Proof. By Equation (2.1) our claim (2.2) holds for $k=1$ .

We assume that Equation (2.2) holds for $k\ge 1$ and will deduce it for $k+1$ . So

(2.3)

as .

We evaluate $\Psi _{p,j}$ at $(X_1^{p^k},\ldots ,X_D^{p^k})$ and use Equation (2.1) to find

(2.4) $$ \begin{align} \begin{aligned} e_j(X_1^{p^k},\ldots,X_D^{p^k})^p = e_j(X_1^{p^{k+1}},\ldots,X_D^{p^{k+1}}) + p \Psi_{p,j}(X_1^{p^k},\ldots,X_D^{p^k}). \end{aligned} \end{align} $$

Finally, $\Psi _{p,j}(X_1^{p^k},\ldots ,X_D^{p^k}) \in \Psi _{p,j}^{p^k} + p\mathbb {Z}[X_1,\ldots ,X_D]$ and hence from Equation (2.4). Thus, Equation (2.2) for $k+1$ follows from Equation (2.3).

Proof. An elementary symmetric polynomial in $R[X_1,\ldots ,X_D]$ lies in $R[e_1,\ldots ,e_D]$ . We conclude (i) as $e_1(x_1,\ldots ,x_D),\ldots ,e_D(x_1,\ldots ,x_D)\in R$ by hypothesis.

For (ii), observe that $\Psi _{p,j}$ is symmetric so $\Psi _{p,j}(x_1,\ldots ,x_D)\in R$ by part (i). For the same reason, $e_j(x_1^{p^k},\ldots ,x_D^{p^k})\in R$ and $e_j(x_1^{p^l},\ldots ,x_D^{p^l})\in R$ .

By the hypothesis in (ii), we conclude . Lemma 2.1 implies

(2.5)

Raising to the p-th power gives for all $a\in R$ . In particular, $e_j(x_1,\ldots ,x_D)^{p^k}$ and $e_j(x_1,\ldots ,x_D)^{p^l}$ are equivalent modulo $p^2$ . Part (ii) now follows from Equation (2.5). $\Box $

Proof. As $f^{(0)}(0)=0$ , the claim holds for $k=1$ . Say $k\ge 2$ . By induction on k, we have for some $g\in R[T]$ . So

and the lemma follows from $f^{(k-1)}(0) = f^{(k-2)}(0)^p+c$ .

Proof. If $k=1$ , then $0=f^{(k-1)}(0)=f^{(l-1)}(0)$ . In this case, Lemma 2.3 implies with $g = T+c$ . Observe that $f^{(k)} = f = g(T^p) = g(T^{p^k})$ .

If $k\ge 2$ , Lemma 2.3 implies and , where $g = (T+f^{(k-1)}(0))^p+c$ .

We can summarize both cases by and ; we sometimes drop the reference to base ring in below.

Let $j\in \{0,\ldots ,D\}$ ; the polynomial $e_j(g(X_1),\ldots ,g(X_D))$ is symmetric. So

$$ \begin{align*} e_j(g(X_1),\ldots,g(X_D))= {P_{j}}(e_1,\ldots,e_D) \end{align*} $$

for some ${P_{j}} \in R[X_1,\ldots ,X_D]$ . Next, we replace $X_i$ by $X_i^{p^{k-\delta }}$ to get

$$ \begin{align*} e_j(g(X_1^{p^{k-\delta}}),\ldots,g(X_D^{p^{k-\delta}}))= P_{j}(e_1(X_1^{p^{k-\delta}},\ldots,X_D^{p^{k-\delta}}),\ldots,e_D(X_1^{p^{k-\delta}},\ldots,X_D^{p^{k-\delta}})). \end{align*} $$

Hence,

(2.8)

The same argument for l yields

(2.9)

We factor $A = (X-x_1)\cdots (X-x_D)$ with $x_1,\ldots ,x_D$ inside a fixed splitting field of A. We specialize $(x_1,\ldots ,x_D)$ in Equation (2.8) to get

$$ \begin{align*} e_j\left(f^{(k)}(x_1),\ldots,f^{(k)}(x_D)\right)\in P_{j}\left(e_1(x_1^{p^{k-\delta}},\ldots,x_D^{p^{k-\delta}}),\ldots,e_D(x_1^{p^{k-\delta}},\ldots,x_D^{p^{k-\delta}})\right)+p^2R[x_1,\ldots,x_D]; \end{align*} $$

both elements lie in R by Lemma 2.2(i). As any element in $R[x_1,\ldots ,x_D]$ is integral over R and since R is integrally closed, we conclude

The same statement holds with k replaced by l by using Equation (2.9).

By hypothesis (2.7) and by Lemma 2.2(ii) applied to $k-\delta ,l-\delta $ , we have

and hence

This holds for all j and so , as desired.

Proof. Lemma 2.4 implies $A_k A_l^{p-1} = A_k^p + p^2 B$ for some $ B\in R[X]$ . As $A_k$ and $A_l$ are both monic of degree $D = \deg A$ , leading terms cancel and we find $\deg B \le pD-1$ . Then,

$$ \begin{align*} A_k A_l^{p-1} = \left(1+p^2C\right)A_k^p \end{align*} $$

with $C={B}/{A_k^p}$ . Since $A_k$ is monic, we have $C\in R((1/X))$ . So $\mathrm {ord}_\infty (C) =-\deg B + p D \ge 1$ . Therefore, C is a formal power series in $1/X$ and even lies in the ideal $(1/X)R[[1/X]]$ .

Recall that $\Phi _p(p^2 U) \in \mathbb {Z}[[U]]$ by Equation (2.11). Therefore, $\Phi _p(p^2C ) \in R[[1/X]]$ and

$$ \begin{align*} A_kA_l^{p-1}=\left(\Phi_p(p^2C) A_k\right)^p \end{align*} $$

by the functional equation $\Phi _p^p=1+U$ ; this is an identity in $R((1/X))$ . We conclude the proof by dividing by $A_k^p$ .

Proof. Let us denote the restriction $f|_{f^{-1}(\mathbb {C}\smallsetminus B)} \colon f^{-1}(\mathbb {C}\smallsetminus B)=\mathbb {C} \smallsetminus f^{-1}(B)\rightarrow \mathbb {C}\smallsetminus B$ by $\psi $ . Then $\psi $ is a surjective continuous map. As a nonconstant polynomial induces a proper map, we see that $\psi $ is proper and thus closed. By hypothesis, $f^{-1}(\mathbb {C}\smallsetminus B)$ does not contain any critical point of f. So $\psi $ is a local homeomorphism and thus open. As $\psi $ is closed and open, the image of a connected component of $\mathbb {C}\smallsetminus f^{-1}(B)$ under $\psi $ is all of $\mathbb {C}\smallsetminus B$ . As $\psi $ is the restriction of a polynomial, we conclude that all connected components of $\mathbb {C}\smallsetminus f^{-1}(B)$ are unbounded. The preimage $f^{-1}(B)$ is compact, and so $\mathbb {C} \smallsetminus f^{-1}(B)$ has a only one unbounded connected component. Therefore, $\mathbb {C}\smallsetminus f^{-1}(B)$ is connected and so is $\widehat {\mathbb {C}}\smallsetminus f^{-1}(B)$ .

