GALOIS EMBEDDING PROBLEMS WITH CYCLIC QUOTIENT OF ORDER p

JÁN MINÁČ^{∗†} AND JOHN SWALLOW^{‡}

Abstract. Let K be a cyclic Galois extension of degree p over a field F containing a primitive pth root of unity. We consider Galois embedding problems involving Galois groups with common quotient Gal(K/F) such that corresponding normal subgroups are indecomposable 𝔽_{p}[Gal(K/F)]- modules. For these embedding problems we prove conditions on solvability, formulas for explicit construction, and results on automatic realizability.

Introduction

Let p > 2 be a prime and F a field of characteristic not p containing a primitive pth root of unity ξ_{p}. Suppose that K = F (\sqrt[p]{a}) is a cyclic field extension of F with Galois group Gal(K/F) of order p. In this paper we consider Galois embedding problems involving Kummer extensions of K of degree p^{n} that are Galois over F, and we establish new automatic realizability results, whereby the solvability of one Galois embedding problem implies the solvability of another. (See e. g. [GrSmSw, Section 5] for some automatic realizations of 2-groups as Galois groups.) We restrict ourselves to the case p > 2 because p = 2 is quite simple and it does not lead to new results.

A simple example serves as a motivating introduction to Galois embedding problems of this type. Consider Heisenberg’s group E, a noncommutative group of order p^{3} and exponent p. These conditions determine E up to isomorphism. The center of E is cyclic of order p, and we have the following short exact sequence:

1 → ℤ/pℤ → E → ℤ/pℤ × ℤ/pℤ → 1.  (1)

Now let L/K be an extension Galois over F such that Gal(L/F) ≅ ℤ/pℤ × ℤ/pℤ. Then the exact sequence naturally gives rise to a Galois embedding problem, asking whether L embeds in a Galois extension L̃/F with group E and such that the surjection in the exact sequence is the surjection of Galois theory.

Date: August 3, 2003.

∗Research supported in part by the Natural Sciences and Engineering Research Council of Canada grant R0370A01, as well as by the special Dean of Science Fund at the University of Western Ontario.

†Supported by the Mathematical Sciences Research Institute, Berkeley.

‡Research supported in part by National Security Agency grant MDA904-02-1-0061.

The obstruction to the solvability of this embedding problem may be computed as follows. Since the group μ_{p} of pth roots of unity is contained in F, we may identify the center of E with μ_{p}. Fixing a primitive root ξ_{p} and an element σ ∈ Gal(L/F) such that \sqrt[p]{a}^{σ−1} = ξ_{p}, there must exist τ ∈ Gal(L/F) such that \sqrt[p]{a}^{τ−1} = 1 and such that the commutator of the lifts of σ and τ in E is ξ_{p}. Then there exists b ∈ F^{×} such that \sqrt[p]{b} ∈ L, \sqrt[p]{b}^{σ−1} = 1, and \sqrt[p]{b}^{τ−1} = ξ_{p}. It is well-known that the Galois embedding problem admits a solution if and only if b ∈ N(K^{×}), where N denotes the norm map from K to F. (See, for instance, [JLY, page 161].)

Moreover, if we suppose that ω ∈ K satisfies N(ω) = b, then it has been observed in [JLY, page 161] that all field extensions L̃/F solving the Galois embedding problem may be written L̃ = L(\sqrt[p]{fα}), where f ∈ F^{×} and α = ω^{p−1}σ(ω)^{p−2} ⋯ σ^{p−2}(ω).

In this paper we generalize and motivate both the condition on solvability and the form of the solution. Observe that in the example above L and L̃ are Kummer extensions of K of pth-power degree that are Galois over F, and the Galois groups Gal(L/K) and Gal(L̃/K) are naturally acted upon by Gal(K/F). The appropriate context for our results turns out to be Kummer extensions L of K such that Gal(L/K) is an indecomposable 𝔽_{p}[Gal(K/F)]- module; as we show later in Proposition 2, any Kummer extension of K of degree p^{n} that is Galois over F decomposes into a compositum of extensions L/F of this type.

This example may be generalized, and one basic generalization is as follows. Further generalizations appear in Theorems 2 and 3. Let G = Gal(K/F) ≅ ℤ/pℤ, with generator σ, and let A = ⊕_{j=0}^{p−1}𝔽_{p}τ^{j} be a free 𝔽_{p}[G]-module on the generator τ, where σ acts by multiplication by τ. Let A_{i} be the 𝔽_{p}[G]-submodule generated by (τ −1)^{i}. (See section 1 for details.) Finally let E_{i}, 1 < i ≤ p, denote the following Galois embedding problem:

E_{i} : 1 → A_{1}/A_{i} → (A/A_{i}) ⋊ G → A/A_{1} ⋊ G = Gal(L/F) → 1.

Observe that A/A_{1} ≅ 𝔽_{p}, a trivial 𝔽_{p}[G]-module; hence A/A_{1} ⋊G ≅ ℤ/pℤ× ℤ/pℤ. Further observe that in the case i = 2, A/A_{2} ⋊ G ≅ E. Hence E_{2} is precisely the embedding problem in equation (1) above.

Given such an L/F, let b ∈ F be any element such that F (\sqrt[p]{b}) is the fixed field of 1 ⋊ G in Gal(L/F). In the following theorem we consider the embedding problems E_{i} where i = 2, …, p. We prove:

Theorem 1. The following are equivalent.

(1) Any E_{i} is solvable.

(2) Each E_{i} is solvable.

(3) b ∈ N(K^{×}).

If the conditions hold, let ω ∈ K^{×} satisfy N(ω) = b. Then a solution to E_{i} is given by

L̃ = K(\sqrt[p]{fω^{(σ−1)^{p−i}}}, \sqrt[p]{ω^{(σ−1)^{p−i+1}}}, …, \sqrt[p]{ω^{(σ−1)^{p−1}}}),

f ∈ F^{×}. Moreover, all solutions of E_{i} arise in this way.

