Main Results {#sec:main}
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Proof of Theorem \[thm:genAR1\] {#sec:univar1}
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We use the same idea as in the proof of [@AR Theorem 1] and Lemma \[lem:univAR\]. Indeed, we write the factorisation in linear factors, $$h_1=\prod_{i=1}^{d_{h_1}}(T-\omega_{1,i}),\quad h_2=\prod_{i=1}^{d_{h_2}}(T-\omega_{2,i}),$$ where $\omega_{1,i},\omega_{2,j}\in\C$, $i=1,\ldots,d_{h_1}$, $j=1,\ldots,d_{h_2}$.

Thus, we reduce the problem to estimating the degree of each $$\gcd \(f^n-\omega_{1,i},g^m-\omega_{2,j}\).$$ For simplicity we use the notation $\omega_1$ and $\omega_2$ for any two roots of $h_1$ and $h_2$, respectively, and we denote $$\cD_{n,m}(\omega_1,\omega_2)=\gcd \(f^n-\omega_{1},g^m-\omega_{2}\).$$

Thus, as in Lemma \[lem:univAR\], we need to bound the number of $t\in\C$ such that $f(t)^n=\omega_1$ and $g(t)^m=\omega_2$ for some positive integers $n$ and $m$.

For every $n,m\ge 1$, we fix an element $t_{n,m}\in\C$ such that $$\label{eq:root}
f(t_{n,m})^n=\omega_1,\quad g(t_{n,m})^m=\omega_2$$ (if no such $t_{n,m}$ exists then we immediately have $\deg \cD_{n,m}(\omega_1,\omega_2) =0$). We define new polynomials $$\widetilde{f}_{n,m}(T)=\frac{1}{f(t_{n,m})}f(T)\mand
 \widetilde{g}_{n,m}(T)=\frac{1}{g(t_{n,m})}g(T).$$ As $f$ and $g$ are multiplicatively independent in $\C(T)^*/\C^*$, we obtain that $\widetilde{f}_{n,m}$ and $\widetilde{g}_{n,m}$ are multiplicatively independent for every $n,m$.

Thus, we can apply Lemma \[lem:univAR\] and conclude that $$\deg \gcd\(\widetilde{f}_{n,m}^n-1,\widetilde{g}_{n,m}^m-1\)\le \(11(d_f+d_g)^{2}\)^{\min(d_f,d_g)}.$$

From  and the definition of $\widetilde{f}_{n,m}$ and $\widetilde{g}_{n,m}$, we have $$\deg \cD_{n,m}(\omega_1,\omega_2)=\deg \gcd\(\widetilde{f}_{n,m}^n-1,\widetilde{g}_{n,m}^m-1\),$$ and thus, for every $n,m\ge 1$, we get $$\deg\cD_{n,m}(\omega_1,\omega_2)\le \(11(d_f+d_g)^{2}\)^{\min(d_f,d_g)}.$$ As this holds for any roots $\omega_1,\omega_2$ of $h_1$ and $h_2$, 