The Galois Group of Iterated Polynomial of X^{p^r}-c over Non-archimedean Valued Field

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(1)國立臺灣師範大學理學院數學系碩士論文 Department of Mathematics, College of Science. National Taiwan Normal University Master's Thesis. The Galois Group of Iterated Polynomial of X^{p^r} - c over Non-archimedean Valued Field. 鍾明廷 Zhong, Ming-Ting. 指導教授：夏良忠博士 Advisor: Hsia, Liang-Chung, Ph.D. 中華民國 109 年 7 月 July 2020.

(2) Abstract Let K be a finite extension over Qp the fraction field of p-adic integers. Let f (X) = r. X p − c ∈ K[X] where r ∈ Z≥2 , and let f n (X) be the nth iterated polynomial of f (X). For any a ∈ K, we examine the Galois groups and the ramified index of Kn over K where Kn is the splitting field of f n (X) − a over K. For some v(c), the behavior depends on v(c). But for. −p p−1. −. pr pr −1 (r. − 1) ≤ v(c) <. −p p−1 ,. we haven’t found results.. Keywords— iterated polynomial, arboreal Galois group, infinitely wildly ramified.. i.

(3) Contents Abstract. i. 1 Introduction. 1. 2 Preliminaries on local field. 2. 3 v(c) ≥ 0 4. −p p−1. 22. < v(c) < 0. 5 v(c) =. −p p−1. 6 v(c) <. −p p−1. 26 29. −. pr pr −1 (r. − 1). 30. References. 34. ii.

(4) 1. Introduction. Let K be a field with characteristic 0 and let f (X) be a polynomial over K. Denote the nth iterated polynomial of f (X) by f n (X) = (f ◦ f ◦ · · · ◦ f )(X). Fix a ∈ K, let f −n (a) be the {z } | n times. set of the solutions of f n (X) = a in K, the algebraic closure of K. Let Kn = K(f −n (a)) and K∞ = ∪n≥1 Kn . When K contains a primitive lth root of unity and f (X) = X l − c is separable, the Galois group of Kn over K can be viewed as a subgroup of a wreath product of cyclic groups of order l. It is natural to ask whether or not the Galois group is isomorphic to the wreath product of groups. For f (X) = X 2 + 1, R. W. K. Odoni [Odo88] gives a criterion to check when the Galois group Gal(Q(f −n (0))/Q) is isomorphic to the n-fold wreath product [C2 ]n of the cyclic group of order 2. M. Stoll [Sto92] gives a sufficient condition of this problem. J. Anderson, S. Hamblen, B. Poonen, and L. Walton [AHPW18] consider a similar problem for the polynomial f (X) = X l − c over a finite extension K of Qp . Theorem ([AHPW18, Theorem 1.3]). Let K be a finite extension over Qp and f (X) = X p − c ∈ K[X] where c ∈ K × . 1. If v(c) < 2. If v(c) =. −p p−1 ,. then K∞ /K is a finite extension.. −p p−1 ,. then K∞ /K is an infinite extension, and K∞ /K is finitely ramified if and. only if a lies within the closed unit disk centered at a fixed point of f (X). 3. If v(c) >. −p p−1 ,. then K∞ /K is infinitely ramified.. In this thesis, we consider a more general situation. Specifically, we focus on the polynomial r. f (X) = X p − c ∈ K[X] where r ≥ 2 and c ∈ K × . We will prove the following result. r. Theorem. Let K be a finite extension over Qp , and let f (X) = X p − c. 1. If v(c) ≥. −p p−1 ,. 2. If v(c) <. −p p−1. then K∞ /K is infinitely ramified.. −. pr pr −1 (r. − 1), then K∞ /K is finite extension.. However, we haven’t found results yet for. −p p−1. −. pr pr −1 (r. − 1) ≤ v(c) <. −p p−1 .. The structure of this thesis is as follows. We introduce necessary tools to deal with the problem in Section 2. The case v(c) ≥ and v(c) =. −p p−1 .. −p p−1. is separated into the next three sections: v(c) ≥ 0,. In Section 6, we focus on the case v(c) <. 1. −p p−1. −. pr. pr −1 (r. − 1).. −p p−1. < v(c) < 0.

(5) 2. Preliminaries on local field. Definition 1. A valuation on a field K is a function v : K × → R satisfying the following conditions: 1. v(ab) = v(a) + v(b) and 2. v(a + b) ≥ min{v(a), v(b)} for all a, b ∈ K × . For convenience, we define v(0) = ∞. Proposition 1. Let v be a valuation on a field K, then 1. v(1) = v(−1) = 0, hence v(k) = v(−k) for any k in K. 2. If a and b in K satisfy v(a) 6= v(b), then v(a + b) = min{v(a), v(b)}. Proof.. 1. First v(1) = v(1 ∗ 1) = v(1) + v(1) which implies v(1) = 0. Moreover, 0 = v(1) = v((−1)(−1)) = v(−1) + v(−1) which implies v(−1) = 0.. 2. Without loss of generality, we suppose v(a) > v(b). Then v(a + b) ≥ min{v(a), v(b)} = v(b). Conversely, v(b) = v((a + b) + (−a)) ≥ min{v(a + b), v(−a)} = min{v(a + b), v(a)} = v(a + b).. Definition 2. Let K be a field with a valuation v, the valuation ring of K is OK = {x ∈ K | v(x) ≥ 0}. Proposition 2. The valuation ring OK is a local ring with its maximal ideal mK = {x ∈ K | v(x) > 0}. Proof. We prove that OK is a ring such that for all x ∈ OK − mK is a unit in OK . For any elements a, b ∈ OK , we have the following.. 2.

(6) 1. v(a + b) ≥ min{v(a), v(b)} ≥ 0 which implies a + b ∈ OK . 2. v(−a) = v(a) ≥ 0 by Proposition 1. So −a ∈ OK . 3. v(ab) = v(a) + v(b) ≥ 0 which implies ab ∈ OK . 4. v(1) = 0 by Proposition 1. So 1 ∈ OK . Hence OK is a subring of K. For any x ∈ OK − mk , we have v(x) = 0. Then there exists x−1 ∈ K such that xx−1 = 1 since x is a nonzero element. Thus, 0 = v(1) = v(xx−1 ) = v(x) + v(x−1 ), so x−1 ∈ OK since v(x−1 ) = 0. Hence x is a unit in OK . Definition 3. Let K be a field. A function | · | : K → R≥0 is called a norm if any a, b ∈ K satisfy 1. |ab| = |a||b|, 2. |a + b| ≤ |a| + |b| and 3. |a| = 0 if and only if a = 0. A norm | · | is said to be a non-archimedean norm if it satisfies |a + b| ≤ max{|a|, |b|}. Proposition 3. Let K be a field. Fix a real number c > 1, then there exists a one-to-one correspondence between the set V of valuations on K and the set N of non-archimedean norms on K: V ↔N v 7→ | · | (v(x) = − logc (|x|)). (|x| = c−v(x) ). v ←[ | · |. Definition 4. A norm | · | on K is called a trivial norm if   0 if x = 0, |x| =  1 if x 6= 0. It is clearly that a trivial norm is a non-archimedean norm. According to Proposition 3 there exists a trivial valuation v such that.   ∞ if x = 0, v(x) =  1 if x 6= 0.. 3.