Said another way, solutions of E_{i} are parameterized by ω with N(ω) = b and f ∈ F^{×}.

Note that in 𝔽_{p}[G] we have the identity

(σ − 1)^{p−2} = (p − 1) + (p − 2)σ + ⋯ + σ^{p−2},

so the construction of L̃ in the theorem above is equivalent to that of [JLY, page 161] in the case i = 2.

1. Preliminaries

In what follows F is a field containing a primitive pth root of unity ξ_{p}, K = F (\sqrt[p]{a}), and G = Gal(K/F) ≅ ℤ/pℤ. We let σ denote the generator of G such that \sqrt[p]{a}^{σ−1} = ξ_{p}. All modules and Galois extensions will be acted upon on the left by their respective groups, even though we will use exponential notation to denote Galois action on fields. We denote by F^{×} the multiplicative group of a field F, and we write N = N_{K/F} for the norm map from K to F. For a subset S of an 𝔽_{p}-module V we denote by ⟨S⟩ the 𝔽_{p}-span of S in V.

1.1. 𝔽_{p}[G]-modules.

Let A = ⊕_{j=0}^{p−1}𝔽_{p}τ^{j} be a free 𝔽_{p}[G]-module on the generator τ, where σ acts by multiplication by τ. There are p quotient modules A/A_{i}, i = 1, …, p of A where for i < p,

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

These quotients are all cyclic and together form a complete set of indecomposable 𝔽_{p}[G]-modules. Each A/A_{i} is of dimension i as a vector space over 𝔽_{p}. We call this dimension the length, and denote the length of a cyclic 𝔽_{p}[G]-module M by l(M), because we have the following criterion for l(M), where M is a cyclic 𝔽_{p}[G]-module generated by m: l(M) = i such that (σ − 1)^{i}m = 0, (σ − 1)^{i−1}m ≠ 0. Moreover, such a cyclic module M of length l contains precisely one submodule of each length 1 ≤ i ≤ l: M_{i} = ⟨(σ − 1)^{l−i}m, …, (σ − 1)^{p−1}m⟩.

For each i ∈ \{1, …, p\} we pick a basis \{1, \overline{τ − 1}, …, (\overline{τ − 1})^{i−1}\} of A/A_{i} consisting of images of 1, τ − 1, …, (τ − 1)^{i−1}. We define an 𝔽_{p}-linear map λ : A/A_{i} ⟶ 𝔽_{p} by λ(f_{0}+f_{1}(\overline{τ − 1})+⋯+f_{i−1}(\overline{τ − 1})^{i−1}) = f_{i−1}, where f_{k} ∈ 𝔽_{p}, k = 0, …, i − 1. Observe that ker(λ) contains no nonzero ideal of A/A_{i}. Then B(a, b) := λ(ab) for each a, b ∈ A/A_{i} defines a nonsingular, symmetric bilinear form B : A/A_{i} × A/A_{i} ⟶ 𝔽_{p}. Thus A/A_{i} is a symmetric algebra. (See [Lam, page 442].) Further we have B(a^{σ}, b^{σ^{−1}}) = B(a, b) for each a, b ∈ A/A_{i} and our bilinear form B induces a G-equivariant isomorphism between A/A_{i} and its dual.

1.2. Groups.

In this section we classify the groups of interest in this paper and the surjections among them. For e ∈ 𝔽_{p}, let B_{i,e} be the group extension of A/A_{i} by G with σ̃^{p} = e(\overline{τ − 1})^{i−1}. Here σ̃ is a lift in B_{i,e} of σ ∈ G. Note that for e = 0, B_{i,0} = A/A_{i} ⋊ G. First we consider the equivalence classes of these groups.

Lemma 1. (See [Wat, Theorem 2])

(1) If H is a group with a normal subgroup isomorphic to A/A_{i} as a G-module, with quotient group G, then H = B_{i,e} for some e.

(2) For fixed 1 ≤ i < p, all B_{i,e}, e ≠ 0, are isomorphic, and these groups are not isomorphic to B_{i,0} = A/A_{i} ⋊ G.

(3) For i = p, all B_{i,e} are isomorphic to B_{p,0} ≅ 𝔽_{p}[G] ⋊ G.

The Galois embedding problems in this paper consist of embedding an extension L/F with group B_{j,e′} in an extension with strictly larger group B_{i,e}. We are interested in all surjections B_{i,e} → B_{j,e′} for which the kernel lies in A/A_{i} ⊂ B_{i,e′}. We call these G-surjections.

Lemma 2. The G-surjections in the set of groups {B_{i,e}}_{i≥1} are precisely

B_{i,e} → B_{j,0}, i > j ≥ 1, e ∈ 𝔽_{p}, with kernel A_{j}/A_{i}.

Proof. Considering the dimensions of A/A_{i} and A/A_{j}, if B_{i,e} → B_{j,e′} is a G-surjection then i > j. Now a surjection of G-modules A/A_{i} to A/A_{j} must have as kernel an 𝔽_{p}[G]-submodule of A/A_{i} of 𝔽_{p}-rank i − j. But since A/A_{i} is cyclic, there is precisely one such submodule, namely A_{j}/A_{i}. Hence (\overline{τ − 1})^{k} lies in the kernel for all j ≤ k < i. In particular, the kernel must contain e(\overline{τ − 1})^{i−1}, which is σ̃^{p} in B_{i,e}. Therefore σ̃ ∈ B_{i,e} is sent to some lift σ̂ ∈ B_{j,e′} of σ ∈ G and hence σ̂^{p} = 1 in B_{j,e′}, or e′ = 0. □

Observe that the order of each B_{i,e} is p^{i+1}; the exponent of B_{i,0} is p, and the exponent of B_{i,e}, e ≠ 0 is p^{2}; and the nilpotent index of B_{i,e} is i.

