# Global Weierstrass equations of hyperelliptic curves

Qing Liu  Université de Bordeaux, Institut de Mathématiques de Bordeaux, CNRS
UMR 5251, 33405 Talence, France<EMAIL_ADDRESS>
###### Abstract.

Given a hyperelliptic curve $C$ of genus $g$ over a number field $K$ and a
Weierstrass model $\mathscr{C}$ of $C$ over the ring of integers
$\mathcal{O}_{K}$ (i.e. the hyperelliptic involution of $C$ extends to
$\mathscr{C}$ and the quotient is a smooth model of $\mathbb{P}^{1}_{K}$ over
$\mathcal{O}_{K}$), we give necessary and sometimes sufficient conditions for
$\mathscr{C}$ to be defined by a global Weierstrass equation. In particular,
if $C$ has everywhere good reduction, we prove that it is defined by a global
integral Weierstrass equation with invertible discriminant if the class number
$h_{K}$ is prime to $2(2g+1)$, confirming a conjecture of M. Sadek.

###### Key words and phrases:

Hyperelliptic curves, discriminant, Weierstrass equation, good reduction

###### 2020 Mathematics Subject Classification:

11G30, 11G05, 14D10, 14H25

I would like to thank heartily Mohammad Sadek for asking the question on the
existence of a global minimal equation and for stimulating discussions. I
would like to thank Yuri Bilu, Jean-François Jaulent and Jean Gillibert for
helpful discussions about Lemma 6.5. I am grateful to one of the anonymous
referees for checking thoroughly the first version, finding mistakes and
proposing several important improvements, especially in the even genera case.

Let $C$ be a hyperelliptic curve of genus $g\geq 2$ or an elliptic curve over
a number field $K$. It is natural to try to represent $C$ with an integral
Weierstrass equation (i.e. with coefficients in the ring of integers
$\mathcal{O}_{K}$ of $K$)

(1) $y^{2}+Q(x)y=P(x)$

in a canonical way. For instance we may require the discriminant of the
equation to be _minimal_ in the sense that, at any finite place $v$ of $K$, it
divides the discriminant of any Weierstrass equation of $C$ with coefficients
in the local ring $\mathcal{O}_{v}$ (Definition 2). If $C$ has everywhere good
reduction, this means that we are looking for an integral Weierstrass equation
with invertible discriminant. It is well known that such a minimal equation
exists if $\mathcal{O}_{K}$ is principal, but this is not true in general (see
e.g. [11], Corollary 6 or [12], Theorem 1.7, for elliptic curves).

The aim of this work is to study general conditions under which a Weierstrass
model, not necessarily minimal, is defined by an integral Weierstrass equation
(Definition 2). Let us describe briefly the main results. We work more
generally over a Dedekind domain $R$ with field of fractions $K$. For a
Weierstrass model $\mathscr{C}$ of $C$, the quotient $\mathscr{Z}$ of
$\mathscr{C}$ by the hyperelliptic involution is a projective bundle over
$S={\rm Spec}\kern 2.0ptR$. For $\mathscr{C}$ to be defined by an integral
Weierstrass equation, there are three necessary conditions, see Proposition
2.6:

  1. (a)

$\mathscr{Z}$ must be trivial (i.e. $\mathscr{Z}\simeq\mathbb{P}^{1}_{S}$);

  2. (b)

the discriminant ideal $\Delta_{\mathscr{C}}\subseteq R$ must be principal;

  3. (c)

if $\omega_{\mathscr{C}/S}$ denotes the dualizing sheaf of $\mathscr{C}$ over
$S$, then $\det H^{0}(\mathscr{C},\omega_{\mathscr{C}/S})\simeq R$.

The converse is true when $g$ is not divisible by $4$ (see below), but is
false in general (Proposition 6.1(d)). These three conditions are independent
of each other (see § 6) except when $g=1,2$ or $4$ in which case
$\Delta_{\mathscr{C}}$ is isomorphic to some power of $\det
H^{0}(\mathscr{C},\omega_{\mathscr{C}/S})$ (Corollary 3.10).

Roughly speaking, Condition (a) means there is a rational function $x$ for the
integral Weierstrass equation (1) potentially representing $\mathscr{C}$. This
condition is automatically satisfied if $\mathrm{Pic}(S)$ is finite of odd
order (Proposition 1.4) or if $\mathscr{C}$ is a pointed Weierstrass model
(Proposition 4.3) The existence of $y$ for the equation means that the
integral closure of $R[x]$ in the function fields $K(C)$ is free over $R[x]$,
with a basis $\\{1,y\\}$. We define the _Weierstrass class_
$[\mathfrak{w}]\in\mathrm{Pic}(S)$ associated to $\mathscr{C}$ (Definition 3)
and we prove that the existence of an integral Weierstrass equation
representing $\mathscr{C}$ is equivalent to the triviality of $[\mathfrak{w}]$
and $\mathscr{Z}\simeq\mathbb{P}^{1}_{S}$ (Proposition 3.4).

The main result of this work is the following theorem. Denote by
$f:\mathscr{C}\to S$ the structure morphism.

######  Theorem 0.1 (See Theorem 3.11).

Let $\mathscr{C}$ be a Weierstrass model of $C$ over $S$. Then $\mathscr{C}$
is defined by an integral Weierstrass equation under any of the following
conditions:

  1. (1)

$\Delta_{\mathscr{C}}$ is principal, $\det f_{*}\omega_{\mathscr{C}/S}$ is
free and either

    1. (a)

$g$ is odd, $\mathscr{Z}\simeq\mathbb{P}^{1}_{S}$ or

    2. (b)

$g\equiv 2\mod 4$;

  2. (2)

$\mathrm{Pic}(S)$ is finite of odd order, $\Delta_{\mathscr{C}}$ is principal
and $\det f_{*}\omega_{\mathscr{C}/S}$ is free;

  3. (3)

$\mathrm{Pic}(S)$ is finite of order prime to $2(2g+1)$ and
$\Delta_{\mathscr{C}}$ is principal;

  4. (4)

$\mathrm{Pic}(S)$ is finite of order prime to $2g$ and $\det
f_{*}\omega_{\mathscr{C}/S}$ is free.

In particular, for hyperelliptic curves having everywhere good reduction over
a number field, the next corollary answers positively a conjecture of Mohammad
Sadek [10]. It is well known for elliptic curves.

######  Corollary 0.2 (See Corollary 3.12).

Let $C$ be a hyperelliptic curve of genus $g$ over a number field $K$ having
everywhere good reduction. If the class number of $K$ is prime to $2(2g+1)$,
then $C$ is defined by an integral Weierstrass equation with invertible
discriminant.

Similar results are given in §4 for pointed hyperelliptic curves $(C,P)$ where
$P$ is a rational Weierstrass point of $C$. Finally in §6, we construct
examples of $\mathscr{C}$ satisfying various nice properties but not defined
by integral (or pointed integral) Weierstrass equations.

Settings In all this paper, $R$ is a Dedekind domain, $K=\mathrm{Frac}(R)$,
$S={\rm Spec}\kern 2.0ptR$ and $C$ is a hyperelliptic curve over $K$ (that is,
$C$ is smooth projective geometrically connected over $K$ and admits a finite
separable morphism $C\to\mathbb{P}^{1}_{K}$ of degree $2$) of genus $g\geq 2$
or an elliptic curve over $K$.

##  1\. Twisted projective lines

Let $\phi:\mathscr{Z}\to S$ be a smooth projective scheme with generic fiber
isomorphic to $\mathbb{P}^{1}_{K}$. This is a _twisted $\mathbb{P}^{1}_{S}$_
(see [9], 3.2). It is the projective bundle over $S$ defined by a locally free
rank $2$ vector bundle on $S$.

In this section we study the sections of $\mathscr{Z}$ and also give a
sufficient condition for $\mathscr{Z}$ to be isomorphic to
$\mathbb{P}^{1}_{S}$. We often consider sections of $\mathscr{Z}$ as effective
Cartier divisors on $\mathscr{Z}$. Then $\mathcal{O}_{\mathscr{Z}}(D)$ is a
subsheaf of the function field $K(\mathscr{Z})$.

  * Definition 1.1

Let $D\in\mathscr{Z}(S)$ be a section. We will say that $D$ is _elementary_ if
$\phi_{*}\mathcal{O}_{\mathscr{Z}}(D)$ is free over $\mathcal{O}_{S}$.

######  Lemma 1.2.

Let $D\in\mathscr{Z}(S)$. Let $F=\phi_{*}\mathcal{O}_{\mathscr{Z}}(D)$.

  1. (1)

The invertible sheaf $\mathcal{O}_{\mathscr{Z}}(D)$ is very ample, $F$ is
locally free of rank $2$ over $\mathcal{O}_{S}$, the morphism
$i:\mathscr{Z}\to\mathbb{P}(F)$ associated to $\mathcal{O}_{\mathscr{Z}}(D)$
is an isomorphism and we have

$i^{*}\mathcal{O}_{\mathbb{P}(F)}(1)\simeq\mathcal{O}_{\mathscr{Z}}(D).$

  2. (2)

The following properties are equivalent:

    1. (a)

$D$ is elementary;

    2. (b)

$\det F\simeq\mathcal{O}_{S}$;

    3. (c)

There exists an isomorphism $\theta:\mathscr{Z}\to\mathbb{P}^{1}_{S}$ such
that
$\mathcal{O}_{\mathscr{Z}}(D)\simeq\theta^{*}\mathcal{O}_{\mathbb{P}^{1}_{S}}(1)$.

  3. (3)

Let $D^{\prime}\in\mathscr{Z}(S)$. Then there exists a unique
$M\in\mathrm{Pic}(S)$ such that

$\mathcal{O}_{\mathscr{Z}}(D^{\prime})\simeq\mathcal{O}_{\mathscr{Z}}(D)\otimes\phi^{*}M.$

We then have

$\phi_{*}\mathcal{O}_{\mathscr{Z}}(D^{\prime})\simeq F\otimes
M,\quad\det\phi_{*}\mathcal{O}_{\mathscr{Z}}(D^{\prime})\simeq(\det F)\otimes
M^{\otimes 2}.$

  4. (4)

For any $M\in\mathrm{Pic}(S)$, there is a section
$D^{\prime}\in\mathscr{Z}(S)$ such that

$\mathcal{O}_{\mathscr{Z}}(D^{\prime})\simeq\mathcal{O}_{\mathscr{Z}}(D)\otimes\phi^{*}M.$

###### Proof.

(1) This is true because it is true locally on $S$.

