset{{\cc@style\underline{\mskip 12.0mu}}}{R}}\subset R\\}$.

  2. (ii)

$\phi^{\prime}(x^{\prime},y^{\prime})\in{\mathfrak{C}}^{\prime}$, where
$\phi^{\prime}$ be the polynomial in $x^{\prime}$ and $y^{\prime}$ defined by
$\phi(x,y)=\phi^{\prime}(x^{\prime},y^{\prime})/({x^{\prime}})^{d}$ with
$(x^{\prime},y^{\prime})=(1/x,y/x)$ and $d=\deg\phi$, and
$\mathfrak{C}^{\prime}$ is the conductor of
$R^{\prime}=K[x^{\prime},y^{\prime}]/(F^{\prime})$ with
$F(x,y)=F^{\prime}(x^{\prime},y^{\prime})/({x^{\prime}})^{e}$ in the integral
closure $\accentset{{\cc@style\underline{\mskip 12.0mu}}}{R^{\prime}}$ of
$R^{\prime}$ in $L$.

  3. (iii)

$\phi^{\prime\prime}(x^{\prime\prime},y^{\prime\prime})\in{\mathfrak{C}}^{\prime\prime}$,
where $\phi^{\prime\prime}$ be the polynomial in $x^{\prime}$ and $y^{\prime}$
defined by
$\phi(x,y)=\phi^{\prime\prime}(x^{\prime\prime},y^{\prime\prime})/({y^{\prime}})^{d}$
with $(x^{\prime},y^{\prime})=(x/y,1/y)$ and $d=\deg\phi$, and
$\mathfrak{C}^{\prime\prime}$ is the conductor of
$R^{\prime\prime}=K[x^{\prime\prime},y^{\prime\prime}]/(F^{\prime\prime})$
with
$F(x,y)=F^{\prime\prime}(x^{\prime\prime},y^{\prime\prime})/({y^{\prime\prime}})^{e}$
in the integral closure $\accentset{{\cc@style\underline{\mskip
18.0mu}}}{R^{\prime\prime}}$ of $R^{\prime\prime}$ in $L$.

######  Remark 3.2.

Here, $\phi(x,y)\in{\mathfrak{C}}$ is equivalent to
$\phi(x,y)\in{\mathfrak{C}}_{P}:=\\{z\in
R_{P}\,;\,z{\accentset{{\cc@style\underline{\mskip 18.0mu}}}{R_{P}}}\subset
R_{P}\\}$ for maximal ideal $P$ of $R$, where
${\accentset{{\cc@style\underline{\mskip 18.0mu}}}{R_{P}}}$ is the integral
closure of $R_{P}$ in $L$. Moreover
${\mathfrak{C}}_{P}={\mathfrak{C}}_{P}^{*}\cap R_{P}$, where
${\mathfrak{C}}_{P}^{*}$ is the conductor of $(R_{P})^{*}$ in
$(\accentset{{\cc@style\underline{\mskip 18.0mu}}}{R_{P}})^{*}$, where $*$
means taking the completion (cf. [2, Theorem 2]).

Let us return to our case $F=y^{N}-f(x)$. Let $\mu_{N}$ be the subgroup of
$\accentset{{\cc@style\underline{\mskip 12.0mu}}}{K}^{\times}\\!$ consisting
of $N$-th roots of unity. First let us consider the case that $K$ contains
$\mu_{N}$ and $\\{\lambda_{0},\ldots,\lambda_{r}\\}$. Let $\phi(x,y)$ be an
adjoint element in the sense of the definition in Theorem 3.1. Using
$y^{N}=f(x)$, one can write

$\phi(x,y)=\phi_{0}(x)+\phi_{1}(x)y+\cdots+\phi_{N-1}(x)y^{N-1}$

with $\phi_{j}(x)\in K[x]$. Consider the action on $R$ of the group $\mu_{N}$
consisting of $N$-th roots of unity by $(x,y)\mapsto(x,\zeta y)$ for
$\zeta\in\mu_{N}$. As this action stabilizes $\mathfrak{C}$,
$\mathfrak{C}^{\prime}$ and $\mathfrak{C}^{\prime\prime}$, we have that
$\phi(x,\zeta y)$ is also an adjoint element for all $\zeta\in\mu_{N}$. This
implies that each term $\phi_{j}(x)y^{j}$ is an adjoint element. Clearly
$\phi_{j}(x)$ is uniquely written as
$\varphi(x)\prod_{i=0}^{r}(x-\lambda_{i})^{a_{i}}$, where $\varphi(x)$ is
coprime to $x-\lambda_{i}$ for $i=0,1,\ldots,r$. As we can check
$\phi_{j}(x)y^{j}\in{\mathfrak{C}}$ by looking at whether $\phi_{j}(x)y^{j}$
in ${\mathfrak{C}}_{P}$ for all $P\in\\{P_{0},\ldots,P_{r}\\}$, we conclude
that $\varphi(x)\prod_{i=0}^{r}(x-\lambda_{i})^{a_{i}}y^{j}\in{\mathfrak{C}}$
if and only if $\prod_{i=0}^{r}(x-\lambda_{i})^{a_{i}}y^{j}\in{\mathfrak{C}}$.
Also at an infinite place, the condition (ii) and (iii) of Theorem 3.1 is
described as the degree of $\phi_{j}$ is less than or equal to a certain
constant depending only on $C$ (cf. the proof of Proposition 3.5 below).
Hence, if $\varphi(x)\prod_{i=0}^{r}(x-\lambda_{i})^{a_{i}}y^{j}$ is an
adjoint element, then
$\prod_{i=0}^{r}(x-\lambda_{i})^{a_{i}}y^{j}\in{\mathfrak{C}}$ is an adjoint
element. Thus we conclude

######  Lemma 3.3.

The regular differential module $\varOmega[X]$ has a basis consisting of
elements of the form

$\omega_{(s,\bm{a})}:=\frac{\prod_{i=0}^{r}(x-\lambda_{i})^{a_{i}}}{y^{s}}dx,\quad
0\leq s\leq N-1$

where $\bm{a}=(a_{0},\ldots,a_{r})$ with $a_{i}\geq 0$.

Now $\pi^{*}\omega_{(s,\bm{a})}$ is regular except at
$Q_{i}\in\pi^{-1}(\\{P_{i}\\})$, however the regularity at $Q_{i}$ is not
ensured.

######  Proposition 3.4.

For $j\in\\{0,\ldots,r\\}$, the pull-back
$\pi^{*}\omega_{(s,\bm{a})}\in\varOmega(X)$ is regular at
$Q_{j}\in\pi^{-1}(\\{P_{j}\\})$ if and only if

$a_{j}\geq\frac{sA_{j}+(N,A_{j})}{N}-1.$

###### Proof.

The equations defining $X_{j}$ gives other equations

$\displaystyle{N}_{j}z^{{N}_{j}-1}dz$
$\displaystyle=u^{m_{j}}dx+m_{j}(x-\lambda_{j})u^{m_{j}-1}du,$ $\displaystyle
g_{j}u^{g_{j}-1}du$ $\displaystyle=(df_{j}/dx)(x)dx.$

Now recall that $\pi^{*}(x)=x$ and
$\pi^{*}(y)=u^{n_{j}}z^{{A^{\prime}}_{\\!j}}$, then we obtain

$\pi^{*}\omega_{(s,\bm{a})}=\frac{Nu^{g_{j}-sn_{j}-(1+a_{j})m_{j}}z^{-s{A^{\prime}}_{\\!j}+(1+a_{j}){N}_{i}-1}\prod(x-\lambda_{i})^{a_{i}}}{g_{j}f_{j}(x)+m_{j}(x-\lambda_{j})(df_{j}/dx)(x)}dz$
(3.1)

since $x-\lambda_{j}=u^{-m_{j}}z^{{N}_{i}}$. Hence
$\pi^{*}\omega_{(s,\bm{a})}$ is regular at $Q_{j}$ if and only if
$-s{A^{\prime}}_{\\!j}+(1+a_{j}){N}_{i}-1\geq 0$. ∎

Similarly, the regularity at the fiber of the infinity is described as below:

######  Proposition 3.5.

The pull-back $\pi^{*}\omega_{(s,\bm{a})}\in\varOmega(X)$ is regular at
$Q_{\infty}\in\pi^{-1}(\\{P_{\infty}\\})$ in Cases 1 and 2 and at
$Q_{\infty}\in\pi^{-1}(\\{(0:1:\zeta);\zeta^{N}=1\\})$ in Case 3 if and only
if

$\sum_{k=0}^{r}a_{k}\leq\frac{s\sum{A_{k}}-(N,N-\sum{A_{k}})}{N}-1.$

###### Proof.

In each case, we can write
$\omega_{(s,\bm{a})}={x_{0}}^{s-\sum{a_{k}}-2}{x_{2}}^{-s}\prod(x_{1}-\lambda_{i}x_{0})^{a_{i}}(x_{0}dx_{1}-x_{1}dx_{0})$.

  * •

Case 1: $N-\sum{A_{k}}>0$. In this case, recall that $x={x_{0}}/{x_{1}}$ and
$y={x_{2}}/{x_{1}}$;

$\omega_{(s,\bm{a})}=x^{s-\sum{a_{k}}-2}y^{-s}\prod(1-\lambda_{i}x)^{a_{i}}dx.$

The equations defining $X_{\infty}$ gives other equations

$\displaystyle N_{\infty}z^{{N}_{\scalebox{0.6}[0.65]{$\infty$}}-1}dz$
$\displaystyle=u^{m_{\scalebox{0.6}[0.65]{$\infty$}}}dx+m_{\infty}xu^{m_{\scalebox{0.6}[0.65]{$\infty$}}-1}du,$
$\displaystyle g_{\infty}u^{g_{\scalebox{0.6}[0.65]{$\infty$}}-1}du$
$\displaystyle=(df_{\infty}/dx)(x)dx.$

Now recall that $\pi^{*}(x)=x$ and
$\pi^{*}(y)=u^{n_{\scalebox{0.6}[0.65]{$\infty$}}}z^{{A^{\prime}}_{\\!\scalebox{0.6}[0.65]{$\infty$}}}$,
then we obtain

$\pi^{*}\omega_{(s,\bm{a})}=\frac{Nu^{g_{\scalebox{0.6}[0.65]{$\infty$}}-s(m_{\scalebox{0.6}[0.65]{$\infty$}}+n_{\scalebox{0.6}[0.65]{$\infty$}})+(1+\sum{a_{k}})m_{\scalebox{0.6}[0.65]{$\infty$}}}z^{s({N}_{\scalebox{0.6}[0.65]{$\infty$}}-{A^{\prime}}_{\\!\scalebox{0.6}[0.65]{$\infty$}})-(1+\sum{a_{k}}){N}_{\scalebox{0.6}[0.65]{$\infty$}}-1}\prod(1-\lambda_{i}x)^{a_{i}}}{g_{\infty}f_{\infty}(x)+m_{\infty}x(df_{\infty}/dx)(x)}dz.$
(3.2)

Hence $\pi^{*}\omega_{(s,\bm{a})}$ is regular at $Q_{\infty}$ if and only if
$s({N}_{\infty}-{A^{\prime}}_{\\!\infty})-(1+\sum{a_{k}}){N}_{\infty}-1\geq
0$.

