Lower Bounds for the Canonical Height of a Unicritical Polynomial and Capacity

Abstract

In a recent breakthrough, Dimitrov [Dim] solved the Schinzel–Zassenhaus conjecture. We follow his approach and adapt it to certain dynamical systems arising from polynomials of the form $T^p+c$, where p is a prime number and where the orbit of $0$ is finite. For example, if $p=2$ and $0$ is periodic under $T^2+c$ with $c\in \mathbb {R}$, we prove a lower bound for the local canonical height of a wandering algebraic integer that is inversely proportional to the field degree. From this, we are able to deduce a lower bound for the canonical height of a wandering point that decays like the inverse square of the field degree. For these f, our method has application to the irreducibility of polynomials. Indeed, say y is preperiodic under f but not periodic. Then any iteration of f minus y is irreducible in $\mathbb {Q}(y)[T]$.

1. Introduction

Let K be a field, and suppose $f\in K[T]$ has degree $\deg f\ge 2$ . For $n\in \mathbb {N}=\{1,2,3,\ldots \}$ , we write $f^{(n)}\in K[T]$ for the n-fold iterate of f and define $f^{(0)}=T$ . Let $x\in K$ . We call x an f-periodic point if there exists $n\in \mathbb {N}$ with $f^{(n)}(x)=x$ . We call x an f-preperiodic point if there exists an integer $m\ge 0$ such that $f^{(m)}(x)$ is f-periodic. Suppose x is f-preperiodic. The preperiod length of x is

$$ \begin{align*} \mathrm{preper}(x) = \min \{ m\ge 0 : f^{(m)}(x) \text{ is } f\text{-periodic}\} \ge 0 \end{align*} $$

and the minimum period of x is

$$ \begin{align*} \mathrm{per}(x) = \min \{ n\ge 1 : f^{(n+\mathrm{preper}(x))}(x) = f^{(\mathrm{preper}(x))}(x)\} \ge 1. \end{align*} $$

An element of K is called an f-wandering point if it is not f-preperiodic. Equivalently, $x\in K$ is an f-wandering point if and only if $\{f^{(n)}(x) : n\in \mathbb {N}\}$ is infinite.

Suppose for the moment that K is a subfield of $\mathbb {C}$ . We call f post-critically finite if its post-critical set

$$ \begin{align*} \mathrm{PCO}^{+}(f) = \{f^{(m)}(x) : m\in\mathbb{N},x\in\mathbb{C}, \text{ and }f'(x)=0\} \end{align*} $$

is finite. For example, $T^2-1$ is post-critically finite as its post-critical set equals $\{-1,0\}$ .

Let $M_{\mathbb {Q}}$ denote the set consisting of the v-adic absolute value on $\mathbb {Q}$ for all prime numbers v together with the Archimedean absolute value. For each prime number v, let $\mathbb {C}_v$ denote a completion of an algebraic closure of the valued field of v-adic numbers. We also set $\mathbb {C}_v=\mathbb {C}$ if v is Archimedean.

For a real number $t\ge 0$ , we set $\log ^{+}t = \log \max \{1,t\}$ . Let $v\in M_{\mathbb {Q}}$ , and suppose $f\in \mathbb {C}_v[T]$ has degree $d\ge 2$ . For all $x\in \mathbb {C}_v$ , the limit

(1.1) $$ \begin{align} \lambda_{f,v}(x) = \lim_{n\rightarrow\infty}\frac{\log^{+} |f^{(n)}(x)|_v}{d^n} \end{align} $$

exists and satisfies $\lambda _{f,v}(f(x)) = d\lambda _{f,v}(x)$ ; see Chapter 3 and in particular Theorem 3.27 or Exercise 3.24 [Sil07] for details. For example, $\lambda _{T^2,v}(x) = \log ^{+} |x|_v$ .

If v is Archimedean, we often abbreviate $\lambda _{f,v} = \lambda _f$ . The local canonical height differs from $z\mapsto \log ^{+}|z|$ by a bounded function on $\mathbb {C}$ . We define

(1.2) $$ \begin{align} \mathsf{C}(f) = \max\left\{1,\sup_{z\in\mathbb{C}} \frac{\max\{1,|z|\}}{e^{\lambda_f(z)}}\right\}<\infty. \end{align} $$

Thus, the filled Julia set attached to f is contained in the closed disk of radius $\mathsf {C}(f)$ centered at $0$ .

Next, we define the canonical height of an algebraic number with respect to f. Let K be a number field and $f\in K[T]$ be of degree $\ge 2$ . Suppose x lies in a finite extension F of K. Let $v\in M_{\mathbb {Q}}$ , and for a ring homomorphism $\sigma \in \mathrm {Hom}(F,\mathbb {C}_v)$ we let $\sigma (f)$ be the polynomial obtained by applying $\sigma $ to each coefficient of f. The canonical or Call–Silverman height of x (with respect to f) is

(1.3) $$ \begin{align} \hat h_f(x) = \frac{1}{[F:\mathbb{Q}]} \sum_{v\in M_{\mathbb{Q}}} \sum_{\sigma \in\mathrm{Hom}(F,\mathbb{C}_v)} \lambda_{\sigma(f),v}(\sigma(x)). \end{align} $$

This value does not depend on the number field $F\supseteq K$ containing x. Let $\overline K$ denote an algebraic closure of K, we obtain a well-defined map $\hat h_f\colon \overline K\rightarrow [0,\infty )$ . For $x\in F$ , we define

$$ \begin{align*} \lambda^{\mathrm{max}}_f(x) = \max\left\{ \lambda_{\sigma(f)}(\sigma(x)) : \sigma\in \mathrm{Hom}( F,\mathbb{C})\right\}. \end{align*} $$

To formulate our results, we require Dimitrov’s notion of a hedgehog. A quill is a line segment $[0,1] z=\{tz : t\in [0,1]\}\subseteq \mathbb {C}$ , where $z\in \mathbb {C}^\times = \mathbb {C}\smallsetminus \{0\}$ . A hedgehog with at most q quills is a finite union of quills

$$ \begin{align*} \mathcal{H}(z_1,\ldots,z_q) = \bigcup_{i=1}^q [0,1] z_i, \end{align*} $$

where $z_1,\ldots ,z_q\in \mathbb {C}^\times $ ; we also define $\mathcal {H}() = \{0\}$ to be the quill-less hedgehog. A hedgehog is a compact, path connected, topological tree.

