GALOIS MODULE STRUCTURE OF pTH-POWER CLASSES OF EXTENSIONS OF DEGREE p

JÁN MINÁČ ^{∗†} AND JOHN SWALLOW

Abstract. For fields F of characteristic not p containing a primitive pth root of unity, we determine the Galois module structure of the group of pth-power classes of K for all cyclic extensions K/F of degree p.

The foundation of the study of the maximal p-extensions of fields K containing a primitive pth root of unity is a group of the pth-power classes of the field: by Kummer theory this group describes elementary p-abelian quotients of a maximal p-extension. The size of the group controls the number of generators of the Galois group of the maximal p-extension. Furthermore, as K varies among the Galois extensions of the base field F, the Galois module structure of pth-power classes plays a fundamental role in the investigation of the Galois group of the maximal p-extension of F. In particular, in the nineteen-sixties Borevič and Faddeev closely studied the Galois module structure of the pth-power classes of K for cyclic extensions K/F of degree p in the case when F is a local field, and in the seventies Miki used this structure in the service of Galois embedding problems (see [Bo], [F], and [Mi]).

From a cohomological point of view, the Galois module structure of pth-power classes is especially important. These Galois modules may be naturally identified with the first cohomology groups of maximal p-extensions with 𝔽_{p}-coefficients. (See [S, Chapter 1].) It is moreover conjectured that the entire Galois cohomology ring of K with coefficients in 𝔽_{p} of degree at least one is generated by its first cohomology group. (See [V] for a proof in the case of p = 2 as well as comments on this conjecture which was initially considered by Beilinson, Bloch-Kato, Lichtenbaum, and Milnor.) As a result, for the investigation of the action of the Galois group Gal(K/F) on the Galois cohomology of K with coefficients in 𝔽_{p}, it should be sufficient to understand the Galois module structure of the pth-power classes of K. Furthermore this is necessary for the study of a Galois cohomology ring of F with coefficients in 𝔽_{p} using the Lyndon-Hochschild-Serre spectral sequence attached to a group extension of an absolute Galois group of K by Gal(K/F).

∗Research supported in part by the Natural Sciences and Engineering Research Council of Canada and by the special Dean of Science Fund at the University of Western Ontario.

†Supported by the Mathematical Sciences Research Institute, Berkeley.

Hence the question of determining the Galois module structure of the pth-power classes of K is a fundamental question related to both the structure of the Galois cohomology of F and the structure of the maximal p-extensions of F.

It is surprising that one can determine this Galois module structure for all cyclic field extensions of degree p, when the base field contains a primitive pth root of unity, and that the proofs are elementary. Just as in the case of local fields, the structure depends upon only the norm subgroup, its intersection with the subgroup of pth roots of unity, and the multiplicative subgroup of the base field. When we first approached these results, the main tools employed were Waterhouse’s elegant treatment of Galois closures of certain Kummer extensions, as well as a criterion for the realization of a nonabelian Galois group of order p^{3} and exponent p as a Galois group over our base field. We were able, however, to simplify our proofs considerably. In particular we require in this paper only Hilbert’s Theorem 90, basic Kummer theory, and linear algebra over 𝔽_{p}[Gal(K/F)]. Although Galois embedding problems do not appear in our exposition, our results are closely related to those in [Ma], [Ma-Ng], and [Wat], and this relationship remains worthy of precise formulation. This will be a topic in subsequent work.

It is clear that one can generalize the main theorems to a larger family of field extensions K/F than we are considering here. Also, these results may be applied to calculations in Galois cohomology, and the study of Galois groups of maximal p-extensions of fields. We concentrate here, however, on the short exposition of a solution of a main significant problem.

1. Main Theorems

Let F be a field of characteristic not p containing a primitive pth root of unity ξ_{p}. Let K be a cyclic extension K = F (\sqrt[p]{a}), K^{×} = K \ {0} and J = K^{×}/K^{×}^{p}.

GALOIS MODULE STRUCTURE OF pTH-POWER CLASSES 3

Denote Gal(K/F) = ⟨σ⟩ by G, where σ ∈ Gal(K/F) such that \sqrt[p]{a}^{σ−1} = ξ_{p}. For G-modules V, denote by V^{G} the fixed submodule of V.

Maps related to the norm will play an important role, and as a result we use N in several different ways. As an element of 𝔽_{p}[G] we let N := 1 + σ + ⋯ + σ^{p−1}. We also view N as the resulting Ghomomorphism V ⟶ V. Then we write N(J) for the image of J, and N(K^{×}) for the image of K^{×}.

The following two theorems classify all modules J using only information about the extension K/F encoded in the multiplicative subgroup of F.

Theorem 1. Suppose p > 2. As an 𝔽_{p}[G]-module, J decomposes as

J = X ⊕ Y ⊕ Z

where

(1) X is an indecomposable 𝔽_{p}[G]-submodule, of dimension 1 if ξ_{p} ∈ N(K^{×}) and of dimension 2 if ξ_{p} ∉ N(K^{×});

(2) Y is a free 𝔽_{p}[G]-submodule with Y^{G} = N(J); and

(3) Z is a trivial 𝔽_{p}[G]-module.

