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 $\mathbb{F}_{q^n}$ 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
In light of the difficult open problem of giving an efficient algorithm for constructing primitive elements in finite fields, Gao [8] has given a method for the more modest task of 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 it was shown that a 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 [16] improves the lower bound on the order to where d = 2 log q n and t = ⌊log d n⌋.
Gao, as a by-product of his method in [8], has also proved a theorem on the multiplicative independence of compositions of polynomials over finite fields, which we consider of independent interest. Our main task is to generalise these results to rational functions, and moreover to general fields, not necessarily finite.
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 define the iterates of f by f (0) (X) = X; f (k) (X) = f • f (k−1) (X) for k ≥ 1, and say that they are multiplicatively independent, if for integers n ≥ 1, k 1 , . . . , k n , we have (f (1) (X)) k1 . . . (f (n) (X)) kn = 1 if and only if k 1 = . . . = k n = 0. Otherwise, we say that they are multiplicatively dependent. In [8], Gao proves that if f ∈ F q [X] is not of the form aX d , or aX p ℓ + b, then the iterates of f are multiplicatively independent.
In generalising this to rational functions, we encounter a few additional exceptional cases. Recall that two rational functions φ, φ ′ ∈ F(X) are linearly conjugate if there exists a rational function ψ ∈ F(X) of degree 1 such that φ ′ = ψ −1 • φ • ψ. We have the following. Theorem 1.1. Suppose that f = g/h ∈ F(X) has degree d ≥ 2, is not a monomial, nor a binomial of the form aX p ℓ + b, and is not linearly conjugate to 1/X d . Let n ≥ 1, and write Let e be the smallest positive integer k such that 0 is a zero of f (k) (we say e = ∞ if f (k) (0) = 0 for all k ≥ 1). Then we have (i) If f ∈ F[X], then Ψ(n) ≥ d n if n ≤ e, and Ψ(n) ≥ d n−e otherwise.
where L ∈ F(X) has degree 1, then there exists an integer j ≥ 0 such that Ψ(n) ≥ d n if n ≤ j, and Ψ(n) ≥ d n−j otherwise.
It is easy to show that the above result implies the multiplicative independence of iterates of f . Corollary 1.2. Suppose f = g/h ∈ F(X) has degree at least 2, is not a monomial nor a binomial of the form aX p ℓ + b, and is not linearly conjugate to 1/X d . If deg g = deg h, or f is separable, or not of the form L(X p ℓ ), where L ∈ F(X) has degree 1, then the iterates f (1) , . . . , f (n) are multiplicatively independent.
We use this in the following extension of the main theorem in [8], with the improved bound from [16].
be coprime with deg h, deg g ≤ d = 2 log q n and suppose f = g/h satisfies the conditions from Corollary 1.2. Suppose that α ∈ F q n has degree n and is a root of X m h(X) − g(X), where m =n = q ⌈log q n⌉ . Then for 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 1.4. Suppose F is a field of characteristic zero, and there are at most en(E+2dn−1)+E+dn E α monic polynomials u ∈ F[X] such that are multiplicatively dependent, and each 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 , written as a polynomial in Y , has degree at most e n , with each coefficient having degree not exceeding d n , we have for i = j, deg Res Y (G i , G j ) ≤ (e n + e n )d n = 2d n e n . Thus, counting n(n−1) 2 distinct pairs {i, j}, we obtain E ≤ 4n(n − 1)d n e n . Theorem 1.4 can be applied to the particular scenario of shifting a given set of polynomials by a polynomial u, giving a analogue of results from [3] and [6], for algebraic numbers. Then there are at most 2C+3dn−1 C monic polynomials u ∈ F[X] such that . . , f n + u are multiplicatively dependent, and each has degree not exceeding C + 2d n − 1.
Therefore E ≤ n(n−1) 2 d n = C, and the result follows, noting that e n = 1.

Proof of Theorem 1.1
To prove Theorem 1.1, we need some facts about the composition of certain classes of rational functions. Let u = v/w, F = G/H ∈ F(X) be in lowest terms over F, chosen so H is monic and G has leading coefficient A, and write where Note that a composition of rational functions in lowest terms is itself in lowest terms ([5, 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 (2), whenever deg G = deg H we have Moreover, when deg G = deg H, the coefficient of X lD is v(A) in P , and w(A) in Q. These can't both be zero as gcd(v, w) = 1, so in all cases we have We can use these facts to obtain results about which zeros and poles are common to different iterates of f , beginning by extending a result of Gao [8, Lemma 2.2].
