# EXAMPLES OF EFFECTIVITY FOR INTEGRAL POINTS ON CERTAIN CURVES OF GENUS 2

Pietro Corvaja, Umberto Zannier

##  1 Introduction

This short article concerns a method to obtain effectivity for the search of
integral points on certain (sets of) curves of genus $2$.

As a preamble, it will be useful to mention that this draft is intended merely
to be a brief and very sketchy exposition of the said method, actually
limiting the statements to a few examples. Not only we shall not give complete
proofs, but often we shall omit even partial arguments. We plan to prepare in
a near future a paper with more extensive statements and complete detail, also
in collaboration with Davide Lombardo, who already performed several
verifications used in this draft, and whom we thank.

However we hope that the present exposition at least will indicate the main
lines of our method and the type of results that can be so obtained.

More precisely, we wish to illustrate just an example of application of a
criterion of Bilu [4], to derive effectivity for integral points on certain
sets of affine curves $X/\overline{{\mathbb{Q}}}$ of genus $2$, with smooth
projective model $\tilde{X}$, so that $X=\tilde{X}-\\{q_{0}\\}$, $q_{0}$ being
some (non-special) point on $\tilde{X}$. As said, future work will contain
details and more general applications (however of the same shape, in
particular again with curves of genus $2$).

The results will be far from giving effectivity for all such curves $X$.
However they shall apply to a set of curves (and points $q_{0}$) which, for
instance, is topologically complex-dense in the whole moduli space for such
data. (See Theorem 1.1 below for a flavour of the statements that we can
obtain. Also, see in particular the Notes at the end of §2.1 and at the end of
§3.1.)

We shall construct morphisms from these curves to ${{\mathbb{G}}}_{\rm
m}^{2}$, with image of increasing degrees. We note that as the degree
increases, we may say that the examples become ‘more interesting’, since they
cannot be derived by substitution from a universal family. As a counterpart,
there is the negative feature in that the relevant curves will somewhat have
increasing fields of definition.

On the other hand, to our knowledge this type of result did not previously
appear in the literature. Actually, the effective known cases of this type
seem to be sporadic; for instance, Serre’s book [13] explicitly inquires about
possible effectiveness for curves of this shape (see the remarks at the end of
p. 116). In fact, this shape is somewhat significant, being in a sense the
simplest for which effectivity is not generally known.111Indeed these may be
realized as affine quartic curves of genus 2 with one point at infinity,
whereas effectivity is known for all affine curves of genus $\leq 1$, in
particular for cubics.

Further, we do not know whether the mentioned kind of limitation is intrinsic,
e.g. of the method. In this direction, we point out that recent work of
Landesman-Poonen [10] indeed suggests that Bilu’s criterion is subject to
strong obstructions.222We note that the results in [10] refer to curves over
${{\mathbb{C}}}$. In principle by some kind of specialization pattern one is
led to expect that the same should hold over $\overline{{\mathbb{Q}}}$;
however some care is needed in taking for granted such a type of inference, as
is shown for instance by remarkable examples like Belyi’s Theorem.

We recall that any curve $\tilde{X}$ as above of genus $2$ may be obtained by
adding two points at infinity to an affine curve defined by an equation
$y^{2}=f(x)$, where $f$ is a polynomial of degree $6$ without multiple roots.
This exhibits $\tilde{X}$ as a double cover of ${{\mathbb{P}}}_{1}$ ramified
at $6$ points (called special points or Weierstrass points). The datum of
these points up to action of $PGL_{2}$ determines $\tilde{X}$ up to
isomorphism.333Here and in the sequel we shall work with
$\overline{{\mathbb{Q}}}$ as ground field. (There are fairly complicated Igusa
invariants for the resulting quotient space.)

There is a natural (unique) involution on $\tilde{X}$ with quotient
${{\mathbb{P}}}_{1}$, defined by $(x,y)\mapsto(x,-y)$, and this fixes the six
said special points. If $q_{0}$ is a special point then $X$ has a model in
${{\mathbb{A}}}^{2}$ given by an equation $y^{2}=f^{*}(u)$ where $f^{*}$ has
degree $5$, and then the integral points may be found effectively with well-
known methods. 444The criterion of Bilu’s that we shall use may be read as
including these cases, however with a construction of a cover which goes back
to Siegel and differs from the one that we shall introduce. The same
effectivity is possible for an affine curve defined in ${{\mathbb{A}}}^{2}$ by
$y^{2}=f(x)$, with $\deg f=6$ (actually any $f$), however in this last affine
model we are removing two points from $\tilde{X}$, moreover related by the
said involution.

Instead, our examples will involve a single non-special point $q_{0}$.

The (complex) algebraic curves of genus $2$ are parametrized by a quasi-
projective variety denoted usually $M_{2}$. This is a rational variety of
dimension $3$. In the sequel the terminology ‘moduli space for curves of genus
$2$’ will refer to some quasi-projective variety in birational correspondence
with $M_{2}$. So generically a point of such variety will represent a curve of
genus $2$ and every curve of genus $2$, with the exception of a family at most
of dimension $2$, will correspond to such a point. We shall also refer to
$M_{2,1}$ as a quasi-projective variety parametrizing the pairs $(X,p)$ where
$X$ is a curve of genus $2$ and $p$ is a point in $X$; it has dimension $4$.

An instance of the results is represented by the following

######  Theorem 1.1.

Let $S$ be a finite cover of a dense Z.-open subspace of a moduli space for
curves of genus $2$. Let $\tilde{X}_{s}$ be the curve corresponding to $s\in
S$. Suppose we are given a prescribed (non-special) point
$q_{s}\in\tilde{X}_{s}$ varying rationally as $s\in S$. Then there is a
complex-dense subset $\Sigma\subset S$ of algebraic points on $S$ such that
for each $s\in\Sigma$ the integral points on $\tilde{X}_{s}-q_{s}$ are
effectively computable, over EVERY number field.

NOTE: (i) The set $\Sigma$ itself is ‘effective’, so to say, for instance in
the sense that we may compute some element of $\Sigma$ in any prescribed open
disk in $S$. And we may establish whether a given algebraic point of $S$
belongs to $\Sigma$. The set $\Sigma$ is described by denumerably many
explicit algebraic equations. The nature of these equations will be clearer
from the arguments. We may say that the points of $\Sigma$ correspond to
points of $S$ where a certain section of an abelian scheme on (a finite cover
$S^{\prime}$ of) $S$, assumes values which are torsion in the appropriate
fiber.

(ii) We can see $S$ as a finite-degree cover of an open affine subvariety of
$M_{2,1}$ whose projection to $M_{2}$ is of finite degree.

We again stress that the present draft will not contain a fully detailed proof
of this theorem; however we shall give many hints on it. A more complete
treatment will follow in §3, where we consider only a pencil of curves inside
the moduli space $S$. There we shall also indicate other types of results
which are in the range of this method.

Before explaining the principles of our method, we pause to give some general
considerations, to which we shall refer in the sequel.

###  1.1 Some explicit examples

Each of the mentioned curves may be also seen as a quartic in
${{\mathbb{A}}}^{2}$, with a single (smooth) point at infinity and precisely
one singular point (see below for a very simple argument). This shape phrases
the problem in classical diophantine terms. Let us see some examples of such
quartics.

Consider the family, of genus $2$, given by equations

$y^{4}+ay^{2}-xy-x^{3}+bx^{2}=0,$

with parameters $a,b$. This is found to be (generically) singular (only) at
the origin, and to have a single (smooth) point at infinity, in a closure in
${{\mathbb{P}}}_{2}$. The genus is $2$ by the formula $g=(4-1)(4-2)/2$ for a
smooth quartic, but the singular point subtracts $1$.

We will treat this family in the mentioned forthcoming paper, performing the
necessary (computational) verifications so as to check that it satisfies the
same conclusion as in Theorem 1.1. But here we limit ourselves to point out a
subfamily capable of effective treatment (and another problem in diophantine
effectivity).

AN EFFECTIVE CASE: Once in a certain expository paper (by Masser-Zannier),
seeking maximum simplicity, the example $y^{4}-axy-x^{3}=0$ was proposed. But
the authors missed to note that the existence of an automorphism of order $5$
($x\to\theta x,y\to\theta^{2}y$) leads to the affine model $y^{5}=w(w+a)^{2}$
where $w=x^{2}/y$. This last of course falls into the usual treatments.
(Bogdan Grechuk pointed out this case to Masser and Zannier, with a different
and much longer series of substitutions.)