Moreover, the rank of any maximal free 𝔽_{p}[G]-submodule of J is equal to the rank of N(J) as an 𝔽_{p}-module.

Theorem 2. Suppose p = 2. As an 𝔽_{2}[G]-module, J decomposes as

J = X ⊕ Y ⊕ Z

where

(1) X is of dimension 1 if −1 ∈ N(K^{×}) and is {0} otherwise;

(2) Y is a free 𝔽_{2}[G]-submodule with Y^{G} = N(J); and

(3) Z is a trivial 𝔽_{2}[G]-module.

Moreover, the rank of any maximal free 𝔽_{2}[G]-submodule of J is equal to the rank of N(J) as an 𝔽_{2}-module.

The results in Theorem 1 and Theorem 2 imply that each 𝔽_{p}[G]- module J can be written as a sum of cyclic 𝔽_{p}[G]-modules. Because each cyclic 𝔽_{p}[G]-module is an indecomposable module, and each ring of the 𝔽_{p}[G]-endomorphisms of such a module is a local ring, we can conclude that our decompositions are unique. This follows from the Krull-Schmidt-Remak-Azumaya theorem. (See [Fa, page 24].)

In the next theorem, which is a supplement to both Theorems 1 and 2, we determine the multiplicity of each cyclic 𝔽_{p}[G]-module in a decomposition of an 𝔽_{p}[G]-module J using the arithmetical invariants associated with a field extension K/F. We provide a direct proof of this theorem, and in particular we do not use the Krull-Schmidt-Remak-Azumaya theorem quoted above. We use the same notation as above, and we also set Υ(K) = 1 or 0 depending upon whether ξ_{p} ∈ N(K^{×}) or ξ_{p} ∉ N(K^{×}). For each set A we denote by |A| the cardinal number of A. Observe that each nonzero cyclic 𝔽_{p}[G]-module M is characterized by its dim_{𝔽_{p}} M ∈ {1, 2, …, p}.

Theorem 3. Let J ≅ ⨁_{i=1}^{p} (⨁_{j∈𝔎_{i}} M_{i,j}) be a decomposition of J into 𝔽_{p}[G]-cyclic modules M_{i,j}, where dim_{𝔽_{p}} M_{i,j} = i and 𝔎_{i} are index sets for each i ∈ {1, 2, …, p}.

Then we have

(1) |𝔎_{1}| + 1 = 2Υ(K) + dim_{𝔽_{p}}(F^{×}/N(K^{×}));

(2) If p > 2 then |𝔎_{2}| = 1 − Υ(K);

( i ) If 3 ≤ i ≤ (p − 1) then 𝔎_{i} = ∅; and

(p) |𝔎_{p}| = dim_{𝔽_{p}} N(J).

The proofs of Theorems 1, 2 and 3 are presented in Section 3.

As a consequence of these theorems and their proofs, we have the following results on extremal cases. A short proof of each is given in the last section of this paper.

Recall that if p > 2 then a ∈ F \ F^{p} is p-rigid if, whenever the cyclic algebra (\frac{a,b}{F,ξ_{p}}) is of trivial class in Br(F), then [b] ∈ ⟨[a]⟩ in F^{×}/F^{×}^{p}. For a discussion of p-rigidity and G-invariant modules J, see [War]. For the basic properties of cyclic algebras and Brauer groups, see [P, Chapters 14 and 15].

Corollary 1.

(1) J is a free 𝔽_{p}[G]-module precisely when p = 2, −1 ∉ N(K^{×}), and F^{×} = N(K^{×}) ∪ −N(K^{×}).

(2) If p > 2 then J contains no free direct summand precisely when a is p-rigid. If p = 2 then J contains no free direct summand precisely when N(K^{×})/F^{×2} ⊂ ⟨[a]⟩ ⊂ F^{×}/F^{×2}. (In this case it follows that \sqrt{−1} ∈ K^{×}.)

(3) If p > 2 then J is G-invariant precisely when a is p-rigid and ξ_{p^{2}} ∈ K^{×}. If p = 2 then J is G-invariant precisely when J contains no free summand.

It is worth pointing out that in the corollary above, we can replace “free direct summand of J” by “free submodule of J”, as it is well known that each free submodule of an 𝔽_{p}[G]-module is in fact its summand. (See [C, Theorem 11.2].)

Observe that if J is an 𝔽_{p}[G]-module as above, and φ: H ⟶ G is any isomorphism, we can consider J to be an 𝔽_{p}[H]-module and the isomorphism type of this module is independent of the choice of φ. In particular, if we have two 𝔽_{p}[G_{i}]-modules J_{i}, i = 1, 2 with G ≅ G_{i}, i = 1, 2, we may consider both modules J_{i} as modules over the same group ring 𝔽_{p}[G]. We shall apply this remark to the following situation.

Suppose p > 2 and K_{1} = F_{1}(\sqrt[p]{a}) and K_{2} = F_{2}(\sqrt[p]{a}) are two cyclic extensions of degree p. We also set T_{i} = K_{i}^{×}/K_{i}^{×p}, G_{i} = Gal(K_{i}/F_{i}) for i = 1, 2, and using any isomorphisms φ_{i}: G ⟶ G_{i} as above we consider both modules T_{1} and T_{2} as modules over 𝔽_{p}[G]. (We write here T_{i} instead of J_{i} so that we avoid possible confusion with our notation for the socle series employed later in this paper.)

