3.2. Proof of Theorem 1·1

Let $\alpha \in K$ be such that there exist integers $k_1,\ldots, k_n$ , not all zero such that

$$f_1(\alpha)^{k_1} \cdots f_n(\alpha)^{k_n} \in\Gamma^{{\textrm{div}}}_\varepsilon.$$

By Lemma 2·1 there exists $\beta \in O_{S_{{\textbf{f}},\Gamma}}^*$ and $\eta \in K^*$ with ${\textrm{h}}(\eta)\ll_{\varepsilon,\Gamma} 1$ such that

$$f_1(\alpha)^{k_1} \cdots f_n(\alpha)^{k_n}= \beta\eta.$$

Since $\eta \in K^*$ is of bounded height depending only on $\varepsilon$ and $\Gamma$ , by Northcott’s theorem there are only finitely many such $\eta$ . Thus we can enlarge the set $S_{{\textbf{f}},\Gamma}$ to include all prime ideals that divide the finitely many elements $\eta$ . We also include in this larger set the prime ideals outside $S_{{\textbf{f}},\Gamma}$ that divide the product $\displaystyle\prod_{1\le i \neq j\le n}$ Res $(f_i,f_j)$ of all the resultants of $f_i$ and $f_j$ for $i\ne j$ (we recall that all ${\textrm{Res}} (f_i,f_j)$ are $S_{{\textbf{f}},\Gamma}$ -integers), which are only finitely many. We denote the new set by $S_{{\textbf{f}},\Gamma,\varepsilon}$ and we note that $S_{{\textbf{f}},\Gamma,\varepsilon}$ is still a finite set.

By the construction of the set $S_{{\textbf{f}},\Gamma, \varepsilon}$ , we have

$$K^* \cap \Gamma^{{\textrm{div}}}_\varepsilon \subseteq O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*.$$

Thus, it suffices to prove the desired result by replacing $\Gamma^{{\textrm{div}}}_\varepsilon$ with $O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*$ .

Now, we write

(3·2) \begin{equation}f_1(\alpha)^{k_1} \cdots f_n(\alpha)^{k_n}= \gamma,\qquad \gamma = \beta\eta \in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*.\end{equation}

If $n = 1$ , since $f_1$ is a polynomial having at least two distinct roots and $O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*$ is a finitely generated subgroup, we see that applying Lemma 2·2 to $f_1$ and $O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*$ gives the desired finiteness result. We thus suppose that $n \geq 2$ , and that the result is valid for $n-1$ , in order to apply an induction.

We note that if some $k_i=0$ , then the desired finiteness of $\alpha \in K$ satisfying (3·2) follows directly from the induction hypothesis. Hence, we can assume from now on that $k_1 \cdots$ $k_n \ne 0$ .

We now complete the proof case by case.

In this case, since $\alpha \in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}$ and $f_i\in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}[X]$ for any $i=1,\ldots,n$ , we have

$$f_1(\alpha), \ldots, f_n(\alpha) \in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}.$$

We note that if $f_i(\alpha)\in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*$ for some $i \in \{1,\ldots,n\}$ , then Lemma 2·2 implies the finiteness of such $\alpha \in K$ satisfying (3·2).

Thus, we now assume that $f_1(\alpha),\ldots, f_n(\alpha)\not\in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*$ . This implies that there exists a prime ideal ${\mathfrak{p}} \not\in S_{{\textbf{f}},\Gamma,\varepsilon}$ in K such that $v_{\mathfrak{p}}(f_1(\alpha))\gt0$ . Moreover, since $f_i(\alpha)\in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}$ , we have $v_{\mathfrak{p}}(f_i(\alpha))\ge 0$ for each $i=2, \ldots, n$ .

If $k_1k_i \gt 0$ for each $i =2,\ldots, n$ , without loss of generality we can assume that $k_1,k_2,\ldots, k_n \gt 0$ . Then, equation (3·2) implies

$$k_1v_{\mathfrak{p}}(f_1(\alpha))+ \cdots +k_nv_{\mathfrak{p}}(f_n(\alpha)) = 0.$$

Since $v_{\mathfrak{p}}(f_1(\alpha))\gt0$ and $v_{\mathfrak{p}}(f_i(\alpha))\ge 0$ for each $i =2, \ldots, n$ , we obtain a contradiction.