  * •

Case 2: $N-\sum{A_{k}}<0$. In this case, recall that $x={x_{0}}/{x_{2}}$ and
$y={x_{1}}/{x_{2}}$;

$\omega_{(s,\bm{a})}=x^{s-\sum{a_{k}}-2}\prod(y-\lambda_{i}x)^{a_{i}}(xdy-
ydx).$

The equations defining $X_{\infty}$ gives other equations

$\displaystyle{A^{\prime}}_{\\!\infty}u^{{A^{\prime}}_{\\!\scalebox{0.6}[0.65]{$\infty$}}-1}du$
$\displaystyle={N}_{\infty}v^{{N}_{\scalebox{0.6}[0.65]{$\infty$}}-1}wdv+v^{{N}_{\scalebox{0.6}[0.65]{$\infty$}}}dw,$
$\displaystyle{A^{\prime}}_{\\!\infty}z^{{A^{\prime}}_{\\!\scalebox{0.6}[0.65]{$\infty$}}-1}dz$
$\displaystyle=w^{n_{\scalebox{0.6}[0.65]{$\infty$}}}dv+nvw^{n_{\scalebox{0.6}[0.65]{$\infty$}}-1}dw,$
$\displaystyle g_{\infty}w^{g_{\scalebox{0.6}[0.65]{$\infty$}}-1}dw$
$\displaystyle=(df_{\infty}/du)(u)du$

since
$v^{{N}_{\scalebox{0.6}[0.65]{$\infty$}}-1}w^{1-n_{\scalebox{0.6}[0.65]{$\infty$}}}z^{{N}_{\scalebox{0.6}[0.65]{$\infty$}}}=v^{{N}_{\scalebox{0.6}[0.65]{$\infty$}}}w$,
then we have

$\pi^{*}\omega_{(s,\bm{a})}=\frac{Nw^{g_{\scalebox{0.6}[0.65]{$\infty$}}+s(m_{\scalebox{0.6}[0.65]{$\infty$}}-n_{\scalebox{0.6}[0.65]{$\infty$}})-(1+\sum{a_{k}})m_{\scalebox{0.6}[0.65]{$\infty$}}}z^{s({N}_{\scalebox{0.6}[0.65]{$\infty$}}+{A^{\prime}}_{\\!\scalebox{0.6}[0.65]{$\infty$}})-(1+\sum{a_{k}}){N}_{\scalebox{0.6}[0.65]{$\infty$}}-1}\prod(1-\lambda_{i}u)^{a_{i}}}{mu(df_{\infty}/du)(u)-g_{\infty}f_{\infty}(u)}dz.$
(3.3)

Hence $\pi^{*}\omega_{(s,\bm{a})}$ is regular at $Q_{\infty}$ if and only if
${s({N}_{\infty}+{A^{\prime}}_{\\!\infty})-(1+\sum{a_{k}}){N}_{\infty}-1}\geq
0$.

  * •

Case 3: $N-\sum A_{i}=0$. In this case, one can confirm that
$\pi^{*}(\omega_{s,\bm{a}})$ is given as

$\pi^{*}\omega_{(s,\bm{a})}=\frac{z^{s-\sum
a_{i}-2}\prod(1-\lambda_{i}z)^{a_{i}}}{y^{s}}dz$ (3.4)

clearly, so $\pi^{*}(\omega_{s,\bm{a}})$ is regular at $Q_{\infty}$ if and
only if $s-\sum a_{i}-2\geq 0$.

After all, we see that the proposition is true in each case. ∎

We can sum up what has been discussed so far as follows:

######  Theorem 3.6.

Assume that a perfect field $K$ contains $\mu_{N}$ and
$\\{\lambda_{0},\ldots,\lambda_{r}\\}$. For $0\leq s\leq N-1$, let $V_{s}$ be
the subspace of $\varOmega[X]$ having the character $\zeta\mapsto\zeta^{s}$
under the action $(x,y)\to(x,\zeta y)$ of $\mu_{n}$ on $\varOmega[X]$. Put

$\displaystyle d_{s}$
$\displaystyle=\max\,\biggl{\\{}0,\left\lfloor\frac{s\sum{A_{k}}-(N,N-\sum{A_{k}})}{N}\right\rfloor-\sum_{j=0}^{r}\left\lceil\frac{sA_{j}+(N,A_{j})}{N}-1\right\rceil\biggr{\\}},$
$\displaystyle e_{s,j}$
$\displaystyle=\left\lceil\frac{sA_{j}+(N,A_{j})}{N}-1\right\rceil\text{ with
}\bm{e_{s}}=(e_{s,0},e_{s,1},\ldots,e_{s,r}).$

We have $\dim V_{s}=d_{s}$. Moreover, a basis of $V_{s}$ is given by

$x^{m}\omega_{(s,\bm{e_{s}})}=x^{m}\frac{\prod_{j=0}^{r}(x-\lambda_{j})^{e_{s,j}}}{y^{s}}dx$

for $0\leq m\leq d_{s}-1$.

###### Proof.

It is obvious that $x^{m}\omega_{(s,\bm{e_{s}})}\in V_{s}$ for all $0\leq
m\leq d_{s}-1$, since $x^{m}$ is a linear combination of $(x-\lambda_{0})^{k}$
for $0\leq k\leq m$ and for $\bm{a}=\bm{e_{s}}+(k,0,\ldots,0)$, so we have
$\omega_{(s,\bm{a})}\in\varOmega[X]$ by Proposition 3.4 and Proposition 3.5.
For the converse, an arbitrary element of $V_{s}$ is a linear combination of
$\omega_{(s,\bm{a})}$ with $a_{j}\geq e_{s,j}$ for all $j\in\\{0,\ldots,r\\}$
and

$\sum_{k=0}^{r}a_{k}\leq\left\lfloor\frac{s\sum{A_{k}}-(N,N-\sum{A_{k}})}{N}\right\rfloor-1.$

Now rewriting $\omega_{(s,\bm{a})}=\varphi(x)\omega_{(s,\bm{e_{s}})}$ where
$\deg\varphi\leq d_{s}-1$, then $\varphi(x)$ is spanned by
$\\{1,x,\ldots,x^{d_{s}-1}\\}$. Clearly these are linearly independent. ∎

######  Remark 3.7.

If $K$ is a perfect field and contains $\mu_{N}$, then each member of the
basis obtained above is defined over $K$. In fact, since
$\omega_{(s,\bm{e_{s}})}$ is the unique “monic” element having the lowest-
degree, it is stable under the action of ${\rm
Gal}(\accentset{{\cc@style\underline{\mskip 12.0mu}}}{K}/K)$.

##  Space of regular differential forms on $C$

In this section, we consider regular differential forms on $C$. We remark that
$C$ has singularities as we see in Lemma 2.5, then we refer [4, Chapter IV,
$\S 3.9$] to the regular differential forms of singular curves. Let us give a
brief review of it. We set $\varOmega[C]:=\bigcap_{P\in C}\varOmega[C]_{P}$
with

$\varOmega[C]_{P}:=\biggl{\\{}\omega\in\varOmega(X);\raise
4.30554pt\hbox{$\displaystyle\sum_{\pi(Q)=P}$}{\rm
res}_{Q}\bigl{(}\pi^{*}(h)\omega\bigr{)}=0,\text{ for all
}h\in{\mathcal{O}}_{C,P}\biggr{\\}}\vspace{-1mm}$ (4.1)

where $\pi:X\rightarrow C$ is the desingularization map we constructed in
Section 2. Note that $\varOmega[C]_{P}$ is a ${\mathcal{O}}_{C,P}$-module and
that $\varOmega[C]$ is the space of global sections of the sheaf
$U\mapsto\bigcap_{P\in U}\varOmega[C]_{P}$, which is the dualizing sheaf on
$C$.

######  Lemma 4.1.

For $j\in\\{0,\ldots,r\\}$ we set $d\geq 0$ and $e\in\\{1,\ldots,g_{j}\\}$. A
differential form $u^{g_{j}-e}dz/z^{d+1}$ belongs to $\varOmega[C]_{P_{j}}$ if
and only if any pair $(a,b)$ does not satisfy

$\displaystyle aN_{j}+b{A^{\prime}}_{\\!j}=d,$ $\displaystyle-
am_{j}+bn_{j}\equiv e~{}({\rm mod}\,g_{j})$

###### Proof.

A set of invertible elements is generated by $(x-\lambda_{j})$ and $y$, so we
may write $h=(x-\lambda_{j})^{a}y^{b}$ where $a,b\geq 0$. Now let
$Q_{j}=(\lambda_{j},u,0)\in\pi^{-1}(\\{P_{j}\\})$ where $u$ runs among
$g_{j}$-th roots of $f_{j}(\lambda_{j})$, then a sum of residues 0 if the
index of $u$ is not divided by $g_{j}$. ∎

According to Lemma 4.1 and Proposition B.4, we directly obtain below:

  1. (i)

If $d\geq g_{j}{N}_{j}{A^{\prime}}_{\\!j}-{N}_{j}-{A^{\prime}}_{\\!j}+1$, then
$u^{g_{j}-e}dz/z^{d+1}\notin\varOmega[C]_{P_{j}}$ for all
$e\in\\{1,\ldots,g_{j}\\}$.

  2. (ii)

If $d=g_{j}{N}_{j}{A^{\prime}}_{\\!j}-{N}_{j}-{A^{\prime}}_{\\!j},\,e\equiv
m_{j}-n_{j}~{}({\rm mod}\ g_{j})$, then
$u^{g_{j}-e}dz/z^{d+1}\in\varOmega[C]_{P_{j}}$.

Now we describe a generator of $\varOmega[C]_{P_{j}}$ in more detail as
follows. If $u^{g_{j}-e}dz/z^{d+1}\in\varOmega(X)$ is equal to (3.1), the
polynomial $\prod(x-\lambda_{i})^{a_{i}}$ is equal to

$\frac{g_{j}u^{g_{j}}+m_{j}(x-\lambda_{j})(\partial{f_{j}}/\partial{x})(x)}{N}\cdot
u^{sn_{j}+m_{j}-e}z^{-{N}_{j}+s{A^{\prime}}_{\\!j}-d}.\vspace{-1mm}$

Hence $u^{sn_{j}+m_{j}-e}z^{-{N}_{j}+s{A^{\prime}}_{\\!j}-d}$ must be
transformed into the polynomial with respect to $x$. By using the relation
$x-\lambda_{j}=u^{-m_{j}}z^{{N}_{j}}$, we have $s=dm_{j}+e{N}_{j}$.

######  Lemma 4.2.

For $j\in\\{0,\ldots,r\\}$, the pull-back $\pi^{*}(dx/y^{N-1})$ is a generator
of $\varOmega[C]_{P_{j}}$.

###### Proof.

Suppose that $\pi^{*}\omega_{(s,\bm{a})}=u^{g_{j}-e}dz/z^{d+1}$ with
$d=g_{j}{N}_{j}{A^{\prime}}_{\\!j}-{N}_{j}-{A^{\prime}}_{\\!j}$ and
$e=m_{j}-n_{j}+g_{j}k$. From above discussion, we get
$s=dm_{j}+e{N}_{j}=N(m_{j}{A^{\prime}}_{\\!j}+k)-1$ thus this means that
$s=N-1$ and $k=n_{j}N_{j}$ since $0\leq s<N$. Then one can confirm that
$sn_{j}+m_{j}-e=-{N}_{j}+s{A^{\prime}}_{\\!j}-d=0$, hence
$\prod(x-\lambda_{i})^{a_{i}}$ is invertible since
$g_{j}u^{g_{j}}+m_{j}(x-\lambda_{j})(\partial{f_{j}}/\partial{x})(x)$ is not
zero at $Q_{j}\in\pi^{-1}(\\{P_{j}\\})$. This means that $a_{j}=0$. ∎

Next, we describe a generator of
$\varOmega[C]_{\scalebox{0.7}[0.7]{$P_{\infty}\\!$}}$ in Case 1 then we
similarly obtain below:

  1. (i)

If $d\geq
g_{\infty}{N}_{\infty}{A^{\prime}}_{\\!\infty}-{N}_{\infty}-{A^{\prime}}_{\\!\infty}+1$,
then
$u^{g_{\infty}-e}dz/z^{d+1}\notin\varOmega[C]_{\scalebox{0.7}[0.7]{$P_{\infty}\\!$}}$
for all $e\in\\{1,\ldots,g_{\infty}\\}$.