We fix a base point $z_0\in \mathbb {C}\smallsetminus f^{-1}(B)$ . The proper, surjective, local homeomorphism $\psi $ is a topological covering by Lemma 2 [Reference HoHo75]. Then $\pi _1(\mathbb {C} \smallsetminus f^{-1}(B),z_0)$ is isomorphic to a subgroup of $\pi _1(\mathbb {C} \smallsetminus B,p(z_0))$ of finite index and thus cyclic.

Let $U_\infty $ be the complement in $\mathbb {C}$ of a closed disk centered at $0$ containing the compact set $f^{-1}(B)$ . We suppose $z_0\in U_\infty $ . Adding the point at infinity gives us a simply connected domain $U_\infty \cup \{\infty \}\subseteq \widehat {\mathbb {C}}$ and we have $\widehat {\mathbb {C}} \smallsetminus f^{-1}(B) = (U_\infty \cup \{\infty \})\cup (\mathbb {C}\smallsetminus f^{-1}(B))$ . The theorem of Seifert and van Kampen tells us that the inclusion induces a short exact sequence

(3.1) $$ \begin{align} 1\rightarrow \pi_1(U_\infty,z_0)\rightarrow \pi_1(\mathbb{C}\smallsetminus f^{-1}(B),z_0) \rightarrow \pi_1(\widehat{\mathbb{C}}\smallsetminus f^{-1}(B),z_0)\rightarrow 1. \end{align} $$

We may assume $B\not =\emptyset $ . Consider the loop $t\mapsto z_0 e^{2\pi \sqrt {-1} t}$ , its image lies in $U_\infty $ . Fix any $w\in f^{-1}(B)$ . It is well-known that such loop represents a generator of $\pi _1(\mathbb {C} \smallsetminus \{w\},z_0)\cong \mathbb {Z}$ and that the natural homomorphism $\pi _1(U_\infty ,z_0)\rightarrow \pi _1(\mathbb {C}\smallsetminus f^{-1}(B),z_0) \rightarrow \pi _1(\mathbb {C}\smallsetminus \{w\},z_0)$ is an isomorphism. So also the middle group is isomorphic to $\mathbb {Z}$ and the first arrow is a group isomorphism. Therefore, $\widehat {\mathbb {C}}\smallsetminus f^{-1}(B)$ is simply connected by Equation (3.1).

Proof. Suppose $z\in \mathbb {C}\smallsetminus H_f^{(n)}$ is a critical point of $g_n$ . Taking the derivative of Equation (3.2) gives $d^n g_n(z)^{d^n-1}g_n'(z) = ({f^{(n)}})'(z)$ and hence $(f^{(n)})'(z)=0$ . The chain rule yields $f'(f^{(n-1)}(z))\cdots f'(z)=0$ , so $f'$ vanishes at $f^{(j)}(z)$ for some $j\in \{0,\ldots ,n-1\}$ . Hence, $f^{(n)}(z) \in \mathrm {PCO}_{n}^{+}(f)$ . But $\mathrm {PCO}_n^{+}(f) \subseteq I_0$ by our standing hypothesis and so $z\in H_f^{(n)}$ . This is a contradiction. We conclude that $g_n$ has no critical points on its domain $\mathbb {C}\smallsetminus H_f^{(n)}$ and that it is a local homeomorphism.

Let us verify that the map $\mathbb {C}\smallsetminus H_f^{(n)}\rightarrow \mathbb {C}\smallsetminus I_n$ is proper. Indeed, let $\mathcal {K}$ be compact with $\mathcal {K}\subseteq \mathbb {C}\smallsetminus I_n$ . We need to verify that $g_n^{-1}(\mathcal {K})$ is compact. To this end, let $(x_k)_{k\in \mathbb {N}}$ be a sequence in $g_n^{-1}(\mathcal {K})$ . If the sequence is unbounded, then so is $(f^{(n)}(x_k))_{k\in \mathbb {N}}$ . This is impossible as $f^{(n)}(x_k) = g_n(x_k)^{d^n}$ lies in the image of $\mathcal {K}$ in $\mathbb {C}$ under the $d^n$ -th power map, itself a bounded set. So we may replace $(x_k)_{k\in \mathbb {N}}$ by a convergent subsequence with limit $x\in \mathbb {C}$ . To verify our claim, it suffices to check $x\in g_n^{-1}(\mathcal {K})$ . If $x\in \mathbb {C}\smallsetminus H_f^{(n)}$ , then $g_n(x)=\lim _{k\rightarrow \infty } g_n(x_k) \in \mathcal {K}$ and we are done. Let us assume $x\in H_f^{(n)}$ . As $\mathcal {K}$ is compact, we may pass to a subsequence for which $(g_n(x_k))_{k\in \mathbb {N}}$ converges to $y\in \mathcal {K}$ . Now, $y^{d^n}=\lim _{k\rightarrow \infty } f^{(n)}(x_k) = f^{(n)}(x)\in I_0$ using that $f^{(n)}$ is a polynomial. Hence, $y\in I_n$ , but this contradicts $\mathcal {K}\subseteq \mathbb {C}\smallsetminus I_n$ .

The continuous map $\mathbb {C}\smallsetminus H_f^{(n)}\rightarrow \mathbb {C}\smallsetminus I_n$ is a proper local homeomorphism. So it is open and closed, hence surjective as both $\mathbb {C}\smallsetminus H_f^{(n)}$ and $\mathbb {C}\smallsetminus I_n$ are connected. We find that $\mathbb {C}\smallsetminus H_f^{(n)}\rightarrow \mathbb {C}\smallsetminus I_n$ is a topological covering using again Lemma 2 [Reference HoHo75].

Recall that $I_0$ contains all critical values of $z\mapsto z^{d^n}$ and $z\mapsto f^{(n)}(z)$ . The fiber of either map above a point of $\mathbb {C}\smallsetminus I_0$ contains $d^n = \deg f^{(n)}$ elements. So Equation (3.2) implies that $g_n$ is injective.

Proof of Proposition 3.2.

