New York Journal of Mathematics

New York J. Math. 14 (2008) 215–224.

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. Using arithmetic invariants of E/F we determine k_{m}E, the Milnor K-groups K_{m}E modulo p, as 𝔽_{p}[Gal(E/F)]-modules for all m ∈ ℕ. In particular, we show that each indecomposable summand of k_{m}E has 𝔽_{p}-dimension a power of p. That all powers p^{i}, i = 0, 1, …, n, occur for suitable examples is shown in a subsequent paper, Mináč, Schultz and Swallow, 2008, where additionally the main result of this paper becomes an essential induction step in the determination of K_{m}E/p^{s}K_{m}E as (ℤ/p^{s}ℤ)[Gal(E/F)]-modules for all m, s ∈ ℕ.

Contents

1. 𝔽_{p}G-modules 218 2. Milnor k-groups 219 3. Proof of Lemma 4 222 4. Alternative proof 223 References 223

In the 1960s (see [Bo]) Z. I. Borevič and D. K. Faddeev classified possible 𝔽_{p}[Gal(E/F)]-modules k_{1}E where E is a cyclic extension of degree p^{n} of a local field F. In [MS] it was shown that if n = 1, the local field F can be replaced by any field containing a primitive pth-root of unity. In [MSS1] this

Received March 9, 2008; revised June 6, 2008. Mathematics Subject Classification. 19D45, 12F10. Key words and phrases. Milnor K-groups modulo p, cyclic extensions, Galois modules. The second author’s research was supported in part by NSERC grant R3276A01. The third author’s research was supported in part by NSERC grant R0370A01, and by

a Distinguished Research Professorship at the University of Western Ontario.

The fourth author’s research was supported in part by NSA grant MDA904-02-1-0061 and by NSF grant DMS-0600122.

215^{ISSN 1076-9803/08}

classification was extended to all base fields F and all cyclic extensions E of degree p^{n}. In [LMS] the classification of all 𝔽_{p}[Gal(E/F)]-modules k_{m}E, m ∈ ℕ was obtained, provided E is a cyclic extension of degree p and F contains a primitive pth-root of unity. In [BLMS] results from [LMS] were used to obtain a restriction on possible absolute Galois pro-p groups. Thus, only when m = 1 has the case of characteristic p been considered ([MSS1]). However, the case of characteristic p is important not only for its intrinsic value, but also because this case arises naturally in considering the residue fields of valuations on fields of characteristic 0.

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].

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 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 a 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. In [MSS2] it is shown that in fact each indecomposable summand of dimension p^{i}, i = 0, 1, …, n, occurs in suitable examples.

Our decomposition of the 𝔽_{p}[Gal(E/F)]-module k_{m}E into indecomposables in Theorem 2 is not canonical, although the ranks of the summands which are free over the various quotients of Gal(E/F) are invariants of the 𝔽_{p}[Gal(E/F)]-module k_{m}E. Still, our results do have a canonical formulation, stemming from Lemma 4, which is in fact equivalent to Theorem 2. In particular, the filtration of (k_{m}E)^{Gal(E/F)} ≃ k_{m}F, induced by the images of successive powers of the maximal ideal of 𝔽_{p}[Gal(E/F)], is determined by the images of the norm maps k_{m}E_{i} → k_{m}F, where E_{i} runs through the intermediate fields between E and F. It is possible that one could use the results in [J] to obtain alternative proofs, but we preferred instead a more self-contained short derivation.

In [MSS2] our main result is used as a critical first induction step to determine K_{m}E/p^{s}K_{m}E as a (ℤ/p^{s}ℤ)[Gal(E/F)]-module for all m, s ∈ ℕ. The classification problem of ℤ/p^{s}ℤ[G]-modules for cyclic G is nontrivial and in fact still not complete, despite some important results when, for instance, the module is free over ℤ/p^{s}ℤ (see [T] and the references contained therein). However one can describe the (ℤ/p^{s}ℤ)[Gal(E/F)]-module K_{m}E/p^{s}K_{m}E completely and simply.

Galois module structure of Milnor K-theory 217

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, [Ma] and [Mi, IX.1]). If E/F is a Galois extension of fields, let G = Gal(E/F) denote the associated Galois group. When G is a cyclic group we write G = ⟨σ⟩, with a suitable fixed generator σ. For the sake of simplicity we write ρ instead of σ − 1. We write i_{E}: K_{m}F → K_{m}E and N_{E/F} : K_{m}E → K_{m}F for the natural map induced by the inclusion and the 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.