1.3. Extensions and Submodules.

Now let J denote the 𝔽_{p}[G]-module J := K^{×}/K^{×p}. We denote elements of J by [γ], γ ∈ K^{×}. Let J_{i} be the kernel of the endomorphism (σ − 1)^{i} and let M_{γ} be the cyclic submodule of J generated by [γ]. Then [γ] ∈ J_{i} if and only if l(M_{γ}) ≤ i.

We denote by M ↔ L_{M} the Kummer correspondence over K of subspaces M of the 𝔽_{p}-vector space J and abelian exponent p extensions L_{M} of K:

M = (L_{M}^{×p} ∩ K^{×})/K^{×p} ↔ L_{M} = K(\sqrt[p]{γ}: [γ] ∈ M).

Set C = Gal(L_{M}/K). Then M and C are dual G-modules and the canonical duality ⟨m, c⟩ := c(\sqrt[p]{m})/\sqrt[p]{m} of M and C is G-equivariant. (See [Wat, pages 134 and 135].) The following proposition rephrases the results in [Wat, page 135] in our notation.

Then we have the following

Proposition 1. Under the Kummer correspondence above,

(1) L_{M} is Galois over F if and only if M is an 𝔽_{p}[G]-submodule of J.

(2) The following are equivalent:

(a) L_{M} is the Galois closure, over F, of K(\sqrt[p]{γ}) for some γ ∈ K^{×};

(b) M = M_{γ} for some γ ∈ K^{×};

(c) Gal(L_{M}/K) ≅ A/A_{i}, as G-modules, for some i;

(d) Gal(L_{M}/F) ≅ B_{i,e}, as G-extensions, for some i and e.

If these conditions hold, then i = l(M) and

L_{M} = K(\sqrt[p]{γ}, \sqrt[p]{γ^{σ−1}}, …, \sqrt[p]{γ^{(σ−1)^{i−1}}}).

Proof. Because L_{M} is Galois if and only if each automorphism of K extends to an automorphism of L_{M}, item (1) and (2a)⇔(2b) follow. That (2c)⇔(2d) follows from Lemma 1.

Suppose (2b) holds. Then M ≅ A/A_{i} for some i ∈ {1, …, p} and Gal(L_{M}/K) is a G-equivariant dual of M. Since M is a G-equivariant self-dual module, we see that Gal(L_{M}/K) and A/A_{i} are G-isomorphic and (2c) follows.

Suppose now that (2c) holds. Then again using the G-equivariant selfduality of A/A_{i} and Kummer theory, we see that M must be a cyclic module M_{γ} for some γ ∈ K^{×}. Hence (2b) follows.

The presentation of L_{M} follows from the fact that a cyclic 𝔽_{p}[G]-module M generated by m is generated over 𝔽_{p} by {(σ − 1)^{k}(m)}_{k=0}^{l(M)−1}. □

We can now prove

Proposition 2. Let L/K be a Kummer extension of pth-power degree which is Galois over F. Then L is a compositum of finitely many Galois closures, over F, of extensions of the form L_{γ} = K(\sqrt[p]{γ}), γ ∈ K^{×}.

Proof. The extension L/K corresponds to an 𝔽_{p}[G]-submodule M of J. Since M is finite, it is decomposable into a direct sum of finitely many indecomposable 𝔽_{p}[G]-modules M_{j}. Each indecomposable 𝔽_{p}[G]-module M_{j} is isomorphic to some A/A_{i} and is hence cyclic. By Proposition 1 (2), these submodules correspond to Galois closures over F of extensions L_{γ} = K(\sqrt[p]{γ}). The submodule of J generated by each of the indecomposables M_{j} then corresponds to the compositum of the L_{γ}, and we are done. □

1.4. The index.

The following homomorphism appears in a somewhat different form in [Wat, Theorem 3]:

Definition. The index e([γ]) ∈ 𝔽_{p} for [γ] ∈ J_{p−1} is defined by

ξ_{p}^{e([γ])} = (\sqrt[p]{N_{K/F} (γ)})^{(σ−1)} .

The index is well-defined, as follows. First, since

1 + σ + ⋯ + σ^{p−1} = (σ − 1)^{p−1}  (2)

in 𝔽_{p}[G], [N(γ)] = [γ]^{(σ−1)^{p−1}}, which is the trivial class [1] by the assumption [γ] ∈ J_{p−1}, and as a result \sqrt[p]{N(γ)} lies in K and is acted upon by σ. Observe further that e([γ]) depends neither on the representative γ of [γ] nor on the particular pth root of N(γ). Also the index function e above is a group homomorphism from J_{p−1} to 𝔽_{p}. Therefore the restriction of e to any M_{γ} is either trivial or surjective.

We show that the index is trivial for any [γ] in the image of σ − 1:

ξ_{p}^{e([γ]^{(σ−1)})} = \sqrt[p]{N(γ^{σ−1})}^{(σ−1)} = \sqrt[p]{1}^{(σ−1)} = 1,

or e([γ]^{σ−1}) = 0.

Lemma 3. (See [Wat, Theorem 2].) Let [γ] ∈ J and M = M_{γ}.

(1) If l(M) < p and e = e([γ]) then Gal(L_{M}/F) ≅ B_{i,e}.

(2) If l(M) = p then Gal(L_{M}/F) ≅ B_{p,0}.

Proof. The second item follows from Proposition 1 and Lemma 1. The fact that Gal(L_{M}/F) ≅ B_{i,e} for some e ∈ 𝔽_{p} follows in the same manner. Therefore it remains only to show that Gal(L_{M}/F) ≅ B_{i,e([γ])}.

