On multiplicative independence of rational function iterates

We give lower bounds for the degree of multiplicative combinations of iterates of rational functions (with certain exceptions) over a general field, establishing the multiplicative independence of said iterates. This leads to a generalisation of Gao’s method for constructing elements in the finite field Fqn\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb {F}}_{q^n}$$\end{document} whose orders are larger than any polynomial in n when n becomes large. Additionally, we discuss the finiteness of polynomials which translate a given finite set of polynomials to become multiplicatively dependent.


Introduction and main results
We say that n non-zero elements a 1 , . . . , a n of a ring are multiplicatively independent if, for integers k 1 , . . . , k n , we have that a k 1 1 . . . a k n n = 1 if and only if k 1 = · · · = k n = 0. Otherwise we say they are multiplicatively dependent. Multiplicative independence, especially of values of polynomials and rational functions, is being increasingly studied. In [4], Bombieri, Masser and Zannier initiate study of the intersection of algebraic curves with proper algebraic subgroups of the multiplicative group G n m . It turns out (see [3,Corollary 3.2.15]) that each such subgroup of G n m is defined by finitely many equations of the form X k 1 1 . . . X k n n = 1, where k 1 , . . . , k n are integers, not all zero. As such, [4], which leads into the area of "unlikely intersections", really concerns the multiplicative dependence of points on curves. More recently, we see multiplicative independence being studied in the context of arithmetic dynamics. In [18], it is shown that under fairly natural conditions on rational functions f 1 , . . . , f s over a number field K, the values f 1 (α), . . . , f s (α) are multiplicatively independent for all but finitely many α ∈ K ab , where K ab is the maximal abelian extension of K. This leads to results on multiplicative dependence in the orbits of a univariate polynomial dynamical system.
Clearly, to study the multiplicative independence of elements in the orbits of polynomials or rational functions, it is necessary to know when the given functions are multiplicatively dependent, as in this case all their values must be multiplicatively dependent. We study this problem in the context of iterates of rational functions over a field.
Throughout the paper, F will denote a field of characteristic p (zero or prime), and f ∈ F(X ) a non-constant rational function in lowest terms over F. That is, f = g/h with d := deg f = max {deg g, deg h} ≥ 1. Being in "lowest terms" means gcd(g, h) = 1, or equivalently, g and h share no roots in any extension field of F. As such, when referring to zeros and poles of a rational function, we mean roots of its numerator and denominator respectively in an algebraic closure F of F. We recursively define the iterates of f by In [10], Gao considers the multiplicative independence of polynomials over a finite field F q , where q is a prime power, proving that if f ∈ F q [X ] is not a monomial or certain binomial, then the iterates f (1) , . . . , f (n) are multiplicatively independent for n ≥ 1. Gao uses this fact to give a method for constructing elements of "high order" in F q n when q is fixed. That is, elements with order larger than any polynomial in n when n is large. In particular, if we definen = q log q n , and g ∈ F q [X ] is not a monomial or certain binomial, then any root of an irreducible factor of degree n of Xn − g(X ) is an element in F q n of order at least n log q n 4 log q (2 log q n) − 1 2 .
Sharper analysis of the same method by Popovych in [19] improves the lower bound on the order to where d = 2 log q n and t = log d n .
In the case of rational functions over a general field, we also have multiplicative independence of iterates, up to a few exceptional cases. We remark (see Lemma 2.8) that these exceptions are precisely the rational functions which, under iteration, eventually become a monomial. For example, if f (n) (X ) = X k , then f (n) (X ) and f (2n) (X ) = X k 2 are multiplicatively dependent. Note also that the cases of zero and positive characteristic are different. One distinction, of course, is the existence of inseparable maps in fields of positive characteristic. We see in Lemma 2.7, that this corresponds to a difference in which rational functions have an iterate which is a polynomial, let alone a monomial. Moreover, especially in the polynomial case, positive characteristic allows terms in iterates to vanish which would otherwise prevent them from becoming monomials. Theorem 1.1 Suppose that f = g/h ∈ F(X ) has degree d ≥ 2, and is not a monomial of the form a X ±d , nor of the form L(X p ), where L ∈ F(X ) has degree 1 and is a positive integer. Let n ≥ 1, and write Then there exists an integer j ≥ 0 depending only on f such that (n) ≥ d n if n ≤ j, Let e be the least positive integer k such that f (k) (0) = 0, and say that e = ∞ if where We use Corollary 1.2 in the following extension of the main theorem in [10].