Lemma 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.

Theorem 1 (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 following sequence is exact:

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

Observe that Lemma 1 can be used to prove the following stronger variant:

Lemma 1′. Suppose E/F is a Galois extension such that G has order p^{n}, n ∈ ℕ. Then i_{E}: k_{m}F → (k_{m}E)^{G} is an isomorphism.

Proof. We use induction on n. By Lemma 1 our statement is true for n = 1. Assume therefore that n > 1. Let H be a central subgroup of G of order p and D be its fixed field. Then D/F is a Galois extension of degree p^{n−1}. Let L = Gal(D/F). By induction i_{D} : k_{m}F → (k_{m}D)^{L} is an isomorphism. By Lemma 1 we see that i_{D,E} : k_{m}D → (k_{m}E)^{G} is an isomorphism. Hence i_{E} : k_{m}F → (k_{m}E)^{G} is injective.

Let α ∈ (k_{m}E)^{G}. Since (k_{m}E)^{G} ⊆ (k_{m}E)^{H}, by Lemma 1 there exists γ ∈ k_{m}D such that i_{D,E}γ = α. Because α ∈ (k_{m}E)^{G} and i_{D,E} is injective we see that γ ∈ (k_{m}D)^{L}. Hence by our induction hypothesis there exists δ ∈ k_{m}F such that i_{F,D}δ = γ. Thus i_{E}δ = α and we see that i_{E} : k_{m}F → (k_{m}E)^{G} is an isomorphism, as required. □

Now suppose E/F is cyclic of degree p^{n}. For i = 0, 1, …, 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 2. 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, and
• 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}.

Remark 1. Observe that Y_{0} is a trivial 𝔽_{p}G-module of dimension over 𝔽_{p} equal to dim_{𝔽_{p}} k_{m}F/N_{E_{1}/F} k_{m}E_{1}. Moreover, the statements about the ranks of the Y_{i} are immediate consequences of the fact that k_{m}E is a direct sum of indecomposable 𝔽_{p}G-modules whose dimensions over 𝔽_{p} are all powers of p — which is the main point of the theorem (see Section 2, proof of the Reduction to Lemma 4).

Our proof of Theorem 2 relies on a filtration lemma, Lemma 4, 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 key to this lemma. In Section 1 we present technical results on 𝔽_{p}G-modules. In Section 2 we state Lemma 4, prove that Theorem 2 is a corollary to Lemma 4, and prove basic lemmas in preparation for its proof. This proof is finally presented in Section 3, and in Section 4 we present an alternative proof for some cases of Theorem 2.

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 (see [C] and [L]).

Let G be a cyclic group of order p^{n} with generator σ. We denote σ − 1 as ρ. For an 𝔽_{p}G-module U, let U^{G} denote the submodule of U fixed by G. For an arbitrary element u ∈ U, we say that the length l(u) is the dimension over 𝔽_{p} of the 𝔽_{p}G-submodule ⟨u⟩ of U generated by u. Then we have

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

If we want to stress the dependence of length l(u) on G, we write l_{G}(u). 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 ρ^{p^{n}−1} acting on module U. For each k = 0, 1, …, n, let H_{k} be the subgroup of G of order p^{n−k}. Observe that if G = Gal(E/F) then this definition coincides with the previous definition H_{k} = Gal(E/E_{k}).

We use the following general lemma about 𝔽_{p}G-modules. The simple proof of this lemma is omitted.

Lemma 2 (Exclusion Lemma). Let M_{i}, i ∈ I, be a family of 𝔽_{p}G-modules contained in a common 𝔽_{p}G-module N. Suppose that the 𝔽_{p}-vector subspace R of N generated by all M_{i}^{G} has the form R = ⊕_{i∈I}M_{i}^{G}. Then the 𝔽_{p}Gmodule M generated by M_{i}, i ∈ I, has the form M = ⊕_{i∈I}M_{i}.

We also use the following result about the restriction of a cyclic 𝔽_{p}Gmodule to a subgroup.