(2) Obviously (c) $\Rightarrow$ (a) $\Rightarrow$ (b). We have (b)
$\Rightarrow$ (a) because $F\simeq\mathcal{O}_{S}\oplus\det F$, $R$ being a
Dedekind domain.

(a) $\Rightarrow$ (c). Compose the isomorphism
$i:\mathscr{Z}\simeq\mathbb{P}(F)$ with an isomorphism
$\mathbb{P}(F)\simeq\mathbb{P}^{1}_{S}$ to get the desired isomorphism
$\theta$. Then
$\theta^{*}\mathcal{O}_{\mathbb{P}^{1}_{S}}(1)\simeq\mathcal{O}_{\mathscr{Z}}(D)$
by the isomorphism of Part (1).

(3) For any noetherian connected scheme $T$ and any locally free
$\mathcal{O}_{T}$-module $G$ of finite rank, it is known that there is a
canonical isomorphism

(2)
$\mathrm{Pic}(\mathbb{P}(G))\simeq\langle\mathcal{O}_{\mathbb{P}(G)}(1)\rangle\times\mathrm{Pic}(T)$

(see [5], II, Remarque 4.2.7, or [4], Proposition 8.4(a)). This implies the
existence and uniqueness of $M$. The second part comes from the projection
formula ([7], Proposition 5.2.32) and the fact that $F$ is locally free of
rank $2$.

(4) We work over $\mathbb{P}(F)$ with
$\mathcal{O}_{\mathbb{P}(F)}(1)\simeq\mathcal{O}_{\mathbb{P}(F)}(D)$. Let
$H=M^{\vee}$ be the dual of $M$. Using the structure of the projective modules
over Dedekind domains, $H$ can be injected into $F$ with locally free quotient
of rank 1. Then $H$ induces a section $D^{\prime}$ of $\mathbb{P}(F)$ ([5],
II.4.2.3). As an effective Cartier divisor on $\mathbb{P}(F)$, $D^{\prime}$ is
defined by the sheaf of homogeneous ideals $H\mathcal{O}_{\mathbb{P}(F)}$. One
can check that
$\mathcal{O}_{\mathbb{P}(F)}(-D^{\prime})\otimes\mathcal{O}_{\mathbb{P}(F)}(1)\simeq\phi^{*}H$,
thus the desired isomorphism. ∎

  * Example 1.3

    1. (1)

If $R$ is a PID, then all sections of $\mathscr{Z}$ are elementary.

    2. (2)

On $\mathbb{P}^{1}_{S}$ with a parameter $x$, a typical elementary section is
the pole divisor $\mathrm{div}_{\infty}(x)$ of $x$ with
$\mathcal{O}_{\mathbb{P}^{1}_{S}}(\mathrm{div}_{\infty}(x))\simeq\mathcal{O}_{\mathbb{P}^{1}_{S}}(1)$.

In view of Lemma 1.2(2), when $\mathscr{Z}$ has an elementary section $D$, we
will often think about the pair $(\mathscr{Z},D)$ as
$(\mathbb{P}^{1}_{S},\mathrm{div}_{\infty}(x))$ where $x$ is some parameter of
$\mathbb{P}^{1}_{S}$.

######  Proposition 1.4.

We have $\mathscr{Z}\simeq\mathbb{P}^{1}_{S}$ if and only if for some
$D\in\mathscr{Z}(S)$, $\det\phi_{*}\mathcal{O}_{\mathscr{Z}}(D)$ is a square
in $\mathrm{Pic}(S)$. In particular, if the map

$\mathrm{Pic}(S)\to\mathrm{Pic}(S),\quad x\mapsto x^{2}$

is surjective (e.g. if $\mathrm{Pic}(S)$ has finite and odd order), then
$\mathscr{Z}\simeq\mathbb{P}^{1}_{S}$.

###### Proof.

Let $D\in\mathscr{Z}(S)$ and $F=\phi_{*}\mathcal{O}_{\mathscr{Z}}(D)$. We have
$\mathscr{Z}\simeq\mathbb{P}(F)$ by Lemma 1.2(1). By Lemma 1.5 below,
$\mathscr{Z}\simeq\mathbb{P}^{1}_{S}$ if and only if $F\simeq M\oplus M$ for
some $M\in\mathrm{Pic}(S)$. The existence of such an isomorphism implies that
$\det F\simeq M^{\otimes 2}$ is a square in $\mathrm{Pic}(S)$. Conversely, if
$\det F\simeq M^{\otimes 2}$ for some $M\in\mathrm{Pic}(S)$, replacing $F$ by
$F\otimes M^{\vee}$ (which does not change $\mathbb{P}(F)$), we can suppose
that $\det F\simeq\mathcal{O}_{S}$. Therefore $F$ is free (Lemma 1.2(2)) and
$\mathscr{Z}\simeq\mathbb{P}^{1}_{S}$. ∎

######  Lemma 1.5.

Let $F,G$ be two locally free finite rank sheaves on a noetherian scheme $T$.
Then the projective bundles $\mathbb{P}(F)$ and $\mathbb{P}(G)$ are
$T$-isomorphic if and only if $F\simeq G\otimes_{\mathcal{O}_{T}}M$ for some
$M\in\mathrm{Pic}(T)$.

###### Proof.

The if part is a general fact ([5], II.4.1.4). To prove the converse, we can
suppose that $T$ is connected. We have a commutative diagram

$\textstyle{\mathbb{P}(F)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\sigma}$$\scriptstyle{q}$$\textstyle{\mathbb{P}(G)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p}$$\textstyle{T}$

with an isomorphism $\sigma$. Using the structure of
$\mathrm{Pic}(\mathbb{P}(G))$ (see Isomorphism (2) in the proof of Lemma
1.2(3)), there exists a $M\in\mathrm{Pic}(T)$ such that
$\sigma_{*}\mathcal{O}_{\mathbb{P}(F)}(1)=\mathcal{O}_{\mathbb{P}(G)}(1)\otimes
p^{*}M$. Taking $p_{*}$ and using the projection formula ([7], Proposition
5.2.32), we find

$q_{*}\mathcal{O}_{\mathbb{P}(F)}(1)=p_{*}\mathcal{O}_{\mathbb{P}(G)}(1)\otimes
M.$

But $q_{*}\mathcal{O}_{\mathbb{P}(F)}(1)\simeq F$ ([13], Example 27.21.3) and
$p_{*}\mathcal{O}_{\mathbb{P}(G)}(1)\simeq G$, the only if part is proved. ∎

##  2\. Weierstrass models and discriminant ideals

In this section we gather some basic definitions and facts on Weierstrass
equations and Weierstrass models.

A _model of $C$ over $S$_ is a normal projective scheme $f:\mathscr{C}\to S$
whose generic fiber $\mathscr{C}_{K}$ is isomorphic to $C$. A model
$\mathscr{C}$ is a _Weierstrass model_ if the hyperelliptic involution on $C$
extends to an $S$-involution $\sigma$ on $\mathscr{C}$ and if the quotient

$\mathscr{Z}:=\mathscr{C}/\langle\sigma\rangle$

is smooth over $S$. The latter is then a smooth model of $\mathbb{P}^{1}_{K}$
over $S$. Equivalently, a Weierstrass model is the normalization of a smooth
model $\mathscr{Z}$ of $\mathbb{P}^{1}_{K}$ in the canonical degree $2$
morphism $C\to\mathbb{P}^{1}_{K}$.

  * Definition 2.1

A _Weierstrass equation of $C$_ is an equation of the form

(3) $y^{2}+Q(x)y=P(x),\quad P(x),Q(x)\in K[x]$

with $\deg Q(x)\leq g+1,\deg P(x)\leq 2g+2$ such that the corresponding affine
curve is isomorphic to an open subscheme of $C$. Such an equation is said to
be _integral over $R$_ if moreover $P(x),Q(x)\in R[x]$. If an integral
Weierstrass equation (3) of $C$ defines a normal affine scheme

(4) ${\rm Spec}\kern 2.0ptR[x,y]/(y^{2}+Q(x)y-P(x))$

(see [6], §3, Lemme 2 for a criterion of normality), then embedding ${\rm
Spec}\kern 2.0ptR[x]$ in $\mathbb{P}^{1}_{S}$ and taking the normalization of
$\mathbb{P}^{1}_{S}$ in the degree $2$ morphism $C\to\mathbb{P}^{1}_{K}$ will
give a Weierstrass model $\mathscr{C}$. Such a Weierstrass model is said to be
_defined by the integral Weierstrass equation_ (3). The scheme $\mathscr{C}$
is then obtained by glueing the two affine schemes:

${\rm Spec}\kern 2.0ptR[x,y]/(y^{2}+Q(x)y-P(x)),\ {\rm Spec}\kern
2.0ptR[t,z]/(z^{2}+t^{g+1}Q(1/t)z-t^{2g+2}P(1/t))$

via the identifications $t=1/x$, $z=y/x^{g+1}$.

  * Remark 2.2

Roughly speaking, for $\mathscr{C}$ to be defined by an integral Weierstrass
equation, there are two conditions: $\mathscr{Z}\simeq\mathbb{P}^{1}_{S}$
(there exists an $x$) and the quotient by $R[x]$ of the integral closure of
$R[x]$ in $K(C)$ is free (there exists an $y$).

######  Lemma 2.3.

Suppose that $R$ is a PID. Then all Weierstrass models are defined by integral
Weierstrass equations.

###### Proof.

This is well known. One can for instance argue as follows: by Proposition 1.4,
$\mathscr{Z}\simeq\mathbb{P}^{1}_{S}$. Choose any (automatically elementary)
section $D\in\mathscr{Z}(S)$ and argue as in the proof of Proposition 3.4. ∎

  * Definition 2.4

Now we define the discriminant ideal and minimal Weierstrass models. Let
$\mathscr{C}$ be a Weierstrass model over $S$. For any closed point $s\in S$,
$\mathscr{C}\times_{S}{\rm Spec}\kern 2.0pt\mathcal{O}_{S,s}$ is then defined
by a Weierstrass equation over the PID $\mathcal{O}_{S,s}$. The discriminant
of the equation ([6], §2) generates an ideal $\Delta_{s}$ of
$\mathcal{O}_{S,s}$ independent on the choice of the equation. We have
$\Delta_{s}=\mathcal{O}_{S,s}$ if and only if $\mathscr{C}$ is smooth at $s$.
The _discriminant ideal $\Delta_{\mathscr{C}}$ of $\mathscr{C}$_ is the ideal
$\prod_{s}\Delta_{s}$. We have $\Delta_{\mathscr{C}}=\mathcal{O}_{S}$ if and
only if $\mathscr{C}$ is smooth over $S$ (then $C$ has everywhere good
reduction). We say that $\mathscr{C}$ is _a minimal Weierstrass model_ if for
all $s$, $\Delta_{s}$ divides the discriminant of any integral Weierstrass
equation of $C$ over $\mathcal{O}_{S,s}$.

