Proof. Let $\phi(x) = x^d + c$ . Since $\alpha$ is preperiodic for $\phi$ , there exist integers $m < n$ such that $\phi^m(\alpha) = \phi^n(\alpha)$ . If $v(c) \ge 0$ , then $\phi^n(x) - \phi^m(x)$ is monic with v-integral coefficients, so $v(\alpha) \ge 0$ as well. Now suppose that $v(c) < 0$ . If $v(\alpha) < {v(c)}/{d}$ , then

\begin{align*} v(\phi(\alpha)) = v(\alpha^d + c) = dv(\alpha) < v(\alpha). \end{align*}

By induction, $v(\phi^n(\alpha)) = d^nv(\alpha) \to -\infty$ , so $\alpha$ cannot be preperiodic. On the other hand, if $v(\alpha) > {v(c)}/{d}$ , then

\begin{align*} v(\phi(\alpha)) = v(\alpha^d + c) = v(c) < \frac{v(c)}{d}. \end{align*}

By the previous case, $\phi(\alpha)$ cannot be preperiodic, hence the same is true for $\alpha$ .

Proof. For each $i = 1,\ldots,s$ , let $\phi_i(x) = x^{d_i} + c_i$ . Let $\alpha \in \textrm{Orb}_S(0)$ be a finite orbit point for $S = \{\phi_1,\ldots,\phi_s\}$ . Suppose for contradiction that there is some valuation v on K for which at least one of the valuations $v(c_i)$ is negative. Since $\alpha$ is a finite-orbit point, $\alpha$ is preperiodic for each $\phi_i$ , so we also have $v(\alpha) < 0$ by Lemma 3·1. (Note that this implies $\alpha \ne 0$ .) More precisely, we have

\begin{align*} v(c_i) = d_i v(\alpha) \textrm{ for all } i = 1,\ldots,s. \end{align*}

Now let $\gamma = (\phi_{i_1}, \phi_{i_2}, \ldots )$ be any element of $\Phi_S$ . We claim that

\begin{align*} v(\gamma_n(0)) = d_{i_1} \cdots d_{i_n} v(\alpha) \end{align*}

for all $n \ge 1$ . The conclusion of the proposition now follows from the claim: Indeed, since we assumed $\alpha$ was in the orbit of 0, we have $\alpha = \gamma_n(0)$ for some $\gamma \in \Phi_S$ and $n \ge 1$ . But then $v(\alpha) = d_{i_1}\cdots d_{i_n} v(\alpha)$ , contradicting the fact that $v(\alpha) \ne 0$ and $d_i \ge 2$ for all $i = 1,\ldots,s$ .

It remains to prove the claim, which we do by induction on n. For $n = 1$ , we have

\begin{align*} v(\gamma_1(0)) = v(\phi_{i_1}(0)) = v(c_{i_1}) = d_{i_1}v(\alpha) \end{align*}

by Lemma 3·1. Now, for $n > 1$ , we write

\begin{align*} v(\gamma_n(0)) = v\big((\phi_{i_1} \circ \cdots \circ \phi_{i_n})(0)\big) = v\big((\phi_{i_2} \circ \cdots \circ \phi_{i_n})(0)^{d_{i_1}} + c_{i_1}\big). \end{align*}

By our induction hypothesis, we have

\begin{align*} v\big((\phi_{i_2} \circ \cdots \circ \phi_{i_n})(0)\big) = d_{i_2} \cdots d_{i_n} v(\alpha). \end{align*}

Since $v(\alpha) < 0$ and $d_i > 2$ for each $i = 1,\ldots,s$ , we have

\begin{align*} v\big((\phi_{i_2} \circ \cdots \circ \phi_{i_n})(0)^{d_{i_1}}\big) = d_{i_1} \cdot d_{i_2} \cdots d_{i_n} v(\alpha) < d_{i_1} v(\alpha) = v(c_{i_1}), \end{align*}

from which it follows that

\begin{align*} v(\gamma_n(0)) = \min\left\{v\big((\phi_{i_2} \circ \cdots \circ \phi_{i_n})(0)^{d_{i_1}}\big), v(c_{i_1})\right\} = d_{i_1} \cdots d_{i_n} v(\alpha). \end{align*}

Proof. Suppose that $K/k$ has transcendence degree 1; the general case follows by induction. Since every place of K is nonarchimedean, Proposition 3·2 tells us that $v(c_i) \ge 0$ for every place of K and every $i = 1,\ldots,s$ . But by the product formula, this implies that $v(c_i) = 0$ for every place of K. Hence, each $c_i$ is a constant.

Proof. Let $S=\{x^2+c_1, x^2+c_2\}$ for distinct $c_i\in\mathbb{Z}$ and assume that $P\in\mathbb{Q}$ is a finite orbit point for S. Then in particular, P is a preperiodic point for both $\phi_1=x^2+c_1$ and $\phi_2=x^2+c_2$ . Hence, [ Reference MortonMor92 , theorem 9] and [ Reference SilvermanSil07 , exercise 2·20] together imply that P enters a 1 or 2-cycle for both $\phi_1$ and $\phi_2$ (meaning that there is an integer $n_i$ such that $\phi_i^{n_i}(P)$ is a fixed point or a periodic point of exact period 2 for $\phi_i$ ). From here, we proceed in cases:

Case (1): P enters a fixed point for both maps. In particular, both $\phi_1$ and $\phi_2$ have rational fixed points, and (after replacing P with $\phi_1^{n_1}(P)$ for some $n_1$ ) we may assume that a fixed point for $\phi_1$ has finite orbit under S. Hence, the tuple $(c_1,c_2,P)$ satisfies the hypotheses of [ Reference HindesHin19 , lemma 2·2], and therefore the pair $(c_1,c_2)\in\mathbb{Z}\times\mathbb{Z}$ must be of the form

(3) \begin{equation}(c_1,c_2)=\bigg(\frac{1-y^2}{4}, \frac{1-(y+2)^2}{4}\bigg)\;\;\; \textrm{or}\;\;\; (c_1,c_2)=\bigg(\frac{t^4-18t^2+1}{4(t^2-1)^2}, \frac{-3t^4-10t^2-3}{4(t^2-1)^2}\bigg)\end{equation}

for some $y,t\in\mathbb{Q}$ . However, in the case on the left $y\in\mathbb{Z}$ and $y\equiv{\pm1}\ ({mod}\ 4)$ since $c_1$ is integral and $\mathbb{Z}\subseteq\mathbb{Q}$ is integrally closed. In particular, we recover the first family in the conclusion of Lemma 3·5. On the other hand, when $\left(c_1,c_2\right)=\left(\frac{t^4-18t^2+1}{4(t^2-1)^2}, \frac{-3t^4-10t^2-3}{4(t^2-1)^2}\right)$ , let $w=\frac{4t}{t^2-1}$ and $z=\frac{2t^2+2}{t^2-1}$ . Then, we see that

\begin{align*} c_1=\frac{1-w^2}{4},\;\; c_2=\frac{1-z^2}{4},\;\;\textrm{and}\;\; w^2-z^2=-4.\end{align*}

In particular, w and z are both integers since $c_1,c_2\in\mathbb{Z}$ and $\mathbb{Z}\subseteq\mathbb{Q}$ is integrally closed. However, it is straightforward to check that the only integral solutions to $w^2-z^2=-4$ are $w=0$ and $z=\pm{2}$ . But this restriction on $w=4t/(t^2-1)$ forces $t=0$ and $(c_1,c_2)=(1/4,-3/4)$ , contradicting our assumption that $c_1$ and $c_2$ are integers. Hence, the only integral pairs of c’s in this case are given by $(c_1,c_2)=\left(\frac{1-y^2}{4}, \frac{1-(y+2)^2}{4}\right)$ for some $y\in\mathbb{Z}$ and $y\equiv{\pm1}\ (\text{mod}\ 4)$ .

