A large arboreal Galois representation for a cubic postcritically finite polynomial
Robert L. Benedetto, Xander Faber, Benjamin Hutz, Jamie Juul, Yu Yasufuku
AA LARGE ARBOREAL GALOIS REPRESENTATION FOR A CUBICPOSTCRITICALLY FINITE POLYNOMIAL
ROBERT L. BENEDETTO, XANDER FABER, BENJAMIN HUTZ, JAMIE JUUL,AND YU YASUFUKU
Abstract.
We give a complete description of the arboreal Galois representation of a certainpostcritically finite cubic polynomial over a large class of number fields and for a large class ofbasepoints. This is the first such example that is not conjugate to a power map, Chebyshevpolynomial, or Latt`es map. The associated Galois action on an infinite ternary rooted treehas Hausdorff dimension bounded strictly between that of the infinite wreath product ofcyclic groups and that of the infinite wreath product of symmetric groups. We deduce azero-density result for prime divisors in an orbit under this polynomial. We also obtaina zero-density result for the set of places of convergence of Newton’s method for a certaincubic polynomial, thus resolving the first nontrivial case of a conjecture of Faber and Voloch. Introduction
Let K be a field and f ∈ K [ z ] a polynomial of degree d ≥
2. Consider the Galois groupsof polynomials of the form f n ( z ) − x, where x ∈ K , and f n = f ◦· · ·◦ f is the n -th iterate of f (with the convention that f ( z ) = z ).Such groups are called arboreal Galois groups because (under certain hypotheses) they canbe made to act on trees.Let T n be the graph whose vertex set is (cid:71) ≤ i ≤ n f − i ( x ) , and where we draw an edge from α to β if f ( α ) = β . Let G n = Gal ( f n ( z ) − x/K ). Clearly, G n acts faithfully on T n , so that G n (cid:44) → Aut( T n ). Provided there is no critical point of f among the points of the above vertex set, the graph T n is a regular d -ary rooted treewith root x . For such f , Aut( T n ) is isomorphic to the n -fold iterated wreath product [ S d ] n of the symmetric group S d on d letters. Odoni and Juul [9, 12] showed that if char( K )and the degree are not both 2, and if f is chosen generically (in the Zariski sense), then G n ∼ = Aut( T n ) ∼ = [ S d ] n .By contrast, for specific choices of f and x , the corresponding Galois groups may be muchsmaller. (See [6, §
3] for a high-level explanation and [2, 4, 8, 14] for detailed examples.)Consider a polynomial f that is postcritically finite , or PCF for short, meaning that allof its critical points have finite orbit under the iteration of f . The simplest examples ofPCF polynomials are the power maps f ( z ) = z d and the Chebyshev polynomials defined by f ( z + 1 /z ) = z d + 1 /z d . These two examples arise from the d -power endomorphism of thealgebraic group G m , which gives a foothold on the associated arboreal Galois representation. a r X i v : . [ m a t h . N T ] D ec A third type of example, Latt`es maps, arises from an endomorphism of an elliptic curve;however, Latt`es maps are never conjugate to polynomials. See [15, § G n have unbounded index inside Aut( T n ) as n → ∞ . However, their proof does not explicitlydescribe G n . One can give an upper bound for G n inside Aut( T n ) by realizing it as aspecialization of Gal ( f n ( z ) − t/K ( t )), with t transcendental over K . The latter group maybe embedded in the profinite monodromy group π ´et1 ( P K (cid:114) P ), where P is the strict postcriticalorbit; this is precisely the tack taken by Pink [14] in the case of quadratic PCF polynomials.In this paper, we give the first complete calculation of the arboreal Galois group attachedto a PCF polynomial over a number field that is not associated with an endomorphism of analgebraic group. More specifically, we describe the Galois groups G n = Gal ( f n ( z ) − x/K )for the polynomial f ( z ) = − z + 3 z over a number field K , where x is chosen to satisfy a certain local hypothesis at the primes 2and 3. In Section 2 we will define groups E n that fit between Aut( T n ) ∼ = [ S ] n and its Sylow3-subgroup [ C ] n — the iterated wreath product of cyclic groups of order 3. The groups E n are somewhat tricky to handle because their action on the tree lacks a certain rigidityproperty: for m < n , the kernel of the restriction homomorphism E n → E m is not the directproduct of copies of E n − m . That is, in contrast to [ S ] n and [ C ] n , the action of E n on onebranch of the tree above T m is not independent of its action on another branch. Our mainresult is the following. Theorem 1.1.
Let K be a number field, let f ( z ) = − z + 3 z ∈ K [ z ] , and let x ∈ K .Suppose there exist primes p and q lying over and , respectively, such that v q ( x ) = 1 , andeither v p ( x ) = ± or v p (1 − x ) = 1 . Then for each n ≥ , (1) The polynomial f n ( z ) − x is irreducible over K . (2) We have an isomorphism
Gal ( f n ( z ) − x/K ) ∼ = E n ⊂ Aut( T n ) .Let E ∞ = lim ←− E n and Aut( T ∞ ) = lim ←− Aut( T n ) be the corresponding inverse limits. Then theHausdorff dimension of E ∞ in Aut( T ∞ ) is lim n →∞ log | E n | log | Aut( T n ) | = 1 −
13 log 2log 6 ≈ . . (1.1) Remark . In this article, we implicitly work in the category of groups with an action onthe regular rooted tree T n . This applies, for example, to the isomorphism between E n andthe Galois group in the theorem.The Galois group Gal ( f n ( z ) − x/K ) depends a priori on the number field K and thebasepoint x , but Theorem 1.1 shows that many choices of K and x give the same isomorphismtype. One key reason is that the discriminant of the second iterate is a square:For any x , Disc (cid:0) f ( z ) − x (cid:1) = (cid:2) · · x ( x − (cid:3) . (1.2)This observation will be vital for forcing the Galois group of f n ( z ) − x to lie inside E n .To fill out the entire group E n , we utilize ramification above the primes 2 and 3. (See theproof of Proposition 3.6). These two features are the only arithmetic-dynamical inputs tothe theorem; the rest is general theory of groups acting on regular rooted trees.We are also able to deduce that the geometric Galois group has the same structure: orollary 1.3. Let f ( z ) = − z + 3 z . Let t be transcendental over Q . For every n ≥ ,we have Gal (cid:0) f n ( z ) − t/ ¯ Q ( t ) (cid:1) ∼ = E n . For a polynomial g ∈ Q [ x ], there are two profinite monodromy groups: G geom g = lim ← n Gal (cid:0) g n ( z ) − t/ ¯ Q ( t ) (cid:1) (geometric monodromy) G arith g = lim ← n Gal ( g n ( z ) − t/ Q ( t )) (arithmetic monodromy) . In general, one knows that G geom g ⊂ G arith g . Theorem 1.1 and its corollary imply that G geom f = G arith f for our special cubic PCF polynomial f ( z ) = − z + 3 z . By contrast, Pink has shownthat G geom g (cid:40) G arith g for all quadratic PCF polynomials g over the rationals [14, Thm. 2.8.4,Cor. 3.10.6]. Similar statements hold upon replacing Q by essentially any other number field.While E n is not an iterated wreath product, it does satisfy the following self-similarityproperty: the action of E n on the subtree of height n − E n − . This self-similarity is a property of geometric iteratedmonodromy 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 toapplications on the density of prime divisors in certain dynamically defined sequences. (Seealso [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.
