Backward orbits of critical points

TL;DR

This paper investigates the Galois groups of certain polynomial extensions over number fields, analyzing prime distributions related to $p$-adic attracting points, and applies findings to the Dynamical Mordell-Lang Conjecture.

Contribution

It introduces a novel analysis of Galois groups of polynomial iterates and connects this to prime distributions and dynamical conjectures in number theory.

Findings

Determines the structure of Galois groups for polynomial iterates.

Estimates the density of primes with $p$-adic attracting points.

Applies results to cases of the Dynamical Mordell-Lang Conjecture.

Abstract

We examine the Galois groups of the extensions $K((f'\circ f^n)^{-1}(0))/K$ where $K$ is a number field for polynomials $f(x)\in K[x]$. We use our understanding of this group to study the proportion of primes for which $f$ has a $\mathfrak p$-adic attracting periodic point for a "typical" $f$ and apply the result to the split case of the Dynamical Mordell-Lang Conjecture.

Equations

$$\Delta (F(x))=c^2\Delta (B/A)$$

$$\Delta (f^{n+1}(x)-\alpha )=\pm \Delta (f^n(x)-\alpha {)}^d\prod _{f^{\prime }(b)=0}(f^{n+1}(b)-\alpha {)}^{m_b}$$

$${\pi }^{d\ell }(f({\pi }^{-\ell }x)-\gamma )=x^d-{\pi }^{(d-m)\ell }b{x}^m-{\pi }^{d\ell }(x_0+\gamma )$$

$${\pi }^{d\ell }(f({\pi }^{-\ell }x)-\gamma )\equiv x^m(x^{d-m}-c)\mathrm{mod}\mathfrak{P}$$

$${\pi }^{(d-m)\ell }{h}_2({\pi }^{-\ell }x)\equiv x^{d-m}-c\mathrm{mod}\mathfrak{P}.$$

$$\mathrm{Gal}\left(K_{\beta }/K_s\right)\cong \mathrm{Gal}\left(K_{\beta ,t}/K_{s,t}\right)\cong S_{d-1}.$$

$$\mathrm{Gal}(M_{\gamma }/K_{\beta ,t}(\gamma ))\cong S_d.$$

$$\mathrm{Gal}(K_{\beta ,t}({\cup }_j{f}^{-n}({\beta }_j))/\widehat{M_{\gamma }})\cong S_d.$$

$$\left|\mathrm{Gal}(K_n/K_{n-1})\right|=\left|\prod _{\gamma \in {\cup }_j{f}^{-(n-1)}({\beta }_j)}{S}_d\right|=d{!}^{d^{n-1}(d-1)}.$$

$$\mathrm{Gal}\left(K_{\beta ,t}/K_{s,t}\right)\cong S_{d-1}.$$

$$|\mathrm{Gal}\left(K_n/K_{s,t}\right)|$$

$$=d{!}^{d^{n-1}(d-1)}\cdot |S_{d-1}[[S_d{]}^{n-1}]|$$

$$=|S_{d-1}[[S_d{]}^n]|.$$

$$\mathrm{Gal}\left(K\left({\left(f^{\prime }\circ f^n\right)}^{-1}(0)\right)/K\right)\cong S_{d-1}[[S_d{]}^n].$$

$$\mathrm{Gal}\left(K\left({\left(f^{\prime }\circ f^n\right)}^{-1}(0)\right)/K\right)\cong \prod _{d-1}[S_d{]}^n.$$

$$(f^n{)}^{\prime }(P)=f^{\prime }(P)f^{\prime }(f(P))\dots f^{\prime }(f^{n-1}(P))\equiv 0\mathrm{mod}\mathfrak{P}.$$

$$1-\mathrm{FPP}\left(\prod _{d-1}[S_d{]}^n\right)$$

$$=\frac{|\{({\sigma }_1,\dots ,{\sigma }_{d-1})\in {\prod }_{d-1}[S_d{]}^n:{\sigma }_i\text{ does not have a fixed point for all }i\}|}{|{\prod }_{d-1}[S_d{]}^n|}$$

$$=\prod _i\frac{|\{{\sigma }_i\in [S_d{]}^n:{\sigma }_i\text{ does not have a fixed point}\}|}{|[S_d{]}^n|}$$

$$={\left(1-\mathrm{FPP}\left([S_d{]}^n\right)\right)}^{d-1}.$$

$$\mathrm{FPP}\left(\prod _{d-1}[S_d{]}^n\right)=1-{\left(1-\mathrm{FPP}\left([S_d{]}^n\right)\right)}^{d-1}.$$

$$\mathrm{FPP}\left(\prod _{d-1}[S_d{]}^n\right)=1-{\left(1-\mathrm{FPP}\left([S_d{]}^n\right)\right)}^{d-1}\le 1-{\left(1-\frac{2}{n+2}\right)}^{d-1}$$

$$=1-{\left(\frac{n}{n+2}\right)}^{d-1}\le 1-\frac{n^{d-1}}{n^{d-1}+C_d{n}^{d-2}}=\frac{C_d{n}^{d-2}}{n^{d-1}+C_d{n}^{d-2}}\le \frac{C_d{n}^{d-2}}{n^{d-1}}=\frac{C_d}{n}.$$

$$\mathrm{Gal}\left(K\left({\left(f^{\prime }\circ f^{n_0}\right)}^{-1}(0)\right)/K\right)\cong \prod _{d-1}[S_d{]}^n.$$

Full text

1. Introduction

Let $K$ be a number field, $d\in{\mathbb{Z}}_{\geq 2}$, and let $s_{1},\dots,s_{d-1},t$ independent transcendentals over $K$. We are using $t$ for the constant term to differentiate it from the other coefficients as it will play a special role in the proof of the main theorem. Then $f(x)=x^{d}+s_{1}x^{d-1}+\dots+s_{d-1}x+t$ is the generic monic polynomial over $K$. Define $K_{s}:=K(s_{1},\dots,s_{d-1})$ and $K_{s,t}:=K_{s}(t)$. We investigate the action of the absolute Galois group of $K_{s,t}$ on the backward orbits of the critical points of $f(x)$. Giving the elements of the backward orbits the structure of a rooted tree induces a homomorphism from the Galois group to the automorphism group of the tree, called an arboreal Galois representation. In this case, the automorphism group of the tree is isomorphic to $S_{d-1}[[S_{d}]^{n}]$. We prove that the action of the Galois group is as large as possible subject to this constraint.

Theorem 1.1**.**

$\operatorname{Gal}\left(K_{s,t}\left(\left(f^{\prime}\circ f^{n}\right)^{-1}(0)\right)/K_{s,t}\right)\cong S_{d-1}[[S_{d}]^{n}]$.

Note, our set up here differs from iterated monodromy groups where one considers preimages of a transcendental basepoint for a specific polynomial or rational function.

Since $f^{\prime}(x)$ is itself a polynomial with independent transcendental coefficients, we have $\operatorname{Gal}\left(K_{s,t}\left((f^{\prime})^{-1}(0)\right)/K_{s,t}\right)\cong S_{d-1}$ by a classical result. Thus, Theorem 1.1 is equivalent to the following proposition.

