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A large arboreal Galois representation for a cubic postcritically finite polynomial
 Robert L. Benedetto^{1},
 Xander Faber^{2}Email authorView ORCID ID profile,
 Benjamin Hutz^{3},
 Jamie Juul^{1} and
 Yu Yasufuku^{4}
 Received: 13 December 2016
 Accepted: 20 July 2017
 Published: 6 December 2017
Abstract
We give a complete description of the arboreal Galois representation of a certain postcritically finite cubic polynomial over a large class of number fields and for a large class of basepoints. This is the first such example that is not conjugate to a power map, Chebyshev polynomial, or Lattès map. The associated Galois action on an infinite ternary rooted tree has Hausdorff dimension bounded strictly between that of the infinite wreath product of cyclic groups and that of the infinite wreath product of symmetric groups. We deduce a zerodensity result for prime divisors in an orbit under this polynomial. We also obtain a zerodensity result for the set of places of convergence of Newton’s method for a certain cubic polynomial, thus resolving the first nontrivial case of a conjecture of Faber and Voloch.
1 Introduction
By contrast, for specific choices of f and x, the corresponding Galois groups may be much smaller (see [6, §3] for a highlevel explanation and [2, 4, 8, 14] for detailed examples). Consider a polynomial f that is postcritically finite, or PCF for short, meaning that all of its critical points have finite orbit under the iteration of f. The simplest examples of PCF polynomials are the power maps \(f(z) = z^d\) and the Chebyshev polynomials defined by \(f\left( z + 1/z\right) = z^d + 1/z^d\). These two examples arise from the dpower endomorphism of the algebraic group \(\mathbb {G}_m\), which gives a foothold on the associated arboreal Galois representation. (A third type of example, Lattès maps, arises from an endomorphism of an elliptic curve; however, Lattès maps are never conjugate to polynomials. See [15, §6.4].)
Jones and Pink [6, Thm. 3.1] have shown that for PCF maps, the Galois groups \(G_n\) have unbounded index inside \({\text {Aut}}(T_n)\) as \(n\rightarrow \infty \). However, their proof does not explicitly describe \(G_n\). One can give an upper bound for \(G_n\) inside \({\text {Aut}}(T_n)\) by realizing it as a specialization of \({\text {Gal}}\left( f^n(z)  t / K(t)\right) \), with t transcendental over K. The latter group may be embedded in the profinite monodromy group \(\pi _1^{\acute{\mathrm{e}}\mathrm{t}}(\mathbb {P}^1_{K} \smallsetminus P)\), where P is the strict postcritical orbit; this is precisely the tack taken by Pink [14] in the case of quadratic PCF polynomials.
Theorem 1.1
 (1)
The polynomial \(f^n(z)  x\) is irreducible over K.
 (2)
We have an isomorphism \({\text {Gal}}\left( f^n(z)  x / K\right) \cong E_n \subset {\text {Aut}}(T_n)\).
Remark 1.2
In this article, we implicitly work in the category of groups with an action on the regular rooted tree \(T_n\). This applies, for example, to the isomorphism between \(E_n\) and the Galois group in the theorem.