Lemma 3 (Restriction Lemma). Let Y be a cyclic 𝔽_{p}G-module of length l generated by γ. Then l_{H_{k}}(γ) is the unique integer such that

l = (l_{H_{k}}(γ) − 1)p^{k} + r

for some unique integer 1 ≤ r ≤ p^{k}. Moreover,

Y ≃ V_{l_{H_{k}} (γ)}^{r} ⊕ V_{l_{H_{k}} (γ)−1}^{p^{k}−r}

where V_{i} is a cyclic 𝔽_{p}H_{k}-module of length i.

Proof. The formula for l_{H_{k}} follows immediately from the definition of length. Since {ρ^{i}γ | 0 ≤ i ≤ l − 1} is an 𝔽_{p}-basis for Y = 𝔽_{p}Gγ, then

𝔽_{p}Gγ|_{H_{k}} = ⊕_{i=0}^{p^{k}−1}𝔽_{p}H_{k}ρ^{i}γ ≃ V_{l_{H_{k}} (γ)}^{r} ⊕ V_{l_{H_{k}} (γ)−1}^{p^{k}−r}

since l_{H_{k}}(ρ^{j}γ) = l_{H_{k}}(γ) if 0 ≤ j < r and l_{H_{k}}(ρ^{j}γ) = l_{H_{k}}(γ) − 1 if r ≤ j ≤ p^{k} − 1. □

Finally, we need a general structure proposition about 𝔽_{p}G-modules that shows that the structure of an 𝔽_{p}G-module X can be determined from the natural filtration on X^{G} obtained by taking the intersection of X^{G} with the images of X under the successive powers of the augmentation ideal of 𝔽_{p}G. The proposition below is proved for cyclic p-groups as [LMS, Proposition 2]. The generalization to cyclic groups of order p^{n} is automatic.

Proposition 1. Let X be an 𝔽_{p}G-module. Set L_{p^{n}} = ρ^{p^{n}−1}X and for 1 ≤ i < p^{n}, suppose that L_{i} is an 𝔽_{p}-complement of ρ^{i}X∩X^{G} in ρ^{i−1}X∩X^{G}.

Then there exist 𝔽_{p}G-modules X_{i}, i = 1, 2, …, p^{n}, such that:

(1) X = ⨁_{i=1}^{p^{n}} X_{i}.

(2) X_{i}^{G} = L_{i} for i = 1, 2, …, p^{n}.

(3) Each X_{i} is a direct sum of dim_{𝔽_{p}}(L_{i}) cyclic 𝔽_{p}G-modules of length i.

(4) For each i = 1, 2, …, p^{n}, there exists an 𝔽_{p}-submodule Y_{i} of X_{i} with dim_{𝔽_{p}}(Y_{i}) = dim_{𝔽_{p}}(L_{i}) such that (𝔽_{p}G)Y_{i} = X_{i}.

Remark 2. The uniqueness of the form of the decomposition of X into a direct sum of cyclic summands is quite easy to see. Indeed, if X = ⊕_{i=1}^{p^{n}}X_{i}, where X_{i} is a direct sum of cyclic modules of length i, indexed by a set N_{i}, then we have

card N_{i} = dim_{𝔽_{p}}(ρ^{i−1}X ∩ X^{G})/(ρ^{i}X ∩ X^{G}).

The uniqueness of the decomposition of X into a direct sum is also a special case of the Krull–Schmidt–Azumaya Theorem (see [AF, Theorem 12.6]).

2. Milnor k-groups

If α ∈ K_{m}E we write α 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 ⟨γ⟩ of k_{m}E generated by γ. Recall that i_{E}N_{E/F} acts on k_{m}E as 1 + σ + ⋯ + σ^{p^{n}−1} (see [FV, Chapter 1, (3.8) Theorem 1]). Because ρ^{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.

We will prove the following key lemma by induction on n.

Lemma 4. Let 1 ≤ j ≤ p^{n} and let 0 ≤ i ≤ n be minimal such that j ≤ p^{i}. Then

ρ^{j−1}k_{m}E ∩ (k_{m}E)^{G} = i_{E}N_{E_{i}/F} k_{m}E_{i}.

Remark 3. (1) First note that Lemma 1′ proves this result for j = 1,

and note also that for j = p^{n}, the result holds by definition.

