Galois module structure of local unit groups
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1 Galois module structure of local unit groups Romyar T. Sharifi Abstract We study the groups U i in the unit filtration of a finite abelian extension K of Q p, for an odd prime p. We determine explicit generators of the U i as modules over the Z p -group ring of Gal(K/Q p ). We work in eigenspaces for powers of the Teichmüller character, first at the level of the field of norms for the extension of K by p-power roots of unity and then at the level of K. 1 Introduction Fix an odd prime p and a finite unramified extension E of Q p. We use F n to denote the field obtained from E by adjoining to E the p n th roots of unity in an algebraic closure of Q p. The ith unit group in the unit filtration of F n will be denoted by U n,i. The object of this paper is to describe generators of the groups U n,i as modules over the Z p -group ring of G n = Gal(F n /Q p ). We express these generators in terms of generators of the pro-p completion D n of F n as a Galois module. In fact, one consequence of our work is a rather elementary proof of an explicit presentation of D n as such a module, as was proven by Greither [Gr] using Coleman theory. Instead of working with all of D n at once, we find it easier to work with certain eigenspaces of it. For this and several purposes, it will be useful to think of the Galois group G n as a direct product of cyclic subgroups G n = Γ n Φ, where Γ n = Gal(F n /E) with = p 1 and Γ n = p n 1, and Φ is isomorphic to Gal(E/Q p ). We then decompose D n into a direct sum of p 1 eigenspaces for powers of the Teichmüller character ω : Z p. For any integer r, the ω r -eigenspace D n (r) of D n is the subgroup of elements upon which σ acts by left multiplication by ω(σ) r. This definition depends only on r modulo p 1, so we fix r with 2 r p. Note that D n (r) is a module over the group ring A n = Z p [Γ n Φ]. In fact, as we shall see in Section 3.1, the A n -module D n (r) has a generating set with just one element if r p 2, three elements if r = p 1, and two elements if r = p. 1
2 Galois module structure 2 We will be interested in the A n -module structure of the groups It turns out that V (r) n,i V (r) n,i = D (r) n U n,i. V (r) n,i+1 = V (r) n,i+2 = = V (r) n,i+p 1 for all i r mod p 1 (see Lemma 2.1), so we will consider only such i and set V n,i = V (r) n,i. Our main results, Theorems and 4.3.3, provide a small set of at most n + 1 generators of V n,i as an A n -module and state that any proper generating subset of it has cocardinality 1. The elements of this set are written down explicitly as A n -linear combinations of elements of the generators of D n (r). In Section 4.2, elements of a special form are constructed so as to lie as deep in the unit filtration as possible. In Section 4.3, these are refined to elements of the same form that instead lie just deep enough to be in V n,i, which are in turn the generators that we use. It is convenient to work first in the field of norms F of Fontaine-Wintenberger for the tower of extensions F n of E. This is a field of characteristic p, the multiplicative group of which is the inverse limit of the F n. We prove analogues of all of the above-mentioned results first at this infinite level, prior to applying them in descending to the level of F n. The fact that the pth power map is an automorphism of F simplifies some of the computations. Moreover, the structure of the eigenspaces of the pro-p completion of F, which we study in Section 3.1, is somewhat simpler than that of the D n (r). We construct special elements in the eigenspaces of the groups in the unit filtration in Section 3.2, refine them in Section 3.3, and prove generation and a minimality result in Section 3.4. We see a number of interesting potential applications for the results of this paper. To mention just one, it appears to make possible the computation of the conductors of all degree p n Kummer extensions of F n in terms of the Kummer generator of the extension. The problem of making this computation, which was approached by the author in three much earlier papers, has until now seemed beyond close reach in this sort of generality. Acknowledgments. The idea for this paper originated with the author s 1999 Ph.D. thesis, and initial computations were performed on an evening in June 2001 during a visit to the University of Nottingham. The author thanks Ivan Fesenko for his hospitality. A short draft was written in 2002, when the author was supported by an NSF Postdoctoral Research Fellowship. In August 2006, some additions were made, and that work was funded in part by an NSERC Discovery Grant and the Canada Research Chairs program. As the paper tripled in size in the summer of 2011, the author was supported in part by NSF award DMS The author thanks Richard Gottesman for his interest in this work, which inspired him to finish the paper.
3 Galois module structure 3 2 Preliminaries We maintain the notation of the introduction and introduce some more. Recall from [W] that the field of norms F for the extension F = n F n of E is a local field of characteristic p with multiplicative group F = lim F n, the inverse limit being taken with respect to norm maps. Let ζ = (ζ p n) n be a norm compatible sequence of p-power roots of unity, with ζ p n primitive p n th root of unity in F n. Then a λ = 1 ζ = (1 ζ p n) n is a prime element of F. For m n, let N m,n : F m F n be the norm map. Recall that the addition on F is given by (α + β) n = lim m N m,n(α m + β m ) for α = (α n ) n and β = (β n ) n in F. We fix an isomorphism of the residue field of E (and thereby each F n ) with F q, with q the order of the residue field. Using this, the field F q is identified with a subfield of F via the map that takes ξ F q to ( ξ p n ) n F, where ξ is the (q 1)th root of unity in E lifting ξ. The field F may then be identified with the field of Laurent series F q ((λ)). If F is the union of the F n, then G = Gal(F /Q p ) acts as automorphisms on the field F. As with G n, we may decompose G = Gal(F /Q p ) into a direct product of procyclic subgroups G = Γ Φ, where Gal(F /E) = Γ, the group has order p 1, the group Γ is isomorphic to Z p, and Φ is isomorphic to Gal(E/Q p ). Let γ denote the topological generator of Γ such that γ(ζ p n) = ζ 1+p pn for all n. The pro-p completion D of F decomposes into a direct sum of eigenspaces for the powers of the Teichmüller character ω on. For an integer r, we let D (r) = D εr, where ε r is the idempotent ε r = 1 ω(δ) r δ Z p [ ]. p 1 For i 1, let U i denote the ith group in the unit filtration of F. We then set V (r) δ i = U i D (r) and (V (r) i ) = V (r) i V (r) i+1. The following is Lemma 2.3 of [Sh] (with F n replaced by F ).
4 Galois module structure 4 Lemma 2.1. We have V (r) i i r mod p 1. /V (r) i+p 1 = F q for every i 1, and (V (r) i ) if and only if From now on, we set V i = V (r) i and V i = (V (r) i ) if i r mod p 1. As a consequence of Lemma 2.1, an element z V i is determined modulo λ i+p 1 by its expansion z 1 + ξλ i mod λ i+1 (2.1) with ξ F q. The following is Lemma 2.4 of [Sh] (with F n replaced by F ). Lemma 2.2. Let z V i. If p i, then z γ 1 V i+p 1. Otherwise, z γ 1 V i+2(p 1). We identify Λ = Z p [[Γ]] with the power series ring Z p [[T ]] via the continuous, Z p - linear isomorphism that takes γ 1 to T, and we use additive notation to describe the action of Z p [[T ]] on D. Ramification theory would already have told us that T V i V i+p 1 for all i. On the other hand, explicit calculation will yield the following two lemmas and proposition, which provide more precise information on how powers of T move elements of V i. For ξ F q, we let V i (ξ) denote the set of z V i for which z has an expansion of the form in (2.1). We use [k] to denote the smallest nonnegative integer congruent to k Z modulo p. Lemma 2.3. Let z V i (ξ) for some i. Then, for 0 j [i], we have ( ) [i]! T j z V i+j(p 1) ([i] j)! ξ. Proof. Note that λ γ = 1 ζ 1+p = 1 (1 λ)(1 λ p ) = λ + λ p λ p+1. (2.2) Using this, we see, for any i 1, that Hence, (1 + ξλ i ) γ iξλ i+p 1 1 λ 1 + ξλ i mod λi+2p 2. (2.3) Applying (2.4) recursively, we obtain the result. (1 + ξλ i ) γ iξλ i+p 1 mod λ i+p. (2.4) Lemma 2.4. Let z V pi p+1 (ξ) for some i 2. If j is a nonnegative multiple of p 1, then T j+1 z V p(i+j) (ξ).