Case (2): P enters a fixed point for one map and a 2-cycle for the other. Then, without loss of generality, we may assume that P enters a fixed point for $\phi_1$ and a 2-cycle for $\phi_2$ . In particular, $\phi_1$ has a rational fixed point and $\phi_2$ has a rational point of exact period 2. Moreover, after replacing P with $\phi_1^{n_1}(P)$ for some $n_1$ , we may assume that a fixed point for $\phi_1$ has finite orbit under S. Hence, the tuple $(c_1,c_2,P)$ satisfies the hypotheses of [ Reference HindesHin19 , lemma 2·3], and therefore the pair $(c_1,c_2)\in\mathbb{Z}\times\mathbb{Z}$ must be of the form

(4) \begin{equation}(c_1,c_2)=\bigg(\frac{1-y^2}{4}, \frac{-3-y^2}{4}\bigg)\;\;\; \textrm{or}\;\;\; (c_1,c_2)=\bigg(\frac{-15t^4-2t^2+1}{4(t^2-1)^2}, \frac{-3t^4-10t^2-3}{4(t^2-1)^2}\bigg)\end{equation}

for some $y,t\in\mathbb{Q}$ . However, by a similar argument to that given in Case (1), only the left parametrisation produces integral c-values. Moreover, $y\in\mathbb{Z}$ and $y\equiv{\pm1}\ (\text{mod}\ 4)$ in that case.

Case (3): P enters a 2-cycle both maps. In particular, both $\phi_1$ and $\phi_2$ have rational points of exact period 2, and (after replacing P with $\phi_1^{n_1}(P)$ for some $n_1$ ) we may assume that a rational point of exact period 2 for $\phi_1$ has finite orbit under S. Hence, the tuple $(c_1,c_2,P)$ satisfies the hypotheses of [ Reference HindesHin19 , lemma 2·4], and therefore the pair $(c_1,c_2)\in\mathbb{Z}\times\mathbb{Z}$ must be of the form

(5) \begin{equation}(c_1,c_2)=\bigg(\frac{-7t^4-2t^2-7}{4(t^2-1)^2}, \frac{-3t^4-10t^2-3}{4(t^2-1)^2}\bigg)\end{equation}

for some $t\in\mathbb{Q}$ . However, by a similar argument to that given in Case (1), one can show that there are no integral c-values produced by this parametrisation.

Proof. We begin with some notation. For $n\geq1$ define $M_{S,n}=\{\theta_1\circ\dots\circ \theta_n\,:\, \theta_i\in S\}$ and $M_{S,0}=\{\textrm{id}\}$ . Likewise, let $M_{S,n}(0)=\{f(0)\,:\, f\in M_{S,n}\}$ , let $U_n=\max\{|a|\,:\, a\in M_{S,n}(0)\}$ , and let $L_n=\min\{|a|\,:\, a\in M_{S,n}(0)\}$ . Now assume that (6) holds. We prove that

(7) \begin{equation}U_n<L_{n+1}\;\textrm{for $n\geq0$.}\end{equation}

by induction. Note first that the statement above is true for $n=0$ since (6) implies that none of the $c_i$ ’s is 0. Moreover, (7) is exactly (6) for $n=1$ . Now suppose that (6) is true for all $n\leq N$ with $N\geq1$ and let $\ell\in M_{S,N+2}(0)$ be such that $|\ell|=L_{N+2}$ and let $u\in M_{S,N+1}(0)$ be such that $|u|=U_{N+1}$ . Next write $\ell=a^2+c_i$ for some $a\in M_{S,N+1}(0)$ . Then since $|a|\geq L_{N+1}\geq U_N+1$ , we have that

(8) \begin{equation}|\ell|\geq\ell\geq U_N^2+2U_N+1+c_i\geq U_N^2+U_N+1;\end{equation}

here the last inequality follows from the fact that $U_N\geq U_1$ by the induction hypothesis (and that $N\geq1$ ) and that $U_1=\max_{1\leq i\leq s}\{|c_i|\}$ . On the other hand, we may write $u=b^2+c_j$ for some $b\in M_{S,N}(0)$ . Then, since $b\leq U_N$ and $c_j\leq U_1\leq U_N$ , we see that

(9) \begin{equation}|u|\leq b^2+|c_j|\leq U_N^2+U_N.\end{equation}

In particular, (7) follows from combining (8) and (9). But then $\{L_n\}$ is a strictly increasing sequence of integers. Hence, for all B there exists $m=m(B)$ such that $|F(0)|>B$ for all $F\in M_{S,n}$ with $n\geq m$ . This precludes the possibility of $\textrm{Orb}_S(0)$ containing a finite orbit point: if g(0) is a finite orbit point, then $|f(g(0))|\leq B$ for some B and all $f\in M_S$ , a contradiction.

Proof. Let $S=\{x^2+c_1,\dots,x^2+c_s\}$ for some $c_i\in\mathbb{Q}$ and suppose that $\textrm{Orb}_S(0)$ contains a finite orbit point for S. Then Corollary 3·3 implies that each $c_i\in\mathbb{Z}$ and Theorem 3·6 implies that $\#S\leq2$ . If $\#S=1$ , then write $S=\{\phi\}$ . But in this case, if $\textrm{Orb}_S(0)$ contains a finite orbit point, then 0 itself is a finite orbit point for $\phi$ . Hence, $\phi$ is a post-critically finite (PCF) map of the form $\phi=x^2+c$ and $c\in\mathbb{Z}$ . However, it is well known that the only c with this property are $c=0$ , $-1$ , and $-2$ . That is, $S=\{x^2\}$ , $\{x^2-1\}$ , and $\{x^2-2\}$ are the only singleton sets (of the desired form) for which $\textrm{Orb}_S(0)$ contains a finite orbit point.