Proposition 1.2**.**

$\operatorname{Gal}\left(K_{s,t}\left(\bigcup_{f^{\prime}(\beta)=0}f^{-n}(\beta)\right)/K_{s,t}((f^{\prime})^{-1}(0))\right)\cong\prod_{d-1}[S_{d}]^{n}$.

By Hilbert’s irreducibility theorem, the corresponding Galois groups for “most” specializations of $f(x)$, that is specializations of the transcendentals to the number field $K$, will be isomorphic to the groups in Theorem 1.1 and Proposition 1.2 (see Corollaries 4.5 and 4.6). The Chebotarev Density theorem allows us to translate information about these Galois groups to information about certain sets of primes. Specifically, we use Proposition 1.2 to show for a “typical” polynomial defined over a number field $K$ and with critical points also defined over $K$ the density of primes for which the polynomial has an attracting point in $\bar{K_{\mathfrak{p}}}$ is small. Here $\bar{K_{\mathfrak{p}}}$ denotes the algebraic closure of the completion of $K$ at the prime ${\mathfrak{p}}$.

Theorem 1.3**.**

Fix a number field $K$ and $\epsilon>0$. There is a Zariski dense subset ${\mathcal{H}}$ of $K^{d}$ such that for all monic polynomials $f(x)$ with critical points $b_{1},\dots,b_{d-1}$ and constant term $c_{0}$ with $(b_{1},\dots,b_{d-1},c_{0})\in{\mathcal{H}}$, the density of the set of primes ${\mathfrak{p}}$ of $K$ for which $f(x)$ has a finite attracting periodic point in $\bar{K}_{\mathfrak{p}}$ is less than $\epsilon$.

Remark 1.4*.*

Using known estimates on fixed point proportions for iterated wreath products, we can conclude for any $f(x)$ as in Theorem 1.3, the set of such primes is bounded above by $\frac{C_{d}}{n}$, for some constant $C_{d}$ depending on $d$.

We apply our results toward the split case of the Dynamical Mordell-Lang Conjecture. The Dynamical Mordell-Lang conjecture is a dynamical analogue to the cyclic case of the classical Mordell-Lang Conjecture, proven by Faltings [Fal83], Vojta [Voj99], and McQuillan [McQ95]. See [BGT16] for a detailed narrative on the Dynamical Mordell-Lang conjecture.

Conjecture 1.5** (Dynamical Mordell-Lang Conjecture, [GT09]).**

Let $X$ be a quasiprojective variety defined over ${\mathbb{C}}$, $\Phi$ be an endomorphism of $X$, $\alpha\in X({\mathbb{C}})$, and $V$ a subvariety of $X$. Then the set of $n\in{\mathbb{N}}_{0}$ such that $\Phi^{n}(\alpha)\in V({\mathbb{C}})$ is a union of finitely many arithmetic progressions.

Using Theorem 5.1, we are able to prove the Dynamical Mordell-Lang Conjecture holds for “most” split maps which act coordinate-wise by polynomials whose critical points are all defined over the same number field.

Theorem 1.6**.**

Let $K$ be a number field. Fix an integer $g\geq 1$. Let $V$ be a subvariety of $({\mathbb{P}}^{1})^{g}$. Fix $g$ integers $d_{1},\dots,d_{g}\geq 2$. Let $\Phi=(f_{1},\dots,f_{g})$ where $f_{i}(x)$ is a monic polynomial of degree $d_{i}$ with critical points $b_{i,1},\dots,b_{i,d_{i}-1}\in K$ and constant term $c_{i,0}\in K$. Let $\alpha=(\alpha_{1},\dots,\alpha_{g})\in({\mathbb{P}}^{1}(\bar{K}))^{g}$ such that $\alpha_{i}\neq\infty$ for each $i$. There are Zariski dense sets ${\mathcal{H}}(g,d_{i},K)\subseteq K^{d_{i}}$ such that if $(b_{i,1},\dots,b_{i,d_{i}-1},c_{0})\in{\mathcal{H}}(g,d_{i},K)$, then the set of $n\in{\mathbb{N}}_{0}$ such that $\Phi^{n}(\alpha)\in V(\bar{K})$ is a union of finitely many arithmetic progressions.

The proof of this theorem relies on a theorem of Benedetto, Ghioca, Kurlberg, Tucker [BGKT12, Theorem 3.4] which gives a sufficient condition for the conclusion of the Dynamical Mordell-Lang Conjecture for split maps in terms of empty intersection of the orbit with the residue classes of attracting periodic points modulo a prime $p$. Their proof proceeds by showing it is possible to find a $p$-adic parametrization of the orbit under these hypotheses. We show our results imply the existence of a suitable prime. The Galois theoretic approach to investigating Dynamical Mordell-Lang used in this paper was suggested by Bell, Ghioca, and Tucker [BGT16]. Interestingly, heuristics suggest this approach likely will not work for non split maps in dimension greater than 4 since there should be a positive proportion of maps in the moduli space such that the orbit of a point passes through the ramification locus at every prime [BGH*+*13]. We also note a similar result was proved by Fakhruddin [Fak14] for generic degree $d$ endomorphisms of ${\mathbb{P}}^{n}_{K}$, where for the definition of generic used in Fakhruddin’s work, a map being generic implies it is in the complement of a countable union of proper subvarieties of the natural parameter variety.

Before proving the main results of this paper, we introduce arboreal Galois representations in Section 2 and give some preliminary results on ramification in Section 3. In Section 4 we prove Theorem 1.1, Proposition 1.2, and corollaries about specializations. Finally, we explore the application to ${\mathfrak{p}}$-adic attracting periodic points and the connection to the Dynamical Mordell-Lang Conjecture in Section 5.

2. Arboreal Galois Representation

In order to analyze the structure of $\operatorname{Gal}\left(K_{s,t}\left(\left(f^{\prime}\circ f^{n}\right)^{-1}(0)\right)/K_{s,t}\right)$ we consider the arboreal Galois representation of this group. These representations have been studied by a number of authors beginning with the work of Odoni [Odo85]. A survey of the subject can be found in [Jon13].

Let $K$ be any field and $f(x)\in K[x]$ a degree $d$ polynomial. We can give the set containing [math] and the roots of $(f^{\prime}\circ f^{n})$ for $n\geq 0$ the structure of a tree graph rooted at [math] as follows. The first level of the tree consists of the $d-1$ roots of $f^{\prime}(x)$. Each root of $f^{\prime}(f^{n}(x))=0$ lies in the $n+1$-st level of the tree. For each node $\gamma$ at level $n+1$ with $n\geq 1$ we draw an edge to the node $f(\gamma)$ and for each node $\beta$ at the 1-st level of the tree we draw an edge to $0=f^{\prime}(\beta)$. Assume $f^{\prime}\circ f^{n}$ is separable for every $n\geq 0$. This gives us a rooted tree with $d-1$ branches above the root point [math] and $d$ branches above each of the other nodes in tree.