5 Galois module structure 5 Proof. Let us begin by proving slightly finer versions of (2.3) in two congruence classes of exponents modulo p. For any t 1, we have (1 + ξλ pt ) γ 1 = 1 + ξλpt (1 + λ p(p 1) λ p2 ) t 1 + ξλ pt 1 mod λ p(t+p 1) and (1 + ξλ pt+1 ) γ 1 = 1 + ξλ pt+1 pt+1 m=1 ( pt+1 ) m (λ p 1 λ p ) m 1 + ξλ pt ξ(λ p(t+1) λ p(t+1)+1 ) mod (λ p(t+p 1)+1, λ p(2t+1)+1 ), the latter congruence following from the fact that p ( ) pt+1 m for 2 m < p. Via some obvious inequalities, we conclude that (1 + ξλ pt ) γ 1 1 mod λ p(t+2), (2.5) (1 + ξλ pt+1 ) γ 1 (1 + ξλ p(t+1) )(1 ξλ p(t+1)+1 ) mod λ p(t+2). (2.6) Let x = 1 + ξλ pi p+1. Recursively applying (2.5) and (2.6), we see that x (γ 1)k+1 (1 + ( 1) k ξλ p(i+k) )(1 + ( 1) k+1 ξλ p(i+k)+1 ) mod λ p(i+k+1), for any positive integer k, as (2.4) implies that U γ 1 p(i+k) U p(i+k+1). The result now follows by application of ε i, since z 1 x ε i V pi, T j+1 x ε i V p(i+j) (ξ), and T j+1 V pi V p(i+j+1) 1 by Lemma 2.2. Let us use {k} to denote the smallest nonnegative integer congruent to k Z modulo p 1. For i 1 with p i, we define a monotonically-increasing function φ (i) : Z 0 Z by φ (i) (0) = i and φ (i) (a) = pa + (i [i]) + {[i] a} (2.7) for a 1. Proposition 2.5. Let z V i (ξ) for some i 2 with p i. Then, for j 1, we have ( ) [i]! T j z V φ (i) (j) {[i] j}! ξ. Proof. Lemma 2.3 implies that T [i] 1 z V φ (i) ([i] 1)([i]! ξ), and note that φ (i) ([i] 1) 1 mod p. Set k = {[i] j}. Since j + k [i] is divisible by p 1, Lemma 2.4 then implies that T j+k z V φ (i) (j+k)([i]! ξ). (2.8)
6 Galois module structure 6 It follows from (2.7) that φ (i) (j + k) i = p(j + k [i]) + (p 1)[i], and so, given (2.8), Lemma 2.2 forces T l z V for all l j + k. In particular, φ (i) (l) applying Lemma 2.3 with j replaced by k and z replaced by T j z, we see that for (2.8) to hold, T j z must have the stated form. Remark 2.6. The obvious analogues of the results of this section all hold at the level of F n for n 2, with λ replaced by λ n = 1 ζ p n. In fact, Lemmas 2.1 and 2.2 were originally proven in that setting in [Sh]. That the other results hold breaks down to the fact that p is a unit times λ pn 1 (p 1) n in F n, which in particular tells us that (2.2) can be replaced by 3 The infinite level λ γ n λ n + λ p n λ p+1 n 3.1 Structure of the eigenspaces mod λ p(p 1)+1 n. In this subsection, we fix choices of certain elements that will be used throughout the paper. From now on, we let ξ denote an element of F q with Tr Φ ξ = 1, the conjugates of which form a normal basis of F q over F p. Let ϕ Φ denote the Frobenius element. Let N Φ Z p [Φ] denote the norm element. Let ζ = (ζ p n) n be a norm-compatible system of primitive p n th roots of unity as before. Let r be an integer satisfying 2 r p. If 2 r p 2, we simply fix an element u r V r (ξ). In the case that r = p 1, generation of D (p 1) requires one additional element π D (p 1), a non-unit, chosen along with u p 1 V p 1 (ξ) in the lemma which follows. The case of r = p shall require more work, but we will fix elements w V 1 ( ξ) and u p V p (ξ) as in Proposition below. Lemma There exist elements π D (p 1) and u p 1 V p 1 (ξ) such that π ϕ = π and π γ 1 = u N Φ p 1. Proof. Set π = λ ε p 1, which satisfies π ϕ = π and π γ 1 V p 1 (1). Since every unit is a norm in an unramified extension, there exists u p 1 D (p 1) such that (u p 1) N Φ = π γ 1, and such an element must lie in V p 1 (ξ ) for some ξ F q with Tr Φ ξ = 1. Hilbert s Theorem 90 tells us that ξ = ξ + (ϕ 1)η for some η F q. Let z V p 1 (η), and set u p 1 = u p 1z 1 ϕ. In fact, one could have chosen u p 1 V p (ξ) arbitrarily and then taken π to satisfy the relations, as can be seen using the results of the following section.
7 Galois module structure 7 Lemma There exist elements w V 1 ( ξ) and u p V p (ξ) with w N Φ u ϕ 1 p = w γ 1 p. = ζ and Proof. First, local class field theory yields the existence of an element w D (p) with (w ) N Φ = ζ. Since ζ V 1 ( 1), we must have w V 1 ( ξ ) for some ξ F q with Tr Φ ξ = 1. Since ξ = ξ + (ϕ 1)η for some η F q, we choose any y V 1 (η), and then w = w y 1 ϕ V 1 ( ξ) satisfies w N Φ = ζ as well. Next, note that (w γ 1 p ) N Φ = 1, and so Hilbert s Theorem 90 allows us to choose an element u p D (p) with (u p) ϕ 1 = w γ 1 p. A simple computation using (2.4) tells us that w γ 1 p V p (ξ p ξ), and therefore u p V p (ξ + a) for some a F p. We may then choose z V p (a) with z ϕ = z and take u p = u pz 1 V p (ξ). We need slightly finer information on the relationship between w and u p inside the unit filtration, as found in the following proposition. Proposition There exist elements w V 1 ( ξ) and u p V p (ξ) with w N Φ and u ϕ 1 p = w γ 1 p such that the element y = u p w pϕ 1 lies in V 2p 1 ( ξ). = ζ Proof. For now, fix any choices of u p and w as in Lemma We must have u p = (1 + ξλ p ) ε 1 α with α V 2p 1 and w = (1 ξλ) ε 1 β with β V p. Note that (1 + ξλ p ) ϕ (ξ p ξ)λ p mod λ 2p and (1 ξλ) γ 1 p = 1 ξ(λ + ( ζλp ) (1 ξλ)(1 ξ p λ p ) 1 + ξ p ξ 1 λ ) λ p mod λ 2p. 1 ξλ We then have (1 ξλ) γ 1 p ξ(1 ξ) 1 + (1 + ξλ p ) ϕ 1 1 ξλ λp+1 mod λ 2p. (3.1.1) We denote the quantity on the right-hand side of (3.1.1) by θ. By Lemma 2.2, we have β γ 1 p V 3p 2, from which it follows that α ϕ 1 θ ε 1 V 3p 2. On the other hand, by Lemma 2.1, we have yα 1 = (1 + ξλ p ) ε 1 (1 ξλ p ) ε 1 β pϕ 1 V 3p 2, so in fact we have y ϕ 1 θ ε 1 V 3p 2. If we can show that θ ε 1 V 2p 1 (ξ ξ p ), we will then have y V 2p 1 ( ξ + a) for some a F p. As in the proof of Lemma 3.1.2, we can then choose an element z V 2p 1 (a) with z ϕ = z and replace u p by u p z 1 to obtain the result.