(7) Proposition 4. Let K be a field equipped with a non-trivial norm | · |. Then for any M ∈ R>0 , there exist x1 , x2 ∈ K such that |x1 | < M < |x2 |. Proof. Because | · | is a non-trivial norm on K, there exist a nonzero element y in K such that |y| = 6 1. Since 1 = |1| = |yy −1 | = |y||y −1 |, either |y| > 1 or |y −1 | > 1. Without loss of generality, we assume |y| > 1. Then we can take a sufficiently large n ∈ N such that |(y −1 )N | < M < |y N |.. Definition 5. A Cauchy sequence in K under a norm | · | is a sequence {xn }∞ n=0 in K such that for any > 0, there exists an integer N such that |xm − xn | < whenever m, n ≥ N . The sequence {xn }∞ n=0 is said to converge to a point x if for any > 0, there exists an integer N such that |xn − x| < whenever n ≥ N . K is called complete under | · | if any Cauchy sequence converges to a point in K. Theorem 1. Let K be a field equipped with a non-archimedean norm | · |. Then there exists a field ˆ equipped with a non-archimedean norm | · |0 such that K ˆ is complete under | · |0 and |k|0 = |k| for K all k ∈ K. Proof. See [Ked10, Theorem1.1.4] Fix a prime number p. For any r ∈ Q, we can write r=. a a0 = pe ( 0 ) for some a, b, e, a0 , b0 ∈ Z with p - a0 b0 . b b. Define a function vp on Q by vp : Q → Z r 7→ e. From the definition of vp , it is not difficult to show that vp is a non-trivial valuation on Q, so there is a non-trivial non-archimedean norm | · |p = p−v(·) on Q with respect to vp (we often take c equal. 4.

(8) to p in the Proposition 3). Since Q is not complete under | · |p . According to Theorem 1, we can extend | · |p to a non-archimedean norm on a field, which is denoted by Qp . For convenience, we still write | · |p (resp. vp ) for the non-archimedean norm (resp. valuation) on Qp . Since vp (Q× ) = Z, we have that vp (Q× p ) is also equal to Z. We denote the valuation ring of Qp by Zp . Since v(Qp ) = Z, the maximal ideal of Zp is {a ∈ Zp | vp (a) > 0} = pZp . Let K be a finite extension over Qp . We will prove that there exists a unique non-archimedean norm extending | · |p on K. First, we discuss the norm on finite dimensional vector space over Qp since we can view K as a finite dimensional vector space over Qp . Definition 6. Let V be a vector space over Qp . A norm on V is a function k · k : V → R≥0 such that for any x, y ∈ V and α ∈ Qp , the following hold: 1. kxk ≥ 0 and kxk = 0 if and only if x = 0, 2. kαxk = |α|p kxk, 3. kx + yk ≤ kxk + kyk. Definition 7. Let V be a vector space over Qp . Two norms k · k1 and k · k2 defined on V are equivalent if there exist two constants C, D ∈ R>0 such that Ckxk1 ≤ kxk2 ≤ Dkxk1 for all x ∈ V. Theorem 2. Let V be vector space over Qp , then two norms k · k1 and k · k2 defined on V are equivalent if and only if they induce the same topology on V . Proof. Suppose k · k1 and k · k2 are equivalent, so there exist C, D ∈ R>0 such that Ckxk1 ≤ kxk2 ≤ Dkxk1 for all x ∈ V . For any y ∈ V and r ∈ R>0 , we have {x ∈ V | kx − yk1 <. r } ⊆ {x ∈ V | kx − yk2 < r} D. and {x ∈ V | kx − yk2 < Cr} ⊆ {x ∈ V | kx − yk1 < r}. Hence k · k1 and k · k2 induce the same topology on V .. 5.

(9) Conversely, suppose k · k1 and k · k2 induce the same topology on V , then there exists s > 0 such that {x ∈ V | kxk1 < s} ⊆ {x ∈ V | kxk2 < 1}. Since | · |p is a non-trivial norm on Qp , we can take γ ∈ Qp such that |γ|p > 1. Then for any nonzero v ∈ V , there exists an integer n such that 1 (|γ|p )n ≤ kvk1 < (|γ|p )n+1 s Hence k. v. k2 γ n+1. <1. since k. v k1 < s. γ n+1. Then we have kvk2 < |γ|n+1 = |γ|p |γ|np ≤ p. |γ|p kvk1 . s. Similarly, there is a constant C such that kvk1 ≤ Ckvk2 for all v ∈ V . Proposition 5. Let V and W be vector spaces over Qp equipped with norms k · kV and k · kW , respectively, and let θ : V → W be a Qp -linear map. Then θ is continuous if and only if kθk := sup {kθ(v)kW } < ∞. kvkV ≤1. Proof. For any a ∈ V , b ∈ W , and r ∈ R>0 , define BV (a, r) := {z ∈ V | kz − akV < r} and BW (b, r) := {z ∈ W | kz − bkW < r}. Suppose θ is continuous. Then for any r0 > 0, there exists an r > 0 such that θ(BV (0, r)) ⊆ BW (0, r0 ). By Proposition 4, there exists an α in Qp satisfying |α|p < r. Hence for any a ∈ BV (0, 1), we have kθ(αa)kW < r0 since kαakV < r. So we obtain kθ(a)k <. 6. r0 |α|p .. This shows that kθk < ∞..

(10) Conversely, suppose kθk = c < ∞. Then for any x in V and any open neighborhood U of θ(x), there exists an r ∈ R>0 such that BW (θ(x), r) ⊆ U . By Proposition 4, there exists an element α ∈ Qp satisfying |α|p < rc . For any element y in BV (x, |α|p ), we have kθ(α−1 (y − x))kW ≤ kθk = c since kα−1 (y − x)kV ≤ 1. Therefore, θ(y) ∈ BW (θ(x), r) ⊆ U since kθ(y) − θ(x)kW = kθ(y − x)kW ≤ |α|p kθk < r. This shows that θ is continuous. Let V be a finite dimensional vector space over Qp and let B = {b1 , . . . , bn } be an ordered basis n P of V where n = dimQp V . For any v ∈ V , v = ci bi for some ci ∈ Qp . For i = 1, . . . , n, define the i=1. Qp -linear map λi by λi : V → Qp v 7→ ci and a norm k · kB on V by kvkB = max {|ci |p }. 1≤i≤n. Moreover, V is complete with respect to k · kB since Qp is complete with respect to | · |p . Theorem 3. Let V be a finite dimensional vector space over Qp equipped with a norm k · k. Then for any ordered basis B of V , the norms k · k and k · kB are equivalent. Hence all norms on V are equivalent. As a consequence, V is complete with respect to any norm on it. Proof. Write B = {b1 , . . . , bn } where n = dimQp V . Then for any x ∈ V , there exist c1 , . . . , cn ∈ Qp n P such that x = ci bi . So we have i=1. kxk = k. n X i=1. ci bi k ≤ max {kci bi k} ≤ max {|ci |p } max {kbi k} = DkxkB 1≤i≤n. 1≤i≤n. where D = max {kbi k}. 1≤i≤n. 7. 1≤i≤n.