Using this notation we shall formalize our classification of 𝔽_{p}[G]- modules J by means of arithmetic invariants associated with the multiplicative group F^{×} as follows.

Corollary 2. The 𝔽_{p}[G]-modules T_{1} and T_{2} are isomorphic if and only if the following three conditions are valid:

(1) ξ_{p} ∈ N(K_{1}^{×}) ⇔ ξ_{p} ∈ N(K_{2}^{×}).

(2) dim_{𝔽_{p}} N(K_{1}^{×})/F_{1}^{×p} = dim_{𝔽_{p}} N(K_{2}^{×})/F_{2}^{×p}.

(3) dim_{𝔽_{p}} F_{1}^{×}/N(K_{1}^{×}) = dim_{𝔽_{p}} F_{2}^{×}/N(K_{2}^{×}).

For p = 2 we obtain a similar Corollary 3. We use the same notation as in the case of p > 2 above. Moreover we set Υ(K_{i}) = 1 or 0 depending upon whether −1 ∈ N(K_{i}^{×}) or −1 ∉ N(K_{i}^{×}), for both i = 1, 2.

Corollary 3. The 𝔽_{2}[G]-modules T_{1} and T_{2} are isomorphic if and only if the following two conditions are valid:

(1) 2Υ(K_{1}) + dim_{𝔽_{2}} F_{1}^{×}/N(K_{1}^{×}) = 2Υ(K_{2}) + dim_{𝔽_{2}} F_{2}^{×}/N(K_{2}^{×}).

(2) Υ(K_{2}) + dim_{𝔽_{2}} N(K_{1}^{×})/F_{1}^{×2} = Υ(K_{1}) + dim_{𝔽_{2}} N(K_{2}^{×})/F_{2}^{×2}.

2. Cyclic Modules and Fixed Elements of J

This section contains a few basic facts used in our proofs of Theorems 1 and 2. For any subset {x, y, … } of V we denote by ⟨x, y, … ⟩ the 𝔽_{p}-submodule of V spanned by {x, y, … }.

We freely use basic Kummer theory (see [A-T, Chapter 6, Section 2]), and, depending upon the context, we use additive and multiplicative notation for our 𝔽_{p}[G]-modules occurring as submodules of K^{×}/K^{×}^{p}.

Let A = ⊕_{j=0}^{p−1}𝔽_{p}τ^{j} be a free 𝔽_{p}[G]-module on one generator 1, where σ acts by multiplication by τ. For i = 1, …, p − 1, let A_{i} denote the cyclic 𝔽_{p}[G]-submodule

A_{i} = ⟨(τ − 1)^{i}, (τ − 1)^{i+1}, …, (τ − 1)^{p−1}⟩, and set also A_{p} = {0}.

The dimension of A/A_{i} is i, and the 𝔽_{p}[G]-modules A/A_{i} exhaust the isomorphism classes of indecomposable 𝔽_{p}[G]-modules. When describing an indecomposable 𝔽_{p}[G]-module, we will use length synonymously with dimension. The only proper 𝔽_{p}[G]-submodules of A/A_{i} are cyclic and they are the images of A/A_{i} under (σ − 1)^{j} for j = 1, …, i − 1.

The socle series of J will be important for the determination of indecomposable submodules of J. The socle series of J is defined by J_{1} = J^{G} and (J_{i}/J_{i−1}) = (J/J_{i−1})^{G} for i > 1. We have that J_{i} = ker(σ − 1)^{i}, when (σ − 1)^{i} is considered as an endomorphism of J.

Indecomposable submodules of J may therefore be obtained as follows. We write the elements of J as [γ], γ ∈ K^{×}. For [γ] ∈ J_{i} \ J_{i−1} let γ_{j} = γ^{(σ−1)^{j}} for j = 0, 1, …, i − 1. Then the 𝔽_{p}-submodule M_{γ} := ⟨[γ], [γ_{1}], …, [γ_{i−1}]⟩ of J is a cyclic 𝔽_{p}[G]-submodule of dimension i isomorphic to A/A_{i}.

In the proof of Theorem 1 in the next section, we shall construct submodules X, Y and Z of J such that X is a cyclic submodule of J of length 1 or 2, Z is a trivial submodule of J, and Y is a free 𝔽_{p}[G]- submodule of J. An important part of the proof will be to show that X, Y and Z generate J as an 𝔽_{p}[G]-module. In order to do this we need to construct a module Y which is sufficiently large. The following lemma will be used to show that we have enough free cyclic submodules of J to construct our sufficiently large free 𝔽_{p}[G]-submodule Y of J.

Lemma 1. Suppose p > 2 and let l denote the length of M_{γ}.

a) If 3 ≤ l ≤ p, then there exists an element [α] ∈ J such that ⟨N([α])⟩ = M_{γ}^{G}.

b) If l = 2 and γ cannot be written as γ = a^{r/p}γ_{1} for some r ∈ ℤ and [γ_{1}] ∈ J_{1}, then there exists an integer t ∈ ℤ and [α] ∈ J such that ⟨N([α])⟩ = (M_{a^{t/p}γ})^{G}.

