THE CONJUGATE DIMENSION OF ALGEBRAIC NUMBERS

NEIL BERRY, ARTŪRAS DUBICKAS, NOAM D. ELKIES, BJORN POONEN, AND CHRIS SMYTH

Abstract. We find sharp upper and lower bounds for the degree of an algebraic number in terms of the ℚ-dimension of the space spanned by its conjugates. For all but seven nonnegative integers n the largest degree of an algebraic number whose conjugates span a vector space of dimension n is equal to 2^{n}n!. The proof, which covers also the seven exceptional cases, uses a result of Feit on the maximal order of finite subgroups of GL_{n}(ℚ); this result depends on the classification of finite simple groups. In particular, we construct an algebraic number of degree 1152 whose conjugates span a vector space of dimension only 4.

We extend our results in two directions. We consider the problem when ℚ is replaced by an arbitrary field, and prove some general results. In particular, we again obtain sharp bounds when the ground field is a finite field, or a cyclotomic extension ℚ(ω_{ℓ}) of ℚ. Also, we look at a multiplicative version of the problem by considering the analogous rank problem for the multiplicative group generated by the conjugates of an algebraic number.

1. Introduction

Let \overline{ℚ} be an algebraic closure of the field ℚ of rational numbers, and let α ∈ \overline{ℚ}. Let α_{1}, …, α_{d} ∈ \overline{ℚ} be the conjugates of α over ℚ, with α_{1} = α. Then d equals the degree d(α) := [ℚ(α) : ℚ], the dimension of the ℚ-vector space spanned by the powers of α. In contrast, we define the conjugate dimension n = n(α) of α as the dimension of the ℚ-vector space spanned by {α_{1}, …, α_{d}}.

In this paper we compare d(α) and n(α). By linear algebra, n ⩽ d. If α has nonzero trace and has Galois group equal to the full symmetric group S_{d}, then n = d (see [Smy86, Lemma 1]). On the other hand, it is shown in [Dub03] that n can be as small as ⌊log_{2} d⌋. It turns out that n can be even smaller. Our first main result gives the minimum and maximum values of d for fixed n.

Theorem 1. Fix an integer n ⩾ 0. If α ∈ \overline{ℚ} has n(α) = n, then the degree d = d(α) satisfies n ⩽ d ⩽ d_{max}(n), where d_{max}(n) is defined by Table 1, equalling 2^{n}n! for all n ∉ \{2, 4, 6, 7, 8, 9, 10\}. Furthermore, for each n ⩾ 1, there exist α ∈ \overline{ℚ} attaining the lower and upper bounds.

We refer to the n with d_{max}(n) ≠ 2^{n}n! as exceptional. To attain d = d_{max}(n), we will use α for which the extension ℚ(α)/ℚ is Galois with Galois group isomorphic to a maximal-order finite subgroup G of GL_{n}(ℚ) given in Table 1.

Date: December 5, 2003. 2000 Mathematics Subject Classification. Primary 11R06; Secondary 20E28, 20H20. This article has appeared in Quarterly J. Math. 55 (2004), no. 3, 237–252.

2 NEIL BERRY, ARTŪRAS DUBICKAS, NOAM D. ELKIES, BJORN POONEN, AND CHRIS SMYTH

n d_{max}(n)/(2^{n}n!) Maximal-order subgroup G  d_{max}(n) = #G
2  3/2 W (G_{2})  12
4  3 W (F_{4})  1152
6  9/4 ⟨W (E_{6}), −I⟩  103680
7  9/2 W (E_{7})  2903040
8  135/2 W (E_{8})  696729600
9  15/2 W (E_{8}) × W (A_{1})  1393459200
10  9/4 W (E_{8}) × W (G_{2})  8360755200
all other n  1 W (B_{n}) = W (C_{n}) = (ℤ/2ℤ)^{n} ⋊ S_{n}  2^{n}n!