Let σ̃ denote a pullback of σ ∈ G to Gal(L_{M}/F). Then σ̃^{p} lies in Z(Gal(L_{M}/F)) ∩ Gal(L_{M}/K). (Here Z(Gal(L_{M}/F)) means the center of Gal(L_{M}/F).) Recall that using Kummer theory and the G-equivariant selfduality of A/A_{i} we may identify Gal(L_{M}/K) with A/A_{i}. Adopting this identification we pick a basis \{1, \overline{τ − 1}, …, (\overline{τ − 1})^{i−1}\} of the Gal(L_{M}/F) dual with {[γ]^{(σ−1)^{i−1}}, …, [γ]^{σ−1}, [γ]} with respect to Kummer pairing. Under our identification, σ̃^{p} lies in the G-invariant submodule of A/A_{i}, which is ⟨(\overline{τ − 1})^{i−1}⟩. Observe that (\overline{τ − 1})^{i−1} sends \sqrt[p]{γ} to ξ_{p} \sqrt[p]{γ}. If σ̃^{p} = e(\overline{τ − 1})^{i−1} then

(\sqrt[p]{γ})^{(σ̃^{p}−1)} = ξ_{p}^{e}.

Therefore

\sqrt[p]{γ}^{(σ̃^{p}−1)} = \sqrt[p]{γ}^{(1+σ̃+···+σ̃^{p−1})(σ̃−1)} = (\sqrt[p]{N_{K/F} (γ)})^{(σ̃−1)} = ξ_{p}^{e([γ])}.

□

We characterize elements of J fixed by σ and of trivial index with the following

Lemma 4. If [γ] ∈ J_{1} and e([γ]) = 0 then there exists f ∈ F^{×} such that [γ] = [f].

Proof. By [MS, Remark 2], 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 from K to F. Now e([γ]) = 0 implies that [γ] is in the kernel of the surjection N above, and we are done. □

We will also need a lemma on the smallest lengths of cyclic submodules of J generated by an element [γ] with nontrivial index. Let Υ = 1 if ξ_{p} ∈ N(K^{×}) and Υ = 0 otherwise.

Lemma 5.

(1) If Υ = 1 then there exists δ ∈ K^{×} such that [δ] ∈ J_{1} and e([δ]) ≠ 0. These are precisely the δ such that K(\sqrt[p]{δ})/F is a cyclic Galois extension of degree p^{2}.

(2) If Υ = 0 then [\sqrt[p]{a}] ∈ J_{2} \ J_{1}, e([\sqrt[p]{a}]) ≠ 0, and e([γ]) = 0 for all [γ] ∈ J_{1}.

Proof. By [A, Theorem 3], Υ = 1 if and only if K/F embeds in an extension L = K(\sqrt[p]{δ}) Galois over F with group ℤ/p^{2}ℤ ≅ B_{1,e}, e ≠ 0. By Proposition 1 and Lemma 3, then, Υ = 1 if and only if there exists [δ] ∈ J_{1} with e([δ]) ≠ 0. This proves the first statement.

Assume now that Υ = 0. We have [\sqrt[p]{a}] ∈ J_{2}, since [\sqrt[p]{a}]^{σ−1} = [ξ_{p}] ∈ J_{1}, and we calculate e([\sqrt[p]{a}]) = 1. Since Υ = 0, [ξ_{p}] ≠ [1] in J_{1} and therefore [\sqrt[p]{a}] ∉ J_{1}. Now consider any [γ] ∈ J_{1}. Then L = K(\sqrt[p]{γ}) is Galois over F and since Υ = 0 we see from [A, Theorem 3] that Gal(L/F) is B_{1,0} ≅ ℤ/pℤ × ℤ/pℤ. Hence e([γ]) = 0 and the second statement is proved. □

Finally, we introduce a variant of [MS, Lemma 1] for submodules generated by elements with trivial index. This is our key lemma:

Lemma 6. Let [γ] ∈ J. Suppose that 2 ≤ l(M_{γ}) < p and e([γ]) = 0.

Then there exists [γ′] ∈ J such that

(1) l(M_{γ′}) = l(M_{γ}) + 1.

(2) [γ′]^{(σ−1)^{2}} = [γ]^{σ−1}.

(3) The fixed elements M_{γ}^{G} of M_{γ} under G coincide with M_{γ′}^{G}.

(4) If l(M_{γ}) < p − 1 then e([γ′]) is defined and has a value of 0.

Proof. Let c = Nγ. Since l(M_{γ}) < p, we have [c] = [γ]^{(σ−1)^{p−1}} = [1]. Hence c ∈ F^{×} ∩K^{×p}. In fact, c = a^{s}f^{p} for some f ∈ F and s ∈ ℤ, as follows. Since c ∈ K^{×p}, F (\sqrt[p]{c}) ⊂ K. The Kummer extension F (\sqrt[p]{c}) is either F or K; if the former, then c ∈ F^{×p}, while if the latter, then by Kummer theory c also has the desired form.

Thus Nγ = a^{s}f^{p} for some s and f. But e(γ) = 0, so p divides s and we see that Nγ = f^{p} for some f ∈ F^{×}. Since N(γ/f) = 1, by Hilbert’s Theorem 90 there exists a ω ∈ K^{×} such that ω^{σ−1} = γ/f. Then l(M_{ω}) = l(M_{γ}) + 1.

If l(M_{γ}) < p − 1 then let t = e([ω]) and set γ′ = ω/(a^{t/p}); otherwise let t = 0 and set γ′ = ω.

We compute [γ′]^{σ−1} = [ξ_{p}^{−t}γ/f] and, since ξ_{p}, f ∈ F^{×}, [γ′]^{(σ−1)^{2}} = [γ]^{σ−1}, which is nontrivial since 2 ≤ l(M_{γ}). Hence (1) and (2) follow.

