# Computing the space of differential forms of
a plane curve and its Cartier-Manin matrix

Momonari Kudo<EMAIL_ADDRESS>and  Shushi Harashita
<EMAIL_ADDRESS>
###### Abstract.

In this paper, we propose a feasible algorithm to give an explicit basis of
the space of regular differential forms on the nonsingular projective model of
any given plane algebraic curve. The algorithm is demonstrated for concrete
examples, with our implementation over the computer algebra system Magma. As
an application, we also describe the Cartier-Manin matrix of the nonsingular
projective curve with respect to the basis computed by the algorithm.

###### Key words and phrases:

Plane curves, Differential forms, Cartier-Manin matrices

###### 2010 Mathematics Subject Classification:

Primary: 14Q05, 14F40, 14G15, 14G17

Graduate School of Information Science and Technology, The University of Tokyo

Graduate School of Environment and Information Sciences, Yokohama National
University

##  1\. Introduction

###  1.1. Background

Let $p$ be a rational prime, and $k$ a perfect field of characteristic $p$.
Let $C$ be a geometrically irreducible plane curve over $k$ define by
$F(x,y)=0$ with $F\in k[x,y]$, and let ${\tilde{C}}$ denote the
desingularization of the Zariski closure $C^{\prime}$ in ${\mathbb{P}^{2}}$ of
$C$. We denote by $\varOmega_{\tilde{C}}$ the sheaf of regular differential
forms on $\tilde{C}$, and ${\mathcal{V}}$ denotes the Cartier operator on the
cohomology group $H^{0}({\tilde{C}},\varOmega_{\tilde{C}})$. A matrix
representing $\mathcal{V}$ with respect to a suitable basis for
$H^{0}({\tilde{C}},\varOmega_{\tilde{C}})$ is called a Cartier-Manin matrix
for $\tilde{C}$. Computing a basis of
$H^{0}({\tilde{C}},\varOmega_{\tilde{C}})$ and the Cartier-Manin matrix is a
very important task both in theory and computation, since they are used to
compute various invariants such as $a$-number, $p$-rank, and so on for the
classification of curves. Indeed, there are many works on this task, e.g.,
[14], [21], [7], [1], [3], [11], [19], [17]. As in [7], some works were on the
first cohomology $H^{1}(\tilde{C},{\mathcal{O}}_{\tilde{C}})$ of the structure
sheaf $\mathcal{O}_{\tilde{C}}$, which is the dual notion of the space of
regular differential forms and on which the natural action of the Frobenius
with respect to a basis is called the Hasse-Witt matrix. For example, for the
case of genus-5 and trigonal curve $C$, we compute
$H^{1}(\tilde{C},{\mathcal{O}}_{\tilde{C}})$ and a Hasse-Witt matrix on it,
see our preceding paper [13]. The first author also obtained a generalized
algorithm to compute coherent cohomologies of projective schemes and the
Frobenius actions on them, see [12].

Among previous works including those described above, one of the most
important results is Stöhr-Voloch’s beautiful formula [18, Theorem 1.1] (see
(6.1) below for a recall). By their formula, once a basis of
$H^{0}(\tilde{C},\varOmega_{\tilde{C}})$ is given explicitly, the Cartier-
Manin matrix of $\tilde{C}$ with respect to the basis can be also computed. As
for the computation of a basis of $H^{0}(\tilde{C},\varOmega_{\tilde{C}})$,
when $C^{\prime}$ is already nonsingular, it is known that under the
assumptions (A1) and (A2) in Section 4 below, $(x^{i}y^{j}/(\partial
F/\partial y))dx$ with $0\leq i+j\leq\mathrm{deg}(F)-3$ form a basis of
$H^{0}({\tilde{C}},\varOmega_{\tilde{C}})$ (this fact can be viewed as a
particular case of Gorenstein’s result described below). On the other hand,
when $C^{\prime}$ is singular, such explicit bases are found only for
particular cases, e.g., [14], [21] for hyperelliptic curves, [7] for Fermat
curves, [19] for superelliptic curves, and [1], [17] for curves associated to
Appell-Lauricella hypergeometic series. For example, Sutherland recently
constructed an efficient algorithm to compute Cartier-Manin matrices for
superelliptic curves in [19], where he found a basis of
$H^{0}(\tilde{C},\varOmega_{\tilde{C}})$ to apply Stöhr-Voloch’s formula. See
also [3] and [11] for efficient algorithms to compute Cartier-Manin matrices
for hyperelliptic curves.

###  1.2. Main results

In this paper, we shall present an algorithm for finding a basis of
$H^{0}({\tilde{C}},\varOmega_{\tilde{C}})$ for general (i.e., possibly
singular) plane curves, based on the work by Gorenstein [8]. In [8, Theorem
12], the necessary and sufficient condition for a rational differential form
to be regular was given in terms of conductor, but unfortunately in the
statement an assumption on infinite places was added. As Gorenstein remarked,
the assumption can be removed if $k$ is infinite, but it causes a problem when
we want to have a basis over $k$ when $k$ is finite. To resolve the problem,
in Theorem 2.2 below we shall give a more general statement than [8, Theorem
12]. Then our problem is reduced to describing the conductors of the
coordinate rings of two affine covers of the Zariski closure $C^{\prime}$ in
${\mathbb{P}}^{2}$, see Subsection 2.1 below for Gorenstein’s result and this
observation. From our observation, we construct an algorithm to compute a
basis of $H^{0}({\tilde{C}},\varOmega_{\tilde{C}})$, by realizing the
following two steps:

  1. Step A.

Compute an explicit basis of the conductor $\mathfrak{C}_{k[C]}$ as a
$k[x]$-module.

  2. Step B.

Compute the truncation of $\mathfrak{C}_{k[C]}$ by the degree described in
Theorem 2.2.

For Step A, we would like to apply Mňuk’s algorithm [16], whose framework is:

  1. (M1)

Compute a basis $\\{w_{1},\ldots,w_{N}\\}$ of the integral closure
$\overline{k[C]}$ as a $k[x]$-module.

  2. (M2)

By computing the trace of $w_{i}w_{j}$ for each $1\leq i\leq j\leq N$,
construct a basis $\\{w_{1}^{\ast},\ldots,w_{N}^{\ast}\\}$ of the
complementary module $\mathcal{C}_{\overline{k[C]}/k[x]}$ (see Definition 2.4
below for the definition). Note that this basis is dual to
$\\{w_{1},\ldots,w_{N}\\}$, and $\\{(\partial F/\partial y)w_{i}^{\ast}:1\leq
i\leq j\leq N\\}$ is a desired basis of $\mathfrak{C}_{k[C]}$.

However, we could not directly apply Mňuk’s algorithm for Step A, since he
assumed that the characteristic of $k$ is zero. Moreover, he utilized an
algorithm [20] for (M1), but [20] also assumes $\mathrm{char}(k)=0$. Here, our
solutions for realizing Step A are:

  1. (i)

We confirm in Subsection 2.2 that Mňuk’s framework works well even in the
positive characteristic case.

  2. (ii)

We present an alternative approach based on the Gröbner basis computation with
respect to plural module orders (TOP, POT and their product orders). In our
alternative approach, the integral closure $\overline{k[C]}$ is realized as an
affine ring by the normalization method [10], and then its basis will be
constructed mainly by the syzygy computation.

As for Step B, we compute a Gröbner basis and the reduced row echelon form of
a Macaulay matrix, see Step B of Algorithm 4.1 in Subsection 4.1 below for
details.

We implemented our algorithm over a computer algebra system, Magma [2], and
demonstrate it for concrete examples, which will be helpful for the further
use of our algorithm. As an application of giving a basis of the space of
regular differential forms of our curve $\tilde{C}$, in Section 6 we also
describe the Cartier-Manin matrix of $\tilde{C}$ with respect to the basis,
where we use the result by Stöhr and Voloch [18]. Furthermore, we give a
remark on finding a basis of the dual space
$H^{1}({\tilde{C}},{\mathcal{O}}_{\tilde{C}})$.

###  1.3. Organization

The rest of this paper is organized as follows: In Section 2, we review
Gorenstein’s work and basic facts on complementary modules. Section 3 is
devoted to computations of syzygy modules and normalizations. In Section 4, we
present our main algorithm, and in Section 5, we demonstrate the algorithm on
an example with our implementation over Magma [2]. In Section 6, we explain a
way to compute the Cartier-Manin matrix of a given plane curve with respect to
a basis of the differential forms, computed in the previous sections. Section
7 is devoted to concluding remarks.

##  2\. Preliminaries

In this section, we collect fundamental results, which are necessary to
describe our algorithm.

###  2.1. Gorenstein’s work

We start with recalling Gorenstein’s description of the regular differential
forms on the nonsingular projective model of a plane curve.

Let $C$ be a plane curve $\operatorname{Spec}A$, say

$A=k[x,y]/\langle F\rangle$

for an element $F=F(x,y)\in k[x,y]$. We assume that $F$ is irreducible over
the algebraic closure of $k$. Let $N$ be the degree of $F$. Consider the
embedding of ${\mathbb{A}}^{2}=\operatorname{Spec}k[x,y]$ into
${\mathbb{P}}^{2}=\operatorname{Proj}k[X,Y,Z]$ by $x=X/Z$ and $y=Y/Z$. Let
$C^{\prime}$ be the Zariski closure of $C$ in ${\mathbb{P}}^{2}$. Let
$\tilde{C}$ be the desingularization of $C^{\prime}$. In this paper, we
propose an algorithm to get a basis of
$H^{0}({\tilde{C}},\varOmega_{\tilde{C}})$ and to obtain the Cartier-Manin
matrix of ${\tilde{C}}$ with respect to the basis.

Let $k(C)$ be the function field of $C$ ($=$ that of $\tilde{C}$). We may
assume that $k(C)$ is separably generated over $k$: moreover after a
coordinate change, we assume that

  1. $\small\bullet$

$x\in k(C)$ is transcendental over $k$ and

  2. $\small\bullet$

$F(x,y)$ is a monic polynomial in $y$ over $k[x]$ of degree $N$, which is
separable as a polynomial over $k(x)$,

which are equivalent to (A1) and (A2) in Section 4 below.

Recall Gorenstein’s result:

######  Theorem 2.1 ([8, Theorem 12]).

Assume that $C^{\prime}$ is nonsingular at any point of
$C^{\prime}\smallsetminus C$. Then

$H^{0}({\tilde{C}},\varOmega_{\tilde{C}})=\left\\{\left.\frac{\phi(x,y)}{F_{y}}dx\
\right|\ \phi\in{\mathfrak{C}}_{A},\ {\rm deg}(\phi)\leq N-3\right\\}$

with $F_{y}=\frac{\partial F}{\partial y}$, where ${\mathfrak{C}}_{A}$ is the
conductor of $A=k[x,y]/\langle F\rangle$ in its integral closure
$\overline{A}$ in $k(C)$, i.e., ${\mathfrak{C}}_{A}=\\{z\in A\mid
z{\overline{A}}\subset A\\}$.

The aim of this paper is to give an algorithm to give a basis of the space of
$\phi(x,y)$ in the theorem and to consider the case where the assumption in
the theorem is removed. Note that if $k$ is small, then we may not remove the
assumption by taking any $k$-linear coordinate-change (cf. Example 6.1, (3)).
Also we give implementations of these algorithms over a computer algebra
system, Magma [2].

The next theorem can be applicable even if $C^{\prime}$ has a singularity at
$C^{\prime}\smallsetminus C$. The proof has already been done essentially in
the proof of [8, Theorem 12].

######  Theorem 2.2.

With the notation above, we have

$H^{0}({\tilde{C}},\varOmega_{\tilde{C}})=\left\\{\left.\frac{\phi(x,y)}{F_{y}}dx\
\right|\ \phi\in{\mathfrak{C}}_{A},\
\phi^{\prime}\in{\mathfrak{C}}_{A^{\prime}}\right\\},$

where $A^{\prime}=k[x^{\prime},y^{\prime}]/\langle F^{\prime}\rangle$, where
$F^{\prime}(x^{\prime},y^{\prime})=F(x,y)/x^{N}$ with
$(x^{\prime},y^{\prime})=(1/x,y/x)$, and
$\phi^{\prime}(x^{\prime},y^{\prime}):=\phi(x,y)/x^{N-3}$. (Note that
$\phi^{\prime}\in{\mathfrak{C}}_{A^{\prime}}$ especially requires ${\rm
deg}(\phi)\leq N-3$, as ${\mathfrak{C}}_{A^{\prime}}\subset A^{\prime}$.)

###### Proof.

As in the proof of [8, Theorem 12], the differential form
$\omega:=\frac{\phi(x,y)}{F_{y}}dx$ with separating variable $x$ is regular at
every point of $\operatorname{Spec}(A)$ if and only if
$\phi\in{\mathfrak{C}}_{A}$, and the differential form $\omega$ is written as
$-\frac{\phi^{\prime}(x^{\prime},y^{\prime})}{F^{\prime}_{y^{\prime}}}dx^{\prime}$
over the open affine part $U$ of $X\neq 0$, where $\phi^{\prime}$ is as in the
statement of the theorem. Let $F_{0}(X,Y,Z)$ be the homogenization of $F(x,y)$
(i.e., the defining equation of $C^{\prime}$). We claim that $X\neq 0$ at any
point of $C^{\prime}\smallsetminus C$. Indeed by the assumption that $F(x,y)$
contains the term $y^{N}$ with $N={\rm deg}(F)$, if $Z=X=0$ held, then $Y=0$
has to hold. By this claim, it suffices to discuss the regularity of $\omega$
over $U$. This is equivalent to $\phi^{\prime}\in{\mathfrak{C}}_{A^{\prime}}$,
since $x^{\prime}$ is a separating variable of $k(C)$ and
$F^{\prime}(x^{\prime},y^{\prime})$ is a monic polynomial in $y^{\prime}$ over
$k[x^{\prime}]$ of degree $N$. The theorem clearly follows from these facts.
$\Box$

######  Remark 2.3.

  1. (1)

Here, $\phi(x,y)\in{\mathfrak{C}}_{A}$ is equivalent to
$\phi(x,y)\in{\mathfrak{C}}_{A,P}=\\{z\in A_{P}:z{\overline{A_{P}}}\subset
A_{P}\\}$ for any maximal ideal $P$ of $A$, where ${\overline{A_{P}}}$ is the
integral closure of $A_{P}$ in $k(C)$. Moreover
${\mathfrak{C}}_{A,P}={\mathfrak{C}}_{A,P}^{*}\cap A_{P}$, where
${\mathfrak{C}}_{A,P}^{*}$ is the conductor of $A_{P}^{*}$ in
${\overline{A_{P}}}^{*}$, where $*$ means taking the completion (cf. [8,
Theorem 2]). For example, if $C^{\prime}$ has a single singular point, say
$P$, in $C^{\prime}\smallsetminus C$, then in order to see
$\frac{\phi(x,y)}{F_{y}}dx\in H^{0}({\tilde{C}},\varOmega_{\tilde{C}})$ for
$\phi\in{\mathfrak{C}}_{A}$ it suffices to check
$\phi^{\prime}\in{\mathfrak{C}}_{A^{\prime},P}$ instead of checking
$\phi^{\prime}\in{\mathfrak{C}}_{A^{\prime}}$.

  2. (2)

As written in Theorem 2.2, the degree of $\phi$ is at most $N-3$. Similarly
${\rm deg}(\phi^{\prime})\leq N-3$ hold. Hence, the space
$H^{0}({\tilde{C}},\varOmega_{\tilde{C}})$ is the intersection of
${\mathfrak{C}}_{A}^{(N-3)}:=\\{\phi\in{\mathfrak{C}}_{A}\mid{\rm
deg}(\phi)\leq N-3\\}$ and ${\mathfrak{C}}_{A^{\prime}}^{(N-3)}$, where we
identify $\phi\in{\mathfrak{C}}_{A}^{(N-3)}$ and
$\phi^{\prime}\in{\mathfrak{C}}_{A^{\prime}}^{(N-3)}$ with the notation of
Theorem 2.2. Since ${\mathfrak{C}}_{A}^{(N-3)}$ and
${\mathfrak{C}}_{A^{\prime}}^{(N-3)}$ are finite dimensional $k$-vector
spaces, it suffices to give an algorithm to describe
$\mathfrak{C}_{A}^{(N-3)}$, in order to have an algorithm to describe
$H^{0}({\tilde{C}},\varOmega_{\tilde{C}})$.

###  2.2. Complementary modules

As we have seen in Subsection 2.1, regular deferential forms of $\tilde{C}$
are obtained from the conductor $\mathfrak{C}$ of $k[C]$. In this subsection,
we review the notion of complementary modules, which enables us to compute a
basis of $\mathfrak{C}$. Let us start with stating the definition of a
complementary module of an integrally closed domain.

######  Definition 2.4 (Complementary modules).

Let $R$ be an integrally closed domain, and $K$ its field of fractions. Let
$K^{\prime}$ be a finite separable (and thus simple) extension of $K$. Let
$\mathrm{Tr}_{K^{\prime}/K}:K^{\prime}\rightarrow K$ be the trace map for
$K^{\prime}/K$. For an integral extension $\tilde{R}\subset K^{\prime}$ of
$R$, we call an $R$-module

$\mathcal{C}_{\tilde{R}/R}:=\\{z\in
K^{\prime}:\mathrm{Tr}_{K^{\prime}/K}(z\tilde{R})\subset R\\}$

the complementary module of $\tilde{R}$ with respect to $R$.