8 Galois module structure 8 By Proposition 2.5, we see that to show that θ ε 1 V 2p 1 (ξ ξ p ), it suffices to show that θ ε 1(γ 1) p 1 V p 2(ξ p ξ). Since p 2 1 mod p 1, for this, it suffices to show that θ (γ 1)p (ξ p ξ)λ p2 mod λ p2 +1. This is a simple consequence of Lemma 3.1.4, which follows. That is, in the notation of said lemma, Fermat s little theorem and the binomial theorem tell us that d p 1,k = 1 for all positive integers k p 1. Lemma For each positive integer j p 1, one has ( ) ( (γ 1) j j ) ξ(1 ξ) ξλ λp d j,k ξ k (1 ξ) λ (j+1)p mod λ (j+1)p+1, where for positive integers k j. Proof. We make the expansion θ = 1 + d j,k = k=1 k ( ) k ( 1) j+h h j F p h h=1 p 1 ξ(1 ξ) 1 ξλ λp+1 (1 + ξ k (1 ξ)λ p+k ) mod λ 2p. k=1 Since Us γ 1 U s+p 1 for all s, as follows from (2.4), to compute θ (γ 1)j modulo λ (j+1)p+1, it suffices to compute (1 + ξ k (1 ξ)λ p+k ) (γ 1)j modulo λ (j+1)p+1. Fix a positive integer k p 1. We claim that the coefficient of λ (j+1)p in the expansion of (1 + ξ k (1 ξ)λ p+k ) (γ 1)j as a power series in F q [[λ]] is 0 if j < k and ξ k (1 ξ)d j,k if j k. As a consequence of (2.3), one sees that (1 + ξλ t ) γ 1 (1 + tξλ t+p 1 )(1 tξλ t+p ) mod λ t+2p 2 for any t p 1. Using this and the finer congruence (2.6) when possible, an induction yields that the expansion in question is determined by min(j,k) m=0 where (a i ) P j,k,m ( 1 + ξ k k! (1 ξ) (k m)! j m i=1 a i λ (j+1)p+k m ) ( 1)j m mod λ (j+1)p+k+1, P j,k,m = {((a 1, a 2,..., a j m ) Z j m k m a 1 a 2 a j m k}
9 Galois module structure 9 if j > m and P j,k,j = {0}, and we consider the empty product to be 1. In particular, the coefficient in question is indeed 0 for j < k and is ξ k (1 ξ)c j,k for j k, where c j,k = ( 1) j k k! j k a i. (a i ) P j,k,k i=1 It remains to verify that c j,k = d j,k for j k. Let D denote the differential operator x d on F dx p[x]. By the binomial theorem, we have that k ( ) k D j ((1 x) k ) x=1 = ( 1) h h j x h x=1 = ( 1) j d j,k. h h=1 On the other hand, repeated application of the product formula for the derivative yields that min(j,k) D j ((1 x) k ) x=1 = ( 1) k h=1 for all j k, hence the result. k! (k h)! (a i ) P j,h,h i=1 j h a i (x 1) k h x h x=1 = ( 1) j c j,k In the next section, we will obtain the following very slight refinement of what is essentially a result of Greither s [Gr, Sections 2-3] (see also [Sh, Corollary 2.2]). Theorem For r p 2, the A-module D (r) is freely generated by any u r V r (ξ). The A-module D (p 1) has a presentation D (p 1) = π, u p 1 π ϕ = π, u N Φ p 1 = π γ 1, for some u p 1 V p 1 (ξ) and π D (p 1). The A-module D (p) has a presentation D (p) = u p, w w γ 1 p = u ϕ 1 p, for some u p V p (ξ) and w V 1 ( ξ) such that w N Φ = ζ. 3.2 Special elements Fix r with 2 r p, and define φ: Z 0 Z by φ(a) = φ (r) (a) for a 1. Set { 0 if 2 r p 1, δ = 1 if r = p. For all a 1, we have φ(a) = p(a + δ) + {r δ a},
10 Galois module structure 10 so φ(a) is the smallest integer that is at least p(a + δ) and congruent to r modulo p 1. From now on, i will be used solely to denote a positive integer congruent to r modulo p 1. We will write α β to denote that both α and β lie in V i (ξ) for some i and ξ F q. We use additive notation for the action of A = Z p [Φ][[T ]] on D (r). We begin with the following useful lemma. Lemma Let j be a positive integer. a. We have ( ) T j [r]! u r V φ(j) {r δ j}! ξ. b. If j r δ mod (p 1) so that T j u r pz for some z D (r), then T j u r pz V φ(j)+p 1 ( [r]!ξ). Proof. For r < p, part a is a direct consequence of Proposition 2.5 and the fact that u r V r (ξ). For r = p, Proposition 2.5 and the fact that φ = φ (2p 1) on positive integers would tell us more directly that T j 1 y V φ(j) ( ξ) for j 1, for y as in Proposition { j}! Note, however, that T u p = T y pϕ 1 T w T y, since pt w V p 2. This is also the key point of part b. That is, we have as pt z V pφ(j) and Since T (T j u r pz) T j+1 u r φ(j + 1) φ(j) + 2(p 1) < pφ(j). T j+1 u r V φ(j)+2(p 1) ([r]!ξ), a final application of Proposition 2.5 tells us that T j u r pz had to be in the stated group. For a nonnegative integer m, let us define φ m : Z 0 Z 0 by We remark that φ m = p m (φ + 1) 1. pφ m = φ (φ m δ). (3.2.1) From now on, we set ρ = pϕ 1 for brevity of notation. We define special elements in the unit filtration of D (r).
11 Galois module structure 11 Theorem Let m and j be nonnegative integers. Define ( ) α m,j = 1 m {r δ j}!ρ m T j ρ m k T φ k 1(j) δ u r, [r]! unless j = 0 and r = p 1, in which case we replace {r δ j}! with 1 in the formula. Then α m,j V φm(j)(ξ). Furthermore, k=1 (p m bt j + c)u r / V φm(j)+p 1 for all b Z p [Φ] with b 0 mod p and c T j+1 A. Proof. We work by induction, the case of m = 0 being Lemma 3.2.1a, aside from the case j = 0, in which case it is simply the definition of u r. Assume we have proven the first statement for m. Then pα m,j V pφm(j)(ξ p ) and, using Lemma 3.2.1a and (3.2.1), we have Lemma 3.2.1b then tells us that T φm(j) δ u r V pφm(j)( [r]!ξ ). α m+1,j = ρα m,j 1 [r]! T φm(j) δ u r V pφm(j)+p 1(ξ). Now assume the second statement is true for m. (For m = 0, this is a consequence of the fact that the conjugates of ξ are F p -linearly independent.) Suppose that α = (p m+1 bt j + c)u r V i with i φ m+1 (j), b Z p [Φ] pz p [Φ] and c T j+1 A. We write c = (pc + T h ν)u r for some c, ν A with ν 0 mod (p, T ) and h j + 1. By induction, we have that (p m bt j + c )u r / V φm(j)+p 1. Since φ m+1 (j) = pφ m (j) + p 1 and α V φm+1 (j) by assumption, this forces p(p m bt j + c )u r T h νu r, which tells us by Lemma 3.2.1a that φ(h) pφ m (j). On the other hand, Lemma 3.2.1b tells us that α V φ(h)+p 1, which forces i = φ m+1(j). The second statement of Theorem insures, in particular, that au r 0 for all nonzero a A. We therefore have the following corollary.