  2. (ii)

If
$d=g_{\infty}{N}_{\infty}{A^{\prime}}_{\\!\infty}-{N}_{\infty}-{A^{\prime}}_{\\!\infty},\,e=m_{\infty}-n_{\infty}~{}({\rm
mod}\ g_{\infty})$, then
$u^{g_{\infty}-e}dz/z^{d+1}\in\varOmega[C]_{\scalebox{0.7}[0.7]{$P_{\infty}\\!$}}$.

If $u^{g_{\infty}-e}dz/z^{d+1}\in\varOmega(X)$ is equal to (3.2), the
polynomial $\prod(1-\lambda_{i}x)^{a_{i}}$ is equal to

$\frac{g_{\infty}u^{g_{\scalebox{0.6}[0.65]{$\infty$}}}+m_{\infty}x(\partial{f}_{\infty}/\partial{x})(x)}{N}\cdot
u^{s(m_{\scalebox{0.6}[0.65]{$\infty$}}+n_{\scalebox{0.6}[0.65]{$\infty$}})-(1+\sum{a_{k}})m_{\scalebox{0.6}[0.65]{$\infty$}}-e}z^{-s({N}_{\scalebox{0.6}[0.65]{$\infty$}}-{A^{\prime}}_{\\!\scalebox{0.6}[0.65]{$\infty$}})+(1+\sum{a_{k}}){N}_{\scalebox{0.6}[0.65]{$\infty$}}-d}.\vspace{-1mm}$

By using the relation
$x=u^{-m_{\scalebox{0.6}[0.65]{$\infty$}}}z^{{N}_{\scalebox{0.6}[0.65]{$\infty$}}}$,
then we have $s=dm_{\infty}+e{N}_{\infty}$.

######  Lemma 4.3.

In Case 1, the pull-back $\pi^{*}\omega_{(N-1,\bm{a})},\,\sum{a_{k}}=N-3$ is a
generator of $\varOmega[C]_{\scalebox{0.7}[0.7]{$P_{\infty}\\!$}}$.

###### Proof.

We set
$d=g_{\infty}{N}_{\infty}{A^{\prime}}_{\\!\infty}-{N}_{\infty}-{A^{\prime}}_{\\!\infty}$
and $e=m_{\infty}+(N-1)n_{\infty}$, thus $s=dm_{\infty}+e{N}_{\infty}=N-1$. By
substituting them, one can easily check
$-s({N}_{\infty}-{A^{\prime}}_{\\!\infty})+(1+\sum{a_{k}}){N}_{\infty}-d=(3-N+\sum{a_{k}}){N}_{\infty}$,
however it must be equal to 0 since $\prod(1-\lambda_{i}x)^{a_{i}}$ is
invertible. This means that $\sum{a_{k}}=N-3$. ∎

In Case 2, we set
${N^{\prime}}_{\infty}={N}_{\infty}+{A^{\prime}}_{\\!\infty}$ then we
similarly obtain below:

  1. (i)

If $d\geq
g_{\infty}{N^{\prime}}_{\infty}{A^{\prime}}_{\\!\infty}-{N^{\prime}}_{\infty}-{A^{\prime}}_{\\!\infty}+1$,
then
$u^{g_{\infty}-e}dz/z^{d+1}\notin\varOmega[C]_{\scalebox{0.7}[0.7]{$P_{\infty}\\!$}}$
for all $e\in\\{1,\ldots,g_{\infty}\\}$.

  2. (ii)

If
$d=g_{\infty}{N^{\prime}}_{\infty}{A^{\prime}}_{\\!\infty}-{N^{\prime}}_{\infty}-{A^{\prime}}_{\\!\infty},\,e=2n_{\infty}-m_{\infty}~{}({\rm
mod}\ g_{\infty})$, then
$u^{g_{\infty}-e}dz/z^{d+1}\in\varOmega[C]_{\scalebox{0.7}[0.7]{$P_{\infty}\\!$}}$.

If $w^{g_{\scalebox{0.6}[0.65]{$\infty$}}-e}dz/z^{d+1}\in\varOmega(X)$ is
equal to (3.3), the polynomial $\prod(1-\lambda_{i}u)^{a_{i}}$ is equal to

$\frac{m_{\infty}u(\partial{f_{\infty}}/\partial{u})(u)-g_{\infty}w^{g_{\scalebox{0.6}[0.65]{$\infty$}}}}{N}\cdot
w^{-s(m_{\scalebox{0.6}[0.65]{$\infty$}}-n_{\scalebox{0.6}[0.65]{$\infty$}})+(1+\sum{a_{k}})m_{\scalebox{0.6}[0.65]{$\infty$}}-e}z^{-s({N}_{\scalebox{0.6}[0.65]{$\infty$}}+{A^{\prime}}_{\\!\scalebox{0.6}[0.65]{$\infty$}})+(1+\sum{a_{k}}){N}_{\scalebox{0.6}[0.65]{$\infty$}}-d}.\vspace{-1mm}$

Recall that equations
$u^{{A^{\prime}}_{\\!\scalebox{0.6}[0.65]{$\infty$}}}=v^{N_{\scalebox{0.6}[0.65]{$\infty$}}}w,\,z^{{A^{\prime}}_{\\!\scalebox{0.6}[0.65]{$\infty$}}}=vw^{n_{\scalebox{0.6}[0.65]{$\infty$}}}$,
then we obtain the relation
$u=w^{m_{\scalebox{0.6}[0.65]{$\infty$}}}z^{{N}_{\scalebox{0.6}[0.65]{$\infty$}}}$
easily. By using this relation, then we have $s=-dm_{\infty}+e{N}_{\infty}$.

######  Lemma 4.4.

In Case 2, the pull-back
$\pi^{*}\omega_{(2-A_{\scalebox{0.6}[0.65]{$\infty$}},\bm{a})}$ is a generator
of $\varOmega[C]_{\scalebox{0.7}[0.7]{$P_{\infty}\\!$}}$ if and only if
$\sum{a_{k}}=0$.

###### Proof.

We set
$d=g_{\infty}{N^{\prime}}_{\infty}{A^{\prime}}_{\\!\infty}-{N^{\prime}}_{\infty}-{A^{\prime}}_{\\!\infty}$
and $e=n_{\infty}+(A_{\infty}-1)(m_{\infty}-n_{\infty})$ with
${N^{\prime}}_{\infty}={N}_{\infty}+{A^{\prime}}_{\\!\infty}$, thus
$s=-dm_{\infty}+e{N}_{\infty}=2-A_{\infty}$. This may not satisfy $0\leq s<N$,
then use $y^{N}=f(x)$ if necessary. By substituting them, one can easily check
$-s({N}_{\infty}+{A^{\prime}}_{\\!\infty})+(1+\sum{a_{k}}){N}_{\infty}-d=\sum{a_{k}}{N}_{\infty}$,
however it must be equal to 0 since $\prod(1-\lambda_{i}u)^{a_{i}}$ is
invertible. This means that $\sum{a_{k}}=0$. ∎

We obtain the regularity of some rational differential forms on $C$, which
will turn out to make a basis of the space of regular differential forms on
$C$ (cf. Corollary 4.6).

######  Theorem 4.5.

  1. (1)

Assume $N-\sum{A_{k}}\geq 0$. Then, for $0\leq s\leq N-1$, we have
$\pi^{*}\omega_{(s,\bm{a})}\in\varOmega[C]$ if

    1. (i)

$a_{j}\geq 0$ for all $j\in\\{0,\ldots,r\\}$ and

    2. (ii)

$0\leq\sum{a_{k}}\leq s-2$.

  2. (2)

Assume $N-\sum{A_{k}}<0$. Then, for $2-A_{\infty}\leq s\leq N-1$, we have
$\pi^{*}\omega_{(s,\bm{a})}\in\varOmega[C]$ if

    1. (i)

$a_{j}\geq 0$ for all $j\in\\{0,\ldots,r\\}$ and

    2. (ii)

$0\leq\sum{a_{k}}\leq s-2+A_{\infty}$.

###### Proof.

A differential form $dx/y^{N-1}$ and its products of some $x,y$ are regular at
$P_{j}$ for $j\in\\{0,\ldots,r\\}$ by Lemma 4.2. (1) In Case 1, the
differential form $\omega_{(N-1,\bm{a}^{\prime})}$ for
$\sum{a^{\prime}_{k}}=N-3$ and its products of some $1/x,y/x$ are regular at
$P_{\infty}$ by Lemma 4.3. The theorem in this case follows from the fact that
$\pi^{*}\omega_{(s,\bm{a})}$ for $\bm{a}$ satisfying (i) and (ii) is a linear
combination of

$\left(\frac{1}{x}\right)^{i}\left(\frac{y}{x}\right)^{N-1-s}\frac{x^{N-3}dx}{y^{N-1}}$

for $0\leq i\leq s-2$. In Case 3, recall that $P_{\infty}$ are nonsingular
points by Lemma 2.5. Then
$\varOmega[C]_{\scalebox{0.7}[0.7]{$P_{\infty}\\!$}}$ is equal to the set of
differential forms which is regular at
$Q_{\infty}\in\pi^{-1}(\\{P_{\infty}\\})$. So this case follows from (3.4).

(2) In Case 2, then a differential form
$\omega_{(2-A_{\scalebox{0.6}[0.65]{$\infty$}},\bm{a}^{\prime})}$ where $\sum
a^{\prime}_{k}=0$ and its products of some $1/y,x/y$ are regular at
$P_{\infty}$ by Lemma 4.4. The theorem in this case follows from the fact that
$\pi^{*}\omega_{(s,\bm{a})}$ for $\bm{a}$ satisfying (i) and (ii) is a linear
combination of

$\left(\frac{1}{y}\right)^{i}\left(\frac{x}{y}\right)^{s-2+A_{\infty}-i}\frac{dx}{y^{2-A_{\infty}}}$

for $0\leq i\leq s-2+A_{\infty}$. ∎

As an application, we obtain a basis of the regular differential module on
$C$. For each $\zeta\in\mu_{N}$, we have the automorphism of $C$ defined by
$(x,y)\mapsto(x,\zeta y)$, which can be extended to an automorphism of $X$,
say $\iota_{\zeta}$. This induces an action of $\mu_{N}$ on $\varOmega[C]$.
Indeed $\iota_{\zeta}$ stabilizes $P_{j}$ and induces a permutation of
$\pi^{-1}(\\{P_{j}\\})$ and we have ${\rm res}_{Q}(\omega)={\rm
res}_{\iota_{\zeta}(Q)}({\iota_{\zeta}}^{*}\omega)$ for each
$Q\in\pi^{-1}(\\{P_{j}\\})$, thanks to Remark 2.9.

######  Corollary 4.6.

Assume that a perfect field $K$ contains $\mu_{N}$ and
$\\{\lambda_{0},\ldots,\lambda_{r}\\}$. For $0\leq s\leq N-1$, let $W_{s}$ be
the subspace of $\varOmega[C]$ consisting of $\omega\in\varOmega[C]$ on which
$\mu_{N}$ acts by $\omega\mapsto\zeta^{s}\omega$ for all $\zeta\in\mu_{N}$.

