Two remarks on Narkiewicz’s property (P)

Abstract

Due to Narkiewicz a field F has property (P) if for no polynomial $f\in F[x]$ of degree at least two there is an infinite f-invariant subset of F. We present a new example of an algebraic extension of ${\mathbb {Q}}$ satisfying (P). This is the first example in which we can find points of arbitrarily small positive Weil-height. Moreover, we study the possibility of property (P) for the field generated by all symmetric Galois extensions of ${\mathbb {Q}}$. In particular we prove that there are no infinite backward orbits of non linear polynomials in this field.

1 Property (P)

In this paper we deal with a definition due to Narkiewicz [14].

Definition

A field F has property (P), if there is no infinite subset $X\subseteq F$, such that $f(X)=X$ for some polynomial $f\in F[x]$ with $\deg (f)\ge 2$.

A similar definition, due to Liardet, replaces polynomial by rational function in the above definition. In this case the field is said to have property (R). It is not known whether property (P) is equivalent to property (R). This is one of several open questions on property (P) (see [15, 16], and [17, Sects. IX and X]). One of these questions asks for a constructive classification of all fields with property (P). With a view to the known examples, see Examples 1.2 and Corollary 1.6 below, this seems to be the most difficult question on property (P).

This property is obviously closely tied to the theory of dynamical systems. Hence, we recall the basic definitions from this discipline. Let $f\in \overline{{\mathbb {Q}}}(x)$ be a rational function. With $f^n$ we denote the nth iterate of f, with the usual convention $f^0=x$. A point $\alpha \in \overline{{\mathbb {Q}}}$ is called a periodic point of f, if for some integer $n\ge 1$ we have $f^n(\alpha )=\alpha$. The smallest n with this property is called the exact period of $\alpha$. The set of all periodic points of f of exact period n is denoted by ${{\,\mathrm{Per}\,}}_n(f)$ and we set ${{\,\mathrm{Per}\,}}(f)=\cup _{n\ge 1}{{\,\mathrm{Per}\,}}_n(f)$.

Remark 1.1

It follows immediately that a field F has property (P) if and only if F satisfies the two following properties:

• (P1) For every $f\in F[x]$, with $\deg (f)\ge 2$, there is no infinite sequence $\alpha _0,\alpha _1,\ldots$ of pairwise distinct elements $\alpha _i \in F$, such that $f(\alpha _i)=\alpha _{i-1}$ for all $i\in {\mathbb {N}}$.

• (P2) For every $f\in F[x]$, with $\deg (f)\ge 2$, there are only finitely many periodic points of f in F.

In this paper we are only interested in fields with property (P) lying inside a fixed algebraic closure $\overline{{\mathbb {Q}}}$ of the rational numbers. In this case, height functions are an important tool for proving that a certain $F\subseteq \overline{{\mathbb {Q}}}$ has property (P). Denote by h the (absolute logarithmic Weil-)height on $\overline{{\mathbb {Q}}}$. An algebraic extension F of ${\mathbb {Q}}$ is said to have the Northcott property (N) if for every $T\in {\mathbb {R}}$ the set $\{\alpha \in F \vert h(\alpha )\le T\}$ is finite. It is easy to see that property (N) implies property (P), we refer to [4] and [6] for a proof and additional results. In particular, every number field satisfies property (P). There are only few examples known of algebraic extensions of ${\mathbb {Q}}$ of infinite degree with property (P).

Example 1.2

Let K denote an arbitrary number field. The following fields satisfy property (P):

• (I) $F=\cup _{i\in {\mathbb {N}}} K_i$, where $K=K_0\subseteq K_1 \subseteq K_2 \subseteq \ldots$ is a nested sequence of number fields, such that

  $$\begin{aligned} \inf _{K_i \subsetneq M \subseteq K_{i+1}} (N_{K_i /{\mathbb {Q}}}(D_{M/K_i}))^{{1}/{[M:K_0][M:K_i]}} \longrightarrow \infty , \text { as } i\rightarrow \infty , \end{aligned}$$

  where $D_{M/K_i}$ denotes the relative discriminant of the extension $M/K_i$ and $N_{K_i/{\mathbb {Q}}}$ is the norm associated to $K_i/{\mathbb {Q}}$ (see [22, Theorem 3]);

• (II) $F\subseteq K(\mu )$, where $\mu$ is the set of roots of unity, and $L(\mu )\cap F$ and $L(\{\zeta + \zeta ^{-1}\vert \zeta \in \mu \})\cap F$ are finite for all linear polynomials $L\in \overline{{\mathbb {Q}}}[x]$ (see [5, Theorem 2*, Proposition 1]);

• (III) F/K Galois, such that infinitely many local degrees of F are finite (see [20, Theorem 4.3, Lemma 4.2]).

A nice example of fields from part (III) are the fields which are generated over ${\mathbb {Q}}$ by elements of bounded degree (see [20, Corollary 4.5]).

Remark 1.3

The fields from (I) satisfy property (N). Other fields with property (N) are constructed in [2, Theorems 1 and 2]. Note that the set of examples from [2, Theorem 2] is non-empty as shown in [3, Theorem 1.3], but they are already covered by the fields in (III).