This very peculiar feature will not happen for the examples below (and it does
not happen for the general equation displayed above), as is not difficult to
check.

ABOUT TRINOMIALS. The case of trinomial equations appearing in the last
paragraphs, actually is always effectively solvable. Over the usual ring of
integers ${{\mathbb{Z}}}$, this was observed and proved by B. Grechuk, T.
Grechuk, A. Wilcox [9], relying on Runge’s theorem, on Baker’s results on
superelliptic equations, after a rather long argument with comparison of
factorisations, and distinction into cases. In particular, due to the use of
Runge, their proof, as it stands, does not work over general number fields but
only over ${{\mathbb{Z}}}$ (and even the rest of the argument does not take
nontrivial units into account).

Such proof in practice exploits the Newton polygon, so as to express $x,y$
multiplicatively in terms of two other functions. A shorter, different, and
completely general, proof goes as follows. Let $x^{n}+ax^{r}y^{s}+by^{m}=0$ be
the equation, defining a curve $Z$, where the exponents are $\geq 0$ and
$a,b\in k^{*}$. Put $\Delta=|nm-rm-sn|$. If $\Delta=0$ then $u:=x^{n}/y^{m}$
and $v:=x^{r}/y^{m-s}$ are multiplicatively dependent and the equation
$u+av+b=0$ shows that the curve $Z$ is the union of translates of tori in
${{\mathbb{G}}}_{m}^{2}$. If $\Delta\neq 0$, the curve is irreducible, as can
be seen by the substitution $x\mapsto t^{m}x,y\mapsto t^{n}y$ which transforms
the equation into a binomial in $t$. 555 See Schinzel’s book [12] for a
complete and subtle theory of reducibility of trinomials over function fields,
not needed in the simple case considered here. There is an abelian-group
action on the function field $\bar{k}(Z)$ of $Z$, defined as follows. Let
$\theta,\eta$ be roots of unity of order (dividing) $\Delta$ such that
$\theta^{n}=\theta^{r}\eta^{s}=\eta^{m}$; then we have an action of the group
of such pairs $(\theta,\eta)$, defined by $(x,y)\mapsto(\theta x,\eta y)$,
well-defined on $\bar{k}(Z)$. If $\theta=\zeta^{p},\eta=\zeta^{q}$ for a fixed
primitive $\Delta$-th root of unity $\zeta$, then this corresponds to $pn-
qm\equiv pr-q(m-s)\equiv 0\pmod{\Delta}$. Hence the group corresponds to
integer vectors $(p,q)$ satisfying these congruences, which is a subgroup of
${{\mathbb{Z}}}^{2}$ of index $\Delta$, modulo $\Delta{{\mathbb{Z}}}^{2}$. The
group leaves invariant the functions $u,v$. Actually, the field of invariants
is generated by $u$ and $v$ (because the lattice generated by $(n,m)$,
$(r,m-s)$ has index $\Delta$ in ${{\mathbb{Z}}}^{2}$). Since $u,v$ are
linearly related, the field of invariants corresponds to a curve of genus $0$.
Also, the group action induces an action on the points of a smooth complete
model $Z^{\prime}$ of $Z$, which leaves invariant the set of poles of $x$.
Hence the affine curve for which we seek the integral points may be seen as a
fiber product of two cyclic covers of the affine line (possibly deprived of
some points), and thus the known methods allow to determine the integral
points, if their set is finite, or to parametrize them. It turns out that
there may be infinitely many ones only if the group above is cyclic (which
happens if an only if $\gcd(m,n,r,s)=1$). 666The case of general Galois covers
of the affine line is less direct than the special case appearing here. It
still may be treated effectively, as in independent results by Bilu [4] and
Dvornicich-Zannier [8]. Both of these results in fact appeared in previous
publications, very difficult to trace back nowadays. Another independent
argument had been given by Poulakis. Serre interpreted all of these arguments
within a short proof, in a letter to Bertrand of 4 June 1993. See also [14]
for full and self-contained detail for the general case of genus $0$.

HYPERELLIPTIC MODEL: Intersecting the quartic with lines through the singular
point one finds the hyperelliptic genus $2$ involution:

$(x,y)\mapsto(2{x^{4}\over y^{4}}-x,2{x^{3}\over y^{3}}-y),$

whence we have the usual sextic model

$\mu^{2}=a\lambda^{6}+\lambda^{5}+b\lambda^{4}-1,\qquad\lambda={y\over
x},\quad\mu=\lambda^{-4}-x.$

Of course this is to be interpreted as defining the function field of the
relevant curves; indeed, the affine model with such equation has two points at
infinity.

###  1.2 Perfect-powers values of algebraic functions

We pause to point out a connection of the problem of effective integral points
on such curves of genus $2$, with another natural problem in diophantine
theory.

PROBLEM. Let $\pi:Y\to{{\mathbb{A}}}^{1}$ be a (possibly ramified) cover of
(affine) curves over a number field $k$, and suppose we are given a rational
function $f\in k(Y)$. The problem is to find effectively the integral (or
rational) points $m$ on ${{\mathbb{A}}}^{1}$ such that they lift to an
algebraic point $\xi=\xi_{m}\in Y(\bar{k})$ with the property that $f(\xi)$ is
a perfect $p$-th power in $k(\xi)$.

This integer $p\geq 2$ may be fixed or variable, and the function $f$ is
supposed not to be a perfect power in $k(Y)$.

NOTE: Let $Y={{\mathbb{A}}}^{1}$ and $\pi$ be the identity. Now $f\in k(x)$,
and we have $\xi=m\in k$, so we are seeking the $p$-th power values (over $k$)
of a usual rational function. These integral points (i.e. such values at
integers of $k$) may be found effectively (by Baker, as applied by Schinzel-
Tijdeman), even letting $p$ be variable.

In general, we can take the norm of $f$ down to $k(x)$ and obtain a rational
function $F=F(x)$ on the affine line. Then we can often apply Schinzel-
Tijdeman (see for instance the paper by Berczes, Evertse and Gyory [3]). But a
problem arises if $F$ is itself identically a perfect power (for instance a
constant) even if $f$ is not.

For any fixed $p$ the issue may be dealt with through Siegel’s theorem. In
fact, if $f$ is not (identically) a $p$-th power of a rational function on
$Y$, the whole graph of fields (going to a normal closure of
$k(Y)(f^{1/p})/k(x)$) degenerates under specialisation at $\xi_{m}$, and we
have an integral point on a certain curve (a quotient of the normal closure by
a suitable decomposition group), to which Siegel’s theorem may be applied, and
we can decide effectively whether there are infinitely many integral
points.777It seems that for $p>2$ we always have finiteness. Certainly for
$p=2$ there are cases with infinitely many integral points, which can be found
through equations of Pell type. But, motivated by the positive results for
powers, especially by the one of Schinzel-Tijdeman, we still may ask:

QUESTION: is this problem completely effective ? Can we find these integral
points when their set is finite ?

CUBIC ROOTS. There is a link of (a special case of) this problem (case $p=3$)
with the above problem of curves of genus $2$. Indeed, we may see the above
equation as a cubic in $x$ and solve with Cardan’s formulae. For instance take
$b=0$ in the above family so we have an equivalent equation

$x^{3}+xy+y^{4}+ay^{2}=0.$

The formulae give, up to a constant, cubic roots of
$-y^{4}-ay^{2}\pm\sqrt{(y^{4}+ay^{2})^{2}+(4/27)y^{3}}$ so of shape

$\root 3 \of{-y^{4}-ay^{2}\pm y\sqrt{(y^{3}+ay)^{2}+(4/27)y}}.$

Corresponding to an integral point $x=r,y=s$ on the original curve, the
expression for which we have to take a cubic root would produce a value in
${{\mathbb{Q}}}(\sqrt{(s^{3}+as)^{2}+(4/27)s})$.

Then, the issue is that we do not know, and is not necessarily true, that the
square root $\sqrt{(s^{3}+as)^{2}+(4/27)s}$ produces a rational value, i.e.
that $(s^{3}+as)^{2}+(4/27)s$ is a square in ${{\mathbb{Q}}}$. Indeed, this is
a stronger constraint and we could effectively find the integers $s$ for which
this is so.

Note that the norm of the relevant function is a perfect cube, so we fall in
the exceptional cases said above, where the known theory does not provide an
effective answer. These formulae provide an explicit link of this problem with
effectivity for integral points on a curve of genus $2$.

