CYCLIC ALGEBRAS AND CONSTRUCTION OF SOME GALOIS MODULES

JÁN MINÁČ^{∗†}, ANDREW SCHULTZ, AND JOHN SWALLOW

Abstract. Let p be a prime and suppose that K/F is a cyclic extension of degree p^{n} with group G. Let J be the 𝔽_{p}G-module K^{×}/K^{×p} of pth-power classes. In our previous paper we established precise conditions for J to contain an indecomposable direct summand of dimension not a power of p. At most one such summand exists, and its dimension must be p^{i} + 1 for some 0 ≤ i < n. We show that for all primes p and all 0 ≤ i < n, there exists a field extension K/F with a summand of dimension p^{i} + 1.

Let p be a prime and K/F a cyclic extension of fields of degree p^{n} with Galois group G. Let K^{×} be the multiplicative group of nonzero elements of K and J = J(K/F) := K^{×}/K^{×p} be the group of pthpower classes of K. We see that J is naturally an 𝔽_{p}G-module. In our previous paper [MSS] we established the decomposition of J into indecomposables, as follows.

For i ∈ ℕ let ξ_{p^{i}} denote a primitive p^{i}th root of unity, and for 0 ≤ i ≤ n let K_{i}/F be the subextension of degree p^{i}, with G_{i} = Gal(K_{i}/F). We adopt the convention that for all i, {0} is a free 𝔽_{p}G_{i}-module.

Theorem. [MSS, Theorems 1, 2, and 3] Suppose

• F does not contain a primitive pth root of unity or

• p = 2, n = 1, and −1 ∉ N_{K/F} (K^{×}),

then

J ≅ ⨁_{i=0}^{n} Y_{i}

where each Y_{i} is a free 𝔽_{p}G_{i}-module.

Date: October 30, 2004. ∗Research supported in part by the Natural Sciences and Engineering Research

Council of Canada, as well as by the special Dean of Science Fund and a Distinguished Research Professorship during 2004–2005 at the University of Western Ontario.

†Supported by the Mathematical Sciences Research Institute, Berkeley.

Otherwise, let

m = m(K/F) := \begin{cases}−∞, & ξ_{p} ∈ N_{K/F} (K^{×}), \\ min\{s: ξ_{p} ∈ N_{K/K_{s}}(K^{×})\} − 1, & ξ_{p} ∉ N_{K/F} (K^{×}). \end{cases}

Then

J ≅ X ⊕ ⨁_{i=0}^{n} Y_{i}

where Y_{i} is a free 𝔽_{p}G_{i}-module and X is an indecomposable 𝔽_{p}G-module of 𝔽_{p}-dimension p^{m} + 1 if m ≥ 0 and 1 if m = −∞.

It is not difficult to show that the decomposition is unique. (See the well-known result of Azumaya [AF, p. 144].)

From the well-known result of Albert [A] concerning embedding a cyclic extension of degree p^{i} to a cyclic extension of degree p^{i+1}, we see that ξ_{p} ∈ N_{K/K_{s}}(K_{s}^{×}) for all s ∈ {0, 1, …, n} if m = −∞ and ξ_{p} ∈ N_{K/K_{s}}(K^{×}) for all s ∈ {m + 1, …, n} if m > −∞.

The submodules Y_{i} are produced naturally using norms from different layers of the tower of field extensions. However, the remaining submodule X is more mysterious, and we consider a first problem concerning the classification of all 𝔽_{p}G-modules occurring as J(K/F):

Given n ≥ 1 and d an element of the set

{1, p^{0} + 1, …, p^{n−1} + 1},

does there exist a cyclic extension K/F with ξ_{p} ∈ F^{×} such that the exceptional summand X has dimension d?

It turns out that we may answer this question in the affirmative using a construction of cyclic division algebras due to Brauer-Rowen. We remark that in [MS] the full realization problem, realizing all possible isomorphism classes for the 𝔽_{p}G-module J(K/F), has been solved in the case n = 1 and ξ_{p} ∈ F^{×}.

1. Strategy and Main Theorem

Our strategy is to reformulate m(K/F) in terms of cyclic algebras and then to use the construction of Brauer-Rowen of suitable cyclic algebras. We will prove the following theorem:

Theorem. Let n ∈ ℕ and t ∈ {−∞, 0, 1, …, n − 1}. Then there exists a cyclic extension K/F of degree p^{n} with ξ_{p} ∈ F^{×} and m(K/F) = t.