Let K be a number field for which there exists an unramified prime above and above . Let y ∈ K (cid:114) { , , / , − / } , and define the sequence y i = f i ( y ) . Then theset of prime ideals P such that y i ≡ or P ) for some i ≥ has natural density zero. In particular, the set of prime divisors of the sequence ( y i ) hasnatural density zero. Our choice of the polynomial f ( z ) = − z + 3 z was originally motivated by the followingconjecture of Faber and Voloch [3]. Conjecture 1.5 (Newton Approximation Fails for 100% of Primes) . Let g be a polynomialof degree d ≥ with coefficients in a number field K and let y ∈ K . Define the Newtonmap N ( z ) = z − g ( z ) /g (cid:48) ( z ) and, for each n ≥ , set y n +1 = N ( y n ) . Assume the Newtonapproximation sequence ( y n ) is not eventually periodic. Let C ( K, g, y ) be the set of primes P of K for which ( y n ) converges in the completion K P to a root of f . Then the naturaldensity of the set C ( K, g, y ) is zero. Faber and Voloch showed that Conjecture 1.5 holds for any polynomial g with at most 2distinct 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 − z ,whose associated Newton map turns out to be conjugate to our polynomial f ( z ) = − z +3 z .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 and over . Let g ( z ) = z − z . Choose y ∈ K such that the Newton iteration sequence .. ... ... ... ... ... ... ... .... . .. . . 012 n -1 n Figure 1.
The ternary rooted tree T n with n levels y i = N i ( y ) does not encounter a root of g . Then the set of primes P of K for which theNewton sequence ( y i ) converges in K P to a root of g has natural density zero. The first and third authors, in collaboration with several others, obtained a weak form ofTheorem 1.6 for a wide class of polynomials [1, Thm. 4.6]. More precisely, they showed thatthe density of primes as in the theorem has natural density strictly less than 1.The outline of the paper is as follows. In Section 2, we will define and discuss the group E n and compute the Hausdorff dimension of equation (1.1). We will then prove the rest ofTheorem 1.1 in Section 3. Next, we consider the case that K is the field of rational functions¯ Q ( t ), proving Corollary 1.3 in Section 4. In Section 5, we compute the proportion of elementsof E n that fix at least one leaf of the tree T n , and in Section 6, we prove Theorems 1.4 and 1.6.2. Tree automorphisms
Let T n denote the regular ternary rooted tree with n levels, as in Figure 1. Note that T n has 3 n leaves and 1 + 3 + · · · + 3 n vertices. Our results and many arguments will dependon an implicit labeling of the vertices of T n . We will make this labeling explicit now forpurposes of rigor, but we will not comment on it again afterward. • The level of a vertex is its distance from the root. • A vertex at level i is given a label ( (cid:96) , . . . , (cid:96) i ), where (cid:96) j ∈ { , , } . The root is giventhe empty label (). • No two vertices at the same level have the same label. • The unique path from the root to the vertex with label ( (cid:96) , . . . , (cid:96) i ) consists of thevertices with labels () , ( (cid:96) ) , ( (cid:96) , (cid:96) ) , . . . , ( (cid:96) , (cid:96) , . . . , (cid:96) i ).This labeling enables us to identify certain canonical subtrees of T n . For example, for each i ∈ { , , } , we consider the subtree T that is induced by the set of vertices with labels ofthe form ( i, ∗ , . . . , ∗ ); then T is isomorphic to T n − .The automorphism group Aut( T n ) of the regular rooted ternary tree is isomorphic to the n -fold iterated wreath product [ S ] n . Indeed, we may decompose T n as a copy of T (verticesof level at most 1) with 3 copies of T n − attached along the leaves of T , so thatAut( T n ) ∼ = Aut( T n − ) (cid:111) Aut( T ) ∼ = [ S ] n − (cid:111) S = [ S ] n . (2.1) ur labeling has the effect of fixing an isomorphism Aut( T ) ∼ = S . For any elements a , a , a ∈ Aut( T n − ) and any b ∈ Aut( T ), the element (cid:0) ( a , a , a ) , b (cid:1) ∈ Aut( T n − ) (cid:111) Aut( T ) ∼ = Aut( T n )acts on the tree by first acting on the 3 copies of T n − via a , a , and a , respectively, andthen by permuting these T n − ’s via b . More precisely, if we label a vertex y of T n by ( x, i ),where x is a vertex of T n − and i ∈ { , , } is the vertex at level 1 that lies below y , then (cid:0) ( a , a , a ) , b (cid:1) . ( x, i ) = ( a i .x, b.i ) . A labeling of the leaves of T n induces an injection Aut( T n ) (cid:44) → S n . Changing the labelingcorresponds to conjugating by an element of S n , and so each automorphism of T n has awell-defined sign attached to it, corresponding to the parity of the number of transpositionsneeded to represent it. Thus, for each n ≥
1, we have a homomorphismsgn : Aut( T n ) → {± } . Lemma 2.1.
Let g = (cid:0) ( a , a , a ) , b (cid:1) be an element of Aut( T n ) for some n ≥ , where b ∈ Aut( T ) , and a i ∈ Aut( T n − ) for i = 1 , , . Then sgn( g ) = sgn( b ) (cid:89) i =1 sgn( a i ) . Proof.
Partition the leaves of T n into three disjoint sets L , L , L so that the elements of L i lie over leaf i of T , for i = 1 , ,
3. Note that | L i | = 3 n − . With this notation, sgn( a i ) is thesign 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, sgn( g ) = (cid:81) sgn( a i ).Next, consider the case that g = (cid:0) (1 , , , b (cid:1) for arbitrary b ∈ Aut( T ). We have alreadyproven the desired result if b = 1. If b is a 2-cycle — say b = ( ij ) — then the inducedpermutation on the leaves of T n decomposes as a product of 3 n − disjoint 2-cycles ( a i a j ),where a i ∈ L i and a j ∈ L j . Therefore,sgn( g ) = ( − n − = − b ) . Similarly, if b is a 3-cycle, then the induced permutation on the leaves of T n decomposes asa product of 3 n − disjoint 3-cycles. Hence,sgn( g ) = 1 = sgn( b ) . Finally, we consider the general case g = (cid:0) ( a , a , a ) , b (cid:1) . Define h = (cid:0) (1 , , , b − (cid:1) . Then hg = (cid:0) ( a , a , a ) , (cid:1) . The previous two paragraphs show that (cid:89) i =1 sgn( a i ) = sgn( hg ) = sgn( h ) sgn( g ) = sgn( b − ) sgn( g ) . (cid:3) For any m ≤ n , we have a restriction homomorphism π m : Aut( T n ) → Aut( T m ), where T m is the subtree with m levels with the same root vertex as T n . We write sgn m = sgn ◦ π m or the composition of restriction followed by the sign map. Define a sequence of subgroups E n ⊂ Aut( T n ) by the following formula: E n = (cid:40) Aut( T ) if n = 1( E n − (cid:111) Aut( T )) ∩ ker(sgn ) if n ≥ . Here we use the embedding E n − (cid:111) Aut( T ) (cid:44) → Aut( T n − ) (cid:111) Aut( T ) ∼ = Aut( T n )from equation (2.1). Thus, for n ≥
2, writing a given automorphism σ ∈ Aut( T n ) as σ = (cid:0) ( a , a , a ) , b (cid:1) , we have σ = (cid:0) ( a , a , a ) , b (cid:1) ∈ E n if and only if a , a , a ∈ E n − and sgn ( σ ) = 1 . (2.2) Proposition 2.2.