and t = log d n , α has order in F q n at least As an aside we additionally ask, given rational functions F 1 , . . . , F n ∈ F(X , Y ) and polynomial u ∈ F[X ], when F 1 (X , u(X )), . . . , F n (X , u(X )) are multiplicatively dependent. In particular, we find upper bounds on the degree of u such that this is possible, and the number of monic u for which this is the case.

Theorem 1.5 Suppose F is a field of characteristic zero, n is a positive integer, and
where Res Y (P, Q) is the resultant of P, Q ∈ F[X , Y ], considered as polynomials in Y , and set If R i j ≡ 0 for all i = j, then there are finitely many monic polynomials u ∈ F[X ] such that are multiplicatively dependent. In particular, such a u has degree not exceeding E + 2d n − 1.
Recalling that the resultant of two polynomials of respective degrees m and n is a polynomial in the coefficients of degree m + n, and that each G i , H i written as a polynomial in Y , has degree at most e n , with each coefficient having degree not exceeding d n . We have that for i = j, deg Res Y (G i , G j ) ≤ (e n + e n )d n = 2d n e n , and the same bound holds for all the factors of the polynomial R i j defined above. Thus, counting n(n−1) Theorem 1.5 can be applied to the particular scenario of shifting a given set of polynomials by a polynomial u, giving a analogue of results for algebraic numbers from [4] and [7]. Corollary 1.6 Suppose F has characteristic zero, n is a positive integer and f 1 , . . . , f n ∈ F[X ] are distinct polynomials, not all constant, of respective degrees d 1 ≤ · · · ≤ d n and let Then there are at most 2C+3d n −1 C monic polynomials u ∈ F[X ] such that . . , f n + u are multiplicatively dependent. In particular, such a u has degree not exceeding C + 2d n − 1.
The paper is organised with sections corresponding to proofs of the main theorems: In the next section, we collect various results on iterates of rational functions, specifically concerning zeros and poles which are common to different iterates, and the degrees of the numerator and denominator of iterates. We use these results to bound from below the number (counted with multiplicity) of zeros and poles of a given iterate which cannot be found in any of the previous ones. We thus obtain Theorem 1.1. In Sect. 3, we give the proof of a version of [8,Main Theorem], which holds for polynomials over fields of arbitrary characteristic. This is used in conjunction with the general method from Sect. 2 to prove Theorem 1.3. In Sect. 4, we discuss elements of high order in finite fields in a manner analogous to [10,19], but in a slightly more general setting. Finally, in Sect. 5, we use resultants in conjunction with the polynomial ABC-theorem to prove Theorem 1.5.

Proof of Theorem 1.1
To prove Theorem 1.1, we need some facts about the composition of rational functions. Let u = v/w, F = G/H ∈ F(X ) be in lowest terms over F, and write Recall that the degree of a rational function f ∈ F(X ), written is lowest terms, is equal to the degree [F(X ) : F( f (X ))], and hence by the product formula for degrees of extensions, Next, we have where Note that a composition of rational functions in lowest terms is itself in lowest terms ([6, Lemma 2.2] is easily extended to our situation). In particular, G, H , q and r are pairwise relatively prime. This means we need not worry about the possibility of factors cancelling after composition. Hence, from (4), whenever deg G = deg H we have where P/Q is in lowest terms. We can use these facts to obtain results about which zeros and poles are common to different iterates of f . It turns out that these relations depend primarily on the earliest iterates of f to have either 0 or a point at infinity as a zero or pole. We hence set the following notation. Definition 2.1 Write f (k) = g k /h k for the k-th iterate of f in lowest terms, and let e be defined as in Theorem 1.3. Further define , μ and ν to be respectively the smallest positive integers k such that h k (0) = 0, deg g k < deg h k , and deg g k > deg h k . These again take the value ∞ if their respective conditions are not satisfied for any k ≥ 1.