Proof. Let 3 ≤ l ≤ i ≤ p. We show by induction on i that there exists an element α_{i} ∈ K^{×} such that ⟨[α_{i}]^{(σ−1)^{i−1}}⟩ = M_{γ}^{G}. Then we may set α := α_{p} and the proof of the first part of our lemma will be complete. If i = l we set α_{l} = γ. Assume now that l ≤ i < p and that our statement is true for i.

Set c = N(α_{i}). Since [α_{i}]^{(σ−1)^{p−1}} = [c] and i < p we see that [c] = [1] ∈ J. Because c ∈ F^{×} ∩ K^{×p} from Kummer theory, we conclude that c = a^{s}f^{p} for some f ∈ F^{×} and s ∈ ℤ. Then N(α_{i}/fa^{s/p}) = 1. By Hilbert’s Theorem 90 there exists an element ω ∈ K^{×} such that ω^{σ−1} = α_{i}/fa^{s/p}. Then ω^{(σ−1)^{2}} = α_{i}^{(σ−1)}/ξ_{p}^{s}. Therefore ⟨ω^{(σ−1)^{i}}⟩ = M_{γ}^{G} and we can set α_{i+1} := ω.

Now we shall prove the second part of our lemma. Assume l = 2 = i. Proceeding in the same way as above, we see that for α_{2} = γ we have N(α_{2}) = c = a^{s}f^{p} for some f ∈ F^{×} and s ∈ ℤ.

As before we see that there exists an element ω ∈ K^{×} such that ω^{σ−1} = α_{2}/fa^{s/p}. Then ω^{(σ−1)^{2}} = (α_{2}a^{−s/p})^{(σ−1)} = (γa^{−s/p})^{(σ−1)}. Observe that γa^{−s/p} ∉ J_{1} by our hypothesis, and therefore the length of M_{γa^{−s/p}} is again 2. Hence we can set α_{3} := ω. We can then continue by induction on i as above, concluding that there exists an element α := α_{p} such that ⟨N([α])⟩ = (M_{γa^{−s/p}})^{G} as required. □

Remark 1. In the case of ξ_{p} ∈ N(K^{×}) we do not need the adjusting factor a^{t/p} in part b) of Lemma 1. Also in this case no restriction on γ is necessary and γ can be any element of K^{×} such that the length of M_{γ} is 2.

Indeed let β ∈ K^{×} such that N(β) = ξ_{p}. As before we have ω^{(σ−1)} = α_{2}/fa^{s/p} for some ω ∈ K^{×} and s ∈ ℤ. Then set θ = (β^{(σ−1)^{p−3}})^{s} ω. Then θ^{(σ−1)^{2}} = α_{2}^{(σ−1)}. Hence we may set α_{3} := θ. We may then continue by induction on i to conclude that there exists an element α := α_{p} such that ⟨N([α])⟩ = M_{γ}^{G}.

In the next lemma we determine J_{1} in terms of arithmetic invariants of K/F encoded in the multiplicative group of F. We write ϵ(F^{×}) for the subgroup F^{×}K^{×}^{p}/K^{×}^{p} of J.

Lemma 2.

a) If ξ_{p} ∉ N(K^{×}) then J_{1} = ϵ(F^{×}).

b) If ξ_{p} ∈ N(K^{×}) then J_{1} ≅ ϵ(F^{×}) ⊕ ⟨δ⟩ where δ ∈ K^{×}, σ(δ)/δ =

λ^{p} and N(λ) = ξ_{p}.

Proof. Suppose that θ ∈ K^{×} such that [θ] ∈ J_{1}. Then σ(θ)/θ = λ^{p} for some λ ∈ K^{×}, and hence N(λ)^{p} = 1. Hence we see that N(λ) = ξ_{p}^{c} for some c ∈ ℤ. Now consider the case ξ_{p} ∉ N(K^{×}). Then N(λ) = 1, because otherwise ξ_{p} would be a norm of a suitable power of λ. Therefore from Hilbert’s Theorem 90 we see that σ(θ)/θ = σ(k^{p})/k^{p} for some k ∈ K^{×}. We conclude that θ/k^{p} ∈ F^{×} and hence [θ] = [f] for some f ∈ F^{×}. Therefore if ξ_{p} ∉ N(K^{×}) then J_{1} = ϵ(F^{×}) as required.

