Proof Let $\theta \in J$ be such that $v_{{\mathfrak {P}}}\big (\phi (\theta )\big )\geq e({\mathfrak {P}}\mid {\mathfrak {p}})$ and $v_{\mathfrak {P}}\big (f(\theta )\big )>e({\mathfrak {P}}\mid {\mathfrak {p}})$ . We see that $\phi (\theta ), f(\theta )\equiv 0\bmod {\mathfrak {P}}$ . Since $k_{\mathfrak {P}}={\mathcal O}_J/{\mathfrak {P}}$ is an extension of $k_{\mathfrak {p}}$ , we see that $\gcd \big (f(x),\phi (x)\big )\neq 1$ in $k_{\mathfrak {p}}[x]$ . Since $\phi (x)$ is irreducible in $k_{{\mathfrak {p}}}[x]$ , we note that $\phi (x)$ divides $f(x)$ in $k_{\mathfrak {p}}[x]$ . However, since $\phi (x)$ also divides $f'(x)$ in $k_{\mathfrak {p}}[x]$ , we see that $\phi (x)^2$ divides $f(x)$ in $k_{\mathfrak {p}}[x]$ . Thus, we can write $f(x)=\phi (x)^2g(x)+bh(x)$ in ${\mathcal O}_K[x]$ for some $b\in {\mathfrak {p}}$ and some $h(x)\in {\mathcal O}_K[x]$ . We evaluate at $\theta $ to obtain $f(\theta )=\phi (\theta )^2 g(\theta )+ bh(\theta )$ . Noting $v_{\mathfrak {P}}\big (\phi (\theta )^2 g(\theta )\big )\geq 2 e({\mathfrak {P}}\mid {\mathfrak {p}})$ , we find $v_{\mathfrak {P}}\big (bh(\theta )\big )>e({\mathfrak {P}}\mid {\mathfrak {p}})$ . Thus, either $b\in {\mathfrak {p}}^2$ or $v_{\mathfrak {P}}\big (h(\theta )\big )>0$ . In the former case, our desired result is clear. In the latter case, we note that $\gcd \big (h(x),\phi (x)\big )\neq 1$ in $k_{\mathfrak {P}}[x]$ and hence in $k_{\mathfrak {p}}[x]$ as above. Thus, $h(x)=\phi (x)j(x)+c\ell (x)$ with $c\in {\mathfrak {p}}{\mathcal O}_K$ . Therefore,

$$\begin{align*}f(x)=\phi(x)^2g(x)+bh(x)=\phi(x)^2g(x)+b\big(\phi(x)j(x)+c\ell(x)\big),\end{align*}$$

and we have a).

Suppose conversely that $\phi (x)$ divides $f(x)$ in ${\mathcal O}_K/{\mathfrak {p}}^2[x]$ . Since $\phi (x)$ is a factor of $f'(x)$ modulo ${\mathfrak {p}}$ , we have $f(x)=\phi (x)^2g(x)+b\phi (x)h(x)+cj(x),$ where $b\in {\mathfrak {p}}$ and $c\in {\mathfrak {p}}^2$ . Evaluating at any $\theta $ such that $v_{\mathfrak {P}}\big (\phi (\theta )\big )\geq e({\mathfrak {P}}\mid {\mathfrak {p}})$ , we see that $v_{\mathfrak {P}}\big (f(\theta )\big )\geq 2e({\mathfrak {P}}\mid {\mathfrak {p}})$ , yielding b).

Proof Since $\phi (x)$ is irreducible in $k_{\mathfrak {p}}[x]$ and since $f(x)\equiv g(x)^p\bmod {\mathfrak {p}}$ , we see that $\phi (x)$ divides $g(x)$ in $k_{\mathfrak {p}}[x]$ . Thus, we can write $g(x)\equiv \phi (x)^mr(x)\bmod {\mathfrak {p}}$ with $\gcd \big (\phi (x),r(x)\big )=1$ in $k_{\mathfrak {p}}[x]$ . Since $g(x)$ splits in L, we see that there is an algebraic integer $\tau $ that is a root of $g(x)$ and such that $\phi (\tau )\equiv 0\bmod {\mathfrak {P}}$ . We have $0=g(\tau )\equiv \phi (\tau )^m r(\tau )\bmod {\mathfrak {p}}{\mathcal O}_L$ . Since $\gcd \big (\phi (x),r(x)\big )=1$ and since $v_{\mathfrak {P}}\big (\phi (\tau )\big )>0$ , we see $v_{\mathfrak {P}}\big (r(\tau )\big )=0$ and $v_{\mathfrak {P}}\big (\phi (\tau )^m\big )\geq e({\mathfrak {P}}\mid {\mathfrak {p}})$ .