Both classes (II) and (III) contain fields which do not satisfy property (N) (see [5, Theorem 3] for the fields in (II), and [3, Theorem 1.7] for the fields in (III)). However, all fields from Example 1.2 satisfy a gap principle for the height (also known as the Bogomolov property (B)). This is, for every F from Example 1.2 there is a positive constant $c_F$ such that every $\alpha \in F$ satisfies either $h(\alpha )=0$ or $h(\alpha )\ge c_F$. This follows from [2, Theorem 2] for the fields in (III), from [22, Theorem 3] for the fields in (I), and from [1, Theorem 1.1] for the fields in (II).

1.1 The construction of Kubota and Liardet

To the best of my knowledge there is only one further class of examples of fields with property (P), due to Kubota and Liardet [9] (see also [17, Theorem 10.8]):

Construction 1.4

Since $\overline{{\mathbb {Q}}}[x]$ is countable, we enumerate the polynomials of degree $\ge 2$ as $f_1,f_2,\ldots$ Moreover, we choose any infinite sequence $p_1,p_2,\ldots$ of distinct primes. Lastly, we set $E_n = \cup _{j\le n} {{\,\mathrm{Per}\,}}(f_j)$. Now we construct a sequence

$$\begin{aligned} K_0\subseteq K_1 \subseteq K_2 \subseteq \ldots \end{aligned}$$

such that $K_0$ is an arbitrary number field and for each $n\ge 1$ we have $[K_n:K_{n-1}]=p_n$ and $K_n \cap E_n=K_{n-1}\cap E_n$. Then $F=\cup _{n\ge 1} K_n$ has property (P).

Note that the condition $K_n \cap E_n=K_{n-1}\cap E_n$ leads to (P2) for F, and the condition $[K_n:K_{n-1}]=p_n$ leads to (P1) for F. Dvornicich and Zannier [5, p. 535] noticed that this construction is rather indirect, and it seems difficult to detect other properties of the field so obtained. In this note we will use a very similar construction which can produce explicit fields satisfying property (P) not contained in Example 1.2. In these explicit fields we can detect other properties. In particular, we can produce a field which satisfies property (P), but—in contrast to all fields from Example 1.2—does contain points of arbitrarily small positive height. This answers a question of the author [20, Question 4.8] in the negative. The precise formulation of this result is as follows.

Let K be a number field, L/K a finite extension, and v a non-archimedean valuation on K. If $w\mid v$ is an extension of v to L, then we denote by $e_{w\mid v}$, resp. $f_{w \mid v}$, the ramification index, resp. the inertia degree, of $L_w / K_v$. If F/K is any algebraic extension, then we say that F has finite inertia degree over v, if there exists a constant $C_v$ such that for every finite subextension F/L/K there is a $w\mid v$ on L such that $f_{w\mid v}\le C_v$. Equivalently, F/K has finite inertia degree over v, if there is an extension w of v to F such that the residue field of $F_w$ is finite.

Proposition 1.5

Let $F \subseteq \overline{{\mathbb {Q}}}$ be a field. If for infinitely many prime numbers p the extension $F/{\mathbb {Q}}$ has finite inertia degree over p, then F satisfies (P2).

The proof, which follows quite immediately from a result of Morton and Silverman [12, Corollary B] (see also [19, Corollary 2] for a related result), can be found in Sect. 2.

Note that Proposition 1.5 reproves (P2) for all fields in (III). In what follows $\root n \of {a}$ denotes any fixed root of the polynomial $x^n-a$. The proof of the next corollary is the content of Sect. 3.

Corollary 1.6

Let $(p_n)_{n\ge 1}$ be a sequence of pairwise distinct prime numbers. Define $\alpha _0=1$ and $\alpha _{n}=\root p_{n} \of {p_{n}\alpha _{n-1}}$ for all $n\ge 1$. Then

$$\begin{aligned} F={\mathbb {Q}}(\alpha _1,\alpha _2,\ldots ) \end{aligned}$$

has property (P). Moreover, $h(\alpha _n)\rightarrow 0$ as n tends to infinity.

Note that the degree of ${\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)/{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _{n-1})$ is equal to the prime number $p_n$. As in Construction 1.4, this will guarantee that the field F satisfies (P1).

1.2 The field ${\mathbb {Q}}^{\mathrm{sym}}$

A hypothetical possibility to prove (P2) for a field F lies in the structure of the Galois groups of the fields generated by periodic points. Let K be a number field, $f\in K[x]$, and $\alpha \in {{\,\mathrm{Per}\,}}(f)$. Moreover, we denote by $K_{\alpha }$ the Galois closure of $K(\alpha )/K$.

We introduce the following hypothesis on f:

$$\begin{aligned} \begin{aligned}&\text { for all but finitely many } \alpha \in {{\,\mathrm{Per}\,}}(f) \text { the group } {{\,\mathrm{Gal}\,}}(K_{\alpha }/K) \\&\text { has a normal abelian subgroup of exponent } \ge 25. \end{aligned} \end{aligned}$$

(1)

Morton and Patel [11, Theorem 7.4] proved that for a generic polynomial $f\in {\mathbb {Q}}[x]$ of degree $\ge 2$ any $\alpha \in {{\,\mathrm{Per}\,}}_n(f)$ is a Galois conjugate of $f(\alpha )$, and that ${{\,\mathrm{Gal}\,}}(K_{\alpha }/K)$ has an abelian normal subgroup of exponent n. In particular, it should be the regular case that a polynomial satisfies hypothesis (1). We will discuss this in more detail in Sect. 6. Let ${\mathbb {Q}}^{\mathrm{sym}}$ be the compositum of all finite Galois extensions of ${\mathbb {Q}}$ with Galois group isomorphic to some symmetric group $S_n$. We will give some support for the possibility that ${\mathbb {Q}}^{\mathrm{sym}}$ satisfies property (P).