We will also use Dubinin’s theorem [Dub84] which states

$$ \begin{align*}\mathrm{cap}(\mathcal{H}(z_1,\ldots,z_q)) \le 4^{-1/q} \max\{|z_1|,\ldots,|z_q|\}\end{align*} $$

if $q\ge 1$ , where $\mathrm {cap}(\mathcal {K})$ is the capacity, or equivalently the transfinite diameter, of a compact subset $\mathcal {K}\subseteq \mathbb {C}$ . We refer to Section 3 for basic properties of the capacity.

Quill Hypothesis. Let K be a number field, and let $f\in K[T]$ have degree $\ge 2$ . We say that f satisfies the quill hypothesis if there exist $\sigma _0\in \mathrm {Hom}(K,\mathbb {C})$ and an integer $q\ge 0$ such that the post-critical set $\mathrm {PCO}^{+}(\sigma _0(f))$ is contained in a hedgehog with at most q quills and

(1.4) $$ \begin{align} q \log \mathsf{C}(\sigma_0(f)) < \log 4. \end{align} $$

This hypothesis will be only be relevant for polynomials $f = T^p+c$ with p a prime number.

The condition and the constant $4$ are artefacts of Dubinin’s theorem. Our main results require f to satisfy the quill hypothesis, which should be unnecessary conjecturally.

Say $f=T^2-1$ . Then $\mathrm {PCO}^{+}(f)$ is contained in the single quill $[-1,0]$ . We will see in Lemma 3.6 that $\mathsf {C}(T^2-1) \le (\sqrt 5+1)/2<4$ . So $f=T^2-1$ satisfies the quill hypothesis.

We are now ready to state our results.

Theorem 1.1. Let p be a prime number, let K be a number field and let $f=T^p+c\in K[T]$ . Suppose that $0$ is an f-periodic point and that f satisfies the quill hypothesis. Then there exists $\kappa =\kappa (f)>0$ with the following properties.

1. (i) If $x\in \overline K$ is an algebraic integer and also an f-wandering point, then $\lambda _f^{\mathrm {max}} (x) \ge \kappa /[K(x):K]$ .

2. (ii) If $x\in \overline K$ is an f-wandering point, then $\hat h_f (x) \ge \kappa /[K(x):K]^{2}$ .

Let us explain why (i) can be seen as a dynamical Schinzel–Zassenhaus property for f. Indeed, let x be an algebraic integer with $\mathbb {Q}$ -minimal polynomial $(X-x_1)\cdots (X-x_d)$ , where $x_1,\ldots ,x_d\in \mathbb {C}$ . The house of x is $\max _{1\le i\le d} |x_i|$ . We have if $x\not =0$ . Kronecker’s theorem states that vanishes if and only if x is $0$ or a root of unity. Dimitrov [Dim] proved for all remaining algebraic integers x. This is sometimes called the Schinzel–Zassenhaus conjecture [SZ65]. Observe that if $x\not =0$ , then . Moreover, x is $T^2$ -preperiodic if and only if it is $0$ or a root of unity.

Part (ii) of the theorem is conjectured to hold for all rational functions $f\in K(T)$ of degree $\ge 2$ with $[K(x):K]^2$ replaced by $[K(x):K]$ . This is called the dynamical Lehmer conjecture; see Conjecture 3.25 [Sil07] or Conjecture 16.2 [BIJ+19] for a higher-dimensional version. The dynamical Lehmer conjecture is already open in the classical case $f=T^2$ . In this case, $\hat h_{T^2}$ is the absolute logarithmic Weil height. Suppose x is a nonzero algebraic number that is not a root of unity. Dobrowolski’s theorem [Dob79] implies $\hat h_{T^2}(x)\ge \frac {c}{d} \left (\frac {\log \log d}{\log d}\right )^3$ with $d=[\mathbb {Q}(x):\mathbb {Q}]$ and for an absolute constant $c>0$ .

In the dynamical setting, little is known. However, we mention that M. Baker obtained a lower bound for the canonical height for general rational maps, up to an exceptional set. For a polynomial f as above, his Theorem 1.14 [Bak06] states the following. There exist constants $c_1>0$ and $c_2>0$ that depend on K and f with the following property. Let $F/K$ be a finite extension. Then $\hat h_f(x)> \frac {c_1}{[F:\mathbb {Q}]}$ for all $x\in F$ with at most $c_2 [F:\mathbb {Q}] \log (2[F:\mathbb {Q}])$ exceptions.

Next, we exhibit some cases where the quill hypothesis is satisfied.

Lemma 1.2. Let p be a prime number, let K be a number field and let $f=T^p+c\in K[T]$ . Suppose that $f\not = T^2-2$ . Then f satisfies the quill hypothesis in both of the following cases.

1. (i) There exists a field embedding $\sigma _0\colon K\rightarrow \mathbb {R}$ with image in the real numbers such that $\mathrm {PCO}^{+}(\sigma _0(f))$ is bounded.

2. (ii) There exists a field embedding $\sigma _0\colon K\rightarrow \mathbb {C}$ such that $\#\mathrm {PCO}^{+}(\sigma _0(f)) \le 2p-2$ .

Let $n\in \mathbb {N}$ , and let $G_n$ denote the monic polynomial whose roots are precisely those $c\in \mathbb {C}$ for which $0$ is $(T^2+c)$ -periodic with exact period n. Then $G_n\in \mathbb {Z}[X]$ and work of Lutzky [Lut93] implies that $G_n$ has at least one real root. In particular, by Lemma 1.2(i) the quill hypothesis is met for infinitely many quadratic polynomials $T^2+c\in \overline {\mathbb {Q}}[T]$ with periodic critical point. It is a folklore conjecture that $G_n$ is irreducible for all $n\in \mathbb {N}$ . Conditional on this conjecture, any $f=T^2+c$ with $c\not =-2$ for which $0$ is f-periodic satisfies the quill hypothesis.