Since $\phi (x)$ divides $f(x)$ modulo ${\mathfrak {p}}^2$ and since $f(x)\equiv g(x)^p\bmod {\mathfrak {p}}$ , we have $f(x)=\big (\phi (x)^mr(x)\big )^p+bh(x)$ with $b\in {\mathfrak {p}}$ and either $b\in {\mathfrak {p}}^2$ or $\phi (x)$ dividing $h(x)$ modulo ${\mathfrak {p}}$ . Evaluating at $\tau $ and noting $v_{\mathfrak {P}}\big (\phi (\tau )^m\big )\geq e({\mathfrak {P}}\mid {\mathfrak {p}})$ , we have our desired result.

For the converse, let $\tau \in L$ be a root of $g(x)$ such that $v_{\mathfrak {P}}\big (f(\tau )\big )>e({\mathfrak {P}}\mid {\mathfrak {p}})$ . Let $\phi (x)$ be an irreducible factor of $g(x)$ in ${\mathcal O}_K[x]$ such that $\phi (\tau )\equiv 0\bmod {\mathfrak {P}}$ . (Note that we can assume that the number of irreducible factors of $g(x)$ in $k_{\mathfrak {p}}[x]$ is equal to that in ${\mathcal O}_K[x]$ .)

By hypothesis, $f(x)=g(x)^p+bh(x)$ with $b\in {\mathfrak {p}}$ and $h(x)\in {\mathcal O}_K[x]$ . Evaluating at $\tau $ , we see that $v_{\mathfrak {P}}\big (bh(\tau )\big )>e({\mathfrak {P}}\mid {\mathfrak {p}})$ . Thus, either $b\in {\mathfrak {p}}^2$ or $\gcd \big (\phi (x),h(x)\big )>1$ in $k_{\mathfrak {p}}[x]$ . In either case, we have our result.

Proof Combining Lemmas 2.6 and 2.7 with Theorem 2.3, we have the result.

Proof First, we focus on the non-vanishing primes. Let ${\mathfrak {p}}\subset {\mathcal O}_K$ be such a prime. Though there is a natural proof via induction, we proceed directly. Note Lemma 2.6 and our hypothesis immediately imply that no $\phi _i(x)$ divides $f^n(x)-a$ in ${\mathcal O}_K/{\mathfrak {p}}^2[x]$ for any $1\leq n\leq N$ . Fix an n in the range $1\leq n\leq N$ . We wish to show that $f^n(x)-a$ is ${\mathfrak {p}}$ -maximal. We have

(2.1) $$ \begin{align} \frac{d}{dx}\Big(f^{n}(x)-a\Big)=f'\big(f^{n-1}(x)\big)\dots f'(x)\equiv \prod_{j=1}^s \gamma_{j}(x)^{m_j} \bmod {\mathfrak{p}}, \end{align} $$

where the $\gamma _j(x)$ are lifts of the irreducible factors of $\frac {d}{dx}\big (f^{n}(x)-a\big )$ in $k_{\mathfrak {p}}[x]$ to ${\mathcal O}_K[x]$ . Let $J'$ be an extension of J where all the $\gamma _j(x)$ split, and let ${\mathcal P}$ be a prime of $J'$ above ${\mathfrak {P}}$ .