Theorem 1.7

The field ${\mathbb {Q}}^{\mathrm{sym}}$ satisfies (P1). Moreover, every polynomial $f\in {\mathbb {Q}}^{\mathrm{sym}}$ that satisfies hypothesis (1) for some number field K has at most finitely many periodic points in ${\mathbb {Q}}^{\mathrm{sym}}$.

In [7] it was shown that there exists a pseudo algebraically closed (PAC) field which satisfies property (N), and hence (P). Since ${\mathbb {Q}}^{\mathrm{sym}}$ is PAC (see [8, Theorem 18.10.4]), Theorem 1.7 gives a potential explicit example of a PAC field satisfying (P).

Section 5 contains the proof of (P1) for ${\mathbb {Q}}^{\mathrm{sym}}$. The remaining statement of Theorem 1.7 is proven in Sect. 6. We colse this introduction by noting that property (P) can actually be defined on any algebraic variety, when we replace polynomials by endomorphisms and the field F by F-rational points. In this setting, property (R) deals with endomorphisms of ${\mathbb {P}}^1$. Narkiewicz also introduced the case (in modern language) of endomorphisms of ${\mathbb {P}}^N$ for arbitrary N and called this property (SP). The case of property (P) for elliptic curves has recently been studied in [13].

2 Ramification and proof of Proposition 1.5

Let K be a number field with ring of integers ${\mathcal {O}}_K$. We use the usual one-to-one correspondence between non-zero prime ideals in ${\mathcal {O}}_K$ and non-archimedean valuations on K. Let v be the valuation corresponding to the prime ideal ${\mathfrak {P}}$ in ${\mathcal {O}}_K$, and let p be the rational prime with $v\mid p$. The norm of ${\mathfrak {P}}$ is given by

$$\begin{aligned} N({\mathfrak {P}})=\vert {{\mathcal {O}}_K}/{{\mathfrak {P}}}\vert = p^{f_{v\mid p}} \end{aligned}$$

(2)

A polynomial $f(x)=a_dx^d + a_{d-1}x^{d-1} + \cdots +a_0\in K[x]$ has good reduction at v if $v(a_i) \le 1$ for all $i\in \{0,\ldots ,d-1\}$, and $v(a_d)=1$. For the necessary facts from valuation and ramification theory, we refer to the first three chapters of [18] (or the English translation thereof), and for information on reduction of polynomials and rational maps, we refer to [21, Sect. 2.5].

We need the following obvious facts:

• A polynomial in K[x] has good reduction at all but finitely many valuations of K.

• Let L/K be a finite extension with an extension $w\mid v$ of non-archimedean valuations. If $f\in K[x]$ has good reduction at v, then f considered as a polynomial in L[x] has good reduction at w.

Proposition 1.5 will follow immediately from the following result (see [21, Corollary 2.26]), where we have applied Equation (2).

Lemma 2.1

Let K be a number field and let v, w be non-archimedean valuations on K. Assume further that $v \mid p$ and $w\mid q$ for distinct rational primes p and q. If $f \in K[x]$ has good reduction at v and w, and if $\alpha \in {{\,\mathrm{Per}\,}}_n(f)\cap K$, then $n \le (p^{2f_{v\mid p}}-1)(q^{2f_{w\mid p}}-1)$.

Proof of Proposition 1.5

Let F be an algebraic extension of ${\mathbb {Q}}$ with finite inertia degree over infinitely many primes. Let $f\in F[x]$ and $\alpha \in {{\,\mathrm{Per}\,}}_n(f)\cap F$ be arbitrary. There is some number field $K\subseteq F$ with $f \in K[x]$ and $\alpha \in K$. Moreover, there are necessarily two distinct primes p and q, such that

• f has good reduction at all valuations on K lying above p and q, and

• F has finite inertia degree over p and q.

Now, we fix $v\mid p$ and $w\mid q$ in K such that $f_{v\mid p}$ and $f_{w\mid q}$ are bounded from above by constants $C_p$ and $C_q$ only depending on p, q and F. It follows from Lemma 2.1 that $n \le (p^{2C_p}-1)(q^{2C_q}-1)$. Hence, we have bounded the possible period size of an element from $F\cap {{\,\mathrm{Per}\,}}(f)$. In particular, there are at most finitely many periodic points for f in F. This proves Proposition 1.5. $\square$

The next result can be seen as a slightly more effective version of Construction 1.4.

Corollary 2.2

Let $K_0$ be a number field and $(p_n)_{n\ge 1}$ a sequence of distinct prime numbers. Let $K_0\subseteq K_1 \subseteq K_2 \subseteq \ldots$ be a nested sequence of number fields, such that $K_n/K_{n-1}$ is totally ramified at all valuations lying above $p_1,\ldots ,p_n$. Then $F=\cup _{n\ge 0}K_n$ satisfies (P2). If in addition $[K_n:K_{n-1}]=p_n$ for all $n \ge 1$, then F satisfies property (P).