If $\alpha _i\in \mathbb {C}\smallsetminus H_f^{(n)}$ , then the value $g_n(\alpha _i)$ is well-defined. We define a further hedgehog

(3.4) $$ \begin{align} I^{\prime}_n = I_n \cup \mathcal{H}\bigl( g_n(\alpha_i)\bigr)_{\substack{1\le i\le m \\ \alpha_i\not\in H_f^{(n)}}} \end{align} $$

with at most $qd^n+m$ quills; see Equation (3.3).

The hedgehog complement $\widehat {\mathbb {C}}\smallsetminus I^{\prime }_n$ is connected and simply connected, and therefore so is its homeomorphic image

(3.5) $$ \begin{align} U = h_n(\widehat{\mathbb{C}}\smallsetminus I^{\prime}_n). \end{align} $$

The image U lies open in $\widehat {\mathbb {C}}\smallsetminus H_f^{(n)}$ and thus in $\widehat {\mathbb {C}}$ . Because $\infty \in U$ , we find that $\widehat {\mathbb {C}}\smallsetminus U=\mathbb {C}\smallsetminus U$ is compact; this yields part of the claim of (i).

Recall $0 \in H_f^{(n)}$ , so $0\not \in U$ . Suppose $\alpha _i\in U$ for some i. So $\alpha _i\not \in H_f^{(n)}$ and thus $g_n(\alpha _i) \in I^{\prime }_n$ by construction. But $\alpha _i$ also lies in the image of $h_n$ , that is, $\alpha _i=h_n(\beta )$ for some $\beta \in \mathbb {C}\smallsetminus I^{\prime }_n$ . We deduce $g_n(\alpha _i) = g_n(h_n(\beta ))=\beta $ , which is a contradiction. So no $\alpha _i$ lies in U. This implies the rest of (i).

We apply Theorem 5.2.3 [Reference RansfordRan95] with $K_1 =I^{\prime }_n,K_2=\widehat {\mathbb {C}}\smallsetminus U,D_1=\widehat {\mathbb {C}}\smallsetminus I^{\prime }_n,D_2= U,$ and f as ${h_n}|_{D_1}\colon {D_1}\rightarrow {D_2}$ extended as above to send $\infty \mapsto \infty $ . Hence,

$$ \begin{align*} \mathrm{cap}(\widehat{\mathbb{C}} \smallsetminus U) \le \mathrm{cap}(I^{\prime}_n). \end{align*} $$

Recall that $I^{\prime }_n$ is a hedgehog with at most $qd^n+m$ quills and lengths given in Equations (3.3) and (3.4). Dubinin’s theorem [Reference DubininDub84] implies

$$ \begin{align*} \mathrm{cap}(I^{\prime}_n) \le 4^{-1/(qd^n+m)} \max\{|\gamma_1|^{1/d^n},\ldots,|\gamma_q|^{1/d^n}, |g_n(\alpha_i)| : \alpha_i\in\mathbb{C} \smallsetminus H_f^{(n)} \}. \end{align*} $$

Combining these two estimates and using $|g_n(\alpha _i)| = |f^{(n)}(\alpha _i)|^{1/d^{n}}$ if $\alpha _i\in \mathbb {C}\smallsetminus H_f^{(n)}$ yields

$$ \begin{align*} \mathrm{cap}(\widehat{\mathbb{C}}\smallsetminus U) \le 4^{-1/(qd^n+m)} \max\{|\gamma_1|,\ldots,|\gamma_q|, |f^{(n)}(\alpha_1)| ,\ldots,|f^{(n)}(\alpha_m)|\}^{1/d^n}. \end{align*} $$

This completes the proof of (ii).

Proof. By hypothesis, there are $\gamma _1,\ldots ,\gamma _q\in \mathbb {C}^\times $ with $\mathrm {PCO}_{n}^{+}(f)\subseteq \mathcal {H}(\gamma _1,\ldots ,\gamma _q)$ . By shortening and omitting quills, we may assume that each $\gamma _i$ lies in $\mathrm {PCO}_{n}^{+}(f)$ . In other words, for each i there is $n_i\in \{1,\ldots ,n\}$ with $\gamma _i = f^{(n_i)}(0)$ . By the definition (1.2) of $\mathsf {C}(f)$ , we have $|\gamma _i|\le \mathsf {C}(f) e^{\lambda _f(f^{(n_i)}(0))} \le \mathsf {C}(f) e^{d^{n_i} \lambda _f(0)} \le \mathsf {C}(f) e^{d^n \lambda _f(0)}$ . The same definition also gives $|f^{(n)}(\alpha _i)|\le \mathsf {C}(f) e^{d^n\lambda _f(\alpha _i)}$ for all i. The proof follows from Proposition 3.2.

Proof. Both statements are invariant under rotating c. So we may assume $c\in [0,\infty )$ .

The polynomial $T^d-T-c$ is a convex function on $(0,\infty )$ , and it is negative for small positive arguments. So it has exact one root r in $(0,\infty )$ . We must have $r\ge 1$ as $1^d-1-c\le 0$ .

Let $z\in \mathbb {C}$ . We prove the lemma by showing

(3.7) $$ \begin{align} \frac{\max\{1,|z^d+c|\}}{\max\{1,|z|^d\}} \ge \frac{1}{r^{d-1}} \end{align} $$

for all $z\in \mathbb {C}$ with $|z|\ge r$ . Note that $|z|\ge r\ge 1$ gives $|z|^d-c\ge |z|$ . The triangle inequality implies $|z^d+c|\ge \bigl ||z^d|-c\bigr |\ge |z|\ge 1$ . So the left-hand side of Equation (3.7) equals $|z^d+c|/|z^d|=|1+c/z^d|>0$ . By the minimum modulus principle, we have $|1+c/z^d|\ge 1$ for all z with $|z|\ge r$ , or $|1+c/z^d|$ attains a minimum $<1$ on the boundary $|z|=r$ . In the first case, Equation (3.7) holds true. Finally, if $|z|=r$ and $|1+c/z^d|$ is minimal, then necessarily $z^d = -r^d$ by elementary geometry. In this case, $|1+c/z^d| = |1-c/r^d|=|(r^d-c)/r^d| = r^{1-d}$ and we recover Equation (3.7).