We now assume $k_1k_i \lt 0$ for some $i \in \{2,\ldots, n\}$ . In this case, without loss of generality, we assume $k_1\gt0, \ldots, k_m \gt0$ and $k_{m+1} \lt0 , \ldots, k_n \lt 0$ for some positive integer m. So, equation (3·2) becomes

$$f_1(\alpha)^{k_1} \cdots f_{m}(\alpha)^{k_m} = \gamma f_{m+1}(\alpha)^{-k_{m+1}} \cdots f_{n}(\alpha)^{-k_n}.$$

Then, since $v_{\mathfrak{p}}(f_1(\alpha)) \gt 0$ and $v_{\mathfrak{p}}(f_i(\alpha))\ge 0$ for each $i=2,\ldots, n$ , there must exist some $j \in \{m+1, \ldots, n\}$ such that $v_{\mathfrak{p}}(f_j(\alpha)) \gt 0$ . In other words, we have

$$f_1(\alpha) \equiv f_j(\alpha) \equiv 0 \pmod {\mathfrak{p}}.$$

This allows us to conclude that $v_{\mathfrak{p}}({\textrm{Res}} (f_1,f_j))\gt0$ (notice that, since $f_1,f_{j} \in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}[X]$ and $f_1$ and $f_j$ do not have common roots, we have ${\textrm{Res}} (f_1,f_j)\in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}$ and ${\textrm{Res}} (f_1,f_j) \ne 0$ ). By our construction of the set $S_{{\textbf{f}},\Gamma,\varepsilon}$ , this implies that ${\mathfrak{p}} \in S_{{\textbf{f}},\Gamma,\varepsilon}$ , which is a contradiction with the choice of ${\mathfrak{p}}$ above. This completes the proof of Case I.

In this case, there exists a prime ideal ${\mathfrak{p}}$ of the ring of integers of K such that

$${\mathfrak{p}} \not \in S_{{\textbf{f}},\Gamma,\varepsilon} \quad\text{and}\quad v_{\mathfrak{p}}(\alpha) \lt 0.$$

Let $d_i = \deg f_i, i =1, \ldots,n$ . Then, using the ultrametric inequality of non-Archimedean valuations and noticing (3.1), we directly have

(3·3) \begin{equation} v_{\mathfrak{p}}(f_i(\alpha)) = d_i v_{\mathfrak{p}}(\alpha)\quad\text{for $i=1,\ldots,n$}.\end{equation}

Considering valuations in (3·2) and using (3·3) we obtain (since $v_{\mathfrak{p}}(\gamma)=0$ due to $\gamma\in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*$ )

(3·4) \begin{equation}k_1d_1 + k_2d_2 + \cdots + k_n d_n = 0.\end{equation}

We view the above identity as a linear Diophantine equation with unknowns $k_1, \ldots, k_n$ in ${\mathbb{Z}}$ . Then, we have a basis of the integer solutions $(k_1, k_2, \ldots, k_n)$ of the equation (3·4), say,

$$(t_{i,1}, t_{i,2}, \ldots, t_{i,n}), \quad i = 1, \ldots, n-1.$$

Therefore, $k_1, k_2, \ldots, k_n$ can be expressed as

$$k_j = \sum_{i=1}^{n-1} s_i t_{i,j}, \quad j=1,\ldots,n,$$

for some integers $s_1, \ldots, s_{n-1}$ . Substituting this into the equation (3·2), we obtain

(3·5) \begin{equation} \left( \prod_{j=1}^{n} f_j(\alpha)^{t_{1,j}} \right)^{s_1} \cdots \left( \prod_{j=1}^{n} f_j(\alpha)^{t_{n-1, j}} \right)^{s_{n-1}}= \gamma.\end{equation}

Now, we let

$$F(X) = \prod_{j=1}^{n} f_j(X)^{t_{1,j}},$$

where the exponent vector $(t_{1,1}, \ldots, t_{1,n})$ is non-zero by its choice above.