Proof

Let $n\ge 1$ be arbitrary, and let $v_1^{n-1},\ldots ,v_k^{n-1}$ be the extensions of $p_n$ to $K_{n-1}$. That the extension $K_n/K_{n-1}$ is totally ramified above all $v_1^{n-1},\ldots ,v_k^{n-1}$ implies that for all $v_i^{n-1}$, $i\in \{1,\ldots ,k\}$, there is precisely one extension $v_i^{n}$ to $K_n$, and this satisfies $f_{v_i^{n}\mid v_i^{n-1}}=1$. Inductively it follows that in any $K_N$ with $N\ge n$ the extensions of $p_n$ are given by some $v_1^{N},\ldots ,v_k^{N}$, with $f_{v_i^{N}\mid v_i^{N-1}}=1$ for all $i\in \{1,\ldots ,k\}$. Since the inertia degree is multiplicative, we have $f_{v_i^{N}\mid p_n}=f_{v_i^{n-1}\mid p_n}$. In particular, $F=\cup _{n\ge 0}K_n$ has finite inertia degree above all primes $p_1,p_2,\ldots$. Hence, by Proposition 1.5 the field F satisfies (P2).

Now we assume that we have $[K_n:K_{n-1}]=p_n$ for all $n \ge 1$. Then (P1) for F follows precisely as in the work of Kubota and Liardet [9]. For completeness we recall the simple argument here. The assumption on the degree implies that there is no proper intermediate field in the extension $K_{n}/K_{n-1}$. Equivalently, the field $K_{n}$ is the smallest proper field extension of $K_{n-1}$ lying in F. Let $f\in F[x]$ be of degree $d\ge 2$, and let $\beta _0 \in F$ be arbitrary. Fix any integer n such that $\beta _0 \in K_n$, $f\in K_n[x]$, and such that $p_n >d$. Construct (if possible) pairwise distinct elements $\beta _1,\ldots ,\beta _k\in K_n$ such that $f(\beta _i)=\beta _{i-1}$ for all $i\in \{1,\ldots ,k\}$. Since any number field satisfies property (P) – and hence (P1) – such a sequence must necessarily be finite. Assume that k is maximal, and that there is some $\beta _{k+1} \in F$, with $f(\beta _{k+1})=\beta _k$. Then $1\ne [K_n(\beta _{k+1}):K_n]\le d < p_n$, which gives a contradiction. Hence, F satisfies (P1), which implies that F satisfies property (P). $\square$

3 Proof of Corollary 1.6

Let $(p_n)_{n\ge 1}$ be a sequence of pairwise distinct primes. We define

$$\begin{aligned} \alpha _0=1 \quad \text { and } \quad \alpha _n = \root p_n \of {p_n\alpha _{n-1}} \text { for } n\ge 1. \end{aligned}$$

Moreover, we set

$$\begin{aligned} F={\mathbb {Q}}(\alpha _1,\alpha _2,\ldots ). \end{aligned}$$

We consider F as the union of number fields. Define $K_0={\mathbb {Q}}$ and $K_n=K_{n-1}(\alpha _n)$ for all $n\ge 1$, then $F=\cup _{n\ge 0}K_n$. We have

$$\begin{aligned} \alpha _n^{p_1\cdots p_n}=(p_1)\cdot (p_2^{p_1})\cdots (p_n^{p_1\cdots p_{n-1}}). \end{aligned}$$

(3)

Hence, $\alpha _n$ is the root of the $p_1$-Eisenstein polynomial $x^{p_1\cdots p_n}-\alpha _n^{p_1\cdots p_n} \in {\mathbb {Z}}[x]$. This implies that $[K_n:{\mathbb {Q}}]=p_1\cdots p_n$, which in turn yields

$$\begin{aligned}{}[K_n:K_{n-1}]=p_n \qquad \text { for all } n\ge 1. \end{aligned}$$

(4)

Note that all primes $p_k$, with $k> n$, are unramified in $K_n$. We need one further result on ramification in radical extensions.

Lemma 3.1

Let K be a number field, $\alpha \in K$, and p a prime number. Assume that the fractional ideal $\alpha {\mathcal {O}}_K$ decomposes as ${\mathfrak {p}}_1^{a_1}\cdots {\mathfrak {p}}_r^{a_r}$ for certain prime ideals ${\mathfrak {p}}_1,\ldots ,{\mathfrak {p}}_r$ in the ring of integers ${\mathcal {O}}_K$. If $p\not \mid a_i$, then ${\mathfrak {p}}_i$ is totally ramified in the extension $K(\root p \of {\alpha })/K$.

Proof

The short proof can be found in [10, p. 132]. $\square$

Proof of Corollary 1.6

In order to prove that F has property (P), thanks to Corollary 2.2 and (4), we are left to prove that the valuations (or equivalently: prime ideals) above $p_1,\ldots ,p_n$ are totally ramified in $K_n/K_{n-1}$.

