1 Introduction
The classical Mordell–Lang conjecture for semiabelian varieties G defined over an algebraically closed field K of characteristic
$0$
(now a theorem due to Laurent [Reference LaurentLau84] for algebraic tori, to Faltings [Reference FaltingsFal91] for abelian varieties, and to Vojta [Reference VojtaVoj96] for the general case of semiabelian varieties) says that the intersection of a subvariety X of G with a finitely generated subgroup
$\Gamma $
of
$G(K)$
is a finite union of cosets of subgroups of
$\Gamma $
. The statement in positive characteristic (i.e., when K is an algebraically closed field of characteristic p) fails; see Hrushovski [Reference HrushovskiHru96]. When G is defined over a finite subfield
$\mathbb F_q$
of K, Moosa and Scanlon [Reference Moosa and ScanlonMS04] described the aforementioned intersection
$X(K)\cap \Gamma $
under the additional assumption that
$\Gamma $
is mapped into itself by a power of the Frobenius endomorphism F of G corresponding to
$\mathbb F_q$
, i.e.,
$\Gamma $
is a finitely generated
${\mathbb Z}[F^\ell ]$
-submodule of
$G(K)$
(for a suitable positive integer
$\ell $
). The precise description of the intersection
$X(K)\cap \Gamma $
is a finite union of F-sets (see Definition 1.2), as given in Theorem 1.3. In [Reference Ghioca and YangGY24], it was shown that if
$\Gamma $
is not a
${\mathbb Z}[F^\ell ]$
-submodule (for any
$\ell \in {\mathbb N}$
), then the structure of the intersection
$X(K)\cap \Gamma $
can be quite wild (see [Reference Ghioca and YangGY24, Section 2]); in particular, [Reference Ghioca and YangGY24, Examples 2.1, 2.2, and 2.3] (see also our Example 1.10) show that the intersection
$X(K)\cap \Gamma $
is no longer a finite union of F-sets (as opposed to what the author claimed erroneously in an earlier paper [Reference GhiocaGhi08]). Furthermore, a geometric description of the intersection
$X(K)\cap \Gamma $
was proven in [Reference Ghioca and YangGY24], even without assuming that G is defined over a finite subfield of K, but with the disadvantage that this description is not intrinsic to the subgroup
$\Gamma $
and instead it relies on the geometry of G. In the present paper, we obtain an explicit description (see Theorem 1.9) of the intersection
$X(K)\cap \Gamma $
, in the spirit of the original description of Moosa–Scanlon [Reference Moosa and ScanlonMS04].
1.1 The intersection of a subvariety of a semiabelian variety defined over a finite field with a finitely generated subgroup invariant under a power of the Frobenius endomorphism
Notation 1.1 Fr om this point on, we fix a semiabelian variety G defined over a finite subfield
$\mathbb F_q$
of an algebraically closed field K. We let F be the Frobenius endomorphism of G corresponding to
$\mathbb F_q$
.
In order to state the result of Moosa–Scanlon [Reference Moosa and ScanlonMS04], we introduce the notion of F-sets.
Definition 1.2 A groupless F-set is any subset of
$G(K)$
of the form:
$$ \begin{align} \left\{\alpha_0+\sum_{i=1}^r F^{k_in_i}(\alpha_i)\colon n_i\in\mathbb{N}\right\}, \end{align} $$
for given integers
$r\ge 0$
and
$k_i\ge 1$
, and given points
$\alpha _0,\alpha _1,\dots , \alpha _r\in G(K)$
. For any finitely generated subgroup
$\Gamma \subset G(K)$
, we define a groupless F-set in
$\Gamma $
(based in
$G(K)$
) as a groupless F-set contained in
$\Gamma $
. Also, an F-set in
$\Gamma $
(based in
$G(K)$
) is any set of the form
$S+B$
, where S is a groupless F-set in
$\Gamma $
and B is a subgroup of
$\Gamma $
(as always, for any two subsets A and B of a given group, we have that
$A+B$
is the set of all elements
$a+b$
where
$a\in A$
and
$b\in B$
).
Next, we state the main result of Moosa–Scanlon [Reference Moosa and ScanlonMS04].
Theorem 1.3 (Moosa–Scanlon [Reference Moosa and ScanlonMS04])
Let G, K,
$\mathbb F_q,$
and F be as in Notation 1.1. Let
$X\subseteq G$
be a subvariety defined over K and let
$\Gamma \subset G(K)$
be a finitely generated subgroup with the property that there exists
$\ell \in {\mathbb N}$
such that
$F^\ell (\Gamma )\subseteq \Gamma $
. Then,
$X(K)\cap \Gamma $
is a finite union of F-sets in
$\Gamma $
.
1.2 The intersection of a subvariety of a semiabelian variety with an arbitrary finitely generated subgroup
In order to state our main result regarding the intersection of a subvariety of G with an arbitrary finitely generated subgroup of G, we introduce a general arithmetic structure associated to any abelian group (see Definition 1.6). But first we define a general class of linear recurrence sequences (see Definition 1.5). We recall that for a linear recurrence equation
$\{a_n\}_{n\ge 1}$
given by the recurrence relation:
$$ \begin{align*}a_{n+m}=\sum_{i=0}^{m-1}c_ia_{n+i},\end{align*} $$
the characteristic equation is
$x^m-c_{m-1}x^{m-1}-\cdots - c_0=0$
; for more details, see [Reference Corvaja, Ghioca, Scanlon and ZannierCGSZ21, Section 2.3].
Definition 1.4 A subset
$S\subseteq \mathbb C^*$
is called powers-closed if for any
$r\in S$
and any non-negative integer n, we have that
$r^n\in S$
. (In particular,
$1\in S$
.)
Definition 1.5 Let
$S\subseteq \mathbb C^*$
be a powers-closed set. A linear recurrence sequence
$\{a_n\}_{n\ge 1}$
of integers is called an S-arithmetic sequence if the characteristic equation for the linear recurrence sequence
$\{a_n\}_{n\ge 1}$
has distinct roots, all contained in S.
So, with the notation as in Definition 1.5, for an S-arithmetic sequence
$\{a_n\}_{n\ge 1}$
, there exist distinct numbers
$r_1,\dots , r_m\in S$
and there exist complex numbers
$d_1,\dots , d_m$
such that
$$ \begin{align} a_n=\sum_{i=1}^m d_ir_i^n \text{ for each }n\ge 1. \end{align} $$
Next, we generalize the notion of S-arithmetic sequences inside an arbitrary abelian group.