Lemma 2.1. Write f (k) = g k /h k for the k-th iterate of f , and let e be defined as in Theorem 1.1. Additionally, let ǫ, µ and ν 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 (again, these take the value ∞ if their respective conditions are not satisfied for any k ≥ 1). Then, for all k > ℓ ≥ 1, (i) A zero of f (ℓ) is a zero of f (k) if and only if e < ∞ and k ≡ ℓ (mod e).
Thus we must have e < ∞, so assume this is the case. Write , which does not have 0 as a zero or pole. If k ≡ ℓ (mod e), say k = ℓ + je where j ≥ 1, then Hence any zero of f (ℓ) is a zero of f (k) . Now, suppose k ≡ ℓ (mod e), say k = ℓ + je + r where u ≥ 0 and 1 ≤ r < e. If f (k) and f (ℓ) have a zero in common then, by the above argument, we have f (r) (0) = 0, contradicting the choice of e. Therefore f (k) and f (ℓ) have no zero in common when k ≡ ℓ (mod e). Writing , the second and third parts follow immediately from (2). (2). If e, ǫ > j, then s = t = 0, and so (3) and (4) give deg g j+µ = deg h j+µ = d j+µ . We thus note that Next, we have , contradicting the choice of ǫ. Hence we have ǫ < e, and by setting with 0 not a zero or pole of ψ. If k ≡ ℓ−µ (mod e), say k = ℓ+je−µ = ℓ+(j−1)e+ǫ, with j ≥ 1, then and so any zero of 0 is also a pole of f ((j−1)e+ǫ) and hence, by part (ii), deg g r > deg h r . This contradicts (7), so we are done.
We may also determine facts about the degrees of iterates of f .
and let S k and T k be respectively the degrees of the lowest order term in g k and h k . We have (i) If ν < µ, then for integer i ≥ 1, deg (iii) Let µ < ν, e < ǫ, and write S e = S. Then, if k = ie + µ for some (iv) Let µ < ν and ǫ < ∞. Recall then, from Lemma 2.1 (iv), that e = ǫ+µ, and 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 (3) and (4). For the first part, we use induction on i. By definition and from (5), we have The second part follows from (7). For the third and fourth parts, . Then, for part (iii), we have e < ǫ, and so T k = 0 for all k by Lemma 2.1 (iv). For part (iv), we set u = f (µ) and F = f (ǫ) so that (2) implies S e = δT . Thus S ie = δ i T i . We similarly obtain T ie+ǫ = δ i T i+1 . Now, suppose ie < k < (i + 1)e. By Lemma 2.1 (iv), we have T k = 0 if and only if k ≡ ie − µ ≡ ǫ (mod e). The results follow.
We hence obtain the following result.
For the second part, by Lemma 2.1 (i) we have that a zero of f (ℓ) is a zero of f (k) if and only if k ≡ ℓ (mod e). Note that this implies that k − ℓ ≥ e = µ + ǫ, and so k − ℓ − µ ≥ ℓ ≥ 1. Then, by Lemma 2.1 (iv), a zero of f (ℓ) is a pole of f (k−µ) if and only if k − µ ≡ ℓ − µ (mod e), which is an equivalent condition. From Lemma 2.1 (iii), a pole of f (ℓ) is a zero of f (k) if and only if deg g k−ℓ < deg h k−ℓ . This occurs precisely when k ≡ ℓ + µ (mod e) by Lemma 2.2 (iv). On the other hand, a pole of f (ℓ) is a pole of f (k−µ) if and only if deg g k−ℓ−µ > deg h k−ℓ−µ . By Lemma 2.2 (iv), this happens exactly when k − µ ≡ ℓ (mod e), which is again equivalent.
Finally, for part (iii), by Lemma 2.1 (iv), a zero of f (ℓ) is a pole of f (k) if and only if k ≡ ℓ − µ (mod e). Since e = µ + ǫ, this is equivalent to k − ǫ ≡ ℓ (mod e), which is the precise condition for a zero of f (ℓ) to be a zero of f (k−ǫ) , by Lemma 2.1 (i). Furthermore, from Lemma 2.1 (ii), a pole of f (ℓ) is a pole of f (k) if and only if deg g k−ℓ > deg h k−ℓ . According to Lemma 2.2 (iv), this is equivalent to k − ℓ being of the form µ + ie + ǫ, which equates to k − ℓ − ǫ = µ + ie. Again by Lemma 2.2 (iv), this is equivalent to having deg g k−ℓ−ǫ < deg h k−ℓ−ǫ , which is in turn equivalent to the given pole of f (ℓ) being a zero of f (k−ǫ) , by Lemma 2.1 (iii).