In two later sections we will examine the relations between m(K/F) and the index of a certain cyclotomic cyclic algebra A defined over F. In particular, we show by example that while these two invariants of K/F are closely related, they are not same.

Before turning to the proof of the theorem, we recall some basic facts about cyclic algebras. If E/F is a cyclic extension of degree r > 1, with Galois group G = Gal(E/F) = ⟨τ⟩, and b ∈ F^{×}, then

B = (E/F, τ, b)

is a central simple algebra such that

B = ⨁_{0≤j<r} u^{j}E,

where u^{−1}du = d^{τ} for all d ∈ E and u^{r} = b. Thus B is an F -algebra of dimension r^{2} over F. We say that deg B := r. If B ≅ M_{s}(D), the matrix algebra containing matrices of size s × s over some division algebra D, then we set ind B = \sqrt{dim_{F} D}. We denote the order of [B] in the Brauer group Br(F) by exp B. Finally, we observe the following important connection:

[B] = 0 in Br(F) ⇔ b ∈ N_{E/F} (E^{×}).

In this case, we say that B splits. For further details on cyclic algebras we refer the reader to [P, Chapter 15] and [R, Chapter 7].

The particular cyclic algebra in which we will be most interested is the cyclotomic cyclic algebra

A := (K/F, σ, ξ_{p}), where G = ⟨σ⟩.

(Recall that we assume ξ_{p} ∈ F for our extensions K/F.)

Proof. We begin with a construction of Brauer-Rowen. (See [Br] for the original construction and see [R, Section 7.3] and [RT, Section 6] for some nice variations of Brauer’s construction.)

First suppose t ≥ 0. Set q = p^{n−t} and let K = ℚ(ξ_{q})(μ_{1}, …, μ_{p^{n}}), where ξ_{q} is a primitive qth root of unity and the μ_{i} are indeterminates over ℚ. Observe that K has an automorphism σ of order p^{n} fixing ℚ(ξ_{q}) and permuting the μ_{i} cyclically.

Let F = K^{⟨σ⟩} be the subfield of K fixed by ⟨σ⟩ and, for each 1 ≤ i ≤ n, K_{i} = K^{⟨σ^{p^{i}}⟩}. Then K/F is a cyclic extension of degree p^{n} satisfying ℚ(ξ_{p}) ⊂ F, and G = ⟨σ⟩ = Gal(K/F). Denote by σ̄ the restriction of σ to the subfield K_{t+1}.

Let A = (K/F, σ, ξ_{p}). Now A is Brauer-equivalent to the cyclic algebra R = (K_{t+1}/F, σ̄, ξ_{q}) by [P, Corollary 15.1b]. On the other hand, the construction of Brauer-Rowen provides that R is a division algebra of degree p^{t+1} and of exponent p [R, Theorem 7.3.8]. Since [A] = [R] ≠ 0, we have ξ_{p} ∉ N_{K/F} (K^{×}).

For all 0 ≤ i ≤ n we have

(K/F, σ, ξ_{p}) ⊗_{F} K_{i} ≅ (K/K_{i}, σ^{p^{i}}, ξ_{p})

by [D, Lemma 6, p. 74]. Therefore, since K_{t+1} is a maximal subfield of R, K_{t+1} splits A:

[A ⊗_{F} K_{t+1}] = 0 ∈ Br(K_{t+1}).

Therefore ξ_{p} ∈ N_{K/K_{t+1}}(K^{×}). Hence m(K/F) ≤ t.

Suppose m = m(K/F) < t. Then [A ⊗_{F} K_{m+1}] = 0 ∈ Br(K_{m+1}), whence K_{m+1}/F splits A. But then p^{t+1} = ind A ≤ [K_{m+1} : F] < p^{t+1}, a contradiction. Hence m(K/F) = t.