Let us briefly address $f=T^2-2$ . Then $\lambda _{f}(w+w^{-1}) = \bigl |\log |w|\bigr |$ for all $w\in \mathbb {C}^\times $ . Moreover, $\hat h_f(w+w^{-1}) = 2h(w)$ for all nonzero algebraic w; here, $h=h_{T^2}$ is the absolute logarithmic Weil height. For $f=T^2-2$ , Theorem 1.1 follows from Dimitrov’s Theorem 1 [Dim]. Moreover, part (ii) with a better exponent in $[K(x):K]$ follows from Dobrowolski’s Theorem 1 [Dob79].

If $0$ is periodic of minimal period larger than 1, then $f = T^p + c$ is not conjugate to a Chebyshev polynomial or $T^p$ . Using the lemma above, we find for each prime p an f not of this form that satisfies the quill hypothesis.

The main technical results of this paper are in Section 4. For example, we also cover some cases where $0$ is an f-preperiodic point. In Theorem 4.4 below, we prove a more general version of Theorem 1.1.

We now draw several corollaries.

Corollary 1.3. Let p be a prime number, let K be a number field and let $f=T^p+c\in K[T]$ . Suppose that $0$ is an f-periodic point and that f satisfies the quill hypothesis. Suppose $y\in K$ .

1. (i) Suppose $K/\mathbb {Q}$ is unramified above p. Let y be an f-periodic point. Let $x\in \overline K$ satisfy $f^{(n)}(x)=y$ with $n\ge 0$ minimal. Then $[K(x):K]\ge p^{n-\mathrm {per}(y)}$ .

2. (ii) Suppose $K/\mathbb {Q}$ is unramified above p. Let y be an f-preperiodic point and not f-periodic. Then $f^{(n)}-y\in K[T]$ is irreducible for all $n\in \mathbb {N}$ .

3. (iii) Let y be an f-wandering point. Then there exists $\kappa =\kappa (f,y)>0$ such that, for all $n\in \mathbb {N}$ , each irreducible factor of $f^{(n)} - y \in K[T]$ has degree at least $\kappa p^n$ .

By Lemma 4.1, the extension $\mathbb {Q}(c)/\mathbb {Q}$ is unramified above p if $0$ is f-preperiodic.

For example, if $f=T^2-1$ , then $f^{(n)}-1$ is irreducible in $\mathbb {Q}[T]$ for all $n\in \mathbb {N}$ . Stoll [Sto92] showed that iterates of certain quadratic polynomials are irreducible over $\mathbb {Q}$ . Part (iii) of Corollary 1.3 was proved in more general form by Jones and Levy [JL17] using different methods.

The method presented here is effective in that it can produce explicit lower bounds for heights. In the next two corollaries, we take a closer look at the quadratic polynomial $f=T^2-1$ .

Corollary 1.4. Let $f = T^2-1$ .

1. (i) Let $x\in \overline {\mathbb {Q}}$ be an algebraic integer and an f-wandering point. Then

   $$ \begin{align*} \lambda^{\mathrm{max}}_{f}(x)\ge \frac{\log(4^{11/10}/(\sqrt 5+1))}{48} \frac{1}{[\mathbb{Q}(x):\mathbb{Q}]}> \frac{0.007}{[\mathbb{Q}(x):\mathbb{Q}]}. \end{align*} $$

2. (ii) Let $x\in \overline {\mathbb {Q}}$ be an f-wandering point. Then

   $$ \begin{align*} \hat h_f(x)\ge \frac{\log(4^{11/10}/(\sqrt 5+1))}{48} \frac{1}{[\mathbb{Q}(x):\mathbb{Q}]^2}> \frac{0.007}{[\mathbb{Q}(x):\mathbb{Q}]^2}. \end{align*} $$

Corollary 1.5. Let $f=T^2-1\in \mathbb {Q}[T]$ , and let $x\in \overline {\mathbb {Q}}$ be an f-preperiodic point. Then

$$ \begin{align*} [\mathbb{Q}(x):\mathbb{Q}] \ge \frac 12 \max\{2^{\mathrm{preper}(x)},\mathrm{per}(x)\}. \end{align*} $$

We now discuss the method of proof. It is an adaptation of Dimitrov’s proof of the Schinzel–Zassenhaus conjecture [Dim].

Suppose we are given f as in Theorem 1.1 and an algebraic integer x such that $\lambda _f^{\mathrm {max}}(x)$ is sufficiently small. For simplicity, say $f=T^p+c\in \mathbb {Q}[T]$ has rational coefficients.

The $\mathbb {Q}$ -minimal polynomial $A\in \mathbb {Q}[X]$ of x factors as $(X-x_1)\cdots (X-x_D)$ with $x_1,\ldots ,x_D\in \mathbb {C}$ . For an integer $k\ge 1$ , we set $A_k = (X-f^{(k)}(x_1))\cdots (X-f^{(k)}(x_D))$ . The basic idea is to consider a p-th root $\phi $ of the rational function $(A_l/A_k)^{p-1}=1+O(1/X)$ for certain integers $1\le k<l$ . A priori, we may consider $\phi $ as a formal power series in $\mathbb {Q}[[1/X]]$ . However, an appropriate choice of k and l will imply $\phi \in \mathbb {Z}[[1/X]]$ . This formal power series is constructed in Section 2 where we also analyze integrality properties via congruence conditions. This step requires that the degree of $T^p+c$ is a prime number. The power series $\phi $ represents a holomorphic function on the complement of a large enough disk. This step is similar as in Dimitrov’s proof.