12 Galois module structure 12 Corollary The A-submodule of D (r) generated by u r is free. In the exceptional case that r = p, we require additional elements. First, we modify the function φ m for this r. For nonnegative integers m and j, we set φ m(j) = φ m (j) unless r = p and j = p l 1 for some l 0, in which case we set φ m(p l 1) = p m+l+1 + p m+1 1 = φ m (p l 1) + p m (p 1). Theorem Let l and m be nonnegative integers. Define ( ) m β m,l = ρ m T pl 1 + ρ m k T φ k 1 (pl 1) 1 u p + ρ m+l+1 w. Then β m,l V φ m (p l 1)( ξ). Moreover, for any j 0, we have k=1 (p m bt j + c)u p + dw / V φ m (j)+p 1 for all b Z p [Φ] pz p [Φ], c T j+1 A and d Z p [Φ]. Proof. The proof is similar to that of Theorem definition of w tell us that Since Lemma 3.2.1a and the T pl 1 u p V p l+1(ξ) and ρ l+1 w V p l+1( ξ), Lemma 3.2.1b yields β 0,l = ρ l+1 w + T pl 1 u p V p l+1 +p 1( ξ). For any m 0, we have By induction and Lemma 3.2.1a, we have β m+1,l = ρβ m,l + T φ m (pl 1) 1 u p. ρβ m,l V pφ m (p l 1)( ξ) and T φ m(p l 1) 1 u p V pφ m (p l 1)(ξ). Since φ m+1(p l 1) = pφ m(p l 1) + p 1, that β m+1,l V φ m+1 (p 1)( ξ) is just another l application of Lemma 3.2.1b. Let j 0, b Z p [Φ] pz p [Φ], c T j+1 A, and d Z p [Φ]. First, suppose that α = (bt j + c)u p + dw V i for some i φ 0(j). Note that (bt j + c)u p V φ(j) by Lemma 3.2.1a, while dw V for some l 0. Since α V p l φ 0 (j), we must have p l φ(j). We then have α V φ(j) unless φ(j) = pl. This occurs if and only if l 1 and j = p l 1 1, in which case φ 0(j) = p l + p 1. For this to hold, we must have (bt j + c)u p dw. Lemma 3.2.1b then implies that α V, so i = p l +p 1 φ 0(j) in all cases.
13 Galois module structure 13 Suppose now that α = (p m+1 bt j + c)u p + dw V i for some i φ m+1(j). Rewrite c as pc + T h ν for some h j + 1 and c, ν A with ν 0 mod (p, T ). If we are to have α V p, we may also write d = pd for some d Z p [Φ]. By induction, we have that (p m bt j + c )u p + d w / V φ m (j)+p 1, and so in order that α V φ m+1 (j), we must have that (p m+1 bt j + pc )u p + dw T h νu p which tells us using Lemma 3.2.1a that φ(h) pφ m(j). On the other hand, Lemma 3.2.1b tells us that α V φ(h)+p 1, so we must have i = φ m+1(j). Theorem may now be proven as a consequence of the description of the above elements and their place in the unit filtration. Proof of Theorem For r p 1, the union of the disjoint images of the functions φ m is exactly the set of positive integers congruent to r modulo p 1. Therefore, Theorem implies that there exists an element of the A-module generated by u r in V i (ξ) for each i r mod p 1. In particular, u r therefore clearly generates V r as an A-module, which equals D (r) for r p 2, and it is free by Corollary Every element of D (p 1) may then be written in the form π m u a p 1 with m Z p and a A, and such an element can clearly only be trivial if m is, and therefore a is as well. Noting that our choices of π and u p 1 as in Lemma satisfy the desired relations, the presentation for r = p 1 is as stated. For r = p, the union of {1} and the images of the functions φ m and φ m is the set of positive integers that are congruent to 1 modulo p 1. Theorem and Theorem imply that there exists an element of the A-module generated by u p and w in V i (ξ) for each i 1 mod p 1. Thus, this A-module is D (p). Our choices of u p and w satisfy the relations of Lemma 3.1.2, and it follows from the second statement of Theorem that if either c A or d Z p [Φ] is nonzero, then so is cu p + dw. 3.3 Refined elements In this section, we provide refinements of the elements constructed in Theorem and Theorem We maintain the notation of Section 3.2. We begin by constructing certain one-sided inverses to the monotonically increasing functions φ and φ m. For any nonnegative integer a and positive integer t, let us set a t = max(a + {t a}, t).
14 Galois module structure 14 Therefore, a t is the smallest integer greater than or equal to t and a and congruent to t modulo p 1. Define ψ : Z 0 Z 0 by a r + 1 ψ(a) = δ p except for r = p 1 and a p 1, in which case we set ψ(a) = 0. For m 0, define ψ m : Z 0 Z 0 by ( ) a + 1 ψ m (a) = ψ 1. Note that ψ 0 = ψ. Lemma We have ψ m (φ m (j)) = j for all nonnegative integers j. Moreover, for all such j and positive integers a, we have φ m (j) a if and only if j ψ m (a). Proof. First, note that φ(j) is congruent to r modulo p 1, so we have φ(j) + 1 p(j + δ) + {r δ j} + 1 ψ(φ(j)) = δ = δ = j p p unless r p 1 and j = 0, but one checks immediately that ψ(φ(0)) = ψ(r) = 0 if r p 1 as well. It follows that we have ( ) p m (φ(j) + 1) ψ m (φ m (j)) = ψ 1 = ψ(φ(j)) = j. p m Therefore, if φ m (j) a, then j = ψ m (φ m (j)) ψ m (a), since ψ m is nondecreasing. To finish the proof, we need only show that φ m (ψ m (a)) a, since φ m is nondecreasing (in fact, strictly increasing). First, note that the definition of ψ is such that p m ψ(a) = ψ( a r ). For i r mod p 1 with i 1, p 1, the value φ(ψ(i)) is the unique integer between p i+1 i+1 and p + p 2 which is congruent to r mod p 1. This implies that p p { a r if a r 1 mod p, or a r = p 1, φ(ψ(a)) = a r + p 1 otherwise. (3.3.1) which is, in particular, at least a. By definition of φ m and ψ m, we then have that a + 1 φ m (ψ m (a)) = p m (φ(ψ m (a)) + 1) 1 p m 1 a. p m