We first handle the prime $p_{n}$ separately. Let ${\mathfrak {P}}$ be any prime ideal in ${\mathcal {O}}_{K_{n-1}}$ lying above $p_{n}$. Since $p_{n}$ is unramified in $K_{n-1}$, we have $e_{{\mathfrak {P}}\vert p_{n}}=1$. Moreover, by (3), ${\mathfrak {P}}$ does not appear as a factor in $\alpha _{n-1}{\mathcal {O}}_{K_{n-1}}$. Hence, the exponent of ${\mathfrak {P}}$ in the ideal $p_{n}\alpha _{n-1} {\mathcal {O}}_{K_{n-1}}$ is equal to $e_{{\mathfrak {P}}\vert p_{n}}=1$. By Lemma 3.1 we conclude that ${\mathfrak {P}}$ is totally ramified in $K_{n}/K_{n-1}$.

Next we prove the desired statement that all prime ideals above $p_1,\ldots ,p_n$ are totally ramified in $K_n/K_{n-1}$. We prove this by induction, where the induction base is the well-known statement that $p_1$ is totally ramified in ${\mathbb {Q}}(\root p_1 \of {p_1})$.

Let $n\ge 1$ be arbitrary and assume that the claim is correct for all positive integers $n'\le n$. Let $i\in \{1,\ldots ,n+1\}$ be arbitrary, and let ${\mathfrak {p}}$ be any prime ideal in ${\mathcal {O}}_{K_{i-1}}$ above $p_i$. We already know that $p_{n+1}$ is totally ramified in $K_{n+1}/K_n$, hence we may assume that $i\le n$. Since $p_i$ is unramified in $K_{i-1}$, we have $e_{{\mathfrak {p}}\mid p_i}=1$. Moreover, by our induction hypothesis, the prime ideal ${\mathfrak {p}}$ is totally ramified in $K_n/K_{i-1}$. Let ${\mathfrak {P}}$ be the unique prime ideal in ${\mathcal {O}}_{K_n}$ above ${\mathfrak {p}}$. Then $e_{{\mathfrak {P}}\mid {\mathfrak {p}}} = [K_n:K_{i-1}]=p_i\cdots p_n$. Combining this with $e_{{\mathfrak {p}}\mid p_i}=1$, gives

$$\begin{aligned} e_{{\mathfrak {P}}\mid p_i}=p_i\cdots p_n. \end{aligned}$$

(5)

Since, by (3), we know

$$\begin{aligned} p_{n+1}\alpha _{n} {\mathcal {O}}_{K_n} \mid (p_{n+1} \cdot p_1 p_2^{p_1}\cdots p_{n}^{p_1\cdots p_{n-1}}){\mathcal {O}}_{K_n}, \end{aligned}$$

the exponent of ${\mathfrak {P}}$ in $p_{n+1}\alpha _{n} {\mathcal {O}}_{K_n}$ must be a divisor of

$$\begin{aligned} p_1\cdots p_{i-1}\cdot e_{{\mathfrak {P}}\mid p_i} \overset{(5)}{=} p_1\cdots p_{n}. \end{aligned}$$

Hence, $p_{n+1}$ is not a divisor of this exponent and by Lemma 3.1 it follows, that ${\mathfrak {P}}$ is totally ramified in $K_{n+1}/K_n$. This concludes the induction and applying Corollary 2.2 yields that F has property (P).

It remains to prove that the height of the elements $\alpha _n$ tends to zero as n tends to infinity. We have $h(\alpha _1)=h(\root p_1 \of {p_1})=\frac{1}{p_1}\log (p_1) <1$, and by elementary properties of the height it follows

$$\begin{aligned} h(\alpha _{n+1})=\frac{1}{p_{n+1}}h(p_{n+1}\alpha _n)\le \frac{1}{p_{n+1}}(\log (p_{n+1})+h(\alpha _n)) \end{aligned}$$

(6)

for all $n \ge 1$. A trivial induction shows that $h(\alpha _n)<1$ for all $n\ge 1$, but this implies by (6) that $h(\alpha _n)$ indeed tends to zero as n tends to infinity. $\square$

4 Some facts about ${\mathbb {Q}}^{\mathrm{sym}}$

Definition

For $k\in {\mathbb {N}}$ we define the field ${\mathbb {Q}}_{k,\mathrm{sym}}$ to be the compositum of all number fields K, such that $K/{\mathbb {Q}}$ is Galois with Galois group isomorphic to some symmetric group $S_n$, with $n\le k$. Then ${\mathbb {Q}}^{\mathrm{sym}}=\cup _{k\in {\mathbb {N}}} {\mathbb {Q}}_{k,\mathrm{sym}}$.

Remark 4.1

The field ${\mathbb {Q}}_{k,\mathrm{sym}}$ is generated by elements of degree bounded from above by k!. Hence, for every k the field ${\mathbb {Q}}_{k,\mathrm{sym}}$ and all its finite extensions have property (P) by Example 1.2 (III).

Lemma 4.2

Let $k\ge 4$ be a rational integer and let $K/{\mathbb {Q}}$ be a finite Galois extension. If ${\mathbb {Q}}(\alpha )/{\mathbb {Q}}$ is Galois, with Galois group isomorphic to $S_n$, then the group ${{\,\mathrm{Gal}\,}}({\mathbb {Q}}_{k,\mathrm{sym}}K(\alpha )/{\mathbb {Q}}_{k,\mathrm{sym}}K)$ is either trivial or isomorphic to the alternating group $A_n$, for $n\ge k$.