For any prime ${\mathfrak{q}} \not\in S_{{\textbf{f}},\Gamma,\varepsilon}$ , if $v_{\mathfrak{q}}(\alpha) \lt 0$ , then (3·3) holds and we have

(3·6) \begin{equation}\begin{split}v_{\mathfrak{q}}(F(\alpha)) & = \sum_{j=1}^{n} t_{1,j}v_{\mathfrak{q}}(f_j(\alpha)) \\& =(t_{1,1}d_1 + t_{1,2}d_2 + \cdots + t_{1,n} d_n) v_{\mathfrak{q}}(\alpha) = 0,\end{split}\end{equation}

since $(t_{1,1}, t_{1,2}, \ldots, t_{1,n})$ is a solution to (3·4).

For any prime ${\mathfrak{q}} \not\in S_{{\textbf{f}},\Gamma,\varepsilon}$ , if $v_{\mathfrak{q}}(\alpha) \ge 0$ , then by (3·1) we have $v_{\mathfrak{q}}(f_i(\alpha)) \ge 0$ for each $i = 1, \ldots, n$ .

If there exists some prime ${\mathfrak{q}} \not\in S_{{\textbf{f}},\Gamma,\varepsilon}$ such that $v_{\mathfrak{q}}(\alpha) \ge 0$ and moreover $v_{\mathfrak{q}}(f_i(\alpha)) \gt 0$ , $v_{\mathfrak{q}}(f_j(\alpha)) \gt 0$ for some $i \ne j$ , then by the same discussion as in the last part of Case I we arrive to a contradiction.

If there exists some prime ${\mathfrak{q}} \not\in S_{{\textbf{f}},\Gamma,\varepsilon}$ such that $v_{\mathfrak{q}}(\alpha) \ge 0$ and moreover $v_{\mathfrak{q}}(f_i(\alpha)) \gt 0$ for exactly one i for $i=1, \ldots, n$ , say $v_{\mathfrak{q}}(f_1(\alpha)) \gt 0$ and $v_{\mathfrak{q}}(f_i(\alpha)) = 0$ for each $i = 2, \ldots, n$ , then by (3·5) we obtain

$$s_1 t_{1,1} + \cdots + s_{n-1} t_{n-1, 1} = 0,$$

which however contradicts with $k_1 \ne 0$ because $k_1 = s_1 t_{1,1} + \cdots + s_{n-1} t_{n-1, 1}$ .

Hence, we may assume that for any prime ${\mathfrak{q}} \not\in S_{{\textbf{f}},\Gamma,\varepsilon}$ , if $v_{\mathfrak{q}}(\alpha) \ge 0$ , then $v_{\mathfrak{q}}(f_i(\alpha)) = 0$ for each $i = 1, \ldots, n$ . In this case, we have $v_{\mathfrak{q}}(F(\alpha)) = 0$ . Combining this with (3·6), we have $v_{\mathfrak{q}}(F(\alpha)) = 0$ for any prime ${\mathfrak{q}} \not\in S_{{\textbf{f}},\Gamma,\varepsilon}$ , and thus $F(\alpha) \in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*$ . Now, the desired result follows directly from Lemma 2·2 (which we can apply, since $f_i$ , $i=1,\ldots,n$ , has at least two distinct roots and they are pairwise coprime, and therefore, F has at least two distinct roots or two distinct poles). This completes the proof.

3.3. Proof of Theorem 1·2

First, we assume that the rational functions $f_1, f_2, \ldots, f_n$ all have no linear factor.

Let $g_1, \ldots, g_m$ be all the distinct monic irreducible factors (over K) in the numerators and denominators of the rational functions $f_{1}, f_{2}, \ldots, f_{n}$ . So, by assumption, the irreducible polynomials $g_1, \ldots, g_m$ are all of degree at least two. Then, for each $f_i, 1\le i \le n$ , we can write

(3·7) \begin{equation} f_i = a_i \prod_{j=1}^{m} g_j^{e_{ij}},\quad a_i\in K^*,\end{equation}

for some integers $e_{i1}, \ldots, e_{im}$ .