(11) We claim that there exist a constant C ∈ R>0 such that kxkB ≤ Ckxk for any x ∈ V. We prove the claim by induction on n, the dimension of V over Qp . For n = 1, write B = {b1 }. Take C ∈ R>0 such that Ckb1 k ≥ 1. Then for any x ∈ V , kxkB = |c1 |p ≤ Ckb1 k|c1 |p = Ckxk where x = c1 b1 . Suppose the claim is true for any (n − 1)-dimensional vector space over Qp . It follows that for any ordered basis B 0 of a (n − 1)-dimensional vector space over Qp equipped with a norm k · k0 , both norms k · k0 and k · kB 0 are equivalent. Then for any x ∈ V , kxkB = max {|λi (x)|p } ≤ kxk max {kλi k} 1≤i≤n. 1≤i≤n. where λi : V → Qp , i = 1, . . . , n, are Qp -linear maps given as above, kλi k = sup {kλi (x)k} and kxkB ≤1. x=. n P. λi (x)bi. i=1. If all the λi are continuous, then according to Proposition 5, we can take C = max {kλi k} < ∞, 1≤i≤n. so the claim follows. Now, we prove that λ1 are continuous. Suppose λ1 is not continuous. Then there exists a sequence {xj }j∈N such that kxj k → 0 but |λ1 (xj )|p 9 0 as j → ∞. Thus, there exists > 0 and a subsequence {x0j } of {xj } such that kx0j k → 0 as j → ∞ but |λ1 (x0j )|p ≥ > 0 for all j. Let yj = and write yj = b1 +. n P. x0j λ1 (x0j ). cij bi for some cij ∈ Qp . Since kyj k → 0, we have. i=2. lim. j→∞. n X. ! cij bi. = −b1 .. i=2. Then b1 ∈ V 0 , the vector space spanned by b2 , . . . , bn , since V 0 is complete with respect to k · k, we get a contradiction by the induction hypothesis. Similarly, we can prove that all the λi are continuous. We complete the proof.. 8.

(12) Theorem 4. Let K be a finite extension over Qp . Then there exists at most one norm extending | · |p on K. Proof. Suppose k · k1 and k · k2 are norms extending | · |p on K. Then k · k1 and k · k2 are norms on K, which is viewed as a vector space over Qp . According to Theorem 3, there exist C, D > 0 such that Ckxk1 ≤ kxk2 ≤ Dkxk1 for any x ∈ K. Replacing x by xn in the preceding inequalities, we can get 1. 1. C n kxk1 ≤ kxk2 ≤ D n kxk1 . 1. 1. Hence C n → 1 and D n → 1 whenever n → ∞, so kxk1 = kxk2 . In the following, we will construct a non-archimedean norm extending | · |p on K. Theorem 5. Let f (X) ∈ Zp [X], suppose there exist g1 (X) and h1 (X) in Zp [X] such that 1. g1 (X) is monic. 2. g1 (X) and h1 (X) are coprime in (Zp /pZp )[X] where g1 (X) = g1 (X) + pZp [X] and h1 (X) = h1 (X) + pZp [X]. 3. f (X) = g1 (X) h1 (X). then there exist polynomials g(X), h(X) ∈ Zp [X] such that 1. g(X) is monic and deg(g(X)) = deg(g1 (X)). 2. g(X) = g1 (X) and h(X) = h1 (X). 3. f (X) = g(X)h(X). Proof. See [Gou97]. Corollary 1. Let f (X) = X n + an−1 X n−1 + · · · + a1 X + a0 be an irreducible polynomial in Qp [X]. If a0 ∈ Zp , then so is ai for all i = 1, . . . , n − 1.. 9.

(13) Proof. Suppose ai ∈ / Zp for some i ∈ {1, . . . , n − 1}. Take r = min{i | |ai |p = max{|a0 |p , . . . , |an−1 |p , |1|p }}. Clearly, r is equal to neither n nor 0. Then the polynomial A(X) := a. since | arj |p =. |aj |p |ar |p. 1 f (X) ∈ Zp [X] ar. ≤ 1 for all j = 0, . . . , n. Since r−1. X ai 1 an−1 ar A(X) = X ( X n−r + r X n−1−r + . . . + ) + X i, ar a ar ar r. i=0. apply Theorem 5 to A(X) with g1 (X) = X r and h1 (X) =. 1 n−r an−1 n−1−r ar X + r X + ... + . ar a ar. Clearly, g1 (X) is monic and g1 (X) and h1 (X) are coprime. By the definition of r, we have A(X) = g1 (X) h1 (X). Hence A(X) is reducible over Qp , so is f (X) which contradicts the condition that f (X) is irreducible over Qp . Lemma 1. Suppose dimQp K = d, and for any x ∈ K, define a Qp -linear map Tx by Tx : K → K k 7→ xk. Then d. det Tx = (−1)d (f (0)) m where f (X) is the minimal polynomial of x over Qp with deg(f (X)) = m. Proof. See [Ash10] Theorem 6. Let K be a finite extension over Qp with [K : Qp ] = d, and define k · kK : K → R≥0 1. x 7→ (| det Tx |p ) d . Then k · kK is a non-archimedean norm extending k · kp on K. Hence there is a unique nonarchimedean norm extending | · |p on K by Theorem 4.. 10.

(14) Proof.. 1. For any x, y ∈ K, Tx ◦ Ty = Txy , so det Tx · det Ty = det Txy . Hence we have 1. 1. 1. 1. kxykK = (| det Txy |p ) d = (| det Tx · det Ty |p ) d = (| det Tx |p ) d (| det Ty |p ) d = kxkK kykK . 2. Let x ∈ K. we claim that if kxkK ≤ 1, then k1 + xkK ≤ 1. Let f (X) = X m + am−1 X m−1 + · · · + a0 be the minimal polynomial of x over Qp . By Lemma 1, we have |a0 |p ≤ 1 since kxkK ≤ 1. By Corollary 1, we have ai ∈ Zp for all i = 0, 1, . . . , m. Since the irreducible polynomial of 1 + x over Qp is f (X − 1), we have k1 + xkK = (|f (−1)|p ). d m. m X d d = (| (−1)i ai |p ) m ≤ ( max {|ai |p }) m ≤ 1. i=0. 0≤i≤m. take any x, y ∈ K, suppose kxkK ≥ kykK , then kx + ykK = kx(1 +. x x )kK = kxkk k1 + kK ≤ kxkK . y y. 3. It is clearly that k0kK = 0. Let x be a nonzero element in K. Then det Tx 6= 0 since Tx has the inverse function Tx−1 . Hence kxkK 6= 0 and this shows that kxkK = 0 if and only if x = 0. 1. 4. For any x ∈ Qp , we have det Tx = xd , so kxkK = (| det Tx |p ) d = |x|p .. Corollary 2. Let K be a finite extension over Qp equipped with a norm k · kK extending | · |p . Then for any field L which is a finite extension over K, there exists a unique norm k · kL extending k · kK on L. Proof. Since L is a finite extension over Qp , there exists the unique norm k · kL extending | · |p on L by Theorem 6. Restrict the norm k · kL to K, then it is a norm extending | · |p on K. According to Theorem 6, it is the same as k · kK , hence k · kL is extending k · kK . For convenience, we denote the norm (resp. valuation) extending | · |p (resp. vp ) on K a finite extension over Qp by k · k (resp. v). Recall Definition 2, we have OK = {x ∈ K | v(x) ≥ 0} and mK = {x ∈ K | v(x) > 0}.. 11.