  * Remark 2.5

The curve $C$ always has minimal Weierstrass models. Indeed, $C$ has a smooth
projective model over a dense open subset $U$ of $S$. Then one can glue this
model with minimal Weierstrass models at the points of $S\setminus U$. Note
that in general the minimal Weierstrass models are not unique (unless $C$ has
good reduction everywhere). However, up to $S$-isomorphisms, there are only
finitely many such models for a given $C$ (use [6], §8, Corollary 4).

Given a Weierstrass model $\mathscr{C}$, there are some natural conditions for
$\mathscr{C}$ to be defined by an integral Weierstrass equation. As
$f:\mathscr{C}\to S$ is locally a complete intersection, its dualizing sheaf
$\omega_{\mathscr{C}/S}$ is invertible and $f_{*}\omega_{\mathscr{C}/S}$ is
locally free of rank $g$ over $S$.

######  Proposition 2.6.

If a Weierstrass model $\mathscr{C}$ is defined by an integral Weierstrass
equation, then

  1. (a)

$\mathscr{Z}\simeq\mathbb{P}^{1}_{S}$;

  2. (b)

$\Delta_{\mathscr{C}}$ is a principal ideal of $R$;

  3. (c)

$\det f_{*}\omega_{\mathscr{C}/S}$ is free.

###### Proof.

The first two conditions are clear. If $\mathscr{C}$ is defined by an integral
Weierstrass equation (3), then one can check exactly as over a ground field
(see e.g. [7], Proposition 7.4.26) that $f_{*}\omega_{\mathscr{C}/S}$ is free
over $R$ with

$\dfrac{x^{i}dx}{2y+Q(x)},\quad 0\leq i\leq g-1$

as a basis. So $\det f_{*}\omega_{\mathscr{C}/S}$ is free. ∎

##  3\. Weierstrass class of a Weierstrass model

We fix a Weierstrass model $\mathscr{C}$ and we look for conditions ensuring
it is defined by an integral Weierstrass equation. Recall that
$f:\mathscr{C}\to S$ is the structure morphism. Denote by
$\pi:\mathscr{C}\to\mathscr{Z}$ the quotient morphism by the hyperelliptic
involution and by $\phi:\mathscr{Z}\to S$ the structure morphism:

$\textstyle{\mathscr{C}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\scriptstyle{f}$$\textstyle{\mathscr{Z}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi}$$\textstyle{S.}$

The morphism $\pi$ is finite, locally free of degree $2$, and we have an exact
sequence

(5)
$0\to\mathcal{O}_{\mathscr{Z}}\to\pi_{*}\mathcal{O}_{\mathscr{C}}\to\mathcal{L}:=\pi_{*}\mathcal{O}_{\mathscr{C}}/\mathcal{O}_{\mathscr{Z}}\to
0.$

######  Lemma 3.1.

Let $D\in\mathscr{Z}(S)$ be a section of $\mathscr{Z}$ over $S$. Let
$\mathcal{L}$ be as above. Then there exists a unique $N\in\mathrm{Pic}(S)$,
depending on $D$, such that

(6) $\mathcal{L}\otimes\mathcal{O}_{\mathscr{Z}}((g+1)D)\simeq\phi^{*}N.$

###### Proof.

For any $s\in S$, we let $U\ni s$ be an affine open neighborhood such that
$\mathscr{C}_{U}$ is defined by an integral Weierstrass equation (Lemma 2.3)

$y^{2}_{U}+Q_{U}(x_{U})y_{U}=P_{U}(x_{U})$

(thus $\mathscr{Z}_{U}\simeq\mathbb{P}^{1}_{U}$ with parameter $x_{U}$) and
that $D|_{\mathscr{Z}_{U}}$ is the pole divisor of $x_{U}$. Consider the
covering of $\mathscr{Z}_{U}$ by

$W:={\rm Spec}\kern 2.0pt\mathcal{O}_{S}(U)[x_{U}],\quad V:={\rm Spec}\kern
2.0pt\mathcal{O}_{S}(U)[1/x_{U}].$

Then

$(\pi_{*}\mathcal{O}_{\mathscr{C}})(W)=\mathcal{O}_{\mathscr{Z}}(W)\oplus\mathcal{O}_{\mathscr{Z}}(W)y_{U}=\mathcal{O}_{\mathscr{Z}}(W)\oplus\mathcal{O}_{\mathscr{Z}}(-(g+1)D)(W)y_{U}$

and

$(\pi_{*}\mathcal{O}_{\mathscr{C}})(V)=\mathcal{O}_{\mathscr{Z}}(V)\oplus\mathcal{O}_{\mathscr{Z}}(V)(y_{U}/x_{U}^{g+1})=\mathcal{O}_{\mathscr{Z}}(V)\oplus\mathcal{O}_{\mathscr{Z}}(-(g+1)D)(V)y_{U}.$

Therefore

$\pi_{*}\mathcal{O}_{\mathscr{C}}|_{\mathscr{Z}_{U}}=\mathcal{O}_{\mathscr{Z}_{U}}\oplus\mathcal{O}_{\mathscr{Z}_{U}}(-(g+1)D)y_{U}$

and

$\mathcal{L}|_{\mathscr{Z}_{U}}\simeq\mathcal{O}_{\mathscr{Z}}(-(g+1)D)|_{\mathscr{Z}_{U}}.$

This proves the existence of $N$. The uniqueness of $N$ up to isomorphism
comes from the fact that $N\simeq\phi_{*}\phi^{*}N$. ∎

  * Remark 3.2 (Dependence on $D$)

Denote by $F=\phi_{*}\mathcal{O}_{\mathscr{Z}}(D)$. Let
$D^{\prime}\in\mathscr{Z}(S)$ be another section with
$F^{\prime}=\phi_{*}\mathcal{O}_{\mathscr{Z}}(D^{\prime})$ and corresponding
$N^{\prime}\in\mathrm{Pic}(S)$ as in Lemma 3.1. Then by Lemma 1.2(3),
$\mathcal{O}_{\mathscr{Z}}(D^{\prime})\simeq\mathcal{O}_{\mathscr{Z}}(D)\otimes\phi^{*}M$
for a unique $M\in\mathrm{Pic}(S)$ and we have

$N^{\prime}\simeq N\otimes M^{\otimes(g+1)},\quad\det F^{\prime}\simeq(\det
F)\otimes M^{\otimes 2}.$

In particular,

$G_{\mathscr{C},2}:=N^{\otimes 2}\otimes(\det
F^{\vee})^{\otimes(g+1)}\in\mathrm{Pic}(S)$

is independent of the choice of $D$. If $g$ is odd, then $G_{\mathscr{C},2}$
is the square of

$G_{\mathscr{C},1}:=N\otimes(\det
F^{\vee})^{\otimes(g+1)/2}\in\mathrm{Pic}(S)$

and the latter is also independent on the choice of $D$.

  * Definition 3.3

Let $\mathscr{C}$ be a Weierstrass model. We define the _Weierstrass class
$[\mathfrak{w}]\in\mathrm{Pic}(S)$ of $\mathscr{C}$_ as the class of
$G_{\mathscr{C},1}$ (resp. $G_{\mathscr{C},2}$) when $g$ is odd (resp. even).

######  Proposition 3.4.

Let $f:\mathscr{C}\to S$ be a Weierstrass model of $C$. Then $\mathscr{C}$ can
be defined by an integral Weierstrass equation if and only if
$\mathscr{Z}\simeq\mathbb{P}^{1}_{S}$ and $[\mathfrak{w}]$ is trivial.

###### Proof.

Step (1) Suppose there exists an elementary section $D\in\mathscr{Z}(S)$ such
that the corresponding $N$ is trivial (Lemma 3.1). Let us show the existence
of an integral Weierstrass equation defining $\mathscr{C}$.

We identify $(\mathscr{Z},D)$ with
$(\mathbb{P}^{1}_{S},\mathrm{div}_{\infty}(x))$ for some parameter $x$ of
$\mathbb{P}^{1}_{S}$. Let $y\in K(C)$ be a basis of
$\mathcal{L}\otimes\mathcal{O}_{\mathscr{Z}}((g+1)D)$. Over $W:={\rm
Spec}\kern 2.0ptR[x]$, we have
$\mathcal{L}\otimes\mathcal{O}_{\mathscr{Z}}((g+1)D)(W)=R[x]y$. This means
that

(7) $\mathcal{O}_{\mathscr{C}}(\pi^{-1}(W))=R[x]\oplus R[x]y.$

So there exist $Q(x),P(x)\in R[x]$ such that $y^{2}+Q(x)y=P(x)$. It remains to
bound the degrees of $Q(x)$ and $P(x)$. Over $V:={\rm Spec}\kern 2.0ptR[1/x]$,
we have

$\mathcal{L}(V)=\mathcal{O}_{\mathscr{Z}}(-(g+1)D)(V)y=R[1/x](y/x^{g+1}),$

which implies that $y/x^{g+1}\in\mathcal{O}_{\mathscr{C}}(\pi^{-1}(V))$, hence
integral over $R[1/x]$. Therefore $Q(x)/x^{g+1},P(x)/x^{2g+2}\in R[1/x]$ and
$\deg Q(x)\leq g+1$, $\deg P(x)\leq 2g+2$. The pair $x,y$ gives an integral
Weierstrass equation for $\mathscr{C}$.

Step (2), “only if” part. Suppose that $\mathscr{C}$ is defined by an integral
Weierstrass equation as Equation (3). We have
$\mathscr{Z}\simeq\mathbb{P}^{1}_{S}$ with parameter $x$. The computations in
the proof of Lemma 3.1 show that for $D=\mathrm{div}_{\infty}(x)$, the
corresponding $N$ is free. As $D$ is elementary, $[\mathfrak{w}]$ is trivial.