15 Galois module structure 15 We actually need a version of Lemma with φ m replaced by φ m and ψ m replaced by an appropriate function ψ m : Z 0 Z 0, which we now define. Set ψ m = ψ m if r p 1 and, if r = p, let { ψm (a) 1 if p m+l+1 + p m a p m+l+1 + p m+1 1 for some l 0, ψ m(a) = ψ m (a) otherwise. Note that ψ m(a) = ψ m (a) 1 for r = p if and only if φ m (p l 1) < a φ m(p l 1) for some l 0, in which case ψ m(a) = p l 1. One then easily checks the following. Corollary We have ψ m(φ m(j)) = j for all nonnegative integers j. Moreover, for all such j and positive integers a, we have φ m(j) a if and only if j ψ m(a). For the rest of this section, we fix a positive integer i with i r mod p 1. Remark Lemma and Theorem (resp., Corollary and Theorem 3.2.4) tell us that each α m,ψm(i) (resp., β m,l with ψ m(i) = p l 1) lies in V i. These elements have the form (p m bt j + c)u r + dw for j = ψ m(i), where b Z p [Φ] pz p [Φ], c T j+1 A, and d Z[Φ], with d = 0 if r p. The same results also show that no such element with j < ψ m(i) can lie in V i. For any m 0, define θ m : Z 1 Z 0 by ( ) a r θ m (a) = ψ. By Lemma 3.2.1a and Lemma 3.3.1, the value θ m (i) for i r mod p 1 is the minimal integer j such that p m T j u r V i. In particular, θ m (i) ψ m (i) for all i. Lemma For all positive integers m and k with k m, we have with equality if and only if p m φ k 1(ψ m(i)) δ θ m k (i) 1, p m k+1 φ k 1(ψ m(i)) < i. (3.3.2) Moreover, we have ψ m(i) θ m (i) 1, with equality if and only if the above equivalent conditions hold for k = 1. Proof. Let us check the case that r = p and ψ m(i) = p l 1 for some l 0 separately. First, suppose that p m+l+1 < i < p m+l+1 +p m+1. In this case, we have ψ m(i) = θ m (i) 1. We also have φ k 1(ψ m(i)) = φ k 1(p l 1) = p k+l + p k 1 = ψ(p k+l+1 + p k+1 ) θ m k (i),
16 Galois module structure 16 with equality if and only if p m k+1 φ k 1(ψ m(i)) = p m+l+1 + p m+1 p m k+1 < i. (3.3.3) Moreover, in the case that p m+l+1 p m+1 + 2p m < i p m+l+1, the values φ k 1 (ψ m(i)) and p m k+1 φ k 1 (ψ m(i)) are the same as in the previous case, while θ m k (i) 1 and i are smaller. So, we may assume from this point forward that r and i are such that ψ m(i) = ψ m (i) and φ k 1 (ψ m(i)) = φ k 1 (ψ m (i)) for all k. We claim that ρ m k T φ k 1(ψ m(i))+1 δ u r lies in V i+p 1 (resp., ρ m k T α k,ψm(i) lies in V i+p 1 ) for all positive (resp., nonnegative) k m. Note that T α m,ψm(i) V i+p 1 as a consequence of Theorem Suppose that ρ m k T α k,ψm(i) V i+p 1 for some positive k m. We then have that ρ m k T φ k 1(ψ m(i))+1 δ u r [r]!ρ m k T α k,ψm(i) V i+p 1, which also forces ρ m k+1 T α k 1,ψm(i) V i+p 1, since ρ m k+1 T α k 1,ψm(i) = ρ m k T α k,ψm(i) + 1 [r]! ρm k T φ k 1(ψ m(i))+1 δ u r, proving the claim. In particular, since ρ m T α 0,ψm(i) V i+p 1, we have ρ m T ψm(i)+1 u r V i+p 1 as well. The definition of θ m k (i) now yields the desired inequalities. Equation (3.3.2) holds for a given k if and only if ρ m k+1 α k 1,ψm(i) / V i. Since [r]!ρα k 1,ψm(i) T φ k 1(ψ m(i)) δ u r, this occurs if and only if ρ m k T φ k 1(ψ m(i)) δ u r / V i and, therefore, if and only if φ k 1 (ψ m (i)) δ θ m k (i) 1, which must then be an equality. Note also that ψ m (i) < θ m (i) if and only if ρ m α 0,ψm(i) / V i, which holds by Lemma 3.2.1a if and only if p m φ(ψ m (i)) < i, the same condition as (3.3.2) for k = 1. From now on, we set i m = i p m for all m 0. Lemma For any pair of positive integers m and k with k m, we have with equality if and only if φ k 1(ψ m(i)) δ θ m k (i) 1, 1. i m+ɛ 0 mod p, or r = p 1 and i m = p, 2. i m+ɛ r + 1 mod p 1, but not r = p 1 and i m = 1, and
17 Galois module structure i j mod p m+ɛ for some 0 < j < p m+1 k, where ɛ = 0 unless r = p and i m+1 = p l + 1 for some l 0, in which case we set ɛ = 1. Moreover, we have ψ m(i) θ m (i) 1, with equality if and only if the above conditions hold with k = 1. Proof. The case that r = p and ψ m(i) = p l 1 for some l 0 follows from the proof of Lemma 3.3.4, noting that if i m+1 = p l + 1, then it is both nonzero modulo p and congruent to p + 1 modulo p 1, and the third condition of the lemma holds exactly when (3.3.3) does. On the other hand, for the remaining i with ψ m(i) = p l 1, we have i m+1 = p l, and the fact that the inequality is strict was shown in the proof of Lemma So, we again assume that r p or i is such that ψ m(i) p l 1 for all l 0. By Lemma 3.3.4, it suffices to determine the precise conditions under which (3.3.2) holds. Let us set a = (i + 1) m. It follows from (3.3.1) that we have p m k+1 φ k 1 (ψ m (i)) = { p m a r+1 p m k+1 if p a r+1, p m a r+1 + p m (p 1) p m k+1 otherwise, (3.3.4) unless r = p 1 and a r+1 = p, in which case p m k+1 φ k 1 (ψ m (i)) = p m+1 p m k+1. Aside from this exceptional case, (3.3.4) implies that p cannot divide a r+1 if (3.3.2) is to hold. Moreover, if a r+1 > a, then again (3.3.2) cannot hold, so for it to hold, we must have a r + 1 mod p 1, but not r = p 1 and a = 1. Assuming that these necessary conditions hold, the condition that p m k+1 φ k 1 (ψ m (i)) = p m a p m k+1 < i is exactly that i j mod p m with 0 < j < p m k+1. For m 0, we will define new elements κ m,i of V i that involve fewer terms and easier-to-compute exponents of powers of T than the expressions for α m,ψm(i) and β m,l. In preparation, set σ(m, i) = log p (p m i m i) for any m 0 such that p m i. Note that 0 σ(m, i) m 1 when it is defined and σ(m + 1, i) is defined and greater than or equal to σ(m, i) whenever σ(m, i) is defined. First, supposing either that r p 1 or that r = p and i m+1 1 is not a power of p, we set κ m,i = ρ m T θm(i) u r (3.3.5) if i m r + 1 mod p 1, p i m, i < p m, or p m i, unless r = p 1 and i m = p, and κ m,i = ρ m T θm(i) 1 a m,i m 1 k=σ(m,i) ρ k T θ k(i) 1 u r (3.3.6)
18 Galois module structure 18 otherwise, where a m,i denotes the least positive residue of the inverse of {r+1 δ θ m (i)}! modulo p unless r = p 1 and θ m (i) = 1, in which case we take a m,i = 1. In the remaining case that r = p and i m+1 1 is a power of p, we set κ m,i = ρ m T θm(i) 1 + m 1 k=σ(m+1,i) ρ k T θ k(i) 1 u r + ρ m+log p (im+1 1)+1 w. (3.3.7) For consistency, we let a m,i = 1 for such m. Note that Lemma tells us that each κ m,i has the form (ρ m T ψ m (i) + c)u r + dw for some c T ψ m (i)+1 A and d Z p [Φ], with d taken to be zero if r p 1. We give two examples for p = 5 and particular values of i. Example Suppose that p = 5, r = 3, and i = Then we have κ 0,i = T 2380 u 3, κ 1,i = ρt 476 u 3, κ 2,i = (ρ 2 T 95 ρt 475 T 2379 )u 3, κ 3,i = (ρ 3 T 19 ρ 2 T 95 )u 3, κ 4,i = ρ 4 T 4 u 3, κ 5,i = (ρ 5 ρ 4 T 3 ρ 3 T 19 )u 3. Example Suppose that p = 5, r = 5, and i = Then we have κ 0,i = T u 5, κ 1,i = (ρt 3708 T )u 5, κ 2,i = ρ 2 T 741 u 5, κ 3,i = (ρ 3 T 147 ρ 2 T 740 ρt 3708 )u 5, κ 4,i = ρ 4 T 29 u 5, κ 5,i = (ρ 5 T 4 + ρ 4 T 28 )u 5 + ρ 7 w, κ 6,i = ρ 6 u 5 + ρ 7 w. Remark It is not hard to see from the definition of σ(m, i) that σ(m, i) k for k < m if and only if p m k i k. Moreover, if for a given k there exists m > k such that σ(m, i) is less than k or not defined, then p i k so κ k,i = ρ k T θ k(i) u r unless r = p and i k+1 1 is a power of p or r = p 1 and i k = p. The previous examples illustrate some of this. Let us show that the κ m,i are actually elements of V i. In the process, we see how they compare to the elements α m,ψm(i) and β m,l previously defined. Proposition The elements κ m,i lie in V i for all nonnegative integers m. Proof. Suppose first that r p or i does not satisfy i m = p l + 1 for any l 0 (and omitting the case r = p 1 and ψ m (i) = 0, for which one should take the fractions in the following two equations to be 1). If ψ m (i) = θ m (i), then we have that κ m,i = [r]! {r δ ψ m (i)}! ρm α 0,ψm(i),