##  2 Our method - Curves of genus $2$

We shall use a criterion of Bilu [4]: 888We apologize for the clash of the
notation $S$ both for our base space and a finite set of places; however the
latter notation is so common that we prefer to forget about this, which is not
likely to cause any confusion.

Bilu’s criterion: If the group of regular morphisms $X\to{{\mathbb{G}}}_{m}$
has rank $\geq 2$ then finding the $S$-integral points on $X$ is effective
over any number field and any finite set $S$ of places.

Further phrasing:

If the group generated in $J_{X}$ by the differences of points at infinity
(i.e. points in $\tilde{X}-X$) has co-rank $\geq 2$, then there is effectivity
for integral points.

There are still other equivalent versions.

This is based on the following result:

The equation $f(x,y)=0$ in $S$-units $x,y$ is effective.

Proof of this: linear forms in logs + Puiseux series (see Bilu’s papers or
[6], Ch. 5, for such an argument avoiding even an appeal to Puiseux theorem).

As Bilu did in some cases (e.g. for modular curves, [4], [5]), one may apply
the criterion after taking a cover unramified outside the points at infinity
(using the Chevalley-Weil Theorem). Then, as formulated explicitly by Bilu, a
sufficient condition for effectivity becomes:

Bilu’s improved statement: If $X$ has an unramified cover $Y$ such that the
differences of the lifted points at infinity in $\tilde{Y}$ have co-rank $\geq
2$ in $Jac(Y)$, then $X({{\mathcal{O}}}_{S})$ is effectively computable (and
actually so is $Y({{\mathcal{O}}}_{S})$).

SIMPLEST CASE OF OUR METHOD

As above, the curve $\tilde{X}$ may be realized as a smooth complete model
with function field given by $v^{2}=f(u)$ where $f$ has degree $6$.

We define $X=\tilde{X}-\\{q_{0}\\}$ for a (non-special) point
$q_{0}\in\tilde{X}$, to be chosen.

We let $J_{X}$ be the Jacobian of $\tilde{X}$.

We then construct an unramified cover $\pi:\tilde{Y}\to\tilde{X}$ cyclic of
degree $3$, with points $p_{1},p_{2},p_{3}$ above $\infty=q_{0}\in X$.999For
Bilu’s criterion it would suffice that $\pi$ is unramified except above
$q_{0}$. However we forget about this freedom in the present approach. The
paper [10] shows that this freedom ‘generically’ is not a significant
improvement.

If the Galois group of $\pi$ is generated by $g$, of order $3$, we may label
the points modulo $3$ so that $p_{i+1}=g(p_{i})=:p_{i}^{g}$.

This cover $\tilde{Y}$ corresponds to taking a cube root of a rational
function on $\tilde{X}$ 101010We suppose that the ground field contains a
cubic root of $1$, actually let us take freely a finite extension of the
ground field in the whole approach.; considering its divisor and the fact that
the cover is unramified, a simple analysis shows that this function
corresponds to a point of order $3$ in $J_{X}$, and there are $80$ (though
half of them yield the same covers). We omit here the relevant verifications,
which will appear in a joint paper also with Lombardo.

Applying Bilu’s criterion: In order to apply Bilu’s criterion, we want that
the $p_{i}-p_{j}$ are torsion in $J_{Y}$.

NOTE: A priori this is too difficult to achieve; in fact, $J_{Y}$ has
dimension $4$, whereas the whole space of curves of genus $2$ has dimension
$3$. The point at infinity yields one more freedom, but still a priori not
sufficient to achieve (many) examples. 111111In fact, see the paper [10] for
grasping the limited hopes of such search.

However there are three peculiar features in our situation, which give more
freedom, which we now illustrate.

A. First of all, $J_{Y}\approx J_{X}\times J^{\prime}$ (i.e. up to isogeny),
since there is a surjective homomorphsm $J_{Y}\to J_{X}$, so $J^{\prime}$ may
be taken as the kernel.

Note that $\pi\circ g=\pi$, so the differences $p_{i}-p_{j}$ map to $0$ on
$J_{X}$, so their classes in $J_{Y}$ (indicated with the same letter here)
belong to $J^{\prime}$, which has dimension $2$. This is already a big gain.

B. These differences (as classes in $J_{Y}$) are found to be not generically
torsion; note however, that if they were, we could apply Bilu’s criterion at
once to the whole set of curves.

Using the cyclic structure it follows that $p_{3}-p_{1}$ and $p_{2}-p_{1}$ are
generically independent over ${{\mathbb{Z}}}$ (i.e. as $s$ and the curve
$\tilde{X}_{s}$ vary). Indeed any linear equivalence relation
$ap_{1}+bp_{2}+cp_{3}\sim 0$, $a,b,c\in{{\mathbb{Z}}}$ with $a+b+c=0$, implies
$ap_{2}+bp_{3}+cp_{1}\sim ap_{3}+bp_{1}+cp_{2}\sim 0$, and it follows that
$ac(p_{2}-p_{1})\sim ab(p_{1}-p_{2})\sim 0$, so the differences would be
torsion.

However they are linearly dependent over $End(J_{Y})$; indeed, $J_{Y}$ too has
an automorphism $g$ of order $3$ coming from the one of $\tilde{Y}$, and
$\\{p_{1}-p_{2},p_{2}-p_{3},p_{3}-p_{1}\\}$ is an orbit, so $p_{3}-p_{2}$ is
the image $g(p_{2}-p_{1})$ of $p_{2}-p_{1}$ (for the previous labelling of the
$p_{i}$).

Hence if one of the differences is torsion, any other one is also torsion.
Therefore it suffices to make one of them torsion to achieve Bilu’s conditions
(in fact $p_{2}-p_{1},p_{3}-p_{1}$ are plainly independent over
${{\mathbb{Z}}}$ as divisors). This is a further gain and (a priori) now we
have 4 or 3 freedoms (that is, depending on whether we vary $q_{0}$ or not)
against only 2 constraints (i.e., the dimension of $J^{\prime}$).

We go ahead by describing a further pleasant feature, which will be exploited
in §3 below.

C. We gain even more simplicity and freedom by the following reason.

It is not difficult to see that the hyperelliptic involution on $\tilde{X}$
lifts to $\tilde{Y}$. Actually it lifts in three ways, conjugate under the
Galois group of the cover. Then $\tilde{Y}$ becomes a Galois (ramified) cover
of ${{\mathbb{P}}}_{1}$ with Galois group $S_{3}$. The curve $\tilde{X}$ is
the quotient of $\tilde{Y}$ by the subgroup of order $3$ in $S_{3}$.

Any of the three quotients of $\tilde{Y}$ by a transposition is found to be a
curve of genus $1$, and these curves are isomorphic because the three
transpositions are conjugate in $S_{3}$. This yields three morphisms from
$J^{\prime}$ to the cube of an elliptic curve $E$; if $\phi$ is one of them,
the other ones are $\phi\circ g,\phi\circ g^{2}$ for a $g$ of order $3$. The
sum of the three maps from $J^{\prime}$ is easily seen to be zero, so we
remain with a morphism of $J^{\prime}$ to a square $E^{2}$. A little
elementary Galois cohomology (by the way for $S_{3}$) yields that
$J^{\prime}\approx E^{2}$ (i.e. up to isogeny). This can be seen also by
checking that two of the maps are independent; indeed if they were not, then
acting with $S_{3}$ we would find that each map is constant, which is absurd.

————–

NOTE: Not all these curves (that is, for other choices of the point of order
$3$) are isomorphic, in fact their $j$-invariants are found to assume $40$
values. We owe these calculations to Lombardo.

QUESTION: are they all isogenous ?

We haven’t checked this. Maybe it is easy to see an answer. In any case, this
does not matter for our application.

NOTE: After this note was written Lombardo has found a negative answer
(details to appear in the said forthcoming paper). He also showed us that in
his joint paper with Garcia, Rizenthaler and Sijsling [11] the decomposition,
up to isogeny, of the jacobian of $\tilde{Y}$ as $J_{X}\times E^{2}$ appears
as a special case of their work on decompositions of jacobians via Galois
covers (see in particular Table 7 therein). ————–

This isogeny decomposition $J_{Y}\approx J_{X}\times E^{2}$, hence
$J^{\prime}\approx E^{2}$, yields further freedom, since we may seek for
instance a section which is torsion identically on one elliptic factor, or
such that its two elliptic components are identically linearly dependent, and
then we have only to prescribe torsion on one of the two factors, which is a
weaker constraint, allowing to lower the dimension of the base. Indeed we
shall later give an example of a 1-dimensional family, based on exactly this
type of construction.