(15) Definition 8. Let K be a finite extension over Qp . The residue degree of K over Qp is f = f (K/Qp ) = [OK /mk : Zp /pZp ] (the degree of field extension), and the ramification index of K over Qp is e = e(K/Qp ) × × = [kK × k : |Q× p |p ] (the index of subgroup |Qp |p in kK k) × × = [v(K × ) : vp (Q× p )] (the index of subgroup vp (Qp ) in v(K )).. Let L be a finite extension over K. By Corollary 2, we can also define the ramified index of L over K by e(L/K) = [v(L× ) : v(K × )], and the residue degree of L over K is f (L/K) = [OL /mL : OK /mK ]. Proposition 6. Let L be a finite extension over K, and let K be a finite extension over Qp , then by definition, we have f (L/Qp ) = f (L/K)f (K/Qp ) and e(L/Qp ) = e(L/K)e(K/Qp ). Theorem 7. Let K be a finite extension over Qp . Then n = [K : Qp ] ≥ e(K/Qp )f (K/Qp ). Proof. Let ω1 , . . . , ωf ∈ OK such that ω1 + mK , . . . , ωf + mk are linearly independent over Zp /pZp , × × × and let π1 , . . . , πe ∈ K × such that the cosets kπ1 k|Q× p |p , . . . , kπe k|Qp |p in kK k/|Qp |p are distinct.. We claim that the set {ωi πj | 1 ≤ i ≤ f, 1 ≤ j ≤ e} is linearly independent over Qp , so that ef ≤ n. First, we prove that if x=. f X. yi ωi where yi ∈ Qp ,. i=1. then kxk = max |yi |p . 1≤i≤f. 12.

(16) It is clearly that if all the yi are zero, so we may assume that |y1 |p is the maximum of the |yi |p , and y1 6= 0. Then yf x y2 = ω1 + ω2 + . . . + ωf ∈ OK y1 y1 y1 since the norm of each term in. x y1. is less than or equal to 1. Moreover, since ω1 + mk , . . . , ωf + mk. are linearly independent over Zp /pZp , we have x + mk 6= 0 + mk . y1 This shows that. x y1. ∈ / mK . Hence kxk = |y1 |p since k yx1 k = 1.. Suppose the set {ωi πj | 1 ≤ i ≤ f, 1 ≤ j ≤ e} is linearly dependent over Qp , so there exist not all zero cij ∈ Qp such that f e X X. cij πj ωi =. j=1 i=1. Let Yj =. f P. f e X X (πj ( cij ωi )) = 0. j=1. i=1. cij ωi , then. i=1. k. e X. πj Yj k = max {kπj Yj k} = 6 0. j=1. 1≤j≤e. since kπj Yj k are in distinct cosets, and this is in contradiction with. e P. πj Yj = 0. Hence we have. j=1. n = [K : Qp ] ≥ e(K/Qp )f (K/Qp ) = ef . Let K be a finite extension over Qp equipped with the valuation v extending vp . Given a polynomial f (X) in K[X], and let α be a root of f (X). Since [K(α) : K] is finite, we can extend the valuation v on K(α) by Corollary 2. If we can calculate v(α) through f (X), then we can get a lower bound of e(K(α)/Qp ), and so is [K(α) : K]. Because our goal is to get a lower bound on [K(α) : K], if f (X) has a trivial root 0, then we just need to consider g(X) ∈ K[X] where f (X) = X r g(X) and g(X) is not divisible by X. The key tool is the Newton polygon which is defined as follows. Definition 9. Let K be a field equipped with a valuation v. For any f (X) = an X n +· · ·+a0 ∈ K[X] with a0 6= 0, the Newton polygon of f (X) is the convex hull of the set of points {(i, v(ai )) | 0 ≤ i ≤ n)} ∪ (0, ∞). 13.

(17) Theorem 8. Let K be a field complete with respect to a valuation v, and let f (X) ∈ K[X] with f (0) 6= 0. Then for each finite side of the Newton polygon of f(X) there corresponds some roots of f (X). The number(counting multiplicities) of roots corresponding to a given side is equal to the length of the projection of that side on the x-axis, and all roots α corresponding to the same side have v(α) = −λ where λ is the slope of the side. Proof. We may assume f (0) = 1, and write f (X) =. n Y. (1 + βi X) = 1 + a1 X + . . . + an X n. i=1. and suppose v(β1 ) ≤ v(β2 ) ≤ . . . ≤ v(βn ). Suppose {v(β1 ), v(β2 ), . . . , v(βn )} = {γ1 , . . . , γl } with γ1 < . . . < γl , and suppose γi appear with multiplicity ni , for 1 ≤ i ≤ l. We must prove that the Newton polygon of f (X) has l non-vertical sides P0 P1 , P1 P2 , . . . , Pl−1 Pl where P0 = (0, 0), P1 = (n1 , n1 γ1 ), P2 = (n1 + n2 , n1 γ1 + n2 γ2 ), . . . , k k n n X X X X Pk = ( ni , ni γi ), . . . , Pn = ( ni , ni γi ). i=1. i=1. i=1. i=1. This is equivalent to prove that v(an1 +...+nk ) =. k X. ni γi. i=1. and for any the integer s satisfying n0 + n1 + . . . + nk < s < n1 + n2 + . . . + nk+1 , then v(as ) ≥. k X. ni γi + γk+1 (s −. i=0. k X. ni ),. i=1. for k ∈ {0, . . . , l − 1} where n0 = γ0 = 0, that is the point (s, v(as )) lies on or above the side Pk Pk+1 . Since as =. X. βi1 βi2 . . . βis ,. 1≤i1 ≤i2 ≤...≤is ≤n. 14.

(18) we have v(as ) ≥ v(β1 β2 . . . βs ) =. s X. v(βi ) =. i=1. Finally, if s =. k P. k X. ni γi + γk+1 (s −. i=0. k X. ni ).. i=1. ni , since v(βj ) > v(βs ) for any j > s, we have. i=1. v(as ) = v(β1 β2 . . . βs ) =. s X. v(αi ) =. i=1. k X. ni γi .. i=1. Example. Let 1 1 f (X) = 57 + 53 X + 5X 3 + X 4 + X 6 + X 7 + 52 X 8 ∈ Q5 [X]. 5 10 Then the Newton polygon of f (X): 7 6 5 4 3 2 1. 1. 2. 3. 4. 5. −1 −2 By Theorem 8, there are 1 root β satisfying v(β) = −. 3−7 = 4, 1−0. 3 roots β satisfying v(β) = −. −1 − 3 4 = , 4−1 3. 15. 6. 7. 8.

(19) 3 roots β satisfying v(β) = −. −1 − (−1) =0 7−4. and 1 root β satisfying v(β) = −. 2 − (−1) = −3. 8−7. Definition 10 ([Lub13]). Let K be a field equipped with a valuation v. For any f (X) = an X n + . . . + a0 ∈ K[X] with a0 6= 0, the Newton copolygon of f (X) is the set of points such that Y − iX − v(ai ) ≤ 0 for all i = 0, . . . , n. Theorem 9. Let K be a field complete with to a valuation v, and let f (X) ∈ K[X] with f (0) 6= 0. A point (c, d) in the Newton coplygon lying on the intersection of two lines corresponds some roots of f (X), the number(counting multiplicities) of roots corresponding (c, d) is equal to the difference of the slope of two lines, and all roots α corresponding (c, d) have v(α) = c. Example. Let 1 1 f (X) = 57 + 53 X + 5X 3 + X 4 + X 6 + X 7 + 52 X 8 ∈ Q5 [X]. 5 10 Then the Newton copolygon of f (X):. 16.