Note in Definition 2.4 that $\tilde{R}\subset\mathcal{C}_{\tilde{R}/R}$ since
$\mathrm{Tr}_{K^{\prime}/K}(r)\in R$ for all $r\in\tilde{R}$. The next theorem
is a consequence of [22, Ch. V, § 11, Theorem 29].

######  Theorem 2.5 ([16], Corollary 3.1).

Let the notation be same as in Definition 2.4. Assume that $R$ is a Dedekind
ring. Let ${\overline{R}}$ be the integral closure of $R$ in $K^{\prime}$. Let
$y\in{\overline{R}}$ be an element such that $K^{\prime}=K(y)$, and $f(Y)$ be
a minimal polynomial of $y$ over $K$. Then we have

$\mathfrak{f}({\overline{R}}/R[y])=f^{\prime}(y)\mathcal{C}_{{\overline{R}}/R},$

where $\mathfrak{f}({\overline{R}}/R[y])$ is the conductor of $R[y]$ in
${\overline{R}}$, that is,

$\displaystyle\mathfrak{f}({\overline{R}}/R[y])$ $\displaystyle=$
$\displaystyle\\{a\in R[y]:\forall b\in{\overline{R}},\
a\cdot(b\bmod{R[y]})=0\bmod{R[y]}\\}$ $\displaystyle=$ $\displaystyle\\{a\in
R[y]:a{\overline{R}}\subset R[y]\\}.$

Note that we can always take an element $y$ as in Theorem 2.5, whihc is proved
in a way similar to the first part of the proof of Lemma 2.8 below.

In the following, we show that both $\overline{R}$ and
$\mathcal{C}_{\overline{R}/R}$ in Theorem 2.5 are finite free $R$-modules of
same rank if $R$ is a principal ideal domain, and that a basis of
$\mathcal{C}_{\overline{R}/R}$ can be computed from that of $\overline{R}$.
For this, we here recall the notion of dual bases in a finite separable field
extension. Proofs of the following lemmas would be well-known, but let us
write all the complete proofs for readers’ convenience:

######  Lemma 2.6.

Let $K^{\prime}/K$ be a finite separable field extension, and
$\\{w_{1},\ldots,w_{N}\\}$ a basis of $K^{\prime}$ over $K$. Put
$w_{i}^{\ast}:=\sum_{j=1}^{N}c_{i,j}w_{j}$ with $c_{i,j}\in K$ for $1\leq
i\leq N$. Then we have the following:

  1. (1)

$(\mathrm{Tr}_{K^{\prime}/K}(w_{i}w_{j}))\in\mathrm{GL}_{N}(K)$.

  2. (2)

By putting $(c_{i,j})=(\mathrm{Tr}_{K^{\prime}/K}(w_{i}w_{j}))^{-1}$, the
basis $\\{w_{1}^{\ast},\ldots,w_{N}^{\ast}\\}$ of $K^{\prime}$ over $K$
satisfies $\mathrm{Tr}_{K^{\prime}/K}(w_{i}w_{j}^{\ast})=\delta_{i,j}$ for
each $1\leq i,j\leq N$, where $\delta_{i,j}$ is the Kronecker delta.
Conversely, if $\mathrm{Tr}_{K^{\prime}/K}(w_{i}w_{j}^{\ast})=\delta_{i,j}$
for each $1\leq i,j\leq N$, then
$(c_{i,j})=(\mathrm{Tr}_{K^{\prime}/K}(w_{i}w_{j}))^{-1}$. In this case,
$\\{w_{1}^{\ast},\ldots,w_{N}^{\ast}\\}$ is the dual basis of
$\\{w_{1},\ldots,w_{N}\\}$.

###### Proof.

(1) Let $\sigma_{1},\ldots,\sigma_{N}$ be all $K$-embeddings from $K^{\prime}$
to $\overline{K}$. Putting $H=(\sigma_{i}(w_{j}))$, one can check that
$(\mathrm{Tr}_{K^{\prime}/K}(w_{i}w_{j}))_{i,j}={}^{t}HH$, where ${}^{t}H$ is
the transpose of $H$. Thus, if
$\mathrm{det}(\mathrm{Tr}_{K^{\prime}/K}(w_{i}w_{j}))=0$, i.e.,
$\mathrm{det}(H)=0$, then the row vectors of $H$ are linearly dependent over
$K$, and hence there exists $(c_{1},\ldots,c_{N})\in
K^{N}\smallsetminus\\{(0,\ldots,0)\\}$ such that
$\sum_{j=1}^{N}c_{j}\sigma_{j}(w_{i})=0$. Since $\\{w_{1},\ldots,w_{N}\\}$
generates $K^{\prime}$ over $K$, we also have
$\sum_{j=1}^{N}c_{j}\sigma_{j}(y)=0$ for all $y\in K^{\prime}$. This means
that $\sigma_{1},\ldots,\sigma_{N}$ are linearly dependent over $K$, which
contradicts Dedekind’s lemma.

(2) It follows from the $K$-linearity of $\mathrm{Tr}_{K^{\prime}/K}$ that

$\sum_{k=1}^{N}c_{i,k}\mathrm{Tr}_{K^{\prime}/K}(w_{k}w_{j})=\mathrm{Tr}_{K^{\prime}/K}\left(\left(\sum_{k=1}^{N}c_{i,k}w_{k}\right)w_{j}\right)=\mathrm{Tr}_{K^{\prime}/K}(w_{i}^{\ast}w_{j}),$

i.e.,
$(c_{ij})\cdot(\mathrm{Tr}_{K^{\prime}/K}(w_{i}w_{j}))=(\mathrm{Tr}_{K^{\prime}/K}(w_{i}^{\ast}w_{j}))$,
and thus the claim holds. $\Box$

######  Lemma 2.7.

Let the notation be same as in Definition 2.4, and suppose that $\tilde{R}$ is
a free $R$-module of finite rank with a basis $\\{w_{1},\ldots,w_{N}\\}$ with
$N:=[K^{\prime}:K]$. Then we have the following:

  1. (1)

$\\{w_{1},\ldots,w_{N}\\}$ is a basis of $K^{\prime}$ over $K$.

  2. (2)

The dual basis $\\{w_{1}^{\ast},\ldots,w_{N}^{\ast}\\}$ given in Lemma 2.6
generates $\mathcal{C}_{\tilde{R}/R}$ over $R$. Hence
$\mathcal{C}_{\tilde{R}/R}$ is a free $R$-module of rank $N$ with a basis
$\mathcal{C}_{\tilde{R}/R}$.

###### Proof.

(1) Assuming $\sum_{i=1}^{N}b_{i}w_{i}=0$ for $b_{i}\in K$, we first prove the
$K$-linear independence of $\\{w_{1},\ldots,w_{N}\\}$. Since $K$ is the field
of fractions for $R$, we can write $b_{i}=p_{i}/q_{i}$ for some
$p_{i},q_{i}\in R$ with $q_{i}\neq 0$, and thus
$\sum_{i=1}^{N}(p_{i}/q_{i})w_{i}=0$. Multiplying the product $q_{1}\cdots
q_{N}$ to the both sides, we have $\sum_{i=1}^{N}\left(p_{i}\prod_{i\neq
k}q_{k}\right)w_{i}=0$. It follows from the $R$-linear independence of
$\\{w_{1},\ldots,w_{N}\\}$ that $p_{i}\prod_{i\neq k}q_{k}=0$ for all $1\leq
i\leq N$. Since $q_{k}\neq 0$ for $1\leq k\leq n$, one has $p_{i}=0$ and thus
$b_{i}=0$. Hence $\\{w_{1},\ldots,w_{N}\\}$ is linearly independent over $K$,
and thus it follows from $[K^{\prime}:K]=N$ that $\\{w_{1},\ldots,w_{N}\\}$
spans $K^{\prime}$ over $K$.

(2) For any element $r^{\prime}=\sum_{i=1}b_{i}w_{i}\in\tilde{R}$ ($b_{i}\in
R$), we have

$\mathrm{Tr}_{K^{\prime}/K}(w_{j}^{\ast}r^{\prime})=\sum_{i=1}^{N}b_{i}\mathrm{Tr}_{K^{\prime}/K}(w_{i}w_{j}^{\ast})=\sum_{i=1}^{N}b_{i}\delta_{j,i}=b_{i}\in
R,$

and thus $w_{j}^{\ast}\in\mathcal{C}_{\tilde{R}/R}$ for each $1\leq j\leq N$.
Next, any element $z^{\prime}\in\mathcal{C}_{\tilde{R}/R}\subset K^{\prime}$
is written as $z^{\prime}=\sum_{j=1}^{N}a_{j}w_{j}^{\ast}\in K^{\prime}$
($a_{j}\in K$) since $\\{w_{1}^{\ast},\ldots,w_{N}^{\ast}\\}$ is a basis of
$K^{\prime}$ over $K$ by (1) together with Lemma 2.6 (2). It suffices to show
that we can take $a_{j}\in R$ for all $1\leq j\leq N$. For each $1\leq i\leq
N$, we have

$\mathrm{Tr}_{K^{\prime}/K}(z^{\prime}w_{i})=\sum_{j=1}^{N}a_{j}\mathrm{Tr}_{K^{\prime}/K}(w_{i}w_{j}^{\ast})=\sum_{j=1}^{N}a_{j}\delta_{j,i}=a_{i},$

and thus it follows from $z^{\prime}\in\mathcal{C}_{\tilde{R}/R}$ and
$w_{i}\in\tilde{R}$ that $a_{i}=\mathrm{Tr}_{K^{\prime}/K}(z^{\prime}w_{i})\in
R$, as desired. $\Box$

######  Lemma 2.8.

Let $R$ be a principal ideal domain (and thus integrally closed) and $K$ its
field of fractions. Let $K^{\prime}$ be a finite separable extension of degree
$N$ of $K$, and ${\overline{R}}$ the integral closure of $R$ in $K^{\prime}$.
Then ${\overline{R}}$ is a free $R$-module of rank $N$.

###### Proof.

Let $\\{w_{1},\ldots,w_{N}\\}$ be a basis of $K^{\prime}$ over $K$. We may
assume $w_{i}\in{\overline{R}}$ for all $1\leq i\leq N$. Indeed, let $w\in
K^{\prime}$, and
$w^{m}+(p_{m-1}/q_{m-1})w^{m-1}+\cdots+(p_{1}/q_{1})w+p_{0}/q_{0}$ its
algebraic relation over $K$, where $p_{i},q_{i}\in R$ with $q_{i}\neq 0$ for
$0\leq i\leq m-1$. Then, multiplying $a^{m}$ with $a:=q_{0}\cdots q_{m-1}\in
R$ to the both sides, we have that $aw$ is integral over $R$, and thus
$aw\in{\overline{R}}$. Thus, there exists $a_{i}\in R\smallsetminus\\{0\\}$
with $a_{i}w_{i}\in{\overline{R}}$ for each $1\leq i\leq N$, and clearly
$\\{a_{1}w_{1},\ldots,a_{N}w_{N}\\}$ is also a basis of $K^{\prime}$ over $K$.

Let ${\tilde{R}}\subset{\overline{R}}$ be the free $R$-module spanned by
$\\{w_{1},\ldots,w_{N}\\}$. Note that ${\tilde{R}}$ is integral over $R$, and
that
${\tilde{R}}\subset{\overline{R}}\subset\mathcal{C}_{{\overline{R}}/R}\subset\mathcal{C}_{{\tilde{R}}/R}$
in $K^{\prime}$. By Lemma 2.7 (2), $\mathcal{C}_{{\tilde{R}}/R}$ is a free
$R$-module of rank $N$. Since $R$ is a principal ideal domain, both
${\overline{R}}$ and $\mathcal{C}_{{\overline{R}}/R}$ are also free
$R$-modules, and have rank $N$. $\Box$

##  3\. Computing syzygies and normalization

In this section, we recall the notion of Gröbner bases of free modules
briefly, and review methods to compute syzygies of polynomials and the
normalization of an affine ring.

###  3.1. Gröbner bases of free modules

Let $S=k[x_{1},\ldots,x_{n}]$ be the polynomial ring of $n$ variables over a
field $k$. For readers’ convenience, we start with recalling the notion of
Gröbner bases in the polynomial ring $S$, and its extension to free
$S$-modules.

A monomial (or term) order on $S$ is a total order $\prec$ on the set of
monomials $\mathrm{Mon}(S):=\\{x_{1}^{\alpha_{1}}\cdots
x_{n}^{\alpha_{n}}:(\alpha_{1},\ldots,\alpha_{n})\in(\mathbb{Z}_{\geq
0})^{n}\\}$ such that (1) $m_{1}\preceq m_{2}$ for $m_{1},m_{2}\in{\rm
Mon}(S)$ implies $m_{1}m_{3}\preceq m_{2}m_{3}$ for all $m_{3}\in{\rm
Mon}(S)$, and (2) $1\preceq m$ for all $m\in{\rm Mon}(S)$. For $f\in
S\smallsetminus\\{0\\}$, we denote by $\mathrm{LT}_{\prec}(f)$,
$\mathrm{LM}_{\prec}(f)$ and $\mathrm{LC}_{\prec}(f)$ the leading term, the
leading monomial, and the leading coefficient of $f$ with respect to $\prec$,
respectively. Note that
$\mathrm{LT}_{\prec}(f)=\mathrm{LC}_{\prec}(f)\cdot\mathrm{LM}_{\prec}(f)$.
For a subset $F\subset S$, we set
$\mathrm{LT}_{\prec}(F):=\\{\mathrm{LT}_{\prec}(f):f\in F\\}$ and
$\mathrm{LM}_{\prec}(F):=\\{\mathrm{LM}_{\prec}(f):f\in F\\}$. A finite subset
$G\subset S$ is called a Gröbner basis for an ideal $I\subset S$ with respect
to $\prec$ if it generates $I$ and if
$\langle\mathrm{LT}_{\prec}(G)\rangle_{S}=\langle\mathrm{LT}_{\prec}(I)\rangle_{S}$.
For simplicity, we denote $\mathrm{LT}_{\prec}$ as $\mathrm{LT}$ and so on, if
$\prec$ is clear from the context. It is well-known that every non-zero ideal
of $S$ has a Gröbner basis with respect to an arbitrary monomial order, and so
far various algorithms for computing Gröbner bases have been proposed, e.g.,
Buchberger’s algorithm [4], Faugère’s $F_{4}$ [5] and $F_{5}$ [6], and so on.

The notion of monomial orders and Gröbner bases is extended to free modules
$S^{r}=\bigoplus_{i=1}^{r}Se_{i}$, where $e_{i}=(0,\ldots,0,1,0,\ldots,0)$
denotes the $i$-th standard basis vector of $S^{r}$ with $1$ at the $i$-th
coordinate and $0$’s elsewhere. A monomial in $S^{r}$ is an element of the
form $me_{i}=(0,\ldots,0,m,0,\ldots,0)\in S^{r}$ with $m\in\mathrm{Mon}(S)$,
and the set of all monomials in $S^{r}$ is denoted by $\mathrm{Mon}(S^{r})$.
For two monomials $me_{i}$ and $m^{\prime}e_{j}$ with
$m,m^{\prime}\in\mathrm{Mon}(S)$, we say that $me_{i}$ is divisible by
$m^{\prime}e_{j}$ if $i=j$ and $m^{\prime}\mid m$. A (module) monomial order
on $S^{r}$ is defined as a total order $\prec$ (we use the same notation as in
the case of polynomial rings) on $\mathrm{Mon}(S^{r})$ such that (1)’
$m_{1}\preceq m_{2}$ for $m_{1},m_{2}\in{\rm Mon}(S^{r})$ implies
$mm_{1}\preceq mm_{2}$ for all $m\in{\rm Mon}(S)$ (!), and (2)’ $e_{i}$ is the
minimum element in the set of all monomials in $Se_{i}$, namely $e_{i}\preceq
me_{i}$ for all $m\in{\rm Mon}(S)$. For an element $f\in S^{r}$, its leading
term, leading monomial, and leading coefficient with respect to $\prec$ are
defined similarly to the case of an element in $S$, and the same notations
$\mathrm{LT}_{\prec}(f)$, $\mathrm{LM}_{\prec}(f)$ and
$\mathrm{LC}_{\prec}(f)$ are used. As for ideals, we also define a Gröbner
basis for a submodule $N\subset S^{r}$: A finite subset $G\subset N$ is called
a Gröbner basis for $N$ with respect to $\prec$ if
$\langle\mathrm{LT}_{\prec}(G)\rangle_{S}=\langle\mathrm{LT}_{\prec}(I)\rangle_{S}$,
equivalently, for every $f\in I$, there exists $g\in G$ such that
$\mathrm{LT}_{\prec}(f)$ is divisible by $\mathrm{LT}_{\prec}(g)$, where we
denote by $\langle F\rangle_{S}$ the $S$-submodule of $S^{r}$ generated by a
subset $F\subset S^{r}$. A Gröbner basis for a submodule $N\subset S^{r}$ with
respect to an arbitrary module monomial order always exists as in the case of
ideals, and it can be computed by extensions of algorithms such as [4], [5]
and [6] over polynomial rings to free modules. In fact, implementations of
such extended algorithms are found in e.g., Macaulay2, Magma and Singular.

In the computation of Gröbner bases in free modules, there are two typical
monomial orders: POT and TOP, where POT (resp. TOP) stands for “position-over-
term” (resp. “term-over-position”). These orders will be used in the syzygy
computation of our main algorithm in Section 4 below. The definition of POT
and TOP orders is as follows: For a monomial order $\succ$ on
$\mathrm{Mon}(S)$, we say

  * •

$me_{i}\succ_{\rm POT}m^{\prime}e_{j}$ with $m,m^{\prime}\in\mathrm{Mon}(S)$
if $i<j$, or if $i=j$ and $m\succ m^{\prime}$.