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. We first look to a result of Silverman [18]. . Then either f ∈ F[X], f is linearly conjugate to 1/X d , or f is inseparable, and f (X) = L(X p ℓ ) for some L ∈ F(X) of degree 1.
Indeed, if no iterate of f is a polynomial, then certainly none can be a monomial. In particular, in the case where f is separable, we have that if f (n) is a polynomial for some n ≥ 1, then already f (2) is a polynomial. This makes it easy to check whether a given rational function becomes a polynomial under iteration. This is not true however, when f is inseparable. For example, if F has characteristic 2, then f (X) = 1 + 1/X 2 satisfies f (2) (X) = 1 X 4 +1 and f (3) (X) = X 8 . Nevertheless, exceptional cases of this type are described completely in the above result. Now, we treat the case where f is a polynomial separately. Note that in the case of characteristic 0, the following can actually be viewed as a corollary of the stronger result [21, Theorem 1], which concerns the number of terms (monomials) of composite polynomials. The results of [21] are further extended to rational functions in [7].
is not a monomial or binomial of the form aX p ℓ + b, with a = 0, b ∈ F, ℓ ∈ N, then f (k) is not a monomial for any k ≥ 1.
Proof. Beginning with the case where 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 has lowest order term a s b ds t X dset = 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 (d1−1)e1+e2 has coefficient d 1 a 1 b d1−1 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 (e1−1)d1+d2 has coefficient e 1 b 1 a e1−1 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 where a 1 , . . . , a t ∈ F \ {0}, b ∈ F, t ≥ 1, ℓ ≥ 0, d 1 > . . . > d t ≥ 1, and p ∤ gcd(d 1 , . . . , d t ).
Here, the degree of f is d = d 1 p ℓ . Denote r = p ℓ and let v(X) = a 1 X d1 + . . . + a t X dt + b, Since r i is a power of p, we have for any i ≥ 1 . . .
Hence we have where c 0 = c r −k ∈ F, and c 0 = 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.
We can now prove Theorem 1.1. Recall that we write f (k) = g k /h k , and define δ, S k , and T k as in Lemma 2.2. Now, where Ψ(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 Ψ(n) counting multiplicity, since k n = 0.
On the other hand, where deg g µ = 0, we set u = f (ǫ) , and F = f (n−ǫ) . If ǫ ≤ µ, then by definition deg g ǫ ≤ deg h ǫ . Otherwise, ǫ = µ + k, with k = ie, ie + ǫ, and so by Lemma 2.2 (iv), we have deg g ǫ − deg h ǫ . Hence, by (2), f (n) = h m−l n−ǫ g −T n−ǫ q/r, where m ≥ l. We thus obtain deg r = deg h n − T deg g n−ǫ . Note that T < d ǫ , as if this were not the case, by Lemma 2.2 (iv) we would have , a contradiction by Lemma 2.4. In particular, this means that d n − T d n−ǫ ≥ d n−ǫ . Hence, if n = µ + ie + ǫ, then n − ǫ = µ + ie, so by Lemma 2.2 (iv), we have Otherwise, once again using Lemma 2.2 (iv), deg h n = d n , and so To conclude, by Lemma 2.3 (iii), we have that gcd(h k , r) = gcd(g k , r) = 1 for all 1 ≤ k < n, and thus Ψ(n) ≥ deg r ≥ d n−ǫ . This completes the proof.

Proof of Theorem 1.3
First recall a lower bound from Lambe [10], on the number of solutions to a linear Diophantine inequality: Now, set m =n. Since α is a root of X m h(X) − g(X), we have α m = f (α). As m is a power of q, applying the Frobenius automorphism iteratively gives (9) α Consider the set We will show that the powers α a , with a ∈ S, are distinct in F q n , so from Lemma 3.1, α has order at least The equation (9) then gives Let Then k 1 (α) = k 2 (α). Since α has degree n and k 1 and k 2 have degree at most i by Corollary 1.2, and hence a = b, a contradiction.
