# Points on a curve with a power on a curve

Gareth J. Boxall<EMAIL_ADDRESS><EMAIL_ADDRESS>Department of
Mathematical Sciences (Mathematics Division), Stellenbosch University, Private
Bag X1, Matieland 7602, South Africa

###### Abstract.

Let $C_{1},C_{2}\subseteq\mathbb{G}_{m}^{N}(\mathbb{C})$ be irreducible closed
algebraic curves, with $N\geq 3$. Suppose $C_{1}$ is not contained in an
algebraic subgroup of $\mathbb{G}_{m}^{N}(\mathbb{C})$ of dimension $1$ and
$C_{1}\cup C_{2}$ is not contained in an algebraic subgroup of
$\mathbb{G}_{m}^{N}(\mathbb{C})$ of dimension $2$. It is a conjecture that at
most finitely many points $x\in C_{1}$ have the property that there is a
positive integer $n$ such that $x^{n}\in C_{2}$ and $[n]C_{1}\nsubseteq
C_{2}$, where $[n]C_{1}=\\{x^{n}:x\in C_{1}\\}$. We prove this in the case
where at least one of the two curves is not defined over
$\overline{\mathbb{Q}}$.

This work is based on the research supported in part by the National Research
Foundation of South Africa (Grant Number 96234).

##  1\. Introduction

We consider the following conjecture.

######  Conjecture 1.1.

Let $C_{1},C_{2}\subseteq\mathbb{G}_{m}^{N}(\mathbb{C})$ be irreducible closed
algebraic curves, with $N\geq 3$. Suppose there does not exist an algebraic
subgroup $G\subseteq\mathbb{G}_{m}^{N}(\mathbb{C})$ of dimension one such that
$C_{1}\subseteq G$ and there does not exist an algebraic subgroup
$H\subseteq\mathbb{G}_{m}^{N}(\mathbb{C})$ of dimension two such that
$C_{1}\cup C_{2}\subseteq H$. Let
$\mathcal{N}=\\{n\in\mathbb{N}:[n]C_{1}\subseteq C_{2}\\}$. Then
$\bigcup\limits_{n\in\mathbb{N}\setminus\mathcal{N}}\\{x\in C_{1}:x^{n}\in
C_{2}\\}$ is finite.

Here and throughout, for subvarieties of $\mathbb{G}_{m}^{N}(F)$,
“irreducible” means “irreducible over $F$”. Addressing a question of Aaron
Levin, Zannier showed in [18] how to deduce Conjecture 1.1 from Zilber’s
conjecture on unlikely intersections with tori (CIT). We prove the following.

######  Theorem 1.2.

Conjecture 1.1 is true under the additional assumption that least one of the
two curves is not defined over $\overline{\mathbb{Q}}$.

This is not dependent on the case where $C_{1}$ and $C_{2}$ are defined over
$\overline{\mathbb{Q}}$, which remains open. To prove Theorem 1.2 we use work
of Bays, Kirby and Wilkie [3] which established an analogue of Schanuel’s
conjecture for the operation of raising to an exponentially transcendental
power. Their reasoning relied on separate work of Kirby [10] on partial
exponential fields which, in turn, made use of Ax’s famous paper [2] on
functional analogues of Schanuel’s conjecture. The main result of [3] is the
following.

######  Theorem 1.3.

Let $F$ be an exponential field and $N$ a positive integer. Let $a\in F^{N}$,
$\lambda\in F$ and $K\subseteq F$ be such that $K$ is a subfield, $a$ is
$\mathbb{Q}$-linearly independent and $\lambda$ is exponentially
transcendental over $K$. The field generated by $(a,\exp(a))$ has
transcendence degree at least $N$ over $K(\lambda)$.

An exponential field is a field of characteristic zero equipped with a
homomorphism $\exp$ from the additive group to the multiplicative group.
Exponential transcendence is a natural analogue, involving $\exp$, of the
usual notion of transcendence. For our argument, we require a slight
generalisation of Theorem 1.3 to the setting of partial exponential fields,
where $\exp$ might not be defined everywhere. Given the results in [10], it is
easy to adapt the reasoning in [3] to obtain this. We apply it in a structure
formed by taking an elementary extension of the complex exponential field,
restricting $\exp$ to the infinitesimal elements and then forming an
ultrapower of the resulting structure. Our exponentially transcendental
exponents are non-standard positive integers.

Aspects of the argument work for higher dimensional subvarieties and we obtain
the following.

######  Theorem 1.4.

Let $V_{1},V_{2}\subseteq\mathbb{G}_{m}^{N}(\mathbb{C})$ be irreducible closed
algebraic subvarieties, with $\dim(V_{1})\geq 1$ and
$\dim(V_{1})+\dim(V_{2})<N$. Suppose that, for any $\mathbb{Q}$-linearly
independent tuples of integers

$(n_{1,1},...,n_{1,N}),...,(n_{k,1},...,n_{k,N}),$

with $k=\dim(V_{1})+1$, and any generic point $(x_{1},...,x_{N})$ of $V_{1}$,
the monomials

$x_{1}^{n_{1,1}}\cdot...\cdot x_{N}^{n_{1,N}},...,x_{1}^{n_{k,1}}\cdot...\cdot
x_{N}^{n_{k,N}}$

are algebraically independent. Let
$\mathcal{N}=\\{n\in\mathbb{N}:[n]V_{1}\subseteq V_{2}\\}$. Then
$\bigcup\limits_{n\in\mathbb{N}\setminus\mathcal{N}}\\{x\in V_{1}:x^{n}\in
V_{2}\\}$ is not Zariski-dense in $V_{1}$.

A similar algebraic independence condition on multiplicative combinations of
coordinates was used by Habegger in [9] to isolate a class of subvarieties for
which he could prove a version of CIT, though there $k$ was equal to the
dimension and the subvariety was assumed to be definable over
$\overline{\mathbb{Q}}$. Masser considered subvarieties satisfying the
assumption for $k=\dim(V)+1$ in positive characteristic in [15].

Interest in CIT motivated an investigation by Bays and Habegger into Levin’s
question in [4]. They considered the case of a single curve $C=C_{1}=C_{2}$
and obtained the following where, with $x=(x_{1},...,x_{N})$,

$\mathbb{T}=\\{x\in\mathbb{G}_{m}^{N}(\mathbb{C}):\left|x_{1}\right|=...=\left|x_{N}\right|=1\\}.$

######  Theorem 1.5.

Let $C\subseteq\mathbb{G}_{m}^{N}(\mathbb{C})$ be an irreducible closed
algebraic curve defined over $\overline{\mathbb{Q}}$, with $N\geq 3$. Suppose
$C$ is not contained in a proper algebraic subgroup of
$\mathbb{G}_{m}^{N}(\mathbb{C})$.

  1. (1)

If $\sigma(C)\cap\mathbb{T}$ is finite, for some automorphism $\sigma$ of
$\mathbb{C}$, then $\bigcup\limits_{n\in\mathbb{N}\setminus\\{1\\}}\\{x\in
C:x^{n}\in C\\}$ is finite.

  2. (2)

If $S$ is a finite set of sufficiently large prime numbers then
$\bigcup\limits_{n\in\mathbb{N}\setminus\\{1\\}}\\{x\in C:x^{n}\in C\\}$
contains at most finitely many $S$-integral points.

As is noted in [4], it follows from known results that
$\\{n\in\mathbb{N}:[n]C\subseteq C\\}=\\{1\\}$ for such $C$. So Theorem 1.5
gives two special cases of a statement that generalises to the two curve
setting as follows.