Our next step will be to construct a sufficiently nice connected domain $U\subseteq \mathbb {C}$ on which $\phi $ represents a holomorphic function. Our choice U will have bounded complement in $\mathbb {C}$ . This construction is done in Section 3, and it is the main new aspect of this paper. Let $\widehat {\mathbb {C}}={\mathbb {C}}\cup \{\infty \}$ denote the extended complex plane. The construction of U depends on a parameter $n\in \mathbb {N}$ . It takes as an input a hedgehog $I_0$ that contains the post-critical set of f (or at least part of it). The critical values of $f^{(n)}$ are contained in the preimage $(f^{(n)})^{-1}(I_0)$ , which one should think of as a finite topological tree. Thus, $f^{(n)}$ has no critical points on $\mathbb {C}\smallsetminus (f^{(n)})^{-1}(I_0)$ . Moreover, adding the point at infinity will make $\mathbb {C}\smallsetminus (f^{(n)})^{-1}(I_0)$ simply connected.

However, this complement is not yet the desired U. We need to ensure that U does not contain the $f^{(k)}(x_i)$ and $f^{(l)}(x_j)$ so that $A_l/A_k$ has neither poles nor zeros on U. This can be done by augmenting $(f^{(n)})^{-1}(I_0)$ to a larger finite topological tree while retaining the property that $U\cup \{\infty \}$ is simply connected. By monodromy, the p-th root $\phi $ extends to a holomorphic function on U.

Dubinin’s theorem is our tool to bound from above the capacity of a suitable hedgehog. We will be able to relate the capacity of $\mathbb {C}\smallsetminus U$ to the capacity of a hedgehog using transformation properties of the capacity under certain holomorphic mappings.

Finally, in Section 4 we prove the results stated in the introduction. As did Dimitrov, we use the Pólya–Bertrandias theorem. It implies that $\phi $ is a rational function if $\mathrm {cap}(\mathbb {C}\smallsetminus U)<1$ . The quill hypothesis, the post-critically finite hypothesis on f, and the fact that $\lambda _f^{\mathrm {max}}(x)$ is small is used to ensure the capacity inequality. Here, we also choose the parameter n that controls the tree and ultimately U. If $\phi $ , whose p-th power is $(A_l/A_k)^{p-1}$ , is a rational function, then we are in one of two cases. The first case is when $A_k$ and $A_l$ have a common root that cancels out in $A_l/A_k$ . Thus, $f^{(k)}(x_i) = f^{(l)}(x_j)$ for some $i,j$ . A standard argument involving the canonical height and $k<l$ shows that x is f-preperiodic. In the second case, $A_k$ has a multiple root, say $f^{(k)}(x_i) = f^{(k)}(x_j)$ for distinct $i,j$ . This collision of Galois conjugates implies that $\mathbb {Q}(f^{(k)}(x))$ is a proper subfield of $\mathbb {Q}(x)$ . We can then proceed by an induction on the field degree. This case study is laid out in detail in Proposition 4.3 which then quickly leads to the results mentioned in the introduction.

2. Constructing the power series

In this section, we construct an auxiliary power series. Our approach largely follows the approach laid out in Section 2 [Dim] and in particular Proposition 2.2 and Lemma 2.5 loc.cit. We adapt it to the dynamical setting. Dimitrov’s crucial observation is that $\sqrt {1+4X}$ is a formal power series in X with integral coefficients.

Throughout this section, let R denote an integrally closed domain of characteristic $0$ . Let p be a prime number.

For $D\in \mathbb {N}$ and $j\in \{0,\ldots ,D\}$ , let $e_j \in \mathbb {Z}[X_1,\ldots ,X_D]$ denote the elementary symmetric polynomial of degree j. We set

$$ \begin{align*} \Psi_{p,j} = \frac 1p \left( e_j(X_1,\ldots,X_D)^p - e_j(X_1^p,\ldots,X_D^p)\right). \end{align*} $$

So $\Psi _{p,j}\in \mathbb {Z}[X_1,\ldots ,X_D]$ by Fermat’s little theorem and

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

If R is subring of a ring S and $a,b\in S$ , we write or to signify $a-b\in pS$ .

Lemma 2.1. For all $k\in \mathbb {N}$ and all $j\in \{0,\ldots ,D\}$ , we have

(2.2)

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).

Lemma 2.2. Let S be a ring of which R is a subring and suppose $x_1,\ldots ,x_D\in S$ such that $(X-x_1)\cdots (X-x_D)\in R[X]$ . Let p be a prime number.

1. (i) The value of a symmetric polynomial in $R[X_1,\ldots ,X_D]$ evaluated at $(x_1,\ldots ,x_D)$ lies in R.

2. (ii) Let $k,l\in \mathbb {N}$ with for all $a\in R$ . For all $j\in \{0,\ldots ,D\}$ , we have

   

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 $

Lemma 2.3. Let $f = T^p+c\in R[T]$ . We have

for all $k\in \mathbb {N}$ .

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 $f^{(k-1)}(0) = f^{(k-2)}(0)^p+c$ .

Let $f\in R[T]$ and $A=\prod _{j=1}^D (X-x_j)\in R[X]$ , where $x_1,\ldots ,x_D$ are in a splitting field of A. For all integers $k\ge 0$ , we set

$$ \begin{align*}A_k = \prod_{j=1}^D (X-f^{(k)}(x_j)).\end{align*} $$

Then $A_k\in R[X]$ by Lemma 2.2(i).

Lemma 2.4. Let $ f= T^p+c\in R[T]$ . Suppose that $k,l\in \mathbb {N}$ satisfy $k\le l$ and $f^{(k-1)}(0) = f^{(l-1)}(0)$ . Set

(2.6) $$ \begin{align} \delta = \left\{ \begin{array}{ll} 0 &: \text{if }k=1,\\ 1 &:\text{if }k\ge 2. \end{array} \right. \end{align} $$

We assume that

(2.7)

Then for all monic $A\in R[X]$ .

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.