Step (3), “if” part. Suppose that $\mathscr{Z}\simeq\mathbb{P}^{1}_{S}$ and
$[\mathfrak{w}]$ is trivial. Let $x$ be a parameter of $\mathbb{P}^{1}_{S}$
and let $D=\mathrm{div}_{\infty}(x)$. If $g$ is odd, then
$N\simeq[\mathfrak{w}]$ is trivial. It follows from Step (1) that
$\mathscr{C}$ is defined by an integral Weierstrass equation.

Suppose now that $g$ is even. Then $N^{\otimes 2}\simeq[\mathfrak{w}]$ is
trivial. By Lemma 1.2(4), there exists a section $D^{\prime}\in\mathscr{Z}(S)$
such that

$\mathcal{O}_{\mathscr{Z}}(D^{\prime})\simeq\mathcal{O}_{\mathscr{Z}}(D)\otimes\phi^{*}N.$

This implies that $\det\phi_{*}\mathcal{O}_{\mathscr{Z}}(D^{\prime})\simeq
N^{\otimes 2}$ and $N^{\prime}\simeq N\otimes N^{\otimes(g+1)}$ are free,
therefore $\mathscr{C}$ is defined by an integral Weierstrass equation, again
by applying Step (1) to $D^{\prime}$. ∎

  * Remark 3.5

In view of this proposition, our Remarque 6 in [6], page 4584, is very likely
to be wrong.

  * Remark 3.6 (Equations defining the same Weierstrass model)

Suppose that $\mathscr{C}$ is defined by an integral Weierstrass equation
$y^{2}+Q(x)y=P(x)$.

    1. (a)

If $g$ is even or if $\mathrm{Pic}(S)[2]=\\{1\\}$, then any other integral
Weierstrass equation $z^{2}+G(u)z=F(u)$ of $\mathscr{C}$ is related to the
previous one by

$u=\frac{ax+b}{cx+d},\quad\text{with}\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\in\mathrm{Gl}_{2}(R)$

and

$z=\frac{ey+H(x)}{(cx+d)^{g+1}},\quad e\in R^{*},H(x)\in R[x],\deg H(x)\leq
g+1.$

Indeed, let $D^{\prime}=\mathrm{div}_{\infty}(u)$ with corresponding
$F^{\prime}=\phi_{*}\mathcal{O}_{\mathscr{Z}}(D^{\prime})\simeq\mathcal{O}_{S}^{2}$
and $N^{\prime}$ as in Lemma 3.1. As seen in Step (2) of the proof of
Proposition 3.4, we have $N^{\prime}\simeq\mathcal{O}_{S}$. Let $M$ be as in
Remark 3, then $M^{\otimes 2}\simeq\mathcal{O}_{S}$ and
$M^{\otimes(g+1)}\simeq\mathcal{O}_{S}$. So $M\simeq\mathcal{O}_{S}$, and this
implies the relation between $x$ and $u$. The condition on $z$ comes from the
fact that $(cx+d)^{g+1}z\in K(C)$ is integral over $R[x]$ and that both
equations give the same discriminant ideal $\Delta_{\mathscr{C}}$.

    2. (b)

On the opposite, suppose that $g$ is odd and that there exists
$M\in\mathrm{Pic}(S)$ of order exactly $2$. We take a section
$D^{\prime}\in\mathscr{Z}(S)$ as in Lemma 1.2(4). Then $D^{\prime}$ is
elementary as $M^{\otimes 2}\simeq\mathcal{O}_{S}$. Moreover the corresponding
$N^{\prime}\simeq N\otimes M^{\otimes g+1}\simeq\mathcal{O}_{S}$. By Step (1)
in the proof of Proposition 3.4, if $D^{\prime}=\mathrm{div}_{\infty}(u)$,
then $\mathscr{C}$ is defined by an integral Weierstrass equation
$z^{2}+G(u)z=F(u)$. But $u,x\in K(\mathscr{Z})$ are not related by the action
of $\mathrm{Gl}_{2}(R)$.

Note that this phenomena does not happen for pointed Weierstrass models
(Definition 4) as we then have
$\mathrm{div}_{\infty}(u)=\mathrm{div}_{\infty}(x)$, so $u=ax+b$ for some
$a\in R^{*}$ and $b\in R$.

Now we relate the Weierstrass class to some other classes attached to
$\mathscr{C}$.

######  Lemma 3.7 (Approximation of local equations).

Let $\mathscr{C}$ be a Weierstrass model of $C$ over $S$. Let
$D\in\mathscr{Z}(S)$. Let us fix

  1. (a)

a Weierstrass equation of $C$ over $K$:

(8) $y^{2}+Q(x)y=P(x)$

with $x$ having its pole at $D_{K}$ and

  2. (b)

a non-empty open subset $U\subset S$.

Then there exist $r\in\mathcal{O}_{S}(U)$, $h(x)\in\mathcal{O}_{S}(U)[x]$ of
degree $\leq g+1$ and, for all $s\in S\setminus U$, an integral Weierstrass
equation

(9) $y_{s}^{2}+Q_{s}(x_{s})y_{s}=P_{s}(x_{s})$

for $\mathscr{C}\times_{S}{\rm Spec}\kern 2.0pt\mathcal{O}_{S,s}$ with

$x-r=a_{s}x_{s},\quad y-h(x)=b_{s}y_{s},\quad a_{s},b_{s}\in K^{*}.$

###### Proof.

For each $s\in S\setminus U$, there exists an integral Weierstrass equation

$y_{s}^{2}+Q_{s}(x_{s})y_{s}=P_{s}(x_{s})$

for $\mathscr{C}\times_{S}{\rm Spec}\kern 2.0pt\mathcal{O}_{S,s}$ with $x_{s}$
having its pole at $D_{K}$. So there exist $a_{s},b_{s}\in K^{*}$, $r_{s}\in
K$, $h_{s}(x)\in K[x]$ of degree $\leq g+1$ such that

$x=a_{s}x_{s}+r_{s},\quad y=b_{s}y_{s}+h_{s}(x).$

By the approximation property in Dedekind domains, there exist
$r\in\mathcal{O}_{S}(U),h(x)\in\mathcal{O}_{S}(U)[x]$ of degree $\leq g+1$
such that

$r-r_{s}\in a_{s}\mathcal{O}_{S,s},\quad h(x)-h_{s}(x)\in
b_{s}\mathcal{O}_{S,s}[x]$

for all $s\in S\setminus U$. We obtain the desired result after replacing
$x_{s}$ with $x_{s}-(r-r_{s})/a_{s}$ and $y_{s}$ with
$y_{s}-(h(x)-h_{s}(x))/b_{s}$. ∎

  * Definition 3.8

Keep the settings of Lemma 3.7. Choose an open subset $U$ such that the
equation (8) has good reduction over $U$ and defines $\mathscr{C}_{U}$ (so
$D|_{\mathscr{Z}_{U}}$ is the pole divisor of $x|_{\mathscr{Z}_{U}}$). Define
$\mathfrak{a}$ as the fractional ideal of $K$ such that
$\mathfrak{a}\otimes\mathcal{O}_{S,s}=a_{s}\mathcal{O}_{S,s}$ for all $s\in
S\setminus U$ and $=\mathcal{O}_{S,s}$ for $s\in U$. Define similarly
$\mathfrak{b}$. These ideals depend on the choice of Equation (8) and on $D$,
but don’t depend on the choice of $U$. Their classes in $\mathrm{Pic}(S)$
don’t depend on the choice of Equation (8) over $K$.

######  Lemma 3.9.

Let $\mathscr{C}$ be a Weierstrass model of $C$ over $S$. Let
$D\in\mathscr{Z}(S)$. Let $F=\phi_{*}\mathcal{O}_{\mathscr{Z}}(D)$ and let $N$
be as in Lemma 3.1.

  1. (1)

We have

$\det F\simeq\mathfrak{a}^{-1},\quad N\simeq\mathfrak{b}^{-1}$

and

$\Delta_{\mathscr{C}}\simeq(\mathfrak{a}^{g+1}\mathfrak{b}^{-2})^{2(2g+1)},\quad\det
f_{*}\omega_{\mathscr{C}/S}\simeq\mathfrak{a}^{-g(g+1)/2}\mathfrak{b}^{g}.$

  2. (2)

Let $[\mathfrak{w}]$ be the Weierstrass class of $\mathscr{C}$ (Definition 3).
Then

$\Delta_{\mathscr{C}}\simeq[\mathfrak{w}]^{\otimes 2(2g+1)},\quad\det
f_{*}\omega_{\mathscr{C}/S}\simeq[\mathfrak{w}]^{\otimes(-g/2)}$

if $g$ is even, and

$\Delta_{\mathscr{C}}\simeq[\mathfrak{w}]^{\otimes 4(2g+1)},\quad\det
f_{*}\omega_{\mathscr{C}/S}\simeq[\mathfrak{w}]^{\otimes(-g)}$

if $g$ is odd.

  3. (3)

In $\mathrm{Pic}(S)$, we have the following equalities of subgroups

$\langle[\mathfrak{w}]^{2}\rangle=\langle\Delta_{\mathscr{C}},\det
f_{*}\omega_{\mathscr{C}/S}\rangle$

if $4\mid g$, and

$\langle[\mathfrak{w}]\rangle=\langle\Delta_{\mathscr{C}},\det
f_{*}\omega_{\mathscr{C}/S}\rangle$

otherwise.

###### Proof.

Part (2) follows immediately from (1) and implies Part (3). Let us prove Part
(1). Keep the notation of Lemma 3.7. We choose the equation (8) with
coefficients in $\mathcal{O}_{S}(U)$ and having good reduction over $U$ for
some dense open subset $U\subseteq S$. We also translate $x$ and $y$ so that
$x=a_{s}x_{s}$ and $y=b_{s}y_{s}$ at $s\in S\setminus U$.