######  Conjecture 1.6.

Let $C_{1},C_{2}\subseteq\mathbb{G}_{m}^{N}(\mathbb{C})$ be irreducible closed
algebraic curves, with $N\geq 3$. Assume there is no proper algebraic subgroup
$H\subseteq\mathbb{G}_{m}^{N}(\mathbb{C})$ such that $C_{1}\subseteq H$. Let
$\mathcal{N}=\\{n\in\mathbb{N}:[n]C_{1}\subseteq C_{2}\\}$. Then
$\bigcup\limits_{n\in\mathbb{N}\setminus\mathcal{N}}\\{x\in C_{1}:x^{n}\in
C_{2}\\}$ is finite.

Conjecture 1.1 follows easily from Conjecture 1.6, given known results. Other
important work towards Conjecture 1.6, for curves defined over
$\overline{\mathbb{Q}}$, was done by Amoroso, Masser and Zannier in [1]. They
showed that the points in
$\bigcup\limits_{n\in\mathbb{N}\setminus\mathcal{N}}\\{x\in C_{1}:x^{n}\in
C_{2}\\}$ will have bounded height. Using this it is possible to extend
Theorem 1.5 to the two curve setting. I shall include that in a subsequent
paper (or an expansion of this one).

This paper is arranged as follows. In Section 2 we recall some known results
concerning unlikely intersections in $\mathbb{G}_{m}^{N}(\mathbb{C})$,
including the dimension one case of CIT and a theorem of Liardet. We also
present a deduction of Conjecture 1.1 from Conjecture 1.6. In Section 3 we
recall some facts about partial exponential fields, from [10], and slightly
modify the proof of the main result in [3] to obtain the generalisation that
we shall want. In Section 4 we use this to prove the case of Conjecture 1.6 in
which $C_{1}$ or $C_{2}$ is not defined over $\overline{\mathbb{Q}}$. By
reasoning presented in Section 2, this implies Theorem 1.2. We also prove
Theorem 1.4.

I am grateful to Martin Bays, Philipp Habegger, Andrew Harrison-Migochi,
Gareth Jones, Sophie Marques, Amador Martin-Pizarro, Harry Schmidt and David
Smith for some helpful comments and conversations while I was working on this
material. I especially thank Gareth for suggesting that some of the reasoning
in the proof of Theorem 1.2 would extend to higher dimensional subvarieties,
which led to Theorem 1.4.

##  2\. CIT for curves and a theorem of Liardet

In this section we discuss some preliminaries. We begin with basic information
about algebraic subgroups of $\mathbb{G}_{m}^{N}(F)$, for an arbitrary field
$F$ of characteristic zero, which can be found in Chapter 3 of [5].

######  Fact 2.1.

Let $N$ be an integer greater than or equal to two. Let $F$ be a field of
characteristic zero.

  1. (1)

A subset $H\subseteq\mathbb{G}_{m}^{N}(F)$ is an algebraic subgroup if and
only if it is the solution set of a finite system of equations of the form
$X_{1}^{j_{1,1}}\cdot...\cdot X_{N}^{j_{1,N}}=1,\ ...,\
X_{1}^{j_{s,1}}\cdot...\cdot X_{N}^{j_{s,N}}=1$, where $j_{i,q}\in\mathbb{Z}$
for all $i\in\\{1,...,s\\}$ and $q\in\\{1,...,N\\}$.

  2. (2)

If $H$ is an algebraic subgroup of $\mathbb{G}_{m}^{N}(F)$ and the exponents
in the equations are as above then $\dim(H)=N-D$, where $D$ is the dimension
of the $\mathbb{Q}$-vector space generated by
$(j_{1,1},...,j_{1,N}),...,(j_{s,1},...,j_{s,N})$.

The following result is the dimension one case of CIT. It was proved by Maurin
[16] for curves defined over $\overline{\mathbb{Q}}$. Bombieri, Masser and
Zannier [6] extended it to curves defined over $\mathbb{C}$.

######  Theorem 2.2.

Let $C\subseteq\mathbb{G}_{m}^{N}(\mathbb{C})$ be an irreducible closed
algebraic curve. Assume $N\geq 2$ and that $C$ is not contained in a proper
algebraic subgroup of $\mathbb{G}_{m}^{N}(\mathbb{C})$. At most finitely many
points $x\in C$ have the property that $x$ is contained in an algebraic
subgroup of $\mathbb{G}_{m}^{N}(\mathbb{C})$ of dimension $N-2$.

The following result was proved by Liardet [13], verifying a conjecture of
Lang and building on work of Lang and others. It is known to be a consequence
of Theorem 2.2, but was established earlier.

######  Theorem 2.3.

Let $C\subseteq\mathbb{G}_{m}^{N}(\mathbb{C})$ be an irreducible closed
algebraic curve, with $N\geq 2$. Let $\Gamma$ be a finite rank subgroup of
$\mathbb{G}_{m}^{N}(\mathbb{C})$. If $C\cap\Gamma$ is infinite then $C$ is a
translate of an algebraic subgroup of $\mathbb{G}_{m}^{N}(\mathbb{C})$.

In some important special cases, the conclusion can be strengthened. The
following is a well-known consequence of Theorem 2.3. We include a proof for
convenience. In fact, the second and third parts of the result were known
before Theorem 2.3 (see [11] and [12]).

######  Corollary 2.4.

Let $C\subseteq\mathbb{G}_{m}^{N}(\mathbb{C})$ be an irreducible closed
algebraic curve, with $N\geq 2$. Assume at least one of the following three
conditions is satisfied.

  1. (1)

There is a non-torsion $x\in\mathbb{G}_{m}^{N}(\mathbb{C})$ such that, for
infinitely many $n\in\mathbb{N}$, $x^{n}\in C$.

  2. (2)

There is a non-torsion $y\in\mathbb{G}_{m}^{N}(\mathbb{C})$ such that, for
infinitely many $n\in\mathbb{N}$, there exists $x\in C$ such that $x^{n}=y$.

  3. (3)

$C$ contains infinitely many torsion points.

Then $C$ is contained in an algebraic subgroup of
$\mathbb{G}_{m}^{N}(\mathbb{C})$ of dimension $1$.

###### Proof.

Assume (1), (2) or (3). By Theorem 2.3, $C$ is a translate of an algebraic
subgroup of $\mathbb{G}_{m}^{N}(\mathbb{C})$ of dimension $1$. So $C$ is
contained in $N-1$ hypersurfaces, defined by equations of the form
$X_{1}^{i_{1}}\cdot...\cdot X_{N}^{i_{N}}=\alpha$, and it is sufficient to
show that, in each case, $\alpha$ is a root of unity. If (1) holds then we
obtain $\beta\in\mathbb{C}\setminus\\{0\\}$ and distinct $n,m\in\mathbb{N}$
such that $\beta^{n}=\alpha=\beta^{m}$. If (2) holds then we obtain
$\beta\in\mathbb{C}\setminus\\{0\\}$ and distinct $n,m\in\mathbb{N}$ such that
$\alpha^{n}=\beta=\alpha^{m}$. If (3) holds then $\alpha$ is a multiplicative
combination of roots of unity. In each case, $\alpha$ is a root of unity. ∎

The following result and its proof are also known (see Section 2.4 of [1]).

######  Corollary 2.5.

Let $C_{1},C_{2}\subseteq\mathbb{G}_{m}^{N}(\mathbb{C})$ be irreducible closed
algebraic curves. Assume $N\geq 2$ and that there is no algebraic subgroup
$H\subseteq\mathbb{G}_{m}^{N}(\mathbb{C})$ of dimension $1$ such that
$C_{1}\subseteq H$. Let $\mathcal{N}=\\{n\in\mathbb{N}:[n]C_{1}\subseteq
C_{2}\\}$. Then $\mathcal{N}$ is finite.

