GALOIS MODULE STRUCTURE OF MILNOR K-THEORY IN CHARACTERISTIC p

GANESH BHANDARI, NICOLE LEMIRE^{⋆}, JÁN MINÁČ^{∗†}, AND JOHN SWALLOW^{‡}

Abstract. Let E be a cyclic extension of degree p^{n} of a field F of characteristic p. We determine k_{m}E, the Milnor K-groups modulo p, as 𝔽_{p}[Gal(E/F)]-modules for all m ∈ ℕ.

Bloch, Gabber, and Kato established an isomorphism between, on one hand, Milnor K-theory modulo p in characteristic p and, on the other, the kernel of the Artin-Schreier operator defined on the exterior algebra on Kähler differentials [BK, K]. In order to better understand Milnor K-theory, it is a natural problem to determine the Galois module structure of Milnor K-theory when the field is a Galois extension. Two significant papers by Izhboldin consider Milnor K-theory in characteristic p and, using the previous work of Bloch, Gabber, and Kato, ascertain its important Galois module properties [I1, I2].

In this paper we establish the remarkable fact that two results of Izhboldin are enough to determine precisely the Galois module structure of the Milnor groups K_{m}E mod p when E is cyclic extension of pth-power degree of a field F of characteristic p. The result is simple and useful: these modules are direct sums of trivial modules and modules free over some quotient of the Galois group. Equivalently, the dimensions over 𝔽_{p} of the indecomposable 𝔽_{p}[Gal(E/F)]-modules occurring as direct summands of K_{m}E/pK_{m}E are all powers of p. Such a description represents a culmination of results of Bloch-Kato, Gabber, and Izhboldin, available since 1987, concerning the Galois module structure of Milnor K-groups mod p.

Date: October 29, 2004.

2000 Mathematics Subject Classification. 19D45, 12F10.

Key words and phrases. Milnor K-groups modulo p, cyclic extensions, Galois modules.

⋆Research supported in part by NSERC grant R3276A01.

∗Research supported in part by NSERC grant R0370A01.

†Supported by the Mathematical Sciences Research Institute, Berkeley.

‡Research supported in part by NSA grant MDA904-02-1-0061.

The precise description of the 𝔽_{p}[Gal(E/F)]-module K_{m}E/pK_{m}E depends on a natural filtration of K_{m}F/pK_{m}F obtained from images of norm maps defined on Milnor K-groups mod p of intermediate fields of E/F. This description, together with the arguments in the proofs, reveals a flavor of abstract class field theory. In place of the axioms of class formations, we use Izhboldin’s theorems below, together with some technical but relatively short consequences of Hilbert 90, which one might call “Hilbert 90 algebra.”

We assume in what follows that all fields have characteristic p and that m is a fixed natural number. For a field F, let K_{m}F denote the mth Milnor K-group of F and k_{m}F = K_{m}F/pK_{m}F. (See, for instance,

[M] and [FV, IX.1].) If E/F is a Galois extension of fields, let G = Gal(E/F) denote the associated Galois group. We write i_{E}: K_{m}F → K_{m}E and N_{E/F} : K_{m}E → K_{m}F for the natural inclusion and norm maps, and we use the same notation for the induced maps modulo p between k_{m}F and k_{m}E. In order to avoid possible confusion, in a few instances we write i_{F,E} instead of i_{E}.

Izhboldin’s results are as follows.

Theorem 1 ([I2, Lemma 2.3]). Suppose E/F is cyclic of degree p. Then i_{E}: k_{m}F → (k_{m}E)^{G} is an isomorphism.

It is useful to observe that Theorem 1 generalizes to cyclic extensions of degree p^{n}. For α ∈ K_{m}E we write ᾱ for the class of α in k_{m}E.

Theorem 2. Suppose that E/F is cyclic of degree p^{n}, n ∈ ℕ. Then i_{E}: k_{m}F → (k_{m}E)^{G} is an isomorphism.

Proof. We use induction on n. By Theorem 1 our statement is true for n = 1. Consider the tower F ⊂ E_{n−1} ⊂ E where E/E_{n−1} is a cyclic extension of degree p. Set H_{n−1} := Gal(E/E_{n−1}) and G_{n−1} := Gal(E_{n−1}/F). Assume by induction that i_{E_{n−1}} : k_{m}F → (k_{m}E_{n−1})^{G_{n−1}} is an isomorphism. From Theorem 1 we see that i_{E_{n−1},E}: k_{m}E_{n−1} → (k_{m}E)^{H_{n−1}} is an isomorphism. Hence i_{E}: k_{m}F → k_{m}E is injective.

Because i_{F,E}(k_{m}F) ⊂ (k_{m}E)^{G} it is sufficient to show that each ᾱ ∈ (k_{m}E)^{G} can be written as i_{F,E}(f̄) for some f ∈ K_{m}F.