We have

$\phi_{*}\mathcal{O}_{\mathscr{Z}}(D)(U)=\mathcal{O}_{S}(U)\oplus\mathcal{O}_{S}(U)x,\quad\phi_{*}\mathcal{O}_{\mathscr{Z}}(D)_{s}=\mathcal{O}_{S,s}\oplus\mathcal{O}_{S,s}x_{s},$

so we have canonically

$(\det F)(U)=\mathcal{O}_{S}(U)(1\wedge x),\quad(\det
F)_{s}=\mathcal{O}_{S,s}(1\wedge x_{s}).$

Then the relations $x=a_{s}x_{s}$ imply that $\det F\simeq\mathfrak{a}^{-1}$.
By the computations in the proof of Lemma 3.1, we see that $N$ has bases $y$
over $\mathcal{O}_{S}(U)$ and $y_{s}$ over $\mathcal{O}_{S,s}$. The relations
$y=b_{s}y_{s}$ then imply that $N\simeq\mathfrak{b}^{-1}$.

Denote by $\Delta$ the discriminant of Equation (8) and by $\Delta_{s}$ that
of Equation (9). Then

$\Delta=b_{s}^{4(2g+1)}a_{s}^{-2(g+1)(2g+1)}\Delta_{s}$

(see [6], bottom of page 4581). This proves that

$\Delta
R=\mathfrak{b}^{4(2g+1)}\mathfrak{a}^{-2(g+1)(2g+1)}\Delta_{\mathscr{C}}$

as fractional ideals of $K$, hence

$\Delta_{\mathscr{C}}\simeq(\mathfrak{a}^{g+1}\mathfrak{b}^{-2})^{2(2g+1)}.$

Now we compute $f_{*}\omega_{\mathscr{C}/S}$. Denote by

$\omega_{U}=\frac{dx}{(2y+Q(x))}\wedge...\wedge\frac{x^{g-1}dx}{(2y+Q(x))}$

and

$\omega_{s}=\frac{dx_{s}}{(2y_{s}+Q_{s}(x_{s}))}\wedge...\wedge\frac{x_{s}^{g-1}dx_{s}}{(2y_{s}+Q(x_{s}))}.$

They are respectively bases of $(\det f_{*}\omega_{\mathscr{C}/S})(U)$ and of
$(\det f_{*}\omega_{\mathscr{C}/S})_{s}$. As $x=a_{s}x_{s}$ and
$y=b_{s}y_{s}$, we have $Q_{s}(x_{s})=Q(x)/b_{s}$. Thus

$\frac{x^{i}dx}{2y+Q(x)}=(a_{s}^{i+1}b_{s}^{-1})\frac{x_{s}^{i}dx_{s}}{2y_{s}+Q(x_{s})}$

and $\omega_{U}=a_{s}^{g(g+1)/2}b_{s}^{-g}\omega_{s}$, so $\det
f_{*}\omega_{\mathscr{C}/S}\simeq\mathfrak{a}^{-g(g+1)/2}\mathfrak{b}^{g}$. ∎

######  Corollary 3.10.

Let $\mathscr{C}$ be a Weierstrass model of $C$ over $S$. Then the following
properties are true.

  1. (1)

If $g=1$, then $\Delta_{\mathscr{C}}\simeq(\det
f_{*}\omega_{\mathscr{C}/S})^{\otimes(-12)}$ (well-known);

  2. (2)

If $g=2$, then $\Delta_{\mathscr{C}/S}\simeq(\det
f_{*}\omega_{\mathscr{C}/S})^{\otimes(-10)}$.

  3. (3)

If $g=4$, then $\Delta_{\mathscr{C}/S}\simeq(\det
f_{*}\omega_{\mathscr{C}/S})^{\otimes(-9)}$.

  4. (4)

In $\mathrm{Pic}(S)$, $\Delta_{\mathscr{C}}$ is a $4(2g+1)$-th power (resp.
$2(2g+1)$-th power) if $g$ is odd (resp. even).

######  Theorem 3.11.

Let $\mathscr{C}$ be a Weierstrass model of $C$ over $S$. Then $\mathscr{C}$
is defined by an integral Weierstrass equation (Definition 2) under any of the
following conditions:

  1. (1)

$\Delta_{\mathscr{C}}$ is principal, $\det f_{*}\omega_{\mathscr{C}/S}$ is
free and either

    1. (a)

$g$ is odd and $\mathscr{Z}\simeq\mathbb{P}^{1}_{S}$ or,

    2. (b)

$g\equiv 2\mod 4$

(so the converse of Proposition 2.6 is true if $4\nmid g$);

  2. (2)

$\mathrm{Pic}(S)$ is finite of odd order, $\Delta_{\mathscr{C}}$ is principal
and $\det f_{*}\omega_{\mathscr{C}/S}$ is free;

  3. (3)

$\mathrm{Pic}(S)$ is finite of order prime to $2(2g+1)$, or $g$ is even and
the torsion elements of $\mathrm{Pic}(S)$ are of order prime to $2(2g+1)$, and
$\Delta_{\mathscr{C}}$ is principal.

  4. (4)

$\mathrm{Pic}(S)$ is finite of order prime to $2g$ and $\det
f_{*}\omega_{\mathscr{C}/S}$ is free.

###### Proof.

We have to show that $\mathscr{Z}\simeq\mathbb{P}^{1}_{S}$ and
$[\mathfrak{w}]$ is trivial under any of the conditions in the theorem.

(1.a) This follows from Lemma 3.9(3).

(1.b) Take any section $D\in\mathscr{Z}(S)$ and consider the fractional ideals
$\mathfrak{a},\mathfrak{b}$ as in Definition 3. Then Lemma 3.9(1) implies that
$\mathfrak{a}^{-g(g+1)/2}\mathfrak{b}^{g}\simeq\mathcal{O}_{S}$, so
$\mathfrak{a}$ is a square, and $\mathscr{Z}\simeq\mathbb{P}^{1}_{S}$. Lemma
3.9(3) implies that $[\mathfrak{w}]=1$.

(2) First we have $\mathscr{Z}\simeq\mathbb{P}^{1}_{S}$ by Proposition 1.4.
Lemma 3.9(3) implies that $[\mathfrak{w}]^{2}=1$, hence $[\mathfrak{w}]=1$.

(3) If $\mathrm{Pic}(S)$ is finite of order prime to $2(2g+1)$, then
$\mathscr{Z}\simeq\mathbb{P}^{1}_{S}$. Therefore $[\mathfrak{w}]^{4(2g+1)}=1$
by Lemma 3.9(2). Thus $[\mathfrak{w}]=1$.

Now suppose $g$ is even and the torsion elements of $\mathrm{Pic}(S)$ are of
order prime to $2(2g+1)$. With the notation of Lemma 3.9, as
$\Delta_{\mathscr{C}}$ is principal, $\mathfrak{a}^{g+1}\mathfrak{b}^{-2}$ is
a torsion element of order dividing $2(2g+1)$, therefore trivial. So
$\mathfrak{a}$ is a square in $\mathrm{Pic}(S)$ as $g$ is even, and
$\mathscr{Z}\simeq\mathbb{P}^{1}_{S}$. We can now argue as in the first part.

(4) Similar to (3). ∎

As an immediate consequence, we proved a conjecture of M. Sadek [10].

######  Corollary 3.12.

Let $K$ be a number field of class number prime to $2(2g+1)$. Let $C$ be a
hyperelliptic curve of genus $g$ having everywhere good reduction over $K$.
Then the smooth model $\mathscr{C}$ of $C$ is defined by an integral
Weierstrass equation.

###### Proof.

By the uniqueness of the smooth projective model, the hyperelliptic involution
of $C$ extends to an involution on $\mathscr{C}$ and the quotient is smooth
because $\mathscr{C}$ is a smooth curve. So $\mathscr{C}$ is a Weierstrass
model of $C$ with $\Delta_{\mathscr{C}}=\mathcal{O}_{K}$. By Theorem 3.11(3),
$\mathscr{C}$ is defined by an integral Weierstrass equation. ∎

##  4\. Pointed integral Weierstrass equations

The usual Weierstrass models of elliptic curves $E$ are pointed Weierstrass
models with respect to the origin $o$ of $E$, in the sense that $o$ extends to
a section contained in the smooth locus of the Weierstrass model. For
hyperelliptic curves $C$ having a rational _Weierstrass point_ $P\in C(K)$
(that is $\dim_{K}H^{0}(C,\mathcal{O}_{C}(2P))=2$), P. Lockhart ([8]) defined
and studied pointed integral Weierstrass equations for $(C,P)$.

  * Definition 4.1

A _pointed Weierstrass model_ of $(C,P)$ over $S$ is a Weierstrass model
$\mathscr{C}$ such that the Zariski closure $\overline{\\{P\\}}$ is contained
in the smooth locus of $\mathscr{C}\to S$. Such a model is said to be _defined
by a pointed integral Weierstrass equation_ of $(C,P)$ if the affine scheme
$\mathscr{C}\setminus\overline{\\{P\\}}$ is defined by

$y^{2}+Q(x)y=P(x)$

with $P(x),Q(x)\in\mathcal{O}_{S}[x]$, $P(x)$ is monic of degree $2g+1$ and
$\deg Q(x)\leq g$.

Note that it may happen that a pointed Weierstrass model $\mathscr{C}$ is
defined by an integral Weierstrass equation, but not by a pointed integral
Weierstrass equation (see Proposition 6.1(e)).

We also have an obvious notion of _minimal pointed Weierstrass model_. Such a
model always exists and is unique up to isomorphism (Corollary 5.2). When
$\mathcal{O}_{S}$ is principal (e.g. a discrete valuation ring), the minimal
pointed Weierstrass model is defined by a pointed integral Weierstrass
equation (see e.g., [8], Proposition 2.8).

Let $\mathscr{C}$ be a pointed Weierstrass model of $(C,P)$. Let us recall the
definition of the _pointed Weierstrass class_
$[\mathfrak{u}]\in\mathrm{Pic}(S)$ of $\mathscr{C}$ similarly to [8],
Definition 2.7. Let

(10) $y^{2}+Q(x)y=P(x)$

be a pointed Weierstrass equation of $(C,P)$ over $K$, of discriminant
$\Delta\in K^{*}$. For any closed point $s\in S$, we have a pointed integral
Weierstrass equation

$y_{s}^{2}+Q_{s}(x_{s})y_{s}=P_{s}(x_{s})$

for $\mathscr{C}\times_{S}{\rm Spec}\kern 2.0pt\mathcal{O}_{S,s}$, with
discriminant $\Delta_{s}\in\mathcal{O}_{S,s}\setminus\\{0\\}$. Using Lemma
3.7, we see that translating, if necessary, $x$ and $x_{s}$ by elements of $K$
and $\mathcal{O}_{S,s}$ respectively and $y,y_{s}$ by polynomials of degree
$\leq g$ in $K[x]$ and in $\mathcal{O}_{S,s}[x]$ respectively (which keeps the
equations as pointed Weierstrass equations), we can suppose that

$x=a_{s}x_{s},\quad y=b_{s}y_{s},\quad a_{s},b_{s}\in K^{*}.$

As $P(x)$ and $P_{s}(x_{s})$ are monic of degree $2g+1$ we find
$a_{s}^{2g+1}=b_{s}^{2}$. Denote by $u_{s}=a_{s}^{g}/b_{s}$, then
$a_{s}=u_{s}^{-2}$ and $b_{s}=u_{s}^{-(2g+1)}$.