Definition 1.6 Let
$(\Gamma _0,+)$
be an abelian group, and let
$S\subseteq \mathbb C^*$
be a powers-closed set. Given an integer
$r\ge 1$
, elements
$P_1,\dots , P_r\in \Gamma _0$
along with finitely many S-arithmetic sequences
$$ \begin{align*}\left\{a^{(1)}_n\right\}_{n\ge 1},\left\{a^{(2)}_n\right\}_{n\ge 1},\cdots,\left\{a^{(r)}_n\right\}_{n\ge 1},\end{align*} $$
we define an S-arithmetic groupless set
$\mathcal {U}$
as a set of the following form:
$$ \begin{align} \mathcal{U}:=\left\{\sum_{i=1}^r a^{(i)}_{n_i}\cdot P_i\colon n_1,\cdots, n_r\ge 1\right\}. \end{align} $$
Given some finitely generated subgroup
$\Gamma \subseteq \Gamma _0$
, we say that
$\mathcal {U}$
is an S-arithmetic groupless set in
$\Gamma $
(based in
$\Gamma _0$
) if
$\mathcal {U}\subseteq \Gamma $
. Furthermore, an S-arithmetic set in
$\Gamma $
(based in
$\Gamma _0$
) is defined as a set of the form
$\mathcal {U}+B$
, where
$\mathcal {U}$
is an S-arithmetic groupless set in
$\Gamma $
, while B is a subgroup of
$\Gamma $
.
The notion of S-arithmetic sets is inspired by the definition of F-sets, as shown by the following notation.
Notation 1.7 With G, K,
$\mathbb F_q,$
and F as in Notation 1.1, then inside the endomorphism ring
$\mathrm {End}(G)$
, we have that F is integral over
${\mathbb Z}$
, i.e., there exists
$m\ge 1$
along with integers
$c_0,\dots , c_{m-1}$
(where
$c_0\ne 0$
) such that
$$ \begin{align} F^m=\sum_{i=0}^{m-1}c_i F^i\text{ in }\mathrm{End}(G). \end{align} $$
The integrality of the Frobenius endomorphism of G follows from the fact that there exists a short exact sequence of algebraic groups
$1\longrightarrow T\longrightarrow G\longrightarrow A\longrightarrow 1$
over
$\overline {\mathbb F_q}$
(where T is an algebraic torus, while A is an abelian variety) and the corresponding Frobenius is integral over both the torus T and the abelian variety A. Furthermore, the equation
$$ \begin{align} x^m-c_{m-1}x^{m-1}-\cdots -c_1x-c_0=0 \end{align} $$
has distinct complex roots (for more details, see [Reference Corvaja, Ghioca, Scanlon and ZannierCGSZ21, Section 2.1]). We let
$S_F$
be the subset of
$\mathbb C^*$
consisting of all complex numbers of the form
$r^m$
for integers
$m\ge 0$
, where r varies among the roots of the equation (1.2.4).
Remark 1.8 Let
$S_F$
be as in Notation 1.7. Then using equation (1.2.3) (see also [Reference Corvaja, Ghioca, Scanlon and ZannierCGSZ21, Section 3]), we obtain that for any finitely generated subgroup
$\Gamma \subset G(K)$
, a groupless F-set in
$\Gamma $
(based in
$G(K)$
) is also an
$S_F$
-arithmetic groupless set in
$\Gamma $
(based in
$G(K)$
). Indeed, there exist m linear recurrence sequences in
$\mathbb {Z}$
:
$\{a^{(1)}_n\}_{n\in \mathbb {N}},\cdots , \{a^{(m)}_n\}_{n\in \mathbb {N}}$
, each one of them satisfying the recurrence relation
$$ \begin{align*}a^{(i)}_{n+m}=\sum_{k=0}^{m-1} c_k\cdot a^{(i)}_{n+k}\text{ for }n\ge 1,\end{align*} $$
such that for any point
$P\in G(K)$
, we have
$$ \begin{align*}F^n(P)=\sum_{i=0}^{m-1}a^{(i+1)}_n\cdot F^i(P).\end{align*} $$
Therefore, an F-set in
$\Gamma $
is an
$S_F$
-arithmetic set in
$\Gamma $
. On the other hand, there are many more
$S_F$
-arithmetic (groupless) sets in
$\Gamma $
, which are not (groupless) F-sets in
$\Gamma $
. Furthermore, our Example 1.10 shows that arbitrary
$S_F$
-arithmetic sets may appear in the intersection of a subvariety
$X\subset G$
with some finitely generated subgroup
$\Gamma \subset G(K)$
; in Theorem 1.9, we prove that
$X\cap \Gamma $
must always be an
$S_F$
-arithmetic set in
$\Gamma $
.
We prove the following main result.
Theorem 1.9 Let G, K,
$\mathbb F_q,$
and F be as in Notation 1.1, and let
$S_F$
be as in Notation 1.7. Let
$X\subseteq G$
be a subvariety defined over K and let
$\Gamma \subset G(K)$
be a finitely generated subgroup. Then, the intersection
$X(K)\cap \Gamma $
is a union of finitely many
$S_F$
-arithmetic sets in
$\Gamma $
(based in
$G(K)$
).
The following example shows that indeed one may obtain intersections
$X(K)\cap \Gamma $
in Theorem 1.9 which are not F-sets, but they are
$S_F$
-arithmetic sets.
Example 1.10 We let A and B be semiabelian varieties defined over a finite subfield
$\mathbb F_q$
of an algebraically closed field K (which we assume to have positive transcendence degree over
$\mathbb F_q$
), let
$G=A\times B$
, and let F be the corresponding Frobenius endomorphism associated to
$\mathbb F_q$
. We let h be the minimal (monic) polynomial with integer coefficients for which
$h(F)=0$
on B. Depending on the abelian part of the semiabelian variety B, the degree m of the polynomial h may be arbitrarily large.
We let
$C\subset B$
be a curve defined over
$\mathbb F_q$
with trivial stabilizer in B and let
$P\in C(K)$
be a non-torsion point; one can even choose C and P so that
$C(K)$
intersects the cyclic
$\mathbb {Z}[F]$
-module
$\Gamma _1$
spanned by P precisely in the orbit of P under the Frobenius endomorphism F. We also let
$Q_1,\dots , Q_m\in A(K)$
be linearly independent points (note that
$A(K)\otimes _{\mathbb {Z}}\mathbb {Q}$
is an infinite dimensional
$\mathbb {Q}$
-vector space). Then, we consider
$X:=A\times C$
and also, consider the group
$\Gamma \subset G(K)$
spanned by the points:
$$ \begin{align*}R_1:=(Q_1,P)\text{; }R_2:=(Q_2,F(P))\text{; }R_3:=\left(Q_3,F^2(P)\right)\text{; }\cdots\text{; }R_m:=\left(Q_m,F^{m-1}(P)\right).\end{align*} $$
Then letting
$\pi _2:G\longrightarrow B$
be the projection of
$G=A\times B$
on the second coordinate, we have that
$\pi _2(\Gamma )=\Gamma _1$
because
$\Gamma _1$
is spanned by the points
$$ \begin{align*}P\text{, }F(P)\text{, }F^2(P),\cdots, F^{m-1}(P)\in B(K)\end{align*} $$
since
$\Gamma _1$
is the cyclic
$\mathbb {Z}[F]$
-module spanned by P and
$h(F)(P)=0$
. So, we can find m sequences
$\left \{a^{(i)}_{n}\right \}_{n\ge 0}$
of integers (for
$i=0,\dots , m-1$
) such that for any
$n\ge 0$
, we have that
$$ \begin{align} F^n(P)=\sum_{i=0}^{m-1}a^{(i)}_n\cdot F^i(P). \end{align} $$
Equation (1.2.5) yields that
$X(K)\cap \Gamma $
is the set:
$$ \begin{align} \left\{\sum_{i=1}^{m} a^{(i-1)}_n\cdot R_i\colon n\ge 0\right\}. \end{align} $$
So, (1.2.6) shows that
$X(K)\cap \Gamma $
is not an F-set, but instead it is an
$S_F$
-arithmetic set.