In light of Theorem 1.3, 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 [8], 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, Conjecture 3.2. Suppose n ≥ 1, and let T be the set of pairs (g, h) ∈ F q [X] 2 of degree not exceeding d := 2 log q n such that f = g/h satisfies the conditions from Corollary 1.2. Then there exists (g, h) ∈ T such that Xnh(X) − g(X) has an irreducible factor of degree n.
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 [8]: Lemma 3.4. Let P q (m, n) be the probability of a random polynomial in F q [X] of degree m ≥ n having at least one irreducible factor of degree n. Then uniformly for q and m ≥ n.
If we model Xnh(X)−g(X) as a random polynomial in F q [X] for each (g, h) ∈ T , Lemma 3.4, in conjunction with (10), 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 1.4
For the following we use the polynomial ABC-theorem (proved first by Stothers [20], then independently by Mason [12] and Silverman [19]). where, for f ∈ F[X], rad(f ) is the product of the distinct monic irreducible factors of f .
For the convenience of having rational function's derivative non-vanishing being equivalent to it being non-constant, we now restrict the field F to having characteristic 0. The results of this section could be extended to characteristic p, given stronger conditions to ensure that our choice of A, B or C has non-vanishing derivative.
We now prove Theorem 1.4. Suppose F 1 (X, u(X)), . . . , F n (X, u(X)) are multiplicatively independent, 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 (11) n ℓ=1 F ℓ (X, u(X)) k ℓ = 1, and hence the proper subset {F ℓ (X, u(X) : 1 ≤ ℓ ≤ n, ℓ = 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 ij (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 ij . In particular, since for all i = j, R ij is not identically zero, we have (12) deg rad and assume, without loss of generality, that g i,ei is not identically zero (if it is, we can replace G i with H i , and g i,ei with h i,ei in the following definitions). For 1 ≤ i < j ≤ n, define D ij (X) = gcd(g i,ei (X)G j (X, u(X)), g j,ei (X)u(X) ej−ei G i (X, u(X))), and set and C = −(A + B). Then A, B, and C are relatively prime polynomials with A + B + C = 0. We have that which is positive if deg u ≥ d n , as e j ≥ 1, and R ij ≡ 0 ensures that A ∤ B. Thus A has non-vanishing derivative. Moreover, in C, the term in u(X) ej cancels out, giving Therefore, we have by Lemma 4.1 and (13), Then, (12) and (14) give 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 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 en(E+2dn−1)+E+dn E from Lemma 3.1. This also gives the number of possibilities for (monic) u if there exists 1 ≤ i ≤ n such that F i ∈ F[X, Y ]. Otherwise, we obtain the same bound for possible H i (X, u(X)), and hence the number of possibilities for F i (X, u(X)), and hence u, does not exceed en(E+2dn−1)+E+dn E 2 .
This completes the proof.

Comments
Considering Theorem 1.1 (i), 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 f . When K/F q is a field extension of degree n, Halter-Koch and Konecná [9] determine the set of all possible cycle lengths in K of polynomials over F q . That is, where N is the number of irreducible monic polynomials of degree n over F q . This bounds finite e above by nN , which equals q (an obvious bound) when K = F q .
In [17], we have the following results by Pezda for a discrete valuation domain of zero characteristic R, with finite residue field of cardinality N (P ), and the special cases of Z p (p-adic integers) and rings of integers in algebraic number fields over the rationals. When e is finite: • If f ∈ R[X], e does not exceed N (P )(N (P ) − 1)p C(p) , where C(p) = 1 + log(ord p) log 2 .
• If f ∈ Z p [X], e does not exceed p 2 .
• If R is the ring of all integers in an algebraic number field of degree n over the rationals and f ∈ R[X], e does not exceed (2 n − 1)2 n+1 . Narkiewicz [13,14,15] also characterised cycle lengths of polynomials in certain rings. Again for f ∈ R[X] with finite e we have • If R is the ring of integers in a cubic field of negative discriminant, then e ≤ 6. • If R is the ring of integers in a quadratic number field, then e ≤ 7.
• If R = Z 1 n for a positive, square-free integer n, then e ≤ 6. In [4], Canci gives an upper bound on the length of finite orbits of rational functions over number fields. Namely, if K is a number field, and S is a finite set of cardinality s of places of K, containing all the archimedean ones, then for rational maps with good reduction outside S, finite e is bounded above by exp(10 12 )(s + 1) 8 (log(5(s + 1))) 8 s .
Bounds on the values of the values of ǫ, µ and ν in the rational function case are similarly of interest.
Also, note that in the case F = C, Theorem 1.4 could 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]).