Let $\alpha \in K$ be such that there exist integers $k_1,\ldots, k_n$ , not all zero such that

$$f_1(\alpha)^{k_1} \cdots f_n(\alpha)^{k_n} \in\Gamma^{{\textrm{div}}}_\varepsilon.$$

As in (3·2), we can write

(3·8) \begin{equation} f_1(\alpha)^{k_1} \cdots f_n(\alpha)^{k_n}= \gamma,\qquad \gamma \in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*,\end{equation}

where the set $S_{{\textbf{f}},\Gamma,\varepsilon}$ is defined as in the proof of Theorem 1·1, however without including the prime ideals outside $S_{{\textbf{f}},\Gamma}$ that divide the product $\displaystyle\prod_{1\le i \neq j\le n}$ Res $(f_i,f_j)$ of all the resultants of $f_i$ and $f_j$ for $i\ne j$ , because $f_i$ and $f_j$ might not be polynomials.

By the discussion in Section 3.1, we know that $a_i \in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*$ for each $i=1,\ldots,n$ . Hence, combining (3·8) with (3·7), we get that for some $\gamma^{\prime} \in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*$ ,

(3·9) \begin{equation} g_1(\alpha)^{k_1e_{11} + \cdots + k_ne_{n1}} \cdots g_m(\alpha)^{k_1e_{1m} + \cdots + k_ne_{nm}} = \gamma^{\prime}.\end{equation}

If for each $1\le j \le m$ , $k_1e_{1j} + \cdots + k_ne_{nj} = 0$ , then this means that $f_1^{k_1} \cdots f_n^{k_n}$ is a constant, which contradicts with the assumption that $f_1, \ldots, f_n$ are multiplicatively independent modulo constants.

So, we must have that $k_1e_{1j} + \cdots + k_ne_{nj} \ne 0$ for some $1 \le j \le m$ . Then, in view of (3·9) and noticing that $g_1, \ldots, g_m$ are pairwise distinct irreducible polynomials of degree at least 2, we obtain directly the desired finiteness result by applying Theorem 1·1 to the polynomials $g_1, \ldots, g_m$ . This completes the proof of the case when $f_1, f_2, \ldots, f_n$ all have no linear factor.

Now, without loss of generality, we assume that for each $f_i, i =1, 2, \ldots, n$ , both its numerator and denominator have linear factors.

Then, for each $f_i, i=1,2,\ldots, n$ , we write

$$f_i = a_i f_{i1} f_{i2}, \quad a_i\in K^*,$$

where $f_{i1} \in K(X)$ is monic and only has linear factors, and $f_{i2} \in K(X)$ is monic and only has irredicible factors of degree at least two; and moreover, we write

$$f_{i1} = \frac{h_{i1}}{h_{i2}}, \quad h_{i1}, h_{i2} \in K[X], \ \gcd(h_{i1}, h_{i2}) = 1.$$

By assumption, for each $i=1,2, \ldots, n$ , both $h_{i1}$ and $h_{i2}$ have at least two distinct linear factors and they only have linear factors. Moreover, since we have assumed that $f_1, f_2, \ldots, f_n$ have distinct linear factors, we know that $h_{11}, h_{12}, \ldots, h_{n1}, h_{n2}$ are pairwise coprime.

Let $g_1, \ldots, g_m$ (assume $m \ge 1$ ) be all the distinct monic irreducible factors (over K) in the numerators and denominators of the rational functions $f_{12}, \ldots, f_{n2}$ .

By assumption, the irreducible polynomials $g_1, \ldots, g_m$ are all of degree at least two. So, the polynomials $h_{11}, h_{12}, \ldots, h_{n1}, h_{n2}$ , $g_1, \ldots, g_m$ are pairwise coprime.

Then, for each $f_i, 1\le i \le n$ , we can write

(3·10) \begin{equation} f_i = a_i h_{i1} h_{i2}^{-1}\prod_{j=1}^{m} g_j^{e_{ij}},\quad a_i\in K^*,\end{equation}

for some integers $e_{i1}, \ldots, e_{im}$ .