Let

(2.10) $$ \begin{align} \Phi_p = \sum_{k=0}^\infty {1/p \choose k} U^k \in \mathbb{Q}[[U]]; \end{align} $$

it satisfies $\Phi _p^p = 1+U$ . If $k\ge 0$ is an integer, then

$$ \begin{align*} {1/p\choose k} p^k = \frac{1}{k!} \frac 1p\left(\frac 1p -1\right)\cdots\left(\frac 1p -k+1\right) p^k \in \frac{1}{k!}\mathbb{Z}. \end{align*} $$

The exponent of the prime p in $k!$ equals the finite sum $\sum _{e=1}^{\infty }[k/p^e]$ which is at most $k\sum _{e=1}^\infty p^{-e}= k/(p-1)\le k$ . So p does not appear in the denominator of ${1/p\choose k} p^{2k}\in \mathbb {Q}$ for all $k\ge 0$ . Moreover, if $\ell $ is a prime distinct from p, then $\ell $ does not appear in the denominator of ${1/p\choose k}$ . Therefore,

(2.11) $$ \begin{align} \Phi_p(p^2U) \in \mathbb{Z}[[U]]. \end{align} $$

Let $R[[1/X]]$ be the ring of formal power series in X. We write $\mathrm {ord}_\infty $ for the natural valuation on $R[[1/X]]$ , that is, $\mathrm {ord}_\infty (1/X) = 1$ . A nonconstant polynomial in $R[X]$ does not lie in $R[[1/X]]$ , rather it lies in the ring of formal Laurent series $R((1/X))$ to which we extend $\mathrm {ord}_\infty $ in the unique fashion. So we will consider $R[X]\subseteq R((1/X))$ . For example, any monic polynomial in $R[X]$ is a unit in $R((1/X))$ .

Proposition 2.5. Let p be a prime number, and let $f=T^p+c\in R[T]$ . Suppose that $k,l\in \mathbb {N}$ satisfy $k\le l$ and $f^{(k-1)}(0) = f^{(l-1)}(0)$ . Let $\delta \in \{0,1\}$ be as in Lemma 2.4, and suppose for all $a\in R$ . If $A\in R[T]$ is monic, there exists $B\in R[X]$ such that $\deg B \le \deg A_k^p - 1$ , $\Phi _p(p^2 B/A_k^p) \in R[[1/X]]$ and $\Phi _p(p^2 B/A_k^p)^p =(A_l/A_k)^{p-1}$ .

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 $A_k^p$ .

3. Constructing simply connected domains

Recall that $\widehat {\mathbb {C}}$ denotes the extended complex plane $\mathbb {C}\cup \{\infty \}$ , that is, the Riemann sphere. This section concerns the construction of a domain $U\subseteq \widehat {\mathbb {C}}$ containing $\infty $ that is related to the discussion in the introduction. It depends on a polynomial $f=T^p+c$ and ultimately on an algebraic integer of small canonical height. The complement $\widehat {\mathbb {C}} \smallsetminus U$ will be a compact subset of $\mathbb {C}$ of capacity strictly less than $1$ . The relevant properties are stated in Proposition 3.4. Later on, we will construct a holomorphic function on U using the power series introduced in Section 2.

We begin with a topological lemma which is most likely well-known. We provide a proof as we could not find a proper reference.

Lemma 3.1. Let $f\in \mathbb {C}[T]\smallsetminus \mathbb {C}$ . Let $B\subseteq \mathbb {C}$ be compact such that $\mathbb {C}\smallsetminus B$ is connected with cyclic fundamental group. Suppose B contains all critical values of f. Then $\mathbb {C}\smallsetminus f^{-1}(B)$ is connected and $\widehat {\mathbb {C}}\smallsetminus f^{-1}(B)$ is simply connected.

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 [Ho75]. 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).

We recall the definition of the transfinite diameter of a nonempty compact subset $\mathcal {K}$ of $\mathbb {C}$ ; see Chapter 5.5 [Ran95] for details. For an integer $n\ge 2$ , we define

$$ \begin{align*} d_n(\mathcal{K}) = \sup \left\{ \prod_{1\le i<j\le n} |z_i-z_j|^{2/(n(n-1))} : z_1,\ldots,z_n\in\mathcal{K}\right\}. \end{align*} $$

The sequence $\bigl (d_n(\mathcal {K})\bigr )_{n\ge 2}$ is nonincreasing, and the transfinite diameter of $\mathcal {K}$ is its limit. The transfinite diameter is known to equal the capacity $\mathrm {cap}(\mathcal {K})$ of $\mathcal {K}$ by the Fekete–Szegö theorem; see Theorem 5.5.2 [Ran95]. It is convenient to define $\mathrm {cap}(\emptyset )=0$ .

For $f\in \mathbb {C}[T]$ and for $n\in \mathbb {N}$ , we introduce the truncated post-critical set

$$ \begin{align*} \mathrm{PCO}^{+}_n(f) = \bigl\{f^{(m)}(x) : m\in\{1,\ldots,n\},x\in\mathbb{C}, \text{ and }f'(x)=0\bigr\} \subseteq \mathrm{PCO}^{+}(f). \end{align*} $$

In the next proposition we will use Dubinin’s theorem.

Proposition 3.2. Let $f\in \mathbb {C}[T]$ be monic of degree $d\ge 2$ such that $f'(0)=0$ . Let $n\in \mathbb {N}$ such that there exist $q\ge 0$ and $\gamma _1,\ldots ,\gamma _q\in \mathbb {C}^\times $ with $\mathrm {PCO}_{n}^{+}(f)\subseteq \mathcal {H}(\gamma _1,\ldots ,\gamma _q)$ . Let $m\ge 1$ , and let $\alpha _1,\ldots ,\alpha _m \in \mathbb {C}$ be arbitrary. There is a simply connected domain $U\subseteq \widehat {\mathbb {C}}$ containing $\infty $ with the following properties.

1. (i) We have $0,\alpha _1,\ldots ,\alpha _m\not \in U$ and $\widehat {\mathbb {C}}\smallsetminus U$ is a compact subset of $\mathbb {C}$ .

2. (ii) We have

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

To prepare for the proof of Proposition 3.2, let $f\in \mathbb {C}[T]$ denote a monic polynomial of degree $d\ge 2$ with $f'(0)=0$ . Let $n\in \mathbb {N}$ , and suppose $\gamma _1,\ldots ,\gamma _q\in \mathbb {C}^\times $ are such that $I_0=\mathcal {H}(\gamma _1,\ldots ,\gamma _q)$ contains $\mathrm {PCO}_{n}^{+}(f)$ .