For a contradiction, suppose $f^n(x)-a$ is not ${\mathfrak {p}}$ -maximal. Thus, Theorem 2.8 shows that there is some $\tau \in J'$ and some $\gamma _j(x)\in {\mathcal O}_K[x]$ such that $v_{\mathcal P}\big (\gamma _j(\tau )\big )\geq e({\mathcal P}\mid {\mathfrak {p}})$ and $v_{\mathcal P}\big (f^n(\tau )-a\big )> e({\mathcal P}\mid {\mathfrak {p}})$ . Since $\gamma _j(x)$ is irreducible in $k_{\mathfrak {p}}[x]$ , we see that $\gamma _j(x)$ divides $f'\big (f^t(x)\big )$ in $k_{\mathfrak {p}}[x]$ for some $0\leq t\leq n-1$ . Hence, $\gamma _j(x)$ divides $\phi _i\big (f^t(x)\big )$ in $k_{\mathfrak {p}}[x]$ for some i. Since $v_{\mathcal P}\big (\gamma _j(\tau )\big )\geq e({\mathcal P}\mid {\mathfrak {p}})$ , we have $v_{\mathcal P}\big (\phi _i\big (f^t(\tau )\big )\big )\geq e({\mathcal P}\mid {\mathfrak {p}})$ . As

$$\begin{align*}v_{\mathcal P}\big(f^n(\tau)-a\big) = v_{\mathcal P}\Big(f^{n-t}\big(f^t(\tau)\big)-a\Big)> e({\mathcal P}\mid {\mathfrak{p}}),\end{align*}$$

Lemma 2.6 shows that $\phi _i(x)$ divides $f^{n-t}(x)-a$ in ${\mathcal O}_K/{\mathfrak {p}}^2[x]$ , yielding the desired contradiction. We have established Condition 1.

Now we turn to the vanishing primes. Suppose $f'(x)\equiv 0\bmod {\mathfrak {p}}$ , so $f(x)-a\equiv g(x)^p\bmod {\mathfrak {p}}$ for some $g(x)\in {\mathcal O}_K[x]$ . Let L and ${\mathfrak {P}}$ be as in the theorem statement. For the monogenicity of $f^n(x)-a$ , Theorem 2.8 shows that it is necessary that $v_{\mathfrak {P}}\big (f(\tau )-a\big )\leq e({\mathfrak {P}}\mid {\mathfrak {p}})$ for each root $\tau $ of $g(x)$ . Suppose this is the case and consider $f^n(x)-a$ with $1\leq n \leq N$ . Since $f'(x)\equiv 0\bmod {\mathfrak {p}}$ , we compute

$$\begin{align*}\frac{d}{dx}\Big(f^{n}(x)-a\Big)=f'\big(f^{n-1}(x)\big)f'\big(f^{n-2}(x)\big)\dots f'(x)\equiv 0\bmod {\mathfrak{p}}.\end{align*}$$

Thus, ${\mathfrak {p}}$ is a vanishing prime for $f^{n}(x)-a$ , and $f^{n}(x)-a\equiv g\big (f^{n-1}(x)\big )^p\bmod {\mathfrak {p}}$ . Let $L'$ be an extension of $L,$ where $g\big (f^{n-1}(x)\big )$ splits completely and let ${\mathcal P}$ be a prime of $L'$ above ${\mathfrak {P}}$ . Theorem 2.8 shows that $f^n(x)-a$ is ${\mathfrak {p}}$ -maximal if and only if for any $\tau '\in L'$ that is a root of $g\big (f^{n-1}(x)\big )$ , one has $v_{\mathcal P}\big (f^n(\tau ')-a\big )\leq e({\mathcal P}\mid {\mathfrak {p}})$ . For a contradiction, suppose $\tau '$ is a root of $g\big (f^{n-1}(x)\big )$ in $L'$ with $v_{\mathcal P}\big (f^n(\tau ')-a\big )> e({\mathcal P}\mid {\mathfrak {p}})$ . Now $f^{n-1}(\tau ')$ is a root of $g(x)$ in L. We have $v_{\mathcal P}\big (f^n(\tau ')-a\big )=e({\mathcal P}\mid {\mathfrak {P}})v_{\mathfrak {P}}\big (f(f^{n-1}(\tau '))-a\big )$ and $e({\mathcal P}\mid {\mathfrak {p}})=e({\mathcal P}\mid {\mathfrak {P}})e({\mathfrak {P}}\mid {\mathfrak {p}})$ , so $v_{\mathfrak {P}}\big (f(f^{n-1}(\tau '))-a\big )> e({\mathfrak {P}}\mid {\mathfrak {p}})$ . This is a contradiction. We conclude that $v_{\mathcal P}\big (f^n(\tau ')-a\big ) \leq e({\mathcal P}\mid {\mathfrak {p}})$ and $f^n(x)-a$ is ${\mathfrak {p}}$ -maximal for all $1\leq n\leq N$ .