Proof. Set $f=T^d+c$ . Observe that by Lemma 3.5 we have $|f(z)|\ge r$ if $|z|\ge r$ . By a simple induction, we find $\log ^{+} |f^{(n)}(z)| \ge -(d-1)\log r + d\log ^{+} |f^{(n-1)}(z)|$ for all $n\ge 1$ if $|z|\ge r$ . Considering the telescoping sum, we get

$$ \begin{align*} \frac{\log^{+}|f^{(n)}(z)|}{d^n} - \log^{+}|z| &= \sum_{j=1}^n \frac{\log^{+}|f^{(j)}(z)|}{d^j} - \frac{d\log^{+} |f^{(j-1)}(z)|}{d^{j}} \\ & \ge \sum_{j=1}^n \frac{-(d-1)\log r}{d^j} \ge \sum_{j=1}^\infty \frac{-(d-1)\log r}{d^j} = -\log r. \end{align*} $$

On taking the limit $n\rightarrow \infty $ , the left-hand side becomes $\lambda _f(z)-\log ^{+}|z|$ by Equation (1.1). So we obtain $\log ^{+}|z| \le \log r + \lambda _f(z)$ if $|z|\ge r$ . Since $\lambda _f$ is nonnegative, this bound also holds if $|z|<r$ . We obtain $\log ^{+}|z|-\lambda _f(z)\le \log r$ for all $z\in \mathbb {C}$ . The lemma follows from the definition (1.2).

Proof. By hypothesis, we have $\lambda _f(c)=\lambda _f(f(0))=0$ . Thus, $|c| \le \mathsf {C}(f)\le r$ by Lemma 3.6, where r is the unique positive real number with $r^d=r+|c|$ . As $T^d-T-|c|$ has no roots on $(0,r)$ by Lemma 3.5 we find that it takes negative values on $(0,r)$ . So $|c|^d \le |c|+|c|$ and therefore $|c|\le t$ , where $t=2^{1/(d-1)}$ . We note $t^d-t-|c|\ge t^d-t-t=0$ and hence $t\ge r$ . We conclude $\mathsf {C}(f)\le r\le t = 2^{1/(d-1)}$ . Part (i) follows.

For the proof of (ii), let us assume $\mathsf {C}(f) = 2^{1/(d-1)}$ and retain the notation from the proof of part (i). Then $r=t$ and hence $2^{1/(d-1)}$ is a root of $T^d-T-|c|$ . This yields $|c|=2^{1/(d-1)}$ after a short calculation.

Suppose $d=2$ . Then it is well-known that the Mandelbrot set meets the circle of radius $2$ with center $0$ in the single point $-2$ . This implies (ii) for $d=2$ . Here is a direct verification that extends to all $d\ge 2$ . It involves $c^d+c$ , the second iterate of $0$ under f. Indeed, its orbit under f is also bounded and so $|c||c^{d-1}+1|=|c^d+c|\le \mathsf {C}(f)=2^{1/(d-1)}$ . But $|c|=2^{1/(d-1)}$ and thus $|c^{d-1}+1|\le 1$ . The circle around $0$ of radius $2$ meets the closed disk around $-1$ of radius $1$ in the single point $-2$ . We conclude $c^{d-1}=-2$ , and this implies (ii).

In (iii), we have that d is odd. We will prove $c^{d-1}\not =-2$ by contradiction; (iii) then follows from (ii). Indeed, if $c^{d-1}=-2$ , then $f^{(2)}(0)=c^d+c=-c$ . So $f^{(3)}(0) = -c^d+c$ has absolute value $|c||c^{d-1}-1| = 3|c|> 2^{1/(d-1)}\ge \mathsf {C}(f)$ . But then $\lambda _f(f^{3}(0))>0$ and this contradicts the hypothesis that the f-orbit of $0$ is bounded.

Proof of Lemma 1.2.

Let $\sigma _0$ be a complex embedding of K as in (i) or (ii). By hypothesis, $\sigma _0(f)\not =T^2-2$ and $d=p$ is a prime. Moreover, $\mathrm {PCO}^{+}(\sigma _0(f))$ is bounded in both (i) and (ii). So Lemma 3.7 yields $\mathsf {C}(\sigma _0(f))<2^{1/(p-1)}\le 2$ ; if $p=2$ we use part (ii), and if $p\ge 3$ we use (iii).

For case (i), the embedding $\sigma _0$ is real. The set $\mathrm {PCO}^{+}(\sigma _0(f))=\{\sigma _0(f)(0),\sigma _0(f)^{(2)}(0),\ldots \}$ lies in $\mathbb {R}$ . So it is contained in the union $[-\alpha ,0]\cup [0,\beta ]$ of $q=2$ quills. Thus, $q\log \mathsf {C}(\sigma _0(f))<\log 4$ and Equation (1.4) is satisfied.

In case (ii), the set $\mathrm {PCO}^{+}(\sigma _0(f))$ is finite and can be covered by $q\le 2p-2$ quills. Now, we find $q\log \mathsf {C}(\sigma _0(f)) < 2(p-1) \log 2^{1/(p-1)} = \log 4$ and Equation (1.4) is again satisfied.

Proof. We shall consider $P=T^p+X$ as polynomial in $\mathbb {Z}[T,X]$ .

We set $P^{(0)}=T,P^{(1)}=P=T^p+X, P^{(2)} = P(P(T,X),X)=(T^p+X)^p+X, P^{(3)}=P(P^{(2)}(T,X),X)=((T^p+X)^p+X)^p+X$ , etc. A simple induction shows that $P^{(k)}$ is monic of degree $p^{k-1}$ in X for all $k\in \mathbb {N}$ .

Suppose $c\in \mathbb {C}$ such that $0$ is f-preperiodic, where $f=T^p+c$ . So there exist integers $k,l$ with $0\le k<l$ such that $P^{(k)}(0,c)=P^{(l)}(0,c)$ . This produces a monic polynomial in integral coefficients of degree $p^{l-1}$ that vanishes at c. Part (i) follows.

For part (ii), we observe that $P^{(k)}=(P^{(k-1)})^p+X$ for $k\in \mathbb {N}$ . The derivative by X satisfies $\frac {\partial }{\partial X} P^{(k)} = pP^{(k-1)} \frac {\partial }{\partial X}P^{(k-1)}+1 \in 1 + p\mathbb {Z}[T,X]$ . We specialize T to $0$ and get $\frac {\partial }{\partial X} P^{(k)}(0,X) \in 1 + p\mathbb {Z}[X]$ .

The determinant of the Sylvester matrix of the pair $P^{(k)}(0,X),\frac {\partial }{\partial X} P^{(k)}(0,X)$ is up to sign the discriminant of $P^{(k)}(0,X)$ . The reduction modulo p of this matrix has upper triangular form with diagonal entries . In particular, the discriminant is not divisible by p.

If $0$ is f-periodic with $f=T^p+c$ , then $f^{(k)}(0)=0$ for some $k\in \mathbb {N}$ . The $\mathbb {Q}$ -minimal polynomial of c has integral coefficients by (i), and it divides $P^{(k)}(0,X)$ . So its discriminant is not divisible by p either. This implies (ii).