Now assume that ξ_{p} ∈ N(K^{×}). Then ξ_{p} = N(λ) for some λ ∈ K^{×} and by Hilbert’s Theorem 90 there exists an element δ ∈ K^{×} such that σ(δ)/δ = λ^{p}. Then the 𝔽_{p}[G]-submodule of J generated by δ and ϵ(F^{×}) is isomorphic with ϵ(F^{×}) ⊕ ⟨δ⟩. Now for each [θ] ∈ J_{1}, σ(θ)/θ = ν^{p}, N(ν) = ξ_{p}^{c}, c ∈ ℤ, and therefore we have [θ] ∈ ϵ(F^{×})[δ]^{c}. Hence J_{1} ≅ ϵ(F^{×}) ⊕ ⟨δ⟩ as required. □

Remark 2. It is worthwhile to observe that we have the following short exact sequence:

0 ⟶ ⟨[a]⟩ ⟶^{i} F^{×}/F^{×p} ⟶^{ϵ} J_{1} ⟶^{N} ⟨[a]⟩,

where ⟨[a]⟩ is the subgroup of F^{×}/F^{×p} generated by [a] ∈ F^{×}/F^{×p}, i is the inclusion map, ϵ is the natural homomorphism induced by the inclusion map F^{×} ⟶ K^{×} and N is the map induced by the norm map N: K^{×} ⟶ F^{×}. Moreover the map N: J_{1} ⟶ ⟨[a]⟩ is surjective if and only if ξ_{p} ∈ N(K^{×}).

The fact that [N(θ)] ∈ ⟨[a]⟩ for each [θ] ∈ J_{1} follows from the fact that N(θ) ∈ F^{×} ∩ K^{×}^{p}. The exactness at F^{×}/F^{×p} follows from Kummer theory. The exactness at J_{1} can be seen as follows.

Suppose that [θ] ∈ J_{1}. Then \sqrt[p]{N(θ)} ∈ K^{×} and [N(θ)] = [1] ∈ F^{×}/F^{×}^{p} if and only if σ(\sqrt[p]{N(θ)})/\sqrt[p]{N(θ)} = 1. Set σ(θ)/θ = λ^{p}, λ ∈ K^{×}. Then the condition σ(\sqrt[p]{N(θ)})/\sqrt[p]{N(θ)} = 1 translates as N(λ) = 1. Using Hilbert’s Theorem 90 as in the proof of Lemma 2, part a) above, we see that N(λ) = 1 if and only if [θ] ∈ ϵ(F^{×}).

Finally if ξ_{p} ∉ N(K^{×}) we know that ϵ: F^{×}/F^{×p} ⟶ J_{1} is surjective and hence N: J_{1} ⟶ ⟨[a]⟩ is trivial. If ξ_{p} ∈ N(K^{×}), then ϵ: F^{×}/F^{×p} ⟶ J_{1} is not surjective and therefore N must be surjective.

3. Proofs

Proof of Theorem 1. In the first part of this proof we shall construct some submodules X, Y and Z of J, and we shall show that they generate a submodule of J isomorphic to X ⊕ Y ⊕ Z. In the second part of this proof we shall show that X, Y and Z generate the full module J.

We first construct X. If ξ_{p} ∈ N(K^{×}), then by Remark 2 at the end of Section 2 we see that there exists an element δ ∈ K^{×} such that N(δ) = a and [δ] ∈ J_{1}. Then we set X = ⟨[δ]⟩.

If ξ_{p} ∉ N(K^{×}), then let δ = \sqrt[p]{a} and X = ⟨[δ], [ξ_{p}]⟩ ⊂ J_{2}. If [ξ_{p}] = [1] then a root of unity ξ_{p^{2}} of order p^{2} lies in K^{×} and we can pick ξ_{p^{2}} such that ξ_{p} = N(ξ_{p^{2}}), contrary to hypothesis. Because [\sqrt[p]{a}]^{σ−1} = [ξ_{p}], X is isomorphic to A/A_{2}.

We proceed to construct Y. Let I be an 𝔽_{p}-basis for N(J). For each [x] ∈ I we construct a free 𝔽_{p}[G]-module M(x), as follows. Choose a representative x ∈ F^{×} for [x], such that x ∈ N(K^{×}). Choose γ ∈ K^{×} such that x = N(γ). Finally set M(x) = M_{γ}.

We claim that the M(x), [x] ∈ I, are independent. First we show by induction on the number of modules that a finite set of modules M(x) is independent. The base case is trivial. Now suppose that W = M(x′) ∩ ∑_{[x′]≠[x]} M(x) ≠ {0}. Then W contains the 1-dimensional submodule V = M(x′) ∩ J_{1}. If M(x′) = M_{ω}, then V = ⟨[N(ω)]⟩ = ⟨[x′]⟩. Since [x′] ∈ W, [x′] is a finite sum ∑_{[x′]≠[x]}[m(x)] where [m(x)] ∈ M(x). By induction the modules M(x) appearing in the sum are independent. Then since [x′] is in J_{1}, each [m(x)] ∈ J_{1} as well. But then each [m(x)] ∈ ⟨[x]⟩, hence [m(x)] ∈ N(J). Since [x′] and the finite number of elements [x] considered above are distinct elements of an 𝔽_{p}-base for N(J), they are independent, and we have a contradiction.

Now if W = M(x′) ∩ ∑_{[x′]≠[x]} M(x) ≠ {0} where the sum is infinite, the same argument holds, since [x′] ∈ V = W ∩ J_{1} is a finite sum of elements [m(x)]. Hence the M(x), [x] ∈ I, are independent.

Let Y = ⊕_{I}M(x). Then Y is a free 𝔽_{p}[G]-module with a generating set in one-to-one correspondence with a generating set for N(J) as an 𝔽_{p}-module. Moreover, Y^{G} = ⊕M(x)^{G} = ⊕⟨[x]⟩ = N(J).