We are also able to deduce that the geometric Galois group has the same structure:
Corollary 1.3
While \(E_n\) is not an iterated wreath product, it does satisfy the following selfsimilarity property: the action of \(E_n\) on the subtree of height \(n1\) stemming from any fixed vertex at level 1 is isomorphic to \(E_{n1}\). This selfsimilarity is a property of geometric iterated monodromy groups [11, Prop. 6.4.2], and \(E_n\) is such a group by Corollary 1.3.
Odoni [13] has shown that descriptions of iterated Galois groups of this sort give rise to applications on the density of prime divisors in certain dynamically defined sequences. (See also [5, 9, 10].) More precisely, after counting elements of \(E_n\) that fix a leaf of the tree \(T_n\), we have the following arithmetic application.
Theorem 1.4
Our choice of the polynomial \(f(z) = 2z^3 + 3z^2\) was originally motivated by the following conjecture of Faber and Voloch [3].
Conjecture 1.5
(Newton Approximation fails for 100% of primes) Let g be a polynomial of degree \(d \ge 2\) with coefficients in a number field K and let \(y_0 \in K\). Define the Newton map \(N(z) = z  g(z) / g'(z)\) and, for each \(n \ge 0 \), set \(y_{n+1} = N(y_n)\). Assume the Newton approximation sequence \((y_n)\) is not eventually periodic. Let \(C(K, g, y_0)\) be the set of primes \(\mathfrak {P}\) of K for which \((y_n)\) converges in the completion \(K_{\mathfrak {P}}\) to a root of f. Then the natural density of the set \(C(K, g, y_0)\) is zero.
Faber and Voloch showed that Conjecture 1.5 holds for any polynomial g with at most 2 distinct roots. Thus, the first nontrivial case of the conjecture is a separable cubic polynomial. For reasons explained in [3, Cor. 1.2], the simplest such cubic polynomial is \(g(z) = z^3  z\), whose associated Newton map turns out to be conjugate to our polynomial \(f(z) = 2z^3 + 3z^2\). Our results therefore yield a proof of the first nontrivial case of the Faber–Voloch conjecture:
Theorem 1.6
Let K be a number field for which there exists an unramified prime over 2 and over 3. Let \(g(z) = z^3  z\). Choose \(y_0 \in K\) such that the Newton iteration sequence \(y_i = N^i(y_0)\) does not encounter a root of g. Then the set of primes \(\mathfrak {P}\) of K for which the Newton sequence \((y_i)\) converges in \(K_\mathfrak {P}\) to a root of g has natural density zero.
The first and third authors, in collaboration with several others, obtained a weak form of Theorem 1.6 for a wide class of polynomials [1, Thm. 4.6]. More precisely, they showed that the density of primes as in the theorem has natural density strictly less than one.
The outline of the paper is as follows. In Sect. 2, we will define and discuss the group \(E_n\) and compute the Hausdorff dimension of Eq. (1). We will then prove the rest of Theorem 1.1 in Sect. 3. Next, we consider the case that K is the field of rational functions \({\bar{\mathbb {Q}}}(t)\), proving Corollary 1.3 in Sect. 4. In Sect. 5, we compute the proportion of elements of \(E_n\) that fix at least one leaf of the tree \(T_n\), and in Sect. 6, we prove Theorems 1.4 and 1.6.
2 Tree automorphisms