19 Galois module structure 19 and this lies in V i by the definition of θ m (i). If ψ m (i) = θ m (i) 1, we claim that κ m,i [r]! {r δ ψ m (i)}! ρσ(m,i) α m σ(m,i),ψm(i). (3.3.8) To see this, note that m σ(m,i) κ m,i = ρ σ(m,i) ρ m σ(m,i) T ψm(i) a m,i ρ m σ(m,i) k T θ m k(i) 1 u r. Lemma tells us that θ m k (i) 1 = φ k 1 (ψ m (i)) δ if and only if p m k+1 > p m i m i, and therefore if k m σ(m, i), proving the claim. (Note that we the reason we do not have actual equality in (3.3.8) is simply that we took a m,i to be an inverse to {r δ ψ m (i)}! modulo p, not in Z p.) Moreover, we have by Theorem that κ m,i V t with t = p σ(m,i) φ m σ(m,i) (ψ m (i)). k=1 Since p σ(m,i) p m i m i, Lemma implies that φ m σ(m,i) (ψ m (i)) δ θ σ(m,i) 1 (i), and Lemma then states that t i. Finally, if r = p and i m+1 = p l +1 for some l 0, then Lemma similarly implies that κ m,i = ρ σ(m+1,i) β m σ(m+1,i),l. By Theorem 3.2.4, we have in this case that κ m,i V t with t = p σ(m+1,i) φ m σ(m+1,i)(p l 1) i, the inequality again following from Lemmas and Generating sets In this subsection, we give explicit minimal generating sets of all of the A-modules V i in terms of the elements κ m,i of the previous section. We begin with generation. Recall that δ {0, 1} is 1 if and only if r = p. Theorem We let S i = {κ m,i 0 m s} for ( ) i + 1 s = log p. r δ(p 1) If 2 r p 1, then S i generates V i as an A-module, while if r = p, then S i {p log p (i) w} generates V i as an A-module.
20 Galois module structure 20 Proof. Let t = (i + 1) m 1. In the case that 2 r p 1, we have ψ m (i) = ψ(t) and ψ(t) > 0 if and only if i+1 > r + 1, or m < log p m p ( i+1 ). The smallest m such that r+1 ψ m (i) = 0 is therefore s. If r = p, then ψ m(i) = ψ(t) ɛ t, where ɛ t {0, 1} is 1 if and only if p l t p l+1 + p 1 for some l 0. In particular, we have ψ(t) > ɛ t if and only if t 2p, so the smallest m such that ψ m(i) = 0 is again s. It suffices to show that the images of our elements generate V i /V i+p 1. Suppose that α = (ρ k bt j + c)u r + dw V i for some j, k 0, b Z p [Φ] pz p [Φ], c T j+1 A, and d Z p [Φ] (with d = 0 if r p). Let m = min(k, s). Then j ψ m(i) by Theorems and and Corollary (and the fact that ψ s(i) = 0), and we set α = α ρ k m bt j ψ m (i) κ m,i V i A(T j+1 u r, w). If r p 1, we may repeat this process recursively until we obtain an element of V i+p 1. If r = p, either κ m,i Au p or κ m,i ρ m+l+1 w + Au p for some l 0 with i < p m+l+1 + p m+1. Since (T, p)p m+l+1 w V p m+l+2, there exists an element α V i (T j+1 Au r + Z p [Φ]w) with α α V i+p 1, and again we may repeat the process until we obtain an element of V i+p 1 plus an element of V i Z p [Φ]w = Z p [Φ]p log p (i) w. We will require the following lemma. Lemma If m 1 is such that θ m (i) 1, then θ m 1 (i) θ m (i) + 2. Proof. First, suppose that θ m (i) 1, and note that i m 1 p(i m 1) + 1. Therefore, 2 + {r im } θ m 1 (i) ψ(p(i m 1) + 1) = i m 1 + δ. (3.4.1) p On the other hand, im {r i m } θ m (i) = δ. (3.4.2) p In particular, we have that θ m (i) = 1 exactly when r + 1 i m r + (δ + 1)(p 1). In this case, θ m 1 (i) r + 1 δ 3 = θ m (i) + 2. In general, (3.4.1) and (3.4.2) tell us that θ m 1 (i) i m 1 δ and i m p + 1 δ θ m(i), and we have i m 1 δ i m p + 3 δ if and only if i m 4 p, which holds for i p 1 m r + p unless i m = 5, r = 2, and p = 3, in which case θ m 1 (i) 5 and θ m (i) = 2.
21 Galois module structure 21 For each m 0, let us set ɛ m (i) = θ m (i) ψ m(i), which lies in {0, 1} by Lemma and the remark before it. The following corollary is useful in understanding the form of our special elements. Corollary For every m 0, we have ψ m(i) ψ m+1(i), with equality if and only if ψ m(i) = 0. Proof. If θ m+1 (i) 1, Lemma and the fact that ɛ k (i) {0, 1} for all k imply that ψ m(i) > ψ m+1(i). Otherwise, ψ m+1(i) = 0, and the inequality holds automatically, with equality exactly if ψ m(i) = 0. We next show that the sets given in Theorem are minimal unless r = p. It is in the proof of this result that the refined elements κ m,i first hold an advantage of ease of use over the elements of Section 3.2. Theorem For r p 1, no proper subset of S i generates V i as an A-module. For r = p, every proper subset of S i {p log p (i) w} that generates V i as an A-module must contain S i. Proof. Assume first that 2 r p 1. Suppose that s c m κ m,i = 0, (3.4.3) m=0 where c m A for m s. We must show that no c m is a unit. We prove the somewhat stronger claim that c m (p, T ɛm(i)+1 ) for each m. Fix an nonnegative integer m s. If ɛ m (i) = 0, then κ m,i = ρ m T θm(i) u r by (3.3.5). If ɛ m (i) = 1, then (3.3.6) tells us that noting Lemma Set κ m,i ρ m T θm(i) 1 u r mod AT θm(i)+1 u r, X m = {k Z m < k s, ɛ k (i) = 1, σ(k, i) m}, (3.4.4) which is actually a set of cardinality at most one, though we do not need this fact. Let k s. If k X m, then (3.3.6) and Lemma together imply that κ k,i a k,i ρ m T θm(i) 1 u r mod (p m+1, T θm(i)+1 )u r, and if k / X m, they and (3.3.5) similarly imply that κ k,i (p m, T θm(i)+1 )u r. Thus, (3.4.3) yields the congruence c m ρ m T ψm(i) c k a k,i ρ m T θm(i) 1 mod (p m+1, T θm+1(i)+1 ). k X m (3.4.5)