1.3 Plan for our paper
We prove Theorem 1.9 as a consequence of a general statement regarding S-arithmetic sets in an abelian group.
Theorem 1.11 Let
$(\Gamma _0,+)$
be an abelian group, let
$\tilde {\Gamma }$
and
$\Gamma $
be finitely generated subgroups of
$\Gamma _0$
, and let
$S\subseteq \mathbb C^*$
be a powers-closed set. Then, the intersection of an S-arithmetic set in
$\tilde {\Gamma }$
(based in
$\Gamma _0$
) with
$\Gamma $
is a finite union of S-arithmetic sets in
$\Gamma $
(based in
$\Gamma _0$
).
In Remark 2.5, we show an example that, if one were to consider an extension of S-arithmetic sets (see Definition 1.6) in an abelian group which allows for linear recurrence sequences in equation (1.2.2) with characteristic roots of higher multiplicity, then the corresponding problem from Theorem 1.11 leads to some deep unknown questions from classical number theory (such as determining the positive integers whose squares have a given number of nonzero digits when written in base-
$2$
).
We prove Theorem 1.11 in Section 2; then Theorem 1.9 follows as an easy corollary of Theorem 1.11 (see Section 3).
2 Proof of Theorem 1.11
We prove Theorem 1.11 over the next several subsections of Section 2; so, throughout the entire Section 2, we work under the hypotheses and notation from Theorem 1.11.
We start with a very simple lemma, which is used repeatedly in our proofs.
Lemma 2.1 Let
$S\subseteq \mathbb C$
be a powers-closed set and let
$\{a_n\}_{n\in \mathbb {N}}$
be an S-arithmetic sequence.
-
(i) Then for each
$u\in \mathbb {N}\cup \{0\}$
and
$v\in \mathbb {N}$
, the sequence
$\{a_{un+v}\}_{n\in \mathbb {N}}$
is also an S-arithmetic sequence. -
(ii) Then for each
$u,v\in {\mathbb Z}$
, the sequence
$\{u\cdot a_n+v\}_{n\in \mathbb {N}}$
is also an S-arithmetic sequence.
Proof The result follows immediately using the general form (1.2.1) of an element in a linear recurrence sequence whose characteristic roots are non-repeated elements of S. Indeed, let
$r_1,\dots , r_m$
be the characteristic roots of
$\{a_n\}_{n\in \mathbb {N}}$
.
For part (i), note that the characteristic roots of the linear recurrence sequence
$\{a_{un+v}\}$
are
$r_i^u$
; since S is powers-closed, then each
$r_i^u$
is also in S.
For part (ii), note that adding a nonzero constant to a linear recurrence sequence leads to another linear recurrence sequence whose characteristic roots are distinct; they form the set
$\{r_1,\dots , r_m\}\cup \{1\}$
(also, note that
$1\in S$
because S is a powers-closed set).
We proceed to proving Theorem 1.11; first, we will sketch the plan for our proof in Section 2.1.
2.1 Plan for our proof of Theorem 1.11
The first step in our proof is to make a couple useful reductions
-
• In Section 2.2, we show that it suffices to prove Theorem 1.11 when
$\Gamma _0=\tilde {\Gamma }$
is a finitely generated group. -
• In Section 2.3, we show that it suffices to assume
$\Gamma _0$
is torsion-free.
So, having reduced the proof of Theorem 1.11 to the case the ambient group
$\Gamma _0$
is a torsion-free, finitely generated group, the next step (established in Section 2.4) is to reformulate our problem as a linear algebra question. Using the analysis from Sections 2.5 and 2.6, we conclude our proof of Theorem 1.11 in Section 2.7.
2.2 It suffices to prove Theorem 1.11 assuming
$\Gamma _0=\tilde {\Gamma }$
is finitely generated
Indeed, we have an S-arithmetic groupless set
$\mathcal {U}$
(contained in
$\tilde {\Gamma }$
) consisting of all elements of
$\Gamma _0$
of the form
$$ \begin{align} \sum_{i=1}^m a^{(i)}_{n_i}\cdot P_i\text{, as we vary }n_i\in{\mathbb N}, \end{align} $$
for some
$m\in {\mathbb N}$
, for some given elements
$P_i\in \Gamma _0$
, and for some given S-arithmetic sequences
$\{a^{(1)}_n\}_{n\ge 1},\cdots , \{a^{(m)}_n\}_{n\ge 1}$
. Then, given a subgroup
$H\subseteq \tilde {\Gamma }$
, our goal is to show that the S-arithmetic set
$$ \begin{align} \mathcal{F}:=\mathcal{U}+H \end{align} $$
intersects
$\Gamma $
in a finite union of S-arithmetic sets (in
$\Gamma $
). At the expense of replacing
$\tilde {\Gamma }$
with a larger subgroup of
$\Gamma _0$
(but still finitely generated), we may assume both that
$\Gamma \subseteq \tilde {\Gamma }$
and that each
$P_i\in \tilde {\Gamma }$
for
$i=1,\dots , m$
. Finally, without loss of generality, we may assume
$\Gamma _0=\tilde {\Gamma }$
is finitely generated.
2.3 It suffices to prove Theorem 1.11 assuming
$\Gamma _0$
is torsion-free
We already reduced proving Theorem 1.11 to the special case
$\Gamma _0=\tilde {\Gamma }$
is finitely generated. Therefore, since
$(\Gamma _0,+)$
is an abelian group, then
$\Gamma _0$
is the direct product of a (finitely generated) free subgroup
$\Gamma _{0,\mathrm {free}}$
with a finite torsion subgroup
$\Gamma _{0, \mathrm {tor}}$
. Hence, both H and
$\Gamma $
(being subgroups of
$\Gamma _0$
) are finite unions of cosets of subgroups of
$\Gamma _{0, \mathrm {free}}$
, i.e.,
$$ \begin{align} H=\bigcup_{i=1}^k \left(h_i+H_{\mathrm{free}}\right)\text{ and }\Gamma=\bigcup_{j=1}^\ell \left(\gamma_{j}+\Gamma_{\mathrm{free}}\right), \end{align} $$
for some
$k,\ell \in \mathbb {N}$
, some elements
$h_{i}$
and
$\gamma _{j}$
in
$\Gamma _{0}$
and some subgroups
$H_{\mathrm {free}}$
and
$\Gamma _{\mathrm {free}}$
of
$\Gamma _{0, \mathrm {free}}$
. So, it suffices to prove that for any given
$i_0\in \{1,\dots , k\}$
and
$j_0\in \{1,\dots , \ell \}$
, the intersection
$$ \begin{align} \left(\mathcal{U}+\left(h_{i_0}+H_{\mathrm{free}}\right)\right)\cap \left(\gamma_{j_0}+\Gamma_{\mathrm{free}}\right) \end{align} $$
is a finite union of S-arithmetic sets in
$\Gamma $
. The following observation will be used repeatedly in our proof.