As in (3·9), combining (3·8) with (3·10) we can get that for some $\gamma^{\prime} \in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*$ ,

(3·11) \begin{equation} \prod_{i=1}^{n} h_{i1}(\alpha)^{k_i} h_{i2}(\alpha)^{-k_i} \cdot \prod_{j=1}^{m}g_j(\alpha)^{k_1e_{1j} + \cdots + k_ne_{nj}}= \gamma^{\prime}.\end{equation}

Then, in view of (3·11) and noticing that the integers $k_1, \ldots, k_n$ are not all zero, we obtain directly the desired finiteness result by applying Theorem 1·1 to the polynomials $h_{11}, h_{12}, \ldots, h_{n1}, h_{n2}$ , $g_1, \ldots, g_m$ . This completes the proof.

3.4. Proof of Corollary 1·4

First, we assume that $f \in K[X]$ and 0 is not a periodic point of f. Since 0 is not a periodic point of f, we have that for any integer $n \ge 1$ , $f^{(n)}(0) \ne 0$ , which means that $f^{(n)}$ has non-zero constant term. So, all the iterates of f are pairwise coprime. In addition, since f has at least two distinct roots, it is easy to see that each iterate of f also has at least two distinct roots. Hence, by Theorem 1·1 we know that there are only finitely many elements $\beta \in K$ such that $f^{(1)}(\beta), \ldots, f^{(n)}(\beta)$ are multiplicatively dependent modulo $\Gamma^{{\textrm{div}}}_\varepsilon$ .

Now, we assume that f has no linear factor. Then, by Lemma 2·5, the iterates $f^{(1)}, \ldots, f^{(n)}$ all have no linear factor. Moreover, it follows directly from Lemma 2·4 that the iterates $f^{(1)}, \ldots, f^{(n)}$ are multiplicatively independent modulo constants. So, using Theorem 1·2 we get that there are only finitely many elements $\beta \in K$ such that $f^{(1)}(\beta), \ldots, f^{(n)}(\beta)$ are multiplicatively dependent modulo $\Gamma^{{\textrm{div}}}_\varepsilon$ .

So, for proving the desired result, it suffices to fix such an element $\beta$ and show that there are only finitely many $\alpha \in K$ such that $f^{(m)}(\alpha) = \beta$ for some integer $m \ge 0$ . Indeed, this follows directly from [Reference Bérczes, Ostafe, Shparlinski and Silverman9, lemma 2·3] and the well-known fact that f has only finitely many preperiodic points lying in K.

3.5 Proof of Theorem 1·5

The proof follows similar ideas as in the proof of [Reference Bérczes, Ostafe, Shparlinski and Silverman9, theorem 1·7].

Let $\alpha \in K$ be such that there exist integers $k_1,k_2$ , not both zero, such that

$$f_1(\alpha)^{k_1}f_2(\alpha)^{k_2}\in\Gamma^{{\textrm{div}}}_\varepsilon.$$

As in the proof of Theorem 1·1, we enlarge the set $S_{{\textbf{f}},\Gamma}$ (in this case ${\textbf{f}}=(f_1,f_2)$ ) to a larger set $S_{{\textbf{f}},\Gamma,\varepsilon}$ such that

(3·12) \begin{equation}f_1(\alpha)^{k_1}f_2(\alpha)^{k_2}= \gamma \in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*.\end{equation}

Also, as in the proof of Theorem 1·1 we can assume that $k_1 k_2 \ne 0$ . From (3·12) and the power saturation of $O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*$ in $K^*$ , we see that

$$ \gamma = \beta^{\gcd(k_1, k_2)} \quad \textrm{for some $\beta \in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*$}.$$

This allows us to take the $\gcd(k_1, k_2)$ -root of (3·12), so without loss of generality we can assume that

$$ \gcd(k_1, k_2)=1.$$

We now complete the proof case by case.

In this case, we have $f_1(\alpha),f_2(\alpha)\in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}$ . We note that if $f_1(\alpha)\in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*$ or $f_2(\alpha)\in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*$ , then Lemma 2·2 implies the finiteness of such $\alpha \in K$ satisfying (3·12).

Thus, we can assume that $f_1(\alpha),f_2(\alpha)\not\in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*$ . This implies that there exists a prime ideal ${\mathfrak{p}}$ of K such that the additive valuation $v_{\mathfrak{p}}(f_1(\alpha))\gt0$ . Moreover, since $f_2(\alpha)\in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}$ , we have $v_{\mathfrak{p}}(f_2(\alpha))\ge 0$ .