Since $f^{\prime}\circ f^{n}$ is separable for every $n$, the extensions $K\left(\left(f^{\prime}\circ f^{n}\right)^{-1}(0)\right)$ are Galois extensions of $K$. Since any element of $\operatorname{Gal}\left(K\left(\left(f^{\prime}\circ f^{n}\right)^{-1}(0)\right)/K\right)$ commutes with $f$ and $f^{\prime}$, $\operatorname{Gal}\left(K\left(\left(f^{\prime}\circ f^{n}\right)^{-1}(0)\right)/K\right)$ acts on the tree up to the $n+1$-st level by permuting the branches and hence is isomorphic to a subgroup of $\operatorname{Aut}(T_{n+1})\cong S_{d-1}[[S_{d}]^{n}]$, where $\operatorname{Aut}(T_{n+1})$ denotes the automorphism group of the tree truncated at the $n+1$-st level and $[S_{d}]^{n}$ denotes the $n$-th iterated wreath product of $S_{d}$ with itself.

3. Preliminary Results

The results in this section will be used in the proof of Theorem 1.1.

Lemma 3.1**.**

Let $A$ be a Dedekind domain with field of fractions $K$, $F(x)\in A[x]$ and $M=K(F^{-1}(0))$. Suppose ${\mathfrak{p}}||\Delta(F(x))$ in $A$, where $\Delta(F(x))$ denotes the polynomial discriminant of $F(x)$. Then for any prime ${\mathfrak{q}}$ of $M$ lying over ${\mathfrak{p}}$, the action of the inertia group $I({\mathfrak{q}}|{\mathfrak{p}})$ on the roots of $F(x)$ consists of a single transposition.

Proof.

Let $\theta$ be a root of $F(x)$, $L=K(\theta)$, and $B$ the integral closure of $A$ in $L$. By [Lan94, Chapter 3, Proposition 13],

[TABLE]

for some $c\in A$ . Hence since ${\mathfrak{p}}||\Delta(F(x))$, we have ${\mathfrak{p}}||\Delta(B/A)$.

Suppose ${\mathfrak{p}}B=\prod{\mathfrak{P}}_{i}^{e_{i}}$. Let $f_{i}=f({\mathfrak{P}}_{i}|{\mathfrak{p}})$, the residue degree of ${\mathfrak{P}}_{i}$ over ${\mathfrak{p}}$, and $e_{i}=e_{i}({\mathfrak{P}}_{i}|{\mathfrak{p}})$, the inertia degree of ${\mathfrak{P}}_{i}$ over ${\mathfrak{p}}$. Then the power of ${\mathfrak{p}}$ in $\Delta(B/A)$ is greater than or equal to $\sum(e_{i}-1)f_{i}$ with equality if ${\mathfrak{p}}$ is tamely ramified, see [Lan94, Chapter 3, Proposition 8] and [Lan94, Chapter 3, Proposition 14]. Hence we must have $e_{i}=2$ for exactly one value of $i$ and $f_{i}=1$ for that $i$.

Now let $D({\mathfrak{q}}|{\mathfrak{p}})$ and $I({\mathfrak{q}}|{\mathfrak{p}})$ denote the decomposition group and inertia group of ${\mathfrak{q}}$ over ${\mathfrak{p}}$ respectively. By [Juu19, Lemma 2.4] and [Juu19, Remark 2.5] there is a bijection between the orbits of the roots of $F(x)$ under the action of $D({\mathfrak{p}}|{\mathfrak{q}})$ and the set of extensions ${\mathfrak{P}}$ of ${\mathfrak{p}}$ to $L$ with the property: if ${\mathfrak{P}}$ corresponds to the orbit $Y$, then the length of $Y$ is $e({\mathfrak{P}}|{\mathfrak{p}})f({\mathfrak{P}}|{\mathfrak{p}})$ and $Y$ decomposes into the union of $f({\mathfrak{P}}|{\mathfrak{p}})$ orbits of length $e({\mathfrak{P}}|{\mathfrak{p}})$ under the action of $I({\mathfrak{q}}|{\mathfrak{p}})$. Therefore by the previous paragraph, $I({\mathfrak{q}}|{\mathfrak{p}})$ acts on the roots of $F(x)$ as a single transposition. ∎

Lemma 3.2** ([AHM05], Proposition 3.2).**

If $f(x)$ is a monic polynomial of degree $d$, then

[TABLE]

where $m_{b}$ is the multiplicity of $c$ as a root of $f^{\prime}(x)$.

Lemma 3.3** ([Odo85], Lemma 2.4).**

Let $I$ be an integrally closed domain with field of fractions $F$ and suppose that $f(x)=a_{0}+\dots+a_{k}x^{k}\in I[x]$ where $k\geq 1$ and $a_{k}\neq 0$. Let $\varphi$ be any morphism mapping $I$ to $K$ where $K$ is a field with $a_{k}\notin\ker\varphi$. Let $\tilde{\varphi}$ be the induced map $I[x]\rightarrow K[x]$. If $\tilde{\varphi}(f(x))$ is separable over $K$ then

1. (i)

$f(x)$* is separable over $F$,* 2. (ii)

$\operatorname{Gal}(K((\tilde{\varphi}(f))^{-1}(0))/K)$* is isomorphic to a subgroup of $\operatorname{Gal}(F(f^{-1}(0))/F)$.*

The following result is similar to [BJ19, Lemma 2.3]. It is modified here to fit our situation. Here and throughout the paper we will use the shorthand of referring to a prime of a number field to mean a prime of the ring of integers of the number field.

Lemma 3.4**.**

Let $K$ be a number field, $d\geq 4$, and $m=d-1$ or $d-2$ with $m\geq 3$. Let $b,x_{0}\in K$, and suppose there is a prime ${\mathfrak{p}}$ of $K$ such that ${\mathfrak{p}}\nmid(d-m)$, $v_{\mathfrak{p}}(b)\in\{-1,-2\}$, $v_{\mathfrak{p}}(x_{0})\geq 1$, $(d-m)\mid v_{\mathfrak{p}}(b)$, $v_{\mathfrak{p}}(x_{0}/b)<m$ and $\gcd(m,v_{\mathfrak{p}}(x_{0}/b))=1$. Let $f(x)=x^{d}-bx^{m}-x_{0}$, let $n\geq 0$, and let $\gamma\in f^{-n}(0)$. Suppose that $f^{n}(x)\in K[x]$ is irreducible over $K$. Then there is a prime ${\mathfrak{P}}$ of $K(\gamma)$ lying above ${\mathfrak{p}}$, and a prime ${\mathfrak{Q}}$ of $K(f^{-1}(\gamma))$ such that

• •

${\mathfrak{P}}$* has ramification index $m^{n}$ over ${\mathfrak{p}}$,*

• •

$m^{n}v_{\mathfrak{P}}((\gamma+x_{0})/b)$* is a positive integer relatively prime to $m$ and $<m^{n+1}$, where $v_{\mathfrak{P}}$ is the ${\mathfrak{P}}$-adic valuation on $K(\gamma)$ extending $v_{\mathfrak{p}}$,*

• •

${\mathfrak{Q}}$* lies above ${\mathfrak{P}}$, and*

• •

the ramification group $I({\mathfrak{Q}}|{\mathfrak{P}})$ acts transitively on $m$ roots of $f(x)-\gamma$ and fixes the other $d-m$ roots.