###### Proof.

Suppose $\mathcal{N}$ is infinite. Let $x\in C_{1}$ be generic over a
countable subfield $K\subseteq\mathbb{C}$ over which both $C_{1}$ and $C_{2}$
are defined. Since $x$ is not a torsion point, it follows by Corollary 2.4
that $C_{2}$ is contained in an algebraic subgroup of
$\mathbb{G}_{m}^{N}(\mathbb{C})$ of dimension $1$. Then the multiplicative
dimension of $x$ is at most $1$. This is a contradiction. ∎

We can deal quickly with one special case of Conjecture 1.6 and it is
convenient to do so. In the proof of the following,
$\pi_{1,2}:\mathbb{G}_{m}^{N}(\mathbb{C})\rightarrow\mathbb{G}_{m}^{2}(\mathbb{C})$
is the projection to the first two coordinates.

######  Proposition 2.6.

Let $C_{1},C_{2}\subseteq\mathbb{G}_{m}^{N}(\mathbb{C})$ be irreducible closed
algebraic curves, with $N\geq 3$. Assume $C_{1}$ is not contained in a proper
algebraic subgroup of $\mathbb{G}_{m}^{N}(\mathbb{C})$. Let
$\mathcal{N}=\\{n\in\mathbb{N}:[n]C_{1}\subseteq C_{2}\\}$. If $C_{1}$ or
$C_{2}$ is contained in a translate of an algebraic subgroup of
$\mathbb{G}_{m}^{N}(\mathbb{C})$ of dimension $N-2$ then
$\bigcup\limits_{n\in\mathbb{N}\setminus\mathcal{N}}\\{x\in C_{1}:x^{n}\in
C_{2}\\}$ is finite.

###### Proof.

Suppose $\bigcup\limits_{n\in\mathbb{N}\setminus\mathcal{N}}\\{x\in
C_{1}:x^{n}\in C_{2}\\}$ is infinite and $C_{1}$ is contained in a translate
of an algebraic subgroup of $\mathbb{G}_{m}^{N}(\mathbb{C})$ of dimension
$N-2$. We may assume there exist $\alpha,\beta\in\mathbb{C}\setminus\\{0\\}$
such that $\pi_{1,2}(x)=(\alpha,\beta)$ for all $x\in C_{1}$. Then
$(\alpha,\beta)$ is not a torsion point. By Corollary 2.4, $\pi_{1,2}(C_{2})$
is contained in an algebraic subgroup of $\mathbb{G}_{m}^{2}(\mathbb{C})$ of
dimension $1$. Then $\alpha$ and $\beta$ are multiplicatively dependent,
contradicting the assumption that $C_{1}$ is not contained in a proper
algebraic subgroup of $\mathbb{G}_{m}^{N}(\mathbb{C})$.

Suppose $\bigcup\limits_{n\in\mathbb{N}\setminus\mathcal{N}}\\{x\in
C_{1}:x^{n}\in C_{2}\\}$ is infinite and $C_{2}$ is contained in a translate
of an algebraic subgroup of $\mathbb{G}_{m}^{N}(\mathbb{C})$ of dimension
$N-2$. We may assume there exist $\alpha,\beta\in\mathbb{C}\setminus\\{0\\}$
such that $\pi_{1,2}(y)=(\alpha,\beta)$ for all $y\in C_{2}$. Given what we
have already proved, we may assume $\pi_{1,2}(C_{1})$ is infinite. Then
$\bigcup\limits_{n\in\mathbb{N}\setminus\mathcal{N}}\\{x\in\pi_{1,2}(C_{1}):x^{n}\in\pi_{1,2}(C_{2})\\}$
is infinite. By Corollary 2.4, $\pi_{1,2}(C_{1})$ is contained in a proper
algebraic subgroup of $\mathbb{G}_{m}^{2}(\mathbb{C})$. Therefore $C_{1}$ is
contained in a proper algebraic subgroup of $\mathbb{G}_{m}^{N}(\mathbb{C})$,
which is a contradiction. ∎

We conclude this section with details of the deduction of Conjecture 1.1 from
the $N=3$ case of Conjecture 1.6.

######  Proposition 2.7.

Suppose it is the case that if
$C^{\prime}_{1},C^{\prime}_{2}\subseteq\mathbb{G}_{m}^{3}(\mathbb{C})$ are
irreducible closed algebraic curves and $C^{\prime}_{1}$ is not contained in a
proper algebraic subgroup of $\mathbb{G}_{m}^{3}(\mathbb{C})$ then, with
$\mathcal{N}^{\prime}=\\{n\in\mathbb{N}:[n]C^{\prime}_{1}\subseteq
C^{\prime}_{2}\\}$, the set
$\bigcup\limits_{n\in\mathbb{N}\setminus\mathcal{N}^{\prime}}\\{x\in
C^{\prime}_{1}:x^{n}\in C^{\prime}_{2}\\}$ is finite. Then it is also the case
that if $C_{1},C_{2}\subseteq\mathbb{G}_{m}^{N}(\mathbb{C})$ are irreducible
closed algebraic curves, with $N\geq 3$, and there does not exist an algebraic
subgroup $G\subseteq\mathbb{G}_{m}^{N}(\mathbb{C})$ of dimension $1$ such that
$C_{1}\subseteq G$ and there does not exist an algebraic subgroup
$H\subseteq\mathbb{G}_{m}^{N}(\mathbb{C})$ of dimension $2$ such that
$C_{1}\cup C_{2}\subseteq H$ then, with
$\mathcal{N}=\\{n\in\mathbb{N}:[n]C_{1}\subseteq C_{2}\\}$, the set
$\bigcup\limits_{n\in\mathbb{N}\setminus\mathcal{N}}\\{x\in C_{1}:x^{n}\in
C_{2}\\}$ is finite.

###### Proof.

Let $N\geq 3$ and let $C_{1},C_{2}\subseteq\mathbb{G}_{m}^{N}(\mathbb{C})$ be
irreducible closed algebraic curves such that $C_{1}$ is not contained in an
algebraic subgroup of $\mathbb{G}_{m}^{N}(\mathbb{C})$ of dimension $1$ and
$C_{1}\cup C_{2}$ is not contained in an algebraic subgroup of
$\mathbb{G}_{m}^{N}(\mathbb{C})$ of dimension $2$. Let
$\mathcal{N}=\\{n\in\mathbb{N}:[n]C_{1}\subseteq C_{2}\\}$. Suppose
$\bigcup\limits_{n\in\mathbb{N}\setminus\mathcal{N}}\\{x\in C_{1}:x^{n}\in
C_{2}\\}$ is infinite. We consider two cases and obtain a contradiction in
each.

Case 1: Suppose $C_{1}$ is not contained in an algebraic subgroup of
$\mathbb{G}_{m}^{N}(\mathbb{C})$ of dimension $2$. Then there is a projection
$\pi:\mathbb{G}_{m}^{N}(\mathbb{C})\rightarrow\mathbb{G}_{m}^{3}(\mathbb{C})$
such that $\pi(C_{1})$ is not contained in a proper algebraic subgroup of
$\mathbb{G}_{m}^{3}(\mathbb{C})$. Let $C^{\prime}_{1}$ and $C^{\prime}_{2}$ be
the Zariski closures of $\pi(C_{1})$ and $\pi(C_{2})$, respectively, in
$\mathbb{G}_{m}^{3}(\mathbb{C})$. These are both irreducible. If either
$C^{\prime}_{1}$ or $C^{\prime}_{2}$ is finite then, using Corollary 2.4 when
one of them is infinite, we obtain the contradictory conclusion that
$C^{\prime}_{1}$ is contained in a proper algebraic subgroup of
$\mathbb{G}_{m}^{3}(\mathbb{C})$. So $C^{\prime}_{1}$ and $C^{\prime}_{2}$ are
curves. Let $\mathcal{N}^{\prime}=\\{n\in\mathbb{N}:[n]C^{\prime}_{1}\subseteq
C^{\prime}_{2}\\}$. By assumption,
$\bigcup\limits_{n\in\mathbb{N}\setminus\mathcal{N}^{\prime}}\\{x\in
C^{\prime}_{1}:x^{n}\in C^{\prime}_{2}\\}$ is finite. Since $C_{1}$ is
irreducible and $C_{1}^{\prime}$ is infinite, it then follows that
$\bigcup\limits_{n\in\mathbb{N}\setminus\mathcal{N}^{\prime}}\\{x\in
C_{1}:x^{n}\in C_{2}\\}$ is finite. By Corollary 2.5, $\mathcal{N}^{\prime}$
is finite. Therefore
$\bigcup\limits_{n\in\mathbb{N}\setminus\mathcal{N}}\\{x\in C_{1}:x^{n}\in
C_{2}\\}$ is finite.