Table 1. Maximal-order finite subgroups of GL_{n}(ℚ)

The groups W (·) are the Weyl groups of classical Lie algebras acting on their maximal tori (see for instance [Hum90]). They are all reflection groups: each is generated by those elements that act on ℚ^{n} by reflection in some hyperplane. For the standard fact that the negative identity matrix −I is not in W (E_{6}), see for instance [Hum90, p. 82]. In particular, W (B_{n}) = W (C_{n}) = (ℤ/2ℤ)^{n} ⋊ S_{n} is better known as the signed permutation group, the group of n × n matrices with entries in {−1, 0, 1} having exactly one nonzero entry in each row and each column.

Feit [Fei96] proved that for each n a subgroup of GL_{n}(ℚ) of maximal finite order is conjugate to the group given in Table 1. (The paper [Fei96] is just a statement of results — no proofs.) Feit’s result uses unpublished work of Weisfeiler depending on the classification theorem for finite simple groups (see also [KP02, p. 185]). See

            http://weisfeiler.com/boris/philinq-8-28-2000.html

for the sad tale of Weisfeiler’s disappearance.

The inequality d ⩽ d_{max}(n) comes from studying the span of {α_{1}, …, α_{d}} as a representation of Gal(ℚ(α_{1}, …, α_{d})/ℚ). To prove the existence of examples where this upper bound is attained, we

(1) observe that if G is one of the maximal-order finite subgroups of GL_{n}(ℚ) listed in Table 1, then the G-invariant subfield ℚ(x_{1}, …, x_{n})^{G} of ℚ(x_{1}, …, x_{n}) is purely transcendental, say ℚ(f_{1}, …, f_{n}) (whence ℚ(x_{1}, …, x_{n})/ℚ(f_{1}, …, f_{n}) is a Galois extension with Galois group G),

(2) apply Hilbert irreducibility to obtain a Galois extension K of ℚ with Galois group G, and

(3) choose α ∈ K generating a suitable subrepresentation of G. Moreover, we give explicit examples for all n except 6, 7, 8, 9, 10, and outline an explicit construction in these remaining five cases.

Many of the arguments work over base fields other than ℚ, so we generalize as appropriate (Theorem 14). In particular, Theorem 15 generalizes Theorem 1 by giving the minimal and maximal degrees over any cyclotomic base field ℚ(ω_{ℓ}). The answers change drastically for base fields of positive characteristic: for instance from Theorem 14(v) there are elements of a separable closure of 𝔽_{q}(t) of conjugate dimension 2 that generate Galois extensions of 𝔽_{q}(t) of arbitrarily large degree. We also give in Section 5 some results on analogous questions concerning the rank of the multiplicative subgroup of \overline{ℚ}^{∗} generated by α_{1}, …, α_{d}, and its generalization over a Hilbertian field.

2. Degree and conjugate dimension over fields in general

2.1. Representations. Let k be a field, and let k^{s} be a separable closure of k. If α ∈ k^{s}, then let d = d(α) be the degree [k(α) : k], and let n = n(α) be the conjugate dimension of α (over k), defined as the dimension of the k-vector space V (α) spanned by the conjugates α_{1}, …, α_{d} of α in k^{s}.

Proposition 2. With notation as above, let K = k(α_{1}, …, α_{d}) and let G = Gal(K/k). Then there exists a faithful n-dimensional k-representation of G.

Proof. Since {α_{1}, …, α_{d}} is G-stable, the G-action on K restricts to a G-action on V (α). If g ∈ G acts trivially on V (α), then g fixes each α_{i}, so g is the identity on K. Thus V (α) is a faithful k-representation of G. Finally, dim_{k} V (α) = n, by definition. □

A partial converse will be given in Proposition 5 below, whose proof relies on the following representation-theoretic result.

Lemma 3. Let k be a field of characteristic 0, and let G be a finite group. Let V be a kG-submodule of the regular representation kG. Assume that G acts faithfully on V. Then V = (kG)α for some α ∈ V with Stab_{G}(α) = {1}.