Since ᾱ ∈ (k_{m}E)^{G} ⊂ (k_{m}E)^{H_{n−1}}, by Theorem 1 we see that there exists γ̄ ∈ k_{m}E_{n−1} such that i_{E_{n−1},E}γ̄ = ᾱ. Because ᾱ ∈ (k_{m}E)^{G} we see also that γ̄ ∈ (k_{m}E_{n−1})^{G_{n−1}} and therefore there exists f ∈ K_{m}F such that i_{F,E_{n−1}}(f̄) = γ̄. Then i_{F,E}(f̄) = ᾱ as required. □

GALOIS MODULE STRUCTURE OF MILNOR K-THEORY 3

Theorem 3 (Hilbert 90 for Milnor K-theory: [I1, Corollary of Proposition 5], [I2, Theorem D]). Suppose E/F is cyclic of pth-power degree. Then the sequence

K_{m}E ⟶^{1−σ} K_{m}E ⟶^{N_{E/F}} K_{m}F

is exact.

Now suppose E/F is cyclic of degree p^{n}. For i = 0, …, n, let E_{i}/F be the subextension of degree p^{i} of E/F, H_{i} = Gal(E/E_{i}), and G_{i} = Gal(E_{i}/F). Our main result is the following

Theorem 4. There exists an isomorphism of 𝔽_{p}[G]-modules k_{m}E ≅ ⊕_{i=0}^{n}Y_{i}, where

• Y_{n} is a free 𝔽_{p}[G]-module of rank dim_{𝔽_{p}} N_{E/F} k_{m}E,

• Y_{i}, 0 < i < n, is a free 𝔽_{p}[G_{i}]-module of rank

dim_{𝔽_{p}} N_{E_{i}/F} k_{m}E_{i}/N_{E_{i+1}/F} k_{m}E_{i+1}, and

• Y_{0} is a trivial 𝔽_{p}[G]-module of rank dim_{𝔽_{p}} k_{m}F/N_{E_{1}/F} k_{m}E_{1}.

Our proof of Theorem 4 proceeds by induction on n. At each stage we use the norm map images N_{E_{i}/F} k_{m}E_{i}, i = 1, …, n, to construct the modules Y_{i}. The most interesting portion is the construction of Y_{n−1}. It is easy to show that the modules Y_{i} are all independent, and we then use Theorems 2 and 3 to show that the sum of the Y_{i} exhausts k_{m}E.

The case n = 1 is section 2, and in section 3 we present the inductive argument. This argument relies on an extension lemma, Lemma 6, which shows that certain elements in k_{m}E are expressible as norms. That the various norm groups N_{E_{i}/F} k_{m}E_{i} contain enough elements is, in fact, the crucial step in our induction. Lemma 6, together with two more basic extension lemmas used in the proof of that lemma, are presented in section 1, along with technical results on 𝔽_{p}[G]-modules.

This work uses some ideas previously developed in investigations of the structure of pth-power classes of field extensions of pth-power degree [MSS].

1. Notation and Lemmas

1.1. 𝔽_{p}[G]-modules. For the reader’s convenience, after introducing some notation, we recall in this section some basic elementary facts about 𝔽_{p}[G]-modules.

Let G be a cyclic group of order p^{n} with generator σ. For an 𝔽_{p}[G]- module U, let U^{G} denote the submodule of U fixed by G. For an arbitrary element u ∈ U, let l(u) denote the dimension over 𝔽_{p} of the 𝔽_{p}[G]-submodule ⟨u⟩ of U generated by u. Then we have

(σ − 1)^{l(u)−1}⟨u⟩ = ⟨u⟩^{G} ≠ {0} and (σ − 1)^{l(u)}⟨u⟩ = {0}.

As usual, if U is a free 𝔽_{p}[G]-module with U = ⊕_{i∈I}𝔽_{p}[G], we say that U is a module of rank |I|. Denote by N the operator (σ −1)^{p^{n}−1} acting on module U.

Lemma 1 (Inclusion Lemma). Let U and V be 𝔽_{p}[G]-modules contained in an 𝔽_{p}[G]-module W. Suppose that (U + V)^{G} ⊂ U and for each w ∈ (U + V) \ (U + V)^{G} there exists u ∈ U such that

(σ − 1)^{l(w)−1}(w) = N(u).

Then V ⊂ U.

Proof. Let {T_{i}}_{i=1}^{s} be the socle series of U + V: T_{1} = (U + V)^{G} and T_{i+1}/T_{i} = ((U + V)/T_{i})^{G}, and let s be the least natural number such that T_{s} = U + V. Observe that since (σ − 1)^{p^{n}} = 0, we have s ≤ p^{n}. We prove the lemma by induction on the socle series.

By hypothesis, T_{1} ⊂ U. Assume now that T_{i} ⊂ U for some i < s. Then for each w ∈ T_{i+1} \T_{i} we have l(w) = i+1 and (σ −1)^{l(w)−1}(w) = N(u) = (σ − 1)^{p^{n}−1}(u) for some u ∈ U. Therefore

(σ − 1)^{l(w)−1}(w − (σ − 1)^{p^{n}−l(w)}(u)) = 0.

Therefore w − (σ − 1)^{p^{n}−l(w)}(u) ∈ T_{i} ⊂ U. Hence w ∈ U and T_{i+1} ⊂ U as required. □

Lemma 2 (Exclusion Lemma). Let U and V be 𝔽_{p}[G]-modules contained in an 𝔽_{p}[G]-module W. Suppose that U^{G} ∩ V^{G} = {0}. Then U + V = U ⊕ V.