  1. (1)

If $N-\sum{A_{k}}>0$, then

$\\{\pi^{*}(x^{i}dx/y^{s})\,;\,0\leq i\leq s-2\\}$ (4.2)

is a basis of $W_{s}$. In particular

$\dim W_{s}=s-1$ (4.3)

with $\dim\varOmega[C]=(N-1)(N-2)/2$.

  2. (2)

If $N-\sum{A_{k}}<0$, then

$\biggl{\\{}\pi^{*}(x^{i}y^{(j-1)N}dx/y^{s})\,;\,0\leq i\leq
s-2-jN+\sum{A_{k}},\ 1\leq j\leq\left\lfloor\frac{s-2+\sum
A_{k}}{N}\right\rfloor\biggr{\\}}$ (4.4)

is a basis of $W_{s}$. In particular

$\dim
W_{s}=\sum_{j=1}^{\left\lfloor\frac{s-2+\sum\\!A_{k}}{N}\right\rfloor}{\left(s-1-jN+\sum{A_{k}}\right)}$
(4.5)

with $\dim\varOmega[C]=(-1+\sum A_{k})(-2+\sum A_{k})/2$.

###### Proof.

(1) The differential forms in (4.2) belong to $W_{s}$ by Theorem 4.6 (1), and
these are linear independent. By using [3, Corollary III.9.10], the arithmetic
genus does not change among fibers of a flat family over a connected
noetherian scheme, and therefore so does the dimension of the space of global
sections of the dualizing sheaf by definition [3, Chapter III, §7]. Consider
the family

$y^{N}=\prod_{j=0}^{r}\prod_{k=1}^{A_{j}}(x-\gamma_{jk}z)\cdot\prod_{l=1}^{A_{\infty}}(-a_{l}x+z)$

which has $C$ as a special fiber (defined by $\gamma_{jk}=\lambda_{j}$ and
$a_{l}=0$). Since its generic fiber has arithmetic genus
$(N-1)(N-2)/2=\sum(s-1)$. This shows that the differential forms in (4.2) span
$W_{s}$.
(2) The differential forms in (4.4) belong to $W_{s}$ by Theorem 4.6 (2), and
these are linear independent. In the same way as in (1), using the fact that
the projective model of $C$:

$y^{N}z^{A_{\infty}}=\prod_{j=1}^{r}(x-\lambda_{j}z)^{A_{j}}$

deforms to a smooth curve of degree $\sum{A_{k}}$, we have
$\dim\varOmega[C]=(-1+\sum A_{k})(-2+\sum A_{k})/2$. Hence, to prove (4.5), it
suffices to show that the sum of the right hand side of (4.5) for
$s=0,\ldots,N-1$ is equal to $(-1+\sum A_{k})(-2+\sum A_{k})/2$. This follows
from the next fact. The set

$\biggl{\\{}s-1-jN+\sum A_{k}\,;\,0\leq s\leq N-1,1\leq
j\leq\left\lfloor\frac{s-2+\sum A_{k}}{N}\right\rfloor\biggr{\\}}$

is equal to $\\{1,\ldots,-2+\sum A_{k}\\}$ for $1\leq s-1-jN+\sum
A_{k}\leq-2+\sum A_{k}$. ∎

######  Example 4.7.

Let $C:y^{3}=x(x-1)^{2}(x-z)^{2}$ and $X$ be the desingularizaiton of $C$,
then

$\Biggl{\\{}\pi^{*}(dx),\pi^{*}\biggl{(}\frac{dx}{y}\biggr{)},\pi^{*}\biggl{(}\frac{xdx}{y}\biggr{)},\pi^{*}\biggl{(}\frac{dx}{y^{2}}\biggr{)},\pi^{*}\biggl{(}\frac{xdx}{y^{2}}\biggr{)},\pi^{*}\biggl{(}\frac{x^{2}dx}{y^{2}}\biggr{)}\Biggr{\\}}$

is a basis of $\varOmega[C]$, while
$\pi^{*}(dx/y),\pi^{*}(x^{2}dx/y^{2})\in\varOmega[X]$.

##  The space of regular differential forms on $\widetilde{C}$

One can consider partial desingularizations of $C$, i.e., the
desingularaization only around a subset of
$\\{P_{0},\ldots,P_{r},P_{\infty}\\}$. Avoiding the general setting, in this
section we consider the desingularization $\widetilde{C}$ only around
$P_{\infty}$ of the curve $C$, which would be the most interesting case. We
can exclude Case 3, since ${\widetilde{C}}=C$ in Case 3. Our aim is to give an
explicit basis of the space of the regular differential forms on
$\widetilde{C}$. Now $\widetilde{C}$ is described as below:

######  Definition 5.1.

Let $C$ be a curve associated to Appell-Lauricella hypergeometric series. We
define $\widetilde{C}$ by gluing $X_{\infty}$ and
$C\smallsetminus\\{P_{\infty}\\}$.

Next we construct the morphism $\widetilde{\pi}:X\rightarrow\widetilde{C}$ as
follows. Recall that $X$ was defined by gluing $X_{i}$ in Section 2, then it
suffices to define the morphism
$\widetilde{\pi}_{i}:X_{i}\rightarrow\widetilde{C}$ for each
$i\in\\{0,\ldots,r,\infty\\}$. We define $\widetilde{\pi}_{\infty}$ to be the
inclusion $X_{\infty}\subset\widetilde{C}$ and $\widetilde{\pi}_{i}$ for
$i=0,\ldots,r$ to be the composition of $\pi_{j}:X_{i}\to
C\smallsetminus\\{P_{0},\overset{\overset{i}{\vee}}{\ldots},P_{r},P_{\infty}\\}$
and the inclusion
$C\smallsetminus\\{P_{0},\overset{\overset{i}{\vee}}{\ldots},P_{r},P_{\infty}\\}\subset
C\smallsetminus\\{P_{\infty}\\}$. Here is a description of the regularity of
differential forms and an explicit basis on $\widetilde{C}$.

######  Theorem 5.2.

We have $\widetilde{\pi}^{*}\omega_{(s,\bm{a})}\in\varOmega[\widetilde{C}]$ if

  1. (i)

$a_{j}\geq 0$ for all $j\in\\{0,\ldots,r\\}$ and

  2. (ii)

$0\leq\sum{a_{k}}\leq\displaystyle\frac{s\sum{A_{k}}-(N,N-\sum{A_{k}})}{N}-1$.

###### Proof.

The fiber $Q_{\infty}\in\pi^{-1}(\\{P_{\infty}\\})$ are not singular points of
$\widetilde{C}$, so $\widetilde{\pi}^{*}\omega_{(s,\bm{a})}$ is an element of
$\varOmega[\widetilde{C}]_{\scalebox{0.7}[0.7]{$Q_{\infty}\\!$}}$ if and only
if $\widetilde{\pi}^{*}\omega_{(s,\bm{a})}\in\varOmega(X)$ is regular at
$Q_{\infty}$. Hence, use Proposition 3.5 and Lemma 4.2. ∎

As with the case of $\varOmega[C]$ in Section 4, the automorphism
$(x,y)\mapsto(x,\zeta y)$ for $\zeta\in\mu_{N}$ induces an action of $\mu_{N}$
on $\varOmega[\widetilde{C}]$. Then, let $\widetilde{W}_{s}$ be the subspace
of $\varOmega[\widetilde{C}]$ consisting of
$\omega\in\varOmega[\widetilde{C}]$ on which $\mu_{N}$ acts by
$\omega\mapsto\zeta^{s}\omega$ for all $\zeta\in\mu_{N}$.

######  Corollary 5.3.

For $0\leq s\leq N-1$, the set of $\widetilde{\pi}^{*}\omega_{s,i}$ with

$\omega_{s,i}:=\frac{x^{i-1}dx}{y^{s}},\quad 1\leq
i\leq\frac{s\sum{A_{k}}-(N,N-\sum{A_{k}})}{N}$ (5.1)

is a basis of $\widetilde{W}_{s}$. In particular

$\dim\widetilde{W}_{s}=\max\left\\{0,\left\lfloor\frac{s\sum{A_{k}}-(N,N-\sum{A_{k}})}{N}\right\rfloor\right\\}.$

###### Proof.

By Theorem 5.2, the differential forms $\omega_{s,i}$ in (5.1) belong to
$\widetilde{W}_{s}$ and they are linearly independent obviously. Let $n=\sum
A_{k}$ and $\mathcal{H}$ be the family

$y^{N}=\prod_{j=0}^{r}\prod_{k=1}^{A_{j}}(x-\gamma_{jk})$

where $(\gamma_{jk})\in\mathbb{A}^{n}(\accentset{{\cc@style\underline{\mskip
12.0mu}}}{K})$. Let $\widetilde{\mathcal{H}}$ be the familiy over
$\mathbb{A}^{n}(\accentset{{\cc@style\underline{\mskip 12.0mu}}}{K})$ obtained
as the fiberwise desingularization only at $\infty$ of $\mathcal{H}$. We
remark that $\varOmega[\widetilde{C}]$ is the space of global secions of the
dualizing sheaf of $\widetilde{C}$. By the similar way as in Corollary 4.6,
the dimension of $\varOmega[\widetilde{\mathcal{H}}]$ is equal to
$\dim\varOmega[{\widetilde{{\mathcal{H}}}}(t)]$ for a smooth fiber
$\widetilde{\mathcal{H}}(t)$ with
$t\in\mathbb{A}^{n}(\accentset{{\cc@style\underline{\mskip 12.0mu}}}{K})$ of
$\widetilde{\mathcal{H}}$. And $\dim\varOmega[{\widetilde{{\mathcal{H}}}}(t)]$
is equal to the dimension we have computed in Theorem 3.6 for $A_{j}=1$ for
$j=1,\ldots,r$, which is

$\sum_{s=0}^{N-1}\max\left\\{0,\left\lfloor\frac{s\sum{A_{k}}-(N,N-\sum{A_{k}})}{N}\right\rfloor\right\\}.$

This implies that $\\{\omega_{s,i}\\}$ has to span $\widetilde{W}_{s}$. ∎

######  Example 5.4.

We consider the curve $C:y^{3}=x(x-1)^{2}(x-z)^{2}$ in Example 4.7, then

$\Biggl{\\{}\pi^{*}\biggl{(}\frac{dx}{y}\biggr{)},\pi^{*}\biggl{(}\frac{dx}{y^{2}}\biggr{)},\pi^{*}\biggl{(}\frac{xdx}{y^{2}}\biggr{)},\pi^{*}\biggl{(}\frac{x^{2}dx}{y^{2}}\biggr{)}\Biggr{\\}}$

is a basis of $\varOmega[\widetilde{C}]$.

##  The modified Cartier operator on the regular differential module

From now on, we consider the case that the characteristic of $K$ is $p>0$. Let

$C:y^{N}=f(x)=x^{A_{0}}(x-\lambda_{1})^{A_{1}}\cdots(x-\lambda_{r})^{A_{r}}$

be a curve associated to Appell-Lauricella hypergeometric series as in
Definition 2.1. Now let us recall [6, Definition 2.1] the definition of the
modified Cartied operator $\mathcal{C}^{\prime}$ on $X$.

######  Definition 6.1.

For all $\omega\in\varOmega(X)$, there exist $\phi,\eta\in K(x,y)$ such that
$\eta^{p}\in K(x^{p},y^{p})$ and

$\omega=d\phi+\eta^{p}x^{p-1}dx.$

Then, the modified Cartier operator
$\mathcal{C}^{\prime}:\varOmega(X)\longrightarrow\varOmega(X)$ is defined as
$\mathcal{C}^{\prime}(\omega)=\eta dx$.