The level of a vertex is its distance from the root.

A vertex at level i is given a label \((\ell _1, \ldots , \ell _i)\), where \(\ell _j \in \{1,2,3\}\). The root is given the empty label ().

No two vertices at the same level have the same label.

The unique path from the root to the vertex with label \((\ell _1, \ldots , \ell _i)\) consists of the vertices with labels \((), (\ell _1), (\ell _1,\ell _2), \ldots , (\ell _1, \ell _2, \ldots , \ell _i)\).
Lemma 2.1
Proof
Partition the leaves of \(T_n\) into three disjoint sets \(L_1, L_2, L_3\) so that the elements of \(L_i\) lie over leaf i of \(T_1\), for \(i=1,2,3\). Note that \(L_i = 3^{n1}\). With this notation, \(\mathrm {sgn}(a_i)\) is the sign of \(a_i\) acting as a permutation on the set \(L_i\).
Consider first the case that \(b = 1\). Then g permutes the elements of each \(L_i\) separately; hence, \(\mathrm {sgn}(g) = \prod \mathrm {sgn}(a_i)\).
Proposition 2.2
For \(n \ge 1\), we have \(\displaystyle E_n = 2^{3^{n1}} \cdot 3^\frac{3^n1}{2}\).
Proof
Our construction of \(E_n\) depends on an identification of \(T_{n1}\) with the subtree of \(T_n\) lying above a vertex at level 1, which in turn depends on the labeling we have assigned to \(T_n\). In other words, \(E_n\) is not normal in \({\text {Aut}}(T_n)\) (for \(n \ge 3\)): a different labeling yields a conjugate subgroup in \({\text {Aut}}(T_n)\).
Proposition 2.3
\(E_n\) is normal in \({\text {Aut}}(T_n)\) if and only if \(n = 1\) or 2.
Proof
We have \(E_1 = {\text {Aut}}(T_1)\), and \(E_2\) has index 2 in \({\text {Aut}}(T_2)\). It remains to show that \(E_n\) is not normal in \({\text {Aut}}(T_n)\) for \(n \ge 3\). To that end, we first construct some special elements of \({\text {Aut}}(T_n)\).
Corollary 2.4
The Hausdorff dimension of \(E_\infty \) in \({\text {Aut}}(T_\infty )\) is given by Eq. (1).
Proof
Corollary 2.5
\(E_\infty \) has infinite index in \({\text {Aut}}(T_\infty )\).
Proposition 2.6
For \(n \ge 1\), \(H_n\) is a Sylow 3subgroup of \(E_n\). It is normal in \(E_n\) if and only if \(n=1\).
Proof
For \(n = 1\), \(H_n\) is an index2 subgroup of \(E_1 = {\text {Aut}}(T_1)\) and, hence, it is normal. Next, for some \(n\ge 1\), suppose we know that \(H_n\) is a subgroup of \(E_n\). By their recursive definitions, to see that \(H_{n+1} \subset E_{n+1}\) it suffices to show that any \(h \in H_{n+1}\) restricts to an even permutation on \(T_2\). The restriction of h to \(T_2\) has the form \(\big ((a_1,a_2,a_3), b\big )\), where each of \(a_1, a_2, a_3\), and b is either trivial or a 3cycle. By Lemma 2.1, we conclude that \(\mathrm {sgn}_2(h) = 1\). Hence, \(h \in E_{n+1}\), as desired.
Since \(H_n\) is the 3power part of \(E_n\), we have proven the first statement of the proposition. It remains to show that \(H_n\) is not normal in \(E_n\) for \(n \ge 2\).
Note that for \(n \ge 2\), we have \(\tau _n^{1} = \big ((1, \tau _{n1}^{1}, 1), (12)\big )\). Note also that for \(n\ge 1\), we have \(\tau _n\not \in H_n\), since the restriction of \(\tau _n\) to \(T_1\) is \((12)\not \in C_3\).
Proposition 2.7
Since \(H_n \subseteq E_n \subseteq {\text {Aut}}(T_n)\), the preceding proposition and Corollary 2.4 show that for large n, \(E_n\) is substantially larger than \(H_n\), but much smaller than \({\text {Aut}}(T_n)\).
Finally, we will need a lemma that constructs certain special elements of \(E_n\):
Lemma 2.8

On the copy of \(T_{n1}\) with the same root as \(T_n\), g acts by the identity.