For n ≥ , we have | E n | = 2 n − · n − .Proof. Since Aut( T ) ∼ = S , the result is clear for n = 1. Suppose it holds for some n ≥ φ be the composition E n (cid:111) Aut( T ) (cid:44) → Aut( T n +1 ) sgn → {± } . By definition, E n +1 = ker( φ ), and φ is onto because ((1 , , , (cid:15) ) (cid:55)→ − (cid:15) of the leaves of T . Thus, | E n +1 | = 12 | E n (cid:111) Aut( T ) | = 12 | Aut( T ) | · | E n | = 12 · · (cid:16) n − · n − (cid:17) = 2 n · n +1 − . (cid:3) Our construction of E n depends on an identification of T n − with the subtree of T n lyingabove a vertex at level 1, which in turn depends on the labeling we have assigned to T n . Inother words, E n is not normal in Aut( T n ) (for n ≥ T n ). Proposition 2.3. E n is normal in Aut( T n ) if and only if n = 1 or .Proof. We have E = Aut( T ), and E has index 2 in Aut( T ). It remains to show that E n is not normal in Aut( T n ) for n ≥
3. To that end, we first construct some special elements ofAut( T n ).Define ν n ∈ Aut( T n ) inductively for n ≥ ν n = (cid:40) (12) n = 1 (cid:0) ( ν n − , , , (cid:1) n ≥ . Thus, ν n transposes two leaves at the n -th level and acts trivially on the rest of T n . Inparticular, sgn( ν n ) = −
1. This yields ν (cid:54)∈ E , and by induction, it follows that ν n (cid:54)∈ E n for n ≥
2. Note further that ν − n = ν n .Next, for fixed n ≥
3, define a = (cid:0) (1 , , , (123) (cid:1) ∈ Aut( T n ). Then a ∈ E n by (2.2).However, ν n a ν − n = (cid:0) ( ν n − , , , (cid:1)(cid:0) (1 , , , (123) (cid:1)(cid:0) ( ν n − , , , (cid:1) = (cid:0) ( ν n − , , ν n − ) , (123) (cid:1) , which does not belong to E n since ν n − (cid:54)∈ E n − for n ≥
3. It follows that E n is not normalin Aut( T n ), as desired. (cid:3) rite T ∞ = (cid:83) n ≥ T n for the infinite ternary rooted tree, which has automorphism groupAut( T ∞ ) = lim ←− Aut( T n ) , where the inverse limit is taken with respect to the restriction homomorphisms π m : Aut( T n ) → Aut( T m ) for m ≤ n. The recursive definition of E n implies that we also have restriction homomorphisms E n → E m for m ≤ n . Passing to the inverse limit gives a subgroup E ∞ = lim ←− E n of Aut( T ∞ ). Corollary 2.4.
The Hausdorff dimension of E ∞ in Aut( T ∞ ) is given by equation (1.1) .Proof. Using the facts that Aut( T n ) ∼ = Aut( T n − ) (cid:111) Aut( T ) and that Aut( T ) ∼ = S , a simpleinduction shows that | Aut( T n ) | = 6 n − . (2.3)Combining this fact with Proposition 2.2 gives the desired result. (cid:3) Corollary 2.5. E ∞ has infinite index in Aut( T ∞ ) . Comparing the cardinalities of E n and Aut( T n ), we see that they share a Sylow 3-subgroup.We can describe one such subgroup explicitly as follows. Let C be the cyclic 3-subgroup ofAut( T ) ∼ = S . Define a sequence of groups H n by the following formula: H n = (cid:40) C if n = 1 H n − (cid:111) C if n ≥ . We identify H n with a subgroup of Aut( T n ) using the embedding H n = H n − (cid:111) C (cid:44) → Aut( T n − ) (cid:111) Aut( T ) ∼ = Aut( T n ) . Evidently, H n ∼ = [ C ] n , the iterated wreath product. By induction, we see that | H n | = 3 n − , (2.4)so that H n is a Sylow 3-subgroup of Aut( T n ). Proposition 2.6.
For n ≥ , H n is a Sylow 3-subgroup of E n . It is normal in E n if andonly if n = 1 .Proof. For n = 1, H n is an index-2 subgroup of E = Aut( T ) and, hence, it is normal.Next, for some n ≥
1, suppose we know that H n is a subgroup of E n . By their recursivedefinitions, to see that H n +1 ⊂ E n +1 it suffices to show that any h ∈ H n +1 restricts to aneven permutation on T . The restriction of h to T has the form (cid:0) ( a , a , a ) , b (cid:1) , where eachof a , a , a , and b is either trivial or a 3-cycle. By Lemma 2.1, we conclude that sgn ( h ) = 1.Hence, h ∈ E n +1 , as desired.Since | H n | is the 3-power 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 ≥ n ≥
1, define τ n ∈ Aut( T n ) inductively as follows: τ n = (cid:40) (12) n = 1 (cid:0) ( τ n − , , , (12) (cid:1) n ≥ . (2.5) e claim that τ n ∈ E n for all n ≥
1. This is clear for n = 1. Suppose that it holds forsome n ≥
1. Then (cid:0) ( τ n , , , (12) (cid:1) ∈ E n +1 if and only if its restriction to T acts by an evenpermutation. The restriction to E is given by (cid:0) ((12) , , , (12) (cid:1) , which has positive sign byLemma 2.1.Note that for n ≥
2, we have τ − n = (cid:0) (1 , τ − n − , , (12) (cid:1) . Note also that for n ≥
1, we have τ n (cid:54)∈ H n , since the restriction of τ n to T is (12) (cid:54)∈ C .Next, for fixed n ≥
2, define a = (cid:0) (1 , , , (123) (cid:1) ∈ H n . Then τ n a τ − n = (cid:0) ( τ n − , , , (12) (cid:1)(cid:0) (1 , , , (123) (cid:1)(cid:0) (1 , τ − n − , , (12) (cid:1) = (cid:0) (1 , τ − n − , τ n − ) , (132) (cid:1) , which does not belong to H n , since τ n − (cid:54)∈ H n − . (cid:3) Proposition 2.7.
The Hausdorff dimension of H ∞ in Aut( T ∞ ) is lim n →∞ log | H n | log | Aut( T n ) | = log 3log 6 ≈ . . Proof.
Immediate from equations (2.3) and (2.4). (cid:3)
Since H n ⊆ E n ⊆ Aut( T n ), the preceding proposition and Corollary 2.4 show that forlarge n , E n is substantially larger than H n , but much smaller than Aut( T n ).Finally, we will need a lemma that constructs certain special elements of E n : Lemma 2.8.
Let n ≥ and let g ∈ Aut( T n ) be any element that acts as follows: • On the copy of T n − with the same root as T n , g acts by the identity. • On each copy of T rooted at a vertex of T n of level n − , g acts by an even permutationof the leaves.Then g ∈ E n .Proof. We proceed by induction on n . For n = 2, the second condition on g implies thatsgn( g ) = 1, so g ∈ E . Suppose that the lemma holds for n −
1, and let g ∈ Aut( T n ) satisfythe given conditions. Let u , u , u be the vertices of T n at level 1. Write T n − ( u i ) for thecopy of T n − inside T n that is rooted at u i . Then g restricts to an element of Aut( T n − ( u i ))that satisfies the two conditions of the lemma. By the induction hypothesis, g | T n − ( u i ) ∈ E n − for i = 1 , ,
3. In addition, g is the identity and, hence, is even, on T . Thus, g ∈ E n by thecriterion of (2.2). (cid:3) Main Theorem
Let K be a field of characteristic zero, and consider the polynomial f ( z ) = − z + 3 z ∈ K [ z ] . The critical points of f in P are 0, 1, and ∞ , all of which are fixed by f . Hence, f is PCFand, in fact, the union of the forward orbits of its critical points is { , , ∞} . Choose a point x ∈ K (cid:114) { , } to be the root of our preimage tree. Note that there is no critical point inthe backward orbit of x — i.e., the set f − ( x ) ∪ f − ( x ) ∪ · · · .For each n ≥
1, define K n = K (cid:0) f − n ( x ) (cid:1) and G n = Gal( K n /K ) . (3.1) Lemma 3.1.