###  2.1 Hints for a proof of Theorem 1.1

Let $S$ be as in the statement, and let us be given a non-special algebraic
section $s\mapsto q_{s}$. So for each smooth complete curve $\tilde{X}$ of
genus $2$ (except possibly for the curves corresponding to finitely many
surfaces in $M_{2}$), we have a finite non-empty set (with uniformly bounded
cardinality) of $s\in S$ such that $\tilde{X}_{s}$ is isomorphic to
$\tilde{X}$, and we have a (generically non-special) point
$q_{s}\in\tilde{X}_{s}$, where $q_{s}$ is given with coordinates which are
rational functions on $S$. (For our purposes here, it suffices to use such
naive notions of moduli space.)

For $s\in S$ we may construct as above a cover $\tilde{Y}_{s}$ of
$\tilde{X}_{s}$, after a choice of a point of order $3$ on $J_{X_{s}}$. This
$\tilde{Y}_{s}$ in fact which will be defined over a suitable finite cover of
$S$ (depending for instance on the choice of such point of order $3$).
121212Explicit formulae can be computed, and part of them will appear below.
Lombardo has performed complete computations, which will appear in a joint
paper with the present authors. The Jacobians of the curves $\tilde{Y}_{s}$
determine an abelian scheme over this cover of $S$.

Replacing $S$ by a further finite cover (and denoting it again by $S$ for
convenience), we obtain two sections from $S$ to this abelian scheme, i.e. the
classes of the differences $p_{2}-p_{1}$, $p_{3}-p_{1}$ (where
$p_{i}=p_{is}$), viewed as points of $J_{Y_{s}}$. As noted above, these
sections are dependent over $End(J_{Y_{s}})$. If they would be dependent over
${{\mathbb{Z}}}$ we would have a nontrivial linear equivalence relation
$a_{1}p_{1}+a_{2}p_{2}+a_{3}p_{3}\sim 0$ with $a_{1}+a_{2}+a_{3}=0$. Acting
with the group $<g>$ the $a_{i}$ get cyclically permuted and we eventually
find that the class $p_{2}-p_{1}$ is identically torsion on $S$. If this were
the case then we could apply Bilu criterion to all members of the family,
leading to a stronger theorem. So let us suppose this is not the case.131313It
is however possible to exclude that this happens on the whole $S$, no matter
the choice of the section. It follows from a general theorem in [7] that this
is true except for a finite number of sections $q_{s}$. However below in §3.2
we shall sketch a proof of the stronger assertion.

Now, even after we have prescribed the family of points at infinity $q_{s}$
(depending on $\tilde{X}$), losing in this way one freedom, we are left with a
base space $S$ of dimension $3$. Then by the theory of the Betti map we can
make one of the sections torsion in a dense set $\Sigma$, so the other section
will be torsion as well by remark (B) above. The former kind of verification
may be generally not easy to achieve (see e.g. the paper [2]), but here the
situation is simpler, since the relevant abelian varieties are (isogenous to)
squares of elliptic curves, by comment (C) just above. This case is easier and
treated in particular in [7]. But we need not even appeal to that paper, the
main points of the argument being as follows:

Sketch of the argument with the Betti map: Suppose we have a doubly elliptic
scheme $E^{2}/Z$ without constant part (i.e., the curve $E$ has not constant
$j$-invariant as a function on $Z$), over a(n affine) surface $Z$, with a
section $\sigma$. Then we have to prove that the points $z\in Z$ when
$\sigma(z)$ is torsion are complex-dense in $S$. Now, writing
$\sigma=(\sigma_{1},\sigma_{2})$, by the elliptic case (which is not difficult
and treated e.g. in [15], see Notes to Ch. 3), there are a complex-dense set
of algebraic curves in $Z$ such that $\sigma_{1}$ is identically torsion on
each of these curves. (In fact, it suffices first to prove this on a ‘general’
curve $C$ in $Z$, which is a known case. Then each point of $C$ where
$\sigma_{1}$ is torsion will yield a full algebraic curve in $Z$ on which
$\sigma_{1}$ will be torsion of the same order.) In turn, for each of these
curves, by the same principle, there is a complex-dense set of points where
$\sigma_{2}$ is torsion, concluding the argument.

NOTES. (i) The torsion orders so obtained will grow to infinity, unless the
section is identically torsion (on the whole base $S$); this last possibility
shall be excluded in §3.2 below (see Proposition 3.1).

(ii) If the curve $E$ is isotrivial, so may be assumed constant, the assertion
is not generally true but remains true (with a very easy proof) provided only
that the section is not (a non-torsion) constant. That this case cannot happen
in our case follows from the fact that the elliptic family coming from the
construction has not constant $j$-invariant on the whole base $M_{2}$ (as
shown by an explicit computation of Lombardo, appearing in a future joint
paper and not inserted here). See also the final considerations at the end of
this article. For the purposes of the construction in §3 below, involving a
pencil, this is immaterial since the section will shown explicitly not to be
constant. (See the step S8 in the construction below, for this specific
issue.)

So the set $\Sigma$ mentioned in the theorem in practice is the set of points
of $S$ where the differences $p_{i}-p_{j}$ become torsion in $Jac(Y_{s})$.

##  3 A pencil with infinitely many effective cases

We give an explicit construction by a direct geometrical process. We start,
for instance, with the Fermat curve of genus $1$ given by
${\mathcal{F}}:u^{3}+v^{3}=1$. In this pencil the relevant elliptic curves
will be all isomorphic to ${\mathcal{F}}$.

We proceed according to the following steps.

S1. We project $E$ from a variable point $(0:\alpha:1)$ (lying on the line
$T=0$ at infinity in homogeneous coordinates $(T:U:V)$) to the line $v=0$,
i.e. the $u$-axis.

S2. There are $6$ branch points for this projection (by Hurwitz), and a
calculation shows that they are given by $v=0$ where

$v^{2}=(\alpha^{3}+1)u^{6}-2u^{3}+1,$

is the equation of a curve of genus $2$, which will define our
$\tilde{X}_{\alpha}$ (where we disregard $\alpha^{3}=-1$). At the moment we
have a scheme over the affine $\alpha$-line.

It may be shown that there are no other automorphisms of $\tilde{X}$ than the
obvious ones (we mean generically in $\alpha$, for instance there is a
nontrivial one for $\alpha=0$), so that we do not fall neither into a Thue nor
in a hyperelliptic case when we take $X$ as $\tilde{X}$ minus a single non-
special point. Indeed, the issue of automorphism is a relevant one: if for
instance there is an automorphism fixing the point at infinity, then the
quotient has genus $0$ or $1$, and our curve is a cyclic cover of it, and
hence there is an effective algorithm by known facts.

It is not too difficult to see as well that the curve-family is non-constant,
that is, the curves in the family are generically non-isomorphic (not even
after extension of the base). We omit these verifications in this preliminary
draft, which seem however not entirely evident and requiring at any rate some
checking.

S3. We have $(\alpha^{3}+1)u^{6}-2u^{3}+1=P^{2}-Q^{3}$ with $P=u^{3}-1$,
$Q=-\alpha u^{2}$.

We use this representation to define a cubic cyclic unramified cover of
$\tilde{X}_{\alpha}$. In general, such a cover can be obtained by choosing a
point of order $3$ on $J_{X_{\alpha}}$; we omit such construction here and
only give the results of the computation (which, as said, will appear
elsewhere). In the present case this cover will continue to be defined on the
$\alpha$-line. Indeed, it may be shown that the said choice of the point of
order $3$ corresponds to a choice of the polynomials $P,Q$ such that
$(\alpha^{3}+1)u^{6}-2u^{3}+1=P^{2}(u)-Q^{3}(u)$. (To be precise, in fact
$P,Q$ yields a bit of more information, immaterial for the present purposes,
we shall make this precise in the mentioned future paper.) In our case this
choice is already implicit in the opening construction; this explains that the
field of definition does not presently increase.

More precisely, omitting the subscript $\alpha$ for notational convenience,
the function field of the cover $\tilde{Y}$ may be obtained by adding to the
function field $k(u,v)$ of $\tilde{X}$ a cube root of the function
$v+P(u)=v+u^{3}-1$, which we denote now $w$ so $w^{3}=v+u^{3}-1$, and the
function field of $\tilde{Y}$ is generated by $u,w$ over
$\overline{{\mathbb{Q}}}(\alpha)$.