We first note that there are restrictions on the possible combinations of the values e, , μ, ν. We make particular use of the next result. Lemma 2.2 Suppose μ < ν and < ∞. Then μ ≤ < e < ∞, and in particular e = + μ.
This contradicts the definition of μ, and so we must have μ ≤ .
, contradicting the choice of . Hence we have < e, and by setting u = f ( j) , F = f ( ) for a positive integer j, (4) gives that 0 is a zero of f ( + j) if and only if deg g j < deg h j . Thus e = + μ.
We have the following extension of a result of Gao [10, Lemma 2.2].

Lemma 2.3 For all integers k
Thus we must have e < ∞, so assume this is the case. If k ≡ (mod e), say k = + je where j ≥ 1, then for any zero β of f ( ) , Hence any zero of f ( ) is a zero of f (k) . Now, suppose k ≡ (mod e), say k = + je + r where u ≥ 0 and 1 ≤ r < e. If f (k) and f ( ) have a zero in common then, by the above argument, f ( je+r contradicting the choice of e. Therefore f (k) and f ( ) have no zero in common when k ≡ (mod e). Writing , the second and third parts follow immediately from (4). Now, suppose that μ < ν. By definition, we have that deg (4). If e, > j ≥ 1, then in (5) and (6), s = t = 0 and so deg g Suppose a zero α of f ( ) is a pole of f (k) . Then we have and so 0 is a pole of f (k− ) . That is, we indeed have < ∞. Thus e = + μ by 0 is also a pole of f (( j−1)e+ ) and hence, by part (ii), deg g r > deg h r . This is a contradiction, since from (7) and the definition of We may also determine facts about the degrees of iterates of f .
and for a positive integer j, let S j and T j be respectively the degrees of the lowest order term in g j and h j . Using the notation from Definition 2.1, we have We moreover have the following.
(b) Suppose < e, and write T = T . Then S = δT . If j = ie + for some integer i ≥ 0, then T j = δ i T i+1 , and otherwise T j = 0.
Proof Throughout the proof, we will write a given iterate f (k) = u • F = P/Q, and infer the degrees of its numerator and denominator via the equations (5) and (6).
For the first part, we proceed by induction on i. By definition, deg g j = deg h j = deg f ( j) = d j for 1 ≤ j < ν, and we have deg g ν = d ν and deg h ν = d ν − δ. This proves the case i = 1. Let i ≥ 1 and suppose that deg as required. The second part follows from (7).
For the third part, setting u = f ( j) and , then no zero of f (e) (in particular 0) is a zero of f ( j) , and so S j = 0.
For the last part of the proof, we make use of Lemma 2.2. If e < , we must have = ∞, and so T j = 0 for all j, proving (a). On the other hand, if < e, then e = + μ. Set u = f (μ) and F = f ( ) so that (4) gives S = δT , and thus S ie = δ i T i . We similarly obtain T ie+ = δ i T i+1 . Finally, if j is not equal to ie + for any integer i ≥ 0, then j ≡ = e − μ (mod e). Thus, by Lemma 2.3 (iv), no zero of f (e) (in particular 0) is a pole of f ( j) , and so T j = 0, proving (b). Proof By definition, we have deg g n = deg h n for n < μ, and deg g n < deg h n for n = μ. Suppose n > μ, and write n = μ+ j, so j = n−μ. Then from Lemma 2.4 (iii), deg g n = d n − δS j and deg h n = d n − δT j . Hence deg g n < deg h n if and only if S j > 0, which occurs precisely when j = n − μ ≡ 0 (mod e) by Lemma 2.4 (iii).