Proof. Since k has characteristic zero, V is a direct summand (and hence a quotient) of the regular representation, so the kG-module V can be generated by one element. An element α ∈ V fails to generate V as a kG-module if and only if {gα : g ∈ G} fails to span V, and this condition can be expressed in terms of the vanishing of certain minors in the coordinates of α with respect to a basis of V. Thus the set Z := { α ∈ V : (kG)α ≠ V } of such elements is contained in the zeros of some nonzero polynomial in the coordinates. Also, for each g ∈ G − {1}, the set V^{g} := {v ∈ V : gv = v} is a proper subspace of V, since V is faithful. Since k is infinite, we can choose α ∈ V outside Z and each V^{g} for g ≠ 1. □

Remark 4. We may also allow k to have characteristic p > 0, as long as p does not divide #G and k is infinite. Then V is still a direct summand and a quotient of kG, and the same proof applies. The hypothesis that k is infinite cannot be removed, however, as the following counterexample shows. Let k be a finite field of characteristic p, let k′/k be a finite extension, and take V = k′. For any subgroup G_{1} of Gal(k′/k), let G be the semidirect product k′^{∗} ⋊ G_{1}, which acts k-linearly on V. Then every nonzero α ∈ V has stabilizer isomorphic to G_{1}. If moreover #G_{1} equals neither 1 nor a multiple of p, then p does not divide #G, and thus V is a submodule of kG since V is multiplicity-free over \overline{k}; but the conclusion of Lemma 3 is false because no α ∈ V has trivial stabilizer.

Proposition 5. Let k be a field of characteristic 0, and let G be a finite group. Suppose that G = Gal(K/k) for some Galois extension K of k, and that there is a faithful ndimensional subrepresentation V of the regular representation of G over k. Then there exists α ∈ K with n(α) = n and d(α) = [K : k] = #G.

Proof. By the Normal Basis Theorem, K, as a representation of G over k, is isomorphic to the regular representation. Hence we may identify V with a subrepresentation of K. Lemma 3 gives an element α ∈ V whose G-orbit has size #G and spans the n-dimensional space V. □

2.2. Invariant subfields.

Proposition 6. Let G be one of the groups in Table 1, viewed as a subgroup of GL_{n}(ℚ). Then for any field k of characteristic 0, the invariant subfield k(x_{1}, …, x_{n})^{G} is purely transcendental over k.

Proof. We may assume k = ℚ. Chevalley [Che55] proved that if G is a finite reflection group, then ℚ[x_{1}, …, x_{n}]^{G} = ℚ[f_{1}, …, f_{n}] for some homogeneous polynomials f_{i} of distinct degrees. In this case, ℚ(x_{1}, …, x_{n})^{G} = ℚ(f_{1}, …, f_{n}) as desired.

The only remaining case is n = 6 and G = ⟨W (E_{6}), −I⟩. Here ℚ(x_{1}, …, x_{6})^{W (E_{6})} = ℚ(I_{2}, I_{5}, I_{6}, I_{8}, I_{9}, I_{12}) where each I_{j} is a homogeneous polynomial of degree j, given explicitly for instance in [Fra51] (see also [Hum90, p. 59]). Moreover −I ∈ G acts on this subfield by I_{j} ↦ (−1)^{j}I_{j}, so ℚ(x_{1}, …, x_{6})^{G} = ℚ(I_{2}, I_{6}, I_{8}, I_{12}, I_{5}^{2}, I_{5}I_{9}). □

Remark 7. Let G be a finite subgroup of GL_{n}(ℝ). Coxeter showed [Cox51] that ℝ[x_{1}, …, x_{n}]^{G} is a polynomial ring over ℝ in n algebraically independent generators if G is a finite reflection group. Shephard and Todd proved that this sufficient condition on G is also necessary ([ST54, Thm. 5.1], see also [Hum90, p. 65]). For example, G = ⟨W (E_{6}), −I⟩ is not a finite reflection group, and the ℝ-algebra ℝ[x_{1}, …, x_{6}]^{G} = ℝ[I_{2}, I_{6}, I_{8}, I_{12}, I_{5}^{2}, I_{5}I_{9}, I_{9}^{2}] cannot be generated by 6 polynomials.