Now if l(M_{γ}) < p − 1 then l(M_{γ′}) = l(M_{γ}) + 1 < p and so γ′ ∈ J_{p−1}. Then e([γ′]) = e([ω]) − t = 0. Therefore (4) is valid.

Finally observe that M_{γ}^{G} is generated by [γ]^{(σ−1)^{l(Mγ)−1}} as well as [γ′]^{(σ−1)^{l(M_{γ′})−1}}, which in turn generates M_{γ′}^{G}. Hence M_{γ}^{G} = M_{γ′}^{G} and therefore (3) follows from (2).

□

2. Embedding Problem Conditions and Solutions

We consider all embedding problems involving groups B_{i,e}, based on the G-surjections determined in Lemma 2, defining the following embedding problems for i > j ≥ 1:

E_{i,j}(L) : 1 → A_{j}/A_{i} → B_{i,0} → A/A_{j} ⋊ G = Gal(L/F) → 1.

and, for any e ≠ 0,

E′_{i,j}(L) : 1 → A_{j}/A_{i} → B_{i,e} → A/A_{j} ⋊ G = Gal(L/F) → 1.

In each case we ask if there exists a Galois extension L̃/F containing L such that Gal(L̃/F) ≅ B_{i,0} or Gal(L̃/F) ≅ B_{i,e}, and under the identification of Gal(L̃/F) with B_{i,0} (or B_{i,e}), the surjection Gal(L̃/F) → Gal(L/F) is identical to the surjection above.

Since B_{p,0} ≅ B_{p,e} as G-extensions for all e, the embedding problems E_{p,i}(L) and E′_{p,i}(L) are identical. Moreover note E_{i,1}(L) = E_{i}(L). (See page 2 for the discussion of E_{i}.)

For each of these problems, by Proposition 1, L is the Galois closure of K(\sqrt[p]{γ}) for some γ ∈ K^{×}. Hence under the Kummer correspondence M_{γ} ↔ L, and by Proposition 1, l(M_{γ}) = j.

Theorem 2. Let p ≥ i > j ≥ 1 and L be the Galois closure of K(\sqrt[p]{γ}) over F.

Then E_{i,j}(L) is solvable if and only if [γ] = [ω]^{(σ−1)^{p−j}} for some ω ∈ K^{×}.

If so, then a solution L̃ to E_{i,j}(L), where i > j + 1, is given by

L̃ = L(\sqrt[p]{fω^{(σ−1)^{p−i}}}, \sqrt[p]{ω^{(σ−1)^{p−i+1}}}, …, \sqrt[p]{ω^{(σ−1)^{p−j−1}}}),

f ∈ F^{×}.

In the case when i = j + 1 a solution L̃ to E_{j+1,j} is given by L̃ = L(\sqrt[p]{fω^{(σ−1)^{p−i}}}), f ∈ F^{×}.

Moreover, all solutions to E_{i,j}(L) arise in this way.

Proof. By Proposition 1, there exists a field L̃ with Gal(L̃/F) ≅ B_{i,e} for some i and e if and only if there exists a cyclic submodule M_{β} of J of length i, and in this case we have M_{β} ↔ L̃ under the Kummer correspondence.

Furthermore, by Lemma 3, if i < p then Gal(L̃/F) ≅ B_{i,e}, where e = e([β]), and if i = p then Gal(L̃/F) ≅ B_{p,0}. Hence if i < p then E_{i,j}(L) is solvable if and only if there exists β ∈ K^{×} with e([β]) = 0, l(M_{β}) = i, and M_{β} ⊃ M_{γ}. If i = p then E_{p,j} is solvable if and only if there exists β ∈ K^{×} with l(M_{β}) = p and M_{β} ⊃ M_{γ}.

Now suppose that [γ] = [ω]^{(σ−1)^{p−j}} for some ω ∈ K^{×}. Then let β = ω^{(σ−1)^{p−i}}. Since l(M_{γ}) = j and [γ] = [β]^{(σ−1)^{i−j}}, l(M_{β}) = i. Now if i = p then the condition of the previous paragraph is satisfied. If i < p then β is in the image of the endomorphism σ − 1, therefore e([β]) = 0 and the condition of the previous paragraph is satisfied.

Going the other way, suppose that there exists β ∈ K^{×} with l(M_{β}) = i, M_{β} ⊃ M_{γ}, and, if i < p, e([β]) = 0. Since M_{γ} is the unique 𝔽_{p}[G]-submodule of M_{β} of length j, M_{γ} = M_{β^{(σ−1)^{i−j}}}. Further since the linear map M_{β} → M_{γ}

defined by [α] ↦ [α]^{(σ−1)^{i−j}} is surjective, there is a [β′] ∈ M_{β} such that we have [β′]^{(σ−1)^{i−j}} = [γ]. Moreover, l(M_{β′}) = l(M_{β}) so M_{β′} = M_{β}. In the case of i < p, because e is trivial on [β], then e is trivial on M_{β} and hence e([β′]) = 0.

If i = p then let ω = β′. Otherwise, by repeated application of Lemma 6, we may find an ω ∈ K^{×} such that [ω]^{(σ−1)^{p−i+1}} = [β′]^{σ−1}. Then [ω]^{(σ−1)^{p−j}} = [β′]^{(σ−1)^{i−j}} = [γ].