It therefore remains to consider the case when $\#S=2$ , say $S=\{x^2+c_1,x^2+c_2\}$ for some distinct $c_i\in\mathbb{Z}$ . Now, since $\textrm{Orb}_S(0)$ contains a finite orbit point for S, Lemma 3·5 implies

\begin{align*} (c_1,c_2)=\Big(\frac{1-y^2}{4}, \frac{1-(y+2)^2}{4}\Big)\;\;\;\textrm{or}\;\;\;(c_1,c_2)=\Big(\frac{1-y^2}{4}, \frac{-3-y^2}{4}\Big) \end{align*}

for some $y\in\mathbb{Z}$ and $y\equiv{\pm1}\ ({mod}\ 4)$ , up to reordering the c’s. Suppose first, without loss of generality, that $(c_1,c_2)=(({1-y^2})/{4}, ({1-(y+2)^2})/{4})$ . Then, after substituting these expressions in for $c_1$ and $c_2$ into Remark 7, we see that at least one of the following inequalities must hold:

\begin{align*}\Big|\frac{1}{16}y^4 - \frac{3}{8}y^2 + \frac{5}{16}\Big|&\leq \Big|-\frac{1}{4}y^2 + \frac{1}{4}\Big|+\Big|-\frac{1}{4}y^2 - y - \frac{3}{4}\Big|, \\[5pt]\Big|\frac{1}{16}y^4 - \frac{3}{8}y^2 - y - \frac{11}{16}\Big|&\leq \Big|-\frac{1}{4}y^2 + \frac{1}{4}\Big|+\Big|-\frac{1}{4}y^2 - y - \frac{3}{4}\Big|, \\[5pt]\Big|\frac{1}{16}y^4 + \frac{1}{2}y^3 + \frac{9}{8}y^2 + \frac{3}{2}y + \frac{13}{16}\Big|&\leq \Big|-\frac{1}{4}y^2 + \frac{1}{4}\Big|+\Big|-\frac{1}{4}y^2 - y - \frac{3}{4}\Big|, \\[5pt]\Big|\frac{1}{16}y^4 + \frac{1}{2}y^3 + \frac{9}{8}y^2 + \frac{1}{2}y - \frac{3}{16} \Big|&\leq \Big|-\frac{1}{4}y^2 + \frac{1}{4}\Big|+\Big|-\frac{1}{4}y^2 - y - \frac{3}{4}\Big|. \\\end{align*}

But each of these inequalities is true only on some bounded, real interval. Moreover, since we have only a single real parameter y, it is a one-variable calculus problem to determine each of these intervals. In particular, it is straightforward to check that as a real number $y\in[-6.8,4.7]$ , otherwise all of the above inequalities fail. On the other hand, $y\in\mathbb{Z}$ and $y\equiv{\pm1}\ ({mod}\ 4)$ so that $y\in\{-5,-3,-1,1,3\}$ . These specific values of y determine the sets $S=\{x^2,x^2-2\}$ and $S=\{x^2-2,x^2-6\}$ . Moreover, among these sets, only $S=\{x^2-2,x^2-6\}$ has the desired property that $\textrm{Orb}_S(0)$ contains a finite orbit point. In this case, $-2\in\textrm{Orb}_S(0)$ and $-2$ is a finite orbit point for S.

Now for the second family from Lemma 3·5. Suppose, without loss of generality, that $(c_1,c_2)=(({1-y^2})/{4}, ({-3-y^2})/{4})$ for some $y\in\mathbb{Z}$ and $y\equiv{\pm1}\ ({mod}\ 4)$ . Then, after substituting these expressions in for $c_1$ and $c_2$ into Remark 7, we see that at least one of the following inequalities must hold:

\begin{align*}\Big|\frac{1}{16}y^4 - \frac{3}{8}y^2 + \frac{5}{16}\Big|&\leq \Big| -\frac{1}{4}y^2 + \frac{1}{4}\Big|+\Big| -\frac{1}{4}y^2 - \frac{3}{4}\Big|, \\[5pt]\Big| \frac{1}{16}y^4 - \frac{3}{8}y^2 - \frac{11}{16} \Big|&\leq \Big| -\frac{1}{4}y^2 + \frac{1}{4}\Big|+\Big| -\frac{1}{4}y^2 - \frac{3}{4}\Big| , \\[5pt]\Big| \frac{1}{16}y^4 + \frac{1}{8}y^2 + \frac{13}{16} \Big|&\leq \Big| -\frac{1}{4}y^2 + \frac{1}{4}\Big|+\Big| -\frac{1}{4}y^2 - \frac{3}{4}\Big| , \\[5pt]\Big| \frac{1}{16}y^4 + \frac{1}{8}y^2 - \frac{3}{16} \Big|&\leq \Big| -\frac{1}{4}y^2 + \frac{1}{4}\Big|+\Big| -\frac{1}{4}y^2 - \frac{3}{4}\Big|. \\\end{align*}

But, as before, each of these inequalities is true only on some bounded, real interval. Hence, it is straightforward to check that as a real number $y\in[-4.9,4.9]$ , otherwise all of the above inequalities fail. On the other hand, $y\in\mathbb{Z}$ and $y\equiv{\pm1}\ ({mod}\ 4)$ so that $y\in\{-3,-1,1,3\}$ . These specific values of y determine the sets $S=\{x^2,x^2-1\}$ and $S= \{x^2-2,x^2-3\}$ . Moreover, both of these sets have the desired property that $\textrm{Orb}_S(0)$ contains a finite orbit point. In the first case, 0 is itself a finite orbit point. While in the second case, $-2\in\textrm{Orb}_S(0)$ is a finite orbit point. This completes the classification in Theorem 3·8.

Proof. We begin with some notation. Write $\phi_1=x^2-2$ and $\phi_2=x^2-3$ , and let $F=x^2$ and $L=x+1$ . Then clearly $\phi_1\equiv F\ ({mod}\ 2)$ , so that $\phi_1$ and $\phi_2$ must commute mod 2; every polynomial commutes with F mod 2. In particular, if $f\in M_S$ , then we can write $f\equiv \phi_1^n\circ\phi_2^m\ ({mod}\ 2)$ for some $n,m\geq0$ . On the other hand, $\phi_2\equiv (x+1)^2\equiv F\circ L\ ({mod}\ 2)$ and $L\circ L\equiv x\ ({mod}\ 2)$ . Therefore, every $f\in M_S$ is of the form

(10) \begin{equation}f\equiv F^n\ ({mod}\ 2)\qquad \textrm{or} \qquad f\equiv F^n\circ L\ ({mod}\ 2)\end{equation}

for some $n\geq0$ . From here we proceed in cases depending on the congruence class of the constant term of f modulo 4. Note that if $f\in M_S$ is not the identity, then $f(0)\not\equiv0\ ({mod}\ 4)$ , since both $x^2-2$ and $x^2-3$ have no roots modulo 4. Hence, we need not consider this case. Suppose first that $f(0)\equiv 2\ ({mod}\ 4)$ . Then, $f(0)\equiv 0\ ({mod}\ 2)$ , and (10) implies that $f\equiv F^n\ ({mod}\ 2)$ . In particular, f satisfies Eisenstein’s irreducibility criterion at the prime $p=2$ in this case. On the other hand, if $f(0)\equiv\pm{1}\ ({mod}\ 4)$ , then $f(0)\equiv 1\ ({mod}\ 2)$ and (10) implies that $f\equiv F^n\circ L\ ({mod}\ 2)$ . Therefore, $f(x+1)\equiv F^n\ ({mod}\ 2)$ . Moreover, the constant term f(1) of $f(x+1)$ is not 0 mod 4, again since both $x^2-2$ and $x^2-3$ have no roots in $\mathbb{Z}/4\mathbb{Z}$ . Therefore, $f(x+1)$ is Eisenstein at the prime $p=2$ as claimed.