  * •

$me_{i}\succ_{\rm TOP}m^{\prime}e_{j}$ with $m,m^{\prime}\in\mathrm{Mon}(S)$
if $m\succ m^{\prime}$, or if $m=m^{\prime}$ and $i<j$.

These two relations are exactly monomial orders on $\mathrm{Mon}(S^{r})$, and
$\succ_{\rm POT}$ (resp. $\succ_{\rm TOP}$) is called a POT (resp. TOP)
extension of $\succ$. Note that we have $e_{1}\succ\cdots\succ e_{r}$ in both
two cases, but other orders such as $e_{1}\prec\cdots\prec e_{r}$ are also
possible; in such a case, we specify the order of $e_{i}$’s.

We will also use a product (or block) order for $S^{r}$, like an elimination
order for polynomial rings: While an elimination order for $S$ divides the
variables $x_{1},\ldots,x_{n}$ into two (or more) blocks, this order divides
the monomials $e_{1},\ldots,e_{r}$ into two (or more) blocks. For two module
monomial orders $\succ_{1}$ on $\bigoplus_{i=1}^{k}Se_{i}$ and $\succ_{2}$ on
$\bigoplus_{i=k+1}^{r}Se_{i}$, their product order
$\succ_{1,2}\;=(\succ_{1},\succ_{2})$ on $S^{r}=\bigoplus_{i=1}^{r}Se_{i}$ is
defined as follows: For two monomials $me_{i}$ and $m^{\prime}e_{j}$ with
$m,m^{\prime}\in\mathrm{Mon}(S)$, we say $me_{i}\succ_{1,2}m^{\prime}e_{j}$ if
$i,j\leq k$ and $me_{i}\succ_{1}m^{\prime}e_{j}$, $i\leq k$ and $k<j$, or
$k<i,j$ and $me_{i}\succ_{2}m^{\prime}e_{j}$. The order $\succ_{1,2}$ has the
following elimination property with respect to $e_{1},\ldots,e_{k}$: $f\in
S^{r}$ with $\mathrm{LT}_{\succ_{1,2}}(f)\in\bigoplus_{i=k+1}^{r}Se_{i}$
satisfies $f\in\bigoplus_{i=k+1}^{r}Se_{i}$. The definition of product order
is clearly extended to arbitrary number of blocks, and the POT order is a
particular case of product orders with the elimination property with respect
to $e_{1},\ldots,e_{k}$ for each $1\leq k\leq r-1$.

###  3.2. Computing syzygy modules

This subsection reviews methods for computing syzygy modules. We use the same
notation as in the previous subsection.

Let $f_{1},\ldots,f_{r}\in S=k[x_{1},\ldots,x_{n}]$. Consider the following
$S$-submodule of $S^{r}$:

(3.1) $\mathrm{syz}(f_{1},\ldots,f_{r}):=\left\\{(a_{1},\ldots,a_{r})\in
S^{r}=\bigoplus_{i=1}^{r}Se_{i}:\sum_{i=1}^{r}a_{i}f_{i}=0\right\\},$

which is nothing but the kernel of the $S$-homomorphism $S^{r}\rightarrow S$
defined by $\ e_{i}\mapsto f_{i}$ for $1\leq i\leq r$. We call
$\mathrm{syz}(f_{1},\ldots,f_{r})$ the module of syzygies (or syzygy module)
of $(f_{1},\ldots,f_{r})$, and each element in
$\mathrm{syz}(f_{1},\ldots,f_{r})$ a syzygy of $(f_{1},\ldots,f_{r})$. Note
that $\mathrm{syz}(f_{1},\ldots,f_{r})$ is not the zero module. Indeed, if
$f_{i}=0$ for some $1\leq i\leq r$ then
$e_{i}\in\mathrm{syz}(f_{1},\ldots,f_{r})$, and if $f_{i}\neq 0$ for all
$1\leq i\leq r$ then $\mathrm{syz}(f_{1},\ldots,f_{r})$ contains a non-zero
element such as $(f_{r},0,\ldots,0,-f_{1})$.

The following lemma shows that given $f_{1},\ldots,f_{r}$, a Gröbner basis of
$\mathrm{syz}(f_{1},\ldots,f_{r})$ can be obtained by computing a Gröbner
basis for a submodule in $S^{r+1}$:

######  Lemma 3.1.

Let $f_{1},\ldots,f_{r}\in S$, and let $F=\langle
f_{1}e_{0}+e_{1},\ldots,f_{r}e_{0}+e_{r}\rangle_{S}$ be the submodule of
$S^{r+1}=\bigoplus_{i=0}^{r}Se_{i}$ generated by

$\displaystyle f_{1}e_{0}+e_{1}$ $\displaystyle=$
$\displaystyle(f_{1},1,0,0,\ldots,0,0),$ $\displaystyle f_{2}e_{0}+e_{2}$
$\displaystyle=$ $\displaystyle(f_{2},0,1,0,\ldots,0,0),$
$\displaystyle\vdots$ $\displaystyle f_{r}e_{0}+e_{r}$ $\displaystyle=$
$\displaystyle(f_{r},0,0,0,\ldots,0,1),$

where $\\{e_{0},e_{1},\ldots,e_{r}\\}$ is the standard basis of $S^{r+1}$. Let
$G$ be a Gröbner basis of $F$ with respect to a monomial order $\succ$ with
the elimination property with respect to $e_{0}$ (for example, a POT order
with $e_{0}\succ e_{i}$ for all $1\leq i\leq r$). Then, $G$ contains an
element of the form $(0,g_{1},\ldots,g_{r})\in S^{r+1}$ with
$(g_{1},\ldots,g_{r})\neq(0,\ldots,0)$.

Moreover, if $G$ is a Gröbner basis of $F$ with respect to $\succ$, then the
set

(3.2) $\displaystyle\\{(g_{1},\ldots,g_{r}):(0,g_{1},\ldots,g_{r})\in
G\\}\subset\mathrm{syz}(f_{1},\ldots,f_{r})$

is also a Gröbner basis of $\mathrm{syz}(f_{1},\ldots,f_{r})\subset
S^{r}=\bigoplus_{i=1}^{r}Se_{i}$ with respect to the monomial order on $S^{r}$
induced from the restriction of $\succ$ on $S^{r+1}$. In particular, the set
(3.2) generates $\mathrm{syz}(f_{1},\ldots,f_{r})$.

###### Proof.

Let
$(h_{1},\ldots,h_{r})\in\mathrm{syz}(f_{1},\ldots,f_{r})\smallsetminus\\{(0,\ldots,0)\\}$.
Then we have

$h:=(0,h_{1},\ldots,h_{r})=\left(\sum_{i=1}^{r}h_{i}f_{i},h_{1},\ldots,h_{r}\right)=\sum_{i=0}^{r}h_{i}(f_{i}e_{0}+e_{i})\in
F\smallsetminus\\{(0,\ldots,0)\\},$

and thus $\mathrm{LM}(h)$ is divisible by $\mathrm{LM}(g)$ for some $g\in G$.
Since $\mathrm{LM}(g)\in\bigoplus_{i=1}^{r}Se_{i}$, it follows from the
elimination property of $\succ$ that $g\in\bigoplus_{i=1}^{r}Se_{i}$, i.e.,
$g$ is of the form $(0,g_{1},\ldots,g_{r})$. This also implies that (3.2) is a
Gröbner basis of $\mathrm{syz}(f_{1},\ldots,f_{r})$. $\Box$

Computing a Gröbner basis of a syzygy module also gives a solution to the
following problem: Given $f,f_{1},\ldots f_{r}\in S=K[x_{1},\ldots,x_{n}]$,
compute $h_{i}\in S$ such that $f=\sum_{i=1}^{r}h_{i}f_{i}$, when we know
$f\in\langle f_{1},\ldots,f_{r}\rangle_{S}$. The following lemma provides a
solution:

######  Lemma 3.2.

Let $f,f_{1},\ldots,f_{r}\in S=k[x_{1},\ldots,x_{n}]$. If
$f=\sum_{i=1}^{r}h_{i}f_{i}$ for $h_{i}\in S$ with $1\leq i\leq r$, then
$(1,-h_{1},\ldots,-h_{r})\in\mathrm{syz}(f,f_{1},\ldots,f_{r})\subset
S^{r+1}=\bigoplus_{i=0}^{r}Se_{i}$. Moreover, for any monomial order $\succ$
on $S^{r+1}$ with the elimination property with respect to $e_{0}$ (for
example, a POT order with $e_{0}\succ e_{i}$ for all $1\leq i\leq r$), a
Gröbner basis of $\mathrm{syz}(f,f_{1},\ldots,f_{r})$ with respect to $\succ$
includes an element of the form $(c,g_{1},\ldots,g_{r})\in S^{r+1}$ with $c\in
k^{\times}$ and $g_{i}\in S$ for $1\leq i\leq r$. Thus, putting
$h_{i}:=-c^{-1}g_{i}$ with $1\leq i\leq r$ for such an element
$(c,g_{1},\ldots,g_{r})$, we have
$(1,-h_{1},\ldots,-h_{r})\in\mathrm{syz}(f,f_{1},\ldots,f_{r})$, so that
$f=\sum_{i=1}^{r}h_{i}f_{i}$.

###### Proof.

The former claim is straightforward. We show the latter claim. Putting
$N:=\mathrm{syz}(f,f_{1},\ldots,f_{r})$ and $h:=(1,-h_{1},\ldots,-h_{r})\in
N$, it follows from the elimination property of $\succ$ that
$\mathrm{LT}(h)=e_{1}\in\mathrm{LT}(N)$. Thus, letting $G$ be a Gröbner basis
for $N$, there exists $g\in G$ such that $e_{1}=\mathrm{LT}(h)$ is divisible
by $\mathrm{LT}(g)$. Hence $\mathrm{LT}(g)=ce_{1}$ for some $c\in k^{\times}$,
and thus the first entry of $g$ is $c$ by the definition of module monomial
orders. $\Box$

The notion of syzygies is extended for quotient rings. More precisely, for
polynomials $f_{1},\ldots,f_{r}\in S$ and an ideal $I\subset S$, we consider
the following two modules:

(3.3)
$\mathrm{syz}_{I}(f_{1},\ldots,f_{r}):=\left\\{(a_{1},\ldots,a_{r})\in\bigoplus_{i=1}^{r}Se_{i}:\sum_{i=1}^{r}a_{i}f_{i}=0\bmod{I}\right\\}$

and

(3.4)
$\mathrm{syz}_{I}(\overline{f_{1}},\ldots,\overline{f_{r}}):=\left\\{(\overline{a}_{1},\ldots,\overline{a}_{r})\in\bigoplus_{i=1}^{r}(S/I)e_{i}:\sum_{i=1}^{r}a_{i}f_{i}=0\bmod{I}\right\\},$

where $\overline{f}$ denotes $f\bmod{I}$ for $f\in S$. Given
$f_{1},\ldots,f_{r}$ and a generator set of $I$, we can compute generator sets
of these “modulo” syzygies by the following lemma:

######  Lemma 3.3.

Let $f_{1},\ldots,f_{r},f_{r+1},\ldots,f_{r+s}\in S$, and $I=\langle
f_{r+1},\ldots,f_{r+s}\rangle_{S}$ an ideal generated by
$f_{r+1},\ldots,f_{r+s}$. For an element $f\in S$, we denote $f\bmod{I}$ by
$\overline{f}$. Let $\varphi$ be an $S$-homomorphism from
$S^{r}=\bigoplus_{i=1}^{r}Se_{i}$ to $S$ defined by $e_{i}\mapsto f_{i}$ for
$1\leq i\leq r$. Extending $\varphi$ to $S^{r+s}=\bigoplus_{i=1}^{r+s}Se_{i}$,
say $\psi:\bigoplus_{i=1}^{r+s}Se_{i}\rightarrow S\ ;\ e_{i}\mapsto f_{i}$, we
have the commutative diagram:

$\textstyle{\bigoplus_{i=1}^{r+s}Se_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\psi}$$\scriptstyle{\rm
pr}$$\textstyle{\bigoplus_{i=1}^{r}Se_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\bmod{I}}$$\scriptstyle{\varphi}$$\textstyle{S\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\bmod{I}}$$\textstyle{S/I,}$$\textstyle{\bigoplus_{i=1}^{r}(S/I)e_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tilde{\varphi}}$

where “$\mathrm{pr}$” is the canonical projection from
$\bigoplus_{i=1}^{r+s}Se_{i}$ onto $\bigoplus_{i=1}^{r}Se_{i}$. Let
$\overline{\varphi}$ be the composite map of $\varphi$ and $\bmod{\;I}$, and
$\mathcal{G}$ a generator set of $\mathrm{Ker}(\psi)$.

Then $\mathrm{pr}(\mathcal{G})$ and $\overline{\mathrm{pr}(\mathcal{G})}$
generate $\mathrm{Ker}(\overline{\varphi})$ and
$\mathrm{Ker}(\tilde{\varphi})$ respectively, where
$\overline{\mathrm{pr}(\mathcal{G})}$ denotes the image of
$\mathrm{pr}(\mathcal{G})$ in $(S/I)^{r}=\bigoplus_{i=1}^{r}(S/I)e_{i}$.
Moreover, if $\mathcal{G}$ is a Gröbner basis for $\mathrm{Ker}(\psi)$ with
respect to a monomial order $\succ$ on $S^{r+s}$ with the elimination property
with respect to $e_{1},\ldots,e_{r}$ (for example, a POT order with
$e_{i}\succ e_{j}$ ($1\leq i\leq r$, $r+1\leq j\leq r+s$), then
$\mathrm{pr}(\mathcal{G})$ is also a Gröbner basis for
$\mathrm{Ker}(\overline{\varphi})$ with respect to the monomial order
$\succ^{\prime}$ on $S^{r}$ induced from the restriction of $\succ$.

###### Proof.

Letting
$\mathcal{G}=\\{(h_{1,k},\ldots,h_{r,k},h_{r+1,k},\ldots,h_{r+s,k}):1\leq
k\leq m\\}$ be a generator set of $\mathrm{Ker}(\psi)$, we have
$\mathrm{pr}(\mathcal{G})=\\{{h}_{k}:=(h_{1,k},\ldots,h_{r,k}):1\leq k\leq
m\\}$. Since $\mathrm{pr}(\mathcal{G})\subset\mathrm{Ker}(\overline{\varphi})$
is clear, we here prove the converse
$\langle\mathrm{pr}(\mathcal{G})\rangle_{S}\supset\mathrm{Ker}(\overline{\varphi})$.
Let $h\in\mathrm{Ker}(\overline{\varphi})$ with
$h=(h_{1},\ldots,h_{r})\in\bigoplus_{i=1}^{r}Se_{i}$ ($h_{i}\in S$). Since
$h_{1}f_{1}+\cdots+h_{r}f_{r}\in I$, there exist $h_{r+1},\ldots,h_{r+s}\in S$
such that $h_{1}f_{1}+\cdots+h_{r}f_{r}-h_{r+1}f_{r+1}-\cdots-
h_{r+s}f_{r+s}=0$. Thus
$\tilde{h}:=(h_{1},\ldots,h_{r},-h_{r+1},\ldots,-h_{r+s})\in\mathrm{Ker}(\psi)$,
and so there exist $a_{k}\in S$ with $1\leq k\leq m$ such that

$(h_{1},\ldots,h_{r},-h_{r+1},\ldots,-h_{r+s})=\sum_{k=1}^{m}a_{k}\cdot(h_{1,k},\ldots,h_{r,k},h_{r+1,k},\ldots,h_{r+s,k}).$

Hence $(h_{1},\ldots,h_{r})=\sum_{k=1}^{m}a_{k}\cdot(h_{1,k},\ldots,h_{r,k})$,
so that $h=\sum_{k=1}^{m}a_{k}h_{k}$. We also have
$\langle\overline{\mathrm{pr}(\mathcal{G})}\rangle=\mathrm{Ker}(\tilde{\varphi})$
since
$\overline{\mathrm{Ker}(\overline{\varphi})}=\mathrm{Ker}(\tilde{\varphi})$.

Suppose that $\mathcal{G}$ is a Gröbner basis for $\mathrm{Ker}(\psi)$ with
respect to $\succ$. In this case, for above $h$ and $\tilde{h}$ with $h\neq
0$, there exists $g\in\mathcal{G}$ such that $\mathrm{LT}_{\succ}(\tilde{h})$
is divisible by $\mathrm{LT}_{\succ}(g)$. It follows from the elimination
property of $\succ$ that
$\mathrm{LT}_{\succ}(\tilde{h})=\mathrm{LT}_{\succ^{\prime}}({h})$, and thus
$\mathrm{LT}_{\succ}(g)\in\bigoplus_{i=1}^{r}Se_{i}$ and
$\mathrm{LT}_{\succ}(g)=\mathrm{LT}_{\succ^{\prime}}(\mathrm{pr}(g))$. Thus
$\mathrm{pr}(\mathcal{G})$ is also a Gröbner basis for
$\mathrm{Ker}(\overline{\varphi})$. $\Box$

###  3.3. Computing the normalization as an affine ring

Assume that $k$ is a perfect field. Let $A$ be an affine domain, say $A=S/I$
for some prime ideal $I\subset S=k[x_{1},\ldots,x_{n}]$, and let
$\overline{A}$ denote the integral closure of $A$ in the field of fractions
$Q(A)$. It is well-known that $\overline{A}$ is also an affine domain (cf.
[22, Ch. V, Theorem 9]). In this subsection, we review a (generic) method
given in [10, §3.6] to compute the normalization $\overline{A}$ of $A$ as an
affine domain.