Case 2: Suppose there is an algebraic subgroup
$H\subseteq\mathbb{G}_{m}^{N}(\mathbb{C})$ of dimension $2$ such that
$C_{1}\subseteq H$. Given that
$\bigcup\limits_{n\in\mathbb{N}\setminus\mathcal{N}}\\{x\in C_{1}:x^{n}\in
C_{2}\\}$ is infinite, there are two subcases to consider.

Subcase 2(a): Assume infinitely many $y\in C_{2}$ have the property that there
exist $x\in C_{1}$ and $n\in\mathbb{N}\setminus\mathcal{N}$ such that
$x^{n}=y$. Then $C_{2}\subseteq H$.

Subcase 2(b): Assume only finitely many $y\in C_{2}$ have this property. Then
there exists $y\in C_{2}$ such that, for infinitely many $x\in C_{1}$, there
is some $n\in\mathbb{N}\setminus\mathcal{N}$ such that $x^{n}=y$. By Corollary
2.4, $C_{1}$ is contained in an algebraic subgroup of
$\mathbb{G}_{m}^{N}(\mathbb{C})$ of dimension $1$.

In either subcase we get a contradiction. ∎

We shall want a relative version of this, given that we are interested in
curves that are not both defined over $\overline{\mathbb{Q}}$.

######  Proposition 2.8.

The case of Conjecture 1.6 in which $N=3$ and at least one of the two curves
is not defined over $\overline{\mathbb{Q}}$ implies the case of Conjecture 1.1
in which $N\geq 3$ and at least one of the two curves is not defined over
$\overline{\mathbb{Q}}$.

###### Proof.

We can use almost exactly the same proof as for Proposition 2.7. We just have
to check that if $C_{1}$ or $C_{2}$ is not defined over
$\overline{\mathbb{Q}}$ then the projection
$\pi:\mathbb{G}_{m}^{N}(\mathbb{C})\rightarrow\mathbb{G}_{m}^{3}(\mathbb{C})$
in Case 1 can be chosen so that $C^{\prime}_{1}$ or $C^{\prime}_{2}$ is not
defined over $\overline{\mathbb{Q}}$. Suppose this is not the case. Let
$\pi^{\prime}:\mathbb{G}_{m}^{N}(\mathbb{C})\rightarrow\mathbb{G}_{m}^{k}(\mathbb{C})$
be a projection such that $\pi^{\prime}(C_{1})$ is not contained in a proper
algebraic subgroup of $\mathbb{G}_{m}^{k}(\mathbb{C})$. Assume $k$ is the
maximum integer for which this is possible. Then $k\geq 3$. By Corollary 2.4,
$\bigcup\limits_{n\in\mathbb{N}\setminus\mathcal{N}}\\{y\in C_{2}:\text{there
exists }x\in C_{1}\text{ such that }x^{n}=y\\}$ is infinite and so, for each
$y\in C_{2}$, the multiplicative dimension of $y$ over $\pi^{\prime}(y)$ is
zero. By our assumption, the Zariski closures of $\pi^{\prime}(C_{1})$ and
$\pi^{\prime}(C_{2})$ in $\mathbb{G}_{m}^{k}(\mathbb{C})$ are both defined
over $\overline{\mathbb{Q}}$. It follows that both $C_{1}$ and $C_{2}$ are
defined over $\overline{\mathbb{Q}}$. This is a contradiction. ∎

##  3\. Exponential transcendence and exponentially transcendental powers

In this section we recall some material from [10] and then slightly modify a
result from [3].

######  Definition 3.1.

A partial exponential field $(F,A,\exp)$ consists of a field $F$ of
characteristic zero, a divisible subgroup $A$ of the additive group of $F$ and
a homomorphism $\exp$ from $A$ to the multiplicative group of $F$.

A total exponential field is a partial exponential field $(F,A,\exp)$ for
which $F=A$. An $E$-ring is a commutative ring with identity, equipped with a
homomorphism $E$ from its additive group to its multiplicative monoid. For
variables $X_{1},...,X_{k}$, $\mathbb{Z}[X_{1},...,X_{k}]^{E}$ denotes the
free $E$-ring generated by $\\{X_{1},...,X_{k}\\}$. If $(F,A,\exp)$ is a
partial exponential field then each $f\in\mathbb{Z}[X_{1},...,X_{k}]^{E}$
determines a partial function $f:F^{k}\rightarrow F$ in the obvious way. The
following definition, which comes from work of Macintyre [14], plays an
important role in [10].

######  Definition 3.2.

Let $(F,A,\exp)$ be a partial exponential field. Let $a\in F$ and $B\subseteq
F$. We say that $a\in\text{ecl}(B)$ if there exist
$f_{1},...,f_{k}\in\mathbb{Z}[X_{1},...,X_{k},Y_{1},...,Y_{m}]^{E}$,
$a_{1},...,a_{k}\in F$ and $b_{1},...,b_{m}\in B$ such that

  1. (1)

$a=a_{1}$,

  2. (2)

for all $i\in\\{1,...,k\\}$, $f_{i}(a_{1},...,a_{k},b_{1},...,b_{m})$ is
defined and equal to zero and

  3. (3)

the determinant of the matrix $\left(\frac{\partial f_{i}}{\partial
X_{j}}(a_{1},...,a_{k},b_{1},...,b_{m})\right)_{i,j\in\\{1,...,k\\}}$ is not
zero.

The determinant of the Jacobian matrix will be defined at
$(a_{1},...,a_{k},b_{1},...,b_{m})$ if each $f_{i}$ is and so there was no
need to include that as an assumption. Kirby also considers another closure
concept in [10]. This is defined in terms of derivations which respect
exponentiation. An exponential derivation $\partial$ on a partial exponential
field $(F,A)$ is a map $\partial:F\rightarrow M$ such that $M$ is an
$F$-vector space and the following conditions are satisfied.

  1. (1)

For all $a,b\in F$, $\partial(a+b)=\partial(a)+\partial(b)$.

  2. (2)

For all $a,b\in F$, $\partial(ab)=a\partial(b)+b\partial(a)$.

  3. (3)

For all $a\in A$, $\partial(\exp(a))=\exp(a)\partial(a)$.

######  Definition 3.3.

Let $(F,A,\exp)$ be a partial exponential field. Let $a\in F$ and $B\subseteq
F$. We say that $a\in\text{ecl}^{\prime}(B)$ if, for every exponential
derivation $\partial$ on $(F,A,\exp)$, if $\partial(b)=0$ for all $b\in B$
then $\partial(a)=0$.

For the main results of [10], it is important to recall that a pregeometry
$(X,\text{cl})$ consists of a set $X$ and a map
$\text{cl}:\mathcal{P}(X)\rightarrow\mathcal{P}(X)$ such that the following
conditions are satisfied.