Proof

Obviously, we have $\alpha \in {\mathbb {Q}}_{k,\mathrm{sym}}$ for $n\le k$. Hence, we assume for the rest of this proof $n>k\ge 4$. It is

$$\begin{aligned} {{\,\mathrm{Gal}\,}}({\mathbb {Q}}_{k,\mathrm{sym}}K(\alpha )/{\mathbb {Q}}_{k,\mathrm{sym}}K) \cong {{\,\mathrm{Gal}\,}}({\mathbb {Q}}(\alpha )/{\mathbb {Q}}(\alpha )\cap {\mathbb {Q}}_{k,\mathrm{sym}}K)=: H \subseteq S_n. \end{aligned}$$

Moreover, $({\mathbb {Q}}(\alpha )\cap {\mathbb {Q}}_{k,\mathrm{sym}}K)/{\mathbb {Q}}$ is Galois and hence H is normal in $S_n$. Since $n> 4$, it follows $H\in \{{{\,\mathrm{id}\,}}, A_n, S_n\}$, and we are left to show $H\ne S_n$.

The field ${\mathbb {Q}}_{k,\mathrm{sym}}$ contains all quadratic field extensions of ${\mathbb {Q}}$. In particular, ${\mathbb {Q}}_{k,\mathrm{sym}}K$ contains ${\mathbb {Q}}(\alpha )^{A_n}$, the field fixed by $A_n \subseteq {{\,\mathrm{Gal}\,}}({\mathbb {Q}}(\alpha )/{\mathbb {Q}})$. Hence, ${\mathbb {Q}}(\alpha )\cap {\mathbb {Q}}_{k,\mathrm{sym}}K\ne {\mathbb {Q}}$ and $H\ne S_n$. $\square$

Proposition 4.3

Let $K/{\mathbb {Q}}$ be a finite Galois extension and let F be a subfield of $K{\mathbb {Q}}^{\mathrm{sym}}$ such that $F/{\mathbb {Q}}_{k,\mathrm{sym}}K$ is also a finite Galois extension for some $k\ge 4$. Then either $F={\mathbb {Q}}_{k,\mathrm{sym}}K$ or

$$\begin{aligned} {{\,\mathrm{Gal}\,}}(F/{\mathbb {Q}}_{k,\mathrm{sym}}K)\cong A_{n_1}\times \cdots \times A_{n_r}, \end{aligned}$$

with $r\in {\mathbb {N}}$ and $n_1,\ldots ,n_r >k$.

Proof

By definition of ${\mathbb {Q}}^{\mathrm{sym}}$ there are $\alpha _1,\ldots ,\alpha _{r'}$ such that ${\mathbb {Q}}(\alpha _i)/{\mathbb {Q}}$ is Galois with group $S_{n_i}$ for all $i\in \{1,\ldots ,r'\}$, and such that $F\subseteq {\mathbb {Q}}_{k,\mathrm{sym}}K(\alpha _1,\ldots ,\alpha _{r'})$.

We may assume that none of the $\alpha _1,\ldots ,\alpha _{r'}$ can be omitted. This implies $n_i>k$ for all $i\in \{1,\ldots ,r'\}$. By Lemma 4.2 we have ${{\,\mathrm{Gal}\,}}({\mathbb {Q}}_{k,\mathrm{sym}}K(\alpha _i)/{\mathbb {Q}}_{k,\mathrm{sym}}K) \cong A_{n_i}$ for all $i\in \{1,\ldots ,r'\}$.

Let $i\in \{1,\ldots ,r'-1\}$ be arbitrary. It is

$$\begin{aligned} ({\mathbb {Q}}_{k,\mathrm{sym}}K(\alpha _1,\ldots ,\alpha _{i})\cap {\mathbb {Q}}_{k,\mathrm{sym}}K(\alpha _{i+1}))/{\mathbb {Q}}_{k,\mathrm{sym}}K \end{aligned}$$

a Galois extension, and the Galois group is a proper (since $\alpha _{i+1}$ cannot be omitted) normal subgroup of ${{\,\mathrm{Gal}\,}}({\mathbb {Q}}_{k,\mathrm{sym}}K(\alpha _{i+1})/{\mathbb {Q}}_{k,\mathrm{sym}}K)\cong A_{n_{i+1}}$ – hence it is trivial and ${\mathbb {Q}}_{k,\mathrm{sym}}K(\alpha _1,\ldots ,\alpha _{i})\cap {\mathbb {Q}}_{k,\mathrm{sym}}K(\alpha _{i+1})={\mathbb {Q}}_{k,\mathrm{sym}}K$. Basic Galois theory gives the result

$$\begin{aligned} {{\,\mathrm{Gal}\,}}({\mathbb {Q}}_{k,\mathrm{sym}}K(\alpha _1,\ldots ,\alpha _{r'})/{\mathbb {Q}}_{k,\mathrm{sym}}K) \cong A_{n_1}\times \cdots \times A_{n_{r'}}. \end{aligned}$$

Therefore, ${{\,\mathrm{Gal}\,}}(F/{\mathbb {Q}}_{k,\mathrm{sym}}K) \trianglelefteq A_{n_1}\times \cdots \times A_{n_{r'}}$. By Goursat’s lemma, the normal subgroup ${{\,\mathrm{Gal}\,}}(F/{\mathbb {Q}}_{k,\mathrm{sym}}K)$ must be either trivial or a direct product of some of the factors of $A_{n_1}\times \cdots \times A_{n_{r'}}$. Hence, after a possible renumbering, we have ${{\,\mathrm{Gal}\,}}(F/{\mathbb {Q}}_{k,\mathrm{sym}}K)\cong A_{n_1}\times \cdots \times A_{n_{r}}$ for some $r\le r'$. $\square$

Corollary 4.4

Let $K/{\mathbb {Q}}$ be a finite Galois extension, and let $K'\subseteq {\mathbb {Q}}^{\mathrm{sym}}$ be a number field such that $K'K/K$ is Galois and abelian. Then it is $K'\subseteq {\mathbb {Q}}_{4,\mathrm{sym}}K$.