Now suppose that t = −∞. Let F be a number field containing ξ_{p}. Then the extension F^{c}/F obtained by adjoining all pth-power roots of unity is the cyclotomic ℤ_{p}-extension of F. Let K/F be the subextension of degree p^{n} of F^{c}/F. Then G = Gal(K/F) is cyclic and K/F embeds in a cyclic extension of F of degree p^{n+1}. Therefore ξ_{p} ∈ N_{K/F} (K^{×}), by a result of Albert [A], and hence m = −∞. □

Remark 1. Observe that the case n = 1 may be handled quite simply. For the case m(K/F) = 0, we set F = ℚ(ξ_{p})(X), where X is a transcendental element over ℚ(ξ_{p}), and K = F (\sqrt[p]{X}). Write G = Gal(K/F) as ⟨σ⟩ with σ(\sqrt[p]{X}) = ξ_{p}\sqrt[p]{X}. Then the cyclic algebra A = (K/F, σ, ξ_{p}) is a symbol algebra A = (\frac{X,ξ_{p}}{F,ξ_{p}} ). (See, for instance, [P, p. 284].) Furthermore,

−[A] = [(\frac{ξ_{p}, X}{F, ξ_{p}} )] = [(E/F, τ, X)] ∈ Br(F),

where E = F (ξ_{p^{2}}) and τ(ξ_{p^{2}}) = ξ_{p}ξ_{p^{2}}. However, it is an easy exercise (solved in [P, p. 380]) that [(E/F, τ, X)] ≠ 0. Hence A is not split, and m = 0 as required.

The m(K/F) = −∞ case follows as in the end of the proof of the theorem. Consider the tower

ℚ(ξ_{p}) ⊂ ℚ(ξ_{p^{2}}) ⊂ ℚ(ξ_{p^{3}}).

By Albert’s result, if F = ℚ(ξ_{p}) and K = ℚ(ξ_{p^{2}}), we have n = 1 and m(K/F) = −∞.

Remark 2. For extensions K/F of local fields one may then deduce that m(K/F) ∈ {−∞, 0}, confirming [B], as follows. If [A] = 0 ∈ Br(F), then m = −∞. Otherwise, since ind A = exp A for local fields (see [P, Corollary 17.10b]), the local invariant inv A of A is s/p with s ∈ ℕ, p ∤ s. Because

inv A ⊗_{F} E = [E : F] inv A

(see [P, Proposition 17.10]), we obtain that inv A ⊗_{F} K_{1} = 0. Hence [A ⊗_{F} K_{1}] = 0 ∈ Br(K_{1}) and m(K/F) = 0, as desired.

2. The Invariants m and ind A

The proof of the theorem turns on the fact that for the particular extension K/F we have ind A = p^{m+1}. It is interesting to ask whether this equality holds generally.

We show in this section that the answer is negative. However, we have an inequality

ind A ≤ p^{m+1},

as follows. Observe that by the definition of m(K/F),

[A ⊗_{F} K_{m(K/F)+1}] = 0 ∈ Br(K_{m(K/F)+1})

for m ≠ −∞. Hence the inequality holds in the case m ≠ −∞. The statement also holds for m = −∞, since A splits if and only if m = −∞. In fact, in this case we obtain an equality.

We show that equality does not always hold by considering the following example in the number field case. Recall first that for number fields exp A = ind A. (See, for instance, [P, Theorem 18.6].) Therefore ind A is either 1 or p since the exponent of A divides p:

[⊗^{p}A] = [(K/F, σ, 1)] = 0 ∈ Br(F).

Hence it is enough to produce a case when m(K/F) > 0.

Let p = 2, c ∈ 4ℤ \ {0}, a = 1 + c^{2} ∉ ℤ^{2}, and d ∈ {1, −1} such that d(a+\sqrt{a}) is not a sum of two squares in ℚ_{2}. (For example, take a = 17 and d = −1.) It is well-known that then

F = ℚ < K_{1} = ℚ(\sqrt{a}) < K_{2} = ℚ (\sqrt{d(a + \sqrt{a})})

is a tower of fields with K_{2}/F cyclic of order 4. (See [JLY, p. 33].)

Let K̂_{i}, i = 1, 2, denote the completion of K_{i} with respect to any valuation v on K_{i} which extends the 2-adic valuation on ℚ. Since 8 | a − 1, we have K̂_{1} = ℚ_{2} and then we may and do assume that K̂_{1} = ℚ_{2} ⊂ K̂_{2}.

Since d(a + \sqrt{a}) is not a sum of two squares in ℚ_{2}, the quaternion algebra (d(a + \sqrt{a}), −1)_{ℚ_{2}} is nonsplit. Hence −1 ∉ N_{K̂_{2}/ℚ_{2}}(K̂_{2}) and therefore −1 ∉ N_{K_{2}/K_{1}}(K_{2}^{×}). (See [P, p. 353].) We obtain then that m(K/F) = 1.