  * Definition 4.2 ([8], Definition 2.7)

The class in $\mathrm{Pic}(S)$ of the fractional ideal $\mathfrak{u}$ of $K$
such that $\mathfrak{u}\otimes\mathcal{O}_{S,s}=u_{s}\mathcal{O}_{S,s}$ for
all $s\in S$ is called the _pointed Weierstrass class 111In fact our
definition is the inverse of that of [8]. of $\mathscr{C}$_. It does not
depend on the choice of the equation (10). With the notation of Definition 3,
if we take for $D$ the image of $\overline{\\{P\\}}$ in $\mathscr{Z}(S)$, we
have $\mathfrak{a}=\mathfrak{u}^{-2}$ and
$\mathfrak{b}=\mathfrak{u}^{-(2g+1)}$.

Note that $[\mathfrak{u}]$ differs from the non-pointed Weierstrass class
(Definition 3). Using Lemma 3.9(1), we see that
$[\mathfrak{w}]=[\mathfrak{u}]^{g}$ if $g$ is odd, and
$[\mathfrak{w}]=[\mathfrak{u}]^{2g}$ if $g$ is even.

######  Proposition 4.3.

Let $\mathscr{C}$ be a pointed Weierstrass model of $(C,P)$. Let
$D\in\mathscr{Z}(S)$ be the image of $\overline{\\{P\\}}\in\mathscr{C}(S)$.

  1. (1)

We have

$\Delta_{\mathscr{C}}\simeq\mathfrak{u}^{4g(2g+1)},\quad\det
f_{*}\omega_{\mathscr{C}/S}\simeq\mathfrak{u}^{-g^{2}}.$

  2. (2)

The model $\mathscr{C}$ is defined by a pointed integral Weierstrass equation
if and only if $\mathfrak{u}$ is principal.

  3. (3)

The quotient $\mathscr{Z}$ is isomorphic to $\mathbb{P}^{1}_{S}$.

###### Proof.

(1) Follows from Lemma 3.9. See also [8], Equality (2.3) and Proposition 1.12.

(2) The only if part is clear. Suppose conversely that $\mathfrak{u}$ is
principal. Then $\mathfrak{a}$ is trivial and $D$ is an elementary section. As
$\mathfrak{b}$ is principal, $\mathscr{C}$ is defined by an integral
Weierstrass equation by Step (1) in the proof of Proposition 3.4.

(3) We use Lemma 3.9. As $\mathfrak{a}=\mathfrak{u}^{-2}$ is a square in
$\mathrm{Pic}(S)$ and
$\det\phi_{*}\mathcal{O}_{\mathscr{Z}}(D)\simeq\mathfrak{a}^{-1}$ by Lemma 3.9
(1), we find $\mathscr{Z}\simeq\mathbb{P}^{1}_{S}$ (Proposition 1.4). ∎

######  Corollary 4.4.

If $\mathrm{Pic}(S)$ has finite order prime to $2g(2g+1)$ and if
$\Delta_{\mathscr{C}}$ is principal (e.g. if $\mathscr{C}$ is smooth over
$S$), then $\mathscr{C}$ is defined by a pointed integral Weierstrass
equation.

The next proposition is a straightforward generalization of [11], Theorem 5.

######  Proposition 4.5.

Let $(C,P)$ be a pointed hyperelliptic curve over a number field $K$. Let
$\mathfrak{u}$ be any fractional ideal of $K$. Then $[\mathfrak{u}]$ is the
pointed Weierstrass class of the minimal pointed Weierstrass model of some
quadratic twist of $(C,P)$.

###### Proof.

The proof is the same as for [11], Theorem 5. The only point we have to check
is that the minimal pointed Weierstrass model commutes with étale base changes
in higher genus. See Proposition 5.4. Note that, in general, this is false for
non-pointed minimal Weierstrass models in higher genus ([6], Proposition 4, p.
4595). ∎

  * Remark 4.6

A referee raised the natural question to study, for each class
$[\mathfrak{a}]$ in ${\mathrm{Cl}}(K)$, the density of the hyperelliptic
curves over $K$ having (pointed or not) Weierstrass class equal to
$[\mathfrak{a}]$, indicating that this question is solved for elliptic curves
in [2]. Recent related works can be found in [3].

##  5\. Complements on minimal pointed Weierstrass models

We prove the uniqueness of the minimal pointed Weierstrass model (see also
Remark 5) and its compatibility with étale base changes. These properties are
local on $S={\rm Spec}\kern 2.0ptR$.

Suppose first that $R$ is a discrete valuation ring with uniformizing element
$t$, residue field $k$ and normalized valuation $v$. Let
$\mathscr{C},\mathscr{C}^{\prime}$ be two distinct Weierstrass models of $C$
over $S$, with respective quotients $\mathscr{Z},\mathscr{Z}^{\prime}$ by the
hyperelliptic involution. Up to actions of $\mathrm{Gl}_{2}(R)$, there are
parameters $x,x^{\prime}$ of $\mathscr{Z},\mathscr{Z}^{\prime}$ respectively
and $d>0$ such that $x=t^{d}x^{\prime}$. Let $p\in\mathscr{C}_{k}$ be a
rational point over the zero of $x$ in $\mathscr{Z}_{k}$.

######  Lemma 5.1.

If $\mathscr{C}$ is regular at $p$, then
$v(\Delta_{\mathscr{C}})<v(\Delta_{\mathscr{C}^{\prime}})$.

###### Proof.

This is a special case of [6], Proposition 3 (iii)-(iv), p. 4594. However, op.
cit. is written under the hypothesis that $k$ is perfect. So we give an ad hoc
proof here.

We suppose $\mathrm{char}(k)=2$. The other cases are similar and easier to
deal with. For $0\leq i\leq d$, denote by $\mathscr{Z}_{i}$ the smooth model
of $\mathbb{P}^{1}_{K}$ with parameter $x_{i}=x/t^{i}$ and by
$\mathscr{C}_{i}$ the normalization of $\mathscr{Z}_{i}$ in $K(C)$. So
$\mathscr{C}=\mathscr{C}_{0}$ and $\mathscr{C}^{\prime}=\mathscr{C}_{d}$. We
are going to prove that $v(\Delta_{\mathscr{C}_{i}})\leq
v(\Delta_{\mathscr{C}_{i+1}})$ for any $0\leq i\leq d-1$ with strict
inequality when $i=0$.

Let

$y^{2}+(a_{1}+a_{3}x+...)y=ta_{0}+a_{2}x+a_{4}x^{2}+...,\quad a_{i}\in R$

be an affine equation of $\mathscr{C}$ with $p$ corresponding to $x=y=0$. As
$\mathscr{C}$ is regular at $p$, at least one of $a_{0},a_{1},a_{2}$ is a unit
in $R$. Using [6], Lemme 2, p. 4582, we verify that the equation

$y^{2}+(a_{1}+ta_{3}x_{1}+...)y=t(a_{0}+a_{2}x_{1}+ta_{4}x_{1}^{2}+...).$

defines an affine normal scheme, hence it is an affine equation of
$\mathscr{C}_{1}$ and we have

$v(\Delta_{\mathscr{C}_{1}})=v(\Delta_{\mathscr{C}})+2(2g+1)(g+1)>v(\Delta_{\mathscr{C}}).$

Suppose that $d\geq 2$. Let $p_{1}\in\mathscr{C}_{1}(k)$ be a point
corresponding to $x_{1}=0$. Then $\mathscr{C}_{1}$ is regular at $p_{1}$ if
either $a_{0}$ or $a_{1}\in R^{*}$ and we can then continue with induction. If
$a_{0},a_{1}\in tR$ and $a_{2}\in R^{*}$, then the above equation becomes

$y_{1}^{2}+(t^{-1}a_{1}+ta_{3}x_{2}+\cdots)y_{1}=t^{-1}a_{0}+a_{2}x_{2}+t^{2}a_{4}x_{2}^{2}+\cdots,$

where $y_{1}=y/t$. Again this defines a normal scheme, hence
$\mathscr{C}_{2}$. We have
$v(\Delta_{\mathscr{C}_{1}})=v(\Delta_{\mathscr{C}_{2}})+2(2g+1)(g-1)\geq
v(\Delta_{\mathscr{C}_{2}})$. As $\mathscr{C}_{2}$ is smooth at $x_{2}=0$, we
can also continue with induction if $d\geq 3$. ∎

######  Corollary 5.2.

Let $C$ be a hyperelliptic curve over $K$ and let $P\in C(K)$ be a Weierstrass
point. Then $(C,P)$ admits a unique minimal pointed Weierstrass model of
$(C,P)$ over $S$.

###### Proof.

The existence is obvious when $S$ is local. In the general case $C$ has good
reduction over a dense open subset $U$ of $S$. Then it is enough to glue local
minimal pointed Weierstrass models over the points of $S\setminus U$ with the
smooth model over $U$.

Now we prove the uniqueness. One can suppose $S$ is local. Let
$\mathscr{C},\mathscr{C}^{\prime}$ be two distinct pointed Weierstrass models
of $(C,P)$. We keep the notation preceding Lemma 5.1. Let $p_{0}$ be the
specialization of $P$ in $\mathscr{C}_{k}$. This is a smooth point by
definition. If $p_{0}=p$, then
$v(\Delta_{\mathscr{C}})<v(\Delta_{\mathscr{C}^{\prime}})$ by the same lemma.
Otherwise, $x(P)\notin tR$ (may be $\infty$). This implies that
$(x^{\prime})^{-1}(P)=t^{d}x(P)^{-1}\in tR$. Hence the specialization of $P$
in $\mathscr{Z}^{\prime}_{k}$ is the zero of $x^{\prime-1}=t^{d}x^{-1}$ and
Lemma 5.1 implies that
$v(\Delta_{\mathscr{C}^{\prime}})<v(\Delta_{\mathscr{C}})$. In other words,
two distinct pointed Weierstrass models always have distinct discriminant
ideals. Therefore the minimal one is unique. ∎

  * Remark 5.3

This corollary also follows from [8], Remark after Definition 2.1 (but whose
proof is left to readers). Note that the non-pointed minimal Weierstrass model
is not unique in general even when $C$ has a rational Weierstrass point.