If $k_1k_2 \gt 0$ , then we can assume that $k_1, k_2 \gt 0$ . In this case, since equation (3·12) implies

$$k_1v_{\mathfrak{p}}(f_1(\alpha))+k_2v_{\mathfrak{p}}(f_2(\alpha))=0,$$

we obtain a contradiction by noticing $v_{\mathfrak{p}}(f_1(\alpha)) \gt 0$ and $v_{\mathfrak{p}}(f_2(\alpha)) \ge 0$ .

We now assume $k_1k_2\lt0$ . Moreover, we can assume $k_1\gt0$ and $k_2\lt0$ (similar discussion applies for $k_1\lt0$ and $k_2\gt0$ ). Since $\gcd(k_1,k_2)=1$ , there exist integers s, t such that

$$sk_1 + tk_2=1.$$

Then, using (3·12), we have

(3·13) \begin{equation}\begin{split}& f_1(\alpha) = f_1(\alpha)^{sk_1 + tk_2} = \gamma^s (f_1(\alpha)^{-t}f_2(\alpha)^s)^{-k_2}, \\& f_2(\alpha) = f_2(\alpha)^{sk_1 + tk_2} = \gamma^t (f_1(\alpha)^{-t}f_2(\alpha)^s)^{k_1}.\end{split}\end{equation}

We note that, since $f_1(\alpha) \in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}, \gamma \in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*$ and $-k_2\gt0$ , we have $f_1(\alpha)^{-t}f_2(\alpha)^s \in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}$ .

If $f_1(\alpha)^{-t}f_2(\alpha)^s \in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*$ , then by (3·13) we obtain that $f_1(\alpha)\in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*$ , which contradicts our assumption above. Thus, $f_1(\alpha)^{-t}f_2(\alpha)^s \not\in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*$ . Then, by Lemma 2·6 (with $y=f_1(\alpha)^{-t}f_2(\alpha)^s$ and noticing $\gamma \in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*$ ), the exponent $-k_2$ is bounded above only in terms of $f_1, f_2, K, \Gamma$ and $\varepsilon$ . Similarly, we obtain that the exponent $k_1$ is also bounded above only in terms of $f_1, f_2, K, \Gamma$ and $\varepsilon$ . Hence, in (3·12) there are only finitely many choices of the two exponents $k_1, k_2$ . Then, fixing $k_1, k_2$ and applying Lemma 2·2 to the rational function $f_1^{k_1}f_2^{k_2}$ , we obtain the desired finiteness result, where we need to use the assumption on $f_1$ and $f_2$ that they can not multiplicatively generate a power of a linear fractional function. This completes the proof of Case I.

In this case, as in the proof of Theorem 1·1, we can choose a prime ideal ${\mathfrak{p}}$ of the ring of integers of K such that

$${\mathfrak{p}}\not \in S_{{\textbf{f}},\Gamma, \varepsilon} \quad\text{and}\quad v_{\mathfrak{p}}(\alpha) \lt 0.$$

Let $d_i = \deg f_i, i =1, 2$ . Then, using the ultrametric inequality of non-Archimedean valuations and noticing (3·1), we directly have

(3·14) \begin{equation} v_{\mathfrak{p}}(f_i(\alpha)) = d_i v_{\mathfrak{p}}(\alpha)\quad\text{for $i=1,2$.}\end{equation}

Considering valuations in (3·12) and using (3.14) we obtain (since $v_{\mathfrak{p}}(\gamma)=0$ due to $\gamma\in O_{S_{{\textbf{f}},\Gamma,\varepsilon}}^*$ )

$$k_1d_1+k_2d_2=0.$$

Since $\gcd(k_1,k_2)=1$ , this implies that $k_1 \mid d_2$ and $k_2 \mid d_1$ . Thus we can assume that both $k_1$ and $k_2$ are fixed. Then, as in Case I, the desired finiteness result follows from Lemma 2·2. This completes the proof.