Proof.

We first prove the first two bullet points by induction on $n$. If $n=0$, then $\gamma=0$ and both statements follow trivially taking ${\mathfrak{P}}={\mathfrak{p}}$.

Now suppose the first two statements hold for $n-1$. Let $\gamma_{n-1}=f(\gamma)\in f^{-(n-1)}(0)$. Since $f^{n}(x)$ is irreducible over $K$, it follows from Capelli’s lemma that $f(x)-\gamma_{n-1}$ is irreducible over $K(\gamma_{n-1})$. Also, by our induction hypothesis we have some prime ${\mathfrak{P}}_{n-1}$ of $K(\gamma_{n-1})$ with ramification index $m^{n-1}$ over ${\mathfrak{p}}$ and $m^{n-1}v_{{\mathfrak{P}}_{n-1}}((\gamma_{n-1}+x_{0})/b)$ is a positive integer relatively prime to $m$ and less than $m^{n}$.

The Newton polygon of $f(x)-\gamma_{n-1}=x^{d}-bx^{m}-(x_{0}+\gamma_{n-1})$ at the prime ${\mathfrak{P}}_{n-1}$ has a line segment of length $m$ with slope $-m^{-1}v_{{\mathfrak{P}}_{n-1}}((\gamma_{n-1}+x_{0})/b)$ and a line segment of slope $-v_{{\mathfrak{P}}_{n-1}}(b)/(d-m)=-v_{\mathfrak{p}}(b)/(d-m)$. This implies $f(x)-\gamma_{n-1}$ factors as a product of two polynomials say $h_{1}$ and $h_{2}$ over $K(\gamma_{n-1})_{{\mathfrak{P}}_{n-1}}$ with $\deg h_{1}=m$ and $\deg h_{2}=d-m$. Also, the Newton polygon for $h_{1}$ is a single line segment of slope $-m^{-1}v_{{\mathfrak{P}}_{n-1}}((\gamma+x_{0})/b)$. Applying the induction hypothesis, we have $-m^{-1}v_{{\mathfrak{P}}_{n-1}}((\gamma_{n-1}+x_{0})/b)=N/m^{n}$ for a positive integer $N$ relatively prime to $m$. Therefore $K(\gamma_{n-1})$ has a prime ${\mathfrak{P}}$ of ramification index $m$ over ${\mathfrak{P}}_{n-1}$. By the induction hypothesis, the ramification index of ${\mathfrak{P}}$ over ${\mathfrak{p}}$ is $m^{n}$, finishing the proof of the first bullet point.

Let $v_{\mathfrak{P}}$ be the ${\mathfrak{P}}$-adic valuation on $K(\gamma)$ extending $v_{{\mathfrak{P}}_{n-1}}$. Applying an element $\sigma\in\operatorname{Gal}(K(f^{-1}(\gamma_{n-1})/K(\gamma_{n-1}))$ if necessary, we may assume $\gamma$ is a root of $h_{1}$. Then $v_{\mathfrak{P}}(\gamma)=N/m^{n}$. Further, since $v_{\mathfrak{P}}(x_{0})=v_{\mathfrak{p}}(x_{0})\geq 1>N/m^{n}$ and $v_{\mathfrak{P}}(b)=v_{\mathfrak{P}}(b)$ is a negative integer, we have $v_{{\mathfrak{P}}}(\gamma+x_{0})=v_{\mathfrak{P}}(\gamma)$ and $m^{n}v_{\mathfrak{P}}((\gamma+x_{0})/b)=m^{n}(N/m^{n})-m^{n}v_{\mathfrak{P}}(b)=N-m^{n}v_{\mathfrak{P}}(b)$ is an integer relatively prime to $m$. Since $N<m^{n}$, $-v_{\mathfrak{P}}(b)=1$ or $2$, and $m\geq 3$, we see $N-m^{n}v_{\mathfrak{P}}(b)<m^{n+1}$. This finishes the proof of the second bullet point.

We now prove the third and fourth bullet points using the first and second. Let ${\mathfrak{P}}$ be a prime of $K(\gamma)$ satisfying the first two bullet points. By the argument above, $f(x)-\gamma$ factors a as a product of two polynomials $h_{1}(x)h_{2}(x)$ over $K(\gamma)_{\mathfrak{P}}$ where the Newton polygon of $h_{1}$ consists of a single line segment of length $m$ and slope $m^{-1}v_{\mathfrak{p}}((\gamma+x_{0})/b)$. Similarly, the Newton polygon of $h_{2}$ consists of a single line segment of length $d-m$ and slope $\ell:=-v_{\mathfrak{P}}(b)/(d-m)$.

Let $\pi$ be a uniformizer for ${\mathfrak{p}}$ in ${\mathcal{O}}_{K}$ and consider the polynomial

[TABLE]

which is defined over $K(\gamma)$ and, since $v_{\mathfrak{p}}(b)=-\ell(d-m)$, has $K(\gamma)_{\mathfrak{P}}$-integral coefficients. Then the monic polynomials $\pi^{m\ell}h_{1}(\pi^{-\ell}x)$ and $\pi^{(d-m)\ell}h_{2}(\pi^{-\ell x})$ have $K(\gamma)_{\mathfrak{P}}$-integral coefficients as well. Reducing modulo ${\mathfrak{P}}$ we have

[TABLE]

where $c\not\equiv 0\mod{\mathfrak{P}}$. Therefore,

[TABLE]

Then since $p\nmid(d-m)$ the splitting field of $h_{2}$ is unramified over ${\mathfrak{P}}$.

Let $L:=K(f^{-1}(\gamma))$ and let ${\mathfrak{Q}}$ be any prime of $L$ lying over ${\mathfrak{P}}$. Let $\delta_{1},\dots,\delta_{d}$ denote the roots of $f(x)-\gamma$ in $L_{\mathfrak{Q}}$. Without loss of generality, we may assume $\delta_{1},\dots,\delta_{m}$ are the roots of $h_{1}(x)$ and $\delta_{m+1},\dots,\delta_{d}$ are the roots of $h_{2}(x)$. Then $\delta_{m+1},\dots,\delta_{d}$ lie in an intermediate unramified extension $L^{\prime}_{\mathfrak{Q}}$ of $K(\gamma)_{\mathfrak{P}}$. The roots $\delta_{1},\dots,\delta_{m}$ correspond to the line segment of length $m$ of the Newton polygon and therefore $g$ is irreducible over $L^{\prime}_{\mathfrak{Q}}$ and the extension $L_{\mathfrak{Q}}/L^{\prime}_{\mathfrak{Q}}$ is totally ramified.

Then since $\operatorname{Gal}(L_{\mathfrak{Q}}/K(\gamma)_{\mathfrak{P}})$ is isomorphic to the decomposition group $D({\mathfrak{Q}}|{\mathfrak{P}})$ and $\operatorname{Gal}(L_{\mathfrak{Q}}/L^{\prime}_{\mathfrak{Q}})$ is isomorphic to the interia group $I({\mathfrak{Q}}|{\mathfrak{P}})$, it follows that $I({\mathfrak{Q}}|{\mathfrak{P}})$ acts transitively on $m$ roots of $f(x)-\gamma$ and fixes the remaining $d-m$. ∎

Lemma 3.5** ([BJ19], Lemma 2.4).**

Let $d\geq 3$, let $m$ be an integer relatively prime to $d$ with $d/2<m<d$ and let $G\subseteq S_{d}$ be a subgroup that contains a transposition, acts transitively on $\{1,2,\dots,d\}$, and has a subgroup that acts trivially on $\{m+1,\dots,d\}$ and transitively on $\{1,2,\dots,m\}$. Then $G=S_{d}$.