(20) 10 Y =7 (4, 7). Y =X −3 5. ( 43 , 13 4 ) Y = 4X − 1. −15. −10. −5. 5 (0, −1) −5. Y = 7X − 1. −10. −15. −20 (−3, 22) Y = 8X + 2 −25. −30 The point (4, 7) corresponds to 1 − 0 = 1 solution β satisfying v(β) = 4 The point ( 43 , 13 4 ) corresponds to 4 − 1 = 3 solutions β satisfying v(β) =. 4 3. The point (0, −1) corresponds to 7 − 4 = 3 solutions β satisfying v(β) = 0. 17. 10. 15.

(21) The point (−3, 22) corresponds to 8 − 7 = 1 solution β satisfying v(β) = −3. Theorem 10. (Krasner’s Lemma) Let K be a finite extension over Qp , and let a ∈ K(an algebraic closure of K). Denote the set of all conjugates of a over K by S. If b ∈ K satisfies kb − ak < min {ka − αk}, α∈S\{a}. then K(a) ⊆ K(b). Proof. Take any b ∈ K such that a ∈ / K(b), let L be the splitting field of the irreducible polynomial of a over K(b) over K(b). Since the characteristic of K is 0, we have L is a Galois extension over K, so there exist a σ ∈ Gal(L/K(b)) such that σ(a) 6= a. Then kb − σ(a)k = kσ(b − a)k = kb − ak, ka − σ(a)k ≤ max{ka − bk, kb − σ(a)k} = kb − ak.. Let K be a finite extension over Qp , fix d, α ∈ K, then we can extend v to K(d, α) by Corollary 2. r. r. And consider the pr roots β1 , . . . , βpr of X p − αp = d. Although we can calculate v(βi ) to get r. r. a lower bound of e(L/K) where L is the splitting field of X p − αp − d over K, we can get more information from v(βi − α). Since βi − α are the roots of r. r. (X + α)p − αp = d, r. r. we can draw the Newton copolygon of (X + α)p − αp = d to compute v(βi − α). For this, we need the following Lemma. r Lemma 2. v( pi ) = r − v(i) where r, i ∈ Z≥0 . Proof. r p pr ! v( ) = v( ) i i!(pr − i)! = v(pr !) − v(i!) − v((pr − i)!) X r r r r r X X p i p −i = − − pj pj pj j=1 j=1 j=1 r r r X X i p −i j = p − + pj pj j=0. j=0. 18.

(22) where bxc is the greatest integer less than or equal to x. Clearly, . i pj. .   r−j p − 1 if pij ∈ / Z, pr − i + = j  pr−j p otherwise. . Hence X v(i) r X pr − i i pr−j + b jc+b j c = p p j=0. j=0. Therefore,. r X. r X pr−j − 1 = pj − (r − v(i)) . j=0. j=v(i)+1. r r r X X p i pr − i j v( )= p − b j c + b j c = r − v(i). i p p j=0. j=0. Since. r. (X + α). pr. −α. pr. p r X p r − d = −d + X i αp −i , i i=1. we consider the lines L0 : Y − v(d) and r p Li : Y − (iX + (p − i)v(α) + v( )) i r. for i = 1, 2, . . . , pr . Clearly, Li passes the point (v(α), pr v(α) + r − v(i)) for i = 1, 2, . . . , pr , and we define Jj := {Li | v(i) = j} = {Li | (v(α), pr v(α) + r − j) ∈ Li }. For i = 1, . . . , pr , we have Li ∈ Jj for some j ∈ {0, 1, . . . , r}. Fix j 6= r and consider the Newton copolygon determined by Jj :. (v(α), pr v(α) + r − j). LM. 19. Lm.

(23) where m = min{i | Li ∈ Jj } = pj and M = max{i | Li ∈ Jj } = pj (pr−j −1). Hence, we only need to consider Li where i = pj or pr − pj for some j ∈ {0, 1, . . . , r}. Since the point (v(α), pr v(α) + r − j) lies on the line X = v(α) for all j ∈ {0, 1, . . . , r}, we consider the Newton copolygon on the left of X = v(α): (v(α), pr v(α) + r). Lpr −1 (v(α), pr v(α) + r − 1). Lpr −p . . . (v(α), pr v(α)). Lpr. Since the slope of Lpr is the largest, the Newton coploygon on the left of X = v(α) is Lpr . Hence the r. r. Newton coploygon of (X + α)p − (α)p = d is determined by {L0 } ∪ {Lpi }ri=0 . Fix i ∈ {0, 1, . . . , r}, for any j ∈ {1, . . . , i}, we have the intersection of Lpi and Lpi−j is (v(α) +. pi. pj j , pr v(α) + + r − i). i−j −p p−1. First, we consider the intersection of Lpr and Lpi for all i ∈ {0, . . . , r − 1}, then we have C. Lpr−1 Lpr. Then, we consider the intersection of Lpr−1 and Lpi for all i ∈ {0, . . . , r − 2} and so on. Hence the Newton copolygon determined by {Lpi }ri=0 : Lp0 L 1 p .. . Lpr−2 Lpr−1 Lpr. 20.

(24) Finally, we consider L0 . We have the following Lemma. Lemma 3. Let K be a finite extension over Qp . Fix d, α ∈ K, and consider the solutions β of r. r. X p − αp = d. 1. If v(d) ≤ pr v(α) +. p p−1 ,. then there are pr solutions β such that v(β − α) =. 2. If v(d) > pr v(α) +. p p−1. v(d) . pr. + r − 1, then there are pr−i − pr−i−1 solutions β such that v(β − α) = v(α) +. 1 pr−i − pr−i−1. for i = 0, 1, . . . , r − 1 and one solution β such that v(β − α) = v(d) − (pr − 1)v(α) − r, moreover, this solution is in K(d, α) by Theorem 10. 3. If pr v(α) +. p p−1. + j < v(d) ≤ pr v(α) +. p p−1. + j + 1 for some j ∈ {0, 1, . . . , r − 2}, then there. are pr−i − pr−i−1 solutions β such that v(β − α) = v(α) +. pr−i. 1 − pr−i−1. for i = 0, 1, . . . , j, and pr−j−1 solutions β such that v(β − α) =. v(d) − (pr − pr−j−1 )v(α) − j − 1 . pr−j−1. 21.