22 Galois module structure 22 If the claim holds for all k > m, then we have c k (p, T 2 ) for each k X m, so c m (p, T ɛm(i)+1 ), as desired. If r = p, a completely analogous argument shows that at most p log p (i) w is unnecessary for generation, if one works modulo Aw = Z p [Φ]w + A(ϕ 1)u p throughout. Here, one should replace X m by X m = {k Z m < k s, ɛ k (i) = 1, σ (k, i) m}, (3.4.6) where we set σ (k, i) = σ(k, i) unless i k+1 = p l + 1 for some l 0, in which case we set σ (k, i) = σ(k + 1, i). For the purpose of completeness, we also give the precise condition on i under which no proper subset of S i {p log p (i) w} generates V i in the case that r = p. Proposition For r = p, the set S i generates V i if and only if i s = p + 1. Proof. To determine whether p log p (i) w is or is not necessary, we work in distinct ranges of i separately. Note that the definition of s forces 2p s < i < 2p s+1. Case 1: 2p s < i p s+1. In this case, all of the elements κ m,i lie in Au p, and therefore p s+1 w is necessary. Case 2: p s+1 < i p s+1 + p s p s 1. In this range, we have κ s,i = ρ s u p + ρ s+1 w and κ s 1,i = ρ s 1 T p 1 u p + ρ s+1 w. Note that so (T p)κ s,i = ρ s (T p)u p + ρ s+1 (ϕ 1)u p = ρ s (T ρ)u p, ρ s T u p ρ s+1 u p mod Aκ s,i and ρ s u p ρ s+1 w mod Aκ s,i. (3.4.7) Applying these to ρκ s 1,i, we obtain ρκ s 1,i ρ s+p 1 u p + ρ s+2 w ( ρ s+p + ρ s+2 )w mod Aκ s,i, which in particular tells us that p s+2 w A(κ s 1,i, κ s,i ). Case 3: p s+1 + p s p s 1 < i p s+1 + p s. In this range, we have κ s,i = ρ s u p + ρ s+1 w and κ s 1,i = ρ s 1 T p 1 + s 2 k=σ(s,i) with σ(s, i) s 2. Moreover, κ m,i Au r for m s 2. ρ k T θ k(i) 1 u p + ρ s+1 w
23 Galois module structure 23 Set ν k,i = ρ k T θ k(i) u r for all nonnegative k. We note that ν m,i A(κ 0,i,..., κ m,i ) for m s 2: if κ m,i ν m,i, which is to say ɛ m (i) = 1, then ν m,i = T κ m,i + a m,i m 1 k=σ(m,i) ν k,i. Let j = θ s 2 (i) p, and note that j p Since T j κ s 1,i ρ s 1 T θ s 2(i) 1 u p + ρ s+1 T j w mod A(ν σ(s,i),i,..., ν s 2,i ) and ρ k+1 T θ k(i) 1 u p Aν k+1,i for all k with σ(s, i) k s 3, we therefore have that (ρ T j )κ s 1,i ρ s T p 1 u p + ρ s+1 (ρ T j )w mod A(ν σ(s,i),i,..., ν s 2,i ). (3.4.8) Using (3.4.7) to reduce (3.4.8), we see that (ρ T j )κ s 1,i ρ s+2 (1 ρ p 2 ρ j 1 )w mod A(ν σ(s,i),i,..., ν s 2,i, κ s,i ), which implies that p s+2 w A(κ 0,i,..., κ s,i ). Case 4: p s+1 + p s < i < 2p s+1. In this case, all of the κ m,i with m s 1 lie in AT p u p, and so for p s+2 w to be unnecessary, there would have to exist c A such that Note that cκ s,i p s+2 w mod AT 2 u p. (3.4.9) cκ s,i c(ρ s u p + ρ s+1 w) mod AT 2 u p, and this forces c T 2 c mod (ϕ 1) for some c A. This means that cκ s,i c p s+3 w mod A(u p, (ϕ 1)w), but p s+2 w / A(p s+3 w, (ϕ 1)w, u p ), so (3.4.9) cannot hold. 4 The finite level 4.1 Norms and eigenspace structure In this section, we explore the consequences of the results of Section 3 for unit groups of actual abelian local fields of characteristic 0. Fix a positive integer n. Recall from the introduction that F n is the field obtained from E by adjoining the p n th roots of unity and that U n,t denotes the tth unit group of F n for t 1. As before, we set Γ n = Gal(F n /F 1 ). For positive integers m n, let N m,n and Tr m,n denote, respectively, the norm and trace from F m to F n. We also let N n denote the restriction map N n : F F n on norm compatible sequences. Recall that λ n = N n (λ) = 1 ζ p n, where ζ p n = N n (ζ) is a primitive p n th root of unity. We will require a few preliminary lemmas.
24 Galois module structure 24 Lemma One has for all k 1 and ɛ {0, 1}. Proof. An easy calculation shows that for every t 0. Since ( ) ( pk ɛ k ɛ = pj j for any j 0, we have the result. Tr n+1,n (λ pk ɛ n+1 ) pλ k ɛ n mod p 3 t p ( ) t Tr n+1,n (λ t n+1) = p ( ζ p n) j pj ) p(k j) s=1 p s j=0 ( 1 + pj ) s ( ) k ɛ mod p 2 j Let e n = p n 1 (p 1) denote the ramification index of F n. In applying Lemma 4.1.1, it is useful to make note of the fact that p λ en n mod λ pn n. (4.1.1) Lemma For t 1 and any unit η in E, one has 1 + η p λ t n mod λ t+1 n if t < p n 1, N n+1,n (1 + ηλ t n+1) 1 + (η p η)λ pn ɛ n 1 ηλ en+k ɛ n mod λ pn +1 ɛ n if t = p n ɛ, ɛ {0, 1}, mod λ en+k+1 ɛ n if t = pk ɛ > p n, ɛ {0, 1}. Moreover, we have for all t > p n. N n+1,n (1 + ηλ t n+1) 1 mod λ en+ t p n. Proof. The jump in the ramification filtration of Gal(F n+1 /F n ) occurs at p n 1. By [Se, Lemmas V.4 and V.5], we have and N n+1,n (1 + ηλ t n+1) 1 + η Tr n+1,n (λ t n+1) + η p λ t n mod λ en+ 2t/p n Tr n+1,n (λ t n+1) 0 mod λ en+ t/p n. The result is then a corollary of Lemma 4.1.1, upon applying (4.1.1).
25 Galois module structure 25 Let D n be the pro-p completion of F n, and let D n (r) = Dn εr for any r Z. As before, we fix r with 2 r p, and i will always denote a positive integer with i r mod p 1. Let V n,i = U εr n,i = U n,i D n (r) for any such i. These V n,i are all modules over A n = Z p [Γ n Φ]. As in Lemma 2.1, we have isomorphisms V n,i /V n,i+p 1 F q that send 1 + xλ i n for some x in the valuation ring of F n to the element x of F q that is identified with the image of x in the residue field of E under the isomorphism fixed in Section 2. We may then set V n,i = V n,i V n,i+p 1 and define V n,i (η) for η F q as the set of elements 1 + xλ i n with x = η. We have the following consequence of Lemma Lemma For any t 1, we have with equality for t 0. Proof. Note that Lemma tells us that N n+1,n (V n+1,p n +t) V n,p n +t (p 1) t+1 p, N n+1,n (U n+1,p n +pk ɛ) = U n,p n +k ɛ for all k 0 and ɛ {0, 1} with k ɛ, since every element in U n,p n +k ɛ can be written as a product of elements of the form 1 + η t λ pn +t n with t k ɛ and η t F q. (For k = 0 and ɛ = 1, it tells us just that any element of U n+1,p n 1 has a norm in U n,p n 1.) Note that U εr n,p n +k ɛ = V n,p n +k ɛ+{r k+ɛ 1} and U εr n+1,p n +pk ɛ = V n,p n +pk ɛ+{r k+ɛ 1}. For any t 0, we may write t = pk ɛ + {r k + ɛ 1} for some k, ɛ, and r, and we have t + 1 t (p 1) = k ɛ + {r k + ɛ 1}, p hence the result. The next corollary is almost immediate from Lemmas and 4.1.3, so we leave it to the reader. Corollary For any unit η in E, one has V n,i (η p ) if i < p n 1, N n+1,n (V n+1,i (η)) V n,i (η p η) if i = p n 1, V n,p n +k 1( η) if i = p n + pk 1 for some k > 0, with equality if r p 1 or i > p n.