(2) That the right-hand side of the equality is contained in the left-hand side follows from the fact that i_{E_{i}}N_{E_{i}/F} is identical to ρ^{p^{i}−1} on k_{m}E_{i} and also on the image under the injection i_{E_{i},E} into k_{m}E. To show inclusion in the opposite direction, it is enough to show that for each cyclic extension E/F of pth-power degree p^{n}, ρ^{j−1}k_{m}E ∩ (k_{m}E)^{G} ⊆ i_{E}N_{E/F} k_{m}E for all j with p^{n−1} < j < p^{n}, as follows.

Assume that the statement is true. Let j satisfy 1 ≤ j ≤ p^{n} and suppose that i is the smallest nonnegative integer such that j ≤ p^{i}. Applying the statement for E_{i}/F case, together with the observation above that the j = p^{i} case for E_{i}/F holds by definition, and using the injectivity of Lemma 1′, we conclude that

ρ^{j−1}(i_{E_{i},E}(k_{m}E_{i})) ∩ i_{E_{i},E}(k_{m}E_{i})^{G} ⊆ i_{E_{i},E}(i_{E_{i}}(N_{E_{i}/F} k_{m}E_{i}))
= i_{E}(N_{E_{i}/F} (k_{m}E_{i})).

Therefore it is enough to observe that

ρ^{j−1}k_{m}E ∩ (k_{m}E)^{G} = ρ^{j−1}(i_{E_{i},E}(k_{m}E_{i})) ∩ i_{E_{i},E}(k_{m}E_{i})^{G},

and to do this it is enough to prove that

ρ^{j−1}k_{m}E ∩ (k_{m}E)^{G} ⊆ ρ^{j−1}(i_{E_{i},E}(k_{m}E_{i})) ∩ i_{E_{i},E}(k_{m}E_{i})^{G},

as the inclusion in the opposite direction is obvious. Let α = ρ^{j−1}γ ∈ (k_{m}E)^{G} where γ ∈ k_{m}E. Then ρ^{p^{i}}γ = 0 and so σ^{p^{i}}γ = γ. Therefore by Lemma 1′ we see that γ ∈ i_{E_{i},E}k_{m}E_{i}, as was required.

The foregoing thus establishes a reduction step for use in the proof of Lemma 4.

Before proving this lemma, we show:

Reduction to Lemma 4. Theorem 2 holds for cyclic extensions of degree p^{n} for which Lemma 4 holds.

Proof. We apply Proposition 1 to k_{m}E. Set L_{p^{n}} = i_{E}N_{E/F} k_{m}E, and for 0 ≤ i < n, define L_{p^{i}} as an 𝔽_{p}-complement of i_{E}N_{E_{i+1}/F} k_{m}E_{i+1} in i_{E}N_{E_{i}/F} k_{m}E_{i}. We also set L_{k} = {0} for each k ∈ {1, …, p^{n}} which is not a power of p. Note that (k_{m}E)^{G} = i_{E}k_{m}F by Lemma 1′. Then, Lemma 4 shows that the 𝔽_{p}-modules L_{j}, j = 1, …, p^{n}, satisfy the conditions of the proposition. By the proposition, we have a decomposition of k_{m}E into a direct sum of 𝔽_{p}G-modules

k_{m}E = ⊕_{i=0}^{n}Y_{i},

where Y_{i}^{G} = L_{p^{i}} and Y_{i} is a direct sum of

dim_{𝔽_{p}} L_{p^{i}} = dim_{𝔽_{p}} i_{E}N_{E_{i}/F} k_{m}E_{i}/i_{E}N_{E_{i+1}/F} k_{m}E_{i+1}
= dim_{𝔽_{p}} N_{E_{i}/F} k_{m}E_{i}/N_{E_{i+1}/F} k_{m}E_{i+1}

cyclic 𝔽_{p}G-modules of length p^{i}, using Lemma 1 for the last equality. Equivalently, Y_{i} ≃ ⊕_{T_{i}}𝔽_{p}(G/H_{i}), where the cardinal number of T_{i} is dim_{𝔽_{p}} L_{p^{i}}, and therefore Y_{i} is a free 𝔽_{p}G_{i}-module of the same rank. □

We now prove several lemmas as partial results toward the proof of Lemma 4. Note it is only in these lemmas, and in the alternative proof in Section 4, that Theorem 1 is used in this paper.

Lemma 5. Lemma 4 holds for all cyclic extensions E/F of degree p.