(25) 3. v(c) ≥ 0 r. Recall our setting: K is a finite extension over Qp and f (X) = X p − c ∈ K[X] where r ∈ N≥2 and c ∈ K × . Denote the nth iterated polynomial of f (X) by f n (X) = (f ◦ f ◦ · · · ◦ f )(X). | {z } n times. Fix an a in K and consider the field K∞ =. [. Kn. n≥1. where Kn is the splitting field of f n (X) − a over K. Let {αn }n∈N be a sequence in K such that f (α1 ) = a and f (αn+1 ) = αn for n ≥ 1. Definition 11. K∞ /K is called infinitely wildly ramified if for any m ∈ N, there exists a positive integer n such that pm | e(Kn /K). Since vp (Q× p ) = Z, it is easier to check whether or not K∞ /Qp is a infinitely wildly ramified extension. By the definition of ramified index, if K∞ /Qp is infinitely wildly ramified, then so is K∞ /K. Lemma 4 ([AHPW18, Lemma 7.1]). If v(c) ≥ 0, min{v(a), v(c)} = 6 0 and v(a) 6= v(c), then K∞ /K is infinitely wildly ramified. Proof. For any n ∈ N, if v(c) > 0, then min{v(a), v(c)} v(c) ≤ r n < v(c). r n (p ) (p ) If v(c) = 0, then v(a) is negative since min{v(a), v(c)} = 6 0. Therefore, min{v(a), v(c)} v(a) = r n < 0 = v(c). r n (p ) (p ) Hence we have min{v(a), v(c)} < v(c) for any n ∈ N. (pr )n We claim that v(αn ) =. min{v(a), v(c)} 6= 0 for all n ∈ N. (pr )n. 22.

(26) If we have the claim, then K∞ /K is infinitely wildly ramified since the denominator of v(αn ) tends to infinity. r. We prove it by induction on n. First, the equation α1p − c = a implies that v(α1 ) =. v(a + c) min{v(a), v(c)} = 6= 0. r p pr. Suppose v(αn−1 ) =. min{v(a), v(c)} . (pr )n−1. r. Then the equation αnp − c = αn−1 implies that v(αn ) =. min{v(c), v(αn−1 )} v(αn−1 ) min{v(a), v(c)} = = 6= 0. pr pr (pr )n. Since the exponent of p in the denominator of v(αn ) grows with n, K∞ /Qp is infinitely wildly ramified, and so is K∞ /K. Lemma 5. If v(c) > 0 and v(a) = 0, then K∞ /K is infinitely wildly ramified. Proof. Since v(c) > 0 and v(a) = 0, the proof of Lemma 4 shows that v(αn ) = 0 for all n ∈ N. We can choose β1 ∈ f −1 (a) such that β1 6= α1 (because v(a + c) = v(a) = 0 6= ∞), take a sequence {βn }n∈N in K such that f (βn+1 ) = βn for all n ∈ N which implies βn+1 ∈ f −(n+1) (a). Since β1 is a root of r. r. X p − α1p = 0, by Lemma 3, we have 0 < v(β1 − α1 ) =. pr−i. 1 p < for some i. r−i−1 −p p−1. We claim that v(βn − αn ) =. v(β1 − α1 ) 6= 0 for n ≥ 2. (pr )(n−1). If we have the claim, then K∞ /K is infinitely wildly ramified since the denominator of v(βn − αn ) tends to infinity. We prove it by induction on n. Since β2 is a root of r. r. X p − α2p = β1 − α1 and v(β1 − α1 ) <. p p−1 ,. we have v(β2 − α2 ) =. 23. v(β1 − α1 ) pr.

(27) by Lemma 3. Suppose v(βn − αn ) =. v(β1 − α1 ) (pr )(n−1). where n ≥ 2. Since βn+1 is a root of r. r. p X p − αn+1 = β n − αn. and v(βn − αn ) < v(β1 − α1 ) <. p , p−1. by Lemma 3 again, we have v(βn+1 − αn+1 ) =. v(βn − αn ) v(β1 − α1 ) = . r p (pr )n. Since the exponent of p in the denominator of v(βn − αn ) grows with n, K∞ /Qp is infinitely wildly ramified, and so is K∞ /K. Theorem 11. If v(c) ≥ 0, then K∞ /K is infinitely wildly ramified. Proof. Assume v(αn ) = v(c) ≥ 0 for all n ∈ N. Then we have v(β1 − α1 ) = v(c) +. 1 p < pr v(c) + pr−i − pr−i−1 p−1. where β1 ∈ f −1 (a) and β1 6= α1 . By the induction on n, we have v(βn − αn ) =. v(β1 − α1 ) (pr )n−1. where βn ∈ f −1 (βn−1 ). So K∞ /K is infinitely wildly ramified. By Lemma 4 and Lemma 5, we only need to consider v(a) > v(c) = 0 or v(a) = v(c) ≥ 0. We separate the above cases into the following cases: 1. If v(a) = v(c) = 0, then v(αn ) = v(c) = 0 for all n ∈ N, hence K∞ /K is infinitely wildly ramified. 2. If v(a) > v(c) = 0, then v(αn ) = v(c) = 0 for all n ∈ N, hence K∞ /K is infinitely wildly ramified.. 24.

(28) 3. If v(a) = v(c) > 0, if v(αn ) = v(c) for all n ∈ N, then we are done. So suppose there exists a N ∈ N such that v(αN ) 6= v(c). Let E = K(αN ), En = E(f −n (αN )) and E∞ = ∪n≥1 En . So if v(αN ) 6= 0, then apply Lemma 4, we have E∞ /K is infinitely wildly ramified. If v(αN ) = 0, then apply Lemma 5, we have E∞ /K is infinitely wildly ramified. Hence K∞ /K is infinitely wildly ramified since E∞ /E is.. 25.

(29) −p p−1. 4. < v(c) < 0. The condition. −p p−1. < v(c) implies that 0<. v(c) 1 p + r < v(c) + r r−1 p p −p p−1. and v(c) 1 + r > 0. r p p − pr−1 First, we estimate v(αn ) to calculate v(βn − αn ) where βn ∈ f −n (a). r. Lemma 6 ([AHPW18, Lemma 3.2]). Suppose v(c) < 0 and f (X) = X p − c, then for any α ∈ f −n (a), v(α) =. v(c) pr. for n large. Moreover, if v(a) > v(c), then v(α) =. v(c) pr. for all n ∈ N and for. any α ∈ f −n (a). According to the Lemma 6, for any a ∈ K we can find an N ∈ N such that v(αN ) =. v(c) pr. for any. αN ∈ f −N (a). Similarly, let E = K(αN ) and let En = E(f −n (αN )). If E∞ /E is infinitely wildly ramified where E∞ = ∪n≥1 En , then so is K∞ /K. Theorem 12. If. −p p−1. < v(c) < 0, then K∞ /K is infinitely wildly ramified.. Proof. Because of the above reason, we can suppose v(a) = v(αn ) = r. v(c) pr .. By Lemma 6, we have. v(c) for all n ∈ N. pr. r. Since the solutions of X p − α1p = 0 and the set f −1 (a) coincide, there exists a β1 ∈ f −1 (a) such that 0 < v(β1 − α1 ) =. 1 p v(c) + r < v(c) + . r r−1 p p −p p−1. Take a sequence {βn+1 }n∈N in K such that f (βn+1 ) = βn for all n ∈ N. We claim that v(βn − αn ) =. v(β1 − α1 ) for n ≥ 2. (pr )n−1. If we have the claim, then K∞ /K is infinitely wildly ramified since the denominator of v(βn − αn ) tends to infinity. We prove it by induction on n. Since β2 is a root of r. X p − α2 = β1 − α1 ,. 26.