Proof. Let $\phi_1=x^2-2$ , let $\phi_2=x^2-6$ , and let $F=x^2$ . Note that $\phi_1\equiv\phi_2\equiv F\ ({mod}\ 2)$ , and therefore every $f\in M_S$ is of the form $f\equiv F^{n}\ ({mod}\ 2)$ for some $n\geq0$ . Likewise, $\phi_1\equiv\phi_2\ ({mod}\ 4)$ , and hence every $f\in M_S$ is of the form $f\equiv \phi_1^{n}\ ({mod}\ 4)$ for some $n\geq0$ . Moreover, it is straightforward to check that $\phi_1(0)=-2$ and $\phi_1^n(0)=2$ for all $n\geq2$ . In particular, $f(0)\equiv\pm{2}\ ({mod}\ 4)$ for all non-identity $f\in M_S$ . Hence, such f are Eisenstein at $p=2$ . Therefore, every $f\in M_S$ is irreducible over $\mathbb{Q}$ .

Proof. It suffices to show that $z^2+c$ is square-free in $\mathbb{Z}[t]$ (meaning it has no non-constant square factor) to show it’s square-free in $\mathbb{Q}[t]$ by Gauss’ Lemma. Suppose for a contradiction that $z^2+c=y^2\cdot w$ for some non-constant $y\in\mathbb{Z}[t]$ and some $w\in\mathbb{Z}[t]$ . Note that y must have odd leading term since $z^2+c$ has odd leading term. In particular, the mod 2 reduction $\bar{y}\in\mathbb{F}_2[t]$ of y must be non-constant. Now we take the expression $z^2+c=y^2\cdot w$ , reduce it mod 2, and take the derivative of both sides in $\mathbb{F}_2[t]$ :

\begin{align*} 1=\frac{d}{dt}(\bar{c})=\frac{d}{dt}(\bar{z}^2+\bar{c})=\frac{d}{dt}(\bar{y}^2\cdot\bar{w}\,)=\bar{y}^2\cdot \frac{d}{dt}(\bar{w}).\end{align*}

Hence, $\bar{y}^2$ is a unit $\mathbb{F}_2[t]$ and is therefore constant. But this contradicts the previously established fact that $\bar{y}\in\mathbb{F}_2[t]$ is non-constant.

Proof. Write $\theta_i(x)=x^2+b_i\in S$ for each $1\leq i\leq n$ . Then

\begin{align*} \gamma_n(0)=\theta_1(\theta_2(\dots(\theta_{n-1}(\theta_n(0)))))=(((b_n^2+b_{n-1})^2+b_{n-2})^2+\dots b_2)^2+b_1. \end{align*}

Now set $z_0=b_n$ and define

(11) \begin{equation}z_{m}=\theta_{n-m}(z_{m-1})=z_{m-1}^2+b_{n-m}\end{equation}

recursively for $1\leq m\leq n-1$ . Note in particular that $\gamma_n(0)=z_{n-1}$ and it suffices to prove the claim below to prove Lemma 5·2:

Claim. $\deg(z_m)\leq d\cdot 2^{m}$ with equality if $\deg(z_0)=d$ . Moreover when $\deg(z_0)=d$ and $m\geq1$ , the leading term of $z_m$ is the square of the leading term of $z_{m-1}$ . We prove this by induction on m. The base case $m=0$ is obvious. On the other hand, if $m\geq1$ and the claim holds for $m-1$ , then (11) implies that

\begin{align*} \deg(z_m)\leq\max\{2\deg(z_{m-1}),b_{n-m}\}\leq\{2\cdot d\cdot 2^{m-1},d\}=\max\{d\cdot 2^m,d\}=d\cdot 2^m\end{align*}

as desired. Moreover, if $\deg(z_0)=d$ then $\deg(z_{m-1})=d\cdot 2^{m-1}$ by induction. Furthermore, since $2^m\cdot d>d\geq\deg(b_{n-m})$ , it follows from (11) that $\deg(z_m)=d\cdot 2^{m}$ .

Proof. Let $\theta_1(0)=c$ and write $\gamma_n(0)=z^2+c$ where $z=0$ if $n=1$ and $z=\gamma_{n-1}(0)$ if $n\geq2$ . For statement (1), suppose $\pm{\gamma_n(0)}$ is a square in $\mathbb{Q}[t]$ . Then it follows from Gauss’ Lemma that $\pm{\gamma_n(0)}=a\cdot y^2$ for some $a\in\mathbb{Z}$ and some $y\in\mathbb{Z}[t]$ . But then

\begin{align*} 1=\frac{d}{dt}(\bar{z}^2+\bar{c})=\frac{d}{dt}(\overline{\gamma_n(0)})=\frac{d}{dt}(\overline{\pm\gamma_n(0)})=\frac{d}{dt}(\overline{a\cdot y^2})=a\frac{d}{dt}(\bar{y}^2)=0,\end{align*}

a contradiction. Therefore, $\pm{\gamma_n(0)}$ is not a square in $\mathbb{Q}[t]$ for all $n\geq1$ .

As for the second statement, note that Lemma 5·2 implies that the leading term of $\gamma_n(0)$ is odd: it’s a power of the odd leading term of $\theta_n(0)$ . Hence, Lemma 5·1 applied to $\gamma_n(0)=z^2+c$ implies that $\gamma_n(0)$ is square-free in $\mathbb{Q}[t]$ as claimed.

Proof. Since $\gamma_n(0)$ is square-free, $\gamma_n(0)=p_1\cdots p_t$ for some coprime, irreducible polynomials $p_i\in\mathbb{Q}[t]$ . In particular, if the conclusion of Lemma 5·4 is false, then it must be the case that each $p_i\big\vert\gamma_{r_i}(0)$ for some $1\leq r_i\leq n-1$ . Therefore, $\gamma_n(0)\big\vert \gamma_1(0)\cdots \gamma_{n-1}(0)$ . However, it then follows from Lemma 5·2 that $d\cdot2^{n-1}\leq d\cdot(2^{n-1}-1)$ , a contradiction.

Finally, putting Lemmas 5·3 and 5·4 together with Theorem 2·3, we obtain the maximality criterion for sets from the Introduction (which we restate for convenience).

Proof. Suppose $\gamma=(\theta_n)_{n\geq1}\in\Phi_S$ satisfies $\theta_1=\phi_j$ and that a particular index n satisfies $\theta_n=\phi_k$ . Then first, since $\theta_1(0)=c_j$ and $\frac{d}{dt}(\overline{c_j})=1$ in $\mathbb{F}_2[t]$ , Lemma 5·3 part 1 implies that $\pm{\gamma_m(0)}$ is not a square in $\mathbb{Q}[t]$ for all $m\geq1$ . In particular, $\gamma$ is stable over $\mathbb{Q}(t)$ by Proposition 2·1. Here we use also that if $a\in\mathbb{Q}[t]$ is a square in $\mathbb{Q}(t)$ , then it is a square in $\mathbb{Q}[t]$ ; this follows from the fact that $\mathbb{Q}[t]$ is integrally closed in $\mathbb{Q}(t)$ .