On the other hand, deg g n > deg h n if and only if T j > 0, and this happens precisely We hence obtain the following result. As we remarked in the introduction, in order to prove multiplicative independence for the iterates of f , it is clearly necessary to show that no iterate of f is a monomial, that is, of the form f (X ) = a X ±d . We first look to a result of Silverman [ Indeed, if no iterate of f is a polynomial, then certainly none can be a monomial. In fact, in the case where f is separable, we show that a rational function has a monomial iterate if and only if it is itself a monomial. This is not true however, when f is not separable. For example, if F has characteristic 2, then f (X ) = 1 + 1/X 2 satisfies f (2) Note that in the case of characteristic 0, some cases of the following can actually be viewed as a corollary of the stronger result [24, Theorem 1], which concerns the number of terms (monomials) of composite polynomials. The results of [24] are further extended to rational functions in [9].
is neither a monomial, nor of the form L(X p ) for some integer ≥ 0 and L ∈ F(X ) of degree 1, then f (k) is not a monomial for any k ≥ 1.

Proof
We begin with the case where f ∈ F[X ] is a polynomial. First suppose F has zero characteristic. We proceed by induction on k. That is, suppose deg f ≥ 2, and that f is not a monomial. Then the case where k = 1 is trivial. If f (k−1) is not a monomial, we can write Hence we have the following cases: If d s = 0, e t = 0, we have that = a 1 (b 1 X e 1 + · · · + b t X e t ) d 1 + · · · + a s (b 1 X e 1 + · · · + b t X e t ) d s has lowest order term a s b d s t X d s e t = 0, since a s = 0, b t = 0. Finally, when d s = e t = 0, if e 2 > 0, we have In this case, the term in X (d 1 −1)e 1 +e 2 has coefficient d 1 a 1 b d 1 −1 1 b 2 = 0, since we have a 1 , b 1 , b 2 = 0, and F has 0 characteristic. Otherwise, e 2 = 0 and Similarly, the term in X (e 1 −1)d 1 +d 2 has coefficient e 1 b 1 a e 1 −1 1 a 2 = 0. That is, in all cases f (k) is not a monomial, and we are done. Now, suppose F has positive characteristic p, and that f (k) is monomial, say of the form cX d k with c ∈ F \ {0}, for some k > 1. We can write and  p gcd(d 1 , . . . , d t ).
Here, the degree of f is d = d 1 p . Denote r = p and let v(X ) = a 1 X d 1 + · · · + a t X d t + b, Since r i is a power of p, we have for any i ≥ 1 . . .
Hence we have where c 0 = c r −k = 0, since c = 0. Differentiating then gives Since p gcd(d 1 , . . . , d t ), w i = 0 for all i ≥ 1. Thus, the polynomial on the left hand side of (8) is not zero. So p d 1 , as otherwise the right hand side would be zero. Since d k 1 c 0 = 0, the Eq. (8) implies that w 1 (X ) divides X d k 1 −1 . Therefore w 1 is a monomial. Since p d 1 , we must have p | d i for 2 ≤ i ≤ t. Hence and so w 2 (w 1 (X )) = d 1 a −r 2 Now, suppose f / ∈ F[X ], and that f (k) is a monomial for some k ≥ 1. Then in particular, some iterate of f is a polynomial.
If f is separable, then by Lemma 2.7, f is linearly conjugate to 1/X d . That is, f has the form Then f (2) , which is a monomial if and only if a = 0, in which case f is a monomial. Suppose a = 0. Since f is separable, d = p for any > 0, and so, since we have already proved the result for polynomials, no iterate of f (2) is a monomial. That is, f (k) is not a monomial for any even k ≥ 2 unless f is a monomial. Moreover, we have in this case ν = 2 < μ, so by Lemma 2.4 (i), deg g k = deg h k , and so f (k) is not a monomial, for all odd k.
Finally, if f is not separable, then by Lemma 2.7, f (k) is not a polynomial, and hence is not a monomial, for any k ≥ 1 unless f is of the form L(X p ) for some L ∈ F(X ) of degree 1.
We can now prove Theorem 1.1. Recall that we write f (k) = g k /h k in lowest terms, and define δ, S k , and T k as in Lemma 2.4, again setting S = S e and T = T where applicable. Now, where (n) is defined as in (1), noting that F(X ) is a unique factorisation domain, any zeros or poles of f (n) which can not be found in previous iterates will contribute to the value of (n) counting multiplicity, since k n = 0.