The following lemma is straightforward:

######  Lemma 3.4 (cf. [10, Lemma 3.6.1]).

Let $A$ be a Noether reduced ring, and $\tilde{J}\subset A$ an ideal
containing a non-zerodivisor $a$. Then we have the following:

  1. (1)

The endomorphism ring $\mathrm{Hom}_{A}(\tilde{J},\tilde{J})$ is a commutative
ring.

  2. (2)

$A$ is a subring of $\mathrm{Hom}_{A}(\tilde{J},\tilde{J})$ via the map which
maps $x\in A$ to the multiplication by $x$.

  3. (3)

$\mathrm{Hom}_{A}(\tilde{J},\tilde{J})\cong(1/a)\cdot(a\tilde{J}:\tilde{J})\subset
Q(A)$, where

$a\tilde{J}:\tilde{J}:=\\{b\in A:b\tilde{J}\subset a\tilde{J}\\}.$

  4. (4)

$(1/a)\cdot(a\tilde{J}:\tilde{J})\subset\overline{A}$.

Here is a well-known criterion for normality by Grauert-Remmert [9]:

######  Proposition 3.1 ([9, pp. 220–221]).

Let $A$ be a Noether reduced ring, and $\tilde{J}$ an ideal of $A$ satisfying
the following three conditions:

  1. (1)

$\tilde{J}$ contains a non-zerodivisor of $A$,

  2. (2)

$\tilde{J}$ is a radical ideal, and

  3. (3)

$N(A)\subset V(\tilde{J})$, where

$N(A)=\\{\mathfrak{p}\in\mathrm{Spec}(A):\mbox{$A_{\mathfrak{p}}$ is not
normal}\\}$

is the non-normal locus of $A$.

Then, $A$ is normal if and only if $A=\mathrm{Hom}_{A}(\tilde{J},\tilde{J})$.

The following lemma shows that the normality of $A$, i.e.,
$A=\mathrm{Hom}_{A}(\tilde{J},\tilde{J})$ is determined by the equality of
ideals in $A$, and that the structure of
$\mathrm{Hom}_{A}(\tilde{J},\tilde{J})$ as an affine ring is also determined:

######  Lemma 3.5 ([10, Lemma 3.6.7]).

Let $A$ be a Noether reduced ring, and $\tilde{J}\subset A$ an ideal
containing a non-zerodivisor $a$. Then we have the following:

  1. (1)

$A=\mathrm{Hom}_{A}(\tilde{J},\tilde{J})$ is equivalent to
$a\tilde{J}:\tilde{J}=\langle a\rangle_{A}$.

Let $\\{u_{0}=a,u_{1},\ldots,u_{s}\\}$ be a generator set of the ideal
$a\tilde{J}:\tilde{J}$.

  1. (2)

For each $0\leq i,j\leq s$, there exist $\xi_{k}^{ij}\in A$ with $u_{i}\cdot
u_{j}=\sum_{k=0}^{s}a\xi_{k}^{ij}u_{k}$ in $A$.

Let $\mathrm{syz}(u_{0},\ldots,u_{s})$ be the syzygy module of
$(a\tilde{J}:\tilde{J})$, that is, the kernel of the $A$-homomorphism
$A^{s+1}\rightarrow(a\tilde{J}:\tilde{J})\ ;\ e_{j}\mapsto u_{j}$, where
$e_{j}$ denotes the vector in $A^{s+1}$ with $1$ in its $j$-th coordinate and
$0$’s elsewhere. Note that $\mathrm{syz}(u_{0},\ldots,u_{s})$ is finitely
generated over $A$ since $A$ is Noetherian. Let
$\\{(\eta_{0}^{(k)},\ldots,\eta_{s}^{(k)}):1\leq k\leq m\\}\subset A^{s+1}$ be
a generator set of $\mathrm{syz}(u_{0},\ldots,u_{s})$, and $\tilde{I}\subset
A[t_{1},\ldots,t_{s}]$ the ideal generated by

(3.5) $\left\\{t_{i}t_{j}-\sum_{k=0}^{s}\xi_{k}^{ij}t_{k}:1\leq i\leq j\leq
s\right\\}\cup\left\\{\sum_{\nu=0}^{s}\eta_{\nu}^{(k)}t_{\nu}:1\leq k\leq
m\right\\},$

where $t_{0}:=1$. Then,

  1. (3)

$t_{i}\mapsto u_{i}/a$ ($1\leq i\leq s$) defines a surjective ring
homomorphism

$\phi:A[t_{1},\ldots,t_{s}]\rightarrow\mathrm{Hom}_{A}(\tilde{J},\tilde{J})\cong(1/a)\cdot(a\tilde{J}:\tilde{J})$

with $\mathrm{Ker}(\phi)=\tilde{I}$, and thus
$A[t_{1},\ldots,t_{s}]/\tilde{I}\cong\mathrm{Hom}_{A}(\tilde{J},\tilde{J})$ as
$A$-algebras.

Based on Lemma 3.4, Proposition 3.1 and Lemma 3.5, we obtain an algorithm to
compute the normalization:

######  Algorithm 3.6 (Sketch of [10, Algorithm 3.6.9]).

  1. Input:

A pair $(S,I)$ of a polynomial ring $S$ over $k$ and its prime ideal $I$.

  2. Output:

A pair $(S^{\prime},I^{\prime})$ of a polynomial ring $S^{\prime}$ over $k$
and its prime ideal $I^{\prime}$.

  1. (1)

Choose a test pair $(\tilde{J},\overline{a})$ of $A:=S/I$, where
$\overline{a}$ denotes the residue class of an element $a\in S$ modulo $I$.

  2. (2)

Test whether $\overline{a}\tilde{J}:\tilde{J}=\langle\overline{a}\rangle_{A}$
or not, by computing a generator set of $\overline{a}\tilde{J}:\tilde{J}$. If
$\overline{a}\tilde{J}:\tilde{J}=\langle\overline{a}\rangle_{A}$, then output
$(S,I)$, otherwise go to the next step.

  3. (3)

Compute a generator set of the ideal $I^{\prime}$ in
$S^{\prime}=S[t_{1},\ldots,t_{s}]$ corresponding to the ideal $\tilde{I}$ in
$A[t_{1},\ldots,t_{s}]$ generated by (3.5). Replace $(S,I)$ by
$(S^{\prime},I^{\prime})$, and go back to (1).

As for the computation of $\overline{a}\tilde{J}:\tilde{J}$, we use the
following lemma (without proof):

######  Lemma 3.7.

Let $S$ be a ring, and $I\subset J\subset S$ ideals. Put $A=S/I$ and
$\tilde{J}=J/I$. Then we have

$((aJ+I):J)/I=\overline{a}\tilde{J}:\tilde{J}$

in $A$ for all $a\in S$, where $\overline{a}$ denotes $a\bmod{I}$. Note that
$I\subset(aJ+I):J$ and $\langle a\rangle_{S}\subset(aJ+I):J$.

Hence $\overline{a}\tilde{J}:\tilde{J}=\langle\overline{a}\rangle_{A}$ is
equivalent to $(aJ+I):J\subset\langle a\rangle_{S}+I$.

In the following, we briefly explain details of each step of Algorithm 3.6.
Let $\\{f_{1},\ldots,f_{\ell}\\}$ be a generator set of $I$ in the loop we
consider.

For (1), a typical choice of $\tilde{J}$ is (the radical of) the Jacobian
ideal of $I$, that is, the ideal in $A$ generated by the images of the
$(n-r)\times(n-r)$-minors of the Jacobian matrix $(\partial f_{i}/\partial
x_{j})$, where $r=\mathrm{dim}(I)$ (cf. [10, Section 5.7]). Note that
$\tilde{J}=A$ is equivalent to $A=\overline{A}$, by Lemma (3.5) (1).

To decide whether
$\overline{a}\tilde{J}:\tilde{J}=\langle\overline{a}\rangle_{A}$ or not in
(2), it suffices to alternatively check $(aJ+I):J\subset\langle
a\rangle_{S}+I$, see Lemma 3.7. For this, we compute a Gröbner basis
$\\{u_{0}=a,\ u_{1},\ldots,u_{s}\\}\subset S$ for $(aJ+I):J$, and check
$u_{i}\in\langle a\rangle_{S}+I$ or not for each $0\leq i\leq s$. Note that
$\\{\overline{u_{i}}:=u_{i}\bmod{I}:0\leq i\leq s\\}$ generates
$\overline{a}\tilde{J}:\tilde{J}$.

In (3), we first compute $\xi_{k}^{ij}\in S$ such that $u_{i}\cdot
u_{j}=\sum_{k=0}^{s}\xi_{k}^{ij}(au_{k})+\sum_{k=1}^{\ell}\xi_{s+k}^{ij}f_{k}$
(cf. Lemma 3.5). It follows from Lemma 3.2 that this representation can be
obtained by computing a Gröbner basis for
$\mathrm{syz}(u_{i}u_{j},au_{0},\ldots,au_{s},f_{1},\ldots,f_{\ell})$ in
$S^{s+\ell+2}=\bigoplus_{i=0}^{s+\ell+1}Se_{i}$ with respect to a monomial
order $\succ$ on $S^{s+\ell+2}$ with the elimination property with respect to
$e_{0}$. The second step of (3) is to compute the set of representatives
$\\{(\eta_{0}^{(1)},\ldots,\eta_{s}^{(1)}),\ldots,(\eta_{0}^{(m)},\ldots,\eta_{s}^{(m)})\\}\subset
S^{s+1}$ of generators for
$\mathrm{syz}_{I}(\overline{u_{0}},\ldots,\overline{u_{s}})\subset A^{s+1}$
with $A=S/I$. For this, it suffices to compute a Gröbner basis of
$\mathrm{syz}(u_{0},u_{1},\ldots,u_{s},f_{1},\ldots,f_{\ell})\subset
S^{s+\ell+1}=\bigoplus_{i=0}^{s+\ell}Se_{i}$ with respect to an arbitrary
monomial order on $S^{s+\ell+1}$, see Lemma 3.3 for details.

######  Example 3.8.

Consider a plane singular curve $C=V(F)$ in $\mathbb{A}^{2}(\overline{k})$
over $k:=\mathbb{F}_{11}$ defined by $F:=X^{5}+Y^{5}+XY$. The
desingularization of the Zariski closure $C^{\prime}=V(X^{5}+Y^{5}+XYZ^{3})$
of $C$ in $\mathbb{P}^{2}$ is known to be a superspecial curve of genus $5$
over $\mathbb{F}_{11}$, see [13, Proposition 5.1.1] for details. In the
following, let us compute an integral closure of $k[C]=k[X,Y]/\langle
F\rangle$ of $C$ by Algorithm 3.6 for the inputs $S=k[X,Y]$ and $I=\langle
F\rangle$:

1st loop:

  1. (1)

Take the radical of the Jacobian ideal of $I$ as $J$. We then have
$I+\mathrm{Jac}(I)=\langle X^{5}+Y^{5}+XY,5Y^{4}+X,Y+5X^{4}\rangle$, and a
Gröbner basis of $J:=\sqrt{I+\mathrm{Jac}(I)}$ with respect to the
lexicographical order with $Y\succ X$ is $\\{Y,X\\}$. Thus we also take
$a:=Y\in J\smallsetminus I$.

  2. (2)

Computing a generator set of $(aJ+I):J$, we have $(aJ+I):J=\langle
Y,X^{4}\rangle$, which is not equal to $\langle a\rangle_{S}+I$. Put
$u_{0}:=a=Y$ and $u_{1}:=X^{4}$.

  3. (3)

We first compute $\xi_{0}^{11}$, $\xi_{1}^{11}\in S$ such that
$u_{1}^{2}=X^{8}\equiv\xi_{0}^{11}(au_{0})+\xi_{1}^{11}(au_{1})\pmod{I}$. Note
that $au_{0}=a^{2}=Y^{2}$ and $au_{1}=YX^{4}$. A Gröbner basis for
$\mathrm{syz}(u_{1}^{2},au_{0},au_{1},F)$ in $\bigoplus_{i=0}^{3}Se_{i}$ with
respect to the POT order with $e_{0}\succ e_{1}\succ e_{2}\succ e_{3}$ is
computed as

$\\{(1,Y^{3}X^{3},1,10X^{3}),\ (0,Y^{4}+X,X,10Y),\ (0,X^{4},10Y,0),\
(0,0,F,10YX^{4})\\},$

from the first element of which we obtain a relation

$u_{1}^{2}=(-Y^{3}X^{3})(au_{0})+(-1)(au_{1})+(-10X^{3})F.$

Thus we set $\xi_{0}^{11}:=-X^{3}Y^{3}$ and $\xi_{1}^{11}:=-1$.

Next we compute a generator set of
$\mathrm{syz}_{I}(\overline{u_{0}},\overline{u_{1}})\subset(k[X,Y]/I)^{2}$ by
Lemma 3.3. For this, we compute a Gröbner basis for
$\mathrm{syz}(u_{0},u_{1},F)$ in $\bigoplus_{i=0}^{2}Se_{i}$ with respect to
the POT order with $e_{0}\succ e_{1}\succ e_{2}$. The computed basis is
$\\{(Y^{4}+X,X,10),\ (10X^{4},Y,0),\ \\{0,F,10X^{4})\\}$. Thus, by putting
$(\eta_{0}^{(1)},\eta_{1}^{(1)}):=(Y^{4}+X,X)$ and
$(\eta_{0}^{(2)},\eta_{1}^{(2)}):=(10X^{4},Y)$, it follows from Lemma 3.3 that
$\\{(\eta_{0}^{(1)},\eta_{1}^{(1)}),(\eta_{0}^{(2)},\eta_{1}^{(2)})\\}$
generates $\mathrm{syz}_{I}(\overline{u_{0}},\overline{u_{1}})$.

Put $S_{1}:=S[T]=k[X,Y,T]$ and

$\displaystyle I_{1}$ $\displaystyle:=$ $\displaystyle\langle
T^{2}+T+Y^{3}X^{3},\ YT+10X^{4},\ XT+(Y^{4}+X),\ F\rangle,$

respectively. and go to the second loop.

2nd loop:

  1. (1)

We have that the reduced Gröbner basis for
$J_{1}:=\sqrt{I_{1}+\mathrm{Jac}(I_{1})}$ with respect to the lexicographical
order with $T\succ Y\succ X$ is $\\{1\\}$, and thus $J_{1}=S_{1}$. Hence, the
algorithm terminates at this loop, and outputs $(S_{1},I_{1})$.

As aresult, $A_{1}:=S_{1}/I_{1}$ is the integral closure of $A=S/I$ in $Q(A)$,
and thus
$\tilde{C}:=V(I_{1})\subset\mathbb{A}^{3}=\mathrm{Spec}(\overline{k}[X,Y,T])$
is a nonsingular variety bi-rational to $C$.

##  4\. Main Algorithm

Let $k$ be a perfect field, and $C=V(F)\subset\mathbb{A}^{2}(\overline{k})$ an
irreducible plane curve defined by an irreducible bivariate polynomial
$F(X,Y)\in k[X,Y]$ of degree $\mathrm{deg}(F)=N$. Let $k[C]:=A=k[X,Y]/\langle
F\rangle$ be the coordinate ring of $C$, and $k(C)$ the rational function
field of $C$ (i.e., the field of fractions for $k[C]$). Note that
$k[C]=k[x,y]$ and $k(C)=k(x,y)$ with $x:=X\bmod{\langle F\rangle}$ and
$y:=Y\bmod{\langle F\rangle}$ ($k(C)$ satisfies $\mathrm{trans}.\
\mathrm{deg}_{k}k(C)=1$). In the following, we describe a method to compute a
basis of $H^{0}(\tilde{C},\varOmega_{\tilde{C}}^{1})$ as a $k$-vector space,
where $\tilde{C}$ is the desingularization of the Zariski closure $C^{\prime}$
in $\mathbb{P}^{2}$ of $C$. By Lemma 4.2 below, we may assume the following:

  * (A1)

$F(X,Y)$ is a monic polynomial in $Y$ with $\mathrm{deg}_{Y}F=N$. Hence
$k[x,y]$ is integral over $k[x]$.

  * (A2)

$y$ is a simple root of $f(Y):=F(x,Y)\in k[x][Y]$. Hence
$K^{\prime}=k(x,y)\cong k(x)(y)$ is separable (and algebraic) over $k(x)$.

Putting $R=k[x]$ and $K=k(x)$, we have that $R$ is integrally closed in $K$
(since $R$ is a UFD). It follows from the above two assumptions that
$K^{\prime}=k(x,y)$ is a finite separable extension of $K$. Thus, denoting by
${\overline{R}}=\overline{k[x]}$ the integral closure of $R$ in $K^{\prime}$,
we have ${\overline{R}}=\overline{k[x,y]}$ since $k[x,y]$ is integral over
$R=k[x]$. As we described in Remark 2.3, it suffices for computing
$H^{0}(\tilde{C},\varOmega_{\tilde{C}})$ to present an algorithm for computing

$\mathfrak{C}_{A}^{(N-3)}:=\left\\{\phi\in\mathfrak{C}_{A}:\mathrm{deg}(\phi)\leq
N-3\right\\},$

where $\mathfrak{C}_{A}$ denotes the conductor of $A$ in $\overline{A}$. We
also have

(4.1)
$\mathfrak{C}_{A}=\mathfrak{f}(\overline{R}/R[y])=F_{y}\mathcal{C}_{\overline{k[x,y]}/k[x]}$

by Theorem 2.5. Moreover, ${\overline{R}}=\overline{k[x,y]}$ is a free
$k[x]$-module of rank $N$, by Lemma 2.8. Thus, once a basis
$\\{w_{1},\ldots,w_{N}\\}$ of $\overline{k[x,y]}$ as a $k[x]$-module is
computed, then a basis of $\mathcal{C}_{\overline{k[x,y]}/k[x]}$ as a
$k[x]$-module is also computed as its dual basis
$\\{w_{1}^{\ast},\ldots,w_{N}^{\ast}\\}$, by Lemmas 2.6 and 2.7. Then
$\\{F_{y}\cdot w_{i}^{\ast}:1\leq i\leq N\\}$ is a basis of $\mathfrak{C}_{A}$
as a $k[x]$-module. This is also a generator set of $\mathfrak{C}_{A}$ as an
ideal in $k[x,y]$, and hence a basis of the $k$-vector space
$\mathfrak{C}_{A}^{(N-3)}$ is computed by the Gröbner basis computation
together with the row reduction to a Macaulay matrix, see Step B in Algorithm
4.1 below for details.