Proof. Let G = Gal(E/F). By Remark 3, it suffices to show the left-to-right inclusion. Let γ ∈ K_{m}E such that 0 ≠ ρ^{l−1}γ ∈ ρ^{l−1}k_{m}E ∩ (k_{m}E)^{G} where 1 < l ≤ p. Then l(γ) = l. We show by induction on i, l ≤ i ≤ p, that there exists α_{i} ∈ K_{m}E such that

⟨ρ^{i−1}α_{i}⟩ = ⟨γ⟩^{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}c = Nα_{i} = ρ^{p−1}α_{i} and i < p, we conclude that i_{E}c = 0. By the injectivity of i_{E} from Lemma 1, c = 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 1, there exists ω ∈ K_{m}E such that ρω = α_{i} − i_{E}(f). Hence ρ^{2}ω = ρα_{i}. Since i ≥ 2,

⟨ρ^{i}ω⟩ = ⟨ρ^{i−1}α_{i}⟩ = ⟨γ⟩^{G}

and we can set α_{i+1} = ω. □

Lemma 6. Lemma 4 holds for all cyclic extensions E/F of degree 4.

Proof. If j = 1 then our statement is a special case of Lemma 1′.

Assume now that j = 2. As we observed in Remark 3, i_{E}N_{E_{1}/F} k_{m}E_{1} ⊆ ρk_{m}E ∩ (k_{m}E)^{G}. In order to show the inclusion in the opposite direction, let γ ∈ k_{m}E such that ργ ∈ (k_{m}E)^{G}. Then ρ^{2}γ = 0 and σ^{2}γ = γ. Thus Lemma 1 implies γ = i_{E_{1},E}δ, where δ ∈ k_{m}E_{1}. Hence

ργ = i_{E_{1},E}ρδ
= i_{E_{1},E} i_{F,E_{1}}N_{E_{1}/F} δ
= i_{E}N_{E_{1}/F} δ.

Hence we have the desired inclusion ρk_{m}E ∩ (k_{m}E)^{G} ⊆ i_{E}N_{E_{1}/F} k_{m}E_{1}.

Now suppose that j = 3, and let γ ∈ K_{m}E such that 0 ≠ ρ^{2}γ ∈ ρ^{2}k_{m}E ∩ (k_{m}E)^{G}. Recall that for γ ∈ K_{m}E, we let l(γ) denote the 𝔽_{p}-dimension of the 𝔽_{p}G-submodule ⟨γ⟩ of k_{m}E generated by γ. Then l(γ) = 3, and set β := ργ. We have

(σ^{2} − 1)β = ρ^{3}γ = 0.

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

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

i_{F,E}c = i_{E_{1},E}(σ + 1)b = (σ^{2} − 1)γ = i_{E_{1},E}N_{E/E_{1}}γ,

and since l(γ) = 3, ⟨γ⟩^{G} = ⟨i_{F,E}c⟩. By Lemma 1 we see that i_{F,E_{1}}c = N_{E/E_{1}}γ and therefore N_{E/E_{1}}γ = i_{F,E_{1}}c + 2g for some g ∈ K_{m}E_{1}.

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

N_{E/F} i_{E_{1},E}δ = 2c + 2N_{E_{1}/F} g,

so that N_{E/F} (γ − i_{E_{1},E}δ) = 0. By Theorem 1 there exists an α ∈ K_{m}E with ρα = γ − i_{E_{1},E}δ. Then observing that ρ^{2}(\overline{i_{E_{1},E}δ}) = 0 as δ ∈ K_{m}E_{1} we see that

i_{F,E}N_{E/F} α = ρ^{3}α = ρ^{2}\overline{(γ − i_{E_{1},E}δ)} = ρ^{2}γ,

as required. □

3. Proof of Lemma 4

We prove Lemma 4 and hence, using the Reduction to Lemma 4, Theorem 2. We do so by induction on n, using Lemmas 5 and 6 as base cases and assuming the result for cyclic extensions of degree p^{n−1}. Observe that Lemmas 5 and 6 and the Reduction to Lemma 4 give us that Theorem 2 holds for degree p and, if p = 2, then for degree 4 as well. By Remark 3, it suffices to prove that

ρ^{j−1}k_{m}E ∩ (k_{m}E)^{G} ⊆ i_{E}N_{E/F} k_{m}E

for all p^{n−1} < j < p^{n}.