2.3. Hilbert irreducibility. It is well known that the field ℚ is Hilbertian — see for instance [Ser92, Theorem 3.4.1] (a form of the Hilbert Irreducibility Theorem). This implies that Galois extensions of purely transcendental extensions ℚ(f_{1}, …, f_{n}) can be specialized to Galois extensions of ℚ having the same Galois group [Ser92, Corollary 3.3.2].

Proposition 8. Let k be a Hilbertian field. Let a finite subgroup G of GL_{n}(k) act on k(x_{1}, …, x_{n}) so that the action on the span of the indeterminates x_{i} corresponds to the inclusion of G in GL_{n}(k). If the invariant subfield k(x_{1}, …, x_{n})^{G} is purely transcendental over k, then there exists a finite Galois extension K of k with Galois group G.

Proof. By assumption k(x_{1}, …, x_{n})^{G} = k(f_{1}, …, f_{n}) for some algebraically independent f_{i}. By Galois theory, k(x_{1}, …, x_{n}) is a Galois extension of k(f_{1}, …, f_{n}) with Galois group G. Now use the assumption that k is Hilbertian to specialize. □

Corollary 9. If k is a Hilbertian field, and G is one of the groups in Table 1, then G is realizable as a Galois group over k.

Proof. Combine Propositions 6 and 8. □

For background material on Hilbert irreducibility see [Sch00] or [Ser92].

3. Degree and conjugate dimension over ℚ

3.1. Proof of Theorem 1.

Proof. The inequality n ⩽ d is immediate. Examples with equality exist by Proposition 5 applied to the standard permutation representation S_{n} ↪ GL_{n}(ℚ), since S_{n} is realizable as a Galois group over ℚ (see [Ser92, p. 42], for example).

On the other hand, d ⩽ #G ⩽ d_{max}(n), where G is the Galois group of α over k, because of Proposition 2, since d_{max}(n) is the size of the largest finite subgroup of GL_{n}(ℚ).

Finally, we prove that d = d_{max}(n) is possible for each n ⩾ 1. Let G be a maximal finite subgroup of GL_{n}(ℚ), as in Table 1. The given n-dimensional faithful representation of G is a subrepresentation of the regular representation, since otherwise it would contain some irreducible subrepresentation with multiplicity > 1, which could be removed once to produce a faithful subrepresentation on a lower-dimensional subspace, contradicting the fact that the function d_{max}(n) is strictly increasing. (Alternatively, this could be deduced from the fact that the given representation is irreducible for n ⩽ 8, and is a direct sum of distinct irreducible representations for n = 9 and n = 10.) Moreover, Corollary 9 shows that G is realizable as a Galois group over ℚ. Thus Proposition 5 yields α ∈ \overline{ℚ} with n(α) = n and d(α) = #G = d_{max}(n). □

3.2. Explicit numbers attaining d_{max}(n). In theory, given n ⩾ 1, we can construct explicit α ∈ \overline{ℚ} with n(α) = n and d(α) = d_{max}(n) as follows. Let G be a maximal-order finite subgroup of GL_{n}(ℚ). Take e_{j} to be the column vector in ℤ^{n} having j-th entry 1 and the rest 0, let G_{1} be the stabilizer of e_{1} under the left action of G, and put N = |G : G_{1}|, the size of the orbit of e_{1} under this action. For most of the groups we consider, all of e_{1}, …, e_{n} are in this orbit, and so we denote the whole orbit by e_{1}, …, e_{n}, …, e_{N}. We then find an auxiliary polynomial P_{N} of degree N, irreducible over ℚ, whose splitting field has Galois group G over ℚ. Further, n zeros β_{1}, …, β_{n} of P_{N} can be chosen so that the full list of conjugates β_{1}, …, β_{N} of β_{1} are the (β_{1}, …, β_{n})e_{j} for j = 1, …, N.