We now treat the explicit construction of the solution fields. Let M_{β} be a solution field to an embedding problem E_{i,j}(L). Let β′ and ω be defined as above. Note that [ω]^{(σ−1)^{p−i+y}} = [β′]^{(σ−1)^{y}} for all y ≥ 1. Hence δ = ω^{(σ−1)^{p−i}}/β′ satisfies [δ]^{(σ−1)} = [1]. Now e([δ]) = 0, so by Lemma 4, [δ] = [f] for some f ∈ F^{×}. If we have a solution L̃ to E_{i,j}(L) with M_{β} ↔ L̃, then L̃ = L(\sqrt[p]{θ} : [θ] ∈ M_{β}), or

L̃ = L(\sqrt[p]{β}, \sqrt[p]{β^{(σ−1)}}, …, \sqrt[p]{β^{(σ−1)^{i−j−1}}}),

by Proposition 1. Since M_{β′} = M_{β}, [ω]^{(σ−1)^{p−i+y}} = [β′]^{(σ−1)^{y}} for all y ≥ 1, and [β′] = [ω^{(σ−1)^{p−i}}/f], we have

L̃ = L(\sqrt[p]{f^{−1}ω^{(σ−1)^{p−i}}}, \sqrt[p]{ω^{(σ−1)^{p−i+1}}}, …, \sqrt[p]{ω^{(σ−1)^{p−j−1}}}),

in the case when i > j + 1 and L̃ = L(\sqrt[p]{f^{−1}ω^{(σ−1)^{p−i}}}) if i = j + 1 again by Proposition 1.

Finally observe that if we have a solution L̃ to E_{i,j}(L) with M_{β} ↔ L̃, then for each f ∈ F^{×} a module M_{fβ} also corresponds to a solution of E_{i,j}(L). Hence in our explicit formula for a solution field L̃, any f ∈ F^{×} is eligible. □

Theorem 3. Let p ≥ i > j ≥ 1 and L be the Galois closure of K(\sqrt[p]{γ}) over F.

(1) E′_{i,j}(L), i > j + 1 − Υ or j = p − 1, is solvable if and only if [γ] = [ω]^{(σ−1)^{p−j}} for some ω ∈ K^{×}. If so, then a solution L̃ to E′_{i,j}(L) is given by

L̃ = L(\sqrt[p]{fαω^{′(σ−1)^{p−i}}}, \sqrt[p]{ω^{′(σ−1)^{p−i+1}}}, …, \sqrt[p]{ω^{′(σ−1)^{p−j−1}}}),

f ∈ F^{×}, where in the case Υ = 1, α is any element in K^{×} with K(\sqrt[p]{α})/F cyclic of degree p^{2}, and in the case Υ = 0, α is \sqrt[p]{a}. Furthermore ω′ = ω^{c_{0}+c_{1}(σ−1)+···+c_{i−1}(σ−1)^{i−1}} for suitable c_{i} ∈ ℤ.

(2) E′_{j+1,j}(L), Υ = 0, is solvable if and only if [γ] = [ξ_{p}]^{e}[ω]^{(σ−1)^{p−j}} for some ω ∈ K^{×} and e ≢ 0 mod p. If so, then a solution L̃ to E′_{i,j}(L) is given by

L̃ = L(\sqrt[p]{fa^{e/p}ω^{(σ−1)^{p−j−1}}}), f ∈ F^{×}.

Moreover, all solutions to E′_{i,j} arise in the way described above.

Note that the two parts of the theorem overlap when i = p, j = p − 1, and Υ = 0.

Proof. We begin in the same manner as the previous proof: if i < p then E′_{i,j}(L) is solvable if and only if there exists β ∈ K^{×} with e([β]) ≠ 0, l(M_{β}) = i, and M_{β} ⊃ M_{γ}. If i = p then E′_{p,j} is solvable if and only if there exists β ∈ K^{×} with l(M_{β}) = p and M_{β} ⊃ M_{γ}.

We first treat the conditions on [γ] that are equivalent to solvability. Then we consider the explicit presentations of the solution fields.

In the case i = p, since E′_{p,j} = E_{p,j}, the condition on [γ] is the same as the condition on [γ] for the solvability of E_{p,j} determined in the previous theorem. This gives us the condition in part 1. Now if additionally j = p−1, consider the condition of part 2: [γ] = [ξ_{p}]^{e}[ω′]^{(σ−1)^{p−j}} = [ξ_{p}]^{e}[ω′]^{σ−1} for e ≢ 0 mod p. If this condition holds, ω = a^{−e/p}ω′ satisfies the condition [γ] = [ω]^{(σ−1)^{p−j}} = [ω]^{σ−1} of part 1. Conversely, if the condition of part 1 holds, set ω′ = a^{1/p}ω and observe that the condition of part 2 holds with e = 1.

Now suppose i < p and E′_{i,j}(L) is solvable with field L̃ such that M_{β} ↔ L̃. We show that the specified conditions on [γ] must hold.

If i > j + 1, then M_{β^{σ−1}} ↔ L̄, where, by Lemma 3, L̄ is a solution to E_{i−1,j}(L). Then by the previous theorem the condition [γ] = [ω]^{(σ−1)^{p−j}} is satisfied.

If i = j + 1 and Υ = 1, then let [α] ∈ J_{1} with e([α]) ≠ 0. (See Lemma 5. Observe that K(\sqrt[p]{α})/F is cyclic of degree p^{2}.) Since E′_{j+1,j}(L) is solvable there exists M_{β} ⊃ M_{γ}, l(M_{β}) = j + 1 and e([β]) ≠ 0. Set β′ = β^{e([α])}/α^{e([β])}. Then e([β′]) = 0. Since α ∈ J_{1} we have M_{(β′)^{σ−1}} = M_{β^{σ−1}} = M_{γ}. By Proposition 1 and Lemma 3, M_{β′} ↔ L̄, where L̄ is a solution to E_{j+1,j}(L). By the previous theorem, the condition [γ] = [ω]^{(σ−1)^{p−j}} is satisfied.