Assume that k_{m}E = ⊕_{γ∈Γ}𝔽_{p}Gγ is a decomposition of k_{m}E into cyclic 𝔽_{p}G-modules given by Proposition 1. As we observed in Remark 2, this decomposition is unique in the sense that the multiplicity of the cyclic summands of given dimension of k_{m}E are determined by k_{m}E.

First assume that p > 2. We know by Lemma 5 and by Reduction to Lemma 4 that k_{m}E|_{H_{n−1}} is a direct sum of trivial and free 𝔽_{p}H_{n−1}-modules. Let γ ∈ k_{m}E satisfy p^{n−1} < l(γ) < p^{n}. We want to show that ρ^{l(γ)−1} γ ∈ ρ^{p^{n}−1}k_{m}E = i_{E}k_{m}E. Without loss of generality we may assume that γ generates a cyclic 𝔽_{p}G-summand of k_{m}E. If l(γ) < p^{n}, then by Lemma 3, 𝔽_{p}Gγ|_{H_{n−1}} contributes direct summands which are not free or trivial, a contradiction. Hence a cyclic generator γ with length greater than p^{n−1} must have length p^{n}.

Now assume p = 2. Suppose that γ ∈ Γ satisfies 2^{n−1} < l(γ) < 2^{n}. By Lemma 6 and the Reduction to Lemma 4, k_{m}E|_{H_{n−2}} is a direct sum of cyclic 𝔽_{2}H_{n−2}-modules of lengths 1, 2, and 4, and 𝔽_{2}Gγ is a cyclic summand of k_{m}E. By Lemma 3, we know that 𝔽_{2}Gγ|_{H_{n−2}} contributes cyclic 𝔽_{2}H_{n−2}- summands of length 3, a contradiction. Hence a cyclic generator γ ∈ Γ of length greater than 2^{n−1} must have length 2^{n}.

In both cases, we have shown for p^{n−1} < j < p^{n} that

ρ^{j−1}k_{m}E ∩ (k_{m}E)^{G} = i_{E}N_{E/F} k_{m}E,

as desired. □

4. Alternative proof

The following simple alternate proof of Theorem 2 for cyclic extensions of prime degree was found by the referee, whom we thank for allowing us to reproduce it here. Observe that using this argument we obtain a simpler proof of Theorem 2 for cyclic extensions of degree p^{n} for p > 2. The case p = 2 remains the hardest, as one sees by examining the proof of Lemma 6.

Let E/F be a cyclic extension of degree p > 2, and write k_{m}E = ⊕_{i=1}^{p}X_{i}, where each X_{i} is a direct sum of cyclic 𝔽_{p}G-modules of length i. We show that X_{i} = {0} for all i ∈ {2, …, p − 1}. Suppose that it is not true: there exists a cyclic summand ⟨γ⟩, γ ∈ K_{m}E, of length i, 2 ≤ i ≤ p − 1. Then N(γ) = 0 and there exists f ∈ K_{m}F such that N_{E/F} γ = pf. From Theorem 1 it follows that γ = ρw+f for some w ∈ K_{m}E. Hence γ = ρw+f. Passing to the projection of k_{m}E onto ⟨γ⟩ we see that ⟨γ⟩ embeds into a cyclic module of length i + 1 — a contradiction.

Acknowledgments. We are very grateful to A. Schultz as some of his ideas, developed in [MSS1], proved quite useful to us during the investigations leading to this paper. We also wish to thank M. Rost and J.-P. Serre for their comments after a lecture delivered in Nottingham by the third author in September 2005. These comments led to the improvement of our exposition. Finally, we would like to thank the referee for such a careful reading of our article, resulting in valuable suggestions for improving our text and simplifications of some proofs.

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Department of Mathematics, Physics and Engineering, Mount Royal College, 4825 Mount Royal Gate SW, Calgary, Alberta T3E 6K6, CANADA gbhandari@mtroyal.ca

Department of Mathematics, Middlesex College, University of Western Ontario, London, Ontario N6A 5B7, CANADA nlemire@uwo.ca minac@uwo.ca

Department of Mathematics, Davidson College, Box 7046, Davidson, North Carolina 28035-7046, USA joswallow@davidson.edu

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