3. When A is a Division Algebra

Observe that if A is a division algebra, then ind A = p^{n} and the chain of inequalities

p^{n} = ind A ≤ p^{m+1} ≤ p^{n}

force the equality ind A = p^{m+1}. In this section we show how a natural construction gives additional field extensions L_{w}/F_{w} with ind A = p^{m+1} = p^{k} for every k < n. More precisely:

Proposition. Suppose that A is a division algebra. Set F_{i} = F (ξ_{p^{i}}) and L_{i} = K(ξ_{p^{i}}) for each i = 1, 2, …, n. Further set A_{i} = A ⊗_{F} F_{i}.

Then

ind A_{i} = p^{m(L_{i}/F_{i})+1} = p^{n−i+1}, i = 1, 2, …, n.

Proof. We proceed by induction on i. The base i = 1 is simply the case K_{1}/F_{1} = K/F, which follows from the observation at the beginning of the section. Hence we assume that A is a division algebra and, for some i ∈ {1, 2, …, n − 1}, we have [L_{i} : F_{i}] = p^{n}, ind A_{i} = p^{n−i+1}, and m(L_{i}/F_{i}) = n − i.

We claim that ξ_{p^{i+1}} ∉ L_{i}. Otherwise, since F_{i}(ξ_{p^{i+1}})/F_{i} is an extension of degree 1 or p, we deduce that ξ_{p^{i+1}} ∈ F′_{i}, where F′_{i} is the subfield of L_{i}/F_{i} with [L_{i} : F′_{i}] = p^{i}. Without loss of generality we may assume that ξ_{p^{i+1}}^{p^{i}} = ξ_{p}. Then ξ_{p} = N_{L_{i}/F′_{i}}(ξ_{p^{i+1}}), and we obtain m(L_{i}/F_{i}) ≤ n − i − 1, a contradiction.

Hence L_{i}/F_{i} and F_{i+1}/F_{i} are linearly disjoint Galois extensions. Therefore L_{i+1}/F_{i+1} is a Galois extension of degree p^{n} and

G = Gal(L_{i}/F_{i}) ≅ Gal(L_{i+1}/F_{i+1}).

Now let σ_{i+1} ∈ Gal(L_{i+1}/F_{i+1}) such that the restriction of σ_{i+1} to L_{i} is σ_{i}. (We assume that σ_{i} is already defined by induction, where σ_{1} = σ.) Then by [D, Lemma 7, p. 74] we see that

A_{i+1} = A_{i} ⊗_{F_{i}} F_{i+1} ≅ (L_{i+1}/F_{i+1}, σ_{i+1}, ξ_{p}).

We therefore obtain from [P, Proposition 13.4v] that

ind A_{i+1} ≥ \frac{ind A_{i}}{p} = p^{n−i}.

On the other hand, we show that p^{m(L_{i+1}/F_{i+1})+1} ≤ p^{n−i}, as follows. Since ξ_{p^{i+1}} ∈ F_{i+1}^{×}, we have

ξ_{p} ∈ N_{L_{i+1}/F′_{i+1}}(L_{i+1}^{×}),

where F_{i+1} ⊂ F′_{i+1} ⊂ L_{i+1} and [L_{i+1} : F′_{i+1}] = p^{i}. Hence m(L_{i+1}/F_{i+1}) ≤ n − i − 1.

Putting these last two equalities together with the equality of the second section, we reach the following chain:

ind A_{i+1} ≤ p^{m(L_{i+1}/F_{i+1})+1} ≤ p^{n−i} ≤ ind A_{i+1}.

We obtain m(L_{i+1}/F_{i+1}) = n − (i + 1) and p^{m(L_{i+1}/F_{i+1})+1} = ind A_{i+1}, as desired. □

To include m = −∞ in the proposition, it is sufficient to continue the induction one step further. Set F_{n+1} = F (ξ_{p^{n+1}}) and L_{n+1} = K(ξ_{p^{n+1}}). Then again L_{n+1}/F_{n+1} is a cyclic extension of degree p^{n} and A_{n+1} = A ⊗_{F} F_{n+1} splits. We conclude that m(L_{n+1}/F_{n+1}) = −∞.

4. Acknowledgement

We thank John Labute for his interest and for his valuable remarks related to this paper.

References

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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, Building 380, Stanford University, Stanford, California 94305-2125 USA

E-mail address: aschultz@stanford.edu

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

E-mail address: joswallow@davidson.edu