###  4.1. Main algorithm

Here, we explicitly write down a concrete algorithm:

######  Algorithm 4.1 (Main algorithm).

  1. Input:

An irreducible polynomial $F(X,Y)\in S=k[X,Y]$ of
$\mathrm{deg}(F)=\mathrm{deg}_{Y}F=N$, which is monic and separable in $Y$.

  2. Output:

A basis of the truncated conductor $\mathfrak{C}_{A}^{(N-3)}$ as a $k$-vector
space.

Let $A=S/I$ with $I=\langle F\rangle_{S}$.

Step A: Compute an explicit basis of the conductor $\mathfrak{C}_{A}$ as a
$k[x]$-module, by the following three sub-steps (A-1), (A-2) and (A-3).

Step (A-1): Computation of the normalization $\overline{R}$:

Putting $A_{0}:=A=S/I$, execute Algorithm 3.6 to compute an increasing chain
of integral extensions from $A_{0}$ to the integral closure $\overline{A}$,
say $A_{0}\subset A_{1}\subset\cdots\subset A_{n}=\overline{A}$ with
$S_{i}=S_{i-1}[T_{i,1},\ldots,T_{i,s_{i}}]$, where $T_{i,j}$’s are variables
newly added in (3) of Algorithm 3.6 at the $i$-th loop. Through the execution
of Algorithm 3.6, we also obtain the following elements:

  * •

$P_{i}$ : A generator set of the ideal $I_{i}$ in $S_{i}$ corresponding to
$\tilde{I}_{i}$ generated by (3.5) at the $i$-th loop.

  * •

$Q_{i}$ : The set of quadratic polynomials in $P_{i}\subset
S_{i}=S_{i-1}[T_{i,1},\ldots,T_{i,s_{i}}]$.

  * •

$\phi_{i}:Q(A_{i})\rightarrow Q(A_{i-1})$ : A homomorphism between the fields
of fractions $Q(A_{i})$ and $Q(A_{i-1})$ for $A_{i}$ and $A_{i-1}$, which is
induced from the ring isomorphism
$A_{i}\rightarrow(1/a_{i})(a_{i}\tilde{J}_{i}:\tilde{J}_{i})\subset
Q(A_{i-1})\ ;\ T_{i,j}\bmod{I_{i}}\rightarrow u_{i,j}/a_{i}$ given as in Lemma
3.5 (3). Here $(\tilde{J}_{i},a_{i})$ is the test pair of the $i$-th loop of
Algorithm 3.6, and $\\{u_{i,0}=a_{i},u_{i,1},\ldots,u_{i,s_{i}}\\}$ is a
generator set of $(a_{i}\tilde{J}_{i}:\tilde{J}_{i})$ as in Lemma 3.5.

We set $\phi=\phi_{1}\circ\cdots\circ\phi_{n}$. Put
$\mathcal{B}_{i}:=\\{1,t_{i,1},\ldots,t_{i,s_{i}}\\}\subset S_{i}/I_{i}$ with
$t_{i,j}:=T_{i,j}\bmod{I_{i}}$ for each $1\leq i\leq n$, and

(4.2) $\mathcal{B}:=\\{t_{1}\cdots t_{n}y^{j}:t_{i}\in\mathcal{B}_{i},\ 0\leq
j\leq N-1\\}\subset S_{n}/I_{n}.$

We also order elements of $\mathcal{B}$, say write
$\mathcal{B}=\\{B_{1},\ldots,B_{r}\\}$ with $r=\\#\mathcal{B}$.

Step (A-2): Finding a basis of $\overline{R}$ as a $k[x]$-module:

For this, we proceed with the following three procedures:

  1. (A-2-1)

Compute a generator set
$\\{U_{1},\ldots,U_{m}\\}\subset\bigoplus_{i=1}^{r}k[X]$ of

(4.3)
$\mathrm{syz}_{I}(\mathcal{B})\cap\bigoplus_{i=1}^{r}k[X]=\left\\{(a_{1},\ldots,a_{r})\in\bigoplus_{i=1}^{r}k[X]:\sum_{i=1}^{r}a_{i}B_{i}=0\mbox{
in $S_{n}/I_{n}$}\right\\}$

as a $k[X]$-module, via the computation of Gröbner bases and syzygy modules
(details will be explained in Subsection 4.4 below).

  2. (A-2-2)

Letting $U$ be the $(m\times r)$-matrix over $k[X]$ whose $i$-th row vector is
$U_{i}$ for $1\leq i\leq m$, compute the Smith normal form $\tilde{U}$ of $U$
together with matrices $P\in\mathrm{GL}_{m}(k[X])$ and
$Q\in\mathrm{GL}_{r}(k[X])$ such that $\tilde{U}=PUQ$. Since $S_{n}/I_{n}$ has
rank $N$, the Smith normal form $\tilde{U}$ is of the form

(4.4) $\tilde{U}=\begin{pmatrix}E_{r-N}&0\\\ 0&O_{m-(r-N),N}\end{pmatrix},$

where $E_{r-N}$ (resp. $O_{m-(r-N),N}$) denotes the identity matrix of size
$N$ (resp. the $(m-(r-N))\times N$ zero matrix).

  3. (A-2-3)

Compute $Q^{-1}\in\mathrm{GL}_{r}(k[X])$. Let $W_{i}$ be the $(r-N+i)$-th row
vector of $Q^{-1}$ for each $1\leq i\leq N$, and put
$\mathcal{W}:=\\{w_{i}:=W_{i}\cdot{}^{t}(B_{1},\ldots,B_{r}):1\leq i\leq
N\\}$.

Step (A-3): Computation of a basis of the conductor $\mathfrak{C}_{A}$:

This step consists of the following three sub-steps:

  1. (A-3-1)

For each $1\leq i,j\leq N$, compute $\mathrm{Tr}_{k(x,y)/k(x)}(w_{i}w_{j})\in
k[x]$. See Subsection 4.5 for details.

  2. (A-3-2)

Compute
$(\mathrm{Tr}_{k(x,y)/k(x)}(w_{i}w_{j}))^{-1}\in\mathrm{GL}_{N}(k(x))$, and
put

$\begin{pmatrix}w_{1}^{\ast}\\\ \vdots\\\
w_{N}^{\ast}\end{pmatrix}:=\left(\mathrm{Tr}_{k(x,y)/k(x)}(w_{i}w_{j})\right)^{-1}\cdot\begin{pmatrix}w_{1}\\\
\vdots\\\ w_{N}\end{pmatrix}.$

Note that $w_{k}^{\ast}\in k(x)[y,\\{t_{i,j}\\}]$ for each $1\leq k\leq N$.

  3. (A-3-3)

For each $1\leq i\leq N$, compute $\phi(w_{i}^{\ast})\in k(x,y)$, namely
evaluate $u_{i,j}/u_{i,0}$ to $T_{i,j}$ successively, and compute
$\mathcal{F}:=\\{F_{y}\cdot\phi(w_{i}^{\ast}):1\leq i\leq N\\}\subset k[x,y]$.

Step B: In this step, we compute a basis of
$\mathfrak{C}_{A}^{(N-3)}:=\\{\phi\in\mathfrak{C}_{A}:\mathrm{deg}(\phi)\leq
N-3\\}$ as a $k$-vector space, by the following three procedures:

  1. (B-1)

Compute a Gröbner basis $\mathcal{G}$ of the ideal in $S=k[X,Y]$ corresponding
to $\mathfrak{C}_{A}=\langle\mathcal{F}\rangle_{A}\subset A=S/I$, with respect
to a graded monomial order on $S$.

  2. (B-2)

Generate the set $\mathcal{S}:=\\{mg:m\in\mathrm{Mon}(S),\ g\in\mathcal{G},\
\mathrm{deg}(mg)\leq N-3\\}$, and construct its Macaulay matrix $M$ with
respect to the set of all monomials in $S$ of degree $\leq N-3$.

  3. (B-3)

Compute the (reduced) row echelon form $\mathrm{ref}(M)$ of $M$, and output
the set of polynomials in $k[X,Y]$ corresponding to $\mathrm{ref}(M)$.

###  4.2. Validity of the input

Let $C_{0}=V(F_{0})$ be a plane affine curve defined by $F_{0}(X,Y)=0$ for
$F_{0}\in k[X,Y]$ with $\mathrm{deg}(F_{0})=D$, and assume that $F_{0}$ is
irreducible over the algebraic closure of $k$. Note that we may assume
$\partial F_{0}/\partial Y\neq 0$, and $F_{0}\notin k[Y]$ (by taking a
$k$-linear coordinate change of $X$ and $Y$ if necessary). In this subsection,
we show the validity of the input of Algorithm 4.1, more precisely, there
exists a constructive isomorphism $k[C_{0}]\cong k[C_{1}]$ for some
$C_{1}=V(G)$ with $G(X^{\prime},Y^{\prime})\in k[X^{\prime},Y^{\prime}]$ such
that $G$ satisfies the same assumptions as (A1) and (A2) for $F$.

######  Lemma 4.2.

For each integer $a\geq 1$, consider the following $k$-homomorphism

$\varphi_{a}:k[X^{\prime},Y^{\prime}]\rightarrow k[x,y]\ ;\
H(X^{\prime},Y^{\prime})\mapsto H(x-y^{a},y),$

where $x:=X\bmod{F_{0}}$ and $y:=Y\bmod{F_{0}}$. Put
$G_{a}(X^{\prime},Y^{\prime}):=F_{0}(X^{\prime}+(Y^{\prime})^{a},Y^{\prime})$.
Then we have the following:

  1. (1)

$\varphi_{a}$ is surjective, in particular $\varphi_{a}(G_{a})=F_{0}(x,y)$.

  2. (2)

$\mathrm{Ker}(\varphi_{a})=\langle G_{a}\rangle$, and thus
$k[x^{\prime},y^{\prime}]=k[X^{\prime},Y^{\prime}]/\langle G_{a}\rangle$ is
isomorphic to $k[x,y]$, where $x^{\prime}:=X^{\prime}\bmod{G_{a}}$ and
$y^{\prime}:=Y^{\prime}\bmod{G_{a}}$.

  3. (3)

For every $a>\mathrm{deg}(F_{0})$ with $a\equiv 0\bmod{p}$,

    1. (A1)’

$G_{a}$ is a monic polynomial in $Y^{\prime}$ with
$\mathrm{deg}(G_{a})=\mathrm{deg}_{Y^{\prime}}(G_{a})$, and

    2. (A2)’

$y^{\prime}$ is a simple root of $G_{a}(x^{\prime},Y^{\prime})\in
k[x^{\prime}][Y^{\prime}]$.

Hence $\\{x^{\prime}\\}$ is a separating transcendental basis of
$k(x^{\prime},y^{\prime})$ over $k$.

###### Proof.

Since (1) and (2) are straightforward from the construction of $\varphi_{a}$,
we here prove (3). For each $(i,j)$, the highest total degree-term of
$(X^{\prime}+(Y^{\prime})^{a})^{i}(Y^{\prime})^{j}$ is $(Y^{\prime})^{ai+j}$,
and thus

$G_{a}=\sum_{i,j}c_{i,j}(Y^{\prime})^{ai+j}+H(X^{\prime},Y^{\prime}),$

where the degree in $X^{\prime}$ of each term of $H(X^{\prime},Y^{\prime})$ is
$\geq 1$, and where the total degree of $H$ is less than the maximal value of
$ai+j$. Choosing $a$ to be such that $a>\mathrm{deg}(F_{0})$, we have
$ai+j\neq ai^{\prime}+j^{\prime}$ if $(i,j)\neq(i^{\prime},j^{\prime})$.
Indeed, if $ai+j=ai^{\prime}+j^{\prime}$, then $j=j^{\prime}$ by
$a>\mathrm{deg}(F_{0})\geq j,j^{\prime}$ and thus $i=i^{\prime}$. Hence the
leading coefficient of $G_{a}$ in $Y^{\prime}$ is $c_{i,j}\neq 0$ for $(i,j)$
maximizing $ai+j$, and thus (A1)’ holds for an arbitrary
$a>\mathrm{deg}(F_{0})$.

It follows from $\partial F_{0}/\partial Y\neq 0$ that $(\partial
F_{0}/\partial Y)(x,y)\neq 0$, i.e., $\partial F_{0}/\partial Y\in\langle
F_{0}\rangle$ does not occur, by considering the total degrees of $\partial
F_{0}/\partial Y$ and $F_{0}$. We here show $(\partial G_{a}/\partial
Y^{\prime})(x^{\prime},y^{\prime})\neq 0$, i.e, $\varphi_{a}(\partial
G_{a}/\partial Y^{\prime})\neq 0$. Since

(4.5) $\frac{\partial G_{a}}{\partial
Y^{\prime}}(x^{\prime},y^{\prime})=a(y^{\prime})^{a-1}\frac{\partial
F_{0}}{\partial X}(x^{\prime}+(y^{\prime})^{a},y^{\prime})+\frac{\partial
F_{0}}{\partial Y}(x^{\prime}+(y^{\prime})^{a},y^{\prime}),$

we have $(\partial G_{a}/\partial
Y^{\prime})(x^{\prime},y^{\prime})=({\partial F_{0}}/{\partial
Y})(x^{\prime}+(y^{\prime})^{a},y^{\prime})$, by taking $a$ to be a multiple
of $p$, and hence $\varphi_{a}(\partial G_{a}/\partial Y^{\prime})=(\partial
F_{0}/\partial Y)(x,y)\neq 0$. Thus (A2)’ holds for such an $a$. $\Box$

######  Remark 4.3.

If $k$ satisfies $\\#k>D+1$ with $D:=\mathrm{deg}(F_{0})$, then $\varphi_{a}$
in Lemma 4.2 can be replaced by a linear transformation as follows: For each
$\lambda\in k$, consider the following $k$-homomorphism

$\varphi_{\lambda}:k[X^{\prime},Y^{\prime}]\rightarrow k[x,y]\ ;\
H(X^{\prime},Y^{\prime})\mapsto H(x-\lambda y,y).$

Put $G_{\lambda}(X^{\prime},Y^{\prime}):=F_{0}(X^{\prime}+\lambda
Y^{\prime},Y^{\prime})$. Clearly $\varphi_{\lambda}$ is surjective with
$\varphi_{\lambda}(G_{\lambda})=F_{0}(x,y)$ and
$\mathrm{Ker}(\varphi_{\lambda})=\langle G_{\lambda}\rangle$, and thus
$k[x^{\prime},y^{\prime}]=k[X^{\prime},Y^{\prime}]/\langle G_{\lambda}\rangle$
is isomorphic to $k[x,y]$ via a homomorphism canonically induced from
$\varphi_{\lambda}$. Similarly to the case of $G_{a}$, we have

$G_{\lambda}=\left(\sum_{i+j=D}c_{i,j}\lambda^{i}\right)(Y^{\prime})^{D}+H(X^{\prime},Y^{\prime})=\left(\sum_{i=0}^{D}c_{i,D-i}\lambda^{i}\right)(Y^{\prime})^{D}+H(X^{\prime},Y^{\prime}),$

where the degree in $Y^{\prime}$ of each term of $H(X^{\prime},Y^{\prime})$ is
$\leq D-1$, and where the total degree of $H$ is less than or equal to $D$.
Since $\\#k>D$, there exists $\lambda\in k^{\times}$ such that
$\sum_{i=0}^{D}c_{i,D-i}\lambda^{i}\neq 0$, and thus $G_{\lambda}$ is monic in
$Y^{\prime}$ for such a $\lambda$.

We also show that for a suitable $\lambda$, we have $(\partial
G_{\lambda}/\partial Y^{\prime})(x^{\prime},y^{\prime})\neq 0$, that is,
$\varphi_{\lambda}(\partial G_{\lambda}/\partial Y^{\prime})\neq 0$. It
follows from

$(\partial G_{\lambda}/\partial
Y^{\prime})(X^{\prime},Y^{\prime})=\lambda({\partial F_{0}}/{\partial
X})(X^{\prime}+\lambda Y^{\prime},Y^{\prime})+({\partial F_{0}}/{\partial
Y})(X^{\prime}+\lambda Y^{\prime},Y^{\prime})$

that $\varphi_{\lambda}(\partial G_{\lambda}/\partial
Y^{\prime})=\lambda({\partial F_{0}}/{\partial X})(x,y)+({\partial
F_{0}}/{\partial Y})(x,y)$. Thus, the number of $\lambda$ with
$\varphi_{\lambda}(\partial G_{\lambda}/\partial Y^{\prime})=0$ is at most
one. This together with $\\#k>D+1$ implies that there exists $\lambda\in
k^{\times}$ such that $\sum_{i=0}^{D}c_{i,D-i}\lambda^{i}\neq 0$ and
$(\partial G_{\lambda}/\partial Y^{\prime})(x^{\prime},y^{\prime})\neq 0$, as
desired.