For any field K of characteristic zero and any x ∈ K (cid:114) { , } , the Galoisgroup G n of (3.1) is isomorphic to a subgroup of E n . roof. Because there is no critical point in the backward orbit of x , the set f − i ( x ) consistsof exactly 3 i distinct elements for each i ≥
0. Identify the vertices of the ternary rootedtree T n with the set (cid:70) ≤ i ≤ n f − i ( x ), with vertex y lying immediately above y (cid:48) if and only if f ( y ) = y (cid:48) . This identification induces a faithful action of G n on T n and, hence, G n may beidentified with a subgroup of 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 = Aut( T ).Fix n ≥
2, and assume we know the lemma holds for n −
1. Write f − ( x ) = { y , y , y } .For each i = 1 , ,
3, applying the lemma to the field K ( y i ) with root point y i shows thatGal (cid:16) K (cid:0) f − ( n − ( y i ) (cid:1) /K ( y i ) (cid:17) is a subgroup of E n − . (The labeling of T n allows us to identify the portion of T n above y i with T n − .) Since Gal( K /K ) is a subgroup of Aut( T ), it follows that G n is isomorphic toa subgroup B n of E n − (cid:111) Aut( T ).It remains to show that B n ⊆ ker(sgn ). Direct computation shows that the discriminantof the degree-nine polynomial f ( z ) − x is given by equation (1.2). Since this discriminantis a square in K , all elements of B n act as even permutations of the nine points of f − ( x ).Thus, B n ⊆ ker(sgn ), as desired. (cid:3) Our goal is to compute the arboreal Galois groups G n in the case that K is a number fieldand that the basepoint x ∈ K (cid:114) { , } satisfies the following local hypothesis:There exist primes p and q of K lying above 2 and 3, respectively, ( † )such that v q ( x ) = 1, and either v p ( x ) = ± v p (1 − x ) = 1.If this hypothesis holds, we will say that the pair ( K, x ) satisfies property ( † ) (relative to p and q ). Example . If K = Q , then the pairs ( Q ,
3) and ( Q , /
2) both satisfy property ( † ). Thelatter pair will be important for our arithmetic applications. Lemma 3.3.
Suppose that ( K, x ) satisfies ( † ) relative to p and q . Then f n ( z ) − x is Eisen-stein at q for all n ≥ . In particular, f n ( z ) − x is irreducible for all n ≥ .Proof. A simple induction shows that f n ( z ) ≡ z n (mod q ) and f n (0) = 0. Since v q ( x ) = 1,it follows immediately that f n ( z ) − x is Eisenstein at q . (cid:3) Proposition 3.4.
Let K be a number field, and let x ∈ K . Suppose that ( K, x ) satisfiesproperty ( † ) relative to primes p and q . Let n ≥ , and let y ∈ f − n ( x ) . Then: (1) There are primes p (cid:48) and q (cid:48) of K ( y ) lying above p and q , respectively, such that e ( p (cid:48) / p ) = 2 n and e ( q (cid:48) / q ) = 3 n . (2) The pair ( K ( y ) , y ) satisfies property ( † ) relative to p (cid:48) and q (cid:48) .Proof. We proceed by induction on n . The statement is trivial for n = 0. We thereforeassume for the rest of the proof that n ≥ n − y ∈ f − n ( x ), let y (cid:48)(cid:48) = f ( y ) ∈ f − ( n − ( x ). By our inductive hypothesis, there areprimes p (cid:48)(cid:48) and q (cid:48)(cid:48) of K ( y (cid:48)(cid:48) ) lying over p and q satisfying the desired properties for n −
1. Thepolynomial f ( z ) − y (cid:48)(cid:48) = − z + 3 z − y (cid:48)(cid:48) ∈ K ( y (cid:48)(cid:48) )[ z ] s Eisenstein at q (cid:48)(cid:48) . Thus, there is only one prime q (cid:48) of K ( y ) lying above q (cid:48)(cid:48) , with ramificationindex e ( q (cid:48) / q (cid:48)(cid:48) ) = 3, and with v q (cid:48) ( y ) = 1. Moreover, e ( q (cid:48) / q ) = e ( q (cid:48) / q (cid:48)(cid:48) ) · e ( q (cid:48)(cid:48) / q ) = 3 · n − = 3 n . Meanwhile, by statement (2) for n −
1, we have either v p (cid:48)(cid:48) ( y (cid:48)(cid:48) ) = 1, v p (cid:48)(cid:48) (1 − y (cid:48)(cid:48) ) = 1, or v p (cid:48)(cid:48) ( y (cid:48)(cid:48) ) = −
1. We consider these three cases separately.If v p (cid:48)(cid:48) ( y (cid:48)(cid:48) ) = 1, then the Newton polygon of f ( z ) − y (cid:48)(cid:48) at p (cid:48)(cid:48) has a segment of length 2and height 1. Thus, there is a prime p (cid:48) of K ( y ) lying above p (cid:48)(cid:48) , with ramification index e ( p (cid:48) / p (cid:48)(cid:48) ) = 2, and with v p (cid:48) ( y ) = 1. Moreover, e ( p (cid:48) / p ) = e ( p (cid:48) / p (cid:48)(cid:48) ) · e ( p (cid:48)(cid:48) / p ) = 2 · n − = 2 n . If v p (cid:48)(cid:48) (1 − y (cid:48)(cid:48) ) = 1, note that f is self-conjugate via z (cid:55)→ − z ; that is, 1 − f (1 − z ) = f ( z ).Thus, 1 − y ∈ f − (1 − y (cid:48)(cid:48) ), and the previous paragraph applied to 1 − y gives the desiredconclusion.Finally, if v p (cid:48)(cid:48) ( y (cid:48)(cid:48) ) = −
1, then because v p (cid:48)(cid:48) ( − ≥
1, the Newton polygon of f ( z ) − y (cid:48)(cid:48) at p (cid:48)(cid:48) has a segment of length 2 and height −
1. Thus, there is a prime p (cid:48) of K ( y ) lying above p (cid:48)(cid:48) , with ramification index e ( p (cid:48) / p (cid:48)(cid:48) ) = 2, and with v p (cid:48) ( y ) = −
1. Once again, then, we have e ( p (cid:48) / p ) = 2 n . (cid:3) Corollary 3.5.
Let K and x be as in Proposition 3.4. Let n ≥ and let K n be the splittingfield of f n ( z ) − x over K . Then n (cid:12)(cid:12) [ K n : K ] . Proof.
Pick y ∈ f − n ( x ), and let p (cid:48) and q (cid:48) 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 p must be divisible by e ( p (cid:48) / p ) = 2 n . Similarly, the ramification index of some primeof K n over q must be divisible by e ( q (cid:48) / q ) = 3 n . Thus, 6 n | [ K n : K ]. (cid:3) Proposition 3.6.
Let K and x be as in Proposition 3.4. Then Gal (cid:16) K (cid:0) f − ( x ) (cid:1) /K (cid:17) ∼ = E ∼ = S and Gal (cid:16) K (cid:0) f − ( x ) (cid:1) /K (cid:17) ∼ = E . Proof.
Let K = K ( f − ( x )) and K = K ( f − ( x )). Then, as a splitting field of a cubicpolynomial, K is Galois over K with Gal( K /K ) isomorphic to a subgroup of S . ByCorollary 3.5 with n = 1, we have 6 | [ K : K ]. Hence, Gal( K /K ) ∼ = S .By Lemma 3.1, Gal( K /K ) acts on f − ( x ) as a subgroup of E . It suffices to show thatevery element of E is realized in Gal( K /K ).Write f − ( x ) = { u , u , u } , and for each i = 1 , ,
3, write f − ( u i ) = { v i , v i , v i } . Claim 1.