(30) by Lemma 3, we have v(β2 − α2 ) =. v(β1 − α1 ) . pr. v(βn − αn ) =. v(β1 − α1 ) (pr )(n−1). Suppose. where n ≥ 2. Since βn+1 is a root of r. r. p X p − αn+1 = β n − αn. and v(βn − αn ) < v(β1 − α1 ) < v(c) +. p , p−1. by Lemma 3 again, we have v(β1 − α1 ) v(βn − αn ) = . r p (pr )n. v(βn+1 − αn+1 ) =. r. Next, we give a upper bound of [Kn : K] for every n. Let µpr = {ξ | ξ p = 1} ⊆ K. Since µpr ⊆ K1 and Km+1 = Km (Am ) where Am is the set of the pr th roots of the (pr )m numbers αm + c as αm ranges over the element in f −m (a), we have [Km+1 : Km ] ≤ (pr )(p. r )m. for all m ≥ 1. And we. have [K1 : K(µpr )] ≤ pr and [K(µpr ) : K] ≤ pr − 1, so we get a upper bound r. [Kn : K] ≤ (p − 1). n−1 Y. r (pr )i. (p ). r. = (p − 1). i=0. ri. (pr )(p ) .. i=0. Lemma 7. Let ∈ Kn for some n ∈ N. If v() > r. n−1 Y. p p−1. r. + r − 1, then 1 + ∈ (Kn× )p .. r. Proof. Consider the solutions of X p − 1p = , by Lemma 3, we get a solution in Kn . Theorem 13. Suppose v(c) < 0, and let k be a natural number such that v(c) < −(. 1 p + r − 1)( r k−1 ), p−1 (p ) −1. Then there exists a constant C depending on k and v(a) such that r )−k. [Kn : K] ≤ CBn1−(p where Bn =. n−1 Y. ri. (pr )(p ) .. i=0. 27.

(31) Proof. By Lemma 6, there exists m0 ∈ N such that if m ≥ m0 and αm ∈ f −m (a), then v(αm ) ≥ v(c). We claim that if m ≥ m0 and αm ∈ f −m (a), then Y. r. × (αm+k + c) ∈ (Km+k )p .. αm+k ∈f −k (αm ). First, the numbers αm+k + c in the product are the zeros of polynomial f k (X − c) − αm . So, Y. (αm+k + c) = (−1)(p. r )k. (f r (−c) − αm ). αm+k ∈f −k (αm ). = (−1)(p. r )k. r. (tp − c − αm ) c + αm r k−1 r ), = ((−1)(p ) t)p (1 − tpr where t := f k−1 (−c). By induction on k, we have v(t) = v(f k−1 (−c)) = (pr )k−1 v(c) and v(c + αm ) ≥ v(c), so v(. c + αm p ) ≥ v(c) − (pr )k−1 v(c) > + r − 1. r p t p−1. By Kummer theory, we have [Km+k+1 : Km+k ] ≤ (pr )(p. r )m ((pr )k −1). . Hence if n ≥ m0 + k, then. [Kn : K] = [Kn : Km0 +k ][Km0 +k : K] ≤ [Km0 +k : K]. n−1 Y. (pr )(p. s=m0 +r. for some C.. 28. r )s (1−(pr )−k ). r )−k. = CBn1−(p.

(32) v(c) =. 5. Suppose v(a) =. −p p−1 v(c) pr .. Then by Lemma 6, we have v(αn ) =. Theorem 14. If v(c) =. −p p−1 ,. v(c) for all n ∈ N. pr. then K∞ /K is infinitely wildly ramified.. Proof. We claim that there exist a sequence {βn }n∈N in K such that f (β1 ) = a, f (βn+1 ) = βn and v(βn − αn ) =. 1 for all n ∈ N. pnr−n−1. If we have the claim, then we complete the proof. We prove it by induction on n. First, consider the solutions of r. r. X p − α1p = 0, then by Lemma 3, there exists a β1 ∈ f −1 (a) such that v(β1 − α1 ) =. 1 1 v(c) + r−1 = r−2 . r r−2 p p −p p. Suppose there exists βn ∈ f −n (a) such that v(βn − αn ) = r. 1 pnr−n−1. < 1. Consider the solutions of. r. p = βn − αn . X p − αn+1. Since 0 = pr v(αn+1 ) +. p p < v(βn − αn ) < pr v(αn+1 ) + + 1 = 1, p−1 p−1. by Lemma 3, there exists βn+1 ∈ f −1 (βn ) such that v(βn+1 − αn+1 ) =. v(βn − αn ) − (pr − pr−1 )v(αn+1 ) − 1 v(βn − αn ) 1 = = (n+1)r−(n+1)−1 . r−1 r−1 p p p. 29.

(33) 6. v(c) <. r. −p p−1. The condition v(c) <. − prp−1 (r − 1) −p p−1. −. pr pr −1 (r. − 1) implies. 1 p v(c) + r > v(c) + + (r − 1) r r−1 p p −p p−1 and −(pr − 1). v(c) pr − 1 − r > − 1 > 0. pr pr − pr−1. Lemma 8. Suppose v(a) > v(c), then for any n ∈ N and any βn ∈ f −n (a), v(βn − αn ) ≥. v(c) 1 + r . pr p − pr−1. Proof. We prove it by induction on n. First, since v(a) > v(c), we have v(αn ) =. v(c) pr. by Lemma 6.. Take β1 ∈ f −1 (a), then it is a solution of r. r. X p − α1p = 0, so v(β1 − α1 ) = v(c) +. 1 1 ≥ v(c) + r pr−i − pr−i−1 p − pr−1. for some i ∈ {0, . . . , r − 1}. Suppose for any βn ∈ f −n (a), we have v(βn − αn ) ≥. v(c) 1 + r . pr p − pr−1. Then for any βn+1 ∈ f −(n+1) (a), it is a solution of r. r. p = βn − αn for some βn ∈ f −n (a). X p − αn+1. Since v(βn − αn ) ≥. v(c) 1 p + r > v(c) + + (r − 1) r r−1 p p −p p−1. By Lemma 3, we have v(βn+1 − αn+1 ) = v(c) +. pr−i. 1 1 ≥ v(c) + r r−i−1 −p p − pr−1. for some i ∈ {0, . . . , r − 1}, as desired. r. Theorem 15. If v(a) > v(c) and µpr = {ξ ∈ K | ξ p = 1} ⊆ K, then for any natural number n, we have. 30.