  1. (1)

For all $B\subseteq X$, $B\subseteq\text{cl}(B)$ and
$\text{cl}(\text{cl}(B))=\text{cl}(B)$.

  2. (2)

For all $B,C\subseteq X$, if $B\subseteq C$ then
$\text{cl}(B)\subseteq\text{cl}(C)$.

  3. (3)

For all $B\subseteq X$ and $a\in X$, if $a\in\text{cl}(B)$ then there is a
finite subset $B_{0}\subseteq B$ such that $a\in\text{cl}(B_{0})$.

  4. (4)

For all $B\subseteq X$ and $a,c\in X$, if
$a\in\text{cl}(B\cup\\{c\\})\setminus\text{cl}(B)$ then
$c\in\text{cl}(B\cup\\{a\\})$.

The following result is Theorem 1.1 in [10]. It generalises earlier work of
Wilkie [17].

######  Theorem 3.4.

Let $(F,A,\exp)$ be a partial exponential field. For all $B\subseteq F$,
$\text{\em ecl}(B)=\text{\em ecl}^{\prime}(B)$. Furthermore, $(F,\text{\em
ecl})$ is a pregeometry.

It is well-known that any pregeometry $(F,\text{cl})$ induces a notion of
dimension. Let $a=(a_{1},...,a_{k})\in F^{k}$ and $B\subseteq F$. We say that
$a$ is cl-independent over $B$ if, for all $i\in\\{1,...,k\\}$,
$a_{i}\notin\text{cl}(B\cup\\{a_{1},...,a_{i-1},a_{i+1},...,a_{k}\\})$. The
dimension $\dim(a/B)$ of a finite tuple $a$ over a set $B$ is then defined to
be the length of any maximal subtuple of $a$ which is cl-independent over $B$.
We use some standard notational conventions regarding $\dim(a/B)$. For
example, if $a=(a_{1},...,a_{k})$, $b=(b_{1},...,b_{m})$ and
$c=(c_{1},...,c_{n})$ are finite tuples and $A$ and $B$ are sets then
$\dim(a,b/A,B,c)=\dim(d/D)$, where $d=(a_{1},...,a_{k},b_{1},...,b_{m})$ and
$D=A\cup B\cup\\{c_{1},...,c_{n}\\}$. The following property of dimension is
well-known and easy to prove.

######  Fact 3.5.

Let $(F,\text{\em cl})$ be a pregeometry. Let $a\in F^{k}$, $c\in F^{n}$ and
$B\subseteq F$. Then

$\dim(a,c/B)=\dim(c/B)+\dim(a/B,c).$

Given a field $F$ and a subfield $Q$, we have familiar pregeometries
$(F,\text{acl})$ and $(F,\text{span}_{Q})$, where $\text{acl}(B)$ is the
algebraic closure in $F$ of the subfield generated by $B$ and
$\text{span}_{Q}(B)$ is the $Q$-vector subspace of $F$ generated by $B$. The
dimensions associated with these two pregeometries are denoted by td and
$\text{ldim}_{Q}$ respectively. Given a partial exponential field $(F,A)$, the
dimension associated with ecl is denoted by $\dim$. The notation $\dim(a)$,
$\text{td}(a)$ and $\text{ldim}_{Q}(a)$ abbreviates $\dim(a/\emptyset)$,
$\text{td}(a/\emptyset)$ and $\text{ldim}_{Q}(a/\emptyset)$.

Theorem 1.2 in [10] applies to partial exponential fields, as well as total
ones, provided all instances of exponentiation are defined, as can be seen
from the proof. Making this explicit, the statement is as follows.

######  Theorem 3.6.

Let $(F,A,\exp)$ be a partial exponential field. Let $b\in A^{N}$ and
$K\subseteq F$ be such that $\text{\em ecl}(K)=K$. Then

$\text{\em td}(b,\exp(b)/K)-\text{\em ldim}_{\mathbb{Q}}(b/K)\geq\dim(b/K).$

Note that $\exp(b)$ is an abbreviation of $(\exp(b_{1}),...,\exp(b_{N}))$,
where $b=(b_{1},...,b_{N})$. The next result is a slight variant of Theorem
1.2 in [3]. That theorem was stated for total exponential fields. The version
we give here is suitable for our purposes and has almost the same proof. We
work over an exponentially-algebraically closed parameter set $K\subseteq F$,
which Bays, Kirby and Wilkie have in their proof though they omit it from the
statement of the theorem.

######  Theorem 3.7.

Let $(F,A,\exp)$ be a partial exponential field. Let $a=(a_{1},...,a_{N})\in
A^{N}$ and $\lambda\in F$. Let $K\subseteq F$ be such that $K=\text{\em
ecl}(K)$ and $\lambda\notin K$. Suppose $\lambda a=(\lambda a_{1},...,\lambda
a_{N})\in A^{N}$ and assume $\exp(a)$ is multiplicatively independent. Then

$\text{\em td}(\text{\em exp}(a),\text{\em exp}(\lambda a)/K,\lambda)\geq N.$

###### Proof.

To prove this in the case where $F=A$, Bays, Kirby and Wilkie first show that

$\text{td}(\text{exp}(a,\lambda
a)/K,\lambda)+\text{ldim}_{\mathbb{Q}(\lambda)}(a,\lambda
a/\text{ker})-\text{ldim}_{\mathbb{Q}}(a,\lambda a/\text{ker})\geq 0$

where $\text{ker}=\\{c\in F:\exp(c)=1\\}$. Their argument involves applying
Theorem 3.6 (above) with $b=(a,\lambda a,\lambda)$ and so makes use of
$\exp(\lambda)$. In our setting, $\exp(\lambda)$ might not be defined. We
therefore modify the argument slightly so that Theorem 3.6 is applied with
$b=(a,\lambda a)$. Adapting the first part of Section 4 in [3], we reason as
follows.

Let $b=(a,\lambda a)$. By Theorem 3.6,

$\text{td}(b,\exp(b)/K)-\text{ldim}_{\mathbb{Q}}(b/K)\geq 1$

and so

$\text{td}(b/K)+\text{td}(\exp(b)/K,b)-\text{ldim}_{\mathbb{Q}}(b/K)\geq 1.$

Now

$\text{td}(b/K)=\text{td}(b,\lambda/K)=\text{td}(\lambda/K)+\text{td}(b/K,\lambda)=1+\text{td}(b/K,\lambda).$

So then

$\text{td}(b/K,\lambda)+\text{td}(\exp(b)/K,b)-\text{ldim}_{\mathbb{Q}}(b/K)\geq
0.$

We have

$\text{td}(b/K,\lambda)\leq\text{ldim}_{\mathbb{Q}(\lambda)}(b/K)$

and

$\text{td}(\exp(b)/K,b)\leq\text{td}(\exp(b)/K,\lambda).$

Therefore

$\text{td}(\exp(b)/K,\lambda)+\text{ldim}_{\mathbb{Q}(\lambda)}(b/K)-\text{ldim}_{\mathbb{Q}}(b/K)\geq
0.$

Since $K=\text{acl}(K)$ and $\lambda\notin K$, we may deduce, exactly as in
[3], that

$\text{ldim}_{\mathbb{Q}(\lambda)}(b/K)-\text{ldim}_{\mathbb{Q}}(b/K)\leq\text{ldim}_{\mathbb{Q}(\lambda)}(b/\text{ker})-\text{ldim}_{\mathbb{Q}}(b/\text{ker}).$

This is done by applying part (iii) of Lemma 3.2 in [3] and then Lemma 3.3 in
[3]. So we have

$\text{td}(\exp(b)/K,\lambda)+\text{ldim}_{\mathbb{Q}(\lambda)}(b/\text{ker})-\text{ldim}_{\mathbb{Q}}(b/\text{ker})\geq
0.$

It remains to show that

$\text{ldim}_{\mathbb{Q}}(b/\text{ker})-\text{ldim}_{\mathbb{Q}(\lambda)}(b/\text{ker})\geq
N.$

This is done in Section 4 of [3], in the part subtitled “Proof of Theorem
1.2”, using no assumptions other than the ones we have made here. Note that
$x$, $z$ and $n$, in the relevant part of [3], correspond to our $a$, $b$ and
$N$, respectively. ∎

##  4\. Proofs of the main results

The first aim of this section is to prove Theorem 1.2. The full statement is
as follows.