Remark 2.2 Since a constant sequence (in
${\mathbb Z}$
) is itself a linear recurrence sequence whose characteristic root is
$1$
(this is covered by the case
$u=0$
in Lemma 2.1, either parts (i) or (ii)), we note that for any S-arithmetic groupless set
$\mathcal {G}$
and for any
$\delta \in \Gamma _0$
, we have that also
$\delta +\mathcal {G}$
is an S-arithmetic groupless set (note also that S contains
$1$
). Furthermore, if
$\mathcal {G}$
is an S-arithmetic set, then also
$\delta +\mathcal {G}$
is an S-arithmetic set.
We re-write the intersection from (2.3.2) as
$$ \begin{align} \gamma_{j_0}+\left(\left( {(-\gamma_{j_0}+h_{i_0})}+\mathcal{U}+H_{\mathrm{free}}\right)\cap \Gamma_{\mathrm{free}}\right). \end{align} $$
In light of Remark 2.2, it suffices to prove that
$$ \begin{align} \left((-\gamma_{j_0}+h_{i_0})+\mathcal{U}+H_{\mathrm{free}}\right)\cap \Gamma_{\mathrm{free}}\text{ is an}\ S \text{-arithmetic set in}\ \Gamma. \end{align} $$
Now, for
$i\in \{1,\dots , m\}$
, we write each
$P_i$
(see (2.2.1)) as
$P_{i, \mathrm {tor}}+P_{i, \mathrm {free}}$
with
$P_{i, \mathrm {tor}}\in \Gamma _{0, \mathrm {tor}}$
and
$P_{i, \mathrm {free}}\in \Gamma _{0,\mathrm {free}}$
. Also, we write
$-\gamma _{j_0}+h_{i_0}=\eta _{\mathrm {tor}}+\eta _{\mathrm {free}}$
for some
$\eta _{\mathrm {tor}}\in \Gamma _{0,\mathrm {tor}}$
and
$\eta _{\mathrm {free}}\in \Gamma _{0, \mathrm {free}}$
.
We let
$M:=|\Gamma _{0, \mathrm {tor}}|$
. Because each linear recurrence sequence of integers is preperiodic modulo any given integer (and thus, in particular, preperiodic modulo M), we obtain that the set of tuples
$(n_1,\dots , n_m)$
of positive integers satisfying
$$ \begin{align} \eta_{\mathrm{tor}}+\sum_{i=1}^m a^{(i)}_{n_i}\cdot P_{i, \mathrm{tor}}=0\text{ in }\Gamma_0 \end{align} $$
is a finite union of sets of the form
$$ \begin{align*}\left\{(k_1n_1+\ell_1, k_2n_2+\ell_2,\dots, k_mn_m+\ell_m)\colon \text{ for arbitrary } n_1,\dots, n_m\ge 1\right\}\end{align*} $$
for some given
$2m$
-tuples of integers:
$$ \begin{align} (k_1,\ell_1,k_2,\ell_2,\cdots, k_m,\ell_m)\text{ with }k_i\ge 0\text{ and }\ell_i\ge 1. \end{align} $$
Thus, at the expense of replacing each linear recurrence sequence
$\left \{a^{(i)}_n\right \}_{n\ge 1}$
with
$\left \{a^{(i)}_{k_in+\ell _i}\right \}_{n\ge 1}$
(which is still an S-arithmetic sequence, as shown in Lemma 2.1 (i)), we reduce our problem to proving that the intersection with
$\Gamma _{\mathrm {free}}$
of the S-arithmetic set
$$ \begin{align} \mathcal{F}_1:=\mathcal{S}_1+H_{\mathrm{free}}, \end{align} $$
where
$\mathcal {S}_1$
is the S-arithmetic groupless set given by
$$ \begin{align} \mathcal{S}_1:=\left\{\eta_{\mathrm{free}}+\sum_{i=1}^m a^{(i)}_{n_i}\cdot P_{i,\mathrm{free}}\colon n_1,\dots,n_m\ge 1\right\} \end{align} $$
is a finite union of S-arithmetic sets in
$\Gamma $
.
Therefore, from now on, we work under the assumption that
$\Gamma _0$
is torsion-free (see also equations (2.3.7) and (2.3.8)).
2.4 Reduction to a linear algebra question
So, we work under the assumption that
$\Gamma _0$
is a finitely generated, free abelian group; hence it is isomorphic to
$\mathbb {Z}^r$
and we let
$Q_1,\dots , Q_r\in \Gamma _0$
be some fixed generators for
$\Gamma _0$
.
We have a subgroup
$\Gamma \subseteq \Gamma _0$
and also, we have an S-arithmetic set
$\mathcal {F}\subseteq \Gamma _0$
. Furthermore,
$\mathcal {F}=\mathcal {U}+H$
for an S-arithmetic groupless set
$\mathcal {U}$
and for a subgroup
$H\subseteq \Gamma _0$
. Since H is a subgroup of
$\Gamma _0$
, then it is also a finitely generated, free abelian group; thus, we let
$\{R_1,\dots , R_s\}$
be a given
${\mathbb Z}$
-basis for H (where
$s\le r$
). For each
$i\in \{1,\dots , s\}$
, we write
$R_i$
in terms of the basis
$\{Q_1,\dots , Q_r\}$
of
$\Gamma _0$
as
$$ \begin{align} R_i:=\sum_{j=1}^r b_{i,j}Q_j\text{ for some }b_{i,j}\in{\mathbb Z}. \end{align} $$
The S-arithmetic groupless set
$\mathcal {U}$
consists of all elements of the form
$$ \begin{align} \sum_{i=1}^m a^{(i)}_{n_i}\cdot P_i\text{, as we vary }n_i\in{\mathbb N}, \end{align} $$
for some given elements
$P_1,\dots , P_m\in \Gamma _0$
, where
$\{a^{(i)}_n\}_{n\in \mathbb {N}}$
are S-arithmetic sequences (for
$1\le i\le m$
). Then for each
$i\in \{1,\dots , m\}$
, we write
$P_i$
as
$$ \begin{align} P_i:=\sum_{j=1}^r c_{i,j}Q_j\text{ for some }c_{i,j}\in{\mathbb Z}. \end{align} $$
Therefore, a point in
$\mathcal {F}=\mathcal {S}+H$
is of the form
$$ \begin{align} \sum_{k=1}^s y_k\cdot R_k + \sum_{i=1}^m a^{(i)}_{n_i}\cdot P_i, \end{align} $$
for some arbitrary integers
$y_k$
and arbitrary positive integers
$n_i$
. Hence, using (2.4.1) and (2.4.3), we write any point in
$\mathcal {F}$
as:
$$ \begin{align} \sum_{j=1}^r \left(\sum_{k=1}^s b_{k,j}y_k + \sum_{i=1}^m c_{i,j}a^{(i)}_{n_i}\right)\cdot Q_j, \end{align} $$
for some arbitrary
$y_k\in {\mathbb Z}$
and some arbitrary
$n_i\in \mathbb {N}$
.