It is well-known that the modified Cartier operator $\mathcal{C}^{\prime}$
stabilizes $\varOmega[X]$. The next theorem says that this also holds for
$\varOmega[C]$ and $\varOmega[\widetilde{C}]$, where $\widetilde{C}$ is the
partial desingularization of $C$ only at $\infty$ as we constructed in
Definition 5.1.

######  Theorem 6.2.

The modified Cartier operator $\mathcal{C}^{\prime}$ stabilizes $\varOmega[C]$
and $\varOmega[\tilde{C}]$. More generally, $\mathcal{C}^{\prime}$ stabilizes
$\varOmega[C]_{P}$ for any closed point $P\in C$, where $\varOmega[C]_{P}$ is
as defined in (4.1).

###### Proof.

It suffices to show the second assertion. Let $\omega\in\varOmega[C]_{P}$.
Write

$\omega=d\phi+\eta^{p}x^{p-1}dx.$ (6.1)

Let $f$ be an arbitrary element of ${\mathcal{O}}_{C,P}$. Mutiplying (6.1) by
$\pi^{*}(f^{p})$, we get

$\pi^{*}(f^{p})\omega=d(\pi^{*}(f^{p})\phi)+(\pi^{*}(f)\eta)^{p}x^{p-1}dx.$

We have

$\left(\sum_{\pi(Q)=P}{\rm res}_{Q}(\pi^{*}(f)\eta
dx)\right)^{p}=\sum_{\pi(Q)=P}{\rm
res}_{Q}((\pi^{*}(f)\eta)^{p}x^{p-1}dx)=\sum_{\pi(Q)=P}{\rm
res}_{Q}(\pi^{*}(f^{p})\omega).$

The right hand side is zero by $\omega\in\varOmega[C]_{P}$. Hence,
$\sum_{\pi(Q)=P}{\rm res}_{Q}(\pi^{*}(f)\eta dx)=0$ and therefore
$\mathcal{C}^{\prime}(\omega)=\eta dx\in\varOmega[C]_{P}$. ∎

######  Definition 6.3.

The Cartier-Manin matrix $A$ of $X$ (resp. $C$ or $\widetilde{C}$) with
respect to a basis $\\{\omega_{i}\\}$ of $\varOmega[X]$ (resp. $\varOmega[C]$
or $\varOmega[\widetilde{C}]$) is given by $A=(a_{ij})$ with
${\mathcal{C}}^{\prime}\omega_{j}=\sum_{i}a_{ij}^{1/p}\omega_{i}$.

In order to describe a Cartier matrix (i.e, the modified Cartier operator
$\mathcal{C}^{\prime}$ on $\varOmega[X]$, $\varOmega[C]$ or
$\varOmega[\tilde{C}]$), it suffices to describe ${\mathcal{C}}^{\prime}$ on

$\omega_{s,j}=\frac{x^{j-1}}{y^{s}}\ dx$

for $0\leq s\leq N-1$, since our basis of the space of regular differential
forms (obtained in the previous sections) is given by linear combinations of
$\omega_{s,j}$. We shall see that it can be described in terms of the Appell-
Lauricella hypergerometric series. Note that $N$ is not divided by $p$, hence
for $1\leq s\leq N-1$ there uniquely exist integers $m^{\prime}_{s}$ and
$n^{\prime}_{s}$ with $1\leq m^{\prime}_{s}\leq N-1$ and $0\leq
n^{\prime}_{s}<p$ such that ${m^{\prime}_{s}}p-{n^{\prime}_{s}}N=s$ by Lemma
B.5. We rewrite $\omega_{s,j}$ as

$\omega_{s,j}=y^{-s}x^{j-1}dx=y^{-m^{\prime}_{s}p}x^{j-1}y^{m^{\prime}_{s}p-s}dx={(y^{m^{\prime}_{s}})}^{-p}x^{j-1}{f(x)}^{n^{\prime}_{s}}dx.$

Let $\gamma_{s,e}$ be the coefficient of $x^{e}$ in the polynomial
$f(x)^{n^{\prime}_{s}}$, namely

${f(x)}^{n^{\prime}_{s}}=\sum_{e=0}^{n^{\prime}_{s}\deg(f)}\gamma_{s,e}x^{e}.$
(6.2)

Now we have

$\displaystyle\omega_{s,j}$
$\displaystyle=(y^{m^{\prime}_{s}})^{-p}\hbox{$\displaystyle\sum_{j+e\not\equiv
0\,({\rm
mod}\,p)}\gamma_{s,e}x^{j+e-1}dx$}+\sum_{l}\gamma_{s,(l+1)p-j}\frac{x^{(l+1)p}}{y^{m^{\prime}_{s}p}}\frac{dx}{x}$
$\displaystyle=d\biggl{(}y^{-m^{\prime}_{s}p}\raise
4.30554pt\hbox{$\displaystyle\sum_{j+e\not\equiv 0\,({\rm
mod}\,p)}$}\frac{\gamma_{s,e}x^{j+e}}{j+e}\biggr{)}+\sum_{l}\gamma_{s,(l+1)p-j}\frac{x^{lp}}{y^{m^{\prime}_{s}p}}x^{p-1}dx$
(6.3)

where $l$ runs from $\displaystyle\left\lceil\frac{j}{p}-1\right\rceil$ to
$\displaystyle\left\lfloor\frac{n^{\prime}_{s}\deg(f)+j}{p}-1\right\rfloor$.

Let ${\mathcal{F}}(a,b_{2},\ldots,b_{r},c\,;\lambda_{2},\ldots,\lambda_{r})$
be the Appell-Lauricella hypergeometric series associated to
$C:y^{N}=x^{A_{0}}(x-1)^{A_{1}}\prod_{k=2}^{r}(x-\lambda_{k})^{A_{k}}$, namely

$a=-1+\sum_{k=0}^{r}(A_{k}/N),\quad b_{i}=A_{i}/N,\quad
c=a+1-(A_{1}/N).\vspace{-1mm}$

As well as in the case of elliptic curves, we need to introduce a truncation
of hypergeometric series.

######  Definition 6.4 (Truncation of AL-hypergeometric series).

For $(\sigma,\tau_{1},\ldots,\tau_{r})\in{\mathbb{Z}}^{r+1}$, let

${\mathcal{F}}^{(\sigma;\tau_{1},\ldots,\tau_{r})}(a,b_{2},\ldots,b_{r},c\,;\lambda_{2},\ldots,\lambda_{r})$

be the sum of the $\lambda_{2}^{e_{2}}\cdots\lambda_{r}^{e_{r}}$-terms of
${\mathcal{F}}(a,b_{2},\ldots,b_{r},c\,;\lambda_{2},\ldots,\lambda_{r})$ for
$(e_{2},\ldots,e_{r})$ satisfying $e_{j}\leq\tau_{j}$ for $j=2,\ldots,r$ and

$\sigma-\tau_{1}\leq\sum_{k=2}^{r}e_{k}\leq\sigma.$

We call it the truncation of
${\mathcal{F}}(a,b_{2},\ldots,b_{r},c\,;\lambda_{2},\ldots,\lambda_{r})$ with
respect to $(\sigma;\tau_{1},\ldots,\tau_{r})$.

Unfortunately, it is not true in general that one can describe
${\mathcal{C}}^{\prime}$ in terms of the AL-hypergeometric series
${\mathcal{F}}(a,b_{2},\ldots,b_{r},c\,;\lambda_{2},\ldots,\lambda_{r})$
itself of $C$. We shall see that ${\mathcal{C}}^{\prime}$ can be described in
terms of AL-hypergeometric series associated to a deformation of $f(x)$ which
is separable except for the factor of $x$, see $f_{0}(x)$ in (6.4) below for
the explicit form. The description is as follows.

######  Theorem 6.5.

Let $a^{\prime}$ be a positive rational number with $a^{\prime}\equiv
s\deg(f)/N-j\pmod{p}$ and set $c^{\prime}=a^{\prime}+1-s/N$ and
$d^{\prime}=n^{\prime}_{s}\deg(f)-(l+1)p+j$. For
$\left\lceil\frac{j}{p}-1\right\rceil\leq
l\leq\left\lfloor\frac{n^{\prime}_{s}\deg(f)+j}{p}-1\right\rfloor$, we have

$\gamma_{s,(l+1)p-j}=\frac{(c^{\prime};d^{\prime})}{(a^{\prime};d^{\prime})}{\mathcal{F}}^{(d^{\prime};n^{\prime}_{s},\ldots,n^{\prime}_{s})}(a^{\prime},\underbrace{s/N,\ldots,s/N}_{-1+\sum_{k\geq
1}A_{k}},c^{\prime};\underbrace{1,\ldots,1}_{A_{1}-1},\underbrace{\lambda_{2},\ldots,\lambda_{2}}_{A_{2}},\ldots,\underbrace{\lambda_{r},\ldots,\lambda_{r}}_{A_{r}}),$

where the right hand side (a priori belonging to
${\mathbb{Q}}[\lambda_{2},\ldots,\lambda_{r}]$) is considered as a polynomial
over $\mathbb{F}_{p}$ (Note that the denominator of any coefficient is coprime
to $p$).

First we see that it is enough to show the case of
$A_{1}=A_{2}=\cdots=A_{r}=1$.

###### Reduction to the case of $A_{k}=1$ for $k=1,\ldots,r$.

Consider

$f_{0}(x)=x^{A_{0}}(x-1)\prod_{t=1}^{A_{1}-1}(x-\lambda_{1t})\prod_{k=2}^{r}\prod_{t=1}^{A_{k}}(x-\lambda_{kt}).$
(6.4)

Write

$f_{0}(x)^{n^{\prime}_{s}}=\sum_{c}(\delta_{0})_{i,c}x^{c}.$ (6.5)

Then

$\gamma_{s,c}=(\delta_{0})_{s,c}|_{\lambda_{kt}=\lambda_{k}\text{ for
}k=1,\ldots,r\text{ and }t=1,\ldots,A_{k}}$ (6.6)

with $\lambda_{1}=1$ holds. We shall see in Proposition 6.7 below that
$(\delta_{0})_{s,(l+1)p-j}$ is described as

$\frac{(c^{\prime};d^{\prime})}{(a^{\prime};d^{\prime})}{\mathcal{F}}^{(d^{\prime};n^{\prime}_{s},\ldots,n^{\prime}_{s})}(a^{\prime},\underbrace{s/N,\ldots,s/N}_{-1+\sum_{k\geq
1}A_{k}},c^{\prime};\lambda_{11},\ldots,\lambda_{1,A_{1}-1},\lambda_{21},\ldots,\lambda_{2A_{2}},\ldots,\lambda_{r1},\ldots,\lambda_{rA_{r}}),$
(6.7)

which is the result in the case of $A_{k}=1$ for $k=1,\ldots,r$. The theorem
follows from (6.6) and (6.7). ∎

######  Lemma 6.6.

Let $a^{\prime}$ be a positive rational number with $a^{\prime}\equiv
s\deg(f)/N-j\pmod{p}$ and set $c^{\prime}=a^{\prime}+1-s/N$ and
$d^{\prime}=n^{\prime}_{s}\deg(f)-(l+1)p+j$. For a partition
$(d^{\prime}_{1},\ldots,d^{\prime}_{r})$ of $d^{\prime}$ (i.e.,
$d^{\prime}=\sum_{k=1}^{r}d^{\prime}_{k}$) with $0\leq d^{\prime}_{k}<p$, the
following is true:

$(-1)^{d_{k}}\binom{-sA_{k}/N}{d_{k}}=\frac{(sA_{k}/N\,;d_{k})}{(1\,;d_{k})}$

in ${\mathbb{Q}}$ for $k=1,\ldots,r$ and

$(-1)^{d_{1}}\binom{-sA_{1}/N}{d_{1}}=(-1)^{d_{1}}\binom{n^{\prime}_{s}A_{1}}{d_{1}}=\frac{(c^{\prime};d^{\prime})}{(a^{\prime};d^{\prime})}\frac{(a^{\prime};d^{\prime}-d_{1})}{(c^{\prime};d^{\prime}-d_{1})}$

in ${\mathbb{F}}_{p}$.