Proof. This is a special case of Théorème 5.4.6 [Reference AmiceAmi75]. Note that our power series has integral coefficients, so we can take $P_1(K)=\emptyset $ in the reference. Let $\overline \sigma $ denote $\sigma $ composed with complex conjugation. Then we may take $U_{\overline \sigma }$ to be the complex conjugate of $U_{\sigma }$ .

Proof. We begin by observing $c\in \mathcal {O}_K$ by Lemma 4.1(i). Let $A\in K[X]$ be the K-minimal polynomial of x. Then $A\in \mathcal {O}_K[X]$ has degree $D=[K(x):K]$

We apply Proposition 2.5 to the ring $R=\mathcal {O}_K$ . There exists $\phi \in \mathcal {O}_K[[1/X]]$ with

(4.2) $$ \begin{align} \phi^p = (A_l/A_k)^{p-1} \end{align} $$

and

(4.3) $$ \begin{align} \phi = \Phi_p(p^2 B/A_k^p)\quad\text{where}\quad\deg B \le \deg A_k^p - 1 =pD-1; \end{align} $$

we recall Equation (2.10). Then $\sigma (\phi )^p = \sigma (A_l/A_k)^{p-1}$ for all $\sigma \in \mathrm {Hom}(K,\mathbb {C})$ .

We split $\sigma (A) = (X-x_{\sigma ,1})\cdots (X-x_{\sigma ,D})$ , where $x_{\sigma ,1},\ldots ,x_{\sigma ,D}\in \mathbb {C}$ are conjugates of $\sigma (x)$ over $\sigma (K)\subseteq \mathbb {C}$ . Then

$$ \begin{align*} \sigma(\phi)^p =\prod_{j=1}^D \left(\frac{X-\sigma(f)^{(l)}(x_{\sigma,j})}{X-\sigma(f)^{(k)}(x_{\sigma,j})}\right)^{p-1}. \end{align*} $$

Recall the degree condition in Equation (4.3). The formal power series $\sigma (\phi ) \in \mathbb {C}[[1/X]]$ converges for all $z\in \mathbb {C}$ for which $|z|$ is sufficiently large. For these z, we have

(4.4) $$ \begin{align} \sigma(\phi)(z)^p =\prod_{j=1}^D \left(\frac{1-\sigma(f)^{(l)}(x_{\sigma,j})/z}{1-\sigma(f)^{(k)}(x_{\sigma,j})/z}\right)^{p-1}. \end{align} $$

We apply Proposition 3.4 to $\sigma (f) = T^p+\sigma (c)$ and $n_\sigma $ . By hypothesis, $\mathrm {PCO}_{n_\sigma }^{+}(\sigma (f))$ is contained in a hedgehog with at most $q_\sigma \ge 0$ quills. We take $m=2D$ and set

(4.5) $$ \begin{align} \alpha_1 = \sigma(f)^{(k)}(x_{\sigma,1}),\ldots, \alpha_D = \sigma(f)^{(k)}(x_{\sigma,D}), \alpha_{D+1} = \sigma(f)^{(l)}(x_{\sigma,1}),\ldots, \alpha_{2D} = \sigma(f)^{(l)}(x_{\sigma,D}). \end{align} $$

We obtain a simply connected domain $U_\sigma \subseteq \widehat {\mathbb {C}}$ with the two stated properties.

The right-hand side of Equation (4.4) is well-defined and nonzero for all $z\in U_\sigma \smallsetminus \{\infty \}$ ; indeed, none among $0$ and the Equation (4.5) lie in $U_\sigma $ by Proposition 3.4. So the right-hand side extends to a holomorphic function on $U_\sigma $ that never vanishes; the extension maps $\infty $ to $1$ . By the monodromy theorem from complex analysis and since $U_\sigma $ is simply connected, we conclude that $z\mapsto \sigma (\phi )(z)$ , defined a priori only if $|z|$ is large, extends to a holomorphic map $\sigma (\phi )\colon U_\sigma \rightarrow \mathbb {C}$ .

We split up into two cases.

Case 1. First, suppose that $\phi $ is not a rational function.

Theorem 4.2 implies

(4.6) $$ \begin{align} 0 \le \sum_{\sigma\in\mathrm{Hom}(K,\mathbb{C})}\log\mathrm{cap}(\widehat{\mathbb{C}}\smallsetminus U_\sigma). \end{align} $$

Recall that $\lambda _{\sigma (f)}(0)=0$ as f is post-critically finite by hypothesis. This contribution can be omitted from the capacity bound from Proposition 3.4 since the local canonical height is nonnegative. So $k\le l$ implies

(4.7) $$ \begin{align} \sum_{\sigma\in\mathrm{Hom}(K,\mathbb{C})}\log \mathrm{cap}(\widehat{\mathbb{C}}\smallsetminus U_\sigma) \le \sum_{\sigma\in \mathrm{Hom}(K,\mathbb{C})} -\frac{\log 4}{q_\sigma p^{n_\sigma}+2D}+\frac{\log \mathsf{C}(\sigma(f))}{p^{n_\sigma}} +p^l \max_{1\le i\le D} \lambda_{\sigma(f)}(x_{\sigma,i}). \end{align} $$

We compare Equations (4.6) and (4.7) and rearrange to find

$$ \begin{align*} \sum_{\sigma\in \mathrm{Hom}(K,\mathbb{C})} \max_{1\le i\le D} \lambda_{\sigma(f)}(x_{\sigma,i}) \ge \frac{1}{p^{l}}\sum_{\sigma\in\mathrm{Hom}(K,\mathbb{C})} \left(\frac{\log 4}{q_\sigma p^{n_\sigma}+2D} - \frac{\log \mathsf{C}(\sigma(f))}{p^{n_\sigma}}\right). \end{align*} $$

The left-hand side is at most $[K:\mathbb {Q}]\lambda ^{\mathrm {max}}_f(x)$ . So conclusion (A) holds.

Case 2. Second, suppose that $\phi $ is a rational function.

In particular, $\phi $ is meromorphic on $\mathbb {C}$ . It follows from Equation (4.2) that any root of $A_l/A_k$ has vanishing order divisible by p. Let F be a splitting field of A. Then $\{f^{(k)}(\tau (x)) : \tau \in \mathrm {Gal}(F/K) \}$ are the roots of $A_k$ and $\{f^{(l)}(\tau (x)) : \tau \in \mathrm {Gal}(F/K) \}$ are the roots of $A_l$ .

We split up into two subcases.