######  Proposition 5.4 (Étale base change).

Let $(C,P)$ be a pointed hyperelliptic curve over $K$. Let $\mathscr{C}$ be
the minimal pointed Weierstrass model of $(C,P)$ over $S$. Let $S^{\prime}\to
S$ be an étale morphism of affine Dedekind schemes and let
$K^{\prime}=K(S^{\prime})$. Then $\mathscr{C}\times_{S}S^{\prime}$ is the
minimal pointed Weierstrass model of $(C_{K^{\prime}},P)$ over $S^{\prime}$.

###### Proof.

The property is local on $S^{\prime}$, so we can suppose that $S$ and
$S^{\prime}={\rm Spec}\kern 2.0ptR^{\prime}$ are local. The base change
$\mathscr{C}\times_{S}S^{\prime}$ is a normal scheme because $S^{\prime}/S$ is
étale. So $\mathscr{C}\times_{S}S^{\prime}$ is a pointed Weierstrass model of
$(C_{K^{\prime}},P)$.

(1) First suppose that the residue extension of $S^{\prime}/S$ is trivial.
Then $R$ is dense in $R^{\prime}$. Let $\mathscr{C}^{\prime}$ be a pointed
Weierstrass model over $S^{\prime}$. We will show that it is defined over $S$.
Let $\mathscr{Z}^{\prime}$ be its quotient by the hyperelliptic involution.
Let $x^{\prime}$ be a parameter of $\mathscr{Z}^{\prime}$ whose pole divisor
is the Zariski closure $\overline{\\{Q\\}}\subset\mathscr{Z}^{\prime}$, where
$Q$ is the image of $P$ in $\mathbb{P}^{1}_{K}(K)$. As $R$ is dense in
$R^{\prime}$, we can find $x\in K(\mathbb{P}^{1}_{K})$ such that
$x^{\prime}=x+a^{\prime}x^{\prime}+b^{\prime}$ with
$a^{\prime},b^{\prime}\in\mathfrak{m}_{R^{\prime}}$. Thus
$R^{\prime}[x^{\prime}]=R^{\prime}[x]=R[x]\otimes_{R}R^{\prime}$ and
$\mathscr{Z}^{\prime}=\mathscr{Z}\times_{S}S^{\prime}$ for some smooth model
$\mathscr{Z}$ of $\mathbb{P}^{1}_{K}$. The normalization $\mathscr{C}_{0}$ of
$\mathscr{Z}$ in $K(C)$ is a pointed Weierstrass model of $(C,P)$ over $S$ and
$\mathscr{C}_{0}\times_{S}S^{\prime}$ is equal to $\mathscr{C}^{\prime}$
because the former is normal. Therefore

$v(\Delta_{\mathscr{C}^{\prime}})=v(\Delta_{\mathscr{C}_{0}})\geq
v(\Delta_{\mathscr{C}})=v(\Delta_{\mathscr{C}\times_{S}S^{\prime}})$

and $\mathscr{C}\times_{S}S^{\prime}$ is minimal.

(2) In the general case, we can first replace $R$ with its henselization (the
first step shows that $\mathscr{C}$ commutes with henselization of $R$). The
integral closure of $R$ in the Galois closure of $K^{\prime}/K$ is a discrete
valuation ring unramified over $R$. So we can suppose that $S^{\prime}/S$ is
Galois with Galois group $\Gamma$. Let $\mathscr{C}^{\prime}$ be the minimal
pointed Weierstrass model of $(C_{K^{\prime}},P)$ over $S^{\prime}$. As $P$ is
invariant by $\Gamma$ and the minimal pointed Weierstrass model is unique,
$\Gamma$ acts on $\mathscr{C}^{\prime}$ (as $S$-scheme) and on the quotient of
$\mathscr{C}^{\prime}$ by the hyperelliptic involution. The quotient
$\mathscr{C}_{0}:=\mathscr{C}^{\prime}/\Gamma$ is a Weierstrass model of $C$
over $S$. As $S^{\prime}/S$ is étale, $\mathscr{C}_{0}\times_{S}S^{\prime}$ is
normal and $\mathscr{C}^{\prime}\to\mathscr{C}_{0}\times_{S}S^{\prime}$ is an
isomorphism. Hence $\mathscr{C}$ is smooth along the Zariski closure of $P$
and is a pointed Weierstrass model of $(C,P)$. Moreover,
$v(\Delta_{\mathscr{C}^{\prime}})=v(\Delta_{\mathscr{C}_{0}})\geq
v(\Delta_{\mathscr{C}})$. So
$\mathscr{C}^{\prime}\simeq\mathscr{C}\times_{S}S^{\prime}$. ∎

##  6\. Examples

We construct examples of Weierstrass models not defined by integral (or
pointed integral) Weierstrass equations, but with various nice properties. We
will work with number fields $K$. Let us summarize first the examples we will
construct.

######  Proposition 6.1.

There exist number fields $K$, hyperelliptic curves $C$ over $K$ with
Weierstrass model $\mathscr{C}$ over $S={\rm Spec}\kern 2.0pt\mathcal{O}_{K}$,
of any of the following types:

  1. (a)

$\mathscr{Z}\not\simeq\mathbb{P}^{1}_{S},\quad\Delta_{\mathscr{C}}\simeq\mathcal{O}_{K},\quad\det
f_{*}\omega_{\mathscr{C}/S}\simeq\mathcal{O}_{S};$

  2. (b)

$\mathscr{Z}\simeq\mathbb{P}^{1}_{S},\quad\Delta_{\mathscr{C}}\simeq\mathcal{O}_{K},\quad\det
f_{*}\omega_{\mathscr{C}/S}\not\simeq\mathcal{O}_{S};$

  3. (c)

$\mathscr{Z}\simeq\mathbb{P}^{1}_{S},\quad\Delta_{\mathscr{C}}\not\simeq\mathcal{O}_{K},\quad\det
f_{*}\omega_{\mathscr{C}/S}\simeq\mathcal{O}_{S};$

(by Proposition 2.6, in all the above situations, $\mathscr{C}$ is not defined
by an integral Weierstrass equation.)

  4. (d)

$\mathscr{Z}\simeq\mathbb{P}^{1}_{S},\quad\Delta_{\mathscr{C}}\simeq\mathcal{O}_{K},\quad\det
f_{*}\omega_{\mathscr{C}/S}\simeq\mathcal{O}_{S}$

but $\mathscr{C}$ is not defined by an integral Weierstrass equation;

  5. (e)

$\mathscr{C}$ is a minimal pointed Weierstrass model, defined by an integral
Weierstrass equation but not by a pointed integral Weierstrass equation.

  * Remark 6.2

The genus $g$ of the curves $C$ appearing in our construction is not
arbitrary. For (a) and for (b) there are infinitely many possible $g$ (always
odd). If we relax the condition $\mathscr{Z}\simeq\mathbb{P}^{1}_{S}$ in (b),
we find all odd integers $g\geq 5$ except those of the form $2^{n}-1$. For
(c), $g$ can be any integer different from $1,2$ and $4$. For (d), $g$ can be
any integer divisible by $4$. Finally, for (e), $g$ can be any integer $\geq
2$.

  * Example 6.3

Fix $g_{0}\geq 1$. We start with a hyperelliptic curve $C_{0}$ of genus
$g_{0}\geq 1$ over a number field $K$ with everywhere good reduction. Such a
curve always exists by enlarging $K$ if necessary.222For some special values
of $g_{0}$, we can give explicit examples as follows. Let $r\geq 3$ be an odd
square-free integer. Consider the hyperelliptic curve $C_{0}$ over
$\mathbb{Q}$ defined by $y^{2}+y=x^{r}$. If $\xi_{r}\in\mathbb{C}$ is a
primitive $r$-th root of the unity and if $m\in\mathbb{Z}$ satisfies
$4m\equiv-1\mod r^{(r+1)/2}$, then $C_{0}$ has everywhere good reduction over
$K=\mathbb{Q}[m^{1/r},\sqrt{1-\xi_{r}}]$. The corresponding smooth model
$\mathscr{C}_{0}$ is defined by an integral global Weierstrass equation, and
$(1-\xi_{r})^{-1}(x-m^{1/r})$ is a parameter of $\mathscr{Z}_{0}$. Let
$\mathscr{C}_{0}$ be its smooth projective model over $S={\rm Spec}\kern
2.0pt\mathcal{O}_{K}$, and let $\mathscr{Z}_{0}$ be the corresponding
projective bundle over $S$. Extending $K$ if necessary, we can suppose that
$\mathscr{Z}_{0}\simeq\mathbb{P}^{1}_{S}$. Let $B\subset\mathscr{Z}_{0}$ be
the branch locus of $\mathscr{C}_{0}\to\mathscr{Z}_{0}$. This is a closed
subset finite over $S$. By [4], Theorem 7.2, there exists an integral point
$D_{0}$ in $\mathscr{Z}_{0}\setminus B$. Applying again the theorem to
$\mathscr{Z}_{0}\setminus(B\cup D_{0})$, we find an integral point
$D_{\infty}$ such that $B$, $D_{0}$ and $D_{\infty}$ are pairwise disjoint.
Extending $K$ if necessary, we can suppose that $\mathscr{C}_{0}$ is defined
by an integral Weierstrass equation

$y^{2}+Q(x)y=P(x),$

and $D_{0},D_{\infty}$ are respectively the zero and pole divisors of $x$.
Write

$4P(x)+Q(x)^{2}=a_{0}x^{2g_{0}+2}+\cdots+a_{2g_{0}+2}\in\mathcal{O}_{K}[x].$

As $(D_{0}\cup D_{\infty})\cap B=\emptyset$, we have
$a_{0},a_{2g_{0}+2}\in\mathcal{O}_{K}^{*}$.