######  Theorem 4.1.

Let $C_{1},C_{2}\subseteq\mathbb{G}_{m}^{N}(\mathbb{C})$ be irreducible closed
algebraic curves, with $N\geq 3$. Suppose there does not exist an algebraic
subgroup $G\subseteq\mathbb{G}_{m}^{N}(\mathbb{C})$ of dimension $1$ such that
$C_{1}\subseteq G$ and there does not exist an algebraic subgroup
$H\subseteq\mathbb{G}_{m}^{N}(\mathbb{C})$ of dimension $2$ such that
$C_{1}\cup C_{2}\subseteq H$. Further assume that at least one of the two
curves is not defined over $\overline{\mathbb{Q}}$. Let
$\mathcal{N}=\\{n\in\mathbb{N}:[n]C_{1}\subseteq C_{2}\\}$. Then
$\bigcup\limits_{n\in\mathbb{N}\setminus\mathcal{N}}\\{x\in C_{1}:x^{n}\in
C_{2}\\}$ is finite.

Let $\mathbb{R}_{(\mathbb{C},\exp)}$ denote the ordered field of real numbers
expanded to include the usual exponential field structure on
$\mathbb{C}=\mathbb{R}^{2}$. For some $\kappa>2^{\aleph_{0}}$, let
$R_{(F,\exp)}$ be a $\kappa$-saturated and strongly $\kappa$-homogeneous
elementary extension of $\mathbb{R}_{(\mathbb{C},\exp)}$, with $F=R^{2}$ and
$\exp:F\rightarrow F$. Let $A\subseteq F$ be the subgroup of elements with
infinitesimal modulus (including zero), so $A=\\{z\in
F:\left|z\right|<\frac{1}{n}\text{ for all }n\in\mathbb{N}\\}$. Restricting
$\exp$ to $A$ gives a partial exponential field $(F,A,\exp)$. We identify
$\mathbb{C}$ with its image under the elementary embedding, so
$\mathbb{C}\subseteq F$. Let $R_{(F,A,\exp,\mathbb{C})}$ be the expansion of
$R_{(F,\exp)}$ obtained by adding unary predicates for $A$ and $\mathbb{C}$.
Then let
$\widetilde{R}_{(\widetilde{F},\widetilde{A},\exp,\widetilde{\mathbb{C}})}$ be
the ultrapower of this with respect to a non-principal ultrafilter on
$\mathbb{N}$. In particular, $(\widetilde{F},\widetilde{A},\exp)$ is an
elementary extension of the partial exponential field $(F,A,\exp)$. If, in the
above construction, we restrict $\exp$ to $D=\\{z\in F:\left|z\right|\leq
1\\}$, instead of to $A$, then neither $(F,D,\exp)$ nor
$(\widetilde{F},\widetilde{D},\exp)$ is a partial exponential field, since $D$
is not a subgroup, but the structure
$\widetilde{R}_{(\widetilde{F},\widetilde{D},\exp)}$ has the advantage of
being o-minimal (see [8]).

For the rest of this section, acl denotes algebraic closure in the field
$\widetilde{F}$, ecl denotes exponential-algebraic closure in the partial
exponential field $(\widetilde{F},\widetilde{A},\exp)$ and dcl denotes
definable closure in the o-minimal structure
$\widetilde{R}_{(\widetilde{F},\widetilde{D},\exp)}$.

######  Lemma 4.2.

Let $B\subseteq\widetilde{\mathbb{C}}$. Then $\text{\em acl}(B)=\text{\em
ecl}(B)$.

###### Proof.

Let $a\in\widetilde{F}$ and suppose $a\in\text{ecl}(B)$. Then there exist
$f_{1},...,f_{k}\in\mathbb{Z}[X_{1},...,X_{k},Y_{1},...,Y_{m}]^{E}$,
$a_{1},...,a_{k}\in F$ and $b_{1},...,b_{m}\in B$ such that $a=a_{1}$,

$f_{1}(a_{1},...,a_{k},b_{1},...,b_{m})=...=f_{k}(a_{1},...,a_{k},b_{1},...,b_{m})=0$

and $(a_{1},...,a_{k})$ is an isolated solution of the system

$f_{1}(X_{1},...,X_{k},b_{1},...,b_{m})=...=f_{k}(X_{1},...,X_{k},b_{1},...,b_{m})=0.$

Using the fact that $\widetilde{A}\cap\widetilde{\mathbb{C}}=\\{0\\}$, there
are, for each $i\in\\{1,...,k\\}$, polynomials
$f_{i,1},...,f_{i,k_{i}}\in\mathbb{Z}[X_{1},...,X_{k},Y_{1},...,Y_{m}]$ such
that the equation $f_{i}(X_{1},...,X_{k},b_{1},...,b_{m})=0$ is implied by the
system

$f_{i,1}(X_{1},...,X_{k},b_{1},...,b_{m})=...=f_{i,k_{i}}(X_{1},...,X_{k},b_{1},...,b_{m})=0$

and, for any $a_{1}^{\prime},...,a_{k}^{\prime}\in\widetilde{\mathbb{C}}$,
$f_{i}(a_{1}^{\prime},...,a_{k}^{\prime},b_{1},...,b_{m})=0$ if and only if

$f_{i,1}(a_{1}^{\prime},...,a_{k}^{\prime},b_{1},...,b_{m})=...=f_{i,k_{i}}(a_{1}^{\prime},...,a_{k}^{\prime},b_{1},...,b_{m})=0.$

It follows that if our tuple $(a_{1},...,a_{k})$ is in
$\widetilde{\mathbb{C}}^{k}$ then $(a_{1},...,a_{k})$ is an isolated solution
of

$\bigwedge\limits_{i=\\{1,...,k\\}}f_{i,1}(X_{1},...,X_{k},b_{1},...,b_{m})=...=f_{i,k_{i}}(X_{1},...,X_{k},b_{1},...,b_{m})=0$

and so $a\in\text{acl}(B)$.

Suppose $(a_{1},...,a_{k})\notin\widetilde{\mathbb{C}}^{k}$. Then, by basic
o-minimality, $a_{i}\notin(\text{dcl}(B))^{2}$, for some $i\in\\{1,...,k\\}$.
It follows that the complete type of $(a_{1},...,a_{k})$ over $B$, in the
sense of $\widetilde{R}_{(\widetilde{F},\widetilde{D},\exp)}$, has infinitely
many realisations. By o-minimality, this contradicts the fact that
$(a_{1},...,a_{k})$ is an isolated solution of the system

$f_{1}(X_{1},...,X_{k},b_{1},...,b_{m})=...=f_{k}(X_{1},...,X_{k},b_{1},...,b_{m})=0.$

∎

It is worth noting the following.

######  Proposition 4.3.

For each $B\subseteq\widetilde{R}$ and $a\in\widetilde{F}$, considered as a
pair $a=(a_{1},a_{2})\in\widetilde{R}^{2}$, if $a\in\text{\em ecl}(B^{2})$
then $a_{1},a_{2}\in\text{\em dcl}(B)$.