Let $g$ generate the group of automorphisms of $\tilde{Y}$ as a cubic cyclic
cover of $\tilde{X}$. Then $g$ may be chosen so to act as
$(u,w)^{g}:=g(u,w)=(u,\theta w)$ where $\theta$ is a given primitive cubic
root of $1$.

S4. We can then apply suitable calculations (again, omitted here) for
obtaining the curves of genus $1$ mentioned above, as quotients of
$\tilde{Y}_{\alpha}$ by an involution lifted from the canonical one on
$\tilde{X}_{\alpha}$. In this case we find the cubic model of one such curve
given by

$E_{\alpha}:\qquad z^{3}-3Q(u)z-2P(u)=z^{3}+3\alpha zu^{2}-2(u^{3}-1)=0.$

We have used affine coordinates but for $E_{\alpha}$ we implicitly mean a
projective closure of this cubic in ${{\mathbb{P}}}_{2}$, whose equation reads
$Z^{3}+3\alpha ZU^{2}-2U^{3}+2T^{3}=0$, in projective coordinates $(Z:U:T)$.
The discriminant is found to be $\alpha^{3}+1$ up to a constant. This
$E_{\alpha}$ is isomorphic (over some extension of ${{\mathbb{Q}}}(\alpha)$)
to the Fermat cubic chosen at the beginning.141414The shape of these equations
is cubic form$=$ constant. By sending through $PGL_{2}$ the three zeros of the
cubic form to the zeros of a fixed one of them, one obtains explicitly a
projective isomorphism with a constant curve.

These $E_{\alpha}$ are only curves of genus $1$ for the moment, since we have
not yet chosen an origin of $E_{\alpha}$, but we shall do that very soon.

Putting $z:=w+Q(u)/w=w-\alpha u^{2}/w$, we have a map
$\phi=\phi_{\alpha}:\tilde{Y}_{\alpha}\to E_{\alpha}$ given by
$\phi(u,w)=(u,z)$, as is easy to check.

Such construction gives a pencil $\tilde{Y}_{\alpha}$ of curves of genus $4$,
with $\tilde{Y}_{\alpha}$ a(n unramified) cubic cyclic cover of
$\tilde{X}_{\alpha}$, and a morphism $\tilde{Y}_{\alpha}\to E_{\alpha}$ (the
curves $E_{\alpha}$ being all isomorphic to $\mathcal{F}$).

In fact, another morphism to $E_{\alpha}$ is simply obtained by composing with
$g$, so we have a morphism $\sigma:\tilde{Y}_{\alpha}\to E_{\alpha}^{2}$ given
by

$\sigma(p)=(\phi(p),\phi(p^{g})),$

where $p\in\tilde{Y}_{\alpha}$, denoting $p^{g}:=g(p)$, as usual.

Note that we can also use $g^{2}$ in place of $g$, but note that the divisors
on $E_{\alpha}$ given by $\phi(p)+\phi^{(}p^{g})+\phi(p^{g^{2}})$ are all
linearly equivalent, since it is easy to check that they come from
${{\mathbb{P}}}_{1}$. Hence this sum-map is constant for every choice of an
origin and acting with $g^{2}$ does not give anything new.

The map $\sigma$ induces a map from $Jac(\tilde{Y}_{\alpha})$ to
$Jac(E_{\alpha})$, which is surjective; in fact the involution on
$\tilde{X}_{\alpha}$ may be lifted in three ways to $\tilde{Y}_{\alpha}$,
which are conjugate under $g$, and this yields the maps from the Jacobian (as
has been said in the discussion (C) above).

S5. On this family $\tilde{X}_{\alpha}$ we now choose our section
$q_{\alpha}\in\tilde{X}_{\alpha}$ giving the point at infinity (to be removed
so as to obtain the affine curve $X_{\alpha}$). We do this as follows.

Let $p_{1},p_{2},p_{3}$ be the fiber above $q_{\alpha}$ in
$\tilde{Y}_{\alpha}$, where $p_{i}=p_{i\alpha}$, the labelling being modulo
$3$ being chosen so that $p_{i+1}=p_{i}^{g}$.

We prescribe that $p_{2}=p_{2,\alpha}$ is a flex on $E_{\alpha}$, which will
be chosen as origin. For a generic $\alpha$, the $u$-coordinates of the flexes
are found to be defined by the equation

$(\lambda^{3}+3\alpha\lambda-2)u^{3}=-2\quad\hbox{where
\quad$\alpha\lambda^{2}-2\lambda-\alpha^{2}=0$.}$

There are also three flexes at infinity (i.e. with $T=0$). Note that for
$\alpha=0$ the flexes are instead the points with $zu=0$ plus the three points
at infinity.

A flex at a finite point is defined either over ${{\mathbb{Q}}}(2^{1/3})$, if
$\alpha=0$, or over a cubic extension of
${{\mathbb{Q}}}(\sqrt{\alpha^{3}+1})$, that may be made explicit by the
equations just given. This yields a cover $B$ of the $\alpha$-line over which
$q_{\alpha}$ and the $p_{i}$ are defined.

SMALL WARNING: So in fact we should not use $\alpha$ as a subscript, but,
since the beginning, a point $\beta$ of $B$; however we omit this precision
here.

Note that by the above formulae, if $u_{\alpha}$ is the $u$-coordinate of the
point $q_{\alpha}$, then $p_{1},p_{2},p_{3}$ will be sent to the points on
$E_{\alpha}$ with $u$-coordinate $u_{\alpha}$, and they will be distinct as
soon as $\alpha$ is not a zero of the discriminant, i.e. $\alpha^{3}\neq-1$.

So, generically, with the actual choice the point $q_{\alpha}$ will not be a
special point on $\tilde{X}_{\alpha}$.

S6. According to the above settings and equations, the section $p_{2}-p_{1}$
on $Jac(\tilde{Y}_{\alpha})$ is sent by $\sigma$ to the point of
$E_{\alpha}^{2}$ given by

$\sigma(p_{2})-\sigma(p_{1})=(\phi(p_{2})-\phi(p_{1}),\phi(p_{3})-\phi(p_{2})).$

Now, let us note that the two components are equal: in fact, this amounts to
$\phi(p_{1})+\phi(p_{2})+\phi(p_{3})\sim 3\phi(p_{2})$ on $E_{\alpha}$ (or the
corresponding equality with $\phi(p_{2})$ as origin, but that is immaterial
for this). In turn, this linear equivalence (or equation) holds because
$\phi(p_{2})$ is a flex, and the $\phi(p_{i})$ are the intersections of
$E_{\alpha}$ with the line $u=u_{\alpha}$.

S7. We now want to check that the section is not identically torsion. If this
were the case, then (acting with $g$) all the differences
$\phi(p_{i})-\phi(p_{j})$ would be torsion.

Now let us specialize to $\alpha=0$, which is a point of good reduction for
the elliptic family, indeed $E_{\alpha}$ reduces to $Z^{3}-2U^{3}+2T^{3}=0$,
which is a twist of the Fermat curve.161616Note that $\alpha=0$ is not of good
reduction for the family $\tilde{X}_{\alpha}$, but this is immaterial for the
present purpose. For this curve the flexes are precisely the points with
$ZUT=0$. The flex that we have chosen goes to a flex with $U=0$, and all
$\phi(p_{1}),\phi(p_{2}),\phi(p_{3})$ specialise to flexes. Then the
differences have order $3$ in the reduced curve. By general reduction theory,
if the section $\phi(p_{2})-\phi(p_{1})$ would be torsion, the order would
divide $3$. But this is not the case, since generally $\phi(p_{1})$ is not a
flex, and is distinct from $\phi(p_{2})$.

We conclude that the section is non-torsion.

S8. We now check that the section $\phi(p_{2})-\phi(p_{1})$ assumes torsion
values for a complex-dense set of $\alpha$. If the elliptic family
$E_{\alpha}$ would be non isotrivial, this would be a consequence of the
theory of the Betti map, an easy case presented e.g. in [15], Notes to Ch. 3.
But instead all curves $E_{\alpha}$ are isomorphic to the Fermat curve
${\mathcal{F}}:u^{3}+z^{3}=1$ and hence we have a(n iso-)constant family. The
section then leads to a rational map from a dense Z.-open subset of a certain
cover $\tilde{B}$ of ${{\mathbb{P}}}_{1}$ to $\mathcal{F}$; if this map is
non-constant, then its image contains a cofinite set in $E$ and our assertion
follows immediately. On the other hand if the section would be constant, it
would be torsion because it is so for $\alpha=0$; but this contradicts what
has been proved at the previous point, concluding the argument.171717That the
section to $\mathcal{F}$ cannot be identically constant follows also by direct
calculation, on computing an explicit isomorphism of $E_{\alpha}$ with
$\mathcal{F}$; since such an isomorphism may be found sending a flex to a
flex, the argument at bottom is the same as specializing $\alpha=0$.