Proof For ease of notation, let $k=v_{\phi ,{\mathfrak {p}}}\big (h(x)\big )$ and $m=v_{\phi ,{\mathfrak {p}}}\big (f(x)\big )$ . If $v_{\phi ,{\mathfrak {p}}}\big (f(x)\big )=0$ , then the result is clear, so suppose $\phi (x)$ divides $f(x)$ in ${\mathcal O}_K/{\mathfrak {p}}^2[x]$ . Let $\theta $ in M be a critical point such that $\phi (\theta )\equiv 0\bmod {\mathfrak {P}}$ . We have $h(x)=\phi (x)^kr(x) + d$ for some $r(x)\in {\mathcal O}_K[x]$ that is relatively prime to $\phi (x)$ in $k_{\mathfrak {p}}[x]$ and some $d\in {\mathfrak {p}}{\mathcal O}_K[x]$ . Since $h(\theta )=0$ , we see that $v_{\mathfrak {P}}\big (\phi (\theta )^k\big )\geq e({\mathfrak {P}}\mid {\mathfrak {p}})$ . We can write $f(x)=\phi (x)^mg(x)+b\phi (x)+c$ with $b\in {\mathfrak {p}}{\mathcal O}_K[x]$ and $c\in {\mathfrak {p}}^2{\mathcal O}_K[x]$ . Since $m>k$ , we evaluate at $\theta $ to obtain $v_{\mathfrak {P}}\big (f(\theta )\big )>e({\mathfrak {P}}\mid {\mathfrak {p}})$ .

Conversely, suppose $v_{\mathfrak {P}}\big (f(\theta )\big )>e({\mathfrak {P}}\mid {\mathfrak {p}})$ for some critical point $\theta $ in M such that $\phi (\theta )\equiv 0\bmod {\mathfrak {P}}$ . We have $h(x)=\phi (x)^kr(x) + d$ for some $r(x)\in {\mathcal O}_K[x]$ that is relatively prime to $\phi (x)$ in $k_{\mathfrak {p}}[x]$ and some $d\in {\mathfrak {p}}{\mathcal O}_K[x]$ . Since $h(\theta )=0$ , we see that $v_{\mathfrak {P}}\big (\phi (\theta )^k\big )\geq e({\mathfrak {P}}\mid {\mathfrak {p}})$ . We write $f(x)=\phi (x)^mg(x)+bj(x)+c$ with $b\in {\mathfrak {p}}$ and $c\in {\mathfrak {p}}^2{\mathcal O}_K[x]$ . Since $v_{\mathfrak {P}}\big (\phi (\theta )^mg(\theta )\big )>e({\mathfrak {P}}\mid {\mathfrak {p}})$ and $v_{\mathfrak {P}}\big (f(\theta )\big )>e({\mathfrak {P}}\mid {\mathfrak {p}})$ , we see $v_{\mathfrak {P}}\big (bj(\theta )\big )>e({\mathfrak {P}}\mid {\mathfrak {p}})$ . By a now familiar argument, $\phi (x)$ divides $f(x)$ in ${\mathcal O}_K/{\mathfrak {p}}^2[x]$ .

The last statement is obtained by applying Theorem 2.3 and noting that any repeated factor of $f(x)$ in $k_{\mathfrak {p}}[x]$ will be a factor of $f'(x)$ in $k_{\mathfrak {p}}[x]$ .

Condition 2.13 For every irreducible factor $h(x)$ of $f'(x)$ in $K[x]$ , every irreducible factor $\phi (x)$ of $h(x)$ in $k_{\mathfrak {p}}[x]$ , and every $n\in \mathbb {N}$ such that $v_{\phi (x)}(f^n(x)-a)>0$ , we have $v_{\phi (x)}(f^n(x)-a)>v_{\phi (x)}(h(x))$ .