Proof. Let Z = U ∩ V and suppose that y ∈ Z \ {0}. Let

z = (σ − 1)^{l(y)−1}(y) ≠ 0.

Then z ∈ U^{G} ∩ V^{G}, a contradiction. Hence U ∩ V = {0} and U + V = U ⊕ V. □

The following lemma follows from the fact that each free 𝔽_{p}[G]- module is injective. (See [C, Theorem 11.2].) We shall, however, provide a direct proof.

Lemma 3 (Free Complement Lemma). Let V ⊂ U be free 𝔽_{p}[G]- modules. Then there exists a free 𝔽_{p}[G]-submodule Ṽ of U such that V ⊕ Ṽ = U.

Proof. Let Z be a complement of V^{G} in U^{G} as 𝔽_{p}-vector spaces, and let Z be an 𝔽_{p}-base of Z. For each z ∈ Z, there exists u(z) such that z = N(u(z)). Let M(z) be the 𝔽_{p}[G]-submodule of W generated by u(z). Then M(z) is a free 𝔽_{p}[G]-submodule. Moreover, its fixed submodule M(z)^{G} is the 𝔽_{p}-vector subspace generated by z. By the Exclusion Lemma, the set of modules M(z), z ∈ Z, is independent.

Set Ṽ := ⊕_{z∈Z}M(z). Then Ṽ is a free 𝔽_{p}[G]-submodule of U and Ṽ^{G} = Z. By the Exclusion Lemma V + Ṽ = V ⊕ Ṽ and (V ⊕ Ṽ)^{G} = V^{G} ⊕ Ṽ^{G} = U^{G}.

Now let u ∈ U be arbitrary and let M be the cyclic 𝔽_{p}[G]-submodule of U generated by u. Then (M +V +Ṽ)^{G} ⊂ U^{G} ⊂ V +Ṽ. Moreover, for any m ∈ M +V +Ṽ \(M +V +Ṽ)^{G}, (σ−1)^{l(m)−1}(m) ∈ (M +V +Ṽ)^{G} ⊂ U^{G} = (V + Ṽ)^{G} = N(V + Ṽ) by the freeness of V and Ṽ. By the Inclusion Lemma, then, M ⊂ V + Ṽ. Hence U = V ⊕ Ṽ. □

1.2. Milnor k-groups.

Recall that if α ∈ K_{m}E we write ᾱ for the class of α in k_{m}E. For γ ∈ K_{m}E, let l(γ) denote the dimension over 𝔽_{p} of the 𝔽_{p}[G]-submodule ⟨γ̄⟩ of k_{m}E generated by γ̄. Because (σ − 1)^{p^{n}−1} = 1 + σ + ⋯ + σ^{p^{n}−1} in 𝔽_{p}[G], we may use i_{E}N_{E/F} and N interchangeably on k_{m}E.

Our proof of Theorem 4 depends on Lemma 6. This lemma, in turn, requires for its induction two basic extension lemmas, the first for the case p > 2, n = 1, and the second for the case p = 2, n = 2. (The case p = 2, n = 1 is clear.) It is in only these two basic lemmas that Theorem 3 is used in this paper.

Lemma 4. Suppose E/F is a cyclic extension of degree p > 2 and γ ∈ K_{m}E with 2 ≤ l(γ) ≤ p. Then there exists α ∈ K_{m}E such that

⟨Nᾱ⟩ = ⟨γ̄⟩^{G}.

Proof. Let l = l(γ) and suppose 2 ≤ l ≤ i ≤ p. We show by induction on i that there exists α_{i} ∈ K_{m}E such that

⟨(σ − 1)^{i−1}ᾱ_{i}⟩ = ⟨γ̄⟩^{G}.

Then setting α := α_{p}, the proof will be complete. If i = l then α_{i} = γ suffices. Assume now that l ≤ i < p and our statement is true for i.

Set c := N_{E/F} α_{i}. Since i_{E}c̄ = Nᾱ_{i} = (σ − 1)^{p−1}ᾱ_{i} and i < p, we conclude that i_{E}c̄ = 0. By the injectivity of i_{E} from Theorem 1, c̄ = 0. Therefore there exists f ∈ K_{m}F such that c = pf in K_{m}F.

We calculate

N_{E/F} (α_{i} − i_{E}(f)) = c − pf = 0.

By Theorem 3, there exists ω ∈ K_{m}E such that

(σ − 1)ω = α_{i} − i_{E}(f).

Hence (σ − 1)^{2}ω = (σ − 1)α_{i}. Since i ≥ 2,

⟨(σ − 1)^{i}ω̄⟩ = ⟨(σ − 1)^{i−1}ᾱ_{i}⟩ = ⟨γ̄⟩^{G}

and we can set α_{i+1} = ω. □

Lemma 5. Suppose p = 2 and E/F is a cyclic extension of degree 4. Suppose γ ∈ K_{m}E with 3 ≤ l(γ) ≤ 4. Then there exists α ∈ K_{m}E such that

⟨Nᾱ⟩ = ⟨γ̄⟩^{G}.