Let Z be any complement in ϵ(F^{×}) of the 𝔽_{p}-submodule of J generated by N(J) and X ∩ ϵ(F^{×}). Clearly Z is a trivial 𝔽_{p}[G]-module.

We claim that X, Y, and Z are independent 𝔽_{p}[G]-submodules of J. If W = Y ∩ Z ≠ {0}, then W is a submodule of ϵ(F^{×}). Let [x] ∈ W. Then [x] lies in Y ∩J_{1} = N(J), as well as in Z. But Z is a complement of ϵ(F^{×}) of a submodule containing N(J), a contradiction. Therefore the 𝔽_{p}[G]-submodule of J generated by Y and Z is isomorphic with Y ⊕ Z.

Now suppose W = X ∩(Y ⊕Z) ≠ {0}. Considering W ∩J_{1} as above, we find that X is of dimension 2, ξ_{p} ∉ N(K^{×}), and W = ⟨[ξ_{p}]⟩. If [ξ_{p}] ∈ N(J) then ξ_{p}a^{c} ∈ N(K^{×}) for some c ∈ ℤ. However, since N(\sqrt[p]{a}) = a, we have ξ_{p} ∈ N(K^{×}), a contradiction. Therefore W ∩ Y = {0} and [ξ_{p}] = y + z, y ∈ Y, z ∈ Z, with z ≠ 0. Hence y ∈ J_{1} ∩ Y = N(J), and z = [ξ_{p}] − y is the relation in ϵ(F^{×}). But Z is a complement in ϵ(F^{×}) of the submodule generated by N(J) and (X ∩ ϵ(F^{×})) = ⟨[ξ_{p}]⟩, a contradiction.

Let J̃ be the 𝔽_{p}[G]-submodule of J generated by X ∪ Y ∪ Z. As we have shown above, J̃ ≅ X ⊕ Y ⊕ Z. We show that J = J̃ by induction on the socle series of J.

First we show J_{1} ⊂ J̃. From our construction of modules X, Y and Z we see that ϵ(F^{×}) ⊂ J̃. Therefore from Lemma 2a) in Section 2 we see that if ξ_{p} ∉ N(K^{×}) then J_{1} ⊂ J̃. If ξ_{p} ∈ N(K^{×}) then from the definition of X and Lemma 2b), we see that again J_{1} ⊂ J̃.

For the inductive step, assume J_{i} ⊂ J̃, 1 ≤ i < p, and let [γ] ∈ J_{i+1} \ J_{i}.

Let ⟨[b]⟩ = M_{γ}^{G}. From our Lemma 1a) in Section 2 we see that if i ≥ 2 then [b] ∈ N(J). If i = 1 and ξ_{p} ∈ N(K^{×}) we see from Remark 1 in Section 2 that again [b] ∈ N(J). If ξ_{p} ∉ N(K^{×}) then [\sqrt[p]{a}] ∈ X and therefore it is sufficient to show that [a^{t/p}γ] ∈ J̃ for some t ∈ ℤ. Therefore in this case if γ = a^{r/p}γ_{1}, where r ∈ ℤ and [γ_{1}] ∈ J_{1}, we see that [γ] ∈ J̃ and if γ cannot be written in this form, we replace γ by a suitable a^{t/p}γ, and we use Lemma 1b) to conclude that (M_{a^{t/p}γ})^{G} ⊂ N(J). Thus using Lemma 1 we have reduced our investigation to the case when [b] ∈ N(J).

We may write [b] = ∑_{I} c_{x}[x] with c_{x} ∈ 𝔽_{p} and almost all c_{x} = 0. Now for each [x], M(x) = M_{ω} for some ω ∈ K^{×} with N([ω]) = [x]. For each [x], set γ(x) = ω^{(σ−1)^{p−i−1}}, so that [γ(x)] ∈ J_{i+1} and [γ(x)]^{(σ−1)^{i}} = [x]. Now let [Γ] = ∑_{I} c_{x}[γ(x)]. The application of (σ − 1)^{i} to [Γ] results in ∑_{I} c_{x}[x] = [b]. Hence [γ/Γ] lies in the kernel of (σ − 1)^{i}, and [γ/Γ] ∈ J_{i} ⊂ J̃ by induction. Since [Γ] ∈ J̃, [γ] ∈ J̃. □

The proof of Theorem 2 below follows the previous proof with some modifications. Since in the case of p = 2 the length of the socle series of J is at most 2, the proof is simpler. Moreover, we make use of the well-known fact on elements of identical norm in quadratic extensions:

If K = F (\sqrt{a}), a ∈ F^{×} \ F^{×2}, γ_{1}, γ_{2} ∈ K^{×} and N(γ_{1}) = N(γ_{2}) then [γ_{1}] ∈ ϵ(F^{×})[γ_{2}]. Indeed by Hilbert’s Theorem 90 we have [γ_{1}] = [N(k)][γ_{2}] for some k ∈ K^{×}. (See [L, page 202].)