###### Proof.

The former equation is obvious from the definition of the symbol $(*;*)$. To
show the latter, we use the following two equalities

$(-x\,;n)=(-1)^{n}(x-n+1\,;n)$ (6.8)

for arbitrary $x\in\mathbb{R}$, and

$(x\,;n+m)=(x\,;n)(x+n\,;m)$ (6.9)

for arbitrary $x,y\in\mathbb{R}$. By definition,

$a=-1+\sum_{k=0}^{r}(A_{k}/N),\quad b_{j}=A_{j}/N~{}(j=2,\ldots,r),\quad
c=a+1-(A_{1}/N),$

and

$a^{\prime}=sa-j+s,\quad{b^{\prime}}_{\\!k}=sb_{k},\quad
c^{\prime}=sc-j+1,\quad
d^{\prime}={n^{\prime}_{s}}\sum_{k=0}^{r}A_{k}-(l+1)p+j$

so we have

$sA_{1}/N=s(a-c+1)=(sa-j+s)-(sa-j+1)+1=a^{\prime}-c^{\prime}+1,$ (6.10)
$d^{\prime}\equiv(a+1){n^{\prime}_{s}}N+j\equiv(a+1)({m^{\prime}_{s}}p-s)+j\equiv-
sa+j-s\equiv-a^{\prime}~{}({\rm mod}\ p).$ (6.11)

Now using (6.9), then

$\frac{(c^{\prime};d^{\prime})}{(a^{\prime};d^{\prime})}\frac{(a^{\prime};d^{\prime}-d_{1})}{(c^{\prime};d^{\prime}-d_{1})}=\frac{(c^{\prime}+d^{\prime}-d_{1}\,;d_{1})}{(a^{\prime}+d^{\prime}-d_{1}\,;d_{1})}$

and we have

$\frac{(c^{\prime}+d^{\prime}-d_{1}\,;d_{1})}{(a^{\prime}+d^{\prime}-d_{1}\,;d_{1})}\equiv\frac{(c^{\prime}-a^{\prime}-d_{1}\,;d_{1})}{(-d_{1}\,;d_{1})}~{}({\rm
mod}\ p)$

And using (6.8), then

$\frac{(c^{\prime}-a^{\prime}-d_{1}\,;d_{1})}{(-d_{1}\,;d_{1})}=\frac{(-1)^{d_{1}}(a^{\prime}-c^{\prime}+1\,;d_{1})}{(-1)^{d_{1}}(1\,;d_{1})}$

and we have

$\frac{(-1)^{d_{1}}(a^{\prime}-c^{\prime}+1\,;d_{1})}{(-1)^{d_{1}}(1\,;d_{1})}=\frac{(sA_{1}/N\,;d_{1})}{(1\,;d_{1})}.$

Thus we conclude

$\frac{(c^{\prime};d^{\prime})}{(a^{\prime};d^{\prime})}\frac{(a^{\prime};d^{\prime}-d_{1})}{(c^{\prime};d^{\prime}-d_{1})}\equiv\frac{(sA_{1}/N\,;d_{1})}{(1\,;d_{1})}~{}({\rm
mod}\ p).$

Note that the denominator is coprime to $p$ in the above discussion if
$d_{1}<p$. ∎

In order to show Theorem 6.5, we reuse the equation 6:

$\frac{x^{j-1}dx}{y^{s}}=d\biggl{(}y^{-{m^{\prime}_{s}}p}\raise
4.30554pt\hbox{$\displaystyle\sum_{j+e\not\equiv 0\,({\rm
mod}\,p)}$}\frac{\gamma_{s,e}x^{j+e}}{j+e}\biggr{)}+\sum_{l}\gamma_{s,(l+1)p-j}\frac{x^{lp}}{y^{{m^{\prime}_{s}}p}}x^{p-1}dx.$
(6.12)

Expanding the left hand side, we have

$\displaystyle\frac{x^{j-1}dx}{y^{s}}$ $\displaystyle=$
$\displaystyle\frac{x^{j-1-\frac{sA_{0}}{N}}dx}{\sqrt[N]{(x-1)^{A_{1}}(x-\lambda_{2})^{A_{2}}\cdots(x-\lambda_{r})^{A_{r}}}^{s}}$
$\displaystyle=$
$\displaystyle\sum_{e_{1},\ldots,e_{r}}\prod_{k=1}^{r}\binom{-sA_{k}/N}{e_{k}}(-\lambda_{k})^{-sA_{k}/N-e_{k}}x^{j-1-\frac{sA_{0}}{N}}x^{e_{1}+\cdots+e_{r}}dx$

with $\lambda_{1}=1$. The right hand side is

$\displaystyle\sum_{l=0}^{\left\lfloor\frac{n^{\prime}_{s}\sum
A_{k}+j}{p}-1\right\rfloor}\gamma_{s,(l+1)p-j}\frac{x^{lp}}{y^{{m^{\prime}_{s}}p}}x^{p-1}dx$
$\displaystyle=$
$\displaystyle\sum_{l}\gamma_{s,(l+1)p-j}\sum_{e^{\prime}_{1},\ldots,e^{\prime}_{r}}\prod_{k=1}^{r}\binom{-m^{\prime}_{s}A_{k}/N}{e^{\prime}_{k}}(-\lambda_{k})^{-m^{\prime}_{s}pA_{k}/N-pe^{\prime}_{k}}x^{(l+1)p-1-\frac{m^{\prime}_{s}pA_{0}}{N}}x^{pe^{\prime}_{1}+\cdots+pe^{\prime}_{r}}dx$

Comparing the coefficients of $x^{pM+p-1-\frac{m^{\prime}_{s}pA_{0}}{N}}$
($M=0,1,2,\ldots$), we have

$\displaystyle\sum_{l=0}^{\min\left(M,\left\lfloor\frac{n^{\prime}_{s}\sum
A_{k}+j}{p}-1\right\rfloor\right)}\gamma_{s,(l+1)p-j}\sum_{e^{\prime}_{1}+\cdots+e^{\prime}_{r}=M-l}\prod_{k=1}^{r}\binom{-m^{\prime}_{s}A_{k}/N}{e^{\prime}_{k}}(-\lambda_{k})^{-m^{\prime}_{s}pA_{k}/N-pe^{\prime}_{k}}$
$\displaystyle=\sum_{e_{1}+\cdots+e_{r}=pM+p-j-n^{\prime}_{s}A_{0}}\prod_{k=1}^{r}\binom{-sA_{k}/N}{e_{k}}(-\lambda_{k})^{-sA_{k}/N-e_{k}}.$

Now recall that $\gamma_{s,e}$ are polynomials in
$\lambda_{2},\ldots,\lambda_{r}$ and the degree about $\lambda_{j}$ is equal
to ${n^{\prime}_{s}}A_{j}$. Hence we can write

$\gamma_{s,(l+1)p-j}=\sum_{d_{2}=0}^{{n^{\prime}_{s}}A_{2}}\cdots\sum_{d_{r}=0}^{{n^{\prime}_{s}}A_{r}}\alpha_{(l,\bm{d})}{\lambda_{2}}^{d_{2}}\cdots{\lambda_{r}}^{d_{r}}$

where $\bm{d}=(d_{2},\ldots,d_{r})$.

Hence

$\displaystyle\sum_{l}\sum_{d_{2}=0}^{{n^{\prime}_{s}}A_{2}}\cdots\sum_{d_{r}=0}^{{n^{\prime}_{s}}A_{r}}\alpha_{(l,\bm{d})}{\lambda_{2}}^{d_{2}}\cdots{\lambda_{r}}^{d_{r}}\sum_{e^{\prime}_{1}+\ldots+e^{\prime}_{r}=M-l}\prod_{k=1}^{r}\binom{-m^{\prime}_{s}A_{k}/N}{e^{\prime}_{k}}(-\lambda_{k})^{-m^{\prime}_{s}pA_{k}/N-pe^{\prime}_{k}}$
$\displaystyle=\sum_{e_{1}+\cdots+e_{r}=pM+p-j-n^{\prime}_{s}A_{0}}\prod_{k=1}^{r}\binom{-sA_{k}/N}{e_{k}}(-\lambda_{k})^{-sA_{k}/N-e_{k}}$

Muliplying $\prod_{k}(-\lambda_{k})^{m^{\prime}_{s}pA_{k}/N}$, we have

$\displaystyle\sum_{l}\sum_{d_{2}=0}^{{n^{\prime}_{s}}A_{2}}\cdots\sum_{d_{r}=0}^{{n^{\prime}_{s}}A_{r}}\alpha_{(l,\bm{d})}{\lambda_{2}}^{d_{2}}\cdots{\lambda_{r}}^{d_{r}}\sum_{e^{\prime}_{1}+\cdots+e^{\prime}_{r}=M-l}\prod_{k=1}^{r}\binom{-m^{\prime}_{s}A_{k}/N}{e^{\prime}_{k}}(-\lambda_{k})^{-pe^{\prime}_{k}}$
$\displaystyle=\sum_{e_{1}+\cdots+e_{r}=pM+p-j-n^{\prime}_{s}A_{0}}\prod_{k=1}^{r}\binom{-sA_{k}/N}{e_{k}}(-\lambda_{k})^{n^{\prime}_{s}A_{k}-e_{k}}$

Comparing the coefficients of
$\lambda_{1}^{d^{\prime}_{1}}\lambda_{2}^{d^{\prime}_{2}}\cdots\lambda_{r}^{d^{\prime}_{r}}$
(for $d^{\prime}_{k}\geq-pe^{\prime}_{k}$), we have

$\sum_{l}\sum_{\small\begin{matrix}d_{1},\ldots,d_{r}\\\
d_{k}-d^{\prime}_{k}\in p{\mathbb{Z}}_{\geq 0}\\\ \sum d_{k}=p(M-l)+\sum
d^{\prime}_{k}\end{matrix}}\alpha_{(l,\bm{d})}\prod_{k=1}^{r}\binom{-m^{\prime}_{s}A_{k}/N}{(d_{k}-d^{\prime}_{k})/p}(-1)^{d_{k}-d^{\prime}_{k}}=\prod_{k=1}^{r}\binom{-sA_{k}/N}{n^{\prime}_{s}A_{k}-d^{\prime}_{k}}(-1)^{d^{\prime}_{k}},$

In LHS we set $d^{\prime}_{k}=d_{k}-pe^{\prime}_{k}$ and in RHS we set
$d^{\prime}_{k}=n^{\prime}_{s}A_{k}-e_{k}$ and therefore
$e_{1}+\cdots+e_{r}=n^{\prime}_{s}\sum A_{k}-\sum
d^{\prime}_{k}=pM+p-j-n^{\prime}_{s}A_{0}$, whence

$M=\frac{n^{\prime}_{s}(\sum_{k=0}^{r}A_{k})-(\sum d^{\prime}_{k})-p+j}{p},$
(6.13)

i.e., $\sum d^{\prime}_{k}$ has to be congruent modulo $p$ to
$j+n^{\prime}_{s}(\sum_{k=0}^{r}A_{k})$. Hence we have

$\sum_{l=0}^{\min\left(M,\left\lfloor\frac{n^{\prime}_{s}\sum
A_{k}+j}{p}-1\right\rfloor\right)}\sum_{{\small\begin{matrix}d_{1},\ldots,d_{r}\\\
d_{k}-d^{\prime}_{k}\in p{\mathbb{Z}}_{\geq 0}\\\ \sum d_{k}=p(M-l)+\sum
d^{\prime}_{k}\end{matrix}}}\alpha_{(l,\bm{d})}\prod_{k=1}^{r}\binom{-m^{\prime}_{s}A_{k}/N}{(d_{k}-d^{\prime}_{k})/p}(-1)^{d_{k}}=\prod_{k=1}^{r}\binom{-sA_{k}/N}{n^{\prime}_{s}A_{k}-d^{\prime}_{k}}$
(6.14)

for each $(d^{\prime}_{k})$, where $M$ is given by (6.13).

