RATIONAL SOLUTIONS OF POLYNOMIAL-EXPONENTIAL
EQUATIONS
AYHAN GÜNAYDIN
Abstract. We give a description of solutions of polynomial-exponential equations where the variables vary over rational numbers. Using this, we also prove
a finiteness result.
1. Introduction
Let X = (X1 , . . . , Xt ) be a tuple of indeterminates. Fix P1 , . . . , Ps ∈ C[X] and
α1 , . . . , αs ∈ Ct . We consider the following equation
s
X
(b)
Pi (X) exp(X · αi ) = 0,
i=1
where exp denotes the usual exponentiation map on complex numbers and X · αi
is short for X1 αi1 + · · · + Xt αit .
A solution a = (a1 , . . . , at ) ∈ Ct of (b) is said to be nondegenerate if
X
Pi (a) exp(a · αi ) 6= 0
i∈I
for every nonempty proper subset I of {1, . . . , s}.
Consider the Q-vector space
V := {q ∈ Qt : q · αi = q · αi0 for all i, i0 }.
Fix a complement V 0 of V in Qt and let π : Qt → V and π 0 : Qt → V 0 denote the
natural projections.
We give a description of nondegenerate rational solutions of (b) as follows.
Theorem 1.1. Given P1 , . . . , Ps ∈ C[X] and α1 , . . . , αs ∈ Ct , there is N ∈ N>0
such that if q ∈ Qt is a nondegenerate solution of (b), then π 0 (q) ∈ ( N1 Z)t .
Note that searching for n = (n1 , . . . , nt ) ∈ Zt satisfying (b) amounts to solving the
classical polynomial-exponential equation
(bb)
s
X
i=1
Pi (n)
t
Y
n
βijj = 0,
j=1
where βij = exp(αij ) (i = 1, . . . , s, j = 1, . . . , t).
Date: March 26, 2012.
2010 Mathematics Subject Classification. 11D61, 11U09.
The author was supported by the FCT grant SFRH/BPD/47661/2008.
1
2
AYHAN GÜNAYDIN
The study of these kind of equations and their relations with linear recurrence
sequences has been a major area of research. For a full account of the area, we
direct the reader to the excellent treatise by Schmidt [11]. Here we only give an
overview of the literature that is relevant to our work.
The first paper to recount the solutions of (bb) is [6] by Laurent. In that paper, he
works with equations where all the data are algebraic and later he generalized his
results to arbitrary complex coefficients in [7]. In order to state his main result, let
H = {n ∈ Zt :
t
Y
j=1
n
βijj =
t
Y
n
βi0 jj for every i, i0 }.
j=1
Theorem 1.2. (Laurent [7]) (i) If Pi is constant for each i, then the set of nondegenerate solutions of (bb) is a finite union of translates of H.
(ii) There are constants a, b ∈ R depending on the polynomials Pi and the numbers
βij such that if n is a nondegenerate solution of (bb), then there is n0 ∈ H with
|n − n0 | < a log(|n|) + b.
(Here and below |(c1 , . . . , cl )| := max{|c1 |, . . . , |cl |}.)
This is a slightly weaker version of Théorème 1 of [7], because there the author
considers not only nondegenerate solutions, but solutions that are ‘maximal’ with
respect to a partition of {1, . . . , s}. Due to the use of specialization arguments,
the constants in this theorem are not effective. However, in the case where all the
coefficients lie in a number field, one can obtain quantitative results. Most notably,
in the papers [9] and [10], Schlickewei and Schmidt gave bounds depending only
on the number of non-zero coefficients of the polynomials Pi and the degree of
the number field. The proofs in those papers rely on the ‘Subspace Theorem’ of
Schmidt. Later Corvaja-Zannier ([1]) and Fuchs ([5]) obtained finer bounds on
the zeros of certain linear recurrence sequences, using the p-adic generalization by
Schlickewei of the Subspace Theorem.
Another situation where one gets effective results is the case of function fields.
In [12], Zannier considers the situation that the coefficients of the polynomials Pi
and the numbers βij vary in a function field K over an algebraically closed field
Qt
n
k. He describes the set of integer tuples n such that Pi (n) j=1 βijj are linearly
dependent over k as a finite union of simpler sets, which play the role of H in
Laurent’s theorem; moreover the number of these simpler sets depend only on s, t
and the degrees of the polynomials Pi .
The cases where the coefficients of the polynomials Pi and the numbers βij are not
algebraic are generally easier to work out due to the fact that no use of Diophantine
approximation is required.