There exists τ ∈ Gal( K /K ) ⊆ Gal( K /K ) that • fixes v j for each j = 1 , , • acts as a 2-cycle on the set f − ( u ) = { v , v , v } , and • acts as a 2-cycle on the set f − ( u ) = { v , v , v } .To prove Claim 1, note that 36 | [ K : K ] by Corollary 3.5 and, hence, 6 | [ K : K ], since[ K : K ] = 6. By Cauchy’s Theorem, there is some τ ∈ Gal( K /K ) of order 2. Since τ fixes each u i , it must act on each set f − ( u i ) as an element of S of order dividing 2. Thus,for each i = 1 , , τ acts as either a 2-cycle or the identity on f − ( u i ). In addition, τ is ven as a permutation on f − ( x ) but (being of order 2) is not the identity. Thus, τ acts asa 2-cycle on the preimages of exactly two of u , u , u , and as the identity on the preimageof the third, u m .Choose γ (cid:48) ∈ Gal( K /K ) with γ (cid:48) ( u ) = u m , and lift γ (cid:48) to γ ∈ Gal( K /K ). Let τ = γ − τ γ .Then τ ( u i ) = u i for each i , and τ satisfies each of the three bulleted properties, provingClaim 1. Claim 2.
There exists ρ ∈ Gal( K /K ) ⊆ Gal( K /K ) that acts as • a 3-cycle on f − ( u ) = { v , v , v } , • either a 3-cycle or the identity on f − ( u ) = { v , v , v } , and • either a 3-cycle or the identity on f − ( u ) = { v , v , v } .To prove Claim 2, we again note that 6 | [ K : K ]. Hence, by Cauchy’s Theorem, there issome ρ ∈ Gal( K /K ) of order 3. We have ρ ( u i ) = u i for each i , with ρ acting as a 3-cycleon either one, two, or all three of f − ( u i ), and as the identity on the others. Lifting someappropriate γ (cid:48) ∈ Gal( K /K ) to γ ∈ Gal( K /K ), we can ensure that ρ = γ − ρ γ satisfies thedesired properties, proving Claim 2.By the conclusions of Claims 1 and 2, the permutation τ ρ acts as • a 3-cycle on f − ( u ) = { v , v , v } , • a 2-cycle on f − ( u ) = { v , v , v } , and • a 2-cycle on f − ( u ) = { v , v , v } .Let σ = ( τ ρ ) . Then σ acts as a 3-cycle on f − ( u ) and as the identity on f − ( u ) ∪ f − ( u ).Conjugating σ by permutations γ as in the proof of Claim 1 and then composing theresulting conjugates with one another, we see that Gal( K /K ) contains each element of E that is the identity on f − ( x ) and is either the identity or a 3-cycle on each f − ( u i ). Thereare 3 = 27 such permutations.Conjugating τ by permutations from the previous paragraph, as well as by permutations γ as in the proof of Claim 1, we also see that Gal( K /K ) contains each element of E thatis the identity on f − ( x ), is a 2-cycle on exactly two of f − ( u ), f − ( u ), and f − ( u ), andis the identity on the third. There are 3 = 27 such permutations.Composing the maps of the previous two paragraphs, we see that Gal( K /K ) containseach element of E that is the identity on f − ( x ), is a 2-cycle on exactly two of f − ( u ), f − ( u ), and f − ( u ), and is a 3-cycle on the third. There are 3 · K : K ] ≥
27 + 27 + 54 = 108, and, hence,[ K : K ] ≥ ·
108 = 648 = | E | . Since Gal( K /K ) is isomorphic to a subgroup of E , we must have Gal( K /K ) ∼ = E . (cid:3) We are now prepared to prove part (b) of Theorem 1.1, which we restate here.
Theorem 3.7.
Let K and x be as in Proposition 3.4. Then for any n ≥ , Gal (cid:16) K (cid:0) f − n ( x ) (cid:1) /K (cid:17) ∼ = E n . 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 , K and x )for a particular n ≥
2, we will now show it for n + 1. u u u ... ... ... ... ... ... ... ... .. . . y . . . 012 n -1 nn +1 Figure 2.
The locations of x , u , u , u and y in T n +1 .Let K n = K (cid:0) f − n ( x ) (cid:1) and K n +1 = K (cid:0) f − ( n +1) ( x ) (cid:1) . Write f − ( x ) = { u , u , u } . By Proposition 3.4, each of the pairs ( K ( u i ) , u i ) satisfies prop-erty ( † ) for i = 1 , ,
3. Thus, our induction hypothesis says that Gal( K n /K ) and all three ofthe Galois groups Gal (cid:16) K (cid:0) f − n ( u i ) (cid:1) /K ( u i ) (cid:17) , for i = 1 , , E n . Pick y ∈ f − ( n − ( u ) ⊆ f − ( n − ( x ) . See Figure 2. Our main goal, which we will achieve at the end of Step 3 below, is to constructan element λ ∈ Gal( K n +1 /K ) that is the identity on f − ( n +1) ( x ) (cid:114) f − ( y ) and on f − ( y ) , but which acts as two disjoint 2-cycles on f − ( y ). Step 1.
Define H = Gal (cid:16) K (cid:0) f − n ( u ) (cid:1) /K ( u ) (cid:17) . By the induction hypothesis, H ∼ = E n . By Lemma 2.8 there is some σ ∈ H that is theidentity on f − n ( u ) (cid:114) f − ( y ) and on f − ( y )and acts as two 2-cycles on f − ( y ). Lift σ to σ ∈ Gal (cid:0) K n +1 /K ( u ) (cid:1) ⊆ Gal( K n +1 /K ) . Thus, σ acts as we would like λ to act on f − n ( u ), but we have no idea how it acts on f − n ( u ) and f − n ( u ). Step 2.
By Lemma 2.8, we may pick τ ∈ Gal( K n /K ) ∼ = E n that is the identity on f − n ( x ) (cid:114) f − ( y )and acts as a 3-cycle on f − ( y ). Lift τ to τ ∈ Gal( K n +1 /K ).Then τ στ − ∈ Gal( K n +1 /K ) acts as • the identity on f − n ( u ) (cid:114) f − ( y ) and f − ( y ), • two 2-cycles on f − ( y ), and • the same as σ on f − ( n − ( { u , u } ), here the two 2-cycles on f − ( y ) for τ στ − do not occur above the same two elements of f − ( y ) as the two 2-cycles for σ .Thus, τ στ − σ − ∈ Gal( K n +1 /K ) acts as • the identity on f − n ( u ) (cid:114) f − ( y ) and f − ( y ), • two 2-cycles and (perhaps) a separate 3-cycle on f − ( y ), and • the identity on f − ( n − ( { u , u } ).Cubing to kill the possible 3-cycle in f − ( y ), we see that ρ = ( τ στ − σ − ) ∈ Gal( K n +1 /K )acts as • the identity on f − n ( x ), • the identity on f − n ( u ) (cid:114) f − ( y ), and • two 2-cycles on f − ( y ). Step 3.
Consider the permutations τ, ρ ∈ Gal( K n +1 /K ) of Step 2. Then τ ρτ − ρ − acts as • the identity on f − n ( x ), • the identity on f − n ( u ) (cid:114) f − ( y ), • two 2-cycles and (perhaps) a separate 3-cycle on f − ( y ), and • for each v ∈ f − ( n − ( { u , u } ), an even permutation of f − ( v ).The even permutations of the last bullet point above are even permutations in S , and,hence, each is either the identity or a 3-cycle. Cubing, we see that λ = ( τ ρτ − ρ − ) ∈ Gal( K n +1 /K )acts as the identity on f − ( n +1) ( x ) (cid:114) f − ( y ) and on f − ( y ) , and it acts as two 2-cycles on f − ( y ), achieving the main goal from the start of the proof. Step 4.