The auxiliary polynomial P_{N} arises, at least generically, as follows: by Proposition 6, we can write ℚ(x_{1}, …, x_{n})^{G} = ℚ(I_{1}, …, I_{n}), where the I_{j} are G-invariant homogeneous polynomials in the x_{i}. Choose c_{1}, …, c_{n} ∈ ℚ, and define a zero-dimensional variety V by the polynomial equations

I_{1}(x_{1}, …, x_{n}) = c_{1},
⋮
I_{n}(x_{1}, …, x_{n}) = c_{n}.

Then successively eliminate x_{n}, x_{n−1}, …, x_{2} to get a monic polynomial R(x_{1}) of degree d_{R} = ∏_{j=1}^{n} deg I_{j}. Clearly xg ∈ V for any x ∈ V and g ∈ G, so the multiset of zeros of R is {xge_{1} | g ∈ G}, which consists of #G_{1} copies of {xe_{j} | j = 1, …, N}. Thus R(x_{1}) = P_{N}(x_{1})^{#G_{1}} for some polynomial P_{N}. For reflection groups and unitary reflection groups we can choose the I_{j} so that d_{R} = #G; in this case P_{N} has degree N. The polynomial P_{N} is our auxiliary polynomial.

Choose b_{1}, …, b_{n} ∈ ℚ such that b_{1}x_{1} + ⋯ + b_{n}x_{n} is not fixed by any g ∈ G except the identity. Then α = b_{1}β_{1} +⋯+b_{n}β_{n} has n(α) = n and degree d_{max}(n), its conjugates being (β_{1}, …, β_{n})g(b_{1}, …, b_{n})^{T} for g ∈ G. (This is the standard “primitive element” construction for the Galois closure of ℚ(β).) For most choices of (c_{1}, …, c_{n}) (that is, for all choices outside a “thin set”, in the sense of [Ser92]), this construction will produce the required α. For small n (such as n = 2, considered in Sections 3.4 and 4.2), this procedure works well. For much larger n, however, the elimination process becomes impractical. Also, it becomes hard to check whether a particular choice of (c_{1}, …, c_{n}) yields a suitable α. The difficulty is to choose c_{1}, …, c_{n} so that not only is P_{N} irreducible, but also it has Galois group G (instead of a subgroup). For this reason, the following sections discuss more practical ways of constructing α, in the nonexceptional case and for n = 4.

For the larger exceptional values of n, even these methods would require special treatment for each value, and the large size of #G (see Table 1) has dissuaded us from trying to do the same for these n. One approach to constructing α ∈ \overline{ℚ} attaining d_{max}(n) for 6 ⩽ n ⩽ 10 is to start with Shioda’s beautiful analysis relating the Weyl groups of E_{6}, E_{7}, E_{8} and their invariant rings with the Mordell-Weil lattices of rational elliptic surfaces with an additive fiber. For instance, in [Shi91, p.484–5] Shioda uses this theory to exhibit a monic polynomial in ℤ[X] with Galois group W (E_{7}), whose roots are the images of the 56 minimal vectors of the E_{7}^{∗} lattice under a ℚ-linear, W (E_{7})-equivariant map from E_{7}^{∗} ⊗ ℚ to \overline{ℚ}. The image under this map of any vector in E_{7}^{∗} ⊗ ℚ with trivial stabilizer in W (E_{7}) (that is, in the interior of a Weyl chamber) is then an α ∈ \overline{ℚ} with n(α) = 7 and d(α) = #W (E_{7}) = d_{max}(7). A similar construction will work for n = 8, and (combined with the analysis of algebraic numbers of conjugate dimension 1, 2) also for n = 9, 10. The case n = 6 will require additional work, because Shioda’s construction, which yields Galois group W (E_{6}), will have to be modified to produce ⟨W (E_{6}), −I⟩.