We prove Theorem 1.1, by reducing the equation (b) to an equation of the form (bb).
We do this by using some earlier results on multiplicative subgroups of fields;
note
that the ‘exponential part’ of the equation, namely the tuple exp(q ·αi ) i varies in
a subgroup of (C× )s of finite rank. The main obstacle in this approach is handling
the roots of unity. A priori for many solutions q ∈ Qt of (b) the exponential part
might be the same, but we go around this by using a theorem of Dvornicich and
Zannier (see Theorem 3.2 below).
RATIONAL SOLUTIONS OF POLYNOMIAL-EXPONENTIAL EQUATIONS
3
We finish this introduction by collecting two consequence of Theorem 1.1. First,
combining it with Laurent’s result, we get a finer understanding of rational solutions.
Corollary 1.3. Given P1 , . . . , Ps ∈ C[X] and α1 , . . . , αs ∈ Ct , there are N ∈ N>0
and cr , dr ∈ R for each r ∈ V such that if q ∈ Qt is a nondegenerate solution
√ of
(b), then π 0 (q) ∈ ( N1 Z)t and there is m ∈ Zt satisfying m · (αi − αi0 ) ∈ Z2N π −1
for each i, i0 such that
m
|π 0 (q) − | < cπ(q) log(|π 0 (q)|) + dπ(q) .
N
We also have the following finiteness result as a consequence of Corollary 1.3.
Corollary 1.4. Let P1 , . . . , Ps ∈ C[X] and α1 , . . . , αs ∈ Ct . Suppose that the set
{βij : i = 1, . . . , s, j = 1, . . . , t} is multiplicatively independent. Then there are only
finitely many nondegenerate solutions q ∈ Qt of (b).
Acknowledgements. The author thanks Amador Martin-Pizarro and Mehmet Haluk
Şengün for very fruitful discussions related to this paper.
2. Linear relations in multiplicative groups
In this section we recall some notations and an earlier result that will be useful in
the rest of the paper and we make the first reduction.
Let K be any field and Γ a subgroup of K × . We consider solutions in Γ of
λ1 x1 + · · · + λk xk = 1,
(*)
where λ1 , . . . , λk ∈
P K. We say that a solution γ = (γ1 , . . . , γk ) in Γ of (*) is
non-degenerate if
λi γi 6= 0 for every nonempty proper subset I of {1, . . . , k}.
i∈I
Definition 2.1. Let G be an abelian group, written multiplicatively. A subgroup
H of G is radical (in G) if for each n > 0 and g ∈ G we have g ∈ H whenever
g n ∈ H.
Given A ⊆ G, we set hAiG to be the smallest radical subgroup of G containing A.
That is,
hAiG = {g ∈ G | g n ∈ [A]G for some n ∈ N}
where [A]G denotes the subgroup generated by A. When G is clear from the context,
we will drop the subscripts and just write hAi and [A].
Also in what follows, U denotes the multiplicative group of roots of unity.
We use the following result from [2].
Lemma 2.2. Let E ⊆ F be fields such that E ∩ U = F ∩ U and G be a radical
subgroup of E × . Then for λ1 , . . . , λn ∈ E × , the equation (*) has the same nondegenerate solutions in G as in hGiF × .
In the rest of the paper, it will be more convenient to consider the following equation
rather than (b)
s
X
(bbb)
Pi (X) exp(X · α0i ) = 0,
i=1
4
AYHAN GÜNAYDIN
where α0i = (αij − α1j )j=1,...,t . Note that q is a nondegenerate solution of (b) if and
only if it is a nondegenerate solution of (bbb). Hence we do not loose any information
by replacing αi with α0i . Let S denote the set of nondegenerate solutions q ∈ Qt
0
of (bbb), and put βij
= exp(αij − α1j ).
Let A be a finite set containing the coefficients of the polynomials Pi and the
0
and put Γ = hAiC× . If q ∈ S, then P1 (q) 6= 0 and the tuple (exp(q·α0i ) :
numbers βij
i = 2, . . . , s) ∈ Γs−1 is a non-degenerate solution of the linear equation
(**)
P2 (q)
Ps (q)
Y2 + · · · +
Ys = 1.
−P1 (q)
−P1 (q)
Note that when q varies in Qt with P1 (q) 6= 0, the coefficients of this equation vary
in the field Q(A).
Let E := Q(U ∪ A) and G := hAiE × . Now by taking C in the place of F in Lemma
2.2, we see that all the possible non-degenerate solutions of the linear equation (**)
in Γ are already in G.