Subcase 2a. Suppose $\#\{f^{(k)}(\tau (x)) : \tau \in \mathrm {Gal}(F/K) \} \le D/p$ . This means $[K(f^{(k)}(x)):K]\le D/p$ , and we are in conclusion (B) because $D=[K(x):K]$ .

Subcase 2b. Suppose $\#\{f^{(k)}(\tau (x)) : \tau \in \mathrm {Gal}(F/K) \}> D/p$ . Recall $D=\deg A_k$ . Then there exists $\tau $ such that the vanishing order of $A_k$ at $f^{(k)}(\tau (x))$ is positive and strictly less than p. But then it must also be a zero of $A_l$ . After replacing $\tau $ , we may assume $f^{(k)}(\tau (x))=f^{(l)}(x)$ . Recall $k<l$ . A simple induction shows $\tau ^e (f^{(k)}(x)) = f^{((l-k)e+k)}(x)$ for all $e\in \mathbb {N}$ . Indeed, for $e\ge 2$ we have

(4.8) $$ \begin{align} \tau^e(f^{(k)}(x))= \tau( \tau^{e-1}(f^{(k)}(x))) =\tau(f^{((l-k)(e-1)+k)}(x)) =f^{((l-k)(e-1)+k)}(\tau(x))=f^{((l-k)e+k)}(x) \end{align} $$

as $f\in K[T]$ .

Take e to be $\#\mathrm {Gal}(F/K)$ . So $\tau ^e(f^{(k)}(x)) = f^{(k)}(x)$ and thus

$$ \begin{align*} f^{(k)}(x) = f^{((l-k)e+k)}(x). \end{align*} $$

In particular, x is f-preperiodic and $f^{(k)}(x)$ is f-periodic. In other words, $\mathrm {preper}(x)\le k$ . So the first two statements in conclusion (C) hold.

By the pigeonhole principle, there are integers e and $e'$ with $0\le e<e'\le [K(x):K]$ with $\tau ^{e}(f^{(k)}(x)) = \tau ^{e'}(f^{(k)}(x))$ . So $f^{((l-k)e+k)}(x)=f^{((l-k)e'+k)}(x)$ by Equation (4.8). Hence, x has minimal period at most $(l-k)(e'-e) \le (l-k)[K(x):K]$ . Therefore, the final statement in conclusion (C) holds.

Proof. We remark that $c\in \mathcal {O}_K$ by Lemma 4.1(i). Let $x\in \overline K$ be f-wandering.

Our proof of (i) is by induction on $[K(x):K]$ . Here, x is an algebraic integer. We will apply Proposition 4.3. Observe that an f-wandering point cannot lead to conclusion (C).

Let $\sigma _0\in \mathrm {Hom}(K,\mathbb {C})$ and $q\ge 0$ be as in the quill hypothesis. We introduce two parameters, $\epsilon _0,\epsilon \in (0,1)$ both are sufficiently small and fixed in terms of f and the other given data such as q and $\sigma _0(f)$ , but independent of x. We will fix $\epsilon $ in function of $\epsilon _0$ .

Let $n_{\sigma _0}\ge 1$ be the unique integer with $p^{n_{\sigma _0}} \ge [K(x):K]/\epsilon _0> p^{n_{\sigma _0}-1}$ . For all $\sigma \in \mathrm {Hom}(K,\mathbb {C})$ with $\sigma \not =\sigma _0$ , we fix the unique $n_{\sigma }\in \mathbb {N}$ with $p^{n_{\sigma }} \ge [K(x):K]/\epsilon> p^{n_{\sigma }-1}$ .

For any $\sigma $ , we set $q_\sigma =\#\mathrm {PCO}^{+}(\sigma (f)) < \infty $ . Then $\mathrm {PCO}^{+}(\sigma (f))$ is contained in a hedgehog with at most $q_\sigma $ quills and therefore so is $\mathrm {PCO}^{+}_{n_{\sigma }}(\sigma (f)) \subseteq \mathrm {PCO}^{+}(\sigma (f))$ .

We have

(4.9) $$ \begin{align} Y&=\sum_{\sigma\in\mathrm{Hom}(K,\mathbb{C})} \frac{1}{p^{n_\sigma}}\left(\frac{\log 4}{q_\sigma + 2[K(x):K]p^{-n_\sigma}} -\log \mathsf{C}(\sigma(f))\right) \\ \nonumber &\ge \frac{1}{p^{n_{\sigma_0}}}\left( \frac{\log 4}{q + 2[K(x):K]p^{-n_{\sigma_0}}} -{\log \mathsf{C}(\sigma_0(f))}\right) - \sum_{\sigma\not=\sigma_0} \frac{\log \mathsf{C}(\sigma(f))}{p^{n_\sigma}} \\ \nonumber &\ge \frac{1}{p^{n_{\sigma_0}}}\left( \frac{\log 4}{q + 2\epsilon_0} -{\log \mathsf{C}(\sigma_0(f))}\right) - \frac{\epsilon}{[K(x):K]}\sum_{\sigma\not=\sigma_0} \log \mathsf{C}(\sigma(f)). \end{align} $$

For $\epsilon _0$ small enough and fixed in function of f, the quill hypothesis (1.4) implies $(\log 4)/(q+2\epsilon _0) - \log \mathsf {C}(\sigma _0(f)) \ge \kappa _1$ where $\kappa _1=\kappa _1(f)> 0$ depends only on f. So

$$ \begin{align*} Y \ge \frac{\kappa_1}{p^{n_{\sigma_0}}} - \frac{\epsilon}{[K(x):K]}\sum_{\sigma\not=\sigma_0} \log \mathsf{C}(\sigma(f)) \ge \frac{1}{ [K(x):K]}\left(\frac{\kappa_1 \epsilon_0}{p} - {\epsilon}\sum_{\sigma\not=\sigma_0} \log \mathsf{C}(\sigma(f))\right). \end{align*} $$

Now, we fix $\epsilon $ small in terms of $\epsilon _0$ and f to achieve $\kappa _1\epsilon _0/(2p) \ge \epsilon \sum _{\sigma \not =\sigma _0} \log \mathsf {C}(\sigma (f))$ . So

(4.10) $$ \begin{align} Y\ge \frac{\kappa_2}{[K(x):K]}\quad\text{with}\quad \kappa_2 = \frac{\kappa_1\epsilon_0}{2p}. \end{align} $$

Suppose we are in conclusion (A) of Proposition 4.3. Then $\lambda _f^{\mathrm {max}}(x) \ge Y [K:\mathbb {Q}]^{-1}p^{-l} \ge \kappa _2p^{-l}/[K(x):\mathbb {Q}]$ . This is stronger than the claim as we may assume $\kappa \le \kappa _2 [K:\mathbb {Q}]^{-1}p^{-l}$ .