Proof To show a), it suffices to prove that $f^n(x)-a$ has no repeated factors of multiplicity divisible by p in $k_{\mathfrak {p}}[x]$ for any $n\in \mathbb {N}$ , since indeed then the multiplicity of any irreducible factor decreases strictly in $\frac {d}{dx}(f^n(x))$ and thus a fortiori in $f'(x)$ . Since the ramification index of a place $x\mapsto \alpha \in \overline {\mathbb {F}_p}$ under the cover $x\mapsto f(x)$ is simply the multiplicity of $x-\alpha $ in $f(x)-f(\alpha )$ , tame ramification translates to saying that $f(x)-a$ has no root of multiplicity divisible by p for any $a\in \overline {\mathbb {F}_p}$ . This shows the claim for $n=1$ . For arbitrary n, the claim simply follows by multiplicativity of ramification indices under composition of covers.

b) follows from the definition: Let $h(x)$ be an irreducible factor of $f'(x)$ in $K[x]$ . A simple factor of $h(x)$ that is a common factor of $f'(x)$ and $f^n(x)-a$ must necessarily be a repeated factor of $f^n(x)-a$ .

Proof Let $1\leq n\leq N$ and consider $\frac {d}{dx}\big (f^{n}(x)-a\big )=f'\big (f^{n-1}(x)\big )\dots f'(x).$

Let $M'$ be an extension of $M,$ where $\frac {d}{dx}(f^n(x)-a)$ splits, and let ${\mathcal P}$ be a prime of $M'$ above ${\mathfrak {P}}$ . If $\tau \in M'$ is a critical point of $f^n(x)-a$ , then $\tau $ is a root of $f'\big (f^t(x)\big )$ for some $0\leq t\leq n-1$ . Thus, $f^t(\tau )$ is a root of $f'(x)$ . From this and the multiplicative nature of valuations, we see $v_{{\mathcal P}}\big (f^n(\tau )-a\big )>e({\mathcal P}\mid {\mathfrak {p}})$ for some $\tau $ that is a critical point of $f^n(x)-a$ if and only if $v_{\mathfrak {P}}\big (f^{n-t}(\theta )-a\big )> e({\mathfrak {P}}\mid {\mathfrak {p}})$ for some $\theta =f^t(\tau )$ that is a critical point of $f(x)-a$ . Lemma 2.12 shows $v_{{\mathcal P}}\big (f^n(\tau )-a\big )>e({\mathcal P}\mid {\mathfrak {p}})$ for some $\tau $ that is a critical point of $f^n(x)-a$ if and only if $f^n(x)-a$ is not ${\mathfrak {p}}$ -maximal.

Proof For the non-vanishing primes, we let M be an extension, where $f'(x)$ splits and let ${\mathfrak {P}}$ be a prime of M above ${\mathfrak {p}}$ . Since $f(\theta )\in {\mathcal O}_K$ for each critical point $\theta $ , we have $v_{\mathfrak {P}}\big (f^n(\theta )-a\big )\leq e({\mathfrak {P}}\mid {\mathfrak {p}})$ if and only if $v_{\mathfrak {p}}\big (f^n(\theta )-a\big )\leq 1$ .

Condition 1 follows from Theorem 2.15.

For the vanishing primes, we let L be an extension, where $g(x)$ splits and let ${\mathfrak {P}}$ be a prime of L above ${\mathfrak {p}}$ . Since $f(\tau )\in {\mathcal O}_K$ for each root $\tau $ of $g(x)$ , we have $v_{\mathfrak {P}}\big (f(\tau )-a\big )\leq e({\mathfrak {P}}\mid {\mathfrak {p}})$ if and only if $v_{\mathfrak {p}}\big (f(\tau )-a\big )\leq 1$ . Condition 2 follows from Theorem 2.9.

Proof The implication 2 $\Rightarrow $ 1 follows directly from Corollary 2.18, so we assume 1.

For every prime ideal $\mathfrak {p}$ of ${\mathcal O}_K$ , let $\Omega _{\mathfrak {p}}$ be the set of all mod- $\mathfrak {p}^2$ reductions of elements of $\Omega $ (excluding those for which the reduction is not defined). It follows by a standard Chinese remainder argument, that the proportion of $a\in {\mathcal O}_K$ such that $(f(x),a)$ fulfills the conclusion of Corollary 2.18 is bounded below by

$$\begin{align*}\prod_{\mathfrak{p} \text{ vanishing}} \frac{1}{N(\mathfrak{p})^2} \cdot \prod_{\mathfrak{p} \text{ non-vanishing}} \left(1-\frac{|\Omega_{\mathfrak{p}}|}{N(\mathfrak{p})^2}\right).\end{align*}$$