By Proposition 2.8, to prove Theorem 4.1 it is sufficient to establish the
following.

######  Theorem 4.4.

Let $C_{1},C_{2}\subseteq\mathbb{G}_{m}^{3}(\mathbb{C})$ be irreducible closed
algebraic curves. Suppose there is no proper algebraic subgroup
$H\subseteq\mathbb{G}_{m}^{3}(\mathbb{C})$ such that $C_{1}\subseteq H$.
Further assume that at least one of the two curves is not defined over
$\overline{\mathbb{Q}}$. Let $\mathcal{N}=\\{n\in\mathbb{N}:[n]C_{1}\subseteq
C_{2}\\}$. Then $\bigcup\limits_{n\in\mathbb{N}\setminus\mathcal{N}}\\{x\in
C_{1}:x^{n}\in C_{2}\\}$ is finite.

###### Proof.

Suppose $\bigcup\limits_{n\in\mathbb{N}\setminus\mathcal{N}}\\{x\in
C_{1}:x^{n}\in C_{2}\\}$ is infinite. Let $e_{1},...,e_{k}\in\mathbb{C}$ be
algebraically independent such that $C_{1}$ and $C_{2}$ are defined over
$K=\text{acl}(e_{1},...,e_{k})$. We may assume that, considered as a
$2k$-tuple in the o-minimal structure $R_{(F,D,\exp)}$, $(e_{1},...,e_{k})$ is
dcl-independent. Let $\sigma$ be an automorphism of $R_{(F,D,\exp)}$ such that
$(\sigma(e_{1}),...,\sigma(e_{k}))$ is dcl-independent over $\mathbb{R}$ and
$\sigma(e_{i})-e_{i}\in A$, for all $i\in\\{1,...,k\\}$. Let
$K^{\prime}=\sigma(K)$, $C_{1}^{\prime}=\sigma(C_{1}(F))$,
$C_{2}^{\prime}=\sigma(C_{2}(F))$ and $e_{i}^{\prime}=\sigma(e_{i})$, for all
$i\in\\{1,...,k\\}$.

Letting $\widetilde{\mathbb{R}}$ denote the canonical extension of
$\mathbb{R}$ to the ultrapower
$\widetilde{R}_{(\widetilde{F},\widetilde{D},\exp)}$, we have that
$(e^{\prime}_{1},...,e^{\prime}_{k})$, as a $2k$-tuple, is dcl-independent
over $\widetilde{\mathbb{R}}$. Therefore, by Proposition 4.3, the $k$-tuple
$(e^{\prime}_{1},...,e^{\prime}_{k})$ is ecl-independent over
$\widetilde{\mathbb{C}}$. Working in the underlying real-closed field of
$R_{(F,D,\exp)}$, it is easy to see that, for each
$n\in\mathbb{N}\setminus\mathcal{N}$ and $x\in C_{1}$, if $x^{n}\in C_{2}$
then there exists $z\in C_{1}^{\prime}$ such that $z^{n}\in C_{2}^{\prime}$,
$x^{-1}z\in\exp(A)^{3}$ and $x^{-n}z^{n}\in\exp(A)^{3}$.

Letting $\widetilde{\mathbb{N}}$ denote the canonical extension of
$\mathbb{N}$ to the ultrapower
$\widetilde{R}_{(\widetilde{F},\widetilde{A},\exp,\widetilde{\mathbb{C}})}$,
we have an infinite $n\in\widetilde{\mathbb{N}}\setminus\mathcal{N}$ and
points $x\in C_{1}(\widetilde{\mathbb{C}})$ and $z\in
C_{1}^{\prime}(\widetilde{F})$ such that $x^{n}\in
C_{2}(\widetilde{\mathbb{C}})$, $z^{n}\in C_{2}^{\prime}(\widetilde{F})$,
$x^{-1}z\in\exp(\widetilde{A})^{3}$, $x^{-n}z^{n}\in\exp(\widetilde{A})^{3}$
and $x$ is not algebraic over $K$. As elements of an ultrapower, $x$, $z$ and
$n$ are really equivalence classes of sequences, so
$x=[(x_{j})_{j\in\mathbb{N}}]$, $z=[(z_{j})_{j\in\mathbb{N}}]$ and
$n=[(n_{j})_{j\in\mathbb{N}}]$. By $x^{n}$ and $z^{n}$ we mean
$[(x_{j}^{n_{j}})_{j\in\mathbb{N}}]$ and $[(z_{j}^{n_{j}})_{j\in\mathbb{N}}]$.

The point $x^{-1}z$ lies on the curve $x^{-1}C_{1}^{\prime}(\widetilde{F})$,
while $x^{-n}z^{n}$ lies on $x^{-n}C_{2}^{\prime}(\widetilde{F})$. These
curves are defined over $K^{\prime}(x,x^{n})$. We consider various cases and
obtain a contradiction in each.

Case 1: Suppose $x^{-1}z$ is multiplicatively independent. Let $u$ be a
maximal subtuple of the 6-tuple $(x,x^{n})$ such that $u$ is acl-independent
over $K\cup\\{n\\}$. It follows by Theorem 3.7 that
$n\in\text{ecl}(K,K^{\prime},u)$. Then $n\in\text{ecl}(K,u)$, since
$e_{1}^{\prime},...,e_{k}^{\prime}$ are ecl-independent over
$\widetilde{\mathbb{C}}$. By Proposition 4.2, $n\in\text{acl}(K,u)$. Since $n$
is infinite, $n\notin\text{acl}(K)$. This contradicts the acl-independence of
$u$ over $K\cup\\{n\\}$.

Case 2: Suppose $x^{-1}z$ has multiplicative dimension $2$. Then there is an
algebraic subgroup $H\subseteq\mathbb{G}_{m}^{3}(\widetilde{F})$ of dimension
$2$ such that $x^{-1}z\in H$ and $x^{-n}z^{n}\in H$. Then $x\in zH$ and
$x^{n}\in z^{n}H$.

Subcase 2a: Suppose both $C_{1}(\widetilde{F})\subseteq zH$ and
$C_{2}(\widetilde{F})\subseteq z^{n}H$. This contradicts the fact that
$\bigcup\limits_{m\in\mathbb{N}\setminus\mathcal{N}}\\{x\in C_{1}:x^{m}\in
C_{2}\\}$ is infinite, unless $zH$ is a translate of $H$ by a torsion point,
which would contradict the fact that $C_{1}$ is not contained in a proper
algebraic subgroup of $\mathbb{G}_{m}^{3}(\mathbb{C})$.

Subcase 2b: Suppose $C_{2}(\widetilde{F})\nsubseteq z^{n}H$. Then
$x^{n}\in\text{acl}(z^{n},K,K^{\prime})$. It follows from Corollary 2.4 and
our assumptions that $x^{n}\notin\text{acl}(K)$. Since
$e^{\prime}_{1},...,e^{\prime}_{k}$ are acl-independent over
$\widetilde{\mathbb{C}}$, $x^{n}\notin\text{acl}(K,K^{\prime})$. Therefore, by
exchange, $z^{n}\in\text{acl}(K,K^{\prime},x^{n})$. As in Case 1, let $u$ be
an acl-basis for $(x,x^{n})$ over $K\cup\\{n\\}$. By Theorem 3.7 and the fact
that $\text{td}(x^{-1}z,x^{-n}z^{n}/K,K^{\prime},u,n)\leq 1$,
$n\in\text{ecl}(K,K^{\prime},u)$. We obtain a contradiction as in Case 1.

Subcase 2c: The subcase where $C_{1}(\widetilde{F})\nsubseteq zH$ is analogous
to the previous one. This time we use the fact that $x\notin\text{acl}(K)$.