As for the claim about maximality, note that Lemma 5·3 part (2) (and the assumed conditions in the theorem) imply that $\gamma_n(0)$ is square-free in $\mathbb{Q}[t]$ . In particular, it follows from Lemma 5·4 that there exists an irreducible polynomial $p\in\mathbb{Q}[t]$ such that $v_p(\gamma_n(0))=1$ and $v_p(\gamma_m(0))=0$ for all $1\leq m<n$ (normally called a primitive prime divisor of $\gamma_n(0)$ appearing to odd valuation). Finally, Theorem 2·3 implies that $K_n(\gamma)/K_{n-1}(\gamma)$ is maximal; here we use that the places of $K=\mathbb{Q}(t)$ correspond to irreducible polynomials in $\mathbb{Q}[t]$ with the usual valuations.

Proof. Let $\phi_i=x^2+c_i$ for each $c_i$ . Then by Theorem 1·4, it suffices to show that

\begin{align*} \mathcal{G}\;:\!=\;\Big\{\gamma=(\theta_n)_{n\geq1}\,:\,\textrm{$\theta_1=\phi_j$ and $\theta_n=\phi_k$ i.o.} \Big\}\end{align*}

has positive $\bar{\nu}$ measure to prove Corollary 5·5. However, note that $\mathcal{G}=\mathcal{G}_1\cap\mathcal{G}_2$ where

\begin{align*} \mathcal{G}_1\;:\!=\;\Big\{\gamma=(\theta_n)_{n\geq1}\,:\,\textrm{$\theta_1=\phi_j$}\Big\}\;\;\;\;\textrm{and}\;\;\;\;\mathcal{G}_2\;:\!=\;\Big\{\gamma=(\theta_n)_{n\geq1}\,:\,\textrm{$\theta_n=\phi_k$ i.o.}\Big\}\end{align*}

respectively. On the other hand, $\bar{\nu}(\mathcal{G}_1)=\nu(\phi_j)>0$ since $\bar{\nu}$ is the product measure of a countable number of copies of $\nu$ on S; see [ Reference Jacod and ProtterJP03 , theorem 10·4]. Moreover, since $\nu(\phi_k)>0$ , the Borel–Cantelli Theorem (specifically, the Monkey and Typewriter problem [ Reference GutGut13 , pp. 96-100]) implies that $\bar{\nu}(\mathcal{G}_2)=1$ . Therefore,

\begin{align*} \bar{\nu}(\mathcal{G})=\bar{\nu}(\mathcal{G}_1\cap\mathcal{G}_2)=\bar{\nu}(\mathcal{G}_1)=\nu(\phi_j)>0\end{align*}

as desired.

Proof. We proceed in cases depending on whether d is even or odd. The key distinction is that in the even case we may choose a single polynomial satisfying conditions (1) and (2) of Theorem 1·4. This is not possible in the odd case, unless $d=1$ .

Case (1): Assume that d is even. We begin by counting polynomials $c\in P_d(B)$ satisfying $\frac{d}{dt}(\bar{c})=1$ in $\mathbb{F}_2[t]$ , $\deg(c)=d$ , and having odd leading term (which we call property ( ^* )). This is clearly true of c if and only if:

(^*) \begin{equation}c=\sum_{i=0}^d a_it^i,\;\,\textrm{$a_d$ is odd,}\; \textrm{$a_1$ is odd},\;\textrm{and}\;\textrm{$a_i$ is even for all odd indices $3\leq i\leq d-1$}.\end{equation}

In particular, after counting even or odd integers with absolute value at most B for each of the $d/2+1$ stipulated coefficients, we see that

(12)

Here we use that $r_d= \big({1}/{2}\big)^{\frac{d}{2}+1}$ for even d. Hence, it follows from Remark 10 that there are $(1-r_d)\cdot {x_B}^{d+1}+O({x_B}^{d})$ polynomials in $P_d(B)$ that do not satisfy property ( ^* ). In particular, there are

\begin{align*} \binom{(1-r_d)\cdot x_B^{d+1}+O({x_B}^{d})}{s}\end{align*}

sets in $\mathcal{S}(d,s,B)$ whose elements all fail to satisfy property ( ^* ). Therefore, there are

(13) \begin{equation}\binom{x_B^{d+1}}{s}-\binom{(1-r_d)\cdot x_B^{d+1}+O(x_B^{d})}{s}=\frac{(1-(1-r_d)^s)}{s!}x_B^{(d+1)s}+O(x_B^{(d+1)s-1})\end{equation}

sets in $\mathcal{S}(d,s,B)$ with at least one element satisfying property ( ^* ). Here we use also that the binomial coefficient $\binom{X}{s}$ is a polynomial of degree s in X with leading coefficient $(s!)^{-1}$ ; see Remark 11.

On the other hand, Corollary 5·5 implies that such sets $\{c_1,\dots,c_s\}\in \mathcal{S}(d,s,B)$ determine sets of quadratic polynomials $S=\{x^2+c_1,\dots,x^2+c_s\}$ which furnish big arboreal representations over $K=\mathbb{Q}(t)$ with positive probability. Hence, (13) and Corollary 5·5 together imply that

\begin{align*} \#\Big\{S\in\mathcal{S}(d,s,B)\,:\,\bar{\nu}_S\big(\textrm{BigArb}(S,K)\big)>0\Big\}\geq \frac{(1-(1-r_d)^s)}{s!}x_B^{(d+1)s}+O(x_B^{(d+1)s-1}).\end{align*}

In particular, after dividing the inequality above by

\begin{align*} \#S(d,s,B)=\binom{{x_B}^{d+1}}{s}=\frac{1}{s!}\cdot{x_B}^{(d+1)s}+O\left(x_B^{(d+1)s-1}\right) \end{align*}

and letting $B\rightarrow\infty$ , we see that

\begin{align*} \liminf_{B\rightarrow\infty}\frac{\#\Big\{S\in\mathcal{S}(d,s,B)\,:\,\bar{\nu}_S\big(\textrm{BigArb}(S,K)\big)>0\Big\}}{\#\mathcal{S}(d,s,B)}\geq 1-(1-r_d)^s. \end{align*}

Therefore, statement (1) of Theorem 1·3 holds for even d and fixed s as claimed.

Before we begin the proof of the case when d is odd, we need the following complementary form of the inclusion-exclusion principle.

Case (2): Assume that d is odd. We begin by counting polynomials $c\in P_d(B)$ satisfying $\frac{d}{dt}(\bar{c})=1$ in $\mathbb{F}_2[t]$ and call this property (I). This is clearly true of c if and only if:

(I) \begin{equation}c=\sum_{i=0}^d a_it^i,\;\, \textrm{$a_1$ is odd,}\;\textrm{and}\;\textrm{$a_i$ is even for all odd indices $3\leq i\leq d$}.\end{equation}

In particular, after counting even or odd integers with absolute value at most B for each of the $(d+1)/2$ stipulated coefficients, we see that

(14)

Here we use that $r_d= \big({1}/{2}\big)^{\frac{d+1}{2}}$ for odd d. On the other hand, we say that $c\in P_d(B)$ has property (II) if it has degree d and odd leading term:

(II) \begin{equation}c=\sum_{i=0}^d a_it^d\;\,\textrm{and $a_d$ is odd.} \end{equation}

Clearly properties (I) and (II) are disjoint events (i.e., there are no polynomials satisfying both properties) by examining the leading coefficient $a_d$ alone, and

(15)

polynomials have property (II).