Let G0 be the complement of U in G. Therefore if q ∈ S, then there are roots of
unity ζq2 , . . . , ζqs and ηq2 , . . . , ηqs ∈ G0 such that
(***)
P1 (q) + P2 (q)ηq2 ζq2 + · · · + Ps (q)ηqs ζqs = 0.
We elaborate on the relation between q and the numbers ηqi later; we first conclude
that there is a bound on the order of ζqi .
3. Roots of unity
We first remark the following easy observation, whose proof follows the lines of
the proof of Lemme 4 of [7] (Note that Laurent considered only finitely generated
Q-algebras, however his conclusion is deeper).
Lemma 3.1. Let R be a subring of Q̄[S], where S is a finite subset of C. Suppose
that b1 , . . . , bq are elements of R and let q 0 be the linear dimension over Q̄ of b.
Then there are ring homomorphisms φ1 , . . . , φq0 from R to Q̄ fixing k := R ∩ Q̄ such
that for every α1 , . . . , αq ∈ k with α1 b1 + · · · + αq bq 6= 0 there is some i ∈ {1, . . . , q 0 }
with φi (α1 b1 + · · · + αq bq ) 6= 0.
Proof. After enlarging R, we may assume that q = q 0 . It suffices then to find
φi : R → Q̄ for each i ∈ {1, . . . , q} fixing k such that the determinant
φ1 (b1 ) · · · φ1 (bq ) ..
..
Dq := .
.
φq (b1 ) · · · φq (bq ) is nonzero.
We proceed by induction on q. For q = 1 take, using Nullstellensatz, a ring homomorphism R[b−1
1 ] → Q̄ that fixes k. Clearly, its restriction to R sends b1 to some
non-zero element.
RATIONAL SOLUTIONS OF POLYNOMIAL-EXPONENTIAL EQUATIONS
Assume now that φ1 , . . . , φq−1 have been
Then the determinant
φ1 (b1 )
..
0
.
Dq := φq−1 (b1 )
b1
5
already constructed such that Dq−1 6= 0.
···
···
···
φq−1 (bq ) bq
φ1 (bq )
..
.
is β1 b1 + · · · + βq bq , where β1 , . . . , βq are algebraic numbers. In particular, by
induction, βq = Dq−1 6= 0. Therefore, since we are assuming that the tuple b is
Q̄-linearly independent, we conclude that Dq0 6= 0. Consider
R0 := R[(Dq0 )−1 ].
Nullstellensatz implies that there is a ring morphism φq from R0 to Q̄ fixing k 0 :=
R0 ∩ Q̄. Its restriction to R has the property that φq Dq0 6= 0 which implies that
Dq 6= 0.
In order to bound the degrees of the roots of unity appearing in (***) we need the
following result.
Theorem 3.2. (Theorem 1 in [3])
Let F be a number field, a0 , a1 , . . . , ak in F and ζ a root of unity of order Q such that
k
P
a0 +
aj ζ nj = 0 with gcd(Q, n1 , . . . , nk ) = 1. Let δ = [F ∩ Q(ζ) : Q] and suppose
j=1
P
that for any nonempty proper subset I of {0, 1, . . . , k} the sum
aj ζ nj 6= 0. Then
j∈I
for each prime p and n > 0, if pn+1 |Q, then pn |2δ and
X
k ≥ dimF (F + F ζ n1 + · · · + F ζ nk ) ≥ 1 +
[
p|Q,p2 -Q
p−1
− 1].
gcd(δ, p − 1)
In particular, the order Q of ζ is bounded by a constant depending on k and δ (and
therefore [F : Q]).
We are ready to prove the following.
Lemma 3.3. There is T ∈ N depending only on the coefficients of the polynomials
Pi and the numbers βij such that if q is a nondegenerate solution of (bbb), then all
the corresponding roots of unity ζqi are of order less than T .
Proof. Let
ζqi
√
lqi 2π −1
),
= exp(
nqi
with lqi < nqi and gcd(lqi , nqi ) = 1. We need to find a bound T on the integers
nqi independent of q.
Let R be the Q̄-algebra generated by elements of A and their inverses. Using Lemma
3.1 with appropriate b1 , . . . , bq , choose a specialization φ such that
φ(P1 (q)) 6= 0.
6
AYHAN GÜNAYDIN
The homomorphism φ transforms (***) into a non-trivial relation
s
X
φ(Pi (q))φ(ηqi ) ζqi = 0.
i=1
So we have a relation
X
aj ζ j = 0,
j∈I
√
exp( 2π T −1 )
with T = lcm(nqi : i = 1, . . . , s) and the aj ’s are algebraic
where ζ =
numbers depending on q and not all zero.