Extending again $K$, we can suppose that the class group $\mathrm{Pic}(S)$ of
$K$ contains a maximal cyclic $2$-subgroup $H_{2}\subseteq\mathrm{Pic}(S)$ of
order $\geq 4$ (see Lemma 6.5 below). Let $[I]\in H_{2}$ (it will be chosen
more precisely later). We will use $[I]$ to construct a hyperelliptic curve
$C$ over $K$ together with a Weierstrass model $\mathscr{C}$ as double
coverings of $\mathscr{C}_{0}$. Let $d$ be the order of $[I]$. It is a power
of $2$. As $I^{d}$ is principal, we identify $\mathbb{P}(\mathcal{O}_{K}\oplus
I^{d})$ with $\mathscr{Z}_{0}$, and the sections corresponding to the
projections of $\mathcal{O}_{K}\oplus I^{d}\to\mathcal{O}_{K}$ and
$\mathcal{O}_{K}\oplus I^{d}\to I^{d}$ respectively to $D_{0}$ and
$D_{\infty}$. Consider now $\mathscr{Z}:=\mathbb{P}(\mathcal{O}_{K}\oplus I)$
and the canonical morphism

$\theta:\ \mathbb{P}(\mathcal{O}_{K}\oplus
I)\to\mathbb{P}(\mathcal{O}_{K}\oplus I^{d})=\mathscr{Z}_{0}.$

Locally on $S$ it corresponds to taking the $d$-th powers of the homogeneous
coordinates. Let

$\mathscr{C}:=\mathscr{C}_{0}\times_{\mathscr{Z}_{0}}\mathscr{Z}.$

Let us study some properties of $\mathscr{C}$. Let $e$ be a basis of $I^{d}$.
For any open subset $U\subseteq S$ such that $I|_{U}$ has a basis $e_{U}$, we
have $e_{U}^{d}=\alpha_{U}^{-1}e$ with $\alpha_{U}\in\mathcal{O}_{S}(U)^{*}$ a
unit. Then $\mathscr{C}_{U}$ is defined by the equation

$y^{2}+Q(\alpha_{U}t^{d})y=P(\alpha_{U}t^{d}).$

By Lemma 6.6 below, this equation defines a hyperelliptic curve $C$ over $K$
of odd genus $g=d(g_{0}+1)-1$ and the discriminant ideal
$\Delta_{\mathscr{C}_{U}}$ defined by this equation satisfies

$\Delta_{\mathscr{C}_{U}}=\pm(a_{0}a_{2g_{0}+2})^{d-1}\alpha_{U}^{(2g_{0}+2)((2g_{0}+2)d-1)}d^{2g+2}\Delta^{d}_{(\mathscr{C}_{0})_{U}}=d^{2g+2}\mathcal{O}_{U}.$

Therefore $\Delta_{\mathscr{C}}=d^{2g+2}\mathcal{O}_{K}$ and $\mathscr{C}\to
S$ is smooth aways from primes dividing $2$. At any prime $\mathfrak{p}$ of
$\mathcal{O}_{K}$ dividing $2$, as $Q(x)$ is non-zero in $k(\mathfrak{p})[x]$,
the fiber $\mathscr{C}_{\mathfrak{p}}$ is reduced, so $\mathscr{C}$ is normal
([7], Lemma 4.1.18) and is a Weierstrass model of $C$ over $S$.

Let $D$ be the section of $\mathscr{Z}$ corresponding to the projection
$\mathcal{O}_{K}\oplus I\to I$. Then
$\det\phi_{*}\mathcal{O}_{\mathscr{Z}}(D)=I$ and $\theta(D)=D_{\infty}$. As
$\mathscr{C}$ is obtained by base change from $\mathscr{C}_{0}$, we have
$N\simeq\mathcal{O}_{K}$ (tensor product the exact sequence (5) corresponding
to $\mathscr{C}_{0}$ by $\mathcal{O}_{\mathscr{Z}}$). By Lemma 3.9(1), we have

$\det f_{*}\omega_{\mathscr{C}/S}\simeq I^{g(g+1)/2}\simeq
I^{d\times\frac{(d(g_{0}+1)-1)(g_{0}+1)}{2}},$

$[\mathfrak{w}]=[I]^{-d(g_{0}+1)/2}$ (and also
$\Delta_{\mathscr{C}}\simeq\mathcal{O}_{K}$, but this is less precise then the
above equality $\Delta_{\mathscr{C}}=d^{2g+2}\mathcal{O}_{K}$). Therefore:

    1. (a)

if we take for $[I]$ a generator of $H_{2}$ and if $g_{0}$ is odd, then $[I]$
is not a square, so

$\mathscr{Z}\not\simeq\mathbb{P}^{1}_{S},\quad\Delta_{\mathscr{C}}\simeq\mathcal{O}_{K},\quad\det
f_{*}\omega_{\mathscr{C}/S}\simeq\mathcal{O}_{S},\quad[\mathfrak{w}]=1;$

    2. (b)

if $[I]$ is the square of some element of $H_{2}$ and if $g_{0}$ is even, then

$\mathscr{Z}\simeq\mathbb{P}^{1}_{S},\quad\Delta_{\mathscr{C}}\simeq\mathcal{O}_{K},\quad\det
f_{*}\omega_{\mathscr{C}/S}\not\simeq\mathcal{O}_{S},\quad[\mathfrak{w}]\neq
1$

  * Example 6.4

Let $g\geq 1$. By [1], there exists a quadratic field $K$ such that
$\mathrm{Cl}(K)$ contains a cyclic subgroup of order $4g^{2}$ (see the
beginning of the proof of Lemma 6.5 below). Let $S={\rm Spec}\kern
2.0pt\mathcal{O}_{K}$ and let $[\mathfrak{u}]\in\mathrm{Pic}(S)$. By
Proposition 4.5, there exists a pointed hyperelliptic curve $(C,P)$ over $K$
such that the pointed Weierstrass class of the minimal pointed Weierstrass
model $\mathscr{C}$ is equal to $[\mathfrak{u}]$.

    1. (c)

If $g\not\equiv 0\mod 4$, we choose $[\mathfrak{u}]$ of order $g^{2}$. Then
$\mathscr{Z}\simeq\mathbb{P}^{1}_{S}$, $\Delta_{\mathscr{C}}$ is not
principal, and $\det f_{*}\omega_{\mathscr{C}/S}$ is free (Proposition
4.3(1)).

    2. (d)

If $g\equiv 0\mod 4$ and if we choose $[\mathfrak{u}]$ of order $4g$, then
$\mathscr{Z}\simeq\mathbb{P}^{1}_{S}$, $\Delta_{\mathscr{C}}$ is principal and
$\det f_{*}\omega_{\mathscr{C}/S}$ is free. Let $[\mathfrak{w}]$ be the
Weierstrass class of $\mathscr{C}$. Then $[\mathfrak{w}]=[\mathfrak{u}]^{2g}$
(Definition 4) is non-trivial. So $\mathscr{C}$ is not defined by an integral
Weierstrass equation (Proposition 3.4).

    3. (e)

If $g\geq 2$, choose $[\mathfrak{u}]$ of order $g$. Then $[\mathfrak{w}]$ is
trivial. As $\mathscr{Z}\simeq\mathbb{P}^{1}_{S}$, the model $\mathscr{C}$ is
defined by an integral Weierstrass equation by Proposition 3.4. But it is not
defined by a pointed integral Weierstrass equation.

Note that in all the above examples except (c), $\Delta_{\mathscr{C}}$ is
principal but not equal to $\mathcal{O}_{K}$ in general.

######  Lemma 6.5.

Let $K$ be a number field. Let $d\geq 2$. Then there exists an extension $L/K$
(of degree at most $2$) such that the class group of $L$ contains a cyclic
subgroup of order $d$.

###### Proof.

For any $n\geq 1$, in the proof of [1], Theorem 1, explicit quadratic
extensions $F/\mathbb{Q}$ whose class groups contain a cyclic subgroup $C_{n}$
of order $n$ are constructed. Fix such an $F$ with $n=d[K:\mathbb{Q}]$. Let
$L=KF$. Using the norm map $\mathrm{Cl}(L)\to\mathrm{Cl}(F)$, we see that the
kernel of the canonical map $\mathrm{Cl}(F)\to\mathrm{Cl}(L)$ is annihilated
by $[K:\mathbb{Q}]$, so the image of $C_{n}$ in $\mathrm{Cl}(L)$ has order
divisible by $d$ and we are done. ∎

######  Lemma 6.6.

Let

$E_{0}:\quad y^{2}+Q(x)y=P(x)$

be a Weierstrass equation of a hyperelliptic curve $C_{0}$ of genus $g_{0}$
over some field $K$ with $4P(x)+Q(x)^{2}=a_{0}x^{2g_{0}+2}+\dots+a_{2g_{0}+2}$
and $a_{0}a_{2g_{0}+2}\neq 0$ (the latter condition means that the covering
$C_{0}\to\mathbb{P}^{1}_{K}$ is unramified above the zero and pole of $x$).
Let $\alpha\in K^{*}$ and let $d\geq 1$ be an integer prime to the
characteristic of $K$. Then

$E_{d}:\quad y^{2}+Q(\alpha x^{d})y=P(\alpha x^{d})$

is a Weierstrass equation of a hyperelliptic curve of genus $g=d(g_{0}+1)-1$,
and the discriminants of the above equations satisfy the relation

$\Delta(E_{d})=\pm(a_{0}a_{2g_{0}+2})^{d-1}\alpha^{(2g_{0}+2)((2g_{0}+2)d-1)}d^{2g+2}\Delta(E_{0})^{d}.$

###### Proof.

(See [6], §2, for the definition of $\Delta(E)$.) This amounts to proving the
following fact: if $F(x)=a_{0}x^{n}+\cdots+a_{n}\in K[x]$ has degree $n$ and
$\alpha\in K^{*}$, then

$\mathrm{disc}(F(\alpha
x^{d}))=\pm(a_{0}a_{n})^{d-1}\alpha^{n(nd-1)}d^{nd}\mathrm{disc}(F(x))^{d}.$

First by substituting $\alpha^{1/d}x$ to $x$, we are reduced to the case
$\alpha=1$. A direct computation of $\mathrm{disc}(F(x^{d}))$ in terms of the
product of the differences of the roots of $F(x^{d})$ in some algebraic
closure of $K$ leads to the desired equality. ∎

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