Now, with Corollary 5·5 and big arboreal representations in mind, we count sets in $\mathcal{S}(d,s,B)$ with at least one element satisfying property (I) and at least one element satisfying property (II). Do do this, let $\mathcal{A}\subseteq \mathcal{S}(d,s,B)$ be the sets in $\mathcal{S}(d,s,B)$ with at least one element satisfying property (I), and let $\mathcal{B}\subseteq \mathcal{S}(d,s,B)$ be the sets in $\mathcal{S}(d,s,B)$ with at least one element satisfying property (II); that is, we want to count $\#\mathcal{A}\cap \mathcal{B}$ . To do this, note first that (14) and Remark 10 imply that there are

\begin{align*} \# P_d(B)-\Big(r_d\cdot x_B^{d+1}+O(x_B^{d})\Big)=(1-r_d)\cdot x_B^{d+1}+O(x_B^{d})\end{align*}

polynomials in $P_d(B)$ that do not have property (I). Therefore, there are

\begin{align*} \binom{(1-r_d)\cdot x_B^{d+1}+O(x_B^d)}{s}\end{align*}

sets in $\mathcal{S}(d,s,B)$ whose elements all fail to satisfy property (I). Equivalently,

(16) \begin{equation}\#\mathcal{A}^c=\binom{(1-r_d)\cdot x_B^{d+1}+O(x_B^d)}{s};\end{equation}

here $\mathcal{A}^c$ denotes the complement of $\mathcal{A}$ in $\mathcal{S}(d,s,B)$ . Likewise, (15) and Remark 10 imply that there are

\begin{align*} \# P_d(B)-\Big(\frac{1}{2}\cdot x_B^{d+1}+O(x_B^{d})\Big)=\frac{1}{2}\cdot x_B^{d+1}+O(x_B^{d})\end{align*}

polynomials in $P_d(B)$ that do not have property (II). Therefore, there are

\begin{align*} \binom{\frac{1}{2}\cdot x_B^{d+1}+O(x_B^d)}{s}\end{align*}

sets in $\mathcal{S}(d,s,B)$ whose elements all fail to satisfy property (II); Equivalently,

(17) \begin{equation}\#\mathcal{B}^c=\binom{\frac{1}{2}\cdot x_B^{d+1}+O(x_B^d)}{s}.\end{equation}

Finally, before we can put all of the pieces above together, it remains to count $\#(\mathcal{A}^c\cap \mathcal{B}^c)$ . However, $\mathcal{A}^c\cap \mathcal{B}^c$ consists precisely of the sets in $\mathcal{S}(d,s,B)$ whose elements all fail to satisfy (I) and all fail to satisfy (II). But, since (I) and (II) are disjoint events, there are exactly

\begin{align*} \# P_d(B)-\Big(r_d\cdot x_B^{d+1}+O(x_B^{d})\Big)-\Big(\frac{1}{2}\cdot x_B^{d+1}+O(x_B^{d})\Big)=\big(1-r_d-\frac{1}{2}\big)\cdot x_B^{d+1}+O(x_B^{d})\end{align*}

polynomials that fail to satisfy (I) and fail to satisfy (II); here we use Remark 10, (14), and (15). Therefore, we see that

(18) \begin{equation}\#(\mathcal{A}^c\cap \mathcal{B}^c)=\binom{(1-r_d-\frac{1}{2})\cdot x_B^{d+1}+O(x_B^d)}{s}.\end{equation}

In particular, (16), (17), (18), and the complementary form of the inclusion-exclusion principle in Remark 12 applied to $\mathcal{X}=\mathcal{S}(d,s,B)$ , $\mathcal{A}$ and $\mathcal{B}$ above imply that there are

\begin{align*} {\binom{x_B^{d+1}}{s}-\binom{(1-r_d)\cdot x_B^{d+1}+O(x_B^d)}{s}-\binom{\frac{1}{2}\cdot x_B^{d+1}+O(x_B^d)}{s}+\binom{(1-r_d-\frac{1}{2})\cdot x_B^{d+1}+O(x_B^d)}{s}} \end{align*}

sets in $\mathcal{S}(d,s,B)$ with at least one element satisfying property (I) and at least one element satisfying property (II) (i.e., the expression above is $\#\mathcal{A}\cap\mathcal{B}$ ). Hence, there are

(19) \begin{equation}\frac{1-(1-r_d)^s-(\frac{1}{2})^s+(1-r_d-\frac{1}{2})^s}{s!}x_B^{(d+1)s}+O(x_B^{(d+1)s-1})\end{equation}

sets in $\mathcal{S}(d,s,B)$ with at least one element satisfying property (I) and at least one element satisfying property (II); see also Remark 11. On the other hand, Corollary 5·5 implies that such sets $\{c_1,\dots,c_s\}\in\mathcal{S}(d,s,B)$ determine sets of quadratic polynomials $S=\{x^2+c_1,\dots,x^2+c_s\}$ which furnish big arboreal representations over $K=\mathbb{Q}(t)$ with positive probability. Hence, (19) and Corollary 5·5 together imply that

\begin{align*} \#\Big\{S\in\mathcal{S}(d,s,B)\,:\,\bar{\nu}_S\big(\textrm{BigArb}(S,K)\big)>0\Big\} & \geq \frac{1-(1-r_d)^s-(\frac{1}{2})^s+(1-r_d-\frac{1}{2})^s}{s!}x_B^{(d+1)s}\\[5pt] &+O(x_B^{(d+1)s-1}) \end{align*}

when d is odd. In particular, after dividing the inequality above by

\begin{align*} \#S(d,s,B)=\binom{{x_B}^{d+1}}{s}=\frac{1}{s!}\cdot{x_B}^{(d+1)s}+O(x_B^{(d+1)s-1}) \end{align*}

and letting $B\rightarrow\infty$ , we see that

\begin{align*} \displaystyle{\liminf_{B\rightarrow\infty}}\frac{\#\Big\{S\in \mathcal{S}(d,s,B)\,:\,\bar{\nu}_S\big(\textrm{BigArb}(S,K)\big)>0\Big\}}{\#\mathcal{S}(d,s,B)}\geq 1-(1-r_{d})^s-\Big(\frac{1}{2}\Big)^s+\Big(1-r_d-\frac{1}{2}\Big)^s\end{align*}

for odd d as claimed in part (2) of Theorem 1·3.