###  4.3. Correctness of the main algorithm

In this subsection, we prove the correctness of Algorithm 4.1 by examining
that of each step.

Step (A-1): Computation of the normalization $\overline{R}$:

This step just executes Algorithm 3.6 for the input $(S,I)$ with $S=k[X,Y]$
and $I=\langle F\rangle$. It follows from the following lemma that the set
$\mathcal{B}$ given in (4.2) generates $\overline{A}=S_{n}/I_{n}$ with $A=S/I$
over $k[x]$:

######  Lemma 4.4.

With notation same as in Step (A-1) of Algorithm 4.1, each $S_{i}/I_{i}$ is
generated by $\mathcal{B}_{i}:=\\{1,t_{i,1},\ldots,t_{i,s_{i}}\\}$ as an
$S_{i-1}/I_{i-1}$-module, and it is a free $k[x]$-module of rank $N$.

###### Proof.

The former claim follows immediately from the construction of $S_{i}/I_{i}$:
Recall from (3.5) that each $I_{i}$ contains elements of the form
$T_{i,j}T_{i,k}-\ell_{i,j,k}$ for all $1\leq j\leq k\leq s_{i}$, where
$\ell_{i,j,k}\in S_{i-1}[T_{i,1},\ldots,T_{i,s_{i}}]$ is linear in $T_{i,j}$
with $1\leq j\leq s_{i}$. This means that any element in $S_{i}$ modulo
$I_{i}$ has no term whose degree in $T_{i,j}$ with $1\leq j\leq s_{i}$ is
greater than $1$, as desired.

As for the latter claim, we have $S/I\subset S_{i}/I_{i}\subset
S_{n}/I_{n}=\overline{k[x,y]}$, and recall from the beginning of this section
that $\overline{k[x,y]}$ is a free $k[x]$-module of rank $N$. Since both
$S/I=k[x,y]$ and $\overline{k[x,y]}$ are of rank $N$ over the PID $k[x]$, the
intermediate module $S_{i}/I_{i}$ also has rank $N$. $\Box$

Step (A-2): Finding a basis of $\overline{R}$ as a $k[x]$-module:

This step first computes a generator set
$\mathcal{U}=\\{U_{1},\ldots,U_{m}\\}\subset\bigoplus_{i=1}^{r}k[X]$ of (4.3)
as a $k[X]$-module, by Algorithm 4.8 in Subsection 4.4 below. After computing
$\mathcal{U}$, we then construct the $(m\times r)$-matrix $U$ over $k[X]$
whose $i$-th row vector is $U_{i}$ for $1\leq i\leq m$, and compute its Smith
normal form $\tilde{U}$ together with matrices $P\in\mathrm{GL}_{m}(k[X])$ and
$Q\in\mathrm{GL}_{r}(k[X])$ such that $\tilde{U}=PUQ$. We here show the
following:

######  Lemma 4.5.

With notation same as in Step (A-2) of Algorithm 4.1, the set $\mathcal{W}$
computed in (A-2-3) is a basis of $S_{n}/I_{n}$ as a free $k[x]$-module.

###### Proof.

Let $e_{i}$ denote a row vector with $1$ in the $i$-th coordinate and $0$’s
elsewhere. Since the set (4.3) is nothing but the kernel of the
$k[X]$-homomorphism $\varphi_{0}:\bigoplus_{i=1}^{r}k[X]\rightarrow
S_{n}/I_{n}\ ;\ e_{i}\mapsto B_{i}$, we have the following commutative
diagram:

$\textstyle{\bigoplus_{i=1}^{m}k[X]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{U}}$$\textstyle{\bigoplus_{i=1}^{r}k[X]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\varphi_{0}}$$\scriptstyle{f_{Q}}$$\textstyle{S_{n}/I_{n}}$$\textstyle{\bigoplus_{i=1}^{m}k[X]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{\tilde{U}}}$$\scriptstyle{f_{P}}$$\textstyle{\bigoplus_{i=1}^{r}k[X],}$

by considering $k[X]$-homomorphisms defined by right-multiplication by
matrices, e.g., $f_{U}$ denotes the $k[X]$-homomorphism defined by $v\mapsto
vU$ for $v\in\bigoplus_{i=1}^{m}k[X]$.

Here, $\varphi_{0}$ canonically induces a $k[X]$-isomorphism from
$\mathrm{Coker}(f_{U})$ to $S_{n}/I_{n}$, and $\mathrm{Coker}(f_{\tilde{U}})$
is $k[X]$-isomorphic to $\mathrm{Coker}(f_{U})$ via $f_{Q}^{-1}=f_{Q^{-1}}$.
It follows from the form of $\tilde{U}$ given in (4.4) that
$\\{e_{r-N+i}:1\leq i\leq N\\}$ in $\bigoplus_{i=1}^{r}k[X]$ gives rise to a
basis of $\mathrm{Coker}(f_{\tilde{U}})$. Since each $W_{i}$ is the image of
$e_{r-N+i}$ by $f_{Q^{-1}}$, we have that $\\{W_{1},\ldots,W_{N}\\}$ is a
basis of $\mathrm{Coker}(f_{U})$, and thus $\\{w_{i}=\varphi_{0}(W_{i}):1\leq
i\leq N\\}$ is also a basis of $S_{n}/I_{n}$. $\Box$

Step (A-3): Computation of a basis of the conductor $\mathfrak{C}_{A}$:

Since $\\{w_{1},\ldots,w_{N}\\}$ computed in Step (A-2) is a basis of
$\overline{k[C]}=S_{n}/I_{n}$ by Lemma 4.5, it follows from Lemmas 2.6 and 2.7
that $\\{w_{1}^{\ast},\ldots,w_{N}^{\ast}\\}$ is a basis of the complementary
module $\mathcal{C}_{\overline{k[C]}/k[x]}$ as a $k[x]$-module. Thus, by
(4.1), the conductor $\mathfrak{C}_{A}$ as a $k[x]$-module is spanned by
$\mathcal{F}=F_{y}\cdot\\{w_{1}^{\ast},\ldots,w_{N}^{\ast}\\}$.

Step B:

Recall from the previous paragraph that $\mathcal{F}$ computed in Step (A-3)
is a basis of $\mathfrak{C}_{A}\subset k[x,y]$ as a $k[x]$-module, and thus it
is also a generator set of $\mathfrak{C}_{A}$ as an ideal of $k[x,y]$. Hence,
for proving the output to be a $k$-basis of $\mathfrak{C}_{A}^{(N-3)}$, it
suffices to show the following lemma:

######  Lemma 4.6.

Let $I\subset S:=k[x_{1},\ldots,x_{n}]$ be an ideal, and $\mathcal{G}$ its
Gröbner basis with respect to a graded monomial order $\succ$ on
$x_{1},\ldots,x_{n}$, i.e., a monomial order first comparing the total degrees
of two monomials. Then, for each integer $d\geq 0$, the (finite-dimensional)
$k$-vector subspace $I_{\leq d}:=\\{f\in I:\mathrm{deg}(f)\leq d\\}$ of $I$ is
generated by

(4.6) $\\{mg:m\in\mathrm{Mon}(S),\ g\in\mathcal{G},\ \mathrm{deg}(mg)\leq
d\\}$

as a $k$-vector space, where $\mathrm{Mon}(S)$ denotes the set of all
monomials in $S$.

Hence, a basis of $I_{\leq d}$ as a $k$-vector space can be also obtained by
computing the row echelon form of the Macaulay matrix corresponding to (4.6).

###### Proof.

Let $f\in I_{\leq d}$ be an arbitrary element, and put
$\mathcal{G}=\\{g_{1},\ldots,g_{s}\\}$. By the division algorithm together
with the definition of a Gröbner basis, there exist $h_{1},\ldots,h_{s}\in S$
such that $f=\sum_{i=1}^{s}h_{i}g_{i}$, where the multi-degree of each
$h_{i}g_{i}$ with respect to $\succ$ does not exceed that of $f$. In
particular, $\mathrm{deg}(h_{i}g_{i})\leq\mathrm{deg}(f)$ for all $1\leq i\leq
s$. Writing $h_{i}=\sum_{j=1}^{t_{i}}c_{i,j}m_{i,j}$ for $c_{i,j}\in k$ and
$m_{i,j}\in\mathrm{Mon}(S)$ with $m_{i,1}\succ\cdots\succ m_{i,t_{i}}$, we
have
$\mathrm{LM}(m_{i,j}g_{i})\preceq\mathrm{LM}(m_{i,1}g_{i})=\mathrm{LM}(h_{i}g_{i})$,
and thus $\mathrm{deg}(m_{i,j}g_{i})\leq\mathrm{deg}(f)$ for all $1\leq j\leq
t_{i}$. Hence, $f=\sum_{i=1}^{s}\sum_{j=1}^{t_{i}}c_{i,j}(m_{i,j}g_{i})$ with
$\mathrm{deg}(m_{i,j}g_{i})\leq\mathrm{deg}(f)\leq d$, as desired. $\Box$

By Lemma 4.6, one can easily verify that the set of elements in the output
modulo $F$ gives rise to a $k$-basis of $\mathfrak{C}_{A}^{(N-3)}$, where
modulo $F$ means to take the reminder of an element by the division of $F$ as
a polynomial in $Y$.

###  4.4. Details of Step (A-2): Elimination on syzygy modules with modulo

In Step (A-2), we compute a generator set of (4.3). Here we consider a more
general case problem, and give a method to solve it: Let
$S=k[x_{1},\ldots,x_{n}]$, $I\subset S$ an ideal, and
$S^{(\ell)}=k[x_{1},\ldots,x_{\ell}]$, where $1\leq\ell\leq n-1$. Given finite
elements $B_{1},\ldots,B_{r},\ldots,B_{r+1},\ldots,B_{r+s}\in S$ with
$I=\langle B_{r+1},\ldots,B_{r+s}\rangle_{S}$, we would like to compute a
generator set of the $S^{(\ell)}$-module

(4.7)
$\left\\{(a_{1},\ldots,a_{r})\in\bigoplus_{i=1}^{r}S^{(\ell)}e_{i}:\sum_{i=1}^{r}a_{i}B_{i}=0\bmod{I}\right\\},$

which is
$\mathrm{syz}_{I}(B_{1},\ldots,B_{r})\cap\bigoplus_{i=1}^{r}S^{(\ell)}e_{i}$,
namely, the kernel of the composition of the following maps:

(4.8)
$\begin{CD}\bigoplus_{i=1}^{r}S^{(\ell)}e_{i}@>{\iota}>{}>\bigoplus_{i=1}^{r}Se_{i}@>{\varphi}>{}>S@>{{\bmod{I}}}>{}>S/I,\end{CD}$

where $\iota$ is the canonical inclusion and $\varphi$ is defined by
$e_{i}\mapsto B_{i}$ for $1\leq i\leq r$. Here we denote by $e_{i}$ the vector
with $1$ in its $i$-th coordinate and $0$’s elsewhere. We also denote by
$\overline{\varphi}:\bigoplus_{i=1}^{r}Se_{i}\rightarrow S/I$ the composition
map of $\varphi$ and $\bmod{\;I}$. Note that
$\mathrm{syz}_{I}(B_{1},\ldots,B_{r})=\mathrm{Ker}(\overline{\varphi})$, and
that (4.7) is equal to
$\mathrm{Ker}(\overline{\varphi})\cap\bigoplus_{i=1}^{r}S^{(\ell)}e_{i}$.

Recall from Lemma 3.3 that, for given $\\{B_{1},\ldots,B_{r}\\}$ and
$\\{B_{r+1},\ldots,B_{r+s}\\}$, a Gröbner basis $\mathcal{G}$ of
$\mathrm{Ker}(\overline{\varphi})$ with respect to an arbitrary monomial order
is computed. With this $\mathcal{G}$, a generator set of
$\mathrm{Ker}(\overline{\varphi})\cap\bigoplus_{i=1}^{r}S^{(\ell)}e_{i}$ can
be also computed by the following lemma:

######  Lemma 4.7 (Elimination theorem for free modules).

Let $S=k[x_{1},\ldots,x_{n}]$ be the polynomial ring of $n$ variables over a
field $k$. Let $M=\bigoplus_{i=1}^{r}Se_{i}$ be the free module over $S$ of
rank $r$ with standard basis $\\{e_{1},\ldots,e_{r}\\}$. Let $N\subset M$ be a
submodule of $M$ and $\mathcal{G}$ a Gröbner basis of $N$ with respect to
$\succ_{\rm TOP}$ derived from an elimination monomial order $\succ$ on $S$
with $x_{1},\ldots,x_{\ell}\prec x_{\ell+1},\ldots,x_{n}$. Put
$S^{(\ell)}:=k[x_{1},\ldots,x_{\ell}]$ and
$M_{\ell}:=\bigoplus_{i=1}^{r}S^{(\ell)}e_{i}$. Then,
$\mathcal{G}_{\ell}:=\mathcal{G}\cap M_{\ell}$ is a Gröbner basis of
$N_{\ell}:=N\cap M_{\ell}$ with respect to the restriction of $\succ_{\rm
TOP}$ to $M_{\ell}$.

###### Proof.

It suffices to show
$\langle\mathrm{LT}(\mathcal{G}_{\ell})\rangle_{S^{(\ell)}}\supset\langle\mathrm{LT}(N_{\ell})\rangle_{S^{(\ell)}}$.
For this, take $v\in N_{\ell}\subset N$, then there exists $w\in\mathcal{G}$
such that $\mathrm{LT}(w)$ divides $\mathrm{LT}(v)$ since $\mathcal{G}$ is a
Gröbner basis of $N$. Namely, we can write $\mathrm{LT}(v)=m_{1}e_{i}$ and
$\mathrm{LT}(w)=m_{2}e_{i}$ for some monomials $m_{1},m_{2}\in\mathrm{Mon}(S)$
with $m_{2}\mid m_{1}$ and some $1\leq i\leq r$. By $v\in N_{\ell}\subset
M_{\ell}$ we have $m_{1}\in S^{(\ell)}$, so that $m_{2}\in S^{(\ell)}$. It
follows from the definition of the TOP order that any term $m_{3}e_{j}$ in $w$
satisfies $m_{3}\in S^{(\ell)}$, and hence $w\in M_{\ell}$. Thus
$w\in\mathcal{G}_{\ell}$, as desired. $\Box$

Based on Lemma 4.7, we here write down explicit procedures to compute a
generator set of (4.7):

######  Algorithm 4.8 (Computation of a modulo-syzygy with elimination of
variables).

  1. Input:

Finite elements $B_{1},\ldots,B_{r},\ldots,B_{r+1},\ldots,B_{r+s}\in S$ and
$1\leq\ell\leq n-1$.

  2. Output:

A generator set of
$\mathrm{syz}_{I}(B_{1},\ldots,B_{r})\cap\bigoplus_{i=1}^{r}S^{(\ell)}$, where
$I$ is an ideal in $S$ generated by $B_{r+1},\ldots,B_{r+s}$.

  1. (1)

Compute a Gröbner basis $\tilde{\mathcal{G}}$ of the syzygy module

$\mathrm{syz}(B_{1},\ldots,B_{r},B_{r+1},\ldots,B_{r+s})=\left\\{(a_{1},\ldots,a_{r+s})\in\bigoplus_{i=1}^{r+s}Se_{i}:\sum_{i=1}^{r+s}a_{i}B_{i}=0\right\\},$

that is, the kernel of the $S$-homomorphism
$\psi:\bigoplus_{i=1}^{r+s}S\longrightarrow Se_{i}\ ;\ e_{i}\mapsto B_{i}$.
For this, we apply Lemma 3.1, where we use a product order
$\succ_{1,2}=(\succ_{1},\succ_{2})$ on $S^{r+s}=\bigoplus_{i=1}^{r+s}Se_{i}$
with $e_{i}\succ_{1,2}e_{j}$ ($1\leq i\leq r$, $r+1\leq j\leq r+s$):
$\succ_{1}$ is a TOP order on $S^{r}=\bigoplus_{i=1}^{r}Se_{i}$ derived from
an elimination monomial order $\succ$ on $S$ with $x_{1},\ldots,x_{\ell}\prec
x_{\ell+1},\ldots,x_{n}$, and $\succ_{2}$ is an arbitrary monomial order on
$S^{s}=\bigoplus_{i=r+1}^{r+s}Se_{i}$. Letting
$\mathrm{pr}:\bigoplus_{i=1}^{r+s}Se_{i}\longrightarrow\bigoplus_{i=1}^{r}Se_{i}$
be the projection which maps $e_{i}$ to $e_{i}$ (resp. $0$) for $1\leq i\leq
r$ (resp. for $r+1\leq i\leq r+s$), then it follows from Lemma 3.3 that
$\mathcal{G}:=\mathrm{pr}(\tilde{\mathcal{G}})$ is a Gröbner basis of
$\mathrm{syz}_{I}(B_{1},\ldots,B_{r})$ with respect to $\succ_{1}$.

  2. (2)

Put $\mathcal{G}_{\ell}:=\mathcal{G}\cap M_{\ell}$, and output it. Note that
by Lemma 4.7, $\mathcal{G}_{\ell}$ is a Gröbner basis of (4.7) with respect to
the restriction of $\succ_{1}$ to $\bigoplus_{i=1}^{r}S^{(\ell)}e_{i}$.