3.6. Proof of Theorem 1·6

The proof follows directly from Maurin’s result (that is, Lemma 2·3). Indeed, let r be the rank of $\Gamma$ modulo torsion and let $g_1,\ldots,g_r \in \Gamma$ be its generators, and thus, they are multiplicatively independent elements. We define the parametric curve

$${\mathcal{C}}=\{(\alpha, f_1(\alpha),\ldots,f_n(\alpha), g_1,\ldots,g_r) \,:\, \alpha \in {\overline K}\}\subset {\mathbb{G}_{\textrm{m}}}^{n+r+1}.$$

We choose $\varepsilon$ to be half of the size of the real $\varepsilon$ from Lemma 2·3.

For an element $\alpha \in \Gamma^{\textrm{div}}_\varepsilon$ , assume that $f_1(\alpha), \ldots, f_n(\alpha)$ are multiplicatively dependent modulo $\Gamma^{\textrm{div}}_\varepsilon$ . Since $\alpha \in \Gamma^{\textrm{div}}_\varepsilon$ , there exist a non-zero vector $(k_0,\ldots,k_r)\in{\mathbb{Z}}^{r+1}$ , $k_0 \ne 0$ , such that

$$\alpha^{k_0}g_1^{k_{1}}\cdots g_r^{k_{r}}=\beta^{k_0}$$

for some $\beta \in \overline{K}^*$ with $ {\textrm{h}}(\beta) \leq \varepsilon$ , implying that

(3·15) \begin{equation}\dfrac{\alpha^{k_0}}{\beta^{k_0}}g_1^{k_{1}}\cdots g_r^{k_{r}}=1.\end{equation}

Moreover, since $f_1(\alpha), \ldots, f_n(\alpha)$ are multiplicatively dependent modulo $\Gamma^{\textrm{div}}_\varepsilon$ , there exist some positive integer t and a non-zero vector $(\ell_1,\ldots,\ell_{n+r})\in{\mathbb{Z}}^{n+r}$ such that

$$f_1(\alpha)^{t\ell_1}\cdots f_n(\alpha)^{t\ell_n}g_1^{\ell_{n+1}}\cdots g_r^{\ell_{n+r}}=\gamma^t$$

for some $\gamma \in \overline{K}^*$ with ${\textrm{h}}(\gamma) \leq \varepsilon$ , implying that (without loss of generality, we assume $\ell_1 \cdots \ell_n \ne 0$ )

(3·16) \begin{equation} \dfrac{f_1(\alpha)^{t\ell_1}}{\gamma^{t\ell_1/n\ell_1}}\cdots \dfrac{f_n(\alpha)^{t\ell_n}}{\gamma^{t\ell_n/n\ell_n}}g_1^{\ell_{n+1}}\cdots g_r^{\ell_{n+r}}=1. \end{equation}

Therefore, the point

$$\left(\dfrac{\alpha}{\beta},\frac{f_1(\alpha)}{\gamma^{1/n\ell_1}},\ldots,\frac{f_n(\alpha)}{\gamma^{1/n\ell_n}},g_1,\ldots,g_r\right)$$

satisfies the multiplicative dependence relations (3.15) and (3.16), which have linearly independent vectors of exponents. Moreover,

\begin{align*}&\left(\alpha,f_1(\alpha),\ldots,f_n(\alpha),g_1,\ldots,g_r\right)\\&=\left(\dfrac{\alpha}{\beta},\frac{f_1(\alpha)}{\gamma^{1/n\ell_1}},\ldots,\frac{f_n(\alpha)}{\gamma^{1/n\ell_n}},g_1,\ldots,g_r\right) \cdot (\beta,\gamma^{1/n\ell_1},\ldots ,\gamma^{1/n\ell_n},1,\ldots,1)\end{align*}

is a point on $\mathcal{C}$ with

\begin{align*}{\textrm{h}}(\beta,\gamma^{1/n\ell_1},\ldots,\gamma^{1/n\ell_n},1,\ldots,1)& :\!=\, {\textrm{h}}(\beta)+ {\textrm{h}}(\gamma^{1/n\ell_1}) + \cdots + {\textrm{h}}(\gamma^{1/n\ell_n}) \\& \leq {\textrm{h}}(\beta) + {\textrm{h}}(\gamma) \leq 2\varepsilon.\end{align*}

We also note that by assumption, the functions $X, f_1,\ldots,f_n,g_1,\ldots,g_r$ are multiplicatively independent. Hence, the desired result follows directly from Lemma 2·3.