Proof. The proof is very similar to that of statement (1) of Theorem 1·3 (only now the leading terms are fixed). In particular, it is straightforward to check that $\#M_d(B)=x_B^d$ and

(20) \begin{equation}\#\Big\{c\in M_d(B)\,:\,\textrm{$\frac{d}{dt}(\bar{c})=1$ in $\mathbb{F}_2[t]$}\Big\}=\Big(\frac{1}{2}\Big)^{\frac{d}{2}}\cdot x_B^d +O(x_B^{d-1}),\end{equation}

since the c’s above must have even coefficients at every odd-powered term, except for the linear term, which must be odd - that is, there is a parity stipulation on the coefficients of c at every odd-powered term (and $d/2$ such terms). Therefore, similar to Case (1) in the proof of Theorem 1·3, there are

(21) \begin{equation}\binom{x_B^d}{s}-\binom{\big(1-({\frac{1}{2}})^{\frac{d}{2}}\big)\cdot x_B^{d}+O({x_B}^{d-1})}{s}=\frac{\Big(1-\big(1-({\frac{1}{2}})^{\frac{d}{2}}\big)^s\Big)}{s!}\,x_B^{ds}+O(x_B^{ds-1})\end{equation}

sets in $\mathcal{S}^{\textrm{Mon}}(d,s,B)$ with at least one term $c_j\in\mathbb{Z}[t]$ satisfying $\frac{d}{dt}(\bar{c}_j)=1$ in $\mathbb{F}_2[t]$ . On the other hand, such collections determine sets of quadratic polynomials $S=\{x^2+c_1,\dots,x^2+c_s\}$ which furnish surjective arboreal representations over $\mathbb{Q}(t)$ with positive probability by Theorem 1·4: the sequences $\gamma=(\theta_n)_{n\geq1}$ with $\theta_1=x^2+c_j$ produce maximal extensions for every n since condition (2) of Theorem 1·4 is satisfied by all $c\in M_d(B)$ . Therefore, it follows from (21) that

(22) \begin{equation}\#\Big\{S\in\mathcal{S}^{\textrm{Mon}}(d,s,B) \,:\,\bar{\nu}_S\big(\textrm{SurArb}(S)\big)>0\Big\}\geq \frac{\Big(1-\big(1-({\frac{1}{2}})^{\frac{d}{2}}\big)^s\Big)}{s!}\,x_B^{ds}+O(x_B^{ds-1}).\end{equation}

In particular, after dividing both sides of (22) by

\begin{align*} \#\mathcal{S}^{\textrm{Mon}}(d,s,B)=\binom{x_B^d}{s}=\frac{1}{s!}x_B^{ds}+O(x_B^{ds-1})\end{align*}

and letting $B\rightarrow\infty$ , we obtain the lower bound

\begin{align*} \liminf_{B\rightarrow\infty}\frac{\#\Big\{S\in \mathcal{S}^{{Mon}}(d,s,B)\,:\,\bar{\nu}_S\big(\textrm{SurArb}(S)\big)>0\Big\}}{\#\mathcal{S}^{{Mon}}(d,s,B)}\geq 1-\Bigg(1-\Big(\frac{1}{2}\Big)^{\frac{d}{2}}\Bigg)^{\hspace{-.1cm}s}\end{align*}

in Theorem 5·6 as claimed.

Proof. Denote the discriminant of $\gamma_n$ by $\textrm{Disc}(\gamma_n)$ , and for a prime ${\mathfrak{p}}$ of ${\mathcal O}_K$ let $v_{\mathfrak{p}}$ be the ${\mathfrak{p}}$ -adic valuation. Let $B_n$ be the following finite set of primes of ${\mathcal O}_K$ :

and note that if ${\mathfrak{p}} \not\in B_n$ then $\gamma_n$ has good reduction, i.e. reducing $\gamma_n$ coefficient-wise modulo ${\mathfrak{p}}$ yields a polynomial $\widetilde{\gamma_n}$ in ${\mathcal O}_K/{\mathfrak{p}}$ of degree $\deg(\gamma_n)$ . In particular, every root $\alpha$ of $\gamma_n$ satisfies $v_{\mathfrak{p}}(\alpha) \geq 0$ . Fix $n \geq 1$ , and let

Assume that ${\mathfrak{p}} \in \Omega_n$ and take $N > n$ . If $\gamma_N(x) \equiv 0 \bmod{{\mathfrak{p}}}$ has a solution in K, then because $\gamma_N = \gamma_n \circ \theta_{n+1} \circ \cdots \circ \theta_N$ , it follows that $\gamma_n(x) \equiv 0 \bmod{{\mathfrak{p}}}$ has a solution in K, and thus $\Omega_n \subseteq R_n$ . Observe that there are only finitely many prime ${\mathfrak{p}}$ with ${\mathfrak{p}} \mid \gamma_N(a_0)$ and $\gamma_N(a_0) \neq 0$ for some $N < n$ , and this together with the finiteness of $B_n$ imply that $R_n$ and the complement $P(\gamma, a_0)^c$ differ by a finite set of primes, which we denote by $F_n$ .

Assume for a moment that $D(\Omega_n)$ exists and equals $d_n$ . Let $\epsilon > 0$ and take $x_0$ large enough so that if $x \geq x_0$ then both

(25) \begin{equation} \frac{\#\{{\mathfrak{p}} \in \Omega_n \;:\; N({\mathfrak{p}}) \leq x\}}{\#\{{\mathfrak{p}} \;:\; N({\mathfrak{p}}) \leq x\}} > d_n - \epsilon/2 \quad \textrm{and} \quad \frac{\#\{{\mathfrak{p}} \in F_n \;:\; N({\mathfrak{p}}) \leq x\}}{\#\{{\mathfrak{p}} \;:\; N({\mathfrak{p}}) \leq x\}} < \epsilon/2.\end{equation}

The first inequality in (25) implies that ${\#\{{\mathfrak{p}} \in R_n \;:\; N({\mathfrak{p}}) \leq x\}}/{\#\{{\mathfrak{p}} \;:\; N({\mathfrak{p}}) \leq x\}} > d_n - \epsilon/2$ , and together with the second inequality in (25) this gives

\begin{align*}\frac{\#\{{\mathfrak{p}} \in P(\gamma, a_0)^c \;:\; N({\mathfrak{p}}) \leq x\}}{\#\{{\mathfrak{p}} \;:\; N({\mathfrak{p}}) \leq x\}} > d_n - \epsilon.\end{align*}

This implies that ${\#\{{\mathfrak{p}} \in P(\gamma, a_0) \;:\; N({\mathfrak{p}}) \leq x\}}/{\#\{{\mathfrak{p}} \;:\; N({\mathfrak{p}}) \leq x\}} < (1 - d_n) + \epsilon,$ from which we obtain $D^+(P(\gamma, a_0)) < 1-d_n$ , and thus $D^+(P(\gamma, a_0)) \leq \lim_{n \to \infty} (1-d_n)$ .