S9. Since by S6 above the two components of the section are equal, and since
by S8 the first component assumes torsion values on a dense set, the section
itself to $E_{\alpha}^{2}$ assumes torsion values on a complex-dense set in
the relevant cover $B$ of the $\alpha$-line.

###  3.1 Conclusion

Let us go ahead by using points $\beta\in B$ in place of $\alpha$, for more
precision. We have a cover $B$ of the $\alpha$-line with a map
$\delta:B\to{{\mathbb{P}}}_{1}$ where we denoted $\delta(\beta)=\alpha$.

Now we have a family $X_{\beta}=\tilde{X}_{\alpha}-q_{\beta}$ as above. The
construction gives a pencil $\tilde{Y}_{\alpha}$ of curves of genus $4$, with
$\tilde{Y}_{\alpha}$ a(n unramified) cover of $\tilde{X}_{\alpha}$.
Correspondingly we have an affine curve
$Y_{\beta}:=\tilde{Y}_{\alpha}-\\{p_{1},p_{2},p_{3}\\}$ and a map
$Y_{\beta}\to E_{\alpha}$ (the curves $E_{\alpha}$ being all isomorphic to
$E$).

The sections $p_{2}-p_{1},p_{3}-p_{1}$ to $Jac(\tilde{Y}_{\alpha})$ become
both torsion in a complex-dense subset $\Sigma$ of (algebraic) points of $B$.
For $\beta\in B$ Bilu’s criterion applied to $Y_{\beta}$ yields effective
finiteness for the $S$-integral points on $Y_{\beta}$ over any prescribed
number field (and finite set $S$ of places). Since $Y_{\beta}$ is an
unramified cover of $X_{\beta}$, the Chevalley-Weil theorem yields effective
finiteness for the $S$-integral points on $X_{\beta}$ as well.

NOTE: We may represent $X_{\beta}$ as a pencil of quartics in
${{\mathbb{A}}}^{2}$, with a single singular point in ${{\mathbb{A}}}^{2}$ and
a single point at infinity in a smooth model $\tilde{X}_{\beta}$. We do not
perform the computation here, but we reproduce an argument to show that this
representation is possible. (We do not call this an ‘embedding’ since there is
a singular point; however this is quite immaterial for our purposes since the
singular point is not at infinity.)

======================

PAUSE:

Representing an affine curve of genus $2$ : $X=\tilde{X}-q_{0}$, $q_{0}$ non-
special, as a plane quartic with $q_{0}=\infty$.

We simply consider the linear system $L(3q_{0})$ (of rational functions on
$\tilde{X}$ with at most a triple pole at $q_{0}$) which is generated by $1,z$
with $z$ a suitable rational function of degree $3$ (since $q_{0}$ is non-
special). Then $L(4q_{0})$ is generated by $1,z,w$ with a $w$ of degree $4$.
Since $z,w$ have coprime degrees, they generate the function field of $X$.
Since they have a unique pole they provide a map $X\to{{\mathbb{A}}}^{2}$ with
a single point at infinity. The extended map $(1:z:w)$ to ${{\mathbb{P}}}_{2}$
provides the projective plane model, with coordinate functions $1/w$ and $z/w$
at infinity, both vanishing therein; since the latter has a simple zero, and
since these functions have no other common zero, the curve is smooth there.

Let us show that the irreducible polynomial equation $f(z,w)=0$ relating $z,w$
is of total degree $4$. First of all, it has degree $4$ in $z$ and $3$ in $w$
(because $z,w$ have degrees resp. $3,4$ as functions on $\tilde{X}$). Now, the
space $L(15q_{0})$ contains the $15$ terms
$1,z,z^{2},z^{3},z^{4},z^{5},w,wz,wz^{2},wz^{3},w^{2},w^{2}z,w^{2}z^{2},w^{3},w^{3}z$.
By Riemann-Roch they must be linearly dependent, giving a nontrivial
polynomial relation $g(z,w)=0$ where $g$ is a linear combination of the said
terms. This $g$ must be divisible by $f$, proving that there is a relation
without the term $z^{5}$, and of total degree therefore at most $4$ (and hence
exactly $4$) as stated.181818It seems that in this case the usual analysis
using the spaces $L(mq_{0})$, to embed the curve, needs the supplementary
consideration involving $f,g$. Or else one can argue with $w+cz$,
$c\in{{\mathbb{C}}}$, in place of $w$ to deduce that $f$ has total degree $4$.

This quartic model will necessarily have a single singular (nodal) point in
${{\mathbb{A}}}^{2}$ by the genus formula. So in fact we have not truly an
embedding of the smooth curve, but for the matter of integral points this is
immaterial, since the singular point (which is always special) is not at
infinity.

======================

Remark. Note that the conclusions are achieved through a morphism from
$Y_{\beta}$ to an irreducible curve defined by an irreducible equation
$F(u_{\beta},v_{\beta})=0$ in ${{\mathbb{G}}}_{m}^{2}$, not a translate of a
torus, where $u_{\beta},v_{\beta}$ are functions on $Y$ with divisors resp.
$n(p_{2}-p_{1})$, $n(p_{3}-p_{1})$, where $p_{i}$ are the points of $Y$ above
$q_{0}$. Then these functions are multiplicatively independent modulo
constants, and assume $S$-unit values at the integral points of $Y$ which are
lifted by Chevalley-Weil from those on $X$. By Puiseux+linear forms this gives
effective finiteness, as in the proof of the criterion of Bilu.

Degree of the image curve. Let us check that for large torsion order of the
relevant section, the degree of the image curve $Y^{*}:F=0$ also increases.
Suppose then that $n$ is the exact (common) torsion order, so for every $m\geq
2$ the functions $u_{\beta},v_{\beta}$ are not both perfect $m$-th powers in
$k(Y)$, where in the following argument we omit the subscripts $\alpha,\beta$
for convenience.

We have an extension $k(Y)/k(u)$ of degree $n$, totally ramified below $p_{1}$
and $p_{2}$, and we have an intermediate extension $k(u,v)/k(u)$. The
extension $k(Y)/k(u,v)$, say of degree a divisor $m$ of $n$, is totally
ramified below $p_{1},p_{2},p_{3}$ (at least).

Suppose first that the genus of $Y^{*}$, i.e. of $k(u,v)$, is $>1$. Then by
the Hurwitz formula we would have $6=2g(Y)-2\geq 2m+3(m-1)=5m-3$, hence $m=1$,
$k(u,v)=k(Y)$, and $F$ would have degree at least $n$, hence exactly $n$.

Suppose now that $g(Y^{*})=1$. Then again we obtain $6\geq 3m-3$, hence $m\leq
3$ and we have a lower bound $n/3$ for the degree of $F$. 191919In case of
equality we would have a map of degree $3$ from $Y$ to an elliptic curve; then
this curve would be isogenous to $E$, and several deductions would be
possible. We do not go ahead, but maybe further inspection could lead to the
impossibility of this case.

Suppose finally that $g(Y^{*})=0$. Then $k(u,v)=k(z)$ for some function $z\in
k(Y)$. After a homography we could assume that $u=z^{q}$ and
$v=c_{1}(z-c_{2})^{q}$ for constants $c_{1},c_{2}\neq 0$ and $qm=n$. The
divisor of $z$ would satisfy
$q\cdot{{\text{div}}}(z)={{\text{div}}}(u)=n(p_{2}-p_{1})$ so
${{\text{div}}}(z)=(n/q)(p_{2}-p_{1})$. Since we are assuming that $n$ is the
exact torsion order we conclude that $q=1$, and then $c_{3}u+c_{4}v=1$ for
nonzero constants $c_{3},c_{4}$. By the functional $abc$-theorem we deduce
$n\leq 2g(Y)+1=9$, so $n$ is bounded, concluding the argument.

NOTE (a). The integer $n$ may be taken as the common torsion orders of the two
sections, which is also the torsion order of each of them, since they are
dependent over $End(J_{Y})$.

It grows to infinity, so it is not possible to perform this procedure in a
uniform (bounded-degree) way for all curves in the pencil. (In fact, if $n$ is
the precise torsion order of our divisors in $J_{Y}$, the said morphism cannot
be factored nontrivially through a morphism of smaller degree to
${{\mathbb{G}}}_{m}^{2}$.)