3.3. Explicit numbers attaining d_{max}(n) for nonexceptional n.

Proposition 10. Let k be a field of characteristic not 2. Let n ⩾ 2. Suppose f(x) = x^{n} − a_{1}x^{n−1} + ⋯ + (−1)^{n}a_{n} ∈ k[x] is a separable polynomial of degree n with Galois group S_{n} and discriminant Δ. Let r_{1}, …, r_{n} ∈ \overline{k} be the zeros of f(x). Choose a square root \sqrt{r_{i}} of each r_{i}, and let K = k(\sqrt{r_{1}}, …, \sqrt{r_{n}}). If a_{n} ∉ Δ^{ℤ}k^{∗2} and either n is even or r_{1} ∉ k^{∗}k(r_{1})^{∗2}, then [K : k] = 2^{n}n!.

Proof. The action of the group G := Gal(K/k) on \{\sqrt{r_{1}}, −\sqrt{r_{1}}, …, \sqrt{r_{n}}, −\sqrt{r_{n}}\} is faithful and preserves the partition \{\{\sqrt{r_{1}}, −\sqrt{r_{1}}\}, …, \{\sqrt{r_{n}}, −\sqrt{r_{n}}\}\}, so G is a subgroup of the signed permutation group W (B_{n}). Recall that W (B_{n}) is a semidirect product

0 → V → W (B_{n}) → S_{n} → 1

where V as a group with S_{n}-action is the standard permutation representation of S_{n} over 𝔽_{2}. Since f has Galois group S_{n}, the group G surjects onto the quotient S_{n} of W (B_{n}). Considering the conjugation action of G on itself gives a (possibly nonsplit) exact sequence

0 → W → G → S_{n} → 1

for some subrepresentation W of V. The only subrepresentations of V are 0, 𝔽_{2} with trivial S_{n}-action, the sum-zero subspace of V = 𝔽_{2}^{n}, and V itself. If W = V, we are done.

If W is contained in the sum-zero subspace, then W acts trivially on the square root β := \sqrt{r_{1}} … \sqrt{r_{n}} of a_{n}. Hence the action of G on β is given by either the trivial character or the sign character of S_{n}. Thus either β ∈ k or β\sqrt{Δ} ∈ k. Squaring yields a_{n} ∈ Δ^{ℤ}k^{∗2}, contrary to assumption.

The only remaining case is where n is odd and W = 𝔽_{2}. Then W acts trivially on the square root β_{1} := \sqrt{r_{2}}\sqrt{r_{3}} … \sqrt{r_{n}} of r_{2}r_{3} … r_{n} = a_{n}/r_{1}. Hence the action of Gal(K/k(r_{1})) on β_{1} is given by either the trivial character or the sign character of S_{n−1} = Gal(k(r_{1}, …, r_{n})/k(r_{1})). Thus either β_{1} ∈ k(r_{1}) or β_{1}\sqrt{Δ} ∈ k(r_{1}). Squaring shows that r_{1} ∈ k^{∗}k(r_{1})^{∗2}, again contrary to assumption. □

In the situation of Proposition 10, when its hypotheses are satisfied, we can take the auxiliary polynomial to be P_{2n}(x) = f(x^{2}).

The following corollary is needed in Section 3.5.

Corollary 11. Let n ⩾ 2. Suppose f(x) = x^{n} − a_{1}x^{n−1} + ⋯ + (−1)^{n}a_{n} ∈ k[x] is a polynomial of degree n over a field k ⊂ ℝ, with Galois group S_{n}. Suppose that the zeros r_{1}, …, r_{n} of f(x) are real and satisfy r_{1} < 0 < r_{2} < ⋯ < r_{n}. Choose a square root \sqrt{r_{i}} ∈ \overline{k} of each r_{i}. and let K = k(\sqrt{r_{1}}, …, \sqrt{r_{n}}). Then [K : k] = 2^{n}n!.