On each copy of \(T_2\) rooted at a vertex of \(T_n\) of level \(n2\), g acts by an even permutation of the 9 leaves.
Proof
We proceed by induction on n. For \(n=2\), the second condition on g implies that \(\mathrm {sgn}(g) = 1\), so \(g \in E_2\). Suppose that the lemma holds for \(n1\), and let \(g \in {\text {Aut}}(T_n)\) satisfy the given conditions. Let \(u_1,u_2,u_3\) be the vertices of \(T_n\) at level 1. Write \(T_{n1}(u_i)\) for the copy of \(T_{n1}\) inside \(T_n\) that is rooted at \(u_i\). Then g restricts to an element of \({\text {Aut}}(T_{n1}(u_i))\) that satisfies the two conditions of the lemma. By the induction hypothesis, \(g_{T_{n1}(u_i)} \in E_{n1}\) for \(i = 1,2,3\). In addition, g is the identity and, hence, is even, on \(T_2\). Thus, \(g\in E_n\) by the criterion of Eq. (4). \(\square \)
3 Main theorem
Lemma 3.1
For any field K of characteristic zero and any \(x\in K\smallsetminus \{0,1\}\), the Galois group \(G_n\) of Eq. (8) is isomorphic to a subgroup of \(E_n\).
Proof
Because there is no critical point in the backward orbit of x, the set \(f^{i}(x)\) consists of exactly \(3^i\) distinct elements for each \(i\ge 0\). Identify the vertices of the ternary rooted tree \(T_n\) with the set \(\bigsqcup _{0 \le i \le n} f^{i}(x)\), with vertex y lying immediately above \(y'\) if and only if \(f(y)=y'\). This identification induces a faithful action of \(G_n\) on \(T_n\) and, hence, \(G_n\) may be identified with a subgroup of \({\text {Aut}}(T_n)\).
To see that this subgroup \(G_n\) lies inside \(E_n\), we proceed by induction on n. For \(n=1\), this is clear because \(E_1={\text {Aut}}(T_1)\).
It remains to show that \(B_n\subseteq \ker (\mathrm {sgn}_2)\). Direct computation shows that the discriminant of the degreenine polynomial \(f^2(z)  x\) is given by Eq. (2). Since this discriminant is a square in K, all elements of \(B_n\) act as even permutations of the nine points of \(f^{2}(x)\). Thus, \(B_n\subseteq \ker (\mathrm {sgn}_2)\), as desired. \(\square \)
Example 3.2
If \(K = \mathbb {Q}\), then the pairs \((\mathbb {Q},3)\) and \((\mathbb {Q},3/2)\) both satisfy property \((\dagger )\). The latter pair will be important for our arithmetic applications.
Lemma 3.3
Suppose that (K, x) satisfies \((\dagger )\) relative to \(\mathfrak {p}\) and \(\mathfrak {q}\). Then \(f^n(z)  x\) is Eisenstein at \(\mathfrak {q}\) for all \(n \ge 1\). In particular, \(f^n(z)  x\) is irreducible for all \(n \ge 1\).
Proof
A simple induction shows that \(f^n(z) \equiv z^{3^n} \pmod \mathfrak {q}\) and \(f^n(0) = 0\). Since \(v_{\mathfrak {q}}(x) = 1\), it follows immediately that \(f^n(z)  x\) is Eisenstein at \(\mathfrak {q}\). \(\square \)
Proposition 3.4
 (1)There are primes \(\mathfrak {p}'\) and \(\mathfrak {q}'\) of K(y) lying above \(\mathfrak {p}\) and \(\mathfrak {q}\), respectively, such that$$\begin{aligned} e(\mathfrak {p}'/\mathfrak {p}) = 2^n \quad \text {and}\quad e(\mathfrak {q}' / \mathfrak {q}) = 3^n. \end{aligned}$$
 (2)
The pair (K(y), y) satisfies property \((\dagger )\) relative to \(\mathfrak {p}'\) and \(\mathfrak {q}'\).
Proof
We proceed by induction on n. The statement is trivial for \(n=0\). We therefore assume for the rest of the proof that \(n\ge 1\) and that the statement holds for \(n1\).
Finally, if \(v_{\mathfrak {p}''}(y'')=1\), then because \(v_{\mathfrak {p}''}(2)\ge 1\), the Newton polygon of \(f(z)y''\) at \(\mathfrak {p}''\) has a segment of length 2 and height \(1\). Thus, there is a prime \(\mathfrak {p}'\) of K(y) lying above \(\mathfrak {p}''\), with ramification index \(e(\mathfrak {p}' / \mathfrak {p}'') = 2\), and with \(v_{\mathfrak {p}'}(y) = 1\). Once again, then, we have \(e(\mathfrak {p}'/\mathfrak {p})=2^n\). \(\square \)
Corollary 3.5
Proof
Pick \(y\in f^{n}(x)\), and let \(\mathfrak {p}'\) and \(\mathfrak {q}'\) be the primes of K(y) given by Proposition 3.4. Since \(K_n/K\) has intermediate extension K(y) / K, the ramification index of some prime of \(K_n\) over \(\mathfrak {p}\) must be divisible by \(e(\mathfrak {p}'/\mathfrak {p})=2^n\). Similarly, the ramification index of some prime of \(K_n\) over \(\mathfrak {q}\) must be divisible by \(e(\mathfrak {q}'/\mathfrak {q})=3^n\). Thus, \(6^n \mid [K_n:K]\). \(\square \)
Proposition 3.6
Proof
Let \(K_1=K(f^{1}(x))\) and \(K_2= K(f^{2}(x))\). Then, as a splitting field of a cubic polynomial, \(K_1\) is Galois over K with \({\text {Gal}}(K_1/K)\) isomorphic to a subgroup of \(\mathfrak {S}_3\). By Corollary 3.5 with \(n=1\), we have \(6\mid [K_1:K]\). Hence, \({\text {Gal}}(K_1/K)\cong \mathfrak {S}_3\).
By Lemma 3.1, \({\text {Gal}}(K_2/K)\) acts on \(f^{2}(x)\) as a subgroup of \(E_2\). It suffices to show that every element of \(E_2\) is realized in \({\text {Gal}}(K_2/K)\).