Here, the factor $\prod _{\mathfrak {p}} \frac {1}{N(\mathfrak {p})^2}$ is due to the fact that Condition 2 of Corollary 2.18 only depends on the mod- ${\mathfrak {p}}^2$ residue class of a. This factor is positive since there are only finitely many vanishing primes. Furthermore, by Assumption i, we have $|\Omega _{\mathfrak {p}}| \le \min \{|\Omega |, N(\mathfrak {p})^2-1\}$ for all non-vanishing primes $\mathfrak {p}$ . From this, it follows that $\prod _{\mathfrak {p}} \big (1-\frac {|\Omega _{\mathfrak {p}}|}{N(\mathfrak {p})^2}\big )$ is bounded below by a positive constant factor (arising from the finitely many primes $\mathfrak {p}$ with $N(\mathfrak {p})^2-1 < |\Omega |$ ) times $\prod _{\mathfrak {p}} \big (1-\frac {|\Omega |}{N(\mathfrak {p})^2}\big )$ , the latter product obviously being bounded away from zero. In total, the proportion of $a\in {\mathcal O}_K$ such that $(f(x),a)$ fulfills the conclusion of Corollary 2.18 is bounded away from $0$ .

Proof For any field $K\subset F\subset \overline {K}$ , let $\textrm {Mon}_{f,F}:=\textrm {Gal}(f(x)-t/F(t))$ denote the monodromy group of $f(x)$ over F. It is well-known that $\textrm {Mon}_{f,F}$ embeds as a permutation group into the iterated wreath product $\text {Mon}_{f_r,F}\circ \dots \circ \text {Mon}_{f_1,F}$ . Furthermore, for $F=K(\zeta _p)$ the p-th cyclotomic field, one clearly has $\text {Mon}_{f_i,F}\cong C_p$ , so that, in particular, $\text {Mon}_{f,F}\le C_p\wr \dots \wr C_p$ is a p-group. The assertion is now a straightforward consequence of [Reference Benedetto, Ghioca, Juul and TuckerBGJT25, Theorem 1.3].

Proof For the first claim, suppose $f(x)$ has degree d. Note that $f(x)\equiv x^d\bmod {\mathfrak {p}}$ . However, x does not divide $f(x)$ modulo ${\mathfrak {p}}^2$ . Thus, $f(x)$ is ${\mathfrak {p}}$ -maximal. Alternatively, one can use the Montes algorithm/Ore’s theorem to see that the principal x-polygon is one-sided with slope $-\frac {1}{d}$ . This also shows that ${\mathfrak {p}}$ has ramification index d. The fact that iterates of a ${\mathfrak {p}}$ -Eisenstein polynomial are ${\mathfrak {p}}$ -Eisenstein is implied by Lemma 2.3 of [Reference OdoniOdo85], though a direct proof via analysis of the valuations of the coefficients of $f^n(x)$ is possible.

To see that an extension totally ramified over ${\mathfrak {p}}$ is given by a ${\mathfrak {p}}$ -Eisenstein polynomial, suppose we are in that setup, and let ${\mathfrak {P}}$ denote the unique prime of ${\mathcal O}_L$ above ${\mathfrak {p}}$ . Let $\pi _{\mathfrak {P}}\in {\mathcal O}_L$ be an element with ${\mathfrak {P}}$ -adic valuation 1. We see that $K\big (\pi _{\mathfrak {P}}\big )=L$ . Moreover, the minimal polynomial of $\pi _{\mathfrak {P}}$ over K is ${\mathfrak {p}}$ -Eisenstein. This can be generalized a good deal; the interested reader should consult Proposition 12 in Section 6 of Chapter III of [Reference SerreSer79].