We first consider the case where ν < μ. Then deg g k ≥ deg h k for all k by Lemma 2.4 (i). Hence gcd(g n , h k ) = 1 for any k < n by Lemma 2.3 (iii). Moreover, if n ≤ e, then gcd(g n , g k ) = 1 for any k < n, by Lemma 2.3 (i). In this case, we have (n) ≥ deg g n = d n . Suppose e < ∞ and n > e. Then for k < n, a zero of f (k) is a zero of f (n) if and only if k ≡ n (mod e) by Lemma 2.3. In this case we also have k ≡ n − e (mod e), and so such a zero must also be a zero of f (n−e) . Write u = f (e) and F = f (n−e) , so (4) gives g n = g S n−e q, where S > 0 and gcd(q, g n−e ) = 1. Since f (e) is not a monomial by Lemma 2.8, we have S < d e , and so (n) ≥ deg q = d n − Sd n−e ≥ d n−e . Now, suppose μ < ν. If e = = ∞ or e < , then by Lemma 2.4 (ii) and (iii)(a), deg g j ≤ deg h j = d j for all j ≥ 1, and so gcd(h k , h n ) = 1 for all 1 ≤ k < n by Lemma 2.3 (ii). Moreover, gcd(g k , h n ) = 1 for all 1 ≤ k < n by Lemma 2.3 (iv). Hence (n) ≥ deg h n = d n . Suppose < ∞. Then μ < < e < ∞ by Lemma 2.2. Moreover, if n ≤ , gcd(g k , h n ) = gcd(h k , h n ) = 1 by Lemma 2.6 (ii), and thus we again have (n) ≥ deg h n = d n . We hence assume that μ ≤ < n < ∞.
We now split into a further two cases. Firstly, suppose that deg g μ > 0, so that δ < d μ . Since e = μ + > μ, we do not have μ ≡ 0 (mod e), and so S μ = 0 by Lemma 2.4 (iii). Hence, where u = f (μ) and F = f (n−μ) , (4) gives for some polynomial q relatively prime to h n−μ . From Lemma 2.6 (i), any zero or pole of a previous iterate f (k) , 1 ≤ k < n, which is also a zero of f (n) , must be a root of 238 M. Young h n−μ . Hence (n) ≥ deg q, and so we aim to bound deg q from below. If n = μ + ie for some integer i ≥ 1, then n − μ = μ + (i − 1)e + , and so by Lemma 2.4 (iii)(b), Otherwise, n − μ ≡ 0 (mod e), and so deg g n ≥ deg h n by Corollary 2.5. That is, deg g n = d n , and so Hence from (9) On the other hand, where deg g μ = 0 and correspondingly δ = d μ , we set u = f ( ) , and F = f (n− ) . If = μ, then by definition deg g < deg h . Otherwise 0 < − μ < , so − μ ≡ 0, (mod e), and thus deg g = deg h by Corollary 2.5.
That is, where r is a polynomial relatively prime to g n− . From Lemma 2.6 (ii), any zero or pole of a previous iterate f (k) , 1 ≤ k < n, which is also a pole of f (n) , must be a root of g n− . Hence (n) ≥ deg r = deg h n − T deg g n− . Note that T < d , as if this were not the case, by Lemma 2.4 (iii) we would have and S μ+ = S e = δT = d μ d , which implies that f (μ+ ) is a monomial, contradicting Lemma 2.8. In particular, this means that Hence, if n = μ + ie + for some integer i ≥ 0, then n − = μ + ie, so by Lemma 2.4 (iii), (10) and (11), we have Otherwise, n − μ ≡ (mod e), and so deg g n ≤ deg h n by Corollary 2.5. That is, deg h n = d n , and so from (10) and (11) We conclude that (n) ≥ deg r ≥ d n− , completing the proof.

Proof of Theorem 1.3
Recall the polynomial ABC-theorem (proved first by Stothers [23], then independently by Mason [14] and Silverman [22]). F be a field and let A, B, C ∈ F[X ] be relatively prime polynomials such that A + B + C = 0 and not all of A, B and C have vanishing derivative. Then

Lemma 3.1 Let
where, for f ∈ F[X ], rad( f ) is the product of the distinct monic irreducible factors of f .
We use this to obtain a version of part of the main result of [8]. Namely, we give a lower bound for the number of distinct zeros of a composite polynomial.