Proof. The case l(γ) = 4 is clear. Assume then that l(γ) = 3, and set β := (σ − 1)γ. Then

(σ^{2} − 1)β̄ = (σ − 1)^{3}γ̄ = 0.

By Theorem 1 applied to E/E_{1}, β̄ ∈ i_{E}(k_{m}E_{1}). Let b ∈ K_{m}E_{1} such that i_{E}b̄ = β̄.

Set c := N_{E_{1}/F} b. Then

i_{E}c̄ = i_{E}(σ + 1)b̄ = (σ^{2} − 1)γ̄ = i_{E}N_{E/E_{1}} γ̄,

and since l(γ) = 3, ⟨γ̄⟩^{G} = ⟨c̄⟩. Now N_{E/E_{1}}γ = c + 2e for some e ∈ K_{m}E, and e_{1} := 2e ∈ K_{m}E_{1}∩2K_{m}E. Then i_{E}(ē_{1}) = 0 ∈ k_{m}E, and by the injectivity of i_{E} from Theorem 1, e_{1} = 2g for some g ∈ K_{m}E_{1}. Hence

N_{E/E_{1}}γ = c + 2g, g ∈ K_{m}E_{1}.

Now N_{E/F} γ = N_{E_{1}/F} (c + 2g) = 2c + 2N_{E_{1}/F} g. Set δ := b + g. We calculate

N_{E/F} δ = 2c + 2N_{E_{1}/F} g,

so that N_{E/F} (γ − δ) = 0. By Theorem 3 there exists an α ∈ K_{m}E with (σ − 1)α = γ − δ. Then

i_{E}N_{E/F} ᾱ = (σ − 1)^{3}ᾱ = (σ − 1)^{2}\overline{(γ − δ)} =
i_{E}\overline{N_{E/E_{1}}(γ − δ)} = i_{E}\overline{(c + 2g − 2b − 2g)} =
i_{E}c̄ = (σ − 1)^{2}γ̄.

Because ⟨(σ − 1)^{2}γ̄⟩ = ⟨γ̄⟩^{G} our statement follows. □

Our full extension lemma is then

Lemma 6. Suppose that E/F is a cyclic extension of degree p^{n}, n ∈ ℕ, and let γ̄ ∈ k_{m}E \ k_{m}E_{n−1}. Then there exists χ ∈ K_{m}E such that

(σ − 1)^{l(γ)−1}γ̄ = (σ − 1)^{p^{n}−1}χ̄ = i_{E}N_{E/F} χ̄.

Proof. We shall first prove our statement in the case p > 2. If n = 1 then our statement follows from Lemma 4.

Assume therefore that n > 1. Let H := H_{n−1}, and write l_{H}(γ) for the dimension over 𝔽_{p} of the 𝔽_{p}[H]-submodule of k_{m}E generated by γ̄. From Theorem 1 we see that l_{H}(γ) ≥ 2. Therefore Lemma 4 tells us that there exists an element α ∈ K_{m}E such that

(σ − 1)^{p^{n}−p^{n−1}} ᾱ = (σ^{p^{n−1}} − 1)^{p−1} ᾱ = (σ^{p^{n−1}} − 1)^{l_{H}(γ)−1} γ̄ ≠ 0.

Set s := l(γ) − p^{n−1}(l_{H}(γ) − 1). Then

(σ − 1)^{p^{n}−p^{n−1}+s−1}ᾱ = (σ − 1)^{p^{n−1}(l_{H}(γ)−1)+s−1}γ̄ ≠ 0.

Furthermore, this element belongs to (k_{m}E)^{G}. Set λ := (σ − 1)^{s}α. Then

(σ − 1)^{p^{n}−p^{n−1}−1}λ̄ = (σ − 1)^{p^{n}−p^{n−1}+s−1}ᾱ,

whence l(λ) = p^{n} − p^{n−1}.

Now we consider l_{H}(λ). On one hand,

(σ^{p^{n−1}} − 1)^{p−1} λ̄ = 0,

and on the other,

(σ − 1)^{p^{n−1}(p−2)}λ̄ = (σ − 1)^{p^{n}−2p^{n−1}} λ̄ ≠ 0.

We deduce that l_{H}(λ) = p − 1 ≥ 2. Applying Lemma 4 again, there exists χ ∈ K_{m}E with

(σ − 1)^{p^{n}−p^{n−1}} χ̄ = (σ − 1)^{p^{n}−2p^{n−1}} λ̄.

In particular, l(χ) = l(λ) + p^{n−1} = p^{n}.

Summarizing, we have obtained

i_{E}N_{E/F} χ̄ = (σ − 1)^{p^{n}−1}χ̄ = (σ − 1)^{p^{n}−p^{n−1}−1}λ̄ = (σ − 1)^{p^{n}−p^{n−1}+s−1}ᾱ.

Because

(σ − 1)^{p^{n}−p^{n−1}+s−1}ᾱ = (σ − 1)^{p^{n−1}(l_{H}(γ)−1)+s−1}γ̄ = (σ − 1)^{l(γ)−1}γ̄,

we have established our equality in the case p > 2.