Proof

The group ${{\,\mathrm{Gal}\,}}(K'K{\mathbb {Q}}_{4,\mathrm{sym}}/{\mathbb {Q}}_{4,\mathrm{sym}}K)$ is abelian. Hence, Proposition 4.3 implies that this group is trivial, proving the claim. $\square$

5 ${\mathbb {Q}}^{\mathrm{sym}}$ satisfies (P1)

We will prove that ${\mathbb {Q}}^{\mathrm{sym}}$ satisfies (P1). Hence, we pick an arbitrary $f\in {\mathbb {Q}}^{\mathrm{sym}}[x]$ of degree $d \ge 2$. Let $\alpha _0, \alpha _1,\ldots \in \overline{{\mathbb {Q}}}$ be pairwise distinct elements such that $f(\alpha _i)=\alpha _{i-1}$ for all $i\in {\mathbb {N}}$, and such that $\alpha _0 \in {\mathbb {Q}}^{\mathrm{sym}}$. We will prove that this sequence is not contained in ${\mathbb {Q}}^{\mathrm{sym}}$. To this end, let $K/{\mathbb {Q}}$ be a finite Galois extension with $K\subset {\mathbb {Q}}^{\mathrm{sym}}$, $f\in K[x]$, and $\alpha _0 \in K$. Set $k=\max \{\deg (f),4\}$, and consider the field ${\mathbb {Q}}_{k,\mathrm{sym}}K$. Since ${\mathbb {Q}}_{k,\mathrm{sym}}K$ has property (P), there are only finitely many elements from $\{\alpha _i\}_{i\in {\mathbb {N}}_0}$ in ${\mathbb {Q}}_{k,\mathrm{sym}}K$. Let $n\in {\mathbb {N}}_0$ be maximal such that $\alpha _n \in {\mathbb {Q}}_{k,\mathrm{sym}}K$. We will show that $\alpha _{n+1}\notin {\mathbb {Q}}^{\mathrm{sym}}$.

Assume for the sake of contradiction that $\alpha _{n+1}\in {\mathbb {Q}}^{\mathrm{sym}}$. Let F be the Galois closure of ${\mathbb {Q}}_{k,\mathrm{sym}}K(\alpha _{n+1})$ over ${\mathbb {Q}}_{k,\mathrm{sym}}K$. Then $F\subseteq {\mathbb {Q}}^{\mathrm{sym}}$, and by Proposition 4.3 it is

$$\begin{aligned} {{\,\mathrm{Gal}\,}}(F/{\mathbb {Q}}_{k,\mathrm{sym}}K)\cong A_{n_1}\times \cdots \times A_{n_r}, \end{aligned}$$

(7)

with $n_1,\ldots ,n_r > \deg (f)$. However, the minimal polynomial of $\alpha _{n+1}$ over ${\mathbb {Q}}_{k,\mathrm{sym}}K$ divides $f(x) - \alpha _n$. Hence ${{\,\mathrm{Gal}\,}}(F/{\mathbb {Q}}_{k,\mathrm{sym}}K)$ is a subgroup of $S_{\deg (f)}$, contradicting (7).

It follows that $\alpha _{n+1} \notin {\mathbb {Q}}^{\mathrm{sym}}$ and hence the sequence $\alpha _0,\alpha _1,\ldots$ cannot be fully contained in ${\mathbb {Q}}^{\mathrm{sym}}$, proving the proposition.

6 Periodic points in ${\mathbb {Q}}^{\mathrm{sym}}$

Let K be a number field, $f\in K[x]$ of degree $\ge 2$, and $\alpha \in {{\,\mathrm{Per}\,}}_n(f)$. For any $\sigma \in {{\,\mathrm{Gal}\,}}(\overline{{\mathbb {Q}}}/K)$, we have

$$\begin{aligned} \alpha = f^k(\alpha ) \quad \Longleftrightarrow \quad \sigma (\alpha )=\sigma (f^k(\alpha ))=f^k(\sigma (\alpha )), \end{aligned}$$

and hence $\sigma (\alpha )\in {{\,\mathrm{Per}\,}}_n(f)$. Obviously, it is also $f(\alpha )\in {{\,\mathrm{Per}\,}}_n(f)$. It follows that the set of Galois conjugates of $\alpha$ over K is contained in a finite union of periodic orbits of length n. Say

$$\begin{aligned} {{\,\mathrm{Gal}\,}}(\overline{{\mathbb {Q}}}/K)\cdot \alpha \subseteq O_1\cup \ldots \cup O_r, \end{aligned}$$