For our purposes we may assume that no subsum is 0. Then Theorem 3.2 gives
that T depends only on the degree of F and |I| and not on q.
4. Proof of Theorem 1.1
In order to finish the proof of Theorem 1.1, we need the following Kummer theoretic
result from [14].
Proposition 4.1. (Zilber) Let L be a finitely generated extension of Q(U). Then
the quotient group L× /U is a free abelian group.
It follows from this proposition that the group G0 from Section 2 is finitely generated. Indeed G0 can be considered as a subgroup of E × /U, which is a free abelian
group by the proposition and being of finite rank, G0 is actually finitely generated.
Then by using Lemma 3.3, if q ∈ S, then (q ·α02 , . . . , q ·α0s ) is in a finitely generated
subgroup A of
W := {(q · α02 , . . . , q · α0s ) : q ∈ Qt }.
Note that W is a Q-linear subspace of C(s−1) and there is a natural surjective linear
map ψ : Qt → W taking q to (q · α02 , . . . , q · α0s ). Then the kernel of ψ is V from
the Introduction.
The preimage of A under ψ is of the form V ⊕ B where B is a finitely generated
subgroup of V 0 . Note that the rank of B is at most the dimension of V 0 .
Let B = Zr 1 ⊕ · · · ⊕ Zr d , where r 1 , . . . , r d ∈ Qt . For k = 1, . . . , d, write r k as
k1
kt
( abk1
, . . . , abkt
), where akj ∈ Z, bkj ∈ N>0 with gcd(ajk , bjk ) = 1 for each j = 1, . . . , t.
Now let
N := lcm(bkj : j = 1, . . . , t, k = 1, . . . , d).
( N1 Z)t .
Then B ⊆
Theorem 1.1.
Hence if q ∈ S, then π 0 (q) ∈ ( N1 Z)t , finishing the proof of
5. Final remarks
The motivation for this work is to answer the following question affirmatively.
Question. Assume Schanuel’s conjecture. Let p(X, Y ) ∈ C[X, Y ] be irreducible in
which both X and Y appear. Let C be the complex curve defined by p. Is it correct
that for each finite A ⊆ C, there is a generic point on C over Q(A) which is of the
form (α, exp(α))?
RATIONAL SOLUTIONS OF POLYNOMIAL-EXPONENTIAL EQUATIONS
7
Schanuel’s conjecture is the statement that for every Q-linearly independent complex numbers α1 , . . . , αn , we have
trdegQ Q α1 , . . . , αn , exp(α1 ), . . . , exp(αn ) ≥ n.
The question above stems from the model theoretic study of the field of complex
numbers expanded by the usual exponential function, (C, exp). In [13], Zilber gives
an axiomatization of a first order theory and conjectures that (C, exp) is a model
of that theory. Since one of the axioms is Schanuel’s conjecture, it seems like
Zilber’s conjecture is out of reach. However, one can try to reduce it to Schanuel’s
conjecture. An affirmative answer to the question above would be a first step for
such a purpose.
To find such a generic point, it is sufficient to show that for each subfield K of
C of finite transcendence degree, there are only finitely many α ∈ K such that
p(α, exp(α)) = 0 (see [8] for an explanation of this). So let K be such a subfield.
Using Schanuel’s conjecture one can conclude that α ∈ K with p(α, exp(α)) = 0
lie in a Q-linear subspace of K of finite dimension. Then the question reduces to
understanding the rational solutions of certain polynomial-exponential equations.
Unfortunately, Corollary 1.3 in this article does not imply that there are only finitely
many such solutions. However, one can try to make the integer N appearing in
Theorem 1.1 effective in the case that the coefficients of the polynomials Pi and the
numbers exp(αij ) vary in a number field. The uses of Lemma 2.2 and Proposition
4.1 in this paper veil that sort of effectiveness.
An alternative approach to solve this question is using the Wronskian techniques
from the papers [12] and [4]. It might even be possible to make use of the results
there directly to conclude partial results. It would, of course, be remarkable to
find proofs independent of Schanuel’s Conjecture in certain known cases, like when
p ∈ Q[X, Y ] (see [8] for a proof of this assuming Schanuel’s Conjecture.)
In order to realize the difficulty of such a project, one could look at the simplest
case when p is X − Y . In this case, for a given algebraically closed field K of finite
transcendence degree, we need to find a fixed point of exp outside of K and we do
not know how to do this without using Schanuel’s Conjecture.
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E-mail address: [email protected]