Now we shall consider the case p = 2. The case n = 1 is trivial, and the case n = 2 is handled by Lemma 5. Assume therefore that n ≥ 3. Let H := Gal(E/E_{n−2}), and write l_{H}(γ) for the dimension over 𝔽_{2} of the cyclic 𝔽_{2}[H]-submodule of k_{m}E generated by γ̄.

Because γ̄ ∉ k_{m}E_{n−1}, we have l_{H}(γ) ≥ 3. From Lemma 5 we see that there exists α ∈ K_{m}E such that

(σ^{2^{n−2}} − 1)^{3} ᾱ = (σ^{2^{n−2}} − 1)^{l_{H}(γ)−1} γ̄.

Hence

(σ − 1)^{2^{n}−2^{n−2}} ᾱ = (σ − 1)^{2^{n−2}(l_{H}(γ)−1)} γ̄ ≠ 0.

Let s := l(γ) − 2^{n−2} (l_{H}(γ) − 1). Then we have

(σ − 1)^{2^{n}−2^{n−2}+s−1} ᾱ = (σ − 1)^{2^{n−2}(l_{H}(γ)−1)+s−1} γ̄ ≠ 0.

Furthermore, this element belongs to (k_{m}E)^{G}. Set λ := (σ − 1)^{s}α. Then

(σ − 1)^{2^{n}−2^{n−2}−1}λ̄ = (σ − 1)^{2^{n}−2^{n−2}+s−1}ᾱ,

whence l(λ) = 2^{n} − 2^{n−2}.

Now we consider l_{H}(λ). On one hand,

(σ^{2^{n−2}} − 1)^{3} λ̄ = (σ − 1)^{2^{n}−2^{n−2}} λ̄ = 0,

and on the other,

(σ^{2^{n−2}} − 1)^{2} λ̄ ≠ 0.

We deduce that l_{H}(λ) = 3. By Lemma 5 there exists χ ∈ K_{m}E with

(σ^{2^{n−2}} − 1)^{3} χ̄ = (σ^{2^{n−2}} − 1)^{2} λ̄.

Equivalently, (σ − 1)^{2^{n}−2^{n−2}} χ̄ = (σ − 1)^{2^{n−1}} λ̄, and therefore l(χ) = l(λ) + 2^{n−2} = 2^{n}.

Summarizing, we have obtained

i_{E}N_{E/F} χ̄ = (σ − 1)^{2^{n}−1}χ̄ = (σ − 1)^{2^{n}−2^{n−2}−1}λ̄
= (σ − 1)^{2^{n}−2^{n−2}+s−1}ᾱ = (σ − 1)^{l(γ)−1}γ̄

as required. □

Remark. Observe that since ⟨γ̄⟩^{G} is a one-dimensional 𝔽_{p}-vector space and N is additive, the equality ⟨Nᾱ⟩ = ⟨γ̄⟩^{G} holds if and only if any δ̄ ∈ ⟨γ̄⟩^{G} \ {0} is expressible as Nβ̄ for some β̄ ∈ k_{m}E.

Lemma 7 (Submodule-Subfield Lemma). Suppose E/F is a cyclic extension of degree p^{n}, and let U be a free 𝔽_{p}[G]-submodule of k_{m}E. Then for each i in {0, 1, …, n − 1} we have:

U^{H_{i}} = (σ − 1)^{p^{n}−p^{i}}U = U ∩ i_{E}(N_{E/E_{i}}k_{m}E) = U ∩ i_{E}(k_{m}E_{i}).

Proof. Suppose ᾱ ∈ U^{H_{i}}. Then (σ^{p^{i}} − 1)ᾱ = (σ − 1)^{p^{i}} ᾱ = 0, so l(α) ≤ p^{i}. Since U is free, ᾱ = (σ − 1)^{p^{n}−l(α)}β̄ for some β̄ ∈ U. In particular,

ᾱ = (σ − 1)^{p^{n}−p^{i}}(σ − 1)^{p^{i}−l(α)}β̄.

Hence U^{H_{i}} ⊂ (σ − 1)^{p^{n}−p^{i}}U. Now suppose ᾱ = (σ − 1)^{p^{n}−p^{i}} γ̄. Then since

(σ − 1)^{p^{n}−p^{i}} γ̄ = i_{E}N_{E/E_{i}} γ̄,

we have (σ − 1)^{p^{n}−p^{i}}U ⊂ U ∩ i_{E}(N_{E/E_{i}}k_{m}E) ⊂ U ∩ i_{E}(k_{m}E_{i}).

Now suppose that ᾱ ∈ U ∩ i_{E}(k_{m}E_{i}). Then ᾱ ∈ U^{H_{i}} and hence all of the inclusions are equalities. □

Remark. Alternatively one can derive the first equality in the lemma above, as follows. If U is a free 𝔽_{p}[G]-module, then U is also a free 𝔽_{p}[H_{i}]-module. But then H^{2}(H_{i}, U) = {0}. Hence U^{H_{i}} = N_{i}(U) := the image of the norm operator N_{i}. Thus U^{H_{i}} = U^{(σ−1)^{p^{n}−p^{i}}} as required.