Now, following [Reference GhiocaGhi08, Definition 3.5], we define
$\mathbf {C}$
-subsets,
$\mathbf {L}$
-subsets, and
$\mathbf {CL}$
-sets of
${\mathbb Z}^r$
; their notation comes from congruence equation (for
$\mathbf { C}$
-subset), respctively linear equation (for
$\mathbf {L}$
-subset).
Definition 2.3 A
$\mathbf {C}$
-subset of
${\mathbb Z}^m$
is a set
$\mathbf {C}(d_1, \dots , d_r,D)$
, where
$d_1,\dots , d_r,D\in {\mathbb Z}$
(with
$D\ne 0$
), which is the set of all solutions
$(x_1,\dots , x_r)\in {\mathbb Z}^r$
of the congruence equation
$\sum _{i=1}^r d_ix_i\equiv 0\ \pmod {D}$
.
An
$\mathbf {L}$
-subset of
${\mathbb Z}^r$
is a set
$\mathbf {L}(d_1,\dots , d_r)$
, where
$d_1,\dots , d_r\in {\mathbb Z}$
, which is the set of all solutions
$(x_1, \dots , x_r)\in {\mathbb Z}^r$
of the linear equation
$\sum _{i=1}^r d_ix_i=0$
.
A
$\mathbf {CL}$
-subset of
${\mathbb Z}^r$
is either a
$\mathbf {C}$
-subset or an
$\mathbf {L}$
-subset of
${\mathbb Z}^r$
.
Remark 2.4 We note that our
$\mathbf {C}$
-subsets,
$\mathbf {L}$
-subsets, and
$\mathbf {CL}$
-subsets are slightly simpler than the ones defined in [Reference GhiocaGhi08] because we can absorb any coset of a subgroup in our definition of S-arithmetic groupless subsets of
$\Gamma _0$
since the constant sequence is a linear recurrence sequence with unique characteristic root equal to
$1$
(see also Lemma 2.1 and Remark 2.2).
As proven in [Reference GhiocaGhi08, Subclaim 3.6], given a subgroup
$\Gamma $
of
$\Gamma _0$
, then there exist finitely many
$\mathbf {C}$
-subsets
$C_i$
of
${\mathbb Z}^r$
(with the index i varying in some finite set I) and there exist finitely many
$\mathbf {L}$
-subsets
$L_j$
of
${\mathbb Z}^r$
(with the index j varying in some finite set J) such that for any
$(x_1,\dots , x_r)\in {\mathbb Z}^r$
, we have
$$ \begin{align} \sum_{i=1}^r x_iQ_i\in\Gamma \end{align} $$
if and only if
$$ \begin{align} (x_1,\dots, x_r)\in \left(\bigcap_{i\in I}C_i\right)\bigcap \left(\bigcap_{j\in J}L_j\right). \end{align} $$
Combining equations (2.4.5), (2.4.6), and (2.4.7), our problem reduces to analyzing for which integers
$y_1,\dots , y_s$
and for which positive integers
$n_1,\dots , n_m$
, we have that
$$ \begin{align} \left(\sum_{k=1}^s b_{k,j}y_k + \sum_{i=1}^m c_{i,j}a^{(i)}_{n_i}\right)_{1\le j\le r}\in \left(\bigcap_{i\in I}C_i\right)\bigcap \left(\bigcap_{j\in J}L_j\right). \end{align} $$
We analyze the conditions imposed on the tuples
$(y_1,\dots ,y_s,n_1,\dots , n_m)$
so that the left-hand side from equation (2.4.8) belongs to some
$C_i$
, respectively some
$L_j$
. We split our analysis for these two cases over the next two Sections 2.5 and 2.6; there are significant differences between these two cases, one of them being that the case of
$\mathbf {C}$
-subsets is easier than the case of
$\mathbf { L}$
-subsets and it can be treated one congruence equation at a time.
2.5 The case of
$\mathbf {C}$
-subsets of
${\mathbb Z}^r$
In this Section 2.5, we analyze the condition that the left-hand side of equation (2.4.8) belongs to some given
$\mathbf {C}$
-subset
$C\subseteq {\mathbb Z}^r$
. Hence, for given integers
$d_1,\dots , d_r$
and nonzero integer D, we analyze the equation
$$ \begin{align} \sum_{j=1}^r d_j\cdot \left(\sum_{k=1}^s b_{k,j}y_k + \sum_{i=1}^m c_{i,j}a^{(i)}_{n_i}\right)\equiv 0\ \ \pmod{D}. \end{align} $$
We note that in equation (2.5.1), the unknowns are the
$y_k$
’s and the
$n_i$
’s, while D, the
$d_j$
’s, the
$b_{k,j}$
’s and the
$c_{i,j}$
’s are given integers. Since any linear recurrence sequence of integers (such as each
$\{a^{(i)}_n\}_{n\ge 1}$
) is preperiodic modulo any given integer (such as D), we obtain that there exist finitely many tuples of non-negative integers
$$ \begin{align*}(u_1,v_1,u_2,v_2,\cdots, u_{s+m},v_{s+m})\end{align*} $$
such that the set of tuples
$(y_1,\dots , y_s,n_1,\dots , n_m)$
satisfying equation (2.5.1) is a finite union of sets of the form
$$ \begin{align} \left\{(u_1y_1+v_1,\cdots, u_sy_s+v_s,u_{s+1}n_1+v_{s+1},\cdots, u_{s+m}n_m+ v_{s+m})\colon \text{for }y_i\in{\mathbb Z}\text{ and }n_i\in{\mathbb N}\right\}. \end{align} $$
As shown in Lemma 2.1 (i), replacing each recurrence sequence
$\{a^{(i)}_n\}_{n\in \mathbb {N}}$
by
$\{a^{(i)}_{u_{s+i}n+v_{s+i}}\}_{n\in \mathbb {N}}$
(for
$i=1,\dots , m$
) leads to another m linear recurrence sequences with simple characteristic roots that all live in the set S. Furthermore, replacing each
$y_k$
by
$u_ky_k+v_k$
leads to replacing the subgroup H with a coset
$\delta _1+H_1$
of a subgroup
$H_1\subseteq H$
. Then, once again using Remark 2.2, we conclude that replacing each
$y_k$
by
$u_ky_k+v_k$
(for
$1\le k\le s$
) and replacing each
$n_i$
by
$u_{s+i}n_i+v_{s+i}$
(for
$1\le i\le m$
) leads to replacing the S-arithmetic subset
$\mathcal {F}=\mathcal {S}+H$
by another S-arithmetic set
$\mathcal {S}_1+H_1$
, where
$\mathcal {S}_1$
is the S-arithmetic groupless set
$$ \begin{align*}\mathcal{S}_1:=\left\{\sum_{k=1}^s v_kR_k+\sum_{i=1}^m a^{(i)}_{u_{s+i}n_i+v_{s+i}}\cdot P_i\colon n_i\in\mathbb{N}\right\},\end{align*} $$
while
$H_1$
is the subgroup of H spanned by
$u_1R_1,\cdots , u_kR_k$
. Therefore, from now on, we may assume that in the right-hand side of the equation (2.4.8) we only have
$\mathbf {L}$
-subsets
$L_j$
.