This is immaterial for the result itself, but it shows that one cannot get
this proof by universal equations through the family (as would be possible for
instance for the affine curves $y^{2}=f(x)$ with two points at infinity). The
situation somewhat reminds of Belyi theorem…

On the other hand, the same method is flexible in some aspects, and will give
also examples of pencils where the relevant sections are identically torsion.
For such pencils we shall have effectivity for all algebraic points. However
this effectivity will be in a sense less interesting, because it will
correspond to a same type of diophantine equation (in the same way that all
equations of type $y^{2}=f(x)$ may be effectively solved in integers). These
examples will appear in the mentioned paper in preparation with Lombardo.

NOTE (b): The parameter will run on a curve of bounded degree over
$\beta\in{{\mathbb{P}}}_{1}$. The relevant values of $\beta$ will be algebraic
numbers with the following properties:

(i) of degree tending to $\infty$,

(ii) topologically complex-dense, and

(iii) of bounded height.

(iv) Moreover it will be effectively possible to check whether a given value
of $\beta$ is or is not in our set.

###  3.2 Discussion about the non-torsion of the section for the general
family

We have seen in the hints of proof presented in §2.1 that IF a certain section
of a doubly elliptic scheme, associated to the choice of the section $s\mapsto
q_{s}\in\tilde{X}_{s}$ for points at infinity, is identically torsion, then we
have effectivity of integral points for the full set of affine smooth curves
$X_{s}$ of genus $2$ defined by $X_{s}:=\tilde{X}_{s}-\\{q_{s}\\}$. We want to
prove now that this cannot happen, no matter the choice of the section at
infinity $q_{s}$. This represents another limitation of the method.

######  Proposition 3.1.

Notation being as in Theorem 1.1, there is no algebraic family
$\\{q_{s}\\}_{s\in S}$, $q_{s}\in\tilde{X}_{s}$, such that the doubly elliptic
section constructed as above is torsion on the whole $S$.

NOTE: (i) As said, the paper [10] goes in a somewhat similar direction as this
statement, namely providing limitations to Bilu’s criterion. However there are
substantial differences with the present proposition. In a sense, the main
result of [10] is much more general, as it concerns arbitrary covers of
arbitrary curves of genus $>1$; and it is different also because it starts
with a given curve (rather than the whole moduli space) and proves that for a
general choice of the point at infinity, Bilu’s criterion will not work.
Instead, here we work only with cyclic cubic covers, but we let the curves
vary, in genus $2$, and we suppose that the choice of the point at infinity is
given.

Also, the paper [10] deals with objects defined over ${{\mathbb{C}}}$ (see a
previous footnote for a comment on this). Here we may work over
$\overline{{\mathbb{Q}}}$. Note that assuming that the value of the section at
$s$ is torsion for each $s\in S(\overline{{\mathbb{Q}}})$, amounts to assume
that it is torsion at a generic point: indeed, if the generic value is not
torsion, known arguments of Silverman (or see Demianenko-Manin in the
isotrivial case) prove that the set of such $s$ has bounded height.

(ii) By appealing to a general result, we can easily prove that the
Proposition holds up to finitely many choices of $q_{s}$. For completeness
(and possible use in other similar issues) we shall offer this short argument
as well, before the proof of the stronger assertion.

As a preamble, suppose that a family as in the statement exists; this
corresponds to having a subvariety $S$ of $M_{2,1}$ projecting with a map of
finite degree to $M_{2}$, so $S$ is a certain hypersurface in $M_{2,1}$.

Also, the construction that we have described gives an abelian scheme above a
Zariski-open dense subset of a finite cover of $M_{2,1}$, with fibers which
are squares of elliptic curves. Moreover, we have a section of this scheme,
and its image is not contained in any group-subscheme (since the components of
the section are generically independent, as has been remarked several times).

====================

Pause: finiteness of possible exceptions. An algebraic family as in the
statement would correspond to a hypersurface where this section becomes
torsion. However then we may apply Theorem 1.1 of [7], which asserts the
finiteness of such hypersurfaces, and this yields just the desired conclusion
as in Note (ii) just above.

====================

###### Proof of Proposition.

Let us now prove the stronger assertion stated in the proposition, namely that
such sections do not exist. In the notation above, it suffices to prove that
for every choice of the section $q_{s}$ at infinity on $\tilde{X}_{s}$, the
section $\phi(p_{2})-\phi(p_{1})$ to the relevant elliptic curve, is not
identically torsion; recall that these elliptic curves $E_{s}$ are defined by
an (affine) equation

(3.1) $z^{3}-3Q_{s}(u)z-2P_{s}(u)=0,$

where the curve $\tilde{X}_{s}$ of genus $2$ is defined (affinely) by
$v^{2}=f_{s}(u)$, so that $f_{s}(u)=P_{s}^{2}(u)-Q_{s}^{3}(u)$; here the
polynomials $P_{s},Q_{s}$ have respective degrees $3,2$.

We remark that any polynomial $f$ of degree $6$ without multiple zeros is
likewise representable, in finitely many ways, and the representations are in
a certain correspondence with the cyclic covers constructed above (again we
omit such verifications here, but they will appear in the mentioned paper in
progress).

Note also that changing $f_{s}(u)$ with $(cu+d)^{6}f_{s}((au+b)/(cu+d))$
(where the coefficients depend on $s$, and $ad\neq bc$) yields an isomorphic
curve. The same then happens changing $P_{s},Q_{s}$ accordingly with the same
transformation.

By such a transformation (actually we need only a translation on $u$) we may
assume that the $u$-coordinate of $q_{s}$ is $0$. We then perform such a
translation; this is only to simplify the notation and to reduce the number of
involved variables. (Of course assuming that $q_{s}$ is non-special amounts to
$f_{s}(0)\neq 0$ (but this does not matter now.)

It is relevant to recall that the points $\phi(p_{i})$, $i=1,2,3$ (where we
again omit to indicate the dependence on $s$ for notational convenience), are
the three points on $E_{s}$ with the same $u$-coordinate as the point at
infinity $q_{s}$, i.e. the point to be removed from $\tilde{X}_{s}$. Hence
with the present normalization the points $\phi(p_{i})$ have $u=0$ and thus
are defined by $z^{3}-3Q_{s}(0)z-2P_{s}(0)=0$, which has no repeated roots if
$q_{s}$ is non-special.

Going to a subspace with all isomorphic elliptic curves. Let us now assume
that the family is identically torsion on $S$; we have to prove that this is
impossible.

We can fiber the threefold $S$ with surfaces, denoted $S_{c}$, where the
$j$-invariant of $E_{s}$ is equal to $c$. Let us restrict our attention to one
(general enough) such $S_{c}$. Then for $s\in S_{c}$ the cubics defined by
(3.1), or rather their projective shape in homogeneous variables $(T:U:Z)$
(where $u=U/T,z=Z/T$), are all projectively isomorphic, so any two of the
resulting equations in $T,U,Z$ are related by a $3\times 3$ invertible linear
substitution on the homogeneous variables.

Let us now choose a flex on $E_{s}$ such that generically on $S_{c}$ its
$U$-coordinate and $Z$-coordinates are both $\neq 0$. Note that such flex
exists (there are in fact $9$ flexes) and can be sent to $(0:1:1)$ by a
projective transformation depending algebraically on $S_{c}$, represented by a
matrix with rows of the shape $(*,*,0)$, $(0,*,0)$, $(0,0,*)$. Taking a finite
cover of $S$ such that the flex becomes rational on the cover, the
transformation too becomes rational; also, composing with such a
transformation does not destroy the said property of $q_{s}$ (to have
vanishing $U$-coordinate), and the polynomials $P_{s},Q_{s}$ are changed
according to an element of $PGL_{2}$, so the shape of the equation (3.1) is
not changed. This further entails that the curve $\tilde{X}_{s}$ does not
change its isomorphism class. Hence we may assume that all curves $E_{s}$,
$s\in S_{c}$, pass through $(0:1:1)$ having a flex there. We can also agree to
choose that flex as an origin for all such curves, which then become elliptic.

Now, let us fix (arbitrarily) one of the curves corresponding to a given point
$s_{0}\in S_{c}$, say $E^{*}:=E_{s_{0}}$. We have a projective transformation
$L_{s}$ sending $E_{s}$ to $E^{*}$. The line $U=0$ is sent by $L_{s}$ to a
line meeting $E^{*}$ in two torsion points (and hence three). Therefore by
continuity this line must be constant as $s$ varies, so the $L_{s}$ all fix
the line $U=0$. And the $L_{s}$ fix the flex $(0:1:1)$ as well, again by
continuity (since the flex $(0:1:1)$ of $E_{s}$ is sent by $L_{s}$ to a flex
of $E^{*}$, and there are $9$).