We apply Lemma 3.1 to $f^{(n)}$ and $B=I_0$ ; the fundamental group of $\mathbb {C}\smallsetminus I_0$ is cyclic and generated by a large enough loop. We set

$$ \begin{align*}H_f^{(n)} = (f^{(n)})^{-1}(I_0).\end{align*} $$

By Lemma 3.1, the complement $\mathbb {C}\smallsetminus H_f^{(n)}$ is connected and $\widehat {\mathbb {C}} \smallsetminus H_f^{(n)}$ is simply connected. Observe that $0\in H_f^{(n)}$ because $0$ is a critical point of f.

We will show that a branch of $z\mapsto f^{(n)}(z)^{1/d^n}$ admits a holomorphic continuation on $\mathbb {C}\smallsetminus H_f^{(n)}$ . Moreover, this continuation will be a biholomorphic map between $\mathbb {C}\smallsetminus H_f^{(n)}$ and the complement of a hedgehog.

Let $z\in \mathbb {C}$ satisfy $f^{(n)}(z)=0$ . Then $z\in H_f^{(n)}$ as $0\in I_0$ . Therefore, $f^{(n)}$ has no roots in $\mathbb {C}\smallsetminus H_f^{(n)}$ . The association $w\mapsto f^{(n)}(w^{-1})$ is meromorphic on $\mathbb {C} \smallsetminus \tilde H$ with $\tilde H = \{w \in \mathbb {C}^\times : w^{-1} \in H_f^{(n)}\}$ closed in $\mathbb {C}$ . It never vanishes and has only one pole at $0\in \mathbb {C}\smallsetminus \tilde H$ the order of which is $d^n$ . It follows that $\tilde f \colon w\mapsto w^{d^n} f^{(n)}(w^{-1})$ is holomorphic on $\mathbb {C} \smallsetminus \tilde H$ without zeros. Moreover, $\tilde f(0)=1$ as f is monic.

The complements $\widehat {\mathbb {C}} \smallsetminus H_f^{(n)}$ and $\mathbb {C} \smallsetminus \tilde H$ are homeomorphic via the Möbius transformation $z\mapsto z^{-1}$ . In particular, the latter is simply connected. So there exists a holomorphic function $\tilde g_n \colon \mathbb {C}\smallsetminus \tilde H \rightarrow \mathbb {C}$ with $\tilde g_n(w)^{d^n}=\tilde f(w)$ for all $w\in \mathbb {C}\smallsetminus \tilde H$ and with $\tilde g_n(0)=1$ . We resubstitute $z=1/w$ and define $g_n(z) = z\tilde g_n(z^{-1})$ to find

(3.2) $$ \begin{align} g_n(z)^{d^n} = f^{(n)}(z) \end{align} $$

for all $z\in \mathbb {C} \smallsetminus H_f^{(n)}$ . Then $g_n$ is holomorphic and $g_n(z) = z + O(1)$ as $z\rightarrow \infty $ .

Let $z\in \mathbb {C} \smallsetminus H_f^{(n)}$ . Then $g_n(z) \in \mathbb {C}\smallsetminus I_n$ with $I_n = \{z\in \mathbb {C} : z^{d^n} \in I_0\}$ by Equation (3.2). Now,

(3.3) $$ \begin{align} I_n=\mathcal{H}\bigl(e^{2\pi \sqrt{-1}j/d^n}w_i\bigr)_{\substack{1\le i\le q \\ 1\le j \le d^n}} \end{align} $$

is a hedgehog with at most $q d^n$ quills where the $w_i$ satisfy $w_i^{d^n}=\gamma _i$ .

Let us consider two examples.

First, suppose $f = T^2$ . In this case, the post-critical set equals $\{0\}$ . We may take $q=0$ and $I_0 = \mathcal {H}()=\{0\}$ . For all $n\in \mathbb {N}$ , we have $H_f^{(n)} = I_n = \{0\}$ . Moreover, $g_n(z) = z$ for all $z\in \mathbb {C}^\times $ .

Second, say $f = T^2-2$ . Its post-critical set is $\{-2,2\}$ . This time the hedgehog $I_0 = \mathcal {H}(-2,2) = [-2,0]\cup [0,2]$ has $q=2$ quills. As $f^{-1}([-2,2])=[-2,2]$ , we have $H^{(n)}_f = [-2,2]$ for all $n\in \mathbb {N}$ . In particular, $f^{(n)}(z)\not =0$ for all $z\in \mathbb {C} \smallsetminus [-2,2]$ . The polynomial mapping $w\mapsto w^{2^n}f^{(n)}(w^{-1})$ does not vanish on the simply connected domain $\mathbb {C}\smallsetminus ((-\infty ,-1/2]\cup [1/2,\infty ))$ and attains $1$ at $z=0$ . So it has a holomorphic $2^n$ -th root $\tilde g_n$ that maps $0$ to $1$ . As above, we set $g_n(z) = z \tilde g_n(z^{-1})$ for all $z\in \mathbb {C} \smallsetminus [-2,2]$ . Then $g_n$ takes values in $\mathbb {C}\smallsetminus I_n$ . Here, $I_n$ is a hedgehog with $2^{n+1}$ quills ending at $2^{1/2^n} e^{2\pi \sqrt {-1} (k+\delta )/2^n}$ for $k\in \mathbb {Z}$ and $\delta \in \{0,1/2\}$ .

We return to the general case. The map $g_n$ is a key tool in constructing U from Proposition 3.2. Let us first show that it is injective and determine its image.

Lemma 3.3. The holomorphic function $g_n$ induces a bijection $\mathbb {C}\smallsetminus H_f^{(n)}\rightarrow \mathbb {C}\smallsetminus I_n$ .

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 [Ho75].

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.