2.6 The case of
$\mathbf {L}$
-subsets of
${\mathbb Z}^r$
So, with the notation as in equation (2.4.8), letting
$|J|=u$
, we need to find
$(y_1,\dots , y_s,n_1,\dots , n_m)\in {\mathbb Z}^s\times \mathbb {N}^m$
such that
$$ \begin{align} \left(\sum_{k=1}^s b_{k,j}y_k + \sum_{i=1}^m c_{i,j}a^{(i)}_{n_i}\right)_{1\le j\le r}\in \bigcap_{1\le h\le u}L_h, \end{align} $$
where for each
$h\in \{1,\dots , u\}$
, the subset
$L_h\subseteq {\mathbb Z}^r$
is cut out by the linear equation
$$ \begin{align} \sum_{j=1}^r d_{h,j}x_j=0, \end{align} $$
for some given integers
$d_{h,j}$
. Combining equation (2.6.1) with the linear equations (2.6.2) (for
$1\le j\le u$
), we obtain a system of u linear equations:
$$ \begin{align} \sum_{k=1}^s \left(\sum_{j=1}^r b_{k,j}d_{h,j}\right)\cdot y_k=- \sum_{i=1}^m \left( \sum_{j=1}^r d_{h,j}c_{i,j}\right)\cdot a^{(i)}_{n_i}\text{ for }1\le h\le u. \end{align} $$
For each
$h=1,\dots , u$
and for each
$k=1,\dots , s$
, we let
$$ \begin{align*}e_{h,k}:=\sum_{j=1}^r b_{k,j}d_{h,j};\end{align*} $$
also, for each
$h=1,\dots , u$
and for each
$i=1,\dots , m$
, we let
$$ \begin{align*}f_{h,i}:=-\sum_{j=1}^r d_{h,j}c_{i,j}.\end{align*} $$
Clearly,
$e_{h,k}\in {\mathbb Z}$
(for each
$1\le h\le u$
and
$1\le k\le s$
) and
$f_{h,i}\in {\mathbb Z}$
(for each
$1\le h\le u$
and
$1\le i\le m$
); also, we recall that each
$a^{(i)}_{n_i}\in {\mathbb Z}$
for
$1\le i\le m$
and each
$n_i\in \mathbb {N}$
. So, we have a linear system of u equations with unknowns
$y_1,\dots , y_s$
:
$$ \begin{align} \sum_{k=1}^s e_{h,k}y_k = \sum_{i=1}^m f_{h,i}a^{(i)}_{n_i}\text{ for }1\le h\le u. \end{align} $$
In order for the system (2.6.4) be solvable for some
$y_1,\dots , y_s\in {\mathbb Q}$
, there are finitely many linear relations to be satisfied by the right-hand side terms from (2.6.4); since each
$e_{h,k}$
is an integer, these linear relations will have integer coefficients as well. Hence, there are finitely many equations, say w equations (for some
$w\ge 0$
), and there are some given integers
$g_{\ell ,h}$
with
$1\le \ell \le w$
and
$1\le h\le u$
such that:
$$ \begin{align} \sum_{h=1}^u g_{\ell,h}\cdot \sum_{i=1}^m f_{h,i}a^{(i)}_{n_i}=0, \end{align} $$
which need to be satisfied (for
$1\le \ell \le w$
) by the positive integers
$n_i$
in order to find a solution
$(y_1,\dots , y_s)$
(even over the rationals) for the system (2.6.4). Letting
$$ \begin{align*}z_{\ell,i}:=\sum_{h=1}^u g_{\ell,h}\cdot f_{h,i}\text{ for each }\ell\in\{1,\dots, w\}\text{ and for each }i\in\{1,\dots, m\},\end{align*} $$
then equations (2.6.5) translate to equations:
$$ \begin{align} \sum_{i=1}^m z_{\ell,i}\cdot a^{(i)}_{n_i}=0\text{ for }\ell=1,\dots, w. \end{align} $$
Since each sequence
$\{a^{(i)}_n\}_{n\in \mathbb {N}}$
is a linear recurrence sequence with simple characteristic roots
$\tilde {r}_{i,1},\dots , \tilde {r}_{i,m_i}\in S$
(for some
$m_i\in \mathbb {N}$
), then the equations (2.6.6) translate into equations:
$$ \begin{align} \sum_{i=1}^m\sum_{j=1}^{m_i} \tilde{z}_{\ell,i,j} \cdot \tilde{r}_{i,j}^{n_i}=0\text{ for }\ell=1,\dots, w, \end{align} $$
for some constants
$\tilde {z}_{\ell ,i,j}\in \mathbb C$
. The famous theorem of Laurent [Reference LaurentLau84] (which solves the classical Mordell–Lang conjecture for algebraic tori) yields that the set of tuples
$(n_1,\dots , n_m)\in \mathbb {N}^m$
satisfying the equations (2.6.7) is a union of finitely many sets
$\tilde {S}_k$
of the following form. Each set
$\tilde {S}_k$
consists of all tuples
$(n_1,\dots , n_m)\in \mathbb {N}^m$
satisfying finitely many equations of the form:
$$ \begin{align} n_j=\tilde{n}_{0,j}\text{ for some given }\tilde{n}_{0,j}\in\mathbb{N}, \end{align} $$
or
$$ \begin{align} n_j\equiv \tilde{n}_{0,j}\ \ \pmod{\tilde{N}_{0,j}}\text{ for some given }\tilde{n}_{0,j},\tilde{N}_{0,j}\in\mathbb{N}, \end{align} $$
or
$$ \begin{align} \tilde{n}_{0,j}\cdot n_j=\tilde{n}_{0,j_1}\cdot n_{j_1}+b\text{ for some given} b\in\mathbb{Z} \text{and }j\ne j_1\text{ and }\tilde{n}_{0,j},\tilde{n}_{0,j_1}\in\mathbb{N}. \end{align} $$
Remark 2.5 This is the only part of our proof which requires that in the definition of S-arithmetic sets the corresponding linear recurrence sequences have distinct characteristic roots. Indeed, otherwise, the problem becomes very difficult. For example, consider the case when
$\Gamma _0={\mathbb Z}^2$
,
$\Gamma ={\mathbb Z}\times \{0\}$
and
$\mathcal {F}=\mathcal {S}$
is the set of all elements of
${\mathbb Z}^2$
of the form:
$$ \begin{align} n_0^2\cdot (1,1) + \sum_{i=1}^m 2^{n_i}\cdot (1,-1)\text{ for arbitrary }n_0,n_1,\dots, n_m\in\mathbb{N}. \end{align} $$
The set (2.6.11) corresponds to the linear recurrence sequences
$\{n_0^2\}_{n_0\ge 1}$
along with
$\{2^{n_i}\}_{n_i\ge 1}$
(for
$1\le i\le m$
); their characteristic roots live in the powers-closed set
$\{2^s\colon s\ge 0\}$
, but the first of these linear recurrence sequences has
$1$
as a repeated characteristic root. Then, analyzing the counterpart of Theorem 1.11 for this example leads to finding all solutions
$(n_0,n_1,\dots , n_m)\in {\mathbb N}^{m+1}$
for the polynomial–exponential equation:
$$ \begin{align} n_0^2=\sum_{i=1}^m 2^{n_i}. \end{align} $$
Equation (2.6.12) is a very deep Diophantine question, well-beyond the reach of the current known methods; for more details, see [Reference Corvaja, Ghioca, Scanlon and ZannierCGSZ21].