Proof. It suffices to check the hypotheses of Proposition 10. The discriminant Δ satisfies Δ > 0, but a_{n} = r_{1} … r_{n} < 0, so a_{n} ∉ Δ^{ℤ}k^{∗2}.

If r_{1} ∈ k^{∗}k(r_{1})^{∗2}, say r_{1} = cγ_{1}^{2} with c ∈ k^{∗} and γ_{1} ∈ k(r_{1}), then applying an automorphism yields r_{2} = cγ_{2}^{2} with γ_{2} ∈ k(r_{2}). These two equations force c < 0 and c > 0, respectively, a contradiction. □

Proposition 12. For n = 1 let r_{1} = 2, while for n ⩾ 2 let r_{1}, …, r_{n} ∈ \overline{ℚ} be the zeros of f(x) = x^{n} + (−1)^{n}(x − 1). Choose a square root of each r_{i}, and let α = \sqrt{r_{1}} + 2\sqrt{r_{2}} + ⋯ + n\sqrt{r_{n}}. Then n(α) = n and d(α) = 2^{n}n!.

Proof. By [Ser92, p. 42], the polynomial (−1)^{n}f(−x) = x^{n} − x − 1 has Galois group S_{n} over ℚ, so f(x) has Galois group S_{n} over ℚ. Also by [Ser92, p. 42], each inertia group of Gal(ℚ(r_{1}, …, r_{n})/ℚ) is either trivial or generated by a transposition; it follows that the same is true for the Galois group G of f over ℚ(i). The group G has index at most 2 in S_{n}, so G is S_{n} or A_{n}. We claim that G = S_{n}. For n = 2 we check this directly.

Take n ⩾ 3. If G = A_{n}, then as G would contain no transpositions, all the inertia groups in G would be trivial, and ℚ(i) would have an A_{n}-extension unramified at all places. The existence of such an extension contradicts the Minkowski discriminant bound for n ⩾ 4, and contradicts class field theory for 3 ⩽ n ⩽ 4. Thus G = S_{n}.

In particular, if Δ is the discriminant of f(x), then Δ ∉ ℚ(i)^{∗2}, so |Δ| ∉ ℚ^{∗2}. Therefore a_{n} := −1 is not in Δ^{ℤ}ℚ^{∗2}.

We now finish checking the hypotheses in Proposition 10 by showing that the assumptions n odd and r_{1} ∈ ℚ^{∗}ℚ(r_{1})^{∗2} lead to a contradiction. Suppose n is odd, and r_{1} = cγ^{2}, with c ∈ ℚ^{∗} and γ ∈ ℚ(r_{1})^{∗}. Taking N_{ℚ(r_{1})/ℚ} of both sides yields (−1)^{n} ≡ c^{n} (mod ℚ^{∗2}). Since n is odd, c ≡ −1 (mod ℚ^{∗2}). Without loss of generality, c = −1. Since γ generates ℚ(r_{1}), the monic minimal polynomial g(t) ∈ ℚ[t] of γ is of degree n. Write g(t)g(−t) = h(t^{2}) for some polynomial h ∈ ℚ[x]. Substituting t = γ shows that h(−r_{1}) = 0, but h has degree n, so h(x) = f(−x). Thus the polynomial −f(−t^{2}) = t^{2n} − t^{2} − 1 factors as −g(t)g(−t). However, it is known to be irreducible (Ljunggren [Lju60, Theorem 3]).

By Proposition 10, the field K = ℚ(\sqrt{r_{1}}, …, \sqrt{r_{n}}) has degree 2^{n}n!. Each \sqrt{r_{i}} lies outside the field generated by the other square roots over ℚ(r_{1}, …, r_{n}), so \sqrt{r_{1}}, …, \sqrt{r_{n}} are linearly independent over ℚ. The conjugates of α are the numbers of the form ∑_{j=1}^{n} ε_{j}j\sqrt{r_{σ(j)}} where σ ∈ S_{n} and ε_{1}, …, ε_{n} ∈ {±1}. The linear independence of the square roots guarantees that these 2^{n}n! elements are distinct. □

3.4. An explicit number attaining d_{max}(n) for n = 2. For n = 2, we can take P_{6}(x) = x^{6} − 2. Taking one zero β of P_{6}, all zeros are spanned by the two zeros β, ω_{3}β where ω_{3} is a primitive cube root of unity. Then α = β + 3ω_{3}β has n(α) = 2 and d(α) = 12, and minimal polynomial y^{12} + 572y^{6} + 470596.