Suppose we are in conclusion (B). Then we have $[K(f^{(k)}(x)):K] \le [K(x):K]/p<[K(x):K]$ . In particular, this rules out $x\in K$ , so the base case of the induction was handled by conclusion (A) above. Now, $\lambda _f^{\mathrm {max}}(f^{(k)}(x)) = p^k \lambda _f^{\mathrm {max}}(x)$ . Induction on $[K(x):K]$ yields $p^k \lambda _f^{\mathrm {max}}(x) \ge \kappa /[K(f^{(k)}(x)):K]^k \ge \kappa p^k/[K(x):K]^k$ . This completes the proof of (i) as conclusion (C) is impossible in the wandering case.

For (ii), we observe that $\hat h_f(x)$ is a normalized sum of local canonical heights as in Equation (1.3). Moreover, every local canonical height takes nonnegative values. In particular, $\hat h_f(x) \ge \lambda _{f}^{\mathrm {max}}(x)/[\mathbb {Q}(x):\mathbb {Q}]$ . If x is an algebraic integer, then part (i) implies the desired lower bound for $\hat h_f(x)$ after adjusting $\kappa $ . So assume that x is not an algebraic integer. Let F be a number field containing x. There is a non-Archimedean $v\in M_{\mathbb {Q}}$ , which we may identify with a prime number, and a field embedding $\sigma \in \mathrm {Hom}( F, \mathbb {C}_v)$ with $|\sigma (x)|_v>1$ . By anticipating ramification, we find $|\sigma (x)|_v \ge v^{1/[\mathbb {Q}(x):\mathbb {Q}]}$ . Thus, $|\sigma (x^p+c)|_v=|\sigma (x)|_v^p>1$ by the ultrametric triangle inequality and since $c\in \mathcal {O}_K$ . Furthermore, $|\sigma (f^{(n)}(x))|_v = |\sigma (x)|_v^{p^n}$ for all $n\in \mathbb {N}$ . Therefore, $\lambda _{\sigma (f),v}(\sigma (x)) = \log ^{+}|\sigma (x)|_v \ge (\log v)/[\mathbb {Q}(x):\mathbb {Q}]\ge (\log 2)/[\mathbb {Q}(x):\mathbb {Q}]$ . Again, we use that local canonical heights are nonnegative and conclude $\hat h_f(x) \ge (\log 2)/[\mathbb {Q}(x):\mathbb {Q}]^2\ge (\log 2)/[K(x):\mathbb {Q}]^2 = [K:\mathbb {Q}]^{-2}(\log 2)/[K(x):K]^2$ . Thetheorem follows as we may assume $\kappa \le [K:\mathbb {Q}]^{-2}\log 2$ .

Proof. By hypothesis, the f-orbit of $0$ is finite and so is $\mathrm {PCO}^{+}(f)$ . By Lemma 4.1(i), we see that c is an algebraic integer. So x, an f-preperiodic point, is also an algebraic integer. Moreover, $\lambda _f^{\mathrm {max}}(x)=0$ .

Let $\sigma _0\in \mathrm {Hom}(K,\mathbb {C})$ and $q\ge 0$ be as in the quill hypothesis. Let $\epsilon _0,\epsilon \in (0,1),n_{\sigma }\in \mathbb {N},$ and $q_{\sigma }\in \mathbb {N}$ , for $\sigma \in \mathrm {Hom}(K,\mathbb {C})$ , be as in the proof of Theorem 4.4. Recall that $\epsilon ,\epsilon _0$ depend on $f,q,$ and $\sigma _0(f)$ , but they are independent of x. Moreover, $\epsilon $ is small in terms of $\epsilon _0$ . Arguing as in the said proof up to Equation (4.10), we find $Y>0$ with Y as in Equation (4.9). So the right-hand side of Equation (4.1) is strictly positive. Therefore, we can rule out conclusion (A) of Proposition 4.3.

So we are either in conclusion (B) where $[K(f^{(k)}(x)):K]\le [K(x):K]/p$ or conclusion (C) where $\mathrm {preper}(x) \le k$ and $\mathrm {per}(x) \le (l-k)[K(x):K]$ .

Observe that we have $\mathrm {preper}(f(x)) = \max \{0,\mathrm {preper}(x)-1\}$ . Iterating gives $\mathrm {preper}(f^{(m)}(x)) = \max \{0,\mathrm {preper}(x)-m\}$ for all integers $m\ge 0$ .

We prove Equation (4.11) by induction on $\mathrm {preper}(x)$ . The claim is trivial if $\mathrm {preper}(x)\le k$ . So let us assume $\mathrm {preper}(x)>k$ . Then we are in conclusion (B) of Proposition 4.3. Hence, $[K(f^{(k)}(x)):K]\le [K(x):K]/p$ and $\mathrm {preper}(f^{(k)}(x)) = \mathrm {preper}(x)-k < \mathrm {preper}(x)$ . By induction,

$$ \begin{align*} [K(x):K]\ge p [K(f^{(k)}(x)):K] \ge p^{\mathrm{preper}(f^{(k)}(x))/k} =p^{\mathrm{preper}(x)/k-1}, \end{align*} $$

as desired.

The first inequality in Equation (4.12) is immediate. We prove the second inequality in the case where x is f-periodic first. Indeed, then $K(f(x))=K(x)$ and so we are again in conclusion (C). Hence, $[K(x):K]\ge \mathrm {per}(x)/(l-k)$ , as desired. If x is f-preperiodic, then Equation (4.12) is applicable to the f-periodic $f^{(\mathrm {preper}(x))}(x)$ . We find $[K(f^{(\mathrm {preper}(x))}(x)):K] \ge {\mathrm {per}(f^{(\mathrm {preper}(x))}(x))}/(l-k)$ . The theorem follows as $\mathrm {per}(f^{(m)}(x)) = \mathrm {per}(x)$ for all integers $m\ge 0$ .

Proof of Theorem 1.1.

We suppose $f^{(l-1)}(0)=0$ with $l\ge 2$ . We want to apply Theorem 4.4 to $\mathbb {Q}(c)$ . Indeed, it suffices to verify that hypothesis (i) holds true for $k=1$ after possibly adjusting l.

By Lemma 4.1, the parameter c is an algebraic integer and $K/\mathbb {Q}$ is unramified above p.

We write m for the least common multiple of $l-1$ and all residue degrees of all prime ideals of $\mathcal {O}_K$ containing p. So for all $a\in \mathcal {O}_K$ . Moreover, $f^{(l-1)}(0)=0$ implies $f^{(m)}(0)=0$ . Note that $[\mathbb {Q}(c,x):\mathbb {Q}(c)]\le [K(x):K][K:\mathbb {Q}]$ . Theorem 1.1 follows from Theorem 4.4 applied to $(k,l)=(1,m+1)$ ; note $\delta =0$ .