4. Proof of Theorem 1.1

Consider $f^{\prime}(x)=dx^{d}+(d-1)s_{1}x^{d-1}+\dots+s_{d-1}$. Note, except for the leading coefficient, the coefficients of $f^{\prime}(x)$ are independent transcendentals over $K$ (and $K_{t}$), hence $\frac{1}{d}f^{\prime}(x)$ is a generic monic polynomial. Let $\beta_{1},\dots,\beta_{d-1}$ denote the roots of $f^{\prime}(x)$, that is $f^{\prime}(x)=d\prod(x-\beta_{i})$. Define $K_{\beta}:=K(\beta_{1},\dots,\beta_{d-1})$ and $K_{\beta,t}:=K_{\beta}(t)$, the splitting field of $f^{\prime}(x)$ over $K_{s}$ and $K_{s,t}$ respectively. Then by a classical result

[TABLE]

Thus Theorem 1.1 follows from Proposition 1.2.

Lemma 4.1**.**

If $i\neq j$, then $f^{n}(\beta_{i})-\beta_{j}\in K(\beta_{1},\beta_{2},\dots\beta_{i-1},\beta_{i+1},\dots,\beta_{d-1},t)[\beta_{i}]$ is irreducible.

Proof.

Note, if $f^{n}(\beta_{i})-\beta_{j}$ has a nontrivial factorization, then for any specialization of the coefficients, the image of $f^{n}(\beta_{i})-\beta_{j}$ has a nontrivial factorization as well. Specialize $\beta_{\ell}\mapsto 0$ for $\ell\neq i$ and $t$ to some prime $\pi$ of the ring of integers ${\mathcal{O}}_{K}$ of $K$ dividing $d$. The image of $f^{n}(\beta_{i})-\beta_{j}$ is $\bar{f}^{n}(\beta_{i})\in K[\beta_{i}]$ where $\bar{f}(x)=x^{d}-\frac{d}{d-1}\beta_{i}x^{d-1}+\pi$.

We claim $\bar{f}^{n}(\beta_{i})$ is Eisenstein in the variable $\beta_{i}$ at $\pi$. We prove this by induction. First note, $\bar{f}(\beta_{i})=-\frac{1}{d-1}\beta_{i}^{d}+\pi$. Now suppose $\bar{f}^{n}(\beta_{i})=c_{d^{n}}\beta_{i}^{d^{n}}+\dots+c_{1}\beta_{i}+c_{0}$ where $v_{\pi}(c_{d^{n}})=0$, $v_{\pi}(c_{k})\geq 1$ for $1\leq k\leq d^{n}-1$, and $v_{\pi}(c_{0})=1$. Then $\bar{f}^{n+1}(\beta_{i})=(c_{d^{n}}\beta_{i}^{d^{n}}+\dots+c_{1}\beta_{i}+c_{0})^{d}-\frac{d}{d-1}\beta_{i}(c_{d^{n}}\beta_{i}^{d^{n}}+\dots+c_{1}\beta_{i}b+c_{0})^{d-1}+\pi=c_{d^{n}}^{d}\beta_{i}^{d^{n+1}}+\pi\beta_{i}p(\beta_{i})+(c_{0}^{d}+\pi)$ is Eisenstein at $\pi$ as well. ∎

We fix the following notation which will be used throughout the remainder of this section. Let $\gamma\in(f^{\prime}\circ f^{n})^{-1}(0)$. Define $M_{\gamma}=K_{\beta,t}(f^{-1}(\gamma))$ and $\hat{M_{\gamma}}=\prod_{\delta\in(f^{\prime}\circ f^{n})^{-1}(0)\setminus\{\gamma\}}M_{\delta}$.

Proposition 4.2**.**

Fix $n\geq 1$ and let $\gamma\in(f^{\prime}\circ f^{n-1})^{-1}(0)$. There is an automorphism of $K_{\beta,t}(\cup_{j}f^{-{n}}(\beta_{j}))$ that fixes $\hat{M_{\gamma}}$ and acts as a single transposition on the roots of $f(x)-\gamma$.

Proof.

Say $f^{n-1}(\gamma)=\beta_{j}$ and fix some $i\neq j$. Lemma 4.1 shows ${\mathfrak{p}}:=(f^{n}(\beta_{i})-\beta_{j})$ is prime in $K(\beta_{1},\beta_{2},\dots\beta_{i-1},\beta_{i+1},\dots,\beta_{d-1},t)[\beta_{i}]$. Now we note $\deg_{\beta_{i}}(f^{n}(\beta_{i})-\beta_{k})=d^{n}$, $\deg_{\beta_{i}}(f^{n}(\beta_{\ell})-\beta_{k})<d^{n}$ for $\ell\neq i$ and all $k$, and $\deg_{\beta_{i}}(f^{m}(\beta_{\ell})-\beta_{k})<d^{n}$, for all $\ell$ and $m<n$. Thus, $f^{n}(\beta_{i})-\beta_{j}$ does not divide $f^{m}(\beta_{\ell})-\beta_{k}$ for any $m\leq n$ and any $\ell,k$ unless $m=n$, $\ell=i$, and $k=j$.

Now by Lemma 3.2, we see ${\mathfrak{p}}\nmid\Delta(f^{n}(x)-\beta_{k})$ for $k\neq j$ and ${\mathfrak{p}}||\Delta(f^{n}(x)-\beta_{j})$. So ${\mathfrak{p}}$ does not ramify in $K_{\beta,t}(f^{-{n}}(\beta_{k}))$ for $k\neq j$. Lemma 3.1 implies for any prime ${\mathfrak{q}}$ of the integral closure of $K(\beta_{1},\beta_{2},\dots\beta_{i-1},\beta_{i+1},\dots,\beta_{d-1},t)[\beta_{i}]$ in $K_{\beta,t}(f^{-{n}}(\beta_{j}))$ lying over ${\mathfrak{p}}$, $I({\mathfrak{q}}|{\mathfrak{p}})$ consists of a single transposition of the roots of $(f^{-n}(x)-\beta_{j})$.