Proof For (1) $\implies $ (2), note that if $\alpha \in {\mathcal O}_L^\times $ , then there exists $\alpha ^{-1}\in {\mathcal O}_L^\times $ so that $\alpha \cdot \alpha ^{-1}=1$ . Taking norms, we have $\text {Norm}_{L/F}(\alpha )\text {Norm}_{L/F}(\alpha ^{-1})=1$ . Thus, $c_0\in R^\times $ . $\checkmark $

For (2) $\implies $ (3), we have $c_n\alpha ^n+\cdots +c_1\alpha +c_0=0$ , so

$$\begin{align*}1=-c_0^{-1}c_n\alpha^n-\cdots -c_0^{-1}c_1\alpha.\ \text{ Thus } \ \alpha^{-1} = -c_0^{-1}c_n\alpha^{n-1}-\cdots -c_0^{-1}c_1. \ \checkmark\end{align*}$$

For (3) $\implies $ (4), we have $\alpha ^{-1} = b_{n-1}\alpha ^{n-1}+\cdots +b_1\alpha +b_0$ for some $b_i\in R$ . Multiplying by $\alpha ^{1-n}$ and subtracting, we obtain

$$\begin{align*}\alpha^{-n}-b_0\alpha^{1-n}-b_1\alpha^{2-n}-\cdots -b_{n-3}\alpha^{-2}-b_{n-2}\alpha^{-1}-b_{n-1}=0.\end{align*}$$

Since $\alpha ^{-1}$ is a unit and since it generates the same extension field as $\alpha $ , we see that $x^n-b_0x^{n-1}-\cdots -b_{n-2}x-b_{n-1}$ is the minimal polynomial. (1) $\implies $ (2) shows that $b_{n-1}\in R^\times $ . Thus,

$$\begin{align*}b_{n-1}^{-1}\alpha^{1-n}-b_{n-1}^{-1}b_0\alpha^{2-n}-b_{n-1}^{-1}b_1\alpha^{3-n}-\cdots -b_{n-1}^{-1}b_{n-3}\alpha^{-1}-b_{n-1}^{-1}b_{n-2}=\alpha\end{align*}$$

is an R-linear expression of $\alpha $ in powers of $\alpha ^{-1}$ . Hence, $R[\alpha ]=R[\alpha ^{-1}]$ . $\checkmark $

Finally, (4) $\implies $ (1) is clear since $R[\alpha ]=R[\alpha ^{-1}]\subset {\mathcal O}_L$ . $\checkmark $

Proof For $n\in \mathbb {N}$ , let $\alpha _n$ be a root of $f^n(x)-\frac {1+b\gamma }{b}$ , let $c_n\in {\mathcal O}_K$ be such that $f^n(c_n)=\gamma $ (note that such a $c_n$ exists due to $\gamma $ being periodic), and let $\beta _n:=\frac {1}{\alpha _n-c_n}$ . In particular, $\beta _n$ is a root of the polynomial $h_n(x):=\big (-bf^n(\frac {1}{x}+c_n) + 1+b\gamma \big )\cdot x^{\deg (f)^n}$ (i.e., the reciprocal of the polynomial $-bf^n(x+c_n) + 1+b\gamma $ ). We will show that, under the conditions of the theorem, ${\mathcal O}_{K_n}={\mathcal O}_K[\beta _n]$ for all $n\in \mathbb {N}$ .

First, we make sure that $h_n(x)$ is indeed monic. This amounts to verifying that the constant coefficient of $-bf^n(x+c_n) + 1+b\gamma $ equals $1$ , which follows directly from the fact that $f^{n}(c_n)=\gamma $ . Next, since the leading coefficient of $f(x)$ is a unit and $v_{{\mathfrak {p}}}(b)\le 1$ , we see that $h_n(x)$ is a ${\mathfrak {p}}$ -Eisenstein polynomial and hence ${\mathfrak {p}}$ -maximal for all primes ${\mathfrak {p}}\mid b$ by Lemma 3.4. Since b is a non-unit, there is at least one such prime ${\mathfrak {p}}$ with $v_{\mathfrak {p}}(b)=1$ . In particular, $h_n(x)$ is irreducible.