######  Proposition 6.7.

Assume $A_{k}=1$ for $k=1,\ldots,r$. Let $a^{\prime}$ be a positive rational
number with $a^{\prime}\equiv s\deg(f)/N-j\pmod{p}$ and
$c^{\prime}=a^{\prime}+1-s/N$. Set $b^{\prime}_{k}=sb_{k}$ and
$d^{\prime}=n^{\prime}_{s}\deg(f)-(l+1)p+j$. Consider a polynomial in
$\lambda_{2},\ldots,\lambda_{r}$ over $\mathbb{Q}$

$\delta_{s,(l+1)p-j}:=\frac{(c^{\prime}:d^{\prime})}{(a^{\prime}:d^{\prime})}{\mathcal{F}}^{(d^{\prime};\tau_{1},\ldots,\tau_{r})}(a^{\prime},b^{\prime}_{2},\ldots,b^{\prime}_{r},c^{\prime};\lambda_{2},\ldots,\lambda_{r}),$

where $\tau_{k}:={n^{\prime}_{s}}$ for $k=1,\ldots,r$.

  1. (1)

The denominator of every coefficient of $\delta_{s,(l+1)p-j}$ is coprime to
$p$. Hence we can consider it as a polynomial over ${\mathbb{F}}_{p}$, say
$\overline{\delta}_{s,(l+1)p-j}$.

  2. (2)

We have the equality $\gamma_{s,(l+1)p-j}=\overline{\delta}_{s,(l+1)p-j}$.

######  Remark 6.8.

For $a^{\prime}_{0}=s\deg(f)/N-j$ and $c^{\prime}_{0}=s(\deg(f)-1)/N-j+1$, the
AL-hypergeometric series
${\mathcal{F}}(a^{\prime}_{0},b^{\prime}_{2},\ldots,b^{\prime}_{r},c^{\prime}_{0};\lambda_{2},\ldots,\lambda_{r})$
is associated to

$\frac{x^{j-1}}{y^{s}}dx=\frac{1}{x^{sA_{0}/N-j+1}(x-1)^{sA_{1}/N}(x-\lambda_{2})^{sA_{2}/N}\cdots(x-\lambda_{r})^{sA_{r}/N}}.$

with $A_{k}=1$ for $k\geq 1$. Indeed
$a^{\prime}_{0}=-1+(sA_{0}/N-j+1)+\sum_{k=1}^{r}sA_{k}/N=s(-1+\sum
A_{k}/N)-j+s=sa-j+s$ and
$c^{\prime}_{0}=a^{\prime}_{0}+1-(sA_{1}/N)=s(a-A_{1}/N+1)-j+1=sc-j+1$. In the
theorem, we use a positive $a^{\prime}$ instead of possibly non-positive
$a^{\prime}_{0}$ so that $(a^{\prime}:d^{\prime})$ ($\in{\mathbb{Q}}$) and the
denominators ($\in{\mathbb{Q}}$) of coefficients of the hypergeometric series
are not zero.

###### Proof of Proposition 6.7.

Assume $A_{k}=1$ for $k=2,\ldots,r$. For $d^{\prime}_{k}$ with $0\leq
d^{\prime}_{k}\leq n^{\prime}_{s}$, in (6.14) $d_{k}$ has to be equal to
$d^{\prime}_{k}$, since $0\leq d_{k}\leq n^{\prime}_{s}<p$ and
$d_{k}-d^{\prime}_{k}\in p{\mathbb{Z}}_{\geq 0}$. Hence (6.14) reads

$\alpha_{(l,d)}=(-1)^{\sum_{k=1}^{r}d_{k}}\prod_{k=1}^{r}\binom{-sA_{k}/N}{n^{\prime}_{s}A_{k}-d_{k}}$

with

$l=\frac{n^{\prime}_{s}(\sum_{k=0}^{r}A_{k})-(\sum d_{k})-p+j}{p}.$ (6.15)

As $-s/N\equiv n^{\prime}_{s}\pmod{p}$, we have
$\binom{-sA_{k}/N}{n^{\prime}_{s}A_{k}-d_{k}}=\binom{-sA_{k}/N}{d_{k}}$ in
${\mathbb{F}}_{p}$. Hence, by Lemma 6.6 we have

$\alpha_{(l,d)}=\frac{(c^{\prime};d^{\prime})}{(a^{\prime};d^{\prime})}\frac{(a^{\prime};d^{\prime}-d_{1})}{(c^{\prime};d^{\prime}-d_{1})}\prod_{k=2}^{r}\frac{(sA_{k}/N;d_{k})}{(1;d_{k})}.$
(6.16)

By (6.15) we have

$d^{\prime}-d_{1}=n^{\prime}_{s}\sum A_{k}-(l+1)p+j-d_{1}=\sum_{k=2}^{r}d_{k}$

and therefore (6.16) is equal to the
$\lambda_{2}^{d_{2}}\cdots\lambda_{r}^{d_{r}}$-coefficient of
$\displaystyle\frac{(c^{\prime};d^{\prime})}{(a^{\prime};d^{\prime})}{\mathcal{F}}^{\tau}(a^{\prime},b^{\prime}_{2},\ldots,b^{\prime}_{r},c^{\prime};\lambda_{2},\ldots,\lambda_{r})$.
∎

######  Example 6.9.

We consider the case that $C$ is a (nonsingular) hyperelliptic curve of genus
$g\geq 1$. Write $C:y^{2}=f(x):=x(x-1)(x-\lambda_{2})\cdots(x-\lambda_{2g})$
with separable $f(x)$. Let us describe the Cartier operator on
$\varOmega[\widetilde{C}]$. The basis is given by $x^{j-1}/ydx$ for
$j=1,\ldots,g$ (cf. Corollary 5.3). Note that $a=g-1/2$, $b_{i}=1/2$ and
$c=g$. We get $m^{\prime}_{1}=1$ and $n^{\prime}_{1}=(p-1)/2$. Put
$a^{\prime}=(2g+1)/2-j=g+1/2-j$ and $c^{\prime}=g-j+1$, which are positive,
and set $d^{\prime}=\frac{p-1}{2}\deg(f)-ip+j$. Then we have

$\gamma_{1,ip-j}=\frac{(p(2g-2i+1)-1)!!}{(p(2g-2i+1)-2)!!}\frac{(2g-2j-1)!!}{(2g-2j)!!}{\mathcal{F}}^{(d^{\prime};n^{\prime}_{1},\ldots,n^{\prime}_{1})}(a^{\prime},1/2,\ldots,1/2,c^{\prime};\lambda_{2},\ldots,\lambda_{2g})$

for $i=1,\ldots,g$. For example, for $(g,p)=(2,3)$ with
$f(x)=x(x-1)(x-z_{1})(x-z_{2})(x-z_{3})$, the Cartier-Manin matrix
$(\gamma_{1,ip-j})$ is

$\begin{pmatrix}2z_{1}z_{2}z_{3}+2z_{1}z_{2}+2z_{1}z_{3}+2z_{2}z_{3}&z_{1}z_{2}z_{3}\\\
1&2z_{1}+2z_{2}+2z_{3}+2\end{pmatrix}.$

Here, ${\mathcal{F}}(5/2-j,1/2,1/2,1/2,3-j;z_{1},z_{2},z_{3})$ truncated by
$z_{k}$-degree $\leq 1$ with coefficients in ${\mathbb{Q}}$ is

$\displaystyle
1+\frac{3}{20}(z_{1}+z_{2}+z_{3})+\frac{1}{32}(z_{1}z_{2}+z_{1}z_{3}+z_{2}z_{3})+\frac{1}{128}z_{1}z_{2}z_{3}$
$\displaystyle\text{for }j=1,$ $\displaystyle
1+\frac{1}{16}(z_{1}+z_{2}+z_{3})+\frac{3}{320}(z_{1}z_{2}+z_{1}z_{3}+z_{2}z_{3})+\frac{1}{512}z_{1}z_{2}z_{3}$
$\displaystyle\text{for }j=2.$

For further truncations, use $d^{\prime}=5-3i+j$ and $n^{\prime}_{1}=1$ with
Definition 6.4.

In Theorem 6.5, we have shown that $\varOmega[C]$ and
$\varOmega[\widetilde{C}]$ are closed under the modified Cartier operator
$\mathcal{C}^{\prime}$. We can show more as for the subspaces $W_{s}$ (resp.
$\widetilde{W}_{s}$) of $\varOmega[C]$ (resp. $\varOmega[\widetilde{C}]$).

######  Theorem 6.10.

  1. (1)

${\mathcal{C}}^{\prime}$ on $\varOmega[C]$ sends $W_{s}$ to
$W_{m^{\prime}_{s}}$. Moreover, in Case 1 (i.e., $N>\deg(f)$) we have

${\mathcal{C}}^{\prime}\omega_{s,j}=\sum_{l=0}^{m^{\prime}_{s}-2}\gamma_{s,(l+1)p-j}^{1/p}\cdot\omega_{m^{\prime}_{s},l+1}.$

  2. (2)

${\mathcal{C}}^{\prime}$ on $\varOmega[\widetilde{C}]$ sends
$\widetilde{W}_{s}$ to $\widetilde{W}_{m^{\prime}_{s}}$. Moreover, we have

${\mathcal{C}}^{\prime}\omega_{s,j}=\sum_{l=0}^{(m^{\prime}_{s}\deg(f)-(N,\deg(f))/N}\gamma_{s,(l+1)p-j}^{1/p}\cdot\omega_{m^{\prime}_{s},l+1}.$

###### Proof.

Note that $\mathcal{C}^{\prime}$ sends $\omega_{s,j}$ to
$\sum_{l}\gamma_{s,(l+1)p-j}^{1/p}\frac{x^{l}}{y^{m^{\prime}_{s}}}dx$, where
$l$ runs between $\left\lceil\frac{j}{p}-1\right\rceil$ and
$\left\lfloor\frac{n^{\prime}_{s}\deg(f)+j}{p}-1\right\rfloor$. Hence the
space of the character $\zeta\mapsto\zeta^{s}$ for $\zeta\in\mu_{N}$ is sent
by $\mathcal{C}^{\prime}$ to that of the character
$\zeta\mapsto\zeta^{m^{\prime}_{s}}$.