Now let ${\mathfrak{P}}$ be any prime of $M_{\gamma}$ dividing $f(\beta_{i})-\gamma$ and note ${\mathfrak{P}}$ lies over ${\mathfrak{p}}$. Also, $(f(\beta_{i})-\gamma)|\Delta(f(x)-\gamma)$, so ${\mathfrak{P}}$ is ramified over ${\mathfrak{p}}$. Let ${\mathfrak{q}}$ be a prime of $K_{\beta,t}(f^{-{n}}(\beta_{j}))$ lying over ${\mathfrak{P}}$. By the arguments in the previous paragraph $|I({\mathfrak{q}}|{\mathfrak{p}})|=2$. But we also have $|I({\mathfrak{q}}|{\mathfrak{P}})|\cdot|I({\mathfrak{P}}|{\mathfrak{p}})|=|I({\mathfrak{q}}|{\mathfrak{p}})|=2$. Hence $|I({\mathfrak{P}}|{\mathfrak{p}})|=2$ and since the elements of $I({\mathfrak{P}}|{\mathfrak{p}})$ extend to elements of $I({\mathfrak{q}}|{\mathfrak{p}})$, the nontrivial element of $I({\mathfrak{P}}|{\mathfrak{p}})$ and hence $I({\mathfrak{q}}|{\mathfrak{p}})$ must act as a single transposition of the roots of $f(x)-\gamma$.

Finally, let ${\mathfrak{Q}}$ denote any extension of ${\mathfrak{q}}$ to $K_{\beta,t}(\cup_{j}f^{-{n}}(\beta_{j}))$. Note since ${\mathfrak{p}}$ does not ramify in $K_{\beta,t}(f^{-{n}}(\beta_{k}))$ for $k\neq i$ and $I({\mathfrak{Q}}|{\mathfrak{p}})$ acts trivially on all of $\hat{M_{\gamma}}$. ∎

Proposition 4.3**.**

Fix $n\geq 1$ and let $\gamma\in(f^{\prime}\circ f^{n-1})^{-1}(0)$. Then

[TABLE]

Proof.

Let $\beta=f^{n-1}(\gamma)$. We first note $f^{n-1}(x)-\beta$ is irreducible over $K_{\beta,t}$ since it is easy to find a specialization that is irreducible, take for example any specialization to ${\mathcal{O}}_{K}$ that sends $f(x)$ to an Eisenstein polynomial with linear term [math] and $\beta$ to [math]. Then Capelli’s Lemma implies $f(x)-\gamma$ is irreducible over $K_{\beta,t}(\gamma)$. Therefore, $\operatorname{Gal}(M_{\gamma}/K_{\beta,t}(\gamma))$ isomorphic to a transitive subgroup of $S_{d}$. Also, by restricting the action of the transposition from Proposition 4.2 to $M_{\gamma}$, we see $\operatorname{Gal}(M_{\gamma}/K_{\beta,t}(\gamma))$ contains a transposition. This is enough to conclude the result if $d=2$, $d=3$, or $d=5$ (or more generally if $d$ is prime).

When $d=4$ or $d\geq 6$, we show there is a specialization of $f(x)$ satisfying the hypotheses of Lemma 3.4 with $m$ satisfying the hypotheses of Lemma 3.5. The result then follows from Lemma 3.3.

We choose a specialization of $\beta_{1},\dots,\beta_{d-1},t$ to $K$ as follows. If $d$ is even, choose $m=d-1$, ${\mathfrak{p}}$ any prime of ${\mathcal{O}}_{K}$, $b={\mathfrak{q}}/{\mathfrak{p}}$ where ${\mathfrak{q}}$ is any other prime of ${\mathcal{O}}_{K}$, and $x_{0}={\mathfrak{p}}{\mathfrak{q}}$. If $d$ is odd, choose $m=d-2$, ${\mathfrak{p}}$ any prime of ${\mathcal{O}}_{K}$ not dividing $2$, $b={\mathfrak{q}}/{\mathfrak{p}}^{2}$ where ${\mathfrak{q}}$ is any other prime of ${\mathcal{O}}_{K}$, and $x_{0}={\mathfrak{q}}{\mathfrak{p}}^{2}$. Specialize $\beta=f^{n-1}(\gamma)$ to [math], $t$ to $-x_{0}$ and $f(x)$ to $x^{d}-bx^{m}-x_{0}$. In either case, $f(x)$ is Eisenstein at ${\mathfrak{q}}$ and hence so is $f^{m}(x)$ for all $m$ and so $f^{n-1}(x)$ is irreducible over $K$. Further, in either case the other hypotheses of Lemma 3.4 and 3.5 are satisfied, as desired. ∎

Proposition 4.4**.**

Fix $n\geq 1$ and let $\gamma\in(f^{\prime}\circ f^{n-1})^{-1}(0)$. Then

[TABLE]

Proof.

We now show we can extend our result from Proposition 4.3 to the base field $\hat{M_{\gamma}}$. Note, $\operatorname{Gal}(K_{\beta,t}(\cup_{j}f^{-{n}}(\beta_{j}))/\hat{M_{\gamma}})\cong\operatorname{Gal}(M_{\gamma}/M_{\gamma}\cap\hat{M_{\gamma}})$ so it suffices to show $\operatorname{Gal}(M_{\gamma}/M_{\gamma}\cap\hat{M_{\gamma}})\cong S_{d}$. Since $\hat{M_{\gamma}}$ is a Galois extension of $K_{\beta,t}(\gamma)$, $M_{\gamma}\cap\hat{M_{\gamma}}$ is also a Galois extension of $K_{\beta,t}(\gamma)$, so $\operatorname{Gal}(M_{\gamma}/M_{\gamma}\cap\hat{M_{\gamma}})$ is a normal subgroup of $S_{d}$. It also contains a transposition, namely the restriction of the automorphism from Proposition 4.2 to $M_{\gamma}$. Therefore $\operatorname{Gal}(M_{\gamma}/M_{\gamma}\cap\hat{M_{\gamma}})\cong S_{d}$ as desired. ∎

We are now ready to complete the proof of Theorem 1.1 by proving Proposition 1.2.

Proof of Proposition 1.2.

Let $K_{n}=K_{s,t}\left(\left(f^{\prime}\circ f^{n}\right)^{-1}(0)\right)=K_{\beta,t}(\cup_{j}f^{-{n}}(\beta_{j}))$, for $n\geq 0$. Note, with this notation $K_{0}=K_{\beta,t}$. Proposition 4.4 implies for all $n\geq 1$,

[TABLE]

As we observed at the beginning of the section, we have

[TABLE]

Thus by induction on $n$, we have

[TABLE]

Since we know $\operatorname{Gal}\left(K_{s,t}\left(\left(f^{\prime}\circ f^{n}\right)^{-1}(0)\right)/K_{s,t}\right)$ is isomorphic to a subgroup of $S_{d-1}[[S_{d}]^{n}]$, we have $\operatorname{Gal}\left(K_{s,t}\left(\left(f^{\prime}\circ f^{n}\right)^{-1}(0)\right)/K_{s,t}\right)\cong S_{d-1}[[S_{d}]^{n}]$. ∎

Corollary 4.5**.**

Fix integers $d,n$ with $d>1$ and $n\geq 0$. There is a Zariski dense subset ${\mathcal{H}}$ of $K^{d}$ such that for all $f(x)=x^{d}+c_{d-1}x^{d-1}+\dots+c_{0}$ with $(c_{0},\dots,c_{d-1})\in{\mathcal{H}}$

[TABLE]

Corollary 4.6**.**

Fix integers $d,n$ with $d>1$ and $n\geq 0$. There is a Zariski dense subset ${\mathcal{H}}$ of $K^{d}$ such that for all degree $d$ monic polynomials $f(x)$ with critical points $b_{1},\dots,b_{d-1}$ and constant term $c_{0}$ with $(b_{1},\dots,b_{d-1},c_{0})\in{\mathcal{H}}$

[TABLE]

Proof of Corollaries 4.5, 4.6.