Proof of Theorem 2. If −1 ∈ N(K^{×}), then we set X = ⟨[δ]⟩ ⊂ J_{1} with δ such that δ^{σ−1} = β^{2} and N(β) = −1 (observe that then [N(δ)] = [a] ∈ F^{×}/F^{×2}). If −1 ∉ N(K^{×}) then we set X = {0}.

Choose I to be an 𝔽_{2}-basis of N(J). For each [x] ∈ I choose a representative x ∈ F^{×} of [x] and ω ∈ K^{×} such that x = N(ω). Then set M(x) = M_{ω} and let Y be the submodule of J generated by all M(x), x ∈ I. The same argument as in the proof of Theorem 1 shows that Y is a free 𝔽_{2}[G]-submodule of J, and that Y ≅ ⊕M(x). Finally let Z be any complement in ϵ(F^{×}) of the 𝔽_{2}-module N(J). As before we check that X, Y and Z are independent, Z is a trivial 𝔽_{2}[G]-module, and we let J̃ ≅ X ⊕ Y ⊕ Z be the 𝔽_{2}[G]-submodule of J generated by X, Y and Z.

Assume now that [γ] ∈ J_{1}. Then N(γ) ∈ F^{×} ∩ K^{×2} = F^{×2} ∪ aF^{×2}. If N(γ) ∈ F^{×2} then [γ] ∈ ϵ(F^{×}) ⊂ J̃. If N(γ) ∈ aF^{×2} then [γ] ∈ [δ]ϵ(F^{×}) ⊂ J̃. Thus J_{1} ⊂ J̃. Now suppose that [γ] ∈ J \ J_{1}. Let λ = γ^{σ−1} and since σ − 1 = σ + 1 in 𝔽_{2}[G], we see that [λ] = [N(γ)] ≠ [1]. Hence for a suitable Γ ∈ Y we have [γ/Γ] ⊂ J_{1}. Therefore J̃ = J as required. □

Proof of Theorem 3. We shall first show that if 3 ≤ i ≤ p−1 then 𝔎_{i} = ∅. Suppose that this is not true and that there exists M_{i,k} = M, 3 ≤ i ≤ (p − 1), k ∈ 𝔎_{i}, appearing as a summand in our decomposition of J.

Let m be a generator of M. Using the decomposition J = X ⊕Y ⊕Z described in Theorem 1, we can write m = u + v, where u ∈ Y and v ∈ X ⊕Z. Using the fact that v ∈ J_{2} and 3 ≤ i we see that we can find an element n ∈ M \ {0} and y ∈ Y \ Y ∩ J_{p−1} such that n = (σ − 1)^{k}y for a suitable k ∈ ℕ.

Now write y = ∑_{s=1}^{t} m_{s} where m_{s} ∈ M_{a,j}, 1 ≤ a ≤ p and j ∈ 𝔎_{a}. Of course we assume that we choose at most one element from each M_{a,j}. Then from the equality n = ∑_{s=1}^{t}(σ−1)^{k}m_{s} we see that we may assume that n = (σ − 1)^{k}m_{1} and (σ − 1)^{k}m_{s} = 0 for s ∈ {2, …, t}. Further from our equalities y = ∑_{s=1}^{t} m_{s}, (σ − 1)^{k}m_{s} = 0 for s ∈ {2, …, t} and y ∈ J_{p} \ J_{p−1} we see that m_{1} ∈ J_{p} \ J_{p−1}. But this implies that M is a free summand of J - a contradiction. Hence if 3 ≤ i ≤ p − 1 then 𝔎_{i} = ∅ as we claimed.

Now we consider the case of i = p. Since N(J) = ⊕_{j∈𝔎_{p}}N(M_{p,j}) and each N(M_{p,j}) is a 1-dimensional 𝔽_{p}-module, we see that |𝔎_{p}| = dim_{𝔽_{p}} N(J).

Let us now consider the case of i = 2 and p > 2. We have

|𝔎_{2}| = dim_{𝔽_{p}}(((σ − 1)J)^{G}/N(J)).

Therefore we see that |𝔎_{2}| is an invariant of J which does not depend upon any particular choice of the decomposition of J into a sum of cyclic submodules of J. Therefore in order to determine |𝔎_{2}| we may use our decomposition J = X ⊕ Y ⊕ Z, where X, Y and Z are each thought of as a sum of their cyclic submodules. Thus in the case of p > 2, from Theorem 1 we conclude that |𝔎_{2}| = 1 − Υ(K).

Finally observe that if p > 2 then |𝔎_{1}| + |𝔎_{2}| = dim_{𝔽_{p}} J_{1}/N(J), and if p = 2 then |𝔎_{1}| = dim_{𝔽_{2}} J_{1}/N(J). Applying Lemma 2 and observing that if p > 2 then a ∈ N(K^{×}) and if p = 2 then a ∈ N(K^{×}) if and only if −1 ∈ N(K^{×}), we obtain

1 + |𝔎_{1}| = dim_{𝔽_{p}}(F^{×}/N(K^{×})) + 2Υ(K).
□

Proof of Corollary 1.