Case 3: Suppose $x^{-1}z$ has multiplicative dimension $1$. Then there is an
algebraic subgroup $H\subseteq\mathbb{G}_{m}^{3}(\widetilde{F})$ of dimension
$1$ such that $x\in zH$ and $x^{n}\in z^{n}H$. By Proposition 2.6,
$C_{1}(\widetilde{F})\nsubseteq zH$ and $C_{2}(\widetilde{F})\nsubseteq
z^{n}H$. So $x\in\text{acl}(K,K^{\prime},z)$ and
$x^{n}\in\text{acl}(K,K^{\prime},z^{n})$. By reasoning from Subcases 2b and
2c, we then have $z\in\text{acl}(K,K^{\prime},x)$ and
$z^{n}\in\text{acl}(K,K^{\prime},x^{n})$. Let $u$ be an acl-basis for
$(x,x^{n})$ over $K\cup\\{n\\}$. By Theorem 3.7 and the fact that
$\text{td}(x^{-1}z,x^{-n}z^{n}/K,K^{\prime},u,n)=0$,
$n\in\text{ecl}(K,K^{\prime},u)$. We obtain a contradiction as in Case 1.

Case 4: Suppose $x^{-1}z$ has multiplicative dimension zero. Then it is the
point $(1,1,1)$ and the curves $C_{1}(F)$ and $C_{1}^{\prime}(F)$ have
infinitely many points in common. It follows that
$C_{1}(F)=C_{1}^{\prime}(F)$. Similarly, $C_{2}(F)=C_{2}^{\prime}(F)$. Then
$C_{1}(F)$ and $C_{2}(F)$ are both defined over $K\cap
K^{\prime}=\overline{\mathbb{Q}}$, which is a contradiction. ∎

We now prove Theorem 1.4, stated again here for convenience.

######  Theorem 4.5.

Let $V_{1},V_{2}\subseteq\mathbb{G}_{m}^{N}(\mathbb{C})$ be irreducible closed
algebraic subvarieties, with $\dim(V_{1})\geq 1$ and
$\dim(V_{1})+\dim(V_{2})<N$. Suppose that, for any $\mathbb{Q}$-linearly
independent tuples of integers

$(n_{1,1},...,n_{1,N}),...,(n_{k,1},...,n_{k,N}),$

with $k=\dim(V_{1})+1$, and any generic point $(x_{1},...,x_{N})$ of $V_{1}$,
the monomials

$x_{1}^{n_{1,1}}\cdot...\cdot x_{N}^{n_{1,N}},...,x_{1}^{n_{k,1}}\cdot...\cdot
x_{N}^{n_{k,N}}$

are algebraically independent. Let
$\mathcal{N}=\\{n\in\mathbb{N}:[n]V_{1}\subseteq V_{2}\\}$. Then
$\bigcup\limits_{n\in\mathbb{N}\setminus\mathcal{N}}\\{x\in V_{1}:x^{n}\in
V_{2}\\}$ is not Zariski-dense in $V_{1}$.

###### Proof.

Suppose $\bigcup\limits_{n\in\mathbb{N}\setminus\mathcal{N}}\\{x\in
V_{1}:x^{n}\in V_{2}\\}$ is Zariski-dense in $V_{1}$. Let
$e_{1},...,e_{k}\in\mathbb{C}$ and $e^{\prime}_{1},...,e^{\prime}_{k}\in F$ be
as in the previous proof, except with $V_{1}$ in place of $C_{1}$ and $V_{2}$
in place of $C_{2}$. Let $K=\text{acl}(e_{1},...,e_{k})$ and
$K^{\prime}=\text{acl}(e^{\prime}_{1},...,e^{\prime}_{k})$. We obtain
$V^{\prime}_{1}$ and $V^{\prime}_{2}$ analogously to how $C^{\prime}_{1}$ and
$C^{\prime}_{2}$ were obtained.

Let $(x_{j})_{j\in\mathbb{N}}$ and $(n_{j})_{j\in\mathbb{N}}$ be sequences in
$V_{1}$ and $\mathbb{N}\setminus\mathcal{N}$ respectively such that
$x_{j}^{n_{j}}\in V_{2}$, for all $j\in\mathbb{N}$, and
$\\{x_{j}:j\in\mathbb{N}\\}$ is Zariski-dense in $V_{1}$. Then
$(n_{j})_{j\in\mathbb{N}}$ is not eventually constant. We get a generic (over
$K$) point $x=[(x_{j})_{j\in\mathbb{N}}]\in V_{1}(\widetilde{\mathbb{C}})$, an
infinite
$n=[(n_{j})_{j\in\mathbb{N}}]\in\widetilde{\mathbb{N}}\setminus\widetilde{\mathcal{N}}$
and some $z=[(z_{j})_{j\in\mathbb{N}}]\in V_{1}^{\prime}(\widetilde{F})$ such
that $x^{n}\in V_{2}(\widetilde{\mathbb{C}})$, $z^{n}\in
V_{2}^{\prime}(\widetilde{F})$, $x^{-1}z\in\exp(\widetilde{A})^{N}$ and
$x^{-n}z^{n}\in\exp(\widetilde{A})^{N}$. We may assume $z$ is a generic point
of $V^{\prime}_{1}(\widetilde{F})$ over $K^{\prime}$ and that
$\text{td}(z_{j}/K^{\prime})=\text{td}(z_{j}/K,K^{\prime},x_{j})$ for all
$j\in\mathbb{N}$.

Suppose $x^{-1}z$ satisfies $d$ independent multiplicative relations, with
$d\leq\dim(V_{1})+1$. Then there is a surjective algebraic homomorphism
$\phi:\mathbb{G}_{m}^{N}(\widetilde{F})\rightarrow\mathbb{G}_{m}^{d}(\widetilde{F})$
such that $\phi(x)^{-1}=\phi(z)$. It follows that
$\phi(x),\phi(z)\in\mathbb{G}_{m}^{d}(\widetilde{L})$, where
$L=\overline{\mathbb{Q}}$. Suppose $\phi^{-1}(\phi(z))\cap
V_{1}^{\prime}(\widetilde{F})$ has dimension greater than
$\dim(V_{1}^{\prime})-d$. Then the Zariski-closure of
$\phi(V_{1}^{\prime}(\widetilde{F}))$ has dimension less than $d$ and, since
$\mathbb{G}_{m}^{d}(\overline{\mathbb{Q}})$ is Zariski-dense in it, is defined
over $\overline{\mathbb{Q}}$. This contradicts our assumptions. So

$0\leq\dim(\phi^{-1}(\phi(z))\cap
V_{1}^{\prime}(\widetilde{F}))\leq\dim(V_{1}^{\prime})-d=\dim(V_{1})-d.$

Therefore $d\leq\dim(V_{1})$. So $x^{-1}z$ satisfies at most $\dim(V_{1})$
independent multiplicative relations. We may assume $d$ is maximal.

Let $u$ be a maximal subtuple of $(x,x^{n})$ such that $u$ is acl-independent
over $K\cup\\{n\\}$. Then
$\text{td}(x^{-1}z,x^{-n}z^{n}/K,K^{\prime},u,n)\leq\dim(V_{1})-d+\dim(V_{2})$
and the multiplicative dimension of $x^{-1}z$ is at least $N-d$. By Theorem
3.7, $n\in\text{ecl}(K,K^{\prime},u)$ and so
$n\in\text{ecl}(K,u)=\text{acl}(K,u)$. Since $n$ is infinite,
$n\notin\text{acl}(K)$. This contradicts the acl-independence of $u$ over
$K\cup\\{n\\}$. ∎

It follows easily from known results that the set $\mathcal{N}$ in this
theorem is finite.

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