2. E/F Cyclic of Degree p

Proposition. Theorem 4 holds for n = 1.

Proof. We first construct submodules Y_{0} and Y_{1} of k_{m}E. Let I be an 𝔽_{p}-basis for the subspace i_{E}(N_{E/F} k_{m}E). For each basis element ȳ ∈ I, let α_{y} ∈ K_{m}E satisfy i_{E}(N_{E/F} ᾱ_{y}) = ȳ. Then ⟨ᾱ_{y}⟩ is a cyclic submodule of dimension p, hence isomorphic to 𝔽_{p}[G], with

⟨ᾱ_{y}⟩^{G} = (σ − 1)^{p−1}⟨ᾱ_{y}⟩ = ⟨Nᾱ_{y}⟩ = ⟨ȳ⟩.

Set Y_{1} := ∑_{ȳ∈I}⟨ᾱ_{y}⟩. By the Exclusion Lemma, Y_{1} = ⊕_{ȳ∈I}⟨ᾱ_{y}⟩ and so Y_{1} is a free 𝔽_{p}[G]-module. Moreover,

Y_{1}^{G} = (σ − 1)^{p−1}Y_{1} = NY_{1} = ⊕_{ȳ∈I}⟨ȳ⟩ = i_{E}(N_{E/F} k_{m}E).

The rank of Y_{1} is equal to the dimension of i_{E}(N_{E/F} k_{m}E), and by the injectivity of i_{E} from Theorem 1, we conclude

rank Y_{1} = dim_{𝔽_{p}} N_{E/F} k_{m}E.

Now let Z be any 𝔽_{p}-module complement of N_{E/F} k_{m}E in k_{m}F, and set Y_{0} := i_{E}Z. By the injectivity of i_{E} from Theorem 1, we obtain

rank Y_{0} = dim_{𝔽_{p}} k_{m}F/N_{E/F} k_{m}E.

By the Exclusion Lemma, Y_{0} + Y_{1} = Y_{0} ⊕ Y_{1}. Moreover, Y_{0} + Y_{1} contains (k_{m}E)^{G}:

(k_{m}E)^{G} = i_{E}(k_{m}F) = i_{E}((N_{E/F} k_{m}E) ⊕ Z)
= i_{E}(N_{E/F} k_{m}E) ⊕ Y_{0} = Y_{1}^{G} ⊕ Y_{0} ⊂ Y_{0} ⊕ Y_{1}.

If p = 2, we already have enough to prove the proposition, as follows. Let γ̄ ∈ k_{m}E. If (σ − 1)γ̄ = 0 then γ̄ ∈ (k_{m}E)^{G} ⊂ Y_{0} ⊕ Y_{1} as above. Otherwise, observe that since p = 2, the operator σ − 1 is equivalent to i_{E}N_{E/F} on k_{m}E. Let β = i_{E}N_{E/F} γ. Then (σ − 1)γ̄ = β̄ ∈ Y_{1}^{G}. Since Y_{1} is free there exists ᾱ ∈ Y_{1} such that (σ − 1)ᾱ = β̄. Then (σ−1)(γ̄−ᾱ) = 0, so l(γ−α) ≤ 1 and γ̄−ᾱ ∈ (k_{m}E)^{G} ⊂ Y_{1}⊕Y_{0}. Since ᾱ ∈ Y_{1}, we conclude that γ̄ ∈ Y_{1} ⊕ Y_{0} as well. Hence k_{m}E = Y_{1} ⊕ Y_{0}.

If p > 2, we instead prove by induction on l(γ) that each γ̄ ∈ k_{m}E lies in Y_{1}⊕Y_{0}, whence we will obtain the proposition. The case l(γ) = 1 follows from (k_{m}E)^{G} ⊂ Y_{1} ⊕ Y_{0}. Suppose now that for each β̄ ∈ k_{m}E with l(β) ≤ i < p we have β̄ ∈ Y_{1}⊕Y_{0}, and assume that l(γ) = i+1 ≥ 2. By Lemma 4 there exists an α ∈ K_{m}E with

⟨Nᾱ⟩ = ⟨γ̄⟩^{G}.

Hence (σ−1)^{i}γ̄ = cNᾱ for some c ∈ ℤ. Now cNᾱ = i_{E}N_{E/F} (cᾱ) ∈ Y_{1}^{G}. Since Y_{1} is free, there exists ω̄ ∈ Y_{1} such that (σ − 1)^{p−1}ω̄ = cNᾱ. Set λ̄ = (σ − 1)^{p−l(γ)}ω̄ ∈ Y_{1}, so that (σ − 1)^{l(γ)−1}(γ̄ − λ̄) = 0. Hence l(γ − λ) < l(γ) and by induction γ̄ = (γ̄ − λ̄) + λ̄ ∈ Y_{1} ⊕ Y_{0}. Thus again we conclude that k_{m}E = Y_{1} ⊕ Y_{0}. □

3. Proof of Theorem 4

We proceed by induction. The case n = 1 is the preceding proposition. Assume then that Theorem 4 holds when degree of E/F is ≤ p^{n−1}, and suppose that E/F is a cyclic extension of degree p^{n} with n ≥ 2.