Proof of Corollary 1.3.

In cases (i) and (ii), the extension $K/\mathbb {Q}$ is unramified above p. As in the proof of Theorem 1.1, there exists $l\ge 2$ such that the pair $(1,l)$ satisfies the hypothesis of Theorem 4.4.

To prove (i), we assume that y is f-periodic. Let $x\in \overline K$ , and suppose $n\ge 0$ is minimal with $f^{(n)}(x)=y$ . We may assume $n\ge m=\mathrm {per}(y)$ . If $\mathrm {preper}(x) \le n-m$ , then $f^{(n-m)}(x)$ is f-periodic and one among $f^{(n-m)}(x),\ldots ,f^{(n-1)}(x)$ must equal y. But this contradicts the minimality of n and so $\mathrm {preper}(x) \ge n-m+1$ . Theorem 4.5 implies $[K(x):K]\ge p^{\mathrm {preper}(x)-1} \ge p^{n-m}$ . We conclude (i).

The proof of part (ii) is very similar. Let $n\in \mathbb {N}$ , and suppose $x\in \overline K$ satisfies $f^{(n)}(x)=y$ . We use $1\le \mathrm {preper}(y)=\mathrm {preper}(f^{(n)}(x)) = \max \{0,\mathrm {preper}(x)-n\}$ to infer $\mathrm {preper}(x)=n+\mathrm {preper}(y)\ge n+1$ . Theorem 4.5 implies $[K(x):K]\ge p^{\mathrm {preper}(x)-1} \ge p^n$ . But $[K(x):K]\le \deg (f^{(n)}-y)=p^n$ , so part (ii) follows.

For part (iii), we no longer restrict the ramification of $K/\mathbb {Q}$ . But we assume that y is f-wandering. Note that c is an algebraic integer by Lemma 4.1(i). Let again $f^{(n)}(x)=y$ .

We first suppose that y is an algebraic integer. Then x is also an algebraic integer. Then $\lambda _f^{\mathrm {max}}(x) \ge \kappa _1/[K(x):K]$ by Theorem 1.1(i). The functional equation of the local canonical height implies $\lambda _f^{\mathrm {max}}(x) = \lambda _f^{\mathrm {max}}(y)/p^n$ . We combine and rearrange to find $[K(x):K]\lambda _f^{\mathrm {max}}(y) \ge \kappa _1p^n$ . So $\lambda _f^{\mathrm {max}}(y)>0$ and $[K(x):K]\ge \kappa p^n$ with $\kappa = \kappa _1/\lambda _f^{\mathrm {max}}(y)$ .

Second, suppose that y is not an algebraic integer. Then there exists a prime number v and a field embedding $\sigma \in \mathrm {Hom}( K,\mathbb {C}_v)$ with $|\sigma (y)|_v>1$ . The ultrametric triangle inequality implies $|\sigma (x)|_v>1$ and $|\sigma (y)|_v=|\sigma (f^{(n)}(x))|_v = |\sigma (x)|_v^{p^n}$ . So the ramification index of $K(x)/K$ at some prime ideal above v grows like a positive multiple of $p^n$ . In particular, $[K(x):K]\ge \kappa p^n$ for all $n\in \mathbb {N}$ , where $\kappa>0$ is independent of n.

Proof of Corollary 1.4.

We will apply Proposition 4.3 to $p=2,K=\mathbb {Q},$ and $f=T^2-1$ .

Observe that $0=f^{(2)}(0)$ . As for all $a\in \mathbb {Z}$ , we see that $(k,l)=(1,3)$ satisfies hypothesis (i) of Proposition 4.3; here, $\delta =0$ . Note that $\mathrm {PCO}^{+}(f) =\{-1,0\}$ . The single quill $[-1,0]$ suffices to cover $\mathrm {PCO}_n^{+}(f)$ for all $n\in \mathbb {N}$ . Moreover, the root r from Lemma 3.5 applied to $d=2$ and $c=-1$ is the golden ratio $(1+\sqrt 5)/2$ . So $\mathsf {C}(T^2-1)\le (1+\sqrt 5)/2$ by Lemma 3.6.

Let x be as in (i). The proof of part (i) is by induction on $[\mathbb {Q}(x):\mathbb {Q}]$ . We fix $n\in \mathbb {N}$ to be minimal with $2^n \ge 3[\mathbb {Q}(x):\mathbb {Q}]$ , so $2^{n-1}< 3[\mathbb {Q}(x):\mathbb {Q}]$ .

Observe that the f-wandering point x is in conclusion (A) or (B) of Proposition 4.3. If $x\in \mathbb {Q}$ , then we must be in conclusion (A).

Conclusion (A) implies

$$ \begin{align*} \lambda_f^{\mathrm{max}}(x) &\ge \frac{1}{2^{3+n}} \left(\frac{\log 4}{1+2[\mathbb{Q}(x):\mathbb{Q}]2^{-n}} - \log \frac{\sqrt 5+1}{2}\right) \ge \frac{1}{2^{3+n}}\log\left(\frac{4^{3/5}\cdot 2}{\sqrt 5+1}\right) \\ &\ge \frac{1}{48}\log\left(\frac{4^{3/5}\cdot 2}{\sqrt 5+1}\right)\frac{1}{[\mathbb{Q}(x):\mathbb{Q}]}. \end{align*} $$

Part (i) follows in this case and in particular if $x\in \mathbb {Q}$ ; which is the base case.

In conclusion (B), we have $[\mathbb {Q}(f(x)):\mathbb {Q}]\le [\mathbb {Q}(x):\mathbb {Q}]/2$ . Part (i) follows by induction on $[\mathbb {Q}(x):\mathbb {Q}]$ and since $\lambda _f^{\mathrm {max}}(f(x))=2\lambda _f^{\mathrm {max}}(x)$ .

The estimate in part (ii) follows from part (i) in the integral case. In the nonintegral case, we use the same argument as in the proof of Theorem 4.4(ii) and conclude using $\log 2> \log (4^{11/10}/(\sqrt 5+1))/48$ .

Proof of Corollary 1.5.

As in the proof of Corollary 1.4, we set $K=\mathbb {Q}$ and $(k,l)=(1,3)$ . The single quill $[-1,0]$ covers $\mathrm {PCO}^{+}(T^2-1)$ and we have $\mathsf {C}(T^2-1)\le (1+\sqrt {5})/2$ . The numerical condition in the quill hypothesis follows as $(1+\sqrt 5)/2<4$ . The corollary follows from Theorem 4.5.