(1) First, consider Case 1 (i.e., $N-\sum A_{k}>0$) of $\varOmega[C]$. In this
case $\omega_{s,j}=x^{j-1}dx/y^{s}$ for $0\leq s<N$ and $1\leq j<s$ give a
basis of $\varOmega[C]$. We have to show $l+1<m^{\prime}_{s}$. This follows
from

$l+1\leq\left\lfloor\frac{{n^{\prime}_{s}}\deg(f)+j}{p}\right\rfloor\leq\left\lfloor\frac{{n^{\prime}_{s}}N+s-1}{p}\right\rfloor\leq\left\lfloor\frac{m^{\prime}_{s}p-1}{p}\right\rfloor<m^{\prime}_{s}.$

(2) Consider the case of $\varOmega[{\widetilde{C}}]$. The set of
$\widetilde{\pi}^{*}\omega_{s,j}$ with

$\omega_{s,j}=\frac{x^{j-1}dx}{y^{s}},\quad 1\leq s\leq N-1,\ 1\leq
j\leq\frac{s\deg(f)-(N,N-\sum{A_{k}})}{N}$

is a basis of $\varOmega[\widetilde{C}]$ by Corollary 5.3. By Corollary 5.3,
we need to show that

$1\leq m^{\prime}_{s}\leq N-1,\quad 1\leq
l+1\leq\frac{m^{\prime}_{s}\sum{A_{k}}-(N,N-\sum{A_{k}})}{N}.$

The former is clear. In addition $m^{\prime}_{s}p-n^{\prime}_{s}N=s$ and
$jN\leq s\sum{A_{k}}-(N,N-\sum{A_{k}})$, then we have

$\left\lfloor\frac{n^{\prime}_{s}\sum{A_{k}}+j}{p}\right\rfloor\leq\left\lfloor\frac{m^{\prime}_{s}\sum{A_{k}}-(N,N-\sum{A_{k}})}{N}+\frac{(N,N-\sum{A_{k}})}{N}\frac{p-1}{p}\right\rfloor.$

We rewrite the first term of right side as the irreducible fraction, then the
denominator is a divisor of $N/(N,N-\sum{A_{k}})$, while the second term of
right side is strictly smaller than $(N,N-\sum{A_{k}})/N$. Hence we obtain

$\left\lfloor\frac{{n^{\prime}_{s}}\sum{A_{k}}+j}{p}\right\rfloor\leq\frac{{m^{\prime}_{s}}\sum{A_{k}}-(N,N-\sum{A_{k}})}{N}.$

This is the desired conclusion. ∎

######  Example 6.11.

We consider the curve $C:y^{3}=x(x-1)^{2}(x-z)^{2}$ in Examples 4.7 and 5.4,
i.e., $(A_{0},A_{1},A_{2})=(1,2,2)$ and $N=3$. Let us describe the Cartier
operator on $\varOmega[\widetilde{C}]$. The basis is given by
$\frac{x^{j-1}}{y^{s}}dx$ for $(s,j)=(1,1),(2,1),(2,2),(2,3)$, see Example
5.4.

  * •

If $p\equiv 1~{}({\rm mod}\,3)$, we have $m^{\prime}_{s}=s$ and
$n^{\prime}_{s}=(p-1)s/3$.

  * •

If $p\equiv 2~{}({\rm mod}\,3)$, we have $m^{\prime}_{1}=2$ and
$n^{\prime}_{1}=(2p-1)/3$ and $m^{\prime}_{2}=1$ and $n^{\prime}_{2}=(p-2)/3$.

Put $a^{\prime}=5s/3-j$ and $c^{\prime}=4s/3-j+1$, which are positive and set
$d^{\prime}=5n^{\prime}_{s}-ip+j$. We have

$\gamma_{s,ip-j}=\frac{(c^{\prime};d^{\prime})}{(a^{\prime};d^{\prime})}{\mathcal{F}}^{(d^{\prime};n^{\prime}_{s},n^{\prime}_{s},n^{\prime}_{s},n^{\prime}_{s})}(a^{\prime},s/3,s/3,s/3,c^{\prime};1,z,z).$

for $i=1$ if $m^{\prime}_{s}=1$ and $i=1,2,3$ if $m^{\prime}_{s}=2$. For
example for $p=7$, the Cartier-Manin matrix is given by

$\begin{pmatrix}z^{4}+2z^{3}+z^{2}+2z+1&0&0&0\\\ 0&z^{7}&-z^{8}-z^{7}&z^{8}\\\
0&-z^{7}-1&z^{8}+z^{7}+z+1&-z^{8}-z\\\ 0&1&-z-1&z\ \end{pmatrix}.$

For example for $p=5$, the Cartier-Manin matrix is given by

$\begin{pmatrix}0&3z+3&z^{2}+4z+1&3z^{2}+3z\\\ 4z^{6}+4z^{5}&0&0&0\\\
z^{6}+z^{5}+z+1&0&0&0\\\ 4z+4&0&0&0\\\ \end{pmatrix}.$

##  Results from Kummer theory

We review the proposition used in Section 2 for the reader’s convenience. Let
$K$ be an arbitrary field and suppose that $\zeta\in K$, where $\zeta$ denotes
a primitive $n$-th root of unity.

######  Lemma A.1.

Let $a\in K,\ L=K(\sqrt[n]{a})$. Then $L/K$ is a finite Galois extension.

###### Proof.

First of all, we show that $n\geq 1$ is not divided by the characteristic of
$K$ when it is positive. Suppose that $p\,|\,n$ where $p>0$ denotes the
characteristic of $K$, there exists a positive integer $m\geq 1$ satisfying
$n=pm$. Since $\zeta^{n}-1=(\zeta^{m}-1)^{p}$, so we get $\zeta^{m}=1$. This
contradicts that $\zeta$ is a primitive $n$-th root of unity.

Now we set $f(x):=x^{n}-a\in K[x]$ then we have $f^{\prime}(x)=nx^{n-1}\neq 0$
and $(f(x),f^{\prime}(x))=1$, thus $f(x)$ is a separable polynomial over $K$.
And $f(x)=\prod(x-\zeta^{i}\sqrt[n]{a})\in K[x]$, hence $L$ is the smallest
splitting field of $f(x)$ over $K$. Hence $L/K$ is a finite Galois extension.
∎

######  Proposition A.2.

Let $a\in K,\ L=K(\sqrt[n]{a})$ and $m=[L:K]$. Then $m$ is a divisor of $n$,
and is equal to the smallest positive integer such that $(\sqrt[n]{a})^{m}\in
K$.

###### Proof.

For any $\sigma\in{\rm Gal}(L/K)$, we have $f(\sqrt[n]{a})=0\Rightarrow
f((\sqrt[n]{a})^{\sigma})=0$. Thus there exists $\zeta_{\sigma}\in K$ such
that $(\sqrt[n]{a})^{\sigma}=\zeta_{\sigma}\sqrt[n]{a}$, where
$\zeta_{\sigma}$ is an $n$-th root of unity. Now let
$\mu_{n}:=\\{\zeta^{i}\,;\,i=0,\ldots,n-1\\}\subset K$ and $\varphi:{\rm
Gal}(L/K)\longrightarrow\mu_{n}\,;\,\sigma\longmapsto\zeta_{\sigma}$, then one
can confirm that $\varphi$ is an injective homomorphism. This implies that
${\rm Gal}(L/K)$ is isomorphism to a certain subgroup of $\mu_{n}$. Since
$\mu_{n}$ is a finite cyclic group of order $n$, thus the order of ${\rm
Gal}(L/K)$ is a divisor of $n$.

Now let $\sigma$ be a generator of a finite cyclic group ${\rm Gal}(L/K)$,
then it follows from $m=[L:K]$ that the order of $\sigma$ is $m$. Since
$((\sqrt[n]{a})^{m})^{\sigma}=((\sqrt[n]{a})^{\sigma})^{m}=(\zeta_{\sigma}\sqrt[n]{a})^{m}=(\sqrt[n]{a})^{m}$,
thus we obtain $(\sqrt[n]{a})^{m}\in K$. And suppose that
$((\sqrt[n]{a})^{i})^{\sigma}=(\sqrt[n]{a})^{i}$ for $i<m$, then this
contradicts that the order of $\sigma$ is $m$. Hence, the proposition is true.
∎

##  Results from elementary number theory

In this section, we prove some propositions used in Sections 4-6.

######  Lemma B.1.

Let $p,q$ be co-prime positive integers, then all positive integers $d\geq
pq+1$ can be written as $d=pa+qb$ where $a,b\geq 1$.

###### Proof.

Let $r$ be the remainder of $d\geq pq+1$ divided by $q$. Now each remainder of
$p,2p,\ldots,pq$ divided by $q$ is different, so there exists $1\leq a\leq q$
such that the remainder of $pa$ divided by $q$ is $r$. Since $d-pa$ is divided
by $q$, then $d=pa+qb$ when we put its quotient $b$. ∎

######  Lemma B.2.

Let $p,q$ be co-prime positive integers, then all positive integers $d\geq
gpq+1$ can be written as $d=pa+qb$ where $a,b\geq 1$ in different $g$ ways.

###### Proof.

Since $d-(g-1)pq\geq pq+1$, we can write $d-(g-1)pq=pa+qb~{}(a,b\geq 1)$ by
Lemma B.1. Then $d=p(a+iq)+q(b-(i-g+1)p)$ where $i=0,\ldots,g-1$. Now
$a+iq,\,b-(i-g+1)p\geq 1$. ∎

######  Corollary B.3.

Let $p,q$ be co-prime positive integers, then all positive integers $d\geq
gpq-p-q+1$ can be written as $d=pa+qb$ where $a,b\geq 0$ in different $g$
ways.

###### Proof.

If $d\geq gpq-p-q+1$, then $d+p+q\geq gpq+1$ can be written as
$d+p+q=pa+qb~{}(a,b\geq 1)$ in different $g$ ways by Lemma B.2. Thus
$d=p(a-1)+q(b-1)$, so the proposition is true. ∎

######  Proposition B.4.

Suppose that $np+mq=1$, then the followings are true:

  1. (1)

If $d\geq gpq-p-q+1$ is written as $d=pa_{i}+qb_{i}~{}(a_{i},b_{i}\geq 0)$,
then $e_{i}\equiv-ma_{i}+nb_{i}\ ({\rm mod}\ g)$ take all values in
$\\{0,\ldots,g-1\\}$.

  2. (2)

If $d\leq gpq-p-q$ can be written as $d=pa_{i}+qb_{i}~{}(a_{i},b_{i}\geq 0)$,
then $e_{i}\equiv-ma_{i}+nb_{i}\ ({\rm mod}\ g)$ are different values by $i$.

###### Proof.

Let $a_{i}=a_{0}+iq,\ b_{i}=b_{0}-(i-g+1)p$ where $i=0,\ldots,g-1$. Suppose
that $e_{i}\equiv e_{j}\ ({\rm mod}\ g)$ for $i\neq j$, that is there exists
$k\in\mathbb{Z}$ such that $e_{i}=e_{j}+gk$. Then

$\displaystyle
gk=e_{i}-e_{j}=(-ma_{i}+nb_{i})-(-ma_{j}+nb_{j})=-m(a_{i}-a_{j})+n(b_{i}-b_{j})$
$\displaystyle=-mq(i-j)-np(i-j)=-(np+mq)(i-j)=j-i$

So $e_{i}$ take different values for $i=0,\ldots,g-1$. ∎

######  Lemma B.5.

Let $p,q$ be co-prime positive integers and let $d$ be an integer with $0<d<q$
such that $d\not\equiv 0~{}({\rm mod}\ p)$. Then there uniquely exist $a,b$
such that $d=pa-qb$ with $0<a<q,\,0<b<p$.

###### Proof.

There exist $a_{0},b_{0}\in\mathbb{Z}$ such that $pa-qb=d$ by Lemma B.1, then
$(a,b)=(a_{0}+qk,b_{0}+pk)$ satisfy $pa-qb=d$ for all $k\in\mathbb{Z}$. Since
$a_{0}\not\equiv 0~{}({\rm mod}\ q)$, then we can choose $k$ such that
$0<a<q$. Now $-d<qb<pq-d$, then $-d/q<b<p-(d/q)$ and note that $0<d/q<1$. ∎

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