The inverse of a holomorphic bijection is again holomorphic. So the inverse

$$ \begin{align*} h_n \colon \mathbb{C}\smallsetminus I_n\rightarrow \mathbb{C}\smallsetminus H_f^{(n)} \end{align*} $$

of $g_n$ is also holomorphic; the function $h_n$ is a branch of $z\mapsto (f^{(n)})^{-1}(z^{d^n})$ . We have

$$ \begin{align*} h_n(z) = z + O(1)\text{ for } z\rightarrow\infty \end{align*} $$

by the same property of $g_n$ . Thus, $h_n$ extends to a homeomorphism $\widehat {\mathbb {C}}\smallsetminus I_n\rightarrow \widehat {\mathbb {C}}\smallsetminus H_f^{(n)}$ .

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 [Ran95] 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 [Dub84] 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).

We keep the notation of the previous proof and make a final expository remark. The complement of U as in Equation (3.5) is obtained by augmenting $H_f^{(n)}$ . More precisely, we claim that

(3.6) $$ \begin{align} \mathbb{C} \smallsetminus U = H_f^{(n)} \cup h_n\left(I^{\prime}_n\smallsetminus I_n\right). \end{align} $$

Indeed, note that $U = h_n(\widehat {\mathbb {C}}\smallsetminus I^{\prime }_n) \subseteq h_n(\widehat {\mathbb {C}}\smallsetminus I_n) = \widehat {\mathbb {C}}\smallsetminus H^{(n)}_f$ as $I_n\subseteq I^{\prime }_n$ . Taking the complement in $\mathbb {C}$ yields $H_f^{(n)}\subseteq \mathbb {C} \smallsetminus U$ . Suppose $z=h_n(w)$ with $w\in I^{\prime }_n\smallsetminus I_n$ . If $z\in U$ , then $z=h_n(w')$ for some $w\in \widehat {\mathbb {C}}\smallsetminus I^{\prime }_n$ . But then $w=w'$ because $h_n$ is injective; this is a contradiction. So the right-hand side of Equation (3.6) is contained in its left-hand side. To see the converse inclusion, we suppose $z\in \mathbb {C}\smallsetminus U$ and $z\not \in H_f^{(n)}$ . As $h_n$ maps onto $\widehat {\mathbb {C}}\smallsetminus H_f^{(n)}$ , there is $w\in \widehat {\mathbb {C}}\smallsetminus I_n$ with $z=h_n(w)$ . Because $z\not \in U$ , we must have $w\in I^{\prime }_n$ . Thus, $z\in h_n(I^{\prime }_n\smallsetminus I_n)$ . This completes the proof of Equation (3.6).

Next, we make the estimate in Proposition 3.2 more explicit for polynomials of the shape $T^d+c$ .

Proposition 3.4. Let $d\ge 2$ be an integer, and let $f=T^d+c\in \mathbb {C}[T]$ . Let $n\in \mathbb {N}$ such that $\mathrm {PCO}_{n}^{+}(f)$ is contained in a hedgehog with at most $q\ge 0$ quills. Let $m\ge 1$ , and let $\alpha _1,\ldots ,\alpha _m \in \mathbb {C}$ . There is a simply connected domain $U\subseteq \widehat {\mathbb {C}}$ with the following properties.

1. (i) We have $0,\alpha _1,\ldots ,\alpha _m\not \in U$ and $\widehat {\mathbb {C}}\smallsetminus U$ is a compact subset of $\mathbb {C}$ .

2. (ii) We have

   $$ \begin{align*} \mathrm{cap}(\widehat{\mathbb{C}}\smallsetminus U)\le 4^{-1/(qd^n+m)} \mathsf{C}(f)^{1/d^n} e^{\max\{\lambda_f(0),\lambda_f(\alpha_1), \ldots, \lambda_f(\alpha_m)\}}. \end{align*} $$

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.

Our applications require $\lambda _f(0)=0$ in Equation (3.4).

We conclude this section by deducing upper bounds for $\mathsf {C}(f)$ when $f=T^d+c\in \mathbb {C}[T]$ . Ingram has related estimates; see Section 2 [Ing09].

Lemma 3.5. Let $d\ge 2$ be an integer, and let $c\in \mathbb {C}$ . The polynomial $T^d -T- |c|$ has a unique root r in $(0,\infty )$ . Moreover, $r\ge 1$ and $\log ^{+}|z^d+c| \ge -(d-1)\log r + d\log ^{+} |z|$ for all $z\in \mathbb {C}$ with $|z|\ge r$ .

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).

Lemma 3.6. Let $d\ge 2$ be an integer, and let $c\in \mathbb {C}$ . Let $r\ge 1$ be as in Lemma 3.5. Then $\mathsf {C}(T^d+c) \le r$ .

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).

Lemma 3.7. Let $d\ge 2$ be an integer, let $c\in \mathbb {C}$ and let $f=T^d+c$ . Suppose $\mathrm {PCO}^{+}(f)=\{f^{(n)}(0) : n\in \mathbb {N}\}$ is a bounded set.

1. (i) We have $|c|\le 2^{1/(d-1)}$ and $\mathsf {C}(f) \le 2^{1/(d-1)}$ .

2. (ii) Suppose $c^{d-1}\not =-2$ . Then $\mathsf {C}(f) < 2^{1/(d-1)}$ .

3. (iii) Suppose d is odd. Then $\mathsf {C}(f) < 2^{1/(d-1)}$ .

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.

We are now ready to prove Lemma 1.2 from the introduction.

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.

4. Proof of main results

The following lemma is well-known; the second claim goes back to Gleason. See Lemme 2, Exposé XIX [DH84] for $p=2$ , and to Epstein and Poonen [Eps12]. For the reader’s convenience, we give a proof.

Lemma 4.1. Let p be a prime number and $f=T^p+c \in \mathbb {C}[T]$ .

1. (i) Suppose $0$ is f-preperiodic. Then c is an algebraic integer.

2. (ii) Suppose $0$ is f-periodic. Then $\mathbb {Q}(c)/\mathbb {Q}$ is unramified above p.

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).

The following result is a special case of the Pólya–Bertrandias theorem; it is crucial for our application and is also at the core of Dimitrov’s proof of the Schinzel–Zassenhaus conjecture.