Thus, it suffices to prove that for each such set
$\tilde {S}:=\tilde {S}_k\subseteq \mathbb {N}^m$
(satisfying finitely many equations of the form (2.6.8), (2.6.9), and (2.6.10)), the corresponding set of points (2.4.5) lying in
$\mathcal {F}\cap \Gamma $
is an S-arithmetic subset in
$\Gamma $
.
On the other hand, due to the simple form of the equations (2.6.8), (2.6.9), and (2.6.10), the set
$$ \begin{align} \left\{\sum_{i=1}^m a^{(i)}_{n_i}\cdot P_i\colon (n_1,\dots, n_m)\in\tilde{S}\right\} \end{align} $$
is still an S-arithmetic groupless set (see Lemma 2.1 (i)). Therefore, at the expense of replacing the original S-arithmetic groupless set
$\mathcal {U}$
by the set from (2.6.13), we may assume that the linear system of equations (2.6.4) is solvable (in
${\mathbb Q}$
) for each
$(n_1,\dots ,n_m)\in \mathbb {N}^m$
.
So, at the expense of re-shuffling our variables
$(y_1,\dots ,y_s)$
(which amounts to re-labelling the points
$R_1,\dots , R_s$
), we may assume that
$y_1,\dots , y_t$
(with
$t\le s$
) are free variables for the linear system (2.6.4) and the general solution in
${\mathbb Q}$
to the system (2.6.4) consists of tuples
$(y_1,\dots , y_s)$
with the property that
$y_1,\dots , y_t$
are arbitrary rational numbers, while for
$j=1,\dots ,s-t$
, we have
$$ \begin{align} y_{t+j}=\sum_{k=1}^t \tilde{a}_{j,k}\cdot y_k + \sum_{\substack{1\le i\le m\\1\le h\le u}} \tilde{b}_{j,h}\cdot f_{h,i}\cdot a^{(i)}_{n_i}, \end{align} $$
for some given constants
$\tilde {a}_{j,k},\tilde {b}_{j,h}\in {\mathbb Q}$
, while the
$n_i$
’s from (2.6.14) are arbitrary positive integers. For each
$j=1,\dots , s-t$
and each
$i=1,\dots , m$
, we let
$$ \begin{align*}\tilde{c}_{j,i}:=\sum_{h=1}^u \tilde{b}_{j,h}f_{h,i};\end{align*} $$
so, we can re-write equation (2.6.14) as follows:
$$ \begin{align} y_{t+j}=\sum_{k=1}^t \tilde{a}_{j,k}\cdot y_k + \sum_{i=1}^m \tilde{c}_{j,i}\cdot a^{(i)}_{n_i}. \end{align} $$
2.7 Conclusion for our proof of Theorem 1.11
Knowing that the constants
$\tilde {a}_{j,k}$
and
$\tilde {c}_{j,i}$
from equation (2.6.15) are rational numbers, coupled with the fact that each linear recurrence sequence of integers is preperiodic with respect to any given moduli, then
$y_{t+j}\in {\mathbb Z}$
for each
$j=1,\dots , s-t$
in equation (2.6.15) yields that the corresponding set of all tuples
$(y_1,\dots , y_t,n_1,\dots ,n_{m})\in {\mathbb Z}^t\times \mathbb {N}^{m}$
consists of finitely many sets of the form
$$ \begin{align} \left\{\left(u_1y_1+v_1,\cdots, u_ty_t+v_t,u_{t+1}n_1+v_{t+1},\cdots , u_{t+m}n_{m}+v_{t+m}\right)\colon \text{ with } y_i\in{\mathbb Z}\text{ and }n_j\in\mathbb{N} \right\} \end{align} $$
for some given integers
$u_j,v_j$
for
$1\le j\le t+m$
(note that the
$u_i$
and
$v_j$
’s are fixed in (2.7.1), while the
$y_i$
and
$n_j$
vary in
$\mathbb {Z}$
, respectively in
$\mathbb {N}$
). So, we consider now a given choice of integers
$u_j$
and
$v_j$
as in equation (2.7.1) so that for each
$(y_1,\dots , y_t,n_1,\dots ,n_{m})\in {\mathbb Z}^t\times \mathbb {N}^{m}$
, we have that
$y_{t+1},\dots , y_s$
given as in (2.6.15) provide a solution over the integers to the system (2.6.4).