Recall H = Gal( K ( f − n ( u )) /K ( u )) ∼ = E n . Pick w ∈ f − ( y ). By Lemma 2.8, wecan pick γ ∈ H that is the identity on f − n ( u ) (cid:114) f − ( w )and acts as a 3-cycle on f − ( w ). Lift γ to γ ∈ Gal( K n +1 /K ).Conjugating the permutation λ (of Step 3) by various products of γ and the permutation τ (of Step 2), we see that Gal( K n +1 /K ) contains each of the 27 permutations that is theidentity on Y = (cid:0) f − ( n +1) ( x ) (cid:114) f − ( y ) (cid:1) ∪ f − ( y )and acts as two disjoint 2-cycles on f − ( y ). In addition, taking products of pairs of suchpermutations, Gal( K n +1 /K ) contains all 27 permutations that are the identity on Y andproducts of disjoint 3-cycles on f − ( y ). Still taking products of pairs, Gal( K n +1 /K ) alsocontains all 54 permutations that are the identity on Y and the product of a disjoint 3-cycleand two 2-cycles on f − ( y ). Together, then, Gal( K n +1 /K ) contains a subgroup H y that actstrivially on Y , with | H y | = 108 = 2 · .Since f n − ( z ) − x is irreducible over K , for each root y (cid:48) ∈ f − ( n − ( x ) there is some δ ∈ Gal( K n − /K ) ∼ = E n − ith δ ( y (cid:48) ) = y . Lift δ to δ ∈ Gal( K n +1 /K ). Then, H y (cid:48) = δ − H y δ is a 108-element subgroup of Gal( K n +1 /K ) that acts trivially on Y (cid:48) = (cid:0) f − ( n +1) ( x ) (cid:114) f − ( y (cid:48) ) (cid:1) ∪ f − ( y (cid:48) ) . There are 3 n − choices for y (cid:48) , and any two of the resulting subgroups H y (cid:48) act nontrivially ondisjoint portions of the preimage tree. In addition, they all act trivially on f − n ( x ) and, hence,trivially on K n . Thus, the product of all of them forms a subgroup B ⊆ Gal( K n +1 /K n ) oforder (cid:0) · (cid:1) ( n − = | E n +1 || E n | , by Proposition 2.2. Hence, (cid:12)(cid:12) Gal( K n +1 /K ) (cid:12)(cid:12) = (cid:12)(cid:12) Gal( K n +1 /K n ) (cid:12)(cid:12) · (cid:12)(cid:12) Gal( K n /K ) (cid:12)(cid:12) ≥ | B | · | E n | = | E n +1 | . Since Gal( K n +1 /K ) is isomorphic to a subgroup of the finite group E n +1 , we must thereforehave Gal( K n +1 /K ) ∼ = E n +1 . (cid:3) The geometric representation
Let L be a number field, and consider the rational function field K = L ( t ). Since t ∈ L ( t ) (cid:114) { , } , Lemma 3.1 states that the Galois group of L ( f − n ( t )) over L ( t ) is a subgroupof E n . In fact, we have the following much stronger statement. Proposition 4.1.
Let L be a number field and t a transcendental element over L . Then Gal (cid:16) L (cid:0) f − n ( t ) (cid:1) /L ( t ) (cid:17) ∼ = E n . Proof.
Let G n = Gal( L ( f − n ( t )) /L ( t )). We can choose x ∈ L such that the pair ( L, x )satisfies property ( † ) of Section 3. By Theorem 3.7, we haveGal (cid:16) L (cid:0) f − n ( x ) (cid:1) /L ( x ) (cid:17) = Gal (cid:16) L (cid:0) f − n ( x ) (cid:1) /L (cid:17) ∼ = E n . Therefore, the specialization lemma of [12, Lem. 2.4] implies that G n has a subgroup iso-morphic to E n . On the other hand, applying Lemma 3.1 to G n shows that G n is isomorphicto a subgroup of E n . Since E n is finite, G n must be isomorphic to E n . (cid:3) Corollary 4.2.
Gal (cid:16) ¯ Q (cid:0) f − n ( t ) (cid:1) / ¯ Q ( t ) (cid:17) ∼ = E n .Proof. Since the previous proposition holds for any number field, Q must be algebraicallyclosed in Q ( f − n ( t )). Hence,Gal (cid:16) ¯ Q (cid:0) f − n ( t ) (cid:1) / ¯ Q ( t ) (cid:17) ∼ = Gal (cid:16) Q (cid:0) f − n ( t ) (cid:1) / Q ( t ) (cid:17) ∼ = E n . (cid:3) . 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 alreadyseen that E ∞ = lim ←− E n is the geometric monodromy group of the PCF polynomial f ( z ) = − z + 3 z . 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 inthe next section, we are able to obtain a more refined statement by working directly withthe group structure of E n : Theorem 5.1.
The proportion of elements of E n that fix a leaf of T n is | E n, fix || E n | = 2 n (cid:18) O (cid:18) log nn (cid:19)(cid:19) as n → ∞ . Remark . The proportion of elements of Aut( T n ) that fix a leaf of T n obeys the sameasymptotic as for E n [12, § H n ∼ = [ C ] n , the Sylow 3-subgroup 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 : n (cid:0) O (cid:0) log nn (cid:1)(cid:1) .We begin by finding a recursive formula for | E n, fix | in terms of certain auxiliary quantities.For n ≥ i ∈ { , , } , we define the following: A n,i = (cid:12)(cid:12)(cid:12)(cid:8) s ∈ E n (cid:12)(cid:12) s acts as an element of order i on T (cid:9)(cid:12)(cid:12)(cid:12) A (cid:48) n,i = (cid:12)(cid:12)(cid:12)(cid:8) s ∈ E n | s acts as an element of order i on T and fixes a leaf of T n (cid:9)(cid:12)(cid:12)(cid:12) . For example, if n = 1, we have A , = 1 A (cid:48) , = 1 A , = 3 A (cid:48) , = 3 A , = 2 A (cid:48) , = 0 . For any n ≥
1, note that A (cid:48) n, = 0, because an element s that permutes the leaves of T by a 3-cycle cannot fix a leaf of T n . It follows that | E n, fix | = A (cid:48) n, + A (cid:48) n, and | E n | = A n, + A n, + A n, . Lemma 5.3.
For n ≥ , we have A n, = 3 A n, , A n, = 2 A n, , | E n | = 6 A n, , and | E n +1 | = 3 | E n | . Proof.
The restriction homomorphism π : E n → E ∼ = S is onto since π (cid:0) (1 , , , (123) (cid:1) =(123) and π ( τ n ) = (12), where τ n was defined in (2.5). The first three equalities follow fromthe fact that A n, = | ker( π ) | . For the final equality, apply Proposition 2.2. (cid:3) Lemma 5.4.
For n ≥ , we have A (cid:48) n +1 , = 54 A n, (cid:0) A (cid:48) n, + A (cid:48) n, (cid:1) − A n, (cid:0) A (cid:48) n, + A (cid:48) n, (cid:1) + 3 A (cid:48) n, (cid:0) A (cid:48) n, (cid:1) + (cid:0) A (cid:48) n, (cid:1) . Proof.
Let s ∈ E n +1 be an element that acts as the identity on T . Then the restriction of s to T is of the form (cid:0) ( a , a , a ) , (cid:1) for some a , a , a ∈ Aut( T ). By Lemma 2.1, the factthat this element lies in E means 1 = (cid:81) sgn( a i ). So among the a i ’s, there are either zero2-cycles or exactly two 2-cycles. We treat these cases separately. ase 1: Zero 2-cycles. As an element of E n +1 , we have s = (cid:0) (˜ a , ˜ a , ˜ a ) , (cid:1) , where each˜ a i ∈ E n restricts to either the identity or a 3-cycle on T . The total number of elements ˜ a i ofthis shape is A n, + A n, , while the number that do not fix a leaf of T n is (cid:0) A n, + A n, − A (cid:48) n, (cid:1) .Thus, the number of elements of E n +1 , fix that act as the identity on T but with no 2-cycleon T is( A n, + A n, ) − (cid:0) A n, + A n, − A (cid:48) n, (cid:1) = 27 A n, A (cid:48) n, − A n, (cid:0) A (cid:48) n, (cid:1) + (cid:0) A (cid:48) n, (cid:1) , (5.1)where we applied Lemma 5.3 when we expanded the two expressions. Case 2: Two 2-cycles.