3.1. Constructing the Y_{i}.

Let I be an 𝔽_{p}-basis for i_{E}(N_{E/F} k_{m}E). For each basis element ȳ ∈ I construct a free 𝔽_{p}[G]-module ⟨ᾱ_{y}⟩, such that i_{E}N_{E/F} ᾱ_{y} = ȳ. We see by the Exclusion Lemma that the modules ⟨ᾱ_{y}⟩, ȳ ∈ I, are independent. Set Y_{n} := ⊕_{y∈I}⟨ᾱ_{y}⟩. Hence Y_{n} is a direct sum of cyclic 𝔽_{p}[G]-modules of dimension p^{n}, and Y_{n}^{G} = i_{E}(N_{E/F} k_{m}E). By the injectivity of i_{E} from Theorem 2,

rank_{𝔽_{p}[G]} Y_{n} = dim_{𝔽_{p}} Y_{n}^{G} = dim_{𝔽_{p}} N_{E/F} k_{m}E.

By induction, we have a 𝔽_{p}[G_{n−1}]-module decomposition

k_{m}E_{n−1} = Ỹ_{n−1} ⊕ Ỹ_{n−2} ⊕ ⋯ ⊕ Ỹ_{0}

into direct sums Ỹ_{i} of cyclic 𝔽_{p}[G_{n−1}]-modules of dimension p^{i}, i = 0, 1, …, n−1. Let σ̄ denote the image of σ under the natural projection G → G_{n−1}. Because i_{E_{n−1}}N_{E_{n−1}/F} acts on k_{m}E_{n−1} as (σ̄ − 1)^{p^{n−1}−1} and i_{E_{n−1}} is injective, we see that N_{E_{n−1}/F} annihilates the sum Ỹ_{n−2} ⊕ ⋯ ⊕ Ỹ_{0}. Therefore

i_{E_{n−1}}(N_{E_{n−1}/F} k_{m}E_{n−1}) = Ỹ_{n−1}^{G_{n−1}}.

Now consider k_{m}E as an 𝔽_{p}[H_{n−1}]-module. By the injectivity of i_{E} from Theorem 1, the images of Ỹ_{i}, i = 0, …, n − 1, under i_{E} are direct sums of cyclic modules of dimension p^{i} and are independent. Because the modules Ỹ_{i} are direct sums of cyclic 𝔽_{p}[G_{n−1}]-modules, the images i_{E}Ỹ_{i} are direct sums of cyclic 𝔽_{p}[G]-modules. Set Y_{i} := i_{E}Ỹ_{i} for i < n − 1. Observe that rank_{𝔽_{p}[G_{i}]} Y_{i} = rank_{𝔽_{p}[G_{i}]} Ỹ_{i}.

Set W := Y_{n}^{H_{n−1}}. By the Submodule-Subfield Lemma,

W = (σ − 1)^{p^{n}−p^{n−1}}Y_{n} = Y_{n} ∩ i_{E}k_{m}E_{n−1}.

Since Y_{n} is a direct sum of cyclic 𝔽_{p}[G]-modules of dimension p^{n}, W is a direct sum of cyclic modules of dimension p^{n−1} and hence is free as an 𝔽_{p}[G_{n−1}]-module. Because W ⊂ i_{E}k_{m}E_{n−1}, we may consider the image P of the projection map pr: W → i_{E}Ỹ_{n−1} from W to the summand i_{E}Ỹ_{n−1} in the decomposition

i_{E}k_{m}E_{n−1} = i_{E}Ỹ_{n−1} ⊕ Y_{n−2} ⊕ ⋯ ⊕ Y_{0}.

Observe that W ≅ P as 𝔽_{p}[G_{n−1}]-modules. Indeed, since W is a free 𝔽_{p}[G_{n−1}]-module, each w ∈ W \ {0} may be written as (σ̄ − 1)^{s}w̃ for some 0 ≤ s ≤ p^{n−1} − 1 and w̃ ∈ W with l(w̃) = p^{n−1}. We have

(σ̄ − 1)^{p^{n−1}−1} pr(w̃) = (σ̄ − 1)^{p^{n−1}−1}w̃ ≠ 0,

since all other components of w̃ are killed by (σ̄ − 1)^{p^{n−1}−1}. (Since n ≥ 2, p^{n−1} − 1 ≥ p^{n−2}.) Therefore (σ̄ − 1)^{s} pr(w̃) = pr(w) ≠ 0. We conclude that the kernel of the projection map is (0), as required.

Since M^{G_{n−1}} = (σ − 1)^{p^{n−1}−1}M for free 𝔽_{p}[G_{n−1}]-modules, we have further obtained that W^{G_{n−1}} = P^{G_{n−1}}. Observe that

W^{G_{n−1}} = (σ − 1)^{p^{n−1}−1}W ⊂ i_{E}(N_{E_{n−1}/F} k_{m}E_{n−1}) = (i_{E}Ỹ_{n−1})^{G}.