Let K be a number field, and let $\mathcal {O}_K$ denote the ring of algebraic integers of K. For $\phi = \sum _{j\ge 0} \phi _j/X^{j}\in K[[1/X]]$ and $\sigma \in \mathrm {Hom}(K,\mathbb {C})$ , we define $\sigma (\phi ) = \sum _{j\ge 0} \sigma (\phi _j)/X^{j}\in \mathbb {C}[[1/X]]$ . Recall that $\mathrm {cap}(\mathcal {K})$ is the transfinite diameter of a nonempty compact subset $\mathcal {K}\subseteq \mathbb {C}$ .

Theorem 4.2 (Pólya–Bertrandias).

Let K be a number field and $\phi \in \mathcal {O}_K[[1/X]]$ . Suppose that for each $\sigma \in \mathrm {Hom}(K,\mathbb {C})$ there exists a connected open subset $U_\sigma \subseteq \mathbb {C}$ with the following properties.

1. (i) The complement $\mathbb {C}\smallsetminus U_\sigma $ is compact.

2. (ii) The formal power series $\sigma (\phi )$ converges at all $z\in \mathbb {C}$ with $|z|$ sufficiently large.

3. (iii) The holomorphic function induced by $\sigma (\phi )$ as in (ii) extends to a holomorphic function on $U_\sigma $ .

If $\prod _{\sigma \in \mathrm {Hom}(K,\mathbb {C})} \mathrm {cap}(\mathbb {C}\smallsetminus U_\sigma )<1$ , then $\phi $ is a rational function.

Proof. This is a special case of Théorème 5.4.6 [Ami75]. 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 $U_{\sigma }$ .

We come to the main technical result of our paper.

Proposition 4.3. Let p be a prime number, let K be a number field and let $f=T^p+c\in K[T]$ . Suppose further that there are $k,l\in \mathbb {N}$ with $k<l, f^{(k-1)}(0)=f^{(l-1)}(0)$ and the following properties.

1. (i) Let $\delta $ be as in Equation (2.6), and assume for all $a\in \mathcal {O}_K$ .

2. (ii) For all field embeddings $\sigma \in \mathrm {Hom}( K,\mathbb {C})$ , we assume that there exists $n_\sigma \in \mathbb {N}$ such that $\mathrm {PCO}_{n_\sigma }^{+}(\sigma (f))$ is contained in a hedgehog with at most $q_\sigma \ge 0$ quills.

Let $x\in \overline K$ be an algebraic integer. Then one of the following conclusions holds true:

1. (A) We have

   (4.1) $$ \begin{align} \lambda^{\mathrm{max}}_{f}(x) \ge \frac{1}{[K:\mathbb{Q}]}\sum_{\sigma\in\mathrm{Hom}(K,\mathbb{C})} \frac{1}{p^{l+n_{\sigma}}}\left(\frac{\log 4}{q_\sigma+2[K(x):K]p^{-n_\sigma}}-{\log \mathsf{C}(\sigma(f))}\right). \end{align} $$

2. (B) We have $[K(f^{(k)}(x)):K]\le [K(x):K]/p$ .

3. (C) The point x is f-preperiodic, $\mathrm {preper}(x) \le k$ , and $\mathrm {per}(x) \le (l-k)[K(x):K]$ .

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.

Theorem 4.4. Let p be a prime number, let K be a number field and let $f=T^p+c\in K[T]$ . Suppose further that there are $k,l\in \mathbb {N}$ with $k<l, f^{(k-1)}(0)=f^{(l-1)}(0),$ and the following properties.

1. (i) Let $\delta $ be as in Equation (2.6), and assume for all $a\in \mathcal {O}_K$ .

2. (ii) We suppose hat f satisfies the quill hypothesis.

Then there exists a constant $\kappa>0$ depending on the data above with the following properties.

1. (i) Let $x\in \overline K$ be an algebraic integer and f-wandering. Then $\lambda _f^{\mathrm {max}} (x) \ge \kappa /[K(x):K]^k$ .

2. (ii) Let $x\in \overline K$ be f-wandering. Then $\hat h_f (x) \ge \kappa /[K(x):K]^{k+1}$ .

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 $\kappa \le [K:\mathbb {Q}]^{-2}\log 2$ .

We consider an example before moving on. Let $c=-1.543689\ldots $ be the real root of $T^3+2T^2+2T+2$ . Then $[K:\mathbb {Q}]=3$ , where $K=\mathbb {Q}(c)$ . Moreover, $f^{(3)}(0)=f^{(4)}(0)$ for $f=T^2+c$ and $\mathrm {PCO}^{+}(f)\subseteq \mathbb {R}$ is contained in a union of $2$ quills. Let $\wp $ be a prime ideal of $\mathcal {O}_K$ containing $2$ . Then $\wp ^3= 2\mathcal {O}_K$ follows from a pari/gp computation. We claim that for all $a\in \mathcal {O}_K$ . Indeed, $a^4(a^4-1)\in 2\mathcal {O}_K$ if $a\in \wp $ . If $a\not \in \wp $ , then a is a unit modulo $2\mathcal {O}_K$ . So $a^4 -1 \in 2\mathcal {O}_K$ as the unit group $(\mathcal {O}_K/2\mathcal {O}_K)^\times $ has order $4$ . So hypothesis (i) in Theorem 4.4 is satisfied with $(k,l)=(4,5)$ ; note $\delta =1$ . The quill hypothesis is met since $\mathsf {C}(f)<2$ by Lemma 3.7(ii).

Theorem 4.5. Let p be a prime number, let K be a number field and let $f=T^p+c\in K[T]$ . Suppose k and l are as in Theorem 4.4. Suppose furthermore that f satisfies the quill hypothesis. Let $x\in \overline K$ be f-preperiodic. Then

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

and

(4.12) $$ \begin{align} [K(x):K] \ge [K(f^{(\mathrm{preper}(x))}(x)):K] \ge \frac{\mathrm{per}(x)}{l-k}. \end{align} $$

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.