As an element of E n +1 , we have s = (cid:0) (˜ a , ˜ a , ˜ a ) , (cid:1) , where twoof the ˜ a i ∈ E n restrict to 2-cycles on T , and the remaining ˜ a i restricts to the identity or a3-cycle. There are 3 choices for the index i such that ˜ a i is the identity or a 3-cycle. For agiven choice of i , there are( A (cid:48) n, + A (cid:48) n, ) A n, = A (cid:48) n, A n, = 9 A n, A (cid:48) n, choices of triples (˜ a , ˜ a , ˜ a ) such that s fixes a leaf of the copy of T n above the i -leaf of T .For s not to fix a leaf of the T n above the i -leaf, at least one of the other two ˜ a i ’s (eachof which acts as a 2-cycle on T ) must fix a leaf of T n . By inclusion-exclusion, the numberof choices for this pair of ˜ a i ’s is2 A n, A (cid:48) n, − (cid:0) A (cid:48) n, (cid:1) = (cid:0) A n, − A (cid:48) n, (cid:1) A (cid:48) n, . For i = i , the element ˜ a i ∈ E n acts as the identity or as a 3-cycle on T but fixes no leaf of T n . The number of such elements of E n is A n, + A n, − A (cid:48) n, = 3 A n, − A (cid:48) n, . (Again, we have applied Lemma 5.3 in all three displayed equations above.)Thus, the number of elements s ∈ E n +1 , fix that act as the identity on T and as two 2-cycleson the leaves of T is3 (cid:104) A n, A (cid:48) n, + (cid:0) A n, − A (cid:48) n, (cid:1) (cid:0) A n, − A (cid:48) n, (cid:1) A (cid:48) n, (cid:105) . (5.2)Adding (5.1) and (5.2) and expanding yields the desired result. (cid:3) Using the same counting technique as in the preceding proof, one obtains:
Lemma 5.5.
For n ≥ , we have A (cid:48) n +1 , = 54 A n, (cid:0) A (cid:48) n, + A (cid:48) n, (cid:1) . Lemma 5.6.
Let φ ( t ) = t − t + t . Then φ is increasing on (0 , , and φ n (1) = 2 n (cid:18) O (cid:18) log nn (cid:19)(cid:19) as n → ∞ . Proof.
The first statement is evident from looking at the derivative. For the second, we set ψ ( z ) = 1 φ ( z − ) = z + 12 − z + 22(6 z − z + 2) . Let R ( z ) = z +22(6 z − z +2) be the final term. Then, by induction, we have ψ n ( z ) = 1 φ n ( z − ) = z + n − n − (cid:88) i =0 R (cid:0) ψ i ( z ) (cid:1) . ince φ n (1) = 1 /ψ n (1), to complete the proof it suffices to show that (cid:80) n − i =0 R ( ψ i (1)) = O (log n ).By elementary algebra, one verifies that R ( z ) ≤ z for all z >
0. Now we show, byinduction, that ψ n (1) ≥ ( n + 5) / n ≥
1. Both sides equal 6 / n = 1. Given n ≥ R ( ψ n (1)) ≤ ψ n (1) ≤ n +15 , and hence ψ n +1 (1) = ψ n (1) + 12 − R ( ψ n (1)) ≥ n + 55 + 12 − n + 15= n + 65 + 9 n − n + 5) > n + 65 , completing the induction.Using our inequalities for R ( z ) and ψ n (1), we conclude that0 ≤ n − (cid:88) i =0 R (cid:0) ψ i (1) (cid:1) < n − (cid:88) i =0 ψ i (1) < n − (cid:88) i =0 i + 5 = O (log n ) . (cid:3) One can apply the technique in the previous proof to obtain the following similar result:
Lemma 5.7.
Let ρ ( t ) = t − t . Then ρ is increasing on (0 , , and ρ n (2 /
3) = 2 n (cid:18) O (cid:18) log nn (cid:19)(cid:19) as n → ∞ . Proof of Theorem 5.1.
By adding the terms (cid:0) A (cid:48) n, (cid:1) and 3 A (cid:48) n, (cid:0) A (cid:48) n, (cid:1) to the formula inLemma 5.4, we obtain the estimate A (cid:48) n +1 , ≤ A n, (cid:0) A (cid:48) n, + A (cid:48) n, (cid:1) − A n, (cid:0) A (cid:48) n, + A (cid:48) n, (cid:1) + (cid:0) A (cid:48) n, + A (cid:48) n, (cid:1) . Adding this to the formula in Lemma 5.5, we find that | E n +1 , fix | = A (cid:48) n +1 , + A (cid:48) n +1 , ≤ A n, (cid:0) A (cid:48) n, + A (cid:48) n, (cid:1) − A n, (cid:0) A (cid:48) n, + A (cid:48) n, (cid:1) + (cid:0) A (cid:48) n, + A (cid:48) n, (cid:1) = 3 (6 A n, ) | E n, fix | −
32 (6 A n, ) | E n, fix | + | E n, fix | = 3 | E n | · | E n, fix | − | E n | · | E n, fix | + | E n, fix | , where we have used Lemma 5.3 to write 6 A n, = | E n | . Dividing by | E n +1 | = 3 | E n | , we findthat | E n +1 , fix || E n +1 | ≤ | E n, fix || E n | − (cid:18) | E n, fix || E n | (cid:19) + 13 (cid:18) | E n, fix || E n | (cid:19) = φ (cid:18) | E n, fix || E n | (cid:19) , (5.3)where φ is the polynomial from Lemma 5.6.Similarly, by discarding the final two terms of the formula in Lemma 5.4 and adding theformula from Lemma 5.5, we obtain the estimate | E n +1 , fix | = A (cid:48) n +1 , + A (cid:48) n +1 , ≥ A n, (cid:0) A (cid:48) n, + A (cid:48) n, (cid:1) − A n, (cid:0) A (cid:48) n, + A (cid:48) n, (cid:1) = 3 | E n | · | E n, fix | − | E n | · | E n, fix | . s above, this yields | E n +1 , fix || E n +1 | ≥ | E n, fix || E n | − (cid:18) | E n, fix || E n | (cid:19) = ρ (cid:18) | E n, fix || E n | (cid:19) , (5.4)where ρ is the polynomial from Lemma 5.7.Set x n = | E n, fix || E n | ∈ [0 , ρ ( x n ) ≤ x n +1 ≤ φ ( x n ). As ρ and φ are increasing on (0 , ρ n (2 /
3) = ρ n ( x ) ≤ ρ n − ( x ) ≤ · · · ≤ ρ ( x n ) ≤ x n +1 and x n +1 ≤ φ ( x n ) ≤ φ ( x n − ) ≤ · · · ≤ φ n ( x ) ≤ φ n (1) . By Lemmas 5.6 and 5.7, the first and last quantities have the same asymptotic value, namely n (cid:0) O (cid:0) log nn (cid:1)(cid:1) , and hence, so does x n +1 . The proof is complete since this asymptotic isunchanged upon replacing n with n − (cid:3) 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 P is a prime ideal of the ring of integers of K with residue field k ( P ), there is a surjective reduction map K → k ( P ) ∪ {∞} . We write x ≡ y (mod P )whenever x, y ∈ K have the same reduction. Proposition 6.1.