###  4.5. Details of Step (A-3): Computation of the trace

In this step, we represent $\mathrm{Tr}_{k(x,y)/k(x)}(w_{i}w_{j})$ as an
element of $k[x]$ for each $1\leq i,j\leq N$. We here describe a method to
find such a representation for an arbitrary element $\overline{a}\in
S_{n}/I_{n}=\overline{k[C]}\subset k(x,y)$, where $\overline{a}$ denotes the
residue class of an element $a$ of $S_{n}$ modulo $I_{n}$. We denote by
$\sigma_{a}$ a linear map defined by multiplication by $a$. Recall that the
trace of $a$ is computed as the trace of the representation matrix of
$\sigma_{a}$ with respect to the basis $\\{w_{1},\ldots,w_{N}\\}$ for
$k(x,y)$. Thus, it suffices to find $a_{i,j}\in k[x]$ with $1\leq i,j\leq N$
such that $aw_{i}\equiv a_{i,1}w_{1}+\cdots+a_{i,N}w_{N}\bmod{I_{n}}$.

We first recall that $\mathcal{B}=\\{B_{1},\ldots,B_{r}\\}$ defined in (4.2)
generates $S_{n}/I_{n}$ over $k[x]$, and thus there exist $b_{i,j}\in k[x]$
with $1\leq i,j\leq r$ such that
$aw_{i}\equiv\sum_{j=1}^{r}b_{i,j}B_{j}\bmod{I_{n}}$. We can find such
$b_{i,j}$ as follows: Let
$\mathrm{syz}_{I_{n}}(aw_{i},B_{1},\ldots,B_{r})\subset\bigoplus_{i=0}^{r}S_{n}e_{i}$
be a modulo-syzygy defined as in (3.3) whose Gröbner basis
$\mathcal{G}^{\prime}$ with respect to an arbitrary monomial order on
$S_{n}^{r+1}$ can be computed by Lemma 3.3. Compute a Gröbner basis
$\mathcal{G}$ of
$\mathrm{syz}_{I_{n}}(aw_{i},B_{1},\ldots,B_{r})\subset\bigoplus_{i=0}^{r}S_{n}e_{i}$
with respect to a TOP extension of a monomial order on $S$ with the
elimination property with respect to $x$. It follows from Lemma 4.7 that
$\mathcal{G}\cap\bigoplus_{i=0}^{r}k[X]$ is a Gröbner basis of
$\mathrm{syz}_{I_{n}}(aw_{i},B_{1},\ldots,B_{r})\cap\bigoplus_{i=0}^{r}k[X]$
with respect to the TOP order induced from that for
$\bigoplus_{i=0}^{r}S_{n}e_{i}$. Writing
$\mathcal{G}\cap\bigoplus_{i=0}^{r}k[X]=\\{\mathbf{g}_{1},\ldots,\mathbf{g}_{s}\\}$,
it follows from
$(1,b_{i,1},\ldots,b_{i,r})\in\mathrm{syz}_{I_{n}}(aw_{i},B_{1},\ldots,B_{r})\cap\bigoplus_{i=0}^{r}k[X]$
that the first coordinates of $\mathbf{g}_{1},\ldots,\mathbf{g}_{s}$ are
coprime, and hence there exists $h_{1},\ldots,h_{s}\in k[X]$ such that the sum
of the first coordinates of $h_{1}\mathbf{g}_{1},\ldots,h_{s}\mathbf{g}_{s}$
is $1$. Since $k[X]$ is an Euclid domain, we can compute such
$h_{1},\ldots,h_{s}$ by the extended Euclidean algorithm, and
$b_{i,1},\ldots,b_{i,r}$ are obtained by putting
$(1,b_{i,1},\ldots,b_{i,r})=h_{1}\mathbf{g}_{1}+\cdots+h_{s}\mathbf{g}_{s}$.

In the following, we next consider to represent each $B_{j}$ as a
$k[x]$-linear combination of $w_{1},\ldots,w_{N}$, by using objects computed
in Step (A-2). Recall that each $w_{i}$ is computed in Step (A-2) as
$w_{i}=\varphi_{0}(W_{i})$, where

$\begin{pmatrix}W_{1}\\\ \vdots\\\
W_{N}\end{pmatrix}=\begin{pmatrix}e_{r-N+1}\\\ \vdots\\\
e_{r}\end{pmatrix}\cdot Q^{-1}.$

Since $\\{W_{i}^{\prime}:=e_{i}Q^{-1}:1\leq i\leq r-N\\}$ is a basis of
$\mathrm{Im}(f_{U})=\mathrm{Ker}(\varphi_{0})$, we obtain an extended basis

$\begin{pmatrix}W_{1}^{\prime}\\\ \vdots\\\ W_{r-N}^{\prime}\\\ W_{1}\\\
\vdots\\\ W_{N}\end{pmatrix}:=\begin{pmatrix}e_{1}\\\ \vdots\\\
e_{r}\end{pmatrix}\cdot Q^{-1}=Q^{-1}\begin{pmatrix}e_{1}\\\ \vdots\\\
e_{r}\end{pmatrix}$

of $\bigoplus_{i=1}^{r}k[X]$. It follows from $w_{i}=\varphi_{0}(W_{i})$ for
$1\leq i\leq N$ and $B_{i}=\varphi_{0}(e_{i})$ for $1\leq i\leq r$ that

$\begin{pmatrix}0\\\ \vdots\\\ 0\\\ w_{1}\\\ \vdots\\\
w_{N}\end{pmatrix}:=Q^{-1}\begin{pmatrix}B_{1}\\\ \vdots\\\
B_{r}\end{pmatrix},$

where we also used $W_{i}^{\prime}\in\mathrm{Ker}(\varphi_{0})$ for $1\leq
i\leq r-N$. Hence, denoting by $Q^{\prime}$ the matrix obtained from $Q$ by
removing its first $r-N$ columns, we have

$\begin{pmatrix}\sigma_{a}(w_{1})\\\ \vdots\\\
\sigma_{a}(w_{N})\end{pmatrix}=\begin{pmatrix}aw_{1}\\\ \vdots\\\
aw_{N}\end{pmatrix}=\begin{pmatrix}b_{1,1}&\cdots&b_{1,r}\\\ \cdots&&\cdots\\\
b_{N,1}&\cdots&b_{N,r}\end{pmatrix}\begin{pmatrix}B_{1}\\\ \vdots\\\
B_{r}\end{pmatrix}=\begin{pmatrix}b_{1,1}&\cdots&b_{1,r}\\\ \cdots&&\cdots\\\
b_{N,1}&\cdots&b_{N,r}\end{pmatrix}Q^{\prime}\begin{pmatrix}w_{1}\\\ \vdots\\\
w_{N}\end{pmatrix}.$

Therefore, the representation matrix of $\sigma_{a}$ with respect to
$\\{w_{1},\ldots,w_{N}\\}$ is ${}^{t}(BQ^{\prime})$, and hence
$\mathrm{Tr}_{k(x,y)/k(x)}(\overline{a})$ is computed as the trace of the
matrix ${}^{t}(BQ^{\prime})$, by the following procedures:

  1. (1)

For each $1\leq i\leq N$, find $b_{i,j}\in k[x]$ with $1\leq j\leq r$ such
that $aw_{i}\equiv\sum_{j=1}^{r}b_{i,j}B_{j}\bmod{I_{n}}$, by the Gröbner
basis computation for free modules. Let $B$ denote the $N\times r$ matrix over
$k[x]$ whose $(i,j)$ entry is $b_{i,j}$.

  2. (2)

Denoting by $Q^{\prime}$ the matrix obtained from $Q$ by removing its first
$r-N$ columns, compute ${}^{t}(BQ^{\prime})$, and then its trace is nothing
but $\mathrm{Tr}_{k(x,y)/k(x)}(\overline{a})$.

##  5\. Implementation and example

We implemented Algorithm 4.1 over Magma [2] V2.25-3 in its 64bit version, and
the source code will be available at

https://sites.google.com/view/m-kudo-official-website/english/code/rdf.

In the code, the function SpaceOfDifferentialForms implements Algorithm 4.1.
Note that, as for the Gröbner basis computation of our implementation, we
utilize Magma’s built-in functions (e.g., Groebner) where Faugere’s $F_{4}$
algorithm and its variants are efficiently implemented. Here, we demonstrate
Algorithm 4.1 in the following example (Example 5.1), where we used our
implementation.

######  Example 5.1.

Consider a plane singular curve $C=V(F)$ in $\mathbb{A}^{2}(\overline{k})$
over $k:=\mathbb{F}_{11}$ defined by $F:=X^{5}+Y^{5}+XY$, whose normalization
is known to be a superspecial curve of genus $5$ over $\mathbb{F}_{11}$, see
[13] for details. Here let us compute an explicit basis of
$H^{0}(\tilde{C},\varOmega_{\tilde{C}})$ as a $k$-vector space. Note that
$C^{\prime}$ is singular only at $(0:0:1)$, and thus it suffices from Theorem
2.1 to compute a basis of $\mathfrak{C}_{A}^{(N-3)}$ by Algorithm 3.6.

Step (A-1): Computation of the normalization:

We execute Algorithm 3.6 for the input $(S,I)$ with $S=k[X,Y]$ and $I=\langle
F\rangle_{S}$. Recall from Example 3.8 that Algorithm 3.6 terminates at the
first step of its second loop, and an integral closure of $A=S/I$ is obtained
as $A_{1}=S_{1}/I_{1}$, where $S_{1}=k[X,Y,T]$ and $I_{1}=\langle
P_{1}\rangle_{S_{1}}$ with

$\displaystyle P_{1}$ $\displaystyle:=$
$\displaystyle\\{T^{2}+T+Y^{3}X^{3},YT+10X^{4},XT+Y^{4}+X,F\\},$

together with $Q_{1}=\\{T^{2}+T+Y^{3}X^{3}\\}$ and $\phi_{1}:A_{1}\rightarrow
k(x,y)$ mapping $(x,y,t)$ to $(x,y,x^{4}/y)$.

In this case, the generator set (4.2) of $S_{1}/I_{1}$ as a $k[x]$-module is

$\mathcal{B}=\\{ty^{4},ty^{3},ty^{2},ty,t,y^{4},y^{3},y^{2},y,1\\},$

where we set $t:=T\bmod{I_{1}}$. Put
$(B_{1},\ldots,B_{10})=(ty^{4},ty^{3},ty^{2},ty,t,y^{4},y^{3},y^{2},y,1)$.

Step (A-2): Finding a basis of $\overline{R}$ as a $k[x]$-module:

  1. (A-2-1)

Compute a generator set of (4.3) as a $k[X]$-module, say

$\mathrm{syz}_{I_{1}}(\mathcal{B})\cap\bigoplus_{i=1}^{10}k[X]=\left\\{(a_{1},\ldots,a_{10})\in\bigoplus_{i=1}^{10}k[X]:\sum_{i=1}^{10}a_{i}B_{i}=0\bmod{I_{1}}\right\\}.$

For this, we execute Algorithm 4.8 for the inputs $B_{1},\ldots,B_{10}$ as
above, $B_{11}=T^{2}+T+Y^{3}X^{3}$, $B_{12}=YT+10X^{4}$, $B_{13}=XT+Y^{4}+X$,
$B_{14}=F$, and $\ell:=1$. As a result,

$U=\begin{pmatrix}U_{1}\\\ U_{2}\\\ U_{3}\\\ U_{4}\\\ U_{5}\\\
\end{pmatrix}=\begin{pmatrix}10&0&0&0&0&0&X^{4}&0&0&0\\\
0&10&0&0&0&0&0&X^{4}&0&0\\\ 0&0&10&0&0&0&0&0&X^{4}&0\\\
0&0&0&10&0&0&0&0&0&X^{4}\\\ 0&0&0&0&X&1&0&0&0&X\\\ \end{pmatrix}$

is a matrix whose row vectors generate
$\mathrm{syz}_{I_{1}}(\mathcal{B})\cap\bigoplus_{i=1}^{10}k[X]$.

  2. (A-2-2)

The computed Smith normal form $\tilde{U}$ of $U$ and transformation matrices
$P\in\mathrm{GL}_{5}(k[X])$ and $Q\in\mathrm{GL}_{10}(k[X])$ are as follows:
$P=E_{5}$, $\tilde{U}=\begin{pmatrix}E_{5}&O_{5}\end{pmatrix}$, where $E_{5}$
(resp. $O_{5}$) is the identity (resp. zero) matrix of size $5$, and

$Q=\begin{pmatrix}10&0&0&0&0&X^{4}&0&0&0&0\\\ 0&10&0&0&0&0&X^{4}&0&0&0\\\
0&0&10&0&0&0&0&X^{4}&0&0\\\ 0&0&0&10&0&0&0&0&X^{4}&0\\\
0&0&0&0&0&0&0&0&0&10\\\ 0&0&0&0&1&0&0&0&10X&X\\\ 0&0&0&0&0&1&0&0&0&0\\\
0&0&0&0&0&0&1&0&0&0\\\ 0&0&0&0&0&0&0&1&0&0\\\ 0&0&0&0&0&0&0&0&1&0\\\
\end{pmatrix}.$

  3. (A-2-3)

We have

$Q^{-1}=\begin{pmatrix}10&0&0&0&0&0&X^{4}&0&0&0\\\ 0&10&0&0&0&0&0&X^{4}&0&0\\\
0&0&10&0&0&0&0&0&X^{4}&0\\\ 0&0&0&10&0&0&0&0&0&X^{4}\\\ 0&0&0&0&X&1&0&0&0&X\\\
0&0&0&0&0&0&1&0&0&0\\\ 0&0&0&0&0&0&0&1&0&0\\\ 0&0&0&0&0&0&0&0&1&0\\\
0&0&0&0&0&0&0&0&0&1\\\ 0&0&0&0&10&0&0&0&0&0\\\ \end{pmatrix}.$

Thus, putting $(w_{1},w_{2},w_{3},w_{4},w_{5})=(y^{3},y^{2},y,1,10t)$, one
also has that $\mathcal{W}=\\{w_{1},\ldots,w_{5}\\}$ is a basis of
$S_{1}/I_{1}$ as a $k[x]$-module.

Step (A-3): Computation of a basis of the conductor $\mathfrak{C}_{A}$:

  1. (A-3-1)

For each $1\leq i,j\leq 5$, compute $\mathrm{Tr}_{k(x,y)/k(x)}(w_{i}w_{j})\in
k[x]$. We here show details on the computation only for
$\mathrm{Tr}_{k(x,y)/k(x)}(w_{1}^{2})$. By computing a generator set of
$\mathrm{syz}_{I_{1}}(w_{1}^{2},B_{1},\ldots,B_{10})\cap\bigoplus_{i=0}^{10}k[X]$
with a method described in Subsection 4.5, we have a relation

$\begin{pmatrix}w_{1}^{2}w_{1}\\\ w_{1}^{2}w_{2}\\\ w_{1}^{2}w_{3}\\\
w_{1}^{2}w_{4}\\\
w_{1}^{2}w_{5}\end{pmatrix}\equiv\begin{pmatrix}0&0&0&x^{2}&0&10x^{5}&0&0&x^{2}&0\\\
10x&0&0&0&0&10x&0&0&0&0\\\ 0&10x&0&0&0&0&10x&0&0&0\\\
0&0&10x&0&0&0&0&10x&0&0\\\ 0&0&x&x^{5}&0&0&0&0&0&0\\\
\end{pmatrix}\cdot\begin{pmatrix}B_{1}\\\ \vdots\\\ B_{10}\end{pmatrix}$

modulo $I_{1}$. Let $Q^{\prime}$ be the matrix obtained from $Q$ by removing
its first $5$ columns. Since
${}^{t}(B_{1},\ldots,B_{10})=(Q^{\prime}){}^{t}(w_{1},\ldots,w_{5})$, we have

$\begin{pmatrix}w_{1}^{2}w_{1}\\\ w_{1}^{2}w_{2}\\\ w_{1}^{2}w_{3}\\\
w_{1}^{2}w_{4}\\\
w_{1}^{2}w_{5}\end{pmatrix}\equiv\begin{pmatrix}0&0&x^{2}&2x^{6}&10x^{6}\\\
10x^{5}&0&0&x^{2}&10x^{2}\\\ 10x&10x^{5}&0&0&0\\\ 0&10x&10x^{5}&0&0\\\
0&0&x^{5}&x^{9}&0\\\ \end{pmatrix}\begin{pmatrix}w_{1}\\\ w_{2}\\\ w_{3}\\\
w_{4}\\\ w_{5}\end{pmatrix},$

and hence $\mathrm{Tr}_{k(x,y)/k(x)}(w_{1}^{2})=0$.

As a result, the computed matrix of traces is

$\left(\mathrm{Tr}_{k(x,y)/k(x)}(w_{i}w_{j})\right)=\begin{pmatrix}0&6x^{5}&7x&0&0\\\
6x^{5}&7x&0&0&0\\\ 7x&0&0&0&6x^{4}\\\ 0&0&0&5&1\\\
0&0&6x^{4}&1&1\end{pmatrix}.$

  2. (A-3-2)

Computing
$(\mathrm{Tr}_{k(x,y)/k(x)}(w_{i}w_{j}))^{-1}\in\mathrm{GL}_{5}(k(x))$, we
have

$\displaystyle{}^{t}\\!\begin{pmatrix}w_{1}^{\ast}&w_{2}^{\ast}&w_{3}^{\ast}&w_{4}^{\ast}&w_{5}^{\ast}\end{pmatrix}=\left(\mathrm{Tr}_{k(x,y)/k(x)}(w_{i}w_{j})\right)^{-1}\cdot{}^{t}\\!\begin{pmatrix}y^{3}&y^{2}&y&1&10t\end{pmatrix}$
$\displaystyle=$
$\displaystyle\frac{1}{x^{15}+3}\begin{pmatrix}5x^{6}&2x^{10}&\frac{2}{x}&3x^{3}&7x^{3}\\\
2x^{10}&\frac{2}{x}&3x^{3}&10x^{7}&5x^{7}\\\
\frac{2}{x}&3x^{3}&10x^{7}&4x^{11}&2x^{11}\\\
3x^{3}&10x^{7}&4x^{11}&9x^{15}+9&2\\\
7x^{3}&5x^{7}&2x^{11}&2&1\end{pmatrix}\cdot\begin{pmatrix}y^{3}\\\ y^{2}\\\
y\\\ 1\\\ 10t\end{pmatrix},$

where $w_{i}^{\ast}\in k(x)[y,t]$ for each $1\leq i\leq 5$.