The statements follow from Hilbert’s irreducibility theorem for number fields applied to the Galois resolvent of $f^{\prime}\circ f^{n}(x)$ in $K_{s,t}[x]$ and $\prod_{i}(f^{n}(x)-\beta_{i})$ in $K_{\beta,t}[x]$ respectively (see for example [Odo85, Lemma 6.1]). Note if the critical points of $f(x)$ are defined over $K$ as in Corollary 4.6, the polynomial $f^{\prime}\circ f^{n}(x)$ factors as $\prod_{i}(f^{n}(x)-b_{i})$ over $K$ and is a specialization of $\prod_{i}(f^{n}(x)-\beta_{i})\in K_{\beta,t}[x]$. ∎

5. Attracting Periodic Points and the Dynamical Mordell-Lang Conjecture

We now apply the results of the previous section to get a bound on the proportion of primes for which a polynomial $f(x)\in K[x]$ with critical points in $K$ has a ${\mathfrak{p}}$-adic attracting periodic point. We say a periodic point $P\in K_{\mathfrak{p}}$ of exact period $n$ is attracting if $|(f^{n})^{\prime}(P)|_{\mathfrak{p}}<1$. Note, this is equivalent to the existence of a neighborhood $U$ of $P$ such that $\lim_{m\rightarrow\infty}|f^{nm}(Q)-P|_{\mathfrak{p}}=0$ for all $Q\in U$ (see [RL03] or [Ben19]). For a group $G$ acting on a set $X$ we define the fixed point proportion of $G$, denoted $\operatorname{FPP}(G)$ to be $|\{g\in G:g(x)=x\text{ for some }x\in X\}|$.

Theorem 5.1**.**

Let $K$ be a number field and $f(x)\in K[x]$ such that all finite critical points of $f(x)$ are defined over $K$. Then for any positive integer $m$, the density of the set of primes ${\mathfrak{p}}$ for which $f$ has a finite attracting periodic point in $\bar{K_{\mathfrak{p}}}$ is bounded above by $\operatorname{FPP}\left(\operatorname{Gal}(K((f^{\prime}\circ f^{m})^{-1}(0))/K)\right)$, the proportion of elements of the Galois group whose action on the roots of $f^{\prime}\circ f^{m}(x)$ has a fixed point.

Proof.

Let $S$ be the set of primes of good reduction for $f(x)$ such that $f$ has a finite attracting periodic point in $\bar{K}_{\mathfrak{p}}$. Let $\delta(S)$ denote the natural density of $S$. Since there are only finitely many primes of bad reduction it suffices to show $\delta(S)$ is bounded above by $\operatorname{FPP}(\operatorname{Gal}(K((f^{\prime}\circ f^{m})^{-1}(0))/K))$.

We first show a prime if a prime ${\mathfrak{p}}$ of $K$ is in $S$, then the reduction of $f(x)\mod{\mathfrak{p}}$ has a periodic critical point. Suppose $\bar{K_{\mathfrak{p}}}$ is a periodic point of $f$ with exact period $n$. Let $K^{\prime}$ be a finite extension of $K_{\mathfrak{p}}$ containing $P$ and let ${\mathfrak{P}}$ be the prime of $K^{\prime}$ extending ${\mathfrak{p}}$. Then $|(f^{n})^{\prime}(P)|_{\mathfrak{P}}<1$, or equivalently $(f^{n})^{\prime}(P)\equiv 0\mod{\mathfrak{P}}$. Then using the chain rule we have,

[TABLE]

This implies that the orbit of $P\mod{\mathfrak{P}}$ contains a critical point of $f(x)\mod{\mathfrak{P}}$. Since the critical points of $f(x)$ are all defined over $K$, each critical point of $f(x)\mod{\mathfrak{P}}$ is the reduction of a critical point of $f(x)$ and hence there is critical point $\beta\in K$ of $f(x)$ such that $\beta\mod{\mathfrak{p}}$ is a periodic point of $f(x)\mod{\mathfrak{p}}$.

Suppose ${\mathfrak{p}}\in S$, then some critical point $\beta$ of $f(x)$ is periodic modulo ${\mathfrak{p}}$, so there is some $n$ such that $f^{n}(\beta)\equiv\beta\mod{\mathfrak{p}}$. For any positive integer $m$, let $k$ be an integer such that $nk\geq m$. Then, $f^{m}(f^{nk-m}(\beta))\equiv f^{nk}(\beta)\equiv\beta\mod{\mathfrak{p}}$. Hence $f^{nk-m}(\beta)\in K$ is a solution to the congruence$(f^{\prime}\circ f^{m})(x)=\prod_{f^{\prime}(\beta)=0}f^{m}(x)-\beta\equiv 0\mod{\mathfrak{p}}$. Thus, the density of $S$ is bounded above by the density of primes of good reduction for which $f^{\prime}(f^{m}(x))\equiv 0\mod{\mathfrak{p}}$ has a solution for any $m$.

With $m$ fixed, we split $S$ into two sets $S_{1}=\{{\mathfrak{p}}\in S:{\mathfrak{p}}|\Delta(f^{\prime}\circ f^{m}(x))\}$ which contains all primes of $S$ which ramify in $K((f^{\prime}\circ f^{m})^{-1}(0))$ and $S_{2}=S\setminus S_{1}$. The set $S_{1}$ is finite, hence it suffices to show $S_{2}$ has density bounded above by $\operatorname{FPP}(\operatorname{Gal}(K((f^{\prime}\circ f^{m})^{-1}(0))/K))$. If ${\mathfrak{p}}\in S_{2}$, then it does not ramify in $K_{m}:=K((f^{\prime}\circ f^{m})^{-1}(0))$. Let $\operatorname{Frob}\left(\frac{K_{m}/K}{{\mathfrak{p}}}\right)$ denote the Frobenius automorphism. Since ${\mathfrak{p}}\in S$, $f^{\prime}(f^{m}(x))\equiv 0\mod{\mathfrak{p}}$ has a linear factor. This implies $\operatorname{Frob}\left(\frac{K_{m}/K}{{\mathfrak{p}}}\right)$ fixes some root of $f^{\prime}\circ f^{m}(x)$. Hence Frobenius automorphisms for ${\mathfrak{p}}\in S_{2}$ are contained in the set of elements of $\operatorname{Gal}(K((f^{\prime}\circ f^{m})^{-1}(0))/K)$ with at least one fixed point. Then by the Chebotarev density theorem (see [FJ08, Theorem 6.3.1]), $\delta(S)\leq\operatorname{FPP}(\operatorname{Gal}(K((f^{\prime}\circ f^{m})^{-1}(0))/K))$ as desired. ∎

Remark 5.2*.*

For $f(x)$ with all critical points in $K$, the polynomial $f^{\prime}\circ f^{m}(x)$ factors as $\prod_{f^{\prime}(\beta)=0}f^{m}(x)-\beta$. So we have $\operatorname{Gal}(K((f^{\prime}\circ f^{m})^{-1}(0))/K)\subseteq\prod_{d-1}[S_{d}]^{m}$, where $d=\deg f$.