Let K be a number field and x ∈ K an element such that ( K, x ) satisfiesproperty ( † ) of Section 3. Choose y ∈ K (cid:114) { x } and define a sequence ( y i ) i ≥ ⊆ K by y i = f i ( y ) . Then the set of prime ideals P of K such that y i ≡ x (mod P ) for some i ≥ has natural density zero.Proof. Note that for all i ≥
0, we have y i (cid:54) = x . This inequality holds for i = 0 by hypothesis.Furthermore, if it failed for some i > y would be a K -rational root of f i ( z ) − x , which isabsurd since this polynomial is irreducible over K by Lemma 3.3.For each n ≥
1, define S n to be the set of prime ideals P of K such that • x is not integral at P , or • y i ≡ x (mod P ) for some 0 ≤ i ≤ n − S n is finite.Let n ≥
1. Let P (cid:54)∈ S n be a prime ideal of the ring of integers of K such that y i ≡ x (mod P ) for some i ≥ n . Then f n ( y i − n ) ≡ x (mod P ), and therefore the polynomial f n ( z ) − x has a k ( P )-rational root. Write δ ( S ) for the natural density of a set of primeideals S (if it exists). For each n ≥
1, the Chebotarev Density Theorem and the finitenessof S n yield δ (cid:0)(cid:8) P (cid:12)(cid:12) y i ≡ x (mod P ) for some i ≥ (cid:9)(cid:1) ≤ δ (cid:0)(cid:8) P (cid:12)(cid:12) P (cid:54)∈ S n , f n ( z ) − x has a k ( P )-rational root (cid:9)(cid:1) = (cid:12)(cid:12)(cid:12)(cid:110) s ∈ Gal (cid:0) K ( f − n ( x )) /K (cid:1) (cid:12)(cid:12)(cid:12) s fixes some root of f n ( z ) − x (cid:111)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Gal (cid:0) K ( f − n ( x )) /K (cid:1)(cid:12)(cid:12) = | E n, fix || E n | . he final equality uses Theorem 3.7 to identify the Galois group of f n ( z ) − x over K with E n . By Theorem 5.1, this last quantity tends to zero as n → ∞ . (cid:3) Recall that the critical points for f are 0 , , ∞ , and that they are all fixed by f . Corollary 6.2.
Let K be a number field for which there exist unramified primes above andabove . Let y ∈ K (cid:114) { , , / , − / } , and define a sequence ( y i ) i ≥ ⊆ K by y i = f i ( y ) .Then the set of prime ideals P of K such that y i ≡ or P ) for some i ≥ has natural density zero. In particular, the set of prime divisors of the sequence ( y i ) i ≥ hasnatural density zero.Proof. We begin by showing that the set of primes P such that y i ≡ P ) for some i ≥ S be the set of primes P of K for which • P lies above 2, or • y ≡ P ).Notice that since y (cid:54) = 0, S is a finite set, and we may safely ignore primes in S for theremainder of the proof.Suppose now that P (cid:54)∈ S is such that y i ≡ P ) for some i ≥
1. We may assumewithout loss that i is minimal with this property. Then y i = f ( y i − ) = − y i − + 3 y i − = − y i − (cid:18) y i − − (cid:19) , which implies that y i − ≡ (mod P ). We claim that y i − (cid:54) = . This is true by hypothesisif i = 1. If it were to fail for some i >
1, then the polynomial f i − ( z ) − / y as a K -rational root. But our hypothesis on K implies that ( K, /
2) satisfies property ( † ),and so we have a contradiction to Lemma 3.3. It follows that δ (cid:16)(cid:8) P (cid:12)(cid:12) y i ≡ P ) for some i ≥ (cid:9)(cid:17) = δ (cid:18)(cid:110) P (cid:12)(cid:12)(cid:12) y i ≡
32 (mod P ) for some i ≥ (cid:111)(cid:19) . Since ( K, /
2) satisfies property ( † ), the density on the right is zero by Proposition 6.1.Now we show that the density of primes P such that y i ≡ P ) for some i ≥ w i = 1 − y i for i ≥
0. As y (cid:54)∈ { , − / } , we see that w (cid:54)∈ { , / } . Moreover, because 1 − f ( z ) = f (1 − z ), we find that f ( w i ) = f (1 − y i ) = 1 − f ( y i ) = 1 − y i +1 = w i +1 . Thus, we may apply the first part of the proof to the sequence ( w i ) i ≥ to deduce that δ (cid:0) { P (cid:12)(cid:12) w i ≡ P ) for some i ≥ } (cid:1) = 0 . Since w i ≡ P ) ⇔ y i ≡ P ), we are done. (cid:3) We now prove a special case of the Faber-Voloch conjecture. Recall that the
Newton map associated to a polynomial g ( z ) ∈ K [ z ] is the rational function N g ( z ) = z − g ( z ) g (cid:48) ( z ) . he simple roots of g are critical fixed points of N g . Hence, for any completion K v of K , theroots of g are super-attracting fixed points of the map N g , viewed as a dynamical systemacting on P ( K v ). Corollary 6.3.
Let K be a number field for which there exist unramified primes above and above . Let g ( z ) = z − z . Choose y ∈ K such that the Newton iteration sequence y i = N ig ( y ) does not encounter a root of g . Then the set of primes P of K for which theNewton sequence ( y i ) i ≥ converges in K P to a root of g has natural density zero.Proof. The Newton map for g is N g ( z ) = 2 z z − . Let η ( z ) = 1 / (1 − z ). Then η − ◦ N g ◦ η ( z ) = − z + 3 z = f ( z ) . (6.1)For each i ≥
0, define w i = η − ( y i ). Then it is immediate from (6.1) that w i +1 = f ( w i ).Moreover, if we had w ∈ { , , / , − / , ∞} , then the sequence ( w i ) i ≥ would encounter afixed point of f , in which case ( y i ) i ≥ would encounter a fixed point of N g and, hence, a rootof g , contradicting our hypotheses. Thus, w (cid:54)∈ { , , / , − / , ∞} . Corollary 6.2 thereforeshows that the set of prime ideals P for which w i ≡ P ) has density zero.On the other hand, the proof of the main theorem of Faber-Voloch [3] shows that for allbut finitely many prime ideals P of K , the sequence ( y i ) i ≥ converges in K P to a root of g if and only if g ( y i ) ≡ P ) for some i ≥
0. Factoring g , this condition is equivalent tosaying that y i ≡ , ± P ) for some i ≥
0, which in turn is equivalent to saying that w i ≡ , , or ∞ (mod P ). (Here, w ≡ ∞ (mod P ) means w is not integral at P .) Clearly,the set of primes P for which w i ≡ ∞ (mod P ) is zero, since f is a polynomial, and so theproof is complete. (cid:3) Acknowledgments:
This project began at a workshop on “The Galois theory of orbits inarithmetic dynamics” at the American Institute of Mathematics in May 2016. We would liketo thank AIM for its generous support and hospitality, Rafe Jones for his early encouragementto pursue this line of thought, and Clay Petsche for several early discussions. We also thankthe anonymous referees for pointing out several opportunities for improvement. The firstauthor gratefully acknowledges the support of NSF grant DMS-1501766. The third authorgratefully acknowledges the support of NSF grant DMS-1415294.
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Graduate Texts in Mathe-matics . Springer, New York, 2007.(Benedetto)
Amherst College, Amherst, MA
E-mail address , Benedetto: [email protected] (Faber)
Center for Computing Sciences, Institute for Defense Analyses, Bowie, MD
E-mail address , Faber: [email protected] (Hutz)
Saint Louis University, Saint Louis, MO
E-mail address , Hutz: [email protected] (Juul)
Amherst College, Amherst, MA
E-mail address , Juul: [email protected] (Yasufuku)
College of Science and Technology, Nihon University, Tokyo, Japan
E-mail address , Yasufuku: [email protected]@math.cst.nihon-u.ac.jp