, h has non-vanishing derivative, and z(g) ≥ 2. Then where γ is defined as in Theorem 1.3.
Proof If deg h = 1, then clearly z( f ) = z(g). Since z(g) ≥ 2, we have In the characteristic 0 case, the result is [8, Main Theorem (i)]. When the characteristic is positive, we proceed in much the same vein. Write where the α i and β j are respectively the distinct roots of f and g in an algebraic closure of f . Then For β i = β j , the factors h(X ) − β i and h(X ) − β j have no zeros in common, so t ≤ n, and there exists a partition of {1, . . . , n} into disjoint subsets S β 1 , . . . , S β t , such that with l m k m = f m , for every j = 1, . . . , t. Since t = z(g) > 1, we can take 1 ≤ i < j ≤ t, and obtain h(X ) = β i + p i (X ) = β j + p j (X ). That is, where the polynomials on the left-hand side are relatively prime, and in particular, since h has non-vanishing derivative, so does p i . Thus, applying Lemma 3.1, we have We now prove Theorem 1.3. Suppose f ∈ F[X ] has non-vanishing derivative. Then for any positive integer n, We can hence apply Lemma 3.2 to obtain z( f (n) ) ≥ γ ( f )d n−1 + 1. As in the proof of Theorem 1.1, any zeros of f (n) which cannot be found in previous iterates will contribute to the value of Z (n), but this time without multiplicity. If n ≤ e, then gcd( f (k) , f (n) ) = 1 for all 1 ≤ k < n by Lemma 2.3 (i), and so Z (n) ≥ z( f (n) ) ≥ γ ( f )d n−1 + 1. Suppose that e < n < ∞, and write Note that any zeros of f (n) which are common with a previous iterate belong to f (n−e) by Lemma 2.3 (i). Now, If e > 1, then z( f (e) ) ≥ d e−1 + 1 > 2, and otherwise z( f (e) ) > 2 by assumption. Hence z(φ) > 1, and so by Lemma

Proof of Theorem 1.4
If f ∈ F[X ], this is the main result of [19], so assume otherwise, in which case we define s = (n − 1)/2 . Recall the following lower bound from Lambe [12], on the number of solutions to a linear Diophantine inequality: with equality when x 0 = · · · = x r −1 = 1.
As m is a power of q, applying the Frobenius automorphism iteratively gives Consider the set We will show that the powers α a , with a ∈ S, are distinct in F q n , so from Lemma 4.1, α has order at least #S.
Suppose that there exist integers a, b in S such that The equation (12) then gives Then k 1 (α) = k 2 (α). Since α has degree n and k 1 and k 2 have degree at most In light of Theorem 1.4, we wish to determine whether such a pair (g, h) of suitable polynomials always exists for all n. If this is so, we can construct a reliable algorithm for finding elements of high order in F q n . Namely, checking Xnh(X ) − g(X ) for irreducible factors of degree n, for each appropriate pair (g, h) ∈ F q [X ] 2 . The case where h(X ) = 1 is considered in [10], where it is reasonably conjectured, but not proved, that for every n, there exists g ∈ F q [X ] with deg g ≤ 2 log q n, such that Xn − g(X ) has an irreducible factor of degree n.
For our more general situation, we make the following weaker conjecture, To give some evidence for this conjecture, we first obtain a rough lower bound for the order of T . See [2] for the next lemma, regarding the probability that two polynomials in F q [X ] are relatively prime.
Now, consider the following result from [10]: If we model Xnh(X )− g(X ) as a random polynomial in F q [X ] for each (g, h) ∈ T , Lemma 4.4, in conjunction with (13), suggests that for large n, we expect on the order of n 3 pairs (g, h) ∈ T such that Xnh(X ) − g(X ) has an irreducible factor of degree n. Thus it is plausible that at least one such pair exists.

Proof of Theorem 1.5
We now restrict the field F to having characteristic 0. The key tool of this section is Lemma 3.1, and so the results could perhaps be extended to characteristic p, given stronger conditions to ensure that one of the polynomials A, B or C, to which we apply the theorem, has non-vanishing derivative.