By the Free Complement Lemma, there exists a free 𝔽_{p}[G_{n−1}]-module complement Y_{n−1} in i_{E}Ỹ_{n−1} of P. Since W = Y_{n} ∩i_{E}k_{m}E_{n−1}, we obtain Y_{n}^{G} = W^{G_{n−1}} = P^{G_{n−1}}. By the Exclusion Lemma, P^{G_{n−1}} ∩Y_{n−1}^{G_{n−1}} = {0} implies that Y_{n−1}+Y_{n} = Y_{n−1}⊕Y_{n}. Then, since P^{G}+Y_{n−1}^{G} = (i_{E}Ỹ_{n−1})^{G}, we obtain (Y_{n−1} + Y_{n})^{G} = (i_{E}Ỹ_{n−1})^{G}. By induction and the injectivity of i_{E},

rank_{𝔽_{p}[G_{n−1}]} i_{E}Ỹ_{n−1} = dim_{𝔽_{p}}(i_{E}Ỹ_{n−1})^{G} = dim_{𝔽_{p}} N_{E_{n−1}/F} k_{m}E_{n−1}.

Since Y_{n−1}^{G} ⊕ Y_{n}^{G} = (i_{E}Ỹ_{n−1})^{G},

rank_{𝔽_{p}[G_{n−1}]} Y_{n−1} = dim_{𝔽_{p}} Y_{n−1}^{G} = dim_{𝔽_{p}} N_{E_{n−1}/F} k_{m}E_{n−1}/N_{E/F} k_{m}E.

Finally, by the Exclusion Lemma, Y_{n−1}+Y_{n} is independent from Y_{n−2}+ ⋯ + Y_{0}. Hence we have a submodule

Y = Y_{n} ⊕ Y_{n−1} ⊕ ⋯ ⊕ Y_{0} ⊂ k_{m}E

with ranks satisfying the claims of Theorem 4.

3.2. Showing k_{m}E = ⊕Y_{i}.

We prove first that Y^{H_{n−1}} = i_{E}k_{m}E_{n−1}. Theorem 1 tells us that Y^{H_{n−1}} ⊂ i_{E}k_{m}E_{n−1}, and we have the decomposition i_{E}k_{m}E_{n−1} = i_{E}Ỹ_{n−1} ⊕ Y_{n−2} ⊕ ⋯ ⊕ Y_{0}. Therefore it is sufficient to show that

i_{E}Ỹ_{n−1} ⊂ Y^{H_{n−1}} = Y_{n}^{H_{n−1}} + Y_{n−1} + ⋯ + Y_{0}.

Because i_{E}Ỹ_{n−1} = Y_{n−1} + P it is enough to show that

P ⊂ Y_{n}^{H_{n−1}} + Y_{n−1} + ⋯ + Y_{0} = W + Y_{n−1} + ⋯ + Y_{0}.

But by the definition of the projection, P ⊂ W +Y_{n−2} +⋯+Y_{0}. Hence P ⊂ W + Y_{n−1} + ⋯ + Y_{0}, and we conclude that Y_{n}^{H_{n−1}} = i_{E}k_{m}E_{n−1}.

We adapt the proof of the Inclusion Lemma to show that k_{m}E ⊂ Y, by induction on the socle series V_{i} of k_{m}E. Our base case is V_{p^{n−1}}. Observe that V_{p^{n−1}} is the kernel of (σ − 1)^{p^{n−1}} = σ^{p^{n−1}} − 1, which is (k_{m}E)^{H_{n−1}}. Hence V_{p^{n−1}} ⊂ Y.

For the inductive step, assume that V_{i} ⊂ Y for all i < t for some p^{n−1} < t ≤ p^{n}, and let γ̄ ∈ V_{t} \ V_{t−1}. Hence l(γ) = t. Therefore γ̄ ∉ i_{E}k_{m}E_{n−1}, and by Lemma 6 there exists χ ∈ K_{m}E such that

(σ − 1)^{t−1}γ̄ = i_{E}N_{E/F} χ̄ ∈ Y_{n}^{G}.

Since Y_{n} is a free 𝔽_{p}[G]-module, there exists ν̄ ∈ Y_{n} such that

i_{E}N_{E/F} ν̄ = (σ − 1)^{p^{n}−1}ν̄ = (σ − 1)^{p^{n}−1}χ̄ = (σ − 1)^{t−1}γ̄.

Set δ̄:= (σ − 1)^{p^{n}−t}ν̄ ∈ Y_{n} ⊂ Y. Then (σ − 1)^{t−1}(γ̄ − δ̄) = 0 and hence l(γ − δ) < t. By induction γ̄ − δ̄ ∈ Y, and since δ̄ ∈ Y, we see that γ̄ ∈ Y as well. □

Acknowledgment

We are very grateful to Andy Schultz, as some of his ideas, developed in [MSS], proved quite useful to us during the investigations leading to this paper.

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Department of Mathematics, Middlesex College, University of Western Ontario, London, Ontario N6A 5B7 CANADA

E-mail address: gbhandar@uwo.ca

E-mail address: nlemire@uwo.ca

E-mail address: minac@uwo.ca

Department of Mathematics, Davidson College, Box 7046, Davidson, North Carolina 28035-7046 USA

E-mail address: joswallow@davidson.edu