Further, by composing (on the left) with a linear projective transformation
with rows $(1,0,0),(0,\rho,0),(0,0,\rho)$, $\rho\neq 0$, does not destroy any
of the shapes, fixes the flex, and fixes the isomorphism class of
$\tilde{X}_{s}$ (it sends $P_{s}(u),Q_{s}(u)$ resp. to $P_{s}(\rho
u),Q_{s}(\rho u)$). We can now choose $\rho=\rho_{s}$ such that one
intersection of $U=0$ with $E_{s}$ is constant in $s$, say equal to $(1:0:1)$.
(In fact, the roots of $z^{3}-3Q(0)z-2P(0)=0$ are finite and distinct.)

Now, moving on a curve $S^{\prime}_{c}$ inside $S_{c}$ we can achieve that
also another intersection with the line $U=0$ is constant in $s\in
S^{\prime}_{c}$. But now note that the shape of (3.1) is such that the sum of
the three roots in $z$ of the resulting equation with $u=0$ vanishes.
Therefore if two intersections are constant, the third one also is.

Summing up, with these normalizations we have achieved that for $s\in
S^{\prime}_{c}$, the transformation $L_{s}^{-1}$ fixes the three points of
$E^{*}$ on $U=0$ and fixes the line $U=0$; this entails that it fixes
pointwise the line $U=0$. Also, it fixes the flex $(0:1:1)$.

Then, up to a nonzero factor, $L_{s}$, for $s\in S^{\prime}_{c}$ has rows of
the shape $(1,0,0),(0,a,0),(0,a-1,1)$, for some $a=a_{s}$. However now we use
that $L_{s}$ sends the shape (3.1) into the same shape, i.e. with vanishing
coefficient of $z^{2}$. This yields $a=1$, so $L_{s}$ is the identity and all
$\tilde{X}_{s}$ for $s\in S^{\prime}_{c}$, would be isomorphic. This is
impossible, as we are assuming that only finitely many $\tilde{X}_{s}$ are
isomorphic to a given one.

This proves the assertion. ∎

NOTE: Note that the proof yields also that our section cannot be a (torsion or
non-torsion) constant on a subscheme with iso-constant fibers, if the base has
dimension $2$. (It makes sense to speak of a ‘constant’ section if the scheme
is isotrivial.)

######  Remark 3.2.

The above argument allows us to be explicit as to the family with iso-constant
elliptic curves. Note that after base change over $S_{c}$ we obtain a trivial
square-elliptic family $(E^{*})^{2}\times R_{c}$, where $R_{c}$ is a finite
cover of $S_{c}$, and where the given section corresponds to a map from
$R_{c}$ to $(E^{*})^{2}$.

This map will now be shown to be not dominant, and actually with image a
certain curve that we now describe. And the fibers will all be curves in
$R_{c}$. The argument now shows that we can obtain these curves as follows.

First, by a translation on $u=U/T$ we can assume that the section is obtained
by taking two differences (in the elliptic-group sense) between the three
points of the curve on the line $U=0$.

Now take a given one elliptic curve of the family, denoted $E^{*}$, having a
flex at $(0:1:1)$, taken as origin, and passing through $(1:0:1)$, where the
section (to $(E^{*})^{2}$) can be assumed to correspond to two of the points
on the line $U=0$. (Note that there will be three such points if the point at
infinity is non-special, and the sum of the three will be zero.) As in the
argument for Proposition 3.1, this curve $E^{*}$ can be gotten through a
projective transformation fixing the isomorphism class of the curve
$\tilde{X}$ of genus $2$ leading to the said elliptic curve (as in the
description above).

Now take the projective transformations $L_{s}$ fixing both the flex $(0:1:1)$
and the point $(1:0:1)$. Their matrices have rows of the shape
$(a,b,-b),(c,d,-c),(e+b-a,e+c-b,e)$. They form a group $\Gamma$ of
(projective) dimension $4$. This groups acts faithfully on the projective
plane ${{\mathbb{P}}}_{2}$, and also on the set of equations (for instance if
it fixes an equation it induces an automorphism of an elliptic curve, with two
fixed points, thus must induce the identity).

Now there is a subvariety $V\subset\Gamma$ formed by the $L\in\Gamma$ such
that the equation for $L(E^{*})$ has zero coefficients of $Z^{2}$ (there are
two coefficients). Note that $V$ has dimension $\geq 2$, as a nonempty fiber
(above the point $(0,0)\in{{\mathbb{A}}}^{2}$) of a map
$\Gamma\to{{\mathbb{A}}}^{2}$.

But then the dimension is exactly $2$. Indeed it cannot have dimension $\geq
3$, because otherwise there would exist inside $V$ an irreducible curve
corresponding to a single isomorphism class of $\tilde{X}$. Then the
corresponding polynomials $P^{2}-Q^{3}$ would all be related by a homography,
up to a constant factor; but then the same would be true individually of the
pairs $P,Q$ (because there are only finitely many pairs producing a given
$P^{2}-Q^{3}$, as will appear in the mentioned forthcoming paper. 202020This
may be proved directly as follows. Let $F(u)=P^{2}(u)-Q^{3}(u)$, where $F$ is
fixed and without multiple roots. If there are infinitely many solutions in
polynomials $P,Q$ of degrees resp. $3,2$, then there is an algebraic family of
solutions of dimension $1$. In other words we may assume that the coefficients
of $P,Q$ are rational functions on a certain curve $Z$, not all constant. Ler
$\delta$ be a derivation on ${{\mathbb{C}}}(Z)$ with constant field
${{\mathbb{C}}}$, and apply it to the equation, obtaining $2(\delta
P)P=3(\delta Q)Q^{2}$. But $P,Q$ must be coprime, hence $Q^{2}$ divides
$\delta P$ and $P$ divides $\delta Q$, forcing $\delta P,\delta Q$ to vanish.
So after all they are constant.)

In conclusion, we would have a $1$-parameter family of matrices in $\Gamma$
carrying a fixed cubic $z^{3}-3Q(u)z-2P(u)$ into a constant multiple of
$z^{3}-3\rho^{2}(cu+d)^{2}Q(\gamma(u))z-2\rho^{3}(cu+d)^{3}P(\gamma(u))$ for a
nonzero $\rho$ and a $\gamma(u)=(au+b)/(cu+d)\in PGL_{2}$. This would produce
a $1$-dimensional family of automorphisms
$(u,z)\mapsto(\gamma(u),z/\rho(cu+d))$ of the corresponding elliptic curve,
which is impossible (e.g. by general theory, or note that these transformation
would have to stabilise the set of flexes, or else the branch points of the
map $u:E^{*}\to{{\mathbb{P}}}_{1}$).

Naturally it is possible to prove the contention about $\dim V$ also by
explicit calculation. (Or else we could also use the above Proposition; in
fact, if $V$ has dimension $3$, we would get a subvariety of dimension $2$
such that the section is constant on it, contrary to the last remark in the
proof of the Proposition.)

Then, the cubic curves of the family may be taken just the $E_{L}:=L(E^{*})$
for $L\in V$: they are all isomorphic, but they come from generically non-
isomorphic curves $\tilde{X}$.

This also shows that the image of the section, from the above space $R_{c}$,
and represented in $E^{*}\times E^{*}$, is of the shape $x_{c}\times E^{*}$
for a certain $x_{c}\in E^{*}$ (depending also on the family of points at
infinity, but now we are assuming this is given). This $x_{c}$ is represented
by the point $(0:1:0)$ in $E^{*}$ (with the chosen flex as origin).

When $x_{c}$ is not torsion, the method does not yield anything on the space
$S_{c}$. On the contrary, when it is torsion, we have pencils like in the
example presented in §3.

This description also yields a rather complete picture of the whole.

In the paper in progress (also with Lombardo) we plan to provide full detail
for this issue too, and to present all types of examples that can be deduced
from it.

## References

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Pietro Corvaja
Dipartimento di Matematica e Informatica
Università di Udine
Via delle Scienze, 206
33100 Udine
Italy

E-mail address<EMAIL_ADDRESS>

Umberto Zannier
Scuola Normale Superiore
Piazza dei Cavalieri, 7
56126 Pisa
Italy

E-mail address<EMAIL_ADDRESS>