fixes \(v_{1j}\) for each \(j=1,2,3\),

acts as a 2cycle on the set \(f^{1}(u_2) = \{v_{21}, v_{22},v_{23} \}\), and

acts as a 2cycle on the set \(f^{1}(u_3) = \{v_{31}, v_{32},v_{33} \}\).
Choose \(\gamma ' \in {\text {Gal}}(K_1/K)\) with \(\gamma '(u_1)=u_m\), and lift \(\gamma '\) to \(\gamma \in {\text {Gal}}(K_2/K)\). Let \(\tau = \gamma ^{1}\tau _1\gamma \). Then \(\tau (u_i)=u_i\) for each i, and \(\tau \) satisfies each of the three bulleted properties, proving Claim 1.

a 3cycle on \(f^{1}(u_1) = \{v_{11}, v_{12},v_{13} \}\),

either a 3cycle or the identity on \(f^{1}(u_2) = \{v_{21}, v_{22},v_{23} \}\), and

either a 3cycle or the identity on \(f^{1}(u_3) = \{v_{31}, v_{32},v_{33} \}\).

a 3cycle on \(f^{1}(u_1) = \{v_{11}, v_{12},v_{13} \}\),

a 2cycle on \(f^{1}(u_2) = \{v_{21}, v_{22},v_{23} \}\), and

a 2cycle on \(f^{1}(u_3) = \{v_{31}, v_{32},v_{33} \}\).
Conjugating \(\sigma \) by permutations \(\gamma \) as in the proof of Claim 1 and then composing the resulting conjugates with one another, we see that \({\text {Gal}}(K_2/K_1)\) contains each element of \(E_2\) that is the identity on \(f^{1}(x)\) and is either the identity or a 3cycle on each \(f^{1}(u_i)\). There are \(3^3=27\) such permutations.
Conjugating \(\tau \) by permutations from the previous paragraph, as well as by permutations \(\gamma \) as in the proof of Claim 1, we also see that \({\text {Gal}}(K_2/K_1)\) contains each element of \(E_2\) that is the identity on \(f^{1}(x)\), is a 2cycle on exactly two of \(f^{1}(u_1)\), \(f^{1}(u_2)\), and \(f^{1}(u_3)\), and is the identity on the third. There are \(3^3=27\) such permutations.
Composing the maps of the previous two paragraphs, we see that \({\text {Gal}}(K_2/K_1)\) contains each element of \(E_2\) that is the identity on \(f^{1}(x)\), is a 2cycle on exactly two of \(f^{1}(u_1)\), \(f^{1}(u_2)\), and \(f^{1}(u_3)\), and is a 3cycle on the third. There are \(3^3 \cdot 2 = 54\) such permutations.
We are now prepared to prove part (b) of Theorem 1.1, which we restate here.
Theorem 3.7
Proof
By Lemma 3.1, we know that the Galois group is isomorphic to a subgroup of \(E_n\). We must show that this subgroup is \(E_n\) itself. We proceed by induction on n. For \(n=1,2\), we are done by Proposition 3.6. Assuming we know the statement (for any such K and x) for a particular \(n\ge 2\), we will now show it for \(n+1\).