(8)

where $O_i=\{f^{k}(\alpha _i)\vert 0\le k \le n-1\}$ for some $\alpha _i \in {{\,\mathrm{Per}\,}}_n(f)$. Denote by $K_{\alpha }$ the Galois closure of $K(\alpha )/K$. If $\tau \in {{\,\mathrm{Gal}\,}}(K_{\alpha }/K)$ maps $\alpha _i$ to some $f^{k}(\alpha _j)\in O_j$, for some integer k, then $\tau (f^{\ell }(\alpha _i))=f^{\ell }(\tau (\alpha _i))=f^{\ell + k}(\alpha _j)\in O_j$. This means that any element in ${{\,\mathrm{Gal}\,}}(K_{\alpha }/K)$ permutes the orbits $O_1,\ldots ,O_r$. Set

$$\begin{aligned} N_{\alpha ,K}=\{\sigma \in {{\,\mathrm{Gal}\,}}(K_{\alpha }/K) \vert \sigma (O_i)=O_i ~\forall ~i\in \{1,\ldots ,r\}\}. \end{aligned}$$

(9)

Then $N_{\alpha ,K}$ is a normal subgroup of ${{\,\mathrm{Gal}\,}}(K_{\alpha }/K)$, which injects into $\left( {{\mathbb {Z}}}/{n{\mathbb {Z}}}\right) ^r$. This follows precisely as in the proof of [11, Theorem 4.1].

Morton and Patel [11, Theorem 7.4] proved that in the generic situation, there is a $\sigma \in {{\,\mathrm{Gal}\,}}(K_{\alpha }/K)$ such that for all $\beta \in {{\,\mathrm{Gal}\,}}(K_{\alpha }/K)\cdot \alpha$ we have $\sigma (\beta )=f(\beta )$. In particular, this $\sigma$ is an element in $N_{\alpha ,K}$ of order n. Hence, for an arbitrarily given $f\in K[x]$ and an $\alpha \in {{\,\mathrm{Per}\,}}_n(f)$, one expects that the exponent of $N_{\alpha ,K}$ grows linearly in n. The 25 in our hypothesis (1) is particularly chosen for our purposes. We will need below that the exponent of a normal abelian subgroup of ${{\,\mathrm{Gal}\,}}(K_{\alpha }/K)$ is greater than the exponent of ${{\,\mathrm{Gal}\,}}({\mathbb {Q}}_{4,\mathrm{sym}}/{\mathbb {Q}})$, which is precisely $4!=24$.

Remark 6.1

Let $f(x)=x^d$, $d\ge 2$, be a power map, and $\zeta \in {{\,\mathrm{Per}\,}}_n(f)$. Then $\zeta$ is a primitive $(d^{n}-1)$-st root of unity. Now, for every number field $K /{\mathbb {Q}}$ at least one of the elements $\zeta ^d,\ldots ,\zeta ^{d^{[K:{\mathbb {Q}}]}}$ is a Galois conjugate of $\zeta$ over K. This means that one of the maps $f,f^{2},\ldots ,f^{[K:{\mathbb {Q}}]}$ induces a K-automorphism on $K(\zeta )$. In particular, for any $n\ge 25[K:{\mathbb {Q}}]$ and any $\alpha \in {{\,\mathrm{Per}\,}}_n(f)$ the exponent of $N_{\alpha ,K}$ is greater than 25.

The same argument is also valid if f is a Chebyshev polynomial. Hence, every power map and every Chebyshev polynomial satisfies hypothesis (1).

Proposition 6.2

Let $f\in {\mathbb {Q}}^{\mathrm{sym}}[x]$ be of degree $\ge 2$, such that f satisfies hypothesis (1), then ${{\,\mathrm{Per}\,}}(f)\cap {\mathbb {Q}}^{\mathrm{sym}}$ is a finite set.

Proof

Let f be as in the statement and let $K\subseteq {\mathbb {Q}}^{\mathrm{sym}}$ be a number field with $f\in K[x]$. Let $\alpha$ be an element from ${\mathbb {Q}}^{\mathrm{sym}} \cap {{\,\mathrm{Per}\,}}(f)$, such that ${{\,\mathrm{Gal}\,}}(K_{\alpha }/K)$ contains an abelian normal subgroup N with exponent $\ge 25$. Denote with $K_{\alpha }^N$ the fixed field of N. Then $K_\alpha /K_{\alpha }^N$ and $K_\alpha ^N/K$ are Galois extensions, where ${{\,\mathrm{Gal}\,}}(K_\alpha /K_\alpha ^N)\cong N$ is abelian.

By Corollary 4.4 we have $K_\alpha \subseteq {\mathbb {Q}}_{4,\mathrm{sym}}K_\alpha ^N$. Therefore,

$$\begin{aligned} {{\,\mathrm{Gal}\,}}(K_\alpha /K_\alpha ^N)\cong N \trianglelefteq {{\,\mathrm{Gal}\,}}({\mathbb {Q}}_{4,\mathrm{sym}}K_\alpha ^N/K_\alpha ^N). \end{aligned}$$

However, the exponent of ${{\,\mathrm{Gal}\,}}({\mathbb {Q}}_{4,\mathrm{sym}}K_\alpha ^N/K_\alpha ^N)$ is 4!, contradicting the choice of N. If f satisfies hypothesis (1), then we can conclude that f contains at most finitely many periodic points in ${\mathbb {Q}}^{\mathrm{sym}}$, proving the claim. $\square$

Together with Sect. 5 this proves Theorem 1.7.