26 Galois module structure 26 As for the p-power map, we have the following well-known and easy-to-prove fact. Lemma Suppose that i > p n 1. Then the pth power map induces an isomorphism V n,i V n,i+en, and we have V n,i (η) p = V n,i+en ( η) for all η F q. Next, we discuss the restriction map from the field of norms to the finite level. Proposition The map N n induces maps N n : D (r) D n (r) that are surjections for r p 1 and which have procyclic cokernel for r = p 1. For η F q, we have V n,i (η p n ) if i p n 2, N n (V i (η)) V n,p n 1(η p n η p n 1 ) if i = p n 1, V n,p n +k 1( η p n 1 ) if i = p n + pk 1 for some 0 < k < e n. Moreover, we have induced maps V i /V i+1 V n,i /V n,i+1 for all i < p n, and these are isomorphisms for i p n 1. For i p n, we have that V n,i = N n V i if r p 1, and V n,i /N n V i is procyclic if r = p 1. Proof. That the cokernel of N n is trivial (resp., procyclic) if r p 1 (resp., r = p 1) follows easily from local class field theory, but it is also a consequence of the argument that follows. The first jump in the ramification filtration of Gal(F /F n+1 ) is at p n+1 1. In particular, for t less than this value, repeated application of Lemma tells us that N n+1 (1 + ηλ t ) = lim N m,n+1(1 + η p m λ t m) 1 + η p n 1 λ t n+1 mod λ t+1 n+1. m Moreover, repeated application of Corollary followed by two applications of Lemma tells us that N n (V p n+1 1+{r}) V n,p n +e n 1+{r}. An application of Corollary then yields the stated containments. Since η p n and η p n 1 run through all elements of F q as as η F q varies, we obtain V n,i = N n V i + V n,i+p 1 for all i p n but p n 1. Noting Lemma 4.1.5, this implies V n,i+ken = N n V p k i + V n,i+ken+p 1 for p n 1 < i p n with i p n 1 and k 0. Note that every element of every V n,i may be written as an infinite product over j 0 of one element from each of a fixed set of representatives of the V n,i+j(p 1) /V n,i+(j+1)(p 1). Thus, we have that N n V i = V n,i so long as r p 1.
27 Galois module structure 27 If r = p 1, we can choose an element z n of V n,p n 1(ξ) that is not a norm. By the above-proven formula for N n (1 + ηλ pn 1 ) modulo λ pn n, we have that V n,p n +ke n 1 = N n V p k (p n 1) + V n,p n +ke n+p 2 + z pk n for k = 0, and then for all k 0 by taking powers. Therefore, V n,i /N n V i is generated by z n for all i < p n with i 0 mod p 1. The following structural result is again essentially found in [Gr], without the stated congruences. Here, we derive it from more basic principles. Theorem For r p 2, the A n -module D n (r) by an element u n,r V n,r (ξ). The A n -module D n (p 1) D (p 1) n = π n, u n,p 1, v n π ϕ n = π n, π γ 1 n is freely generated as an A n -module has a presentation = u N Φ n,p 1, v γ n = v n, u N Γn n,p 1 = v 1 ϕ n, where v = vn ϕ2 n 1 + pξ mod p 2 is independent of n and u n,p 1 V n,p 1 (ξ) for n 2, while u 1,p 1 V n,p 1 (ξ ξ p 1 ). The A n -module D n (p) has a presentation D (p) n = u n,p, w n w γ 1 p n = u ϕ 1 n,p with u n,p V n,p (ξ) and w n V n,1 ( ξ) such that w N Φ n = ζ p n. Proof. We set u n,r = (N n u r ) ϕn, π n = N n π, and w n = (N n w) ϕn with u r, π, and w as in Theorem It follows from the surjectivity of N n for r p 1 in Proposition that the element u n,r generates D n (r) for r p 2, while the elements w n and u n,r generate D n (p). By Hilbert s Theorem 90, the kernel of N n consists exactly of elements of the form α γpn 1 1 with α D, and therefore it follows that D n (r) is free of rank 1 on u n,r over A n for r p 2 and that D n (p) has the stated presentation. (That u 1,p V 1,p (ξ) requires a simple check using Propositions and ) The elements π n and u n,p 1 automatically satisfy the first two relations in the desired presentation of D n (p 1). In particular, u N Γn Φ n,p 1 = π (γ 1)N Γn n = 1, so Hilbert s Theorem 90 tells us that u N Γn n,p 1 = vn 1 ϕ for some v n in the pro-p completion of E. By Proposition 4.1.6, we have that u N Γn n,p (ξ pn 1 ξ pn 2 )λ p 1 1 mod λ p 1. Noting (4.1.1), we may in fact choose v n 1 + pϕ n 2 (ξ) mod p 2 with v = vn ϕ2 n independent of n. HiIbert s theorem 90 and Theorem tell us that the A n -module generated by u n,p 1 is isomorphic to A n /(N Γn Φ). By Proposition 4.1.6, the cokernel of N n on D (p 1)
28 Galois module structure 28 is isomorphic to Z p. We claim that the image of v topologically generates this cokernel. If this is the case, then clearly D n (p 1) is generated by π n, u n,p 1, and v, and any solution with b, d Z p and c A n to πnu b c n,p 1v d = 1 must satisfy b = d = 0 and c Z p N Γn Φ. It remains only to demonstrate the claim. Suppose by way of contradiction that there exists a A n such that x = vu a n,p 1 is a pth power in D n (p 1). This implies that x γ 1 = u a(γ 1) n,p 1 is a pth power in the A n -module generated by u n,p 1. It follows that a(γ 1) A n (p, N Γn Φ), which forces a(γ 1) 0 mod p, so a A n (p, N Γn ). It then suffices to show that vu bn Γn n,p 1 = v 1+bϕn 2 (1 ϕ) is not a pth power in F n for any b Z p [Φ]. If it were for some b, then v N Φ and hence 1 + p would be a pth power in F n as well, but this is clearly not the case. 4.2 Special elements We assume for the rest of the paper that n 2, the case that n = 1 being slightly exceptional but also completely straightforward. In this subsection, we construct special elements in the groups in the unit filtration of F n. Aside from the case that r = p 1, these arise as restrictions of the elements introduced in Section 3.2. Note that Z p [Γ n ] = Z p [T ]/(f n ), where f n = (T +1) pn 1 1. Of course, we can then speak of the action of T on an element of D n (r). Once again reverting to additive notation, the following is now an immediate corollary of Theorem and Proposition Proposition Let m and j be nonnegative integers with φ m (j) < p n 1. Define ( ) α n,m,j = 1 m {r δ j}!ρ m T j ρ m k T φ k 1(j) δ u n,r, [r]! unless j = 0 and r = p 1, in which case we replace {r δ j}! with 1 in the formula. Then α n,m,j V n,φm(j)(ξ). Furthermore, k=1 for all b Z p [Φ] pz p [Φ] and c T j+1 A n. (p m bt j + c)u n,r / V n,φm(j)+p 1 For nonnegative m n 2, define φ n,m : Z 0 Z 0 by φ n,m(j) = φ m(j) unless r = p 1 and j = e n m 1, in which case we set φ n,m(e n m 1 ) = e n + p m+1 1 = φ m (e n m 1 ) + p m (p 1).
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