the identity on \(f^{n}(u_3) \smallsetminus f^{2}(y)\) and \(f^{1}(y)\),

two 2cycles on \(f^{2}(y)\), and

the same as \(\sigma \) on \(f^{(n1)}(\{ u_1, u_2\})\),

the identity on \(f^{n}(u_3) \smallsetminus f^{2}(y)\) and \(f^{1}(y)\),

two 2cycles and (perhaps) a separate 3cycle on \(f^{2}(y)\), and

the identity on \(f^{(n1)}(\{ u_1, u_2\})\).

the identity on \(f^{n}(x)\),

the identity on \(f^{n}(u_3) \smallsetminus f^{2}(y)\), and

two 2cycles on \(f^{2}(y)\).

the identity on \(f^{n}(x)\),

the identity on \(f^{n}(u_3) \smallsetminus f^{2}(y)\),

two 2cycles and (perhaps) a separate 3cycle on \(f^{2}(y)\), and

for each \(v\in f^{(n1)}(\{u_1,u_2\})\), an even permutation of \(f^{1}(v)\).
4 The geometric representation
Let L be a number field, and consider the rational function field \(K=L(t)\). Since \(t\in L(t) \smallsetminus \{0,1\}\), Lemma 3.1 states that the Galois group of \(L(f^{n}(t))\) over L(t) is a subgroup of \(E_n\). In fact, we have the following much stronger statement.
Proposition 4.1
Proof
Corollary 4.2
\({\text {Gal}}\Big ({\bar{\mathbb {Q}}}\big (f^{n}(t)\big )/{\bar{\mathbb {Q}}}(t)\Big )\cong E_n\).
Proof
5 Counting elements that fix leaves of \(T_n\)
Write \(E_{n,fix }\) for the set of elements of \(E_n\) that fix at least one leaf of \(T_n\). We have already seen that \(E_\infty = \varprojlim E_n\) is the geometric monodromy group of the PCF polynomial \(f(z) = 2z^3 + 3z^2\). Using this fact, one could apply [7, Thm. 1.1] to show that the ratio \(E_{n,fix }/E_n\) tends to zero with n. And while this would be sufficient for the arithmetic applications in the next section, we are able to obtain a more refined statement by working directly with the group structure of \(E_n\):
Theorem 5.1
Remark 5.2
The proportion of elements of \({\text {Aut}}(T_n)\) that fix a leaf of \(T_n\) obeys the same asymptotic as for \(E_n\) [12, §4]. By way of contrast, consider \(H_n \cong [C_3]^n\), the Sylow 3subgroup of \(E_n\) from Proposition 2.6. The proportion of elements of \(H_n\) that fix a leaf of \(T_n\) is half that of \(E_n\): \(\frac{1}{n}\left( 1 + O\left( \frac{\log n}{n}\right) \right) \).
Lemma 5.3
Proof
The restriction homomorphism \(\pi :E_n \rightarrow E_1 \cong \mathfrak {S}_3\) is onto since \(\pi \big ( (1,1,1), (123) \big ) = (123)\) and \(\pi (\tau _n) = (12)\), where \(\tau _n\) was defined in Eq. (7). The first three equalities follow from the fact that \(A_{n,1} = \ker (\pi )\). For the final equality, apply Proposition 2.2. \(\square \)
Lemma 5.4
Proof
Let \(s \in E_{n+1}\) be an element that acts as the identity on \(T_1\). Then the restriction of s to \(T_2\) is of the form \(\big ((a_1,a_2,a_3),1\big )\) for some \(a_1,a_2,a_3\in {\text {Aut}}(T_1)\). By Lemma 2.1, the fact that this element lies in \(E_2\) means \(1 = \prod \mathrm {sgn}(a_i)\). So among the \(a_i\)’s, there are either zero 2cycles or exactly two 2cycles. We treat these cases separately.
Using the same counting technique as in the preceding proof, one obtains:
Lemma 5.5
Lemma 5.6
Proof
\(\square \)
One can apply the technique in the previous proof to obtain the following similar result:
Lemma 5.7
Proof of Theorem 5.1
6 Arithmetic applications
We now prove our applications on density of prime divisors in orbits and Newton’s method. If K is a number field and \(\mathfrak {P}\) is a prime ideal of the ring of integers of K with residue field \(k(\mathfrak {P})\), there is a surjective reduction map \(K \rightarrow k(\mathfrak {P}) \cup \{\infty \}\). We write \(x \equiv y \pmod \mathfrak {P}\) whenever \(x,y \in K\) have the same reduction.
Proposition 6.1
Proof
Note that for all \(i\ge 0\), we have \(y_i \ne x\). This inequality holds for \(i = 0\) by hypothesis. Furthermore, if it failed for some \(i > 0\), \(y_0\) would be a Krational root of \(f^i(z)  x\), which is absurd since this polynomial is irreducible over K by Lemma 3.3.