  3. (A-3-3)

Substituting $x^{4}/y$ into $t$ in $w_{i}^{\ast}$ for $1\leq i\leq 5$, we have

$F_{y}\cdot\begin{pmatrix}w_{1}^{\ast}&\cdots&w_{5}^{\ast}\end{pmatrix}\equiv\begin{pmatrix}y&y^{2}&y^{3}&y^{4}&x\end{pmatrix}\pmod{F}.$

Thus, $\\{y,y^{2},y^{3},y^{4},x\\}$ spans $\mathfrak{C}_{A}$ as a
$k[x]$-module, and hence it also generates $\mathfrak{C}_{A}$ as an ideal
(i.e., as a $k[x,y]$-module).

Step B: Finally, we compute a basis of
$\mathfrak{C}_{A}^{(2)}:=\\{\phi\in\mathfrak{C}_{A}:\mathrm{deg}(\phi)\leq
2\\}$ as a $k$-vector space, by the following three procedures:

  1. (B-1)

A computed Gröbner basis of the ideal $\langle
Y,Y^{2},Y^{3},Y^{4},X,F(X,Y)\rangle_{k[X,Y]}$ with respect to the graded
lexicographic order with $X\succ Y$ is $\mathcal{G}=\\{X,Y\\}$.

  2. (B-2)

Putting $\mathcal{S}:=\\{mg:m\in\mathrm{Mon}(k[X,Y]),\ g\in\mathcal{G},\
\mathrm{deg}(mg)\leq 2\\}$, we have $\mathcal{S}=\\{Y^{2},XY,X^{2},Y,X\\}$.
The Macaulay matrix $M$ corresponding to $\mathcal{S}$ with respect to the set
of monomials of degree $\leq 2$ is as follows:

$\displaystyle\begin{split}M=\hbox{}\vbox{\kern 0.86108pt\hbox{$\kern
0.0pt\kern 2.5pt\kern-5.0pt\left[\kern
0.0pt\kern-2.5pt\kern-5.55557pt\vbox{\kern-0.86108pt\vbox{\vbox{
\halign{\kern\arraycolsep\hfil\@arstrut$\kbcolstyle#$\hfil\kern\arraycolsep&
\kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep&&
\kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep\cr
5.0pt\hfil\@arstrut$\scriptstyle$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle
X^{2}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle XY$\hfil\kern
5.0pt&5.0pt\hfil$\scriptstyle Y^{2}$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle
X$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle Y$\hfil\kern
5.0pt&5.0pt\hfil$\scriptstyle 1\\\Y^{2}$\hfil\kern
5.0pt&5.0pt\hfil$\scriptstyle 0$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle
0$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle 1$\hfil\kern
5.0pt&5.0pt\hfil$\scriptstyle 0$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle
0$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle 0\\\XY$\hfil\kern
5.0pt&5.0pt\hfil$\scriptstyle 0$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle
1$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle 0$\hfil\kern
5.0pt&5.0pt\hfil$\scriptstyle 0$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle
0$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle 0\\\X^{2}$\hfil\kern
5.0pt&5.0pt\hfil$\scriptstyle 1$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle
0$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle 0$\hfil\kern
5.0pt&5.0pt\hfil$\scriptstyle 0$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle
0$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle 0\\\Y$\hfil\kern
5.0pt&5.0pt\hfil$\scriptstyle 0$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle
0$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle 0$\hfil\kern
5.0pt&5.0pt\hfil$\scriptstyle 0$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle
1$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle 0\\\X$\hfil\kern
5.0pt&5.0pt\hfil$\scriptstyle 0$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle
0$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle 0$\hfil\kern
5.0pt&5.0pt\hfil$\scriptstyle 1$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle
0$\hfil\kern 5.0pt&5.0pt\hfil$\scriptstyle 0\\\$\hfil\kern
5.0pt\crcr}}}}\right]$}}.\end{split}$

  3. (B-3)

The reduced row echelon form $\mathrm{ref}(M)$ of $M$ is
$\begin{pmatrix}E_{5}&{\bf 0}\end{pmatrix}$, where $E_{5}$ (resp. ${\bf 0}$)
is the identity matrix of size $5$ (resp. the zero column vector of dimension
$5$). Thus the set of its corresponding polynomials is
$\\{x^{2},xy,y^{2},x,y\\}$, which gives rise to a basis of
$\mathfrak{C}_{A}^{(2)}$ as a $k$-vector space.

Therefore, it follows from Theorem 2.1 that

$\left\\{\frac{x^{2}}{y^{4}+x}dx,\frac{xy}{y^{4}+x}dx,\frac{y^{2}}{y^{4}+x}dx,\frac{x}{y^{4}+x}dx,\frac{y}{y^{4}+x}dx\right\\}$

spans $H^{0}(\tilde{C},\varOmega_{\tilde{C}})$ as a $k$-vector space.

##  6\. Computing Cartier-Manin matrices

Let $C=V(F)$ be a plane curve with ${\rm deg}(F)=N$. Let $C^{\prime}$ be the
the Zariski closure of $C$ in ${\mathbb{P}}^{2}$. Let $\pi:\tilde{C}\to
C^{\prime}$ be the normalization. The aim of this section is to describe the
Cartier operator (Cartier-Manin matrix) on
$H^{0}({\tilde{C}},\varOmega_{\tilde{C}})$ and to give a remark on computation
of a basis of $H^{1}({\tilde{C}},{\mathcal{O}}_{\tilde{C}})$.

In the previous sections, we gave an algorithm to obtain a basis of
$H^{0}({\tilde{C}},\varOmega_{\tilde{C}})$. Let ${\mathcal{V}}$ be the Cartier
operator on $H^{0}({\tilde{C}},\varOmega_{\tilde{C}})$ (usually the symbol
$\mathcal{C}$ is used for the Cartier operator, but this symbol is used for
complementary modules in Section 2.2). We can calculate the Cartier-Manin
matrix of $\tilde{C}$ with respect to the basis. We recall the result [18,
Theorem 1.1] by Stöhr and Voloch:

(6.1)
${\mathcal{V}}\left(\phi\frac{dx}{F_{y}}\right)=\left(\frac{\partial^{2p-2}}{\partial
x^{p-1}\partial y^{p-1}}(F^{p-1}\phi)\right)^{1/p}\frac{dx}{F_{y}}.$

######  Example 6.1.

  1. (1)

Consider $C=V(F)$ with $F=x^{5}+y^{5}+xy$ over ${\mathbb{F}}_{11}$ as in
Example 5.1. As we have seen, as a basis of
$H^{0}({\tilde{C}},\varOmega_{\tilde{C}})$ we have
$\left\\{\left.\phi_{i}\frac{dx}{F_{y}}\ \right|\ i=1,\ldots,5\right\\}$ with
$(\phi_{1},\ldots,\phi_{5})=(x^{2},xy,y^{2},x,y)$. It is straightforward to
see that the Cartier matrix of $\tilde{C}$ is zero, i.e., $\tilde{C}$ is
superspecial.

  2. (2)

Consider $C=V(F)$ with $F=x^{5}+y^{5}+(x+y)^{3}+xy$ over $k={\mathbb{F}}_{2}$.
This curve is singular at $(x,y)=(0,0),(1,0),(0,1)$ and $C^{\prime}$ is
regular at the infinite points. By executing an implementation of our
algorithm in Section 4, $H^{0}({\tilde{C}},\varOmega_{\tilde{C}})$ has a basis
$\left\\{\left.\phi_{i}\frac{dx}{F_{y}}\ \right|\ i=1,2,3\right\\}$ with
$\phi_{1}=x^{2}+x$, $\phi_{2}=xy$ and $\phi_{3}=y^{2}+y$. A straightforward
computation, for example $(\partial^{2}F\phi_{1}/\partial x\partial
y)^{1/2}=(y^{4}+y^{2})^{1/2}=\phi_{3}$, shows that the Cartier-Manin matrix of
$\tilde{C}$ with respect to this basis is

$\begin{pmatrix}0&0&1\\\ 0&0&0\\\ 1&0&0\end{pmatrix}.$

  3. (3)

Consider $C=\operatorname{Spec}(A)$, where $A=\mathbb{F}_{2}[x,y]/\langle
F\rangle$ with $F=y^{7}-x^{2}(x-1)^{2}$. Note that $C^{\prime}$ is singular at
every $\mathbb{F}_{2}$-rational point, which means that any
$\mathbb{F}_{2}$-linear transformation can not make the infinity regular. By
executing an implementation of our algorithm, we see that
${\mathfrak{C}}_{A}^{(4)}$ has a basis

$\\{x^{4}+x,\ x^{3}y+xy,\ x^{2}y^{2}+xy^{2},\ xy^{3},\ y^{4},\ x^{3}+x,\
x^{2}y+xy,\ y^{3},\ x^{2}+x\\}.$

The $\mathbb{F}_{2}$-subspace of ${\mathfrak{C}}_{A}^{(4)}$ where the
associated differential form is regular at the infinite point is generated by
$y^{4}$, $y^{3}$, $x^{2}+x$ (by looking at the conductor at the infinite point
or at ${\mathfrak{C}}_{A^{\prime}}^{(4)}$, cf. Remark 2.3). Thus, a basis of
$H^{0}(\tilde{C},\varOmega_{\tilde{C}})$ can be taken as $1/y^{2}dx$,
$1/y^{3}dx$ and $(x^{2}+x)/y^{6}dx$. The Cartier-Manin matrix of $\tilde{C}$
with respect to this basis is

$\begin{pmatrix}0&0&0\\\ 0&0&1\\\ 0&0&0\end{pmatrix}$

by a straightforward computation, for example
${\mathcal{V}}\left((x^{2}+x)/y^{6}dx\right)=1/y^{3}dx$.

Finally, we mention about $H^{1}(\tilde{C},{\mathcal{O}}_{\tilde{C}})$. This
is just the dual notion of $H^{0}(\tilde{C},\varOmega_{\tilde{C}})$, but it is
important to have an explicit basis of
$H^{1}(\tilde{C},{\mathcal{O}}_{\tilde{C}})$. Let $U_{0},U_{1}$ and $U_{2}$ be
the affine open subshemes of $C^{\prime}$ corresponding to the parts of $X\neq
0$, $Y\neq 0$ and $Z\neq 0$ respectively. The short exact sequence

$\begin{CD}0@>{}>{}>{\mathcal{O}}_{C^{\prime}}@>{}>{}>\pi_{*}{\mathcal{O}}_{\tilde{C}}@>{}>{}>\pi_{*}{\mathcal{O}}_{\tilde{C}}/{\mathcal{O}}_{C^{\prime}}@>{}>{}>0\end{CD}$

induces

$\begin{CD}0@>{}>{}>H^{0}(\pi_{*}{\mathcal{O}}_{\tilde{C}}/{\mathcal{O}}_{C^{\prime}})@>{}>{}>H^{1}(C^{\prime},{\mathcal{O}}_{C^{\prime}}<EMAIL_ADDRESS>
From the short exact sequence

$\begin{CD}0@>{}>{}>{\mathcal{I}}@>{}>{}>{\mathcal{O}}_{{\mathbb{P}}^{2}}@>{}>{}>{\mathcal{O}}_{{\mathcal{C}}^{\prime}}@>{}>{}>0\end{CD}$

with the ideal sheaf ${\mathcal{I}}$ defining $C^{\prime}$, we have

$H^{2}({\mathbb{P}}^{2},\mathcal{O}(-N))\overset{F\times}{\simeq}H^{2}({\mathbb{P}}^{2},{\mathcal{I}})\simeq
H^{1}(C^{\prime},{\mathcal{O}}_{C^{\prime}})\simeq
H^{0}({C^{\prime}},{\mathcal{K}}/{\mathcal{O}}_{C^{\prime}})/K,$

where $\mathcal{K}$ is the constant sheaf of rational functions on
$C^{\prime}$. The class of $1/x^{\ell}y^{m}z^{n}$ with $\ell,m,n\geq 1$ and
$\ell+m+n=N$ in $H^{2}({\mathbb{P}}^{2},\mathcal{O}(-N))$ is sent to the class
of $F/x^{\ell}y^{m}z^{n}$, which can be written as $f_{ij}-f_{ik}+f_{jk}$ for
some $f_{ij}\in\Gamma(U_{i}\cap U_{j},{\mathcal{O}}_{C^{\prime}})$. Then
$\\{f_{ij}\\}$ gives a class of
$H^{1}(C^{\prime},{\mathcal{O}}_{C^{\prime}})$. For a fixed $r_{0}\in
H^{0}(U_{0},{\mathcal{K}}/{\mathcal{O}}_{C^{\prime}})$, we put
$r_{i}:=f_{i0}+r_{0}$; then this defines an element of
$H^{0}(C^{\prime},{\mathcal{K}}/{\mathcal{O}})$, which is called a
repartition. The pairing

$\langle\ ,\ \rangle:\quad H^{1}(C^{\prime},{\mathcal{O}}_{C^{\prime}})\otimes
H^{0}({\tilde{C}},\varOmega_{\tilde{C}})\simeq
H^{0}(C^{\prime},{\mathcal{K}}/{\mathcal{O}})/K\otimes
H^{0}({\tilde{C}},\varOmega_{\tilde{C}})\to k$

is defined by taking their residues, and it is embedded into the commutative
diagram

$\begin{CD}H^{1}(C^{\prime},\mathcal{O}_{C^{\prime}})\otimes
H^{0}(C^{\prime},\omega_{C^{\prime}}^{o})@>{}>{}>k\\\
@V{}V{}V@A{}A{}A\Big{\|}\\\ H^{1}({\tilde{C}},\mathcal{O}_{\tilde{C}})\otimes
H^{0}({\tilde{C}},\varOmega_{\tilde{C}})@>{}>{}>k,\end{CD}$

where $\omega_{C}^{o}$ is the dualizing sheaf on $C$. In particular, the image
of $H^{0}(\pi_{*}{\mathcal{O}}_{\tilde{C}}/{\mathcal{O}}_{C^{\prime}})\to
H^{1}(C^{\prime},{\mathcal{O}}_{C^{\prime}})$ is equal to

$H^{0}({\tilde{C}},\varOmega_{\tilde{C}})^{\perp}:=\left\\{\eta\in
H^{1}(C^{\prime},{\mathcal{O}}_{C^{\prime}})\ \left|\
\langle\eta,\omega\rangle=0\text{ for all }\omega\in
H^{0}({\tilde{C}},\varOmega_{\tilde{C}})\right.\right\\}.$

######  Lemma 6.2.

With identification $F\cdot:H^{2}({\mathbb{P}}^{2},\mathcal{O}(-N))\simeq
H^{1}(C^{\prime},{\mathcal{O}}_{C^{\prime}})$, we have

$H^{1}({\tilde{C}},{\mathcal{O}}_{\tilde{C}})\simeq F\cdot
H^{2}({\mathbb{P}}^{2},\mathcal{O}(-N))/H^{0}({\tilde{C}},\varOmega_{\tilde{C}})^{\perp}.$

This lemma gives us a feasible algorithm to give a basis of
$H^{1}({\tilde{C}},{\mathcal{O}}_{\tilde{C}})$. In a future work, we shall
give an algorithm determining the Ekedahl-Oort type of $\tilde{C}$ based on
these descriptions of $H^{0}({\tilde{C}},\varOmega^{1}_{\tilde{C}})$ and
$H^{1}({\tilde{C}},{\mathcal{O}}_{\tilde{C}})$ using bases on them discussed
in the present paper (cf. see the recent work [15] by Moonen, for a way to
compute the Ekedahl-Oort type of nonsingular complete-intersection curves).

##  7\. Concluding Remarks

In this paper, we proposed an algorithm to compute an explicit basis of the
space of regular differential forms on a general plane curve, whereas most of
previous works study particular cases such as hyperelliptic curves. More
precisely, we reduced the problem into computing a basis of the conductor of
the coordinate ring of the curve, based on Gorenstein’s work [8]. As for the
computation of the conductor, we also confirmed that Mňuk’s framework [16]
works well even in the positive characteristic case, and realized it as an
algorithm via the theory of Gröbner bases for free modules. By our algorithm
together with Stöhr-Voloch’s formula in [18] for computing the Cartier
operator, we can symbolically compute the Cartier-Manin matrix of an arbitrary
plane curve. Our algorithm was also demonstrated for concrete examples, which
will be helpful for the further use of our algorithm.

While our algorithm might derive fruitful applications such as computing
invariants (e.g., Ekedahl-Oort type), the computational complexity has not
been determined yet neither in theory nor in practice, due to the difficulty
of estimating that of computing Gröbner bases for free modules. Let us leave
the complexity analysis of our algorithm as future work, where it is important
to find some mathematical invariants for measuring the complexity.

### Acknowledgments

This work was supported by JSPS Grant-in-Aid for Young Scientists 20K14301,
and JSPS Grantin-Aid for Scientific Research (C) 21K03159. This work was also
supported by JST CREST Grant Number JPMJCR2113, Japan.

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