(34) 1. Gal(Kn /K) is isomorphic to a subgroup of (Z/pr Z)n . 2. For any αn ∈ f −n (a), K(αn ) = Kn . Proof.. 1. Let δ =. v(c) pr. +. 1 p−1 .. For x, y ∈ f −m (a), we write x ∼ y if v(x − y) > δ. By the. definition of the valuation v, it is not difficult to check that ∼ is an equivalent relation. Take arbitrary βm−1 ∈ f −(m−1) (a) and consider the solutions of r. r. p X p − αm = βm−1 − αm−1 .. Since v(βm−1 − αm−1 ) ≥. v(c) 1 p + r > v(c) + + (r − 1), r r−1 p p −p p−1. there is exactly one element βm in f −1 (βm−1 ) satisfying v(βm − αm ) > δ by Lemma 3. Take any element αn−1 ∈ f −(n−1) (a), fix an αn ∈ f −1 (αn−1 ), then f −1 (αn−1 ) = {αn ξ i | 0 ≤ i ≤ pr − 1} where ξ is a primitive (pr )th root of unity. Consider the solutions of r. r. X p − αnp = 0. Since the solutions and f −1 (αn−1 ) coincide, for any 1 ≤ j ≤ pr − 1, we have v(αn ξ j − αn ) = v(αn ) +. v(c) 1 1 = r + r−i ≤δ pr−i − pr−i−1 p p − pr−i−1 r. for some i by Lemma 3 and Lemma 6. Hence f −m (a) = ∪pi=0−1 [αn ξ i ] where [x] is the equivalence class of x. For any σ ∈ Gal(Kn /K), if σ(αn ) ∈ [αn ξ j ] for some j, then for any γn ∈ [αn ], we have v(σ(γn ) − σ(αn )) = v(γn − αn ) > δ. Hence σ([αn ]) ⊆ [αn ξ j ]. For any i, we have σ(αn ξ i ) = ξ i σ(αn ) since µpr ⊆ K. Then v(ξ i σ(αn ) − αn ξ i+j ) = v(ξ i (σ(αn ) − αn ξ j )) = v(ξ i ) + v(σ(αn ) − αn ξ j ) > δ since v(ξ i ) = iv(ξ) = 0. So σ(αn ξ i ) ∈ [αn ξ i+j ]. Similarly, we have σ([αn ξ i ]) ⊆ [αn ξ i+j ]. Define a map φ by φ : Gal(Kn /K) → (Z/pr Z) × Gal(Kn−1 /K) σ 7→ (j, σ|Kn−1 ).. 31.

(35) where σ(αn ) ∈ [αn ξ j ] for any αn ∈ f −n (a). φ is well defined since Kn−1 /K is a Galois extension. We claim that φ is a one to one group homomorphism. For any σ, τ ∈ Gal(Kn /K), suppose φ(σ) = (jσ , σ|Kn−1 ) and φ(τ ) = (jτ , τ |Kn−1 ). Then for any αn ∈ f −n (a) σ(τ (αn )) ∈ σ([αn ξ jτ ]) ⊆ [αn ξ jτ ξ jσ ] = [αn ξ jτ +jσ ]. Hence φ(σ ◦ τ ) = (jσ + jτ , (σ ◦ τ )|Kn−1 ) = (jσ , σ|Kn−1 ) ∗ (jτ , τ |Kn−1 ) = φ(σ) ∗ φ(τ ). Suppose σ ∈ Kerφ, that is φ(σ) = (0, idKn−1 ) where idKn−1 is the identity map on Kn−1 . For any αn ∈ f −n (a), we have σ(αn ) ∈ [αn ] and σ(αn ) ∈ f −1 (f (αn )) since r. r. r. f (σ(αn )) = σ(αn )p − c = σ(αnp − c) = αnp − c = f (αn ). So σ(αn ) = αn , and we have σ = idKn . We prove Gal(Kn /K) is isomorphic to a subgroup of (Z/pr Z)n by induction on n. For any α1 ∈ f −1 (a), since K1 = K(α1 , α1 ξ, α1 ξ 2 , . . . , α1 ξ p. r −1. ) and µpr ⊆ K, we have Gal(Kn /K) is. isomorphic to a subgroup of Z/pr Z. Suppose Gal(Kn−1 /K) is isomorphic to a subgroup of (Z/pr Z)n−1 . Since φ : Gal(Kn /K) → Z/pr Z × Gal(Kn−1 /K) is a one to one group homomorphism, we have Gal(Kn /K) is isomorphic to a subgroup of (Z/pr Z)n . 2. We prove it by induction on n. Since µpr ⊆ K, we have K(α1 ) = K for any α1 ∈ f −1 (a). Suppose for any αn−1 ∈ f −(n−1) (a), we have K(αn−1 ) = Kn−1 . For any αn ∈ f −n (a) and σ ∈ Gal(Kn /K) such that σ(αn ) = αn , then αn−1 is fixed by σ, and so is Kn−1 . For any βn ∈ f −n (a), we have σ(βn ) ∈ [βn ] since σ(αn ) = αn . This implies σ(βn ) = βn since σ fixes Kn−1 pointwise. Hence we have σ = idKn , this implies K(αn ) = Kn .. 32.

(36) Theorem 16. Suppose v(a) ≥ v(c) <. v(c) pr ,. if n ∈ N such that. pr+1 1 r−1 ( + + nr), r+1 2 r+1 1−p + p − np + np − p p − 1 p. then K∞ = Kn+1 . Proof. Let α0 = a, and let αm ∈ f −1 (αm−1 ) maximizing v(αm − αm−1 ). By Lemma 6, we have v(αm ) =. v(c) pr. for all m ≥ 1 since v(a) > v(c) Consider the solutions of r. r. X p − α1p = α1 − a, since v(c) +. p p v(c) + (r − 2) < r ≤ v(c) + + (r − 2) + 1, p−1 p p−1. we have 1 v(c) v(c) v(α2 − α1 ) = ( r − (pr − p) r − (r − 2) − 1). p p p Consider the solutions of r. r. X p − α2p = α2 − α1 , since v(α2 − α1 ) ≥ v(c) +. p + r − 1, p−1. we have v(α3 − α2 ) = v(α2 − α1 ) − (pr − 1). v(c) − r ≥ v(α2 − α1 ). pr. So, we have v(αn+2 − αn+1 ) = v(α2 − α1 ) − n((pr − 1). v(c) + r). pr. The condition v(c) <. 1 r−1 pr+1 ( + + nr), r+1 2 r+1 1−p + p − np + np − p p − 1 p. implies that v(αn+2 − αn+1 ) >. 1 v(c) + . r p p−1. So, by the Theorem 10, we have αn+2 ∈ Kn+1 . By the Theorem 15, we have Kn+2 = K(αn+2 ) ⊆ Kn+1 , hence we have K∞ = Kn+1 . Theorem 17. If v(c) <. −p p−1. −. pr pr −1 (r. − 1), then K∞ is a finite extension over K.. Proof. By Lemma 6, there exist an m such that every αm ∈ f −m (a) satisfies v(αm ) = Theorem 16 on Km with each αm replacing a, then we complete the proof.. 33. v(c) pr .. Apply.

(37) References [AHPW18] J. Anderson, S. Hamblen, B. Poonen, and L. Walton. Local arboreal representations. Int. Math. Res. Not., 2018(19):5974–5994, 2018. doi:10.1093/imrn/rnx054. [Ash10] R. B. Ash. A Course in Algebraic Number Theory. Dover books on mathematics. Courier Corporation, 2010. [Gou97] F. Q. Gouvêa. p-adic Numbers. Universitext. Springer-Verlag, Berlin, second edition, 1997. doi:10.1007/978-3-642-59058-0. [Ked10] K. S. Kedlaya. p-adic Differential Equations, volume 125 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 2010. doi:10.1017/ CBO9780511750922,. [Lub13] J. Lubin. Elementary analytic methods in higher ramification theory. Journal of Number Theory, 133(3):983–999, 2013. doi:10.1016/j.jnt.2012.02.017. [Odo88] R. W. K. Odoni. Realising wreath products of cyclic groups as Galois groups. Mathematika, 35(1):101–113, 1988. doi:10.1112/S002557930000632X. [Sto92] M. Stoll. Galois groups over Q of some iterated polynomials. Arch. Math, 59:239–244, 1992. doi:10.1007/BF01197321.. 34.

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