Now consider the case i = j+1 and Υ = 0. By choosing another generator [β] of M_{β} if necessary, we may assume that [β]^{σ−1} = [γ]. Consider β′ = βa^{−e([β])/p}. Then e([β′]) = 0. Now [a^{−e([β])/p}]^{σ−1} = [ξ_{p}^{−e([β])}], so [β′]^{σ−1} = [ξ_{p}^{−e([β])}γ]. Therefore if l(M_{ξ_{p}^{−e([β])}γ}) is at least 1 there exists a solution to an embedding problem corresponding to M_{β′} and M_{ξ_{p}^{−e([β])}γ}. By the previous

theorem, the condition [ξ_{p}]^{−e([β])}[γ] = [ω]^{(σ−1)^{p−j}} is satisfied. If [ξ_{p}^{−e([β])}γ] = [1] then we can set ω = 1 and again the condition [ξ_{p}]^{−e([β])}[γ] = [ω]^{(σ−1)^{p−j}} is satisfied.

In all cases, then, we have shown that if E′_{i,j}(L) is solvable, the corresponding condition on [γ] holds.

Now suppose that the condition of part 1 holds: [γ] = [ω]^{(σ−1)^{p−j}} for some ω. Here we include the case i = p.

If Υ = 1 then let [α] ∈ J_{1} with e([α]) ≠ 0. (See Lemma 5.) Consider β = αω^{(σ−1)^{p−i}}. If i < p, e([ω]^{(σ−1)^{p−i}}) = 0 and hence e([β]) ≠ 0. Since [α] ∈ J_{1}, l(M_{β}) = l(M_{ω^{(σ−1)^{p−i}}}) = l(M_{γ}) + i − j = i. Moreover, since [β]^{(σ−1)^{i−j}} = [γ], M_{β} ⊃ M_{γ} and we have shown that E′_{i,j}(L) is solvable. In the case when i = p and Υ = 1 we observed above that E′_{p,j}(L) is equivalent with E_{p,j}(L) and also the solvability conditions are the same. Hence by Theorem 2 we see that E′_{p,j}(L) is solvable.

If Υ = 0 and i > j + 1, then let α = \sqrt[p]{a}. Consider β = αω^{(σ−1)^{p−i}}. Again if i < p then e([ω]^{(σ−1)^{p−i}}) = 0 and hence e([β]) ≠ 0. Since [α] ∈ J_{2} (see Lemma 5) and i > j + 1, l(M_{β}) = l(M_{ω^{(σ−1)^{p−i}}}) = l(M_{γ}) + i − j = i. Moreover, since [β]^{(σ−1)^{i−j}} = [γ], M_{β} ⊃ M_{γ} and we have shown that E′_{i,j}(L) is solvable. (Observe that we employed the condition i < p only to ensure that e([β]) ≠ 0 in this case. If i = p then e([β]) plays no role, and therefore we have covered this case in the construction above.)

Now suppose that the condition of part 2 holds: [γ] = [ξ_{p}]^{e}[ω]^{(σ−1)^{p−j}} for some ω ∈ K^{×} and e ≢ 0 mod p. Let β = a^{e/p}ω^{(σ−1)^{p−j−1}}. If j + 1 < p then because e([ω]^{(σ−1)^{p−j−1}}) = 0 we have e([β]) = e ≢ 0 mod p. Moreover, [β]^{σ−1} = [ξ_{p}^{e}][ξ_{p}^{−e}][γ] = [γ], so M_{β} ⊃ M_{γ} and we have shown that E′_{j+1,j}(L) is solvable. Finally observe that if j +1 = p we showed at the beginning of our proof that both embedding problems E′_{p,p−1} and E_{p,p−1} are the same, and that also the conditions in Theorem 2 and Theorem 3 for the existence of a solution of this problem are equivalent. Hence the existence of a solution in this case follows from Theorem 2.

Next we shall derive an explicit form of any solution field L̃ of our embedding problem.

Observe that for any f ∈ F^{×} and J_{p−1} ⊃ M_{β} ⫌ M_{γ}, we have l(M_{fβ}) = l(M_{β}), M_{fβ} ⊃ M_{γ}, and e([fβ]) = e([β]). Recall that in the case Υ = 1, α is any element in K^{×} with K(\sqrt[p]{α})/F cyclic of degree p^{2}, and in the case Υ = 0, α is \sqrt[p]{α}. By Proposition 1, then,

L̃ = L(\sqrt[p]{fβ}, \sqrt[p]{β^{(σ−1)}}, …, \sqrt[p]{β^{(σ−1)^{p−1}}})

is a solution to the appropriate embedding problem for β = αω^{(σ−1)^{p−i}} in the case i > j +1−Υ or j = p−1 and β = a^{e/p}ω^{(σ−1)^{p−i}} in the case i = j +1, Υ = 0, as above.

To show that every solution field L̃ takes this form, suppose that M_{β} ↔ L̃ is a solution to E′_{i,j}(L). Hence M_{β} ⊃ M_{γ}. We consider first the case of part 1 when Υ = 1. If i < p then by Lemma 3, e([β]) ≠ 0; in this case we let c ∈ ℤ be such that e([β^{c}]) = e([α]) and set β′ = β^{c} so that e([β′/α]) = 0. If i = p then let β′ = β. Because i > j ≥ 1 and [α] ∈ J_{1}, l(M_{β′/α}) = l(M_{β′}) = i. Observe that [β′/α]^{(σ−1)} = [β^{c}]^{(σ−1)}, so M_{β′/α} ⊃ M_{(β′/α)^{(σ−1)}} = M_{β^{(σ−1)}} ⊃ M_{γ}, because M_{γ} is properly contained in M_{β} and M_{β^{(σ−1)}} is the maximal proper 𝔽_{p}[G]-submodule of M_{β}. Hence M_{β′/α} ↔ L̄, for L̄ a solution to E_{i,j}(L).