Example 5.3*.*

As an initial example, consider the case where $f(x)$ has a superattracting point $P$ in $K$ of exact period $n$. Then $|(f^{n})^{\prime}(P)|=0$, which implies some critical point $\beta$ of $f(x)$ is periodic. Then $\beta$ is a root of $f^{\prime}\circ f^{m}(x)$ for infinitely many $m$ and since $\beta$ is in the base field $K$, every element of $\operatorname{Gal}(K((f^{\prime}\circ f^{m})^{-1}(0))/K)$ has a fixed point. That is $\operatorname{FPP}(\operatorname{Gal}(K((f^{\prime}\circ f^{m})^{-1}(0))/K))=1$, which is what we expect since $P$ is a ${\mathfrak{p}}$-adic attracting point for all primes ${\mathfrak{p}}$.

Example 5.4*.*

Consider $f(x)=x^{3}+5$ and $K={\mathbb{Q}}$. In this case $f(x)$ has one finite critical point [math], which is not periodic. The map $f:{\mathbb{Q}}_{p}\rightarrow{\mathbb{Q}}_{p}$, has a finite attracting periodic point if and only if [math] is periodic for the map $f:{\mathbb{F}}_{p}\rightarrow{\mathbb{F}}_{p}$. It was shown in [HJM15] that the proportion of primes of ${\mathbb{Z}}$ for which this holds is $\frac{1}{2}=\operatorname{FPP}(\operatorname{Gal}({\mathbb{Q}}(f^{-n}(0))/{\mathbb{Q}})$. The set contains all the $2\mod 3$ primes where the cubing map is a bijection and a set of $1\mod 3$ primes of density [math]. On the other hand if we extend the base field to $K={\mathbb{Q}}(\zeta_{3})$, then $\operatorname{FPP}(\operatorname{Gal}(K(f^{-n}(0))/K))\rightarrow 0$ and $n\rightarrow\infty$ and hence the density of primes for which $K_{\mathfrak{p}}$ has a ${\mathfrak{p}}$-adic attracting point is [math].

In each of these examples $\operatorname{Gal}(K((f^{\prime}\circ f^{n})^{-1}(0))/K)$ is a small subset of $\prod_{d-1}[S_{d}]^{n}$. However, by Corollary 4.6, “most” polynomials $f(x)$ with critical points in $K$ have $\operatorname{Gal}(K((f^{\prime}\circ f^{n})^{-1}(0))/K)\cong\prod_{d-1}[S_{d}]^{n}$. For example, one can show $f(x)=x^{2}+1$ is such an example by [Odo85] and [Sto92].

Proof of Theorem 1.3.

First note

[TABLE]

Thus,

[TABLE]

By [Odo85, Lemma 4.3], $\operatorname{FPP}([S_{d}]^{n})\rightarrow 0$ as $n\rightarrow\infty$ and hence $1-\left(1-\operatorname{FPP}\left([S_{d}]^{n}\right)\right)^{d-1}\rightarrow 0$ as $n\rightarrow\infty$ as well. We can make this more precise, using [Juu21, Proposition 4.5], which says $\operatorname{FPP}([S_{d}]^{n})\leq\frac{2}{n+2}$. Then, using the fact that $(n+2)^{d-1}\leq n^{d-1}+C_{d}n^{d-2}$ for some positive constant $C_{d}$, we have

[TABLE]

Choose $n_{0}$ so that $\operatorname{FPP}\left(\prod_{d-1}[S_{d}]^{n}\right)<\epsilon$ then by Corollary 4.6, there is a Zariski dense subset ${\mathcal{H}}$ of $K^{d}$ such that for all monic polynomials $f(x)$ with critical points $b_{1},\dots,b_{d-1}$ and constant term $c_{0}$ with $(b_{1},\dots,b_{d-1},c_{0})\in{\mathcal{H}}$

[TABLE]

The result then follows from Theorem 5.1. ∎

Finally, we turn to the connection of $p$-adic attracting periodic points to the Dynamical Mordell-Lang Conjecture. The proof of Theorem 1.6 follows quickly from the following theorem of Benedetto, Ghioca, Kurlberg, and Tucker.

Theorem 5.5** (Theorem 3.4, [BGKT12]).**

Let $V$ be a subvariety of $({\mathbb{P}}^{1})^{g}$ defined over ${\mathbb{C}}_{p}$, let $f_{1},\dots,f_{g}\in{\mathbb{C}}_{p}(t)$ be rational functions of good reduction on ${\mathbb{P}}^{1}$, and let $\Phi$ denote the coordinatewise action of $(f_{1},\dots,f_{g})$ on $({\mathbb{P}}^{1})^{g}$. Let ${\mathcal{O}}$ be the $\Phi$-orbit of a point $\alpha=(x_{1},\dots,x_{g})\in({\mathbb{P}}^{1}({\mathbb{C}}_{p}))^{g}$, and suppose that for each $i$, the orbit ${\mathcal{O}}_{f_{i}}(x_{i})$ does not intersect the residue class of any attracting $f_{i}$-periodic point. Then $V({\mathbb{C}}_{p})\cap{\mathcal{O}}$ is a union of at most finitely many orbits of the form $\{\Phi^{nk+\ell}(\alpha)\}_{n\geq 0}$ for nonnegative integers $k$ and $\ell$.

Proof of Theorem 1.6.

Let $\epsilon=\frac{1}{g}$. By Theorem 1.3, for each $d_{i}$ we can find a Zariski dense subset ${\mathcal{H}}(g,d_{i},K)$ of $K^{d}$ such that for all monic polynomials $f(x)$ with critical points $b_{1},\dots,b_{d_{i}-1}$ and constant term $c_{0}$ with $(b_{1},\dots,b_{d_{i}-1},c_{0})\in{\mathcal{H}}(g,d_{i},K)$, the density of the set of primes ${\mathfrak{p}}$ of $K$ for which $f(x)$ has a finite attracting periodic point in $\bar{K}_{\mathfrak{p}}$ is less than $\epsilon$.

Suppose $\Phi=(f_{1},\dots,f_{g})$, where $f_{i}(x)$ is a monic polynomial with critical points $b_{i,1},\dots,b_{i,d-1}$ and constant term $c_{i,0}$ and $(b_{i,1},\dots,b_{i,d-1},c_{0})\in{\mathcal{H}}(g,d_{i},K)$. Let $S_{i}$ be the set of primes for which $f_{i}$ has an attracting periodic point in $K_{\mathfrak{p}}$, then $\delta(S_{i})<\epsilon$. Let $S$ be the union of the $S_{i}$ and the finitely many primes for which some $f_{i}$ has bad reduction, then $\delta(S)\leq\sum_{g}\delta(S_{i})<g\epsilon=1$. Thus, the set of primes for which each $f_{i}$ has good reduction and none of the $f_{i}$ have an attracting periodic point in $\bar{K_{\mathfrak{p}}}$ has positive density and so contains at least one prime. The result now follows from Theorem 5.5. ∎