The proof will be complete once we show that $D(\Omega_n)$ exists and equals $1 - \textrm{FPP}(G_n)$ . If ${\mathfrak{p}} \nmid \textrm{Disc}(\gamma_n)$ , then ${\mathfrak{p}}$ cannot divide the field discriminant of $K_n(\gamma)/K$ ; see, for instance, [ Reference NarkiewiczNar04 , corollary 2, p. 157]. Hence, such ${\mathfrak{p}}$ are unramified in the extension $K_n(\gamma)/K$ . Now $\gamma_n(x) \equiv 0 \pmod{{\mathfrak{p}}}$ having a solution in K is equivalent to $\widetilde{\gamma_n}$ having at least one linear factor in $({\mathcal O}_K/{\mathfrak{p}})[x]$ . Except for possibly finitely many ${\mathfrak{p}}$ , this implies that ${\mathfrak{p}}{\mathcal O}_L = \mathfrak{P}_1 \cdots \mathfrak{P}_r$ , where $L/K$ is obtained by adjoining a root of $\gamma_n$ , ${\mathcal O}_L$ is the ring of integers of L, and at least one of the $\mathfrak{P}_i$ has residue class degree one [ Reference NarkiewiczNar04 , theorem 4·12]. This is equivalent to the disjoint cycle decomposition of the Frobenius conjugacy class at ${\mathfrak{p}}$ having a fixed point (in the natural permutation representation of $G_n$ on the roots $R_n$ of $\gamma_n$ ). From the Chebotarev Density Theorem it follows [ Reference NarkiewiczNar04 , proposition 7·15] that the density of ${\mathfrak{p}}$ with ${\mathfrak{p}}{\mathcal O}_L$ having such a decomposition exists and equals $\textrm{FPP}(G_n)$ . We have thus shown that $D(\Omega_n^c)$ exists and equals $\textrm{FPP}(G_n)$ , and the proof is complete.

Definition 6·2. A stochastic process $(X_0, X_1, X_2, \ldots)$ taking values in $\mathbb{Z}$ is a martingale if for all $n \geq 1$ and any $t_i \in \mathbb{Z}$ ,

\begin{align*} E(X_n \mid X_{0} = t_{0}, X_1 = t_1, \ldots, X_{n-1} = t_{n-1}) = t_{n-1}.\end{align*}

We call $(X_0, X_1, X_2, \ldots)$ an eventual martingale if for some $n_0 \geq 1$ the tail end process $X_{n_0}, X_{n_0 + 1}, X_{n_0 + 2}, \ldots $ is a martingale.

Proof. First note that since K is a number field, the irreducibility of $\gamma_n$ over K implies that $\gamma_n$ is separable over K. From Lemma 2·4 and theorem 2·5 of [ Reference JonesJon08 ] it follows that the present theorem is proved provided that the following holds: there exists $n_0 \geq 1$ such that for all $n \geq n_0$ and every root $\alpha$ of $\gamma_{n-1}$ , the polynomial $\theta_n - \alpha$ is irreducible over $K_{n-1}(\gamma)$ . Let $n_0$ be such that $H_n$ is non-trivial for $n \geq n_0$ , and fix $n \geq n_0$ . Observe that

(27) \begin{equation} \gamma_n = \prod_{\alpha \in \gamma_{n-1}^{-1}(0)} (\theta_n(x) - \alpha).\end{equation}

Because $H_n$ is non-trivial, there must be some $\alpha$ such that $\theta_n - \alpha$ has a root not in $K_{n-1}(\gamma)$ . Because $\theta_n$ has degree 2, this implies that $\theta_n - \alpha$ is irreducible over $K_{n-1}(\gamma)$ . By hypothesis, $\gamma_{n-1}$ is irreducible over K, and hence $G_{n-1}$ acts transitively on the roots of $\gamma_{n-1}$ . It follows that $\theta_n - \alpha$ is irreducible over $K_{n-1}(\gamma)$ for all $\alpha$ , as desired.

Proof. As in the discussion above, write $G_\infty$ for $G_{\gamma, K}$ , the inverse limit of the $G_n$ , and write $H_n$ for the Galois group of $K_n(\gamma)/K_{n-1}(\gamma)$ . We first claim that $H_n$ is non-trivial for all sufficiently large n, which by Theorem 6·5 shows that the fixed point process of $G_\infty$ is an eventual martingale.

Let ${\mathfrak{C}}$ be the partition of $T_n$ into the sets $\{\textrm{roots of $_n(x)$ -}\},$ where $\alpha$ varies over roots of $\theta_n$ , as in (27). Let $\sigma_{{\mathfrak{C}}} \in {\rm Sym} (V_n)$ be the permutation associated to ${\mathfrak{C}}$ , i.e. the unique permutation whose orbits are precisely the sets belonging to ${\mathfrak{C}}$ . We wish to show that $\sigma_{{\mathfrak{C}}} \in G_n$ for n sufficiently large, and then the fact that $\sigma_{\mathfrak{C}}$ only acts non-trivially on fibers of $\theta_n$ implies that $\sigma_{\mathfrak{C}} \in H_n$ , whence $H_n$ is non-trivial.

Because S contains only quadratic polynomials, $G_n$ is a 2-group, and thus has non-trivial center. Let $\delta$ be a nontrivial element of the center of $G_n$ , and ${\mathfrak{D}}$ the corresponding central fiber system, i.e. the collection of orbits of $\delta$ (see [ Reference JonesJon07 , proposition-definition 4·10]). Then by [ Reference JonesJon07 , theorem 4·9] we have either $\sigma_{{\mathfrak{C}}} = \delta$ or $G_n$ is composed entirely of alternating permutations.

In the latter case, $\textrm{Disc}(\gamma_n)$ is a square in K. Now for $n \geq 2$ , the discriminant formula in [ Reference HindesHin , proposition 6·2] and the fact that all elements of S are monic and of even degree imply that $\pm \gamma_n(0)$ must be a square in K. This implies that the either the curve $C^+ \;:\; y^2 = \gamma_2(x)$ or $C^{-} \;:\; y^2 = -\gamma_2(x)$ has a point P with $x(P) = 0$ for $n = 2$ and $x(P) = (\theta_3 \circ \cdots \circ \theta_n)(0)$ for $n \geq 3$ . The coordinates of P must have non-negative ${\mathfrak{p}}$ -adic valuation for every prime ${\mathfrak{p}}$ of ${\mathcal O}_K$ except for the finitely many where $v_{\mathfrak{p}}(c_i) < 0$ for some i.

By hypothesis the degree-4 polynomial $\gamma_2(x)$ is separable, so me may apply Siegel’s Theorem [ Reference Hindry and SilvermanHS00 , p.353] to conclude that both $C^+$ and $C^-$ have only finitely many such points P, and hence $\gamma_n(0)$ is a square in K for only finitely many n. We have thus shown that $\sigma_{{\mathfrak{C}}} = \delta$ for n sufficiently large, and because $\delta \in G_n$ , this implies that $\sigma_{{\mathfrak{C}}} \in G_n$ . Hence $H_n$ is non-trivial for n sufficiently large, and thus the fixed point process of $G_\infty$ is an eventual martingale.

We may now apply both Proposition 6·3 and Lemma 6·4. An argument identical to the proof of theorem 1·3 on p. 1122 of [ Reference JonesJon07 ] shows that

\begin{align*}\lim_{n \to \infty} \textbf{P}(X_n > 0) = 0.\end{align*}

By the definition of the Galois process, this is the same as $\lim_{n \to \infty} \textrm{FPP}(G_n) = 0$ , and so Theorem 6·1 gives $D^+(P(\gamma, a_0)) = 0$ , whence $D(P(\gamma, a_0)) = 0$ .