x is not integral at \(\mathfrak {P}\), or

\(y_i \equiv x \! \pmod \mathfrak {P}\) for some \(0\le i \le n1\).
Recall that the critical points for f are \(0,1,\infty \), and that they are all fixed by f.
Corollary 6.2
Proof

\(\mathfrak {P}\) lies above 2, or

\(y_0 \equiv 0 \pmod \mathfrak {P}\).
Corollary 6.3
Let K be a number field for which there exist unramified primes above 2 and above 3. Let \(g(z) = z^3  z\). Choose \(y_0 \in K\) such that the Newton iteration sequence \(y_i = N_g^i(y_0)\) does not encounter a root of g. Then the set of primes \(\mathfrak {P}\) of K for which the Newton sequence \((y_i)_{i\ge 0}\) converges in \(K_\mathfrak {P}\) to a root of g has natural density zero.
Proof
On the other hand, the proof of the main theorem of Faber–Voloch [3] shows that for all but finitely many prime ideals \(\mathfrak {P}\) of K, the sequence \((y_i)_{i \ge 0}\) converges in \(K_\mathfrak {P}\) to a root of g if and only if \(g(y_i) \equiv 0 \! \pmod \mathfrak {P}\) for some \(i \ge 0\). Factoring g, this condition is equivalent to saying that \(y_i \equiv 0, \pm 1 \!\pmod \mathfrak {P}\) for some \(i \ge 0\), which in turn is equivalent to saying that \(w_i \equiv 0,1, \text { or } \infty \!\pmod \mathfrak {P}\). (Here, \(w\equiv \infty \!\pmod \mathfrak {P}\) means w is not integral at \(\mathfrak {P}\)). Clearly, the set of primes \(\mathfrak {P}\) for which \(w_i \equiv \infty \!\pmod \mathfrak {P}\) is zero, since f is a polynomial, and so the proof is complete. \(\square \)
Declarations
Acknowledgements
This project began at a workshop on “The Galois theory of orbits in arithmetic dynamics” at the American Institute of Mathematics in May 2016. We would like to thank AIM for its generous support and hospitality, Rafe Jones for his early encouragement to pursue this line of thought, and Clay Petsche for several early discussions. We also thank the anonymous referees for pointing out several opportunities for improvement. The first author gratefully acknowledges the support of NSF Grant DMS1501766. The third author gratefully acknowledges the support of NSF Grant DMS1415294.
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