Next, pick a prime ${\mathfrak {p}}$ not dividing b which is a non-vanishing prime for $f(x)$ . We wish to show that the polynomial $g_n(x):=f^{n}(x+c_n)-\frac {1+b\gamma }{b}\in {\mathcal O}_{{\mathfrak {p}}}[x]$ is ${\mathfrak {p}}$ -maximal. Since translating by $c_n\in {\mathcal O}_K$ does not affect whether an element of ${\mathcal O}_{K_n}$ is a monogenerator for ${\mathcal O}_{K_n}\otimes {\mathcal O}_{\mathfrak {p}}$ , we may replace $g_n(x)$ by $f^{n}(x)-\frac {1+b\gamma }{b}$ . Let $\pi _{\mathfrak {p}}$ denote a generator of the maximal ideal of ${\mathcal O}_{\mathfrak {p}}$ . To apply Theorem 1.1, we fix $N\in \mathbb {N}$ and choose $a:=a(N)\in {\mathcal O}_K$ such that $a\equiv \frac {1+b\gamma }{b}\bmod \pi _{\mathfrak {p}}^2$ and such that $f^n(x)-a$ is irreducible for all $1\le n\le N$ . (For a fixed N, there are infinitely many such a by Hilbert’s irreducibility theorem.)

From Remark 2.2, we see that $f^{N}(x)-\frac {1+b\gamma }{b}$ is ${\mathfrak {p}}$ -maximal if and only if $f^{N}(x)-a$ is ${\mathfrak {p}}$ -maximal. Now Assumption ii shows that, for every critical point $\theta $ of $f(x)$ and every $k\in \mathbb {N}$ ,

$$\begin{align*}v_{{\mathfrak{P}}}\left(f^k(\theta)-a\right)=v_{{\mathfrak{P}}}\left(f^k(\theta)-\frac{1+b\gamma}{b}\right) \le e({\mathfrak{P}}\mid {\mathfrak{p}}),\end{align*}$$

where the former equality comes from the fact that $a\equiv \frac {1+b\gamma }{b}\bmod \pi _{\mathfrak {p}}^2$ so $v_{{\mathfrak {P}}}(a-\frac {1+b\gamma }{b})\ge 2e({\mathfrak {P}}\mid {\mathfrak {p}})$ . Thus, Condition 1 of Theorem 1.1 applied to the pair $\big (f(x), a\big )$ shows that $f^{N}(x)-a$ is ${\mathfrak {p}}$ -maximal. Since $N\in \mathbb {N}$ was arbitrary, we have obtained that $f^{n}(x)-\frac {1+b\gamma }{b}$ , and thus $g_n(x)$ , is ${\mathfrak {p}}$ -maximal for all $n\in \mathbb {N}$ . Since $g_n(x)$ has constant coefficient $f^n(c_n)-\frac {1+b\gamma }{b}=-\frac {1}{b}$ , which is a unit in ${\mathcal O}_{K_{\mathfrak {p}}}$ , it follows from Lemma 3.5 that ${\mathfrak {p}}\nmid \big [{\mathcal O}_{K_n}:{\mathcal O}_K[\beta _n]\big ]$ .

Suppose now that ${\mathfrak {p}}$ is a vanishing prime not dividing b. As above, given $N\in \mathbb {N}$ , we take any $a:=a(N)\equiv \frac {1+b\gamma }{b} \bmod ~\pi _{\mathfrak {p}}^2$ such that $f^n(x)-a$ is irreducible for all $1\le n\le N$ . Hence, $f(x)-a\equiv f(x)-\frac {1+b\gamma }{b}\equiv g(x)^p\bmod {\mathfrak {p}}$ . Assumption iii yields

$$\begin{align*}v_{{\mathfrak{P}}}\big(f(\tau)-a\big) = v_{{\mathfrak{P}}}\left(f(\tau)-\frac{1+b\gamma}{b}\right)\leq e({\mathfrak{P}}\mid {\mathfrak{p}}).\end{align*}$$

It follows from Theorem 1.1 that the polynomial $f^N(x)-a(N)$ is ${\mathfrak {p}}$ -maximal. Hence, by Remark 2.2 and since $N\in \mathbb {N}$ was arbitrary, $f^n(x)-\frac {1+b\gamma }{b}$ and $g_n(x)$ are ${\mathfrak {p}}$ -maximal for all $n\in \mathbb {N}$ . Thus, ${\mathfrak {p}}~\nmid ~\big [{\mathcal O}_{K_n}~:~{\mathcal O}_K[\beta _n]\big ]$ .

In conclusion, ${\mathcal O}_K[\beta _n]$ is ${\mathfrak {p}}$ -maximal for all primes ${\mathfrak {p}}$ . Thus, $\beta _n$ is a monogenerator of ${\mathcal O}_{K_n}$ over ${\mathcal O}_K$ .