(1): From Theorem 3 we see that if p > 2 then J is a free 𝔽_{p}[G]- module if and only if 𝔎_{1} = ∅ = 𝔎_{2}, and if p = 2 then J is a free 𝔽_{p}[G]- module if and only if |𝔎_{1}| = ∅. Therefore if J is a free 𝔽_{p}[G]-module, p = 2 and −1 ∉ N(K^{×}). Moreover assuming p = 2, −1 ∉ N(K^{×}) we see that 𝔎_{1} = ∅ if and only if F^{×} = N(K^{×}) ∪ −N(K^{×}).

(2): J contains no free direct summand precisely when N(J) is trivial. If p > 2 then N(J) is trivial precisely when a is p-rigid. Suppose now that p = 2. Then we see that N(J) is trivial if and only if N(K^{×})/F^{×2} ⊂ ⟨[a]⟩ ⊂ F^{×}/F^{×2}. Since [−1] = [−a] ∈ N(J) we see that in this case \sqrt{−1} ∈ K^{×}.

(3): Suppose first that p > 2. If J is G-invariant then J contains no free direct summand. Hence a is p-rigid. If a is p-rigid then N(J) is trivial, and J is G-invariant if additionally ξ_{p} ∈ N(K^{×}). By the definition of p-rigidity, [ξ_{p}] ∈ ⟨[a]⟩ in F^{×}/F^{×}^{p}, which is equivalent with ξ_{p^{2}} ∈ K^{×}. Now if p = 2 then J is G-invariant precisely when J contains no free direct summand. □

Proof of Corollary 2. Suppose first that our three conditions mentioned in this corollary are valid. Then we see that if we use Theorem 1 and its proof, we can write T_{i} = K_{i}^{×}/K_{i}^{×p} and T_{i} = X_{i} ⊕ Y_{i} ⊕ Z_{i}, i = 1, 2, where modules X_{i}, Y_{i} and Z_{i} are described in Theorem 1 and its proof.

From condition 1 of our corollary we see that X_{1} ≅ X_{2}, from condition 2 we see that Y_{1} ≅ Y_{2}, and finally with condition 3 together with our choice of Z_{i} as an 𝔽_{p}-complement of the 𝔽_{p}-submodule of T_{i} generated by X_{i} ∩ ϵ(F_{i}^{×}) and N(T_{i}) in ϵ(F_{i}^{×}) we see that Z_{1} ≅ Z_{2}. Hence if conditions (1), (2) and (3) are valid, then T_{1} ≅ T_{2}.

Now assume that T_{1} = X_{1} ⊕ Y_{1} ⊕ Z_{1} where submodules X_{1}, Y_{1}, Z_{1} of T_{1} are constructed as in the proof of Theorem 1 and ψ is an 𝔽_{p}[G]- isomorphism T_{1} ⟶ T_{2}. Set X_{2} = ψ(X_{1}), Y_{2} = ψ(Y_{1}) and Z_{2} = ψ(Z_{1}). Then T_{2} = X_{2} ⊕ Y_{2} ⊕ Z_{2}. We also see that N(T_{2}) = N(Y_{2}) ≅ N(Y_{1}) = N(T_{1}). Therefore we have dim_{𝔽_{p}} N(K_{1}^{×})/F_{1}^{×p} = dim_{𝔽_{p}} N(K_{2}^{×})/F_{2}^{×p}.

We shall now assume that X_{1}, Y_{1} and Z_{1} are decomposed into their cyclic summands, and we shall apply Theorem 3.

We obtain ξ_{p} ∈ N(K_{1}^{×}) ⇔ 𝔎_{2} ≠ ∅ ⇔ ξ_{p} ∈ N(K_{2}^{×}). Finally from Theorem 3 we see that dim_{𝔽_{p}}(F_{1}^{×}/N(K_{1}^{×})) = dim_{𝔽_{p}}(F_{2}^{×}/N(K_{2}^{×})) as required. □

Finally we shall prove Corollary 3. Observe that this proof is a straightforward corollary of Theorem 3. We write 𝔎_{2,1} for the index set 𝔎_{2} used in Theorem 3 when applied to the field extension K_{1}/F_{1}. Similarly we extend this notation for other index sets under consideration.

Proof of Corollary 3. From Theorem 3 we see that T_{1} ≅ T_{2} if and only if |𝔎_{1,1}| = |𝔎_{1,2}| and |𝔎_{2,1}| = |𝔎_{2,2}|. Rewriting these equalities we obtain T_{1} ≅ T_{2} if and only if we have the following two equalities.

(1) 2Υ(K_{1}) + dim_{𝔽_{2}} F_{1}^{×}/N(K_{1}^{×}) = 2Υ(K_{2}) + dim_{𝔽_{2}} F_{2}^{×}/N(K_{2}^{×}),

(2) Υ(K_{2}) + dim_{𝔽_{2}} N(K_{1}^{×})/F_{1}^{×2} = Υ(K_{1}) + dim_{𝔽_{2}} N(K_{2}^{×})/F_{2}^{×2}.

□

4. Acknowledgments

We gratefully acknowledge the support of the Mathematical Sciences Research Institute in Berkeley, California, where during the fall of 1999 the authors met, for their subsequent discussions led to the results in this paper. The first author is also very appreciative of the stimulating discussions with Adrian Wadsworth.

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

E-mail address: minac@uwo.ca

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

  E-mail address: joswallow@davidson.edu