By Kummer theory and Theorem 2,

M_{β′/α} = ⟨[fω^{(σ−1)^{p−i}}], [ω^{(σ−1)^{p−i+1}}], …, [ω^{(σ−1)^{p−1}}]⟩

for some f ∈ F^{×} and ω ∈ K^{×}. Observe that hence M_{β′/α} = M_{fω^{(σ−1)^{p−i}}}.

Because [β′/α] and [fω^{(σ−1)^{p−i}}] are both 𝔽_{p}[G]-module generators of the same module of length i,

β′/α = (fω^{(σ−1)^{p−i}})^{c_{0}+c_{1}(σ−1)+···+c_{i−1}(σ−1)^{i−1}}

for some c_{i} ∈ 𝔽_{p}. Let f′ = f^{c_{0}} and ω′ = ω^{c_{0}+c_{1}(σ−1)+···+c_{i−1}(σ−1)^{i−1}}. Then β′/α = f′(ω′)^{(σ−1)^{p−i}}, or β′ = f′α(ω′)^{(σ−1)^{p−i}}. Since M_{β′} = M_{β} ↔ L̃, L̃ takes the form

L̃ = L(\sqrt[p]{f′α(ω′)^{(σ−1)^{p−i}}}, \sqrt[p]{(ω′)^{(σ−1)^{p−i+1}}}, …, \sqrt[p]{(ω′)^{(σ−1)^{p−1}}})

by Proposition 1.

The case when Υ = 0 can be treated as above with slight modifications. First in this case instead of [α] ∈ J_{1} we take [\sqrt[p]{a}]. We use our hypothesis i > j +1 to make sure as above that l(M_{β′/ \sqrt[p]{a}}) = l(M_{β′}). Next observe that

M_{β′/ \sqrt[p]{a}} ⊃ M_{(β′/ \sqrt[p]{a})^{(σ−1)^{2}}} = M_{β^{(σ−1)^{2}}} ⊃ M_{γ}

as the 𝔽_{p}[G]-submodules of M_{β} are linearly ordered by inclusion and l(M_{β})− l(M_{γ}) ≥ 2. The rest of the argument for case (1) when Υ = 0 faithfully follows the argument for case (1) when Υ = 1 as above.

In order to show that every solution field L̃ in part 2 takes the specified form, observe that A_{j}/A_{j+1} is in the center of B_{j+1,e}. Therefore, since we have one solution of the embedding problem E′_{j+1,j} of the form L′ = L(\sqrt[p]{a^{e/p}ω^{(σ−1)^{p−j−1}}}), by the well-known theorem on solutions of central embedding problems (see [JLY, Lemma A.1.1]), any other solution L̃ of the embedding problem E′_{j+1,j} takes the form

L̃ = L(\sqrt[p]{fa^{e/p}ω^{(σ−1)^{p−j−1}}}), f ∈ F^{×},

as required. □

Remark. Lemma 2 implies that among our embedding problems, only Galois extensions L/F with Gal(L/F) ≅ B_{j,0} may be solved. This result agrees with our Theorems 2 and 3, as follows. Suppose that L is the Galois closure of K(\sqrt[p]{γ}), γ ∈ K^{×}, and l(M_{γ}) = j. From our solvability conditions we see that if L can be embedded in some extension L̃ such that Gal(L̃/F) ≅ B_{i,e} and p ≥ i > j ≥ 1, e ∈ 𝔽_{p}, then necessarily e([γ]) = 0.

3. Proof of Theorem 1

Proof. The condition on b preceding the statement of the theorem implies that M_{b} ↔ L/K under the Kummer correspondence. Since E_{i} = E_{i,1}, by Theorem 2 we need only show that there exists α with [b] = [α]^{(σ−1)^{p−1}} if and only if there exists ω with N(ω) = b. By equation (2),

[ω]^{(σ−1)^{p−1}} = [ω]^{1+σ+···+σ^{p−1}} = [ω^{1+σ+···+σ^{p−1}}] = [N(ω)] = [b],

so if ω exists satisfying N(ω) = b, then α = ω satisfies [b] = [α]^{(σ−1)^{p−1}}.

Going the other way, suppose that [b] = [α]^{(σ−1)^{p−1}} for some α. By equation (2) again, [b] = [N(α)]. Then N(α) = k^{p}b for some k ∈ K^{×}. Hence N(α)/b ∈ F^{×} ∩ K^{×p}, and by Kummer theory k^{p} = a^{s}f^{p} for some s ∈ ℤ. Choosing ω = α/(a^{s/p}f) we obtain N(ω) = b. □

4. Acknowledgement

It is clear from the content of our paper that we are strongly influenced by the beautiful paper of Waterhouse quoted in the references. We gratefully acknowledge this crucial influence on our work.

References

[A] A. Albert, On cyclic fields, Trans. Amer. Math. Soc. 37 (1935), no. 3, 454–462.

[GrSmSw] H. G. Grundman, T. L. Smith and J. R. Swallow, Groups of order 16 as Galois groups, Expositiones Math. 13 (1995), 289–319.

[JLY] C. Jensen, A. Ledet and N. Yui, Generic polynomials: constructive aspects of the inverse Galois problem, Mathematical Sciences Research Institute Publications 45, Cambridge: Cambridge University Press, 2002.

[Lam] T.-Y. Lam, Lectures on modules and rings, Graduate Texts in Mathematics 189, New York: Springer-Verlag, 1999.

[MS] J. Mináč and J. Swallow, Galois module structure of pth-power classes of extensions of degree p. To appear, Israel J. Math.

[Wat] W. Waterhouse, The normal closures of certain Kummer extensions, Canad. Math. Bull. 37 (1994), no. 1, 133–139.

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

E-mail address: minac@uwo.ca

   URL: http://www.math.uwo.ca/minac.html

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

   E-mail address: joswallow@davidson.edu
   URL: http://www.davidson.edu/math/swallow/