We now prove Theorem 1.5. Suppose F 1 (X , u(X )), . . . , F n (X , u(X )) are multiplicatively dependent, and and assume that no proper subset of these is also multiplicatively dependent, as we can remove functions until this is the case. Then every zero and pole of F i for 1 ≤ i ≤ n must be a zero or pole of F j for some j = i. This is because otherwise we would require k i = 0 in the equation and hence the proper subset {F (X , u(X )) : 1 ≤ ≤ n, = i} would be multiplicatively dependent. Hence, if α is a zero or pole or F i (X , u(X )), there exists j = i such that F i (α, Y ) and F j (α, Y ) have the common zero or pole u(α), giving R i j (α) = 0. Thus, any zero or pole of F i (X , u(X )) for 1 ≤ i ≤ n is a zero of 1≤i< j i< j≤n R i j .
In particular, since for all i = j, R i j is not identically zero, we have deg rad

M. Young
Now, for 1 ≤ i ≤ n, write and assume, without loss of generality, that g i,e i is not identically zero (if it is, we can replace G i with H i , and g i,e i with h i,e i in the following definitions). For 1 ≤ i < j ≤ n, define and D i j (X ) = gcd(P(X ), Q(X )). Then set Then A, B, and C are relatively prime polynomials with A + B + C = 0. Suppose deg u > 2d n . By construction, P and Q have the same degree and same leading coefficient, and hence we have P | Q if and only if P = Q. If P = Q, then Since deg u > 2d n , the term in u(X ) ν in the above expression contains monomials in X of degree between ν deg u and ν deg u + 2d n < (ν + 1) deg u. Thus there can be no cancellation between these terms, and so g i,e i (X )g j,ν (X ) − g j,e j (X )g i,ν−e j +e i (X ) = 0, e j − e i ≤ ν ≤ e j , and g i,e i (X )g j,ν (X ) = 0, 0 ≤ ν < e j − e i . We conclude that in fact but R i j ≡ 0 implies that gcd(G j (X , Y ), G i (X , Y )) = 1, and so we must have This is impossible as G j (X , Y ) has degree e j > e j − e i in Y . Therefore P Q, and so deg D i j < deg P gives (16) Thus A has non-vanishing derivative. Moreover, in C, the term in u(X ) e j cancels out, giving Therefore, we have by Lemma 3.1 and (16), From (15), deg radG i (X , u(X ))G j (X , u(X )) ≤ E, and so (17) gives Therefore, for 1 ≤ i ≤ n, G i (X , u(X )) is a product of at most E distinct irreducible factors, with degree not exceeding e n (E + 2d n − 1) + d n . If w 0 , . . . , w E−1 are the respective multiplicities of said factors, then up to multiplication by a nonzero constant, the number of possibilities for G i (X , u(X )) is at most the number of non-negative integer solutions to the inequality E−1 j=0 w j ≤ e n (E + 2d n − 1) + d n , which is at most e n (E + 2d n − 1) + E + d n E if u is monic then A is uniquely determined. Moreover, we have so there are finitely many possibilities for monic u. For corollary 1.6, let has degree at most d n , and thus The result follows from substituting this into (18), noting that e n = 1 in this case.

Comments
Considering the case ν < μ (which encompasses the polynomial case) of Theorem 1.1, and additionally Theorem 1.3, it is of interest to obtain upper bounds for the value e when it is finite. That is, bounds for the period of 0 under iteration of a polynomial or rational function f . This problem is investigated in various contexts in [5,11,[15][16][17]20]. Bounds on the values of the values of , μ and ν in the rational function case are similarly of interest. Another problem is to generalise Theorem 1.3 to rational functions. Our approach used for the polynomial case can plausibly be extended to the situation where ν ≤ μ, mirroring the proof of the relevant case in Theorem 1.1, but applying an appropriate version of the main theorem in [8]. Such an extension, however, is not immediate for the case μ < ν.
Also, note that in the case F = C, Theorem 1.5 may be able to be generalised to several variables, where F i ∈ C(X 1 , . . . , X m , Y ) and u ∈ C[X 1 , . . . , X m ], using an appropriate analogue of Mason's theorem (for example [1,Theorem 2]).
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