Next, we show that the set of points in
$\mathcal {F}$
corresponding to a subset
$\tilde {E}\subseteq {\mathbb Z}^t\times \mathbb {N}^{m}$
of the form (2.7.1) is an S-arithmetic set. Indeed, we note first that replacing each
$n_i$
(for
$1\le i\le m$
) by
$u_{t+i}\cdot n_i+v_{t+i}$
yields that the corresponding points in
$\mathcal {U}$
:
$$ \begin{align*}\sum_{i=1}^m a^{(i)}_{u_{t+i}n_i+v_{t+i}}\cdot P_i\text{ (as we vary} (n_1,\dots, n_m)\in\mathbb{N}^m)\end{align*} $$
still form an S-arithmetic groupless set (once again, see Lemma 2.1 (i)). Second, we claim that the set of all points in H of the form:
$$ \begin{align} \sum_{k=1}^t (u_ky_k+v_k)\cdot R_k+ \sum_{j=1}^{s-t} y_{t+j}\cdot R_{t+j}, \end{align} $$
with
$y_{t+1},\dots , y_s$
given as in equation (2.6.15) (where each
$y_i$
is replaced by
$u_iy_i+v_i$
for
$i=1,\dots , t$
) is actually an S-arithmetic set. Indeed, the set from (2.7.2) can be re-written as the sum of an S-arithmetic groupless subset (see also Lemma 2.1 and Remark 2.2):
$$ \begin{align} \sum_{k=1}^t v_k\cdot R_k + \sum_{j=1}^{s-t}\left(\sum_{k=1}^t \tilde{a}_{j,k}v_k + \sum_{i=1}^{m} \tilde{c}_{j,i}a^{(i)}_{u_{t+i}n_i+v_{t+i}}\right)\cdot R_{t+j} \end{align} $$
(as we vary
$n_1,\dots , n_m$
freely in
$\mathbb {N}$
) with the subgroup of H consisting of all points:
$$ \begin{align} \sum_{k=1}^t u_ky_k\cdot R_k + \sum_{j=1}^{s-t}\left(\sum_{k=1}^t \tilde{a}_{j,k}u_ky_k\right)\cdot R_{t+j} , \end{align} $$
as we vary
$y_1,\dots , y_t$
freely in
${\mathbb Z}$
.
This concludes our proof of Theorem 1.11.
2.8 Further remarks
As suggested by the anonymous referee, we state here a special case of the result we proved in Section 2.6 since its statement may be of independent interest. As noted by the referee, Ginsburg and Spanier [Reference Ginsburg and SpanierGS66] proved that a subset of
$\mathbb {Z}^d$
is definable in
$\left (\mathbb {Z}, +, 0, 1, <\right )$
if and only if it is a finite union of cosets of subgroups of
$\mathbb {Z}^d$
(they actually showed this result for subsets of
$\mathbb {N}^d$
, but the proof carries to the slightly more general setting of subsets of
$\mathbb {Z}^d$
). So, then it may be of interest both to model theorists and to number theorists the following result.
Proposition 2.6 Let
$d\in \mathbb {N}$
, let
$m\ge 1$
, let
$P_1,\dots , P_m\in \mathbb {Z}^d$
and let
$\{a_{i,n}\}_{n\in \mathbb {Z}}$
be simple linear recurrence sequences of integers, for each
$i\in \{1,\dots , m\}$
. Then for any finite union V of cosets of subgroups of
$\mathbb {Z}^d$
, the set
$$ \begin{align} \left\{(n_1,\dots, n_m)\in\mathbb{Z}^m\colon \sum_{i=1}^m a_{i,n_i}\cdot P_i\in V\right\} \end{align} $$
is a finite union of sets, each one of them satisfying finitely many equations of the form
$$ \begin{align} n_j=\tilde{n}_{0,j}\text{ for some given }\tilde{n}_{0,j}\in\mathbb{Z}, \end{align} $$
(for some
$j=1,\dots , m$
), or
$$ \begin{align} n_j\equiv \tilde{n}_{0,j}\ \ \pmod{\tilde{N}_{0,j}}\text{ for some given }\tilde{n}_{0,j},\tilde{N}_{0,j}\in\mathbb{Z}, \end{align} $$
(for some
$j=1,\dots , m$
), or
$$ \begin{align} \tilde{n}_{0,j}\cdot n_j=\tilde{n}_{0,j_1}\cdot n_{j_1}+b\text{ for some given}\ b\in\mathbb{Z}\ \text{and } \tilde{n}_{0,j},\tilde{n}_{0,j_1}\in\mathbb{Z}, \end{align} $$
for some
$j\ne j_1$
in
$\{1,\dots , m\}$
.
Actually, Proposition 2.6 was first obtained in [Reference GhiocaGhi08] (as part of proving that the intersection between a groupless F-set with a coset of subgroup is a finite union of groupless F-sets), but it was not formally stated there. In our proof of Theorem 1.11 (specifically in Section 2.6), we derived an even stronger result than Proposition 2.6 since we dealt with intersecting subgroups of
$\Gamma $
with an S-arithmetic set, whose components (inside
$\mathbb {Z}^d$
) are more general than a sum of simple linear recurrence sequences; note that an S-arithmetic set inside
${\mathbb Z}^d$
consists of a sum between a subgroup of
${\mathbb Z}^d$
with a set as the one appearing in equation (2.8.1).
3 Proof of Theorem 1.9
Theorem 1.9 follows now as an easy corollary of Theorem 1.11.
So, we let G, K, X,
$\Gamma $
,
$\mathbb F_q$
,
$F,$
and
$S_F$
be as in Theorem 1.9 (see also Notations 1.1 and 1.7). We let
$\tilde {\Gamma }$
be the finitely generated
${\mathbb Z}[F]$
-submodule of
$G(K)$
spanned by
$\Gamma $
; in particular,
$\tilde {\Gamma }$
is also a finitely generated subgroup of
$G(K)$
. Using Theorem 1.3, the intersection
$X(K)\cap \tilde {\Gamma }$
is a finite union of F-sets in
$\tilde {\Gamma }$
(based in
$G(K)$
). Furthermore, according to Remark 1.8, we have that each F-set in
$\tilde {\Gamma }$
is also an
$S_F$
-arithmetic set in
$\tilde {\Gamma }$
(based in
$G(K)$
). Finally, using Theorem 1.11 (with
$\Gamma _0:=G(K)$
) coupled with the fact that
$$ \begin{align*}X(K)\cap\Gamma=\left(X(K)\cap\tilde{\Gamma}\right)\cap \Gamma,\end{align*} $$
we conclude that
$X(K)\cap \Gamma $
is a finite union of
$S_F$
-arithmetic sets in
$\Gamma $
(based in
$G(K)$
), as desired.
Remark 3.1 As remarked by the referee (whom we thank once again for their useful comments), Hieronymi–Schulz [Reference Hieronymi and SchulzHS22] have shown that the first-order theory of
$(\mathbb {N}, 0, 1, +, k^{\mathbb {N}},\ell ^{\mathbb {N}})$
is undecidable whenever k and
$\ell $
are multiplicatively independent. That means there are first-order formulas involving powers of
$2$
and
$3$
(for example) for which we cannot decide whether the given formula is true or false in the standard model of
$\mathbb {N}$
. In particular, this hints that the related question of whether one can decide if
$X(K)\cap \Gamma $
is non-empty in Theorem 1.9 is very difficult when the Frobenius has at least two multiplicatively independent eigenvalues; note that our Theorem 1.11 in fact shows this is related to Skolem’s problem [Reference SkolemSko34] for simple linear recurrence sequences.
Acknowledgment
We thank the anonymous referee for their very useful suggestions and comments.