Remark 13. This example can be produced using the procedure outlined in Section 3.2, as follows. The group W (G_{2}) from Table 1 equals ⟨( \begin{matrix}0 & −1 \\ 1 & 1 \end{matrix} ) , ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} )⟩, and has invariants I_{1} = x_{1}^{2} − x_{1}x_{2} + x_{2}^{2} and I_{2} = (x_{1}x_{2}(x_{1} − x_{2}))^{2}. Taking c_{1} = 0, c_{2} = 2, b_{1} = 1, b_{2} = −3, we get the minimal polynomial of α as the x_{2}-resultant of I_{1}(y+3x_{2}, x_{2}) and I_{2}(y+3x_{2}, x_{2})−2.

3.5. An explicit number attaining d_{max}(n) for n = 4. For n = 4, one maximal-order finite subgroup of GL_{4}(ℚ) is the order-1152 group W (F_{4}) generated by its index-3 subgroup W (B_{4}) (of order 384) and the order-2 matrix

\frac{1}{2} (\begin{matrix}1 & 1 & 1 & 1 \\ 1 & −1 & 1 & −1 \\ 1 & 1 & −1 & −1 \\ 1 & −1 & −1 & 1 \end{matrix}) .

Thus by Galois correspondence we should be able to apply the construction of Section 3.2 for β defined over a suitable cubic extension of ℚ. And indeed, this is possible.

Define s_{2k} = z_{1}^{2k} + z_{2}^{2k} + z_{3}^{2k} + z_{4}^{2k} for k = 1, 2, . . .. Four independent homogeneous invariants for W (F_{4}) are known [Meh88] to be

I_{2k} = (8 − 2^{2k−1})s_{2k} + ∑_{j=1}^{k−1} \binom{2k}{2j}s_{2j}s_{2k−2j}

for k = 1, 3, 4, 6. Using the Newton identities and with the help of Maple these can be written entirely as polynomials in s_{2}, s_{4}, s_{6}, s_{8} as follows:

I_{2} = 6s_{2},  I_{6} = −24 s_{6} + 30 s_{2} s_{4},  I_{8} = −120 s_{8} + 56 s_{2} s_{6} + 70 s_{4}^{2},

\begin{matrix}I_{12} = −540 s_{4} s_{8} + 244 s_{6}^{2} − 1365 s_{2}^{2} s_{8} + \frac{1365}{2} s_{2}^{2} s_{4}^{2} + 255 s_{4}^{3} \\ −710 s_{2}^{4} s_{4} + 1250 s_{2}^{3} s_{6} + \frac{159}{2} s_{2}^{6} + 110 s_{2} s_{4} s_{6}. \end{matrix}

We now use resultants to eliminate s_{4} and s_{6}. This shows that s_{8} is cubic over ℚ(I_{2}, I_{6}, I_{8}, I_{12}), and also that s_{4}, s_{6} ∈ ℚ(I_{2}, I_{6}, I_{8}, I_{12})(s_{8}). Specifically, we take I_{2} = 6s_{2} = 30, I_{6} = 1410, I_{8} = 13670 and I_{12} = 1161749, and then γ := s_{8} (the real root, say) satisfies

γ^{3} + \frac{5735}{32} γ^{2} + \frac{5811288377}{36864} γ − \frac{114051068048293}{6220800} = 0.

Then, with the Newton identities, we compute the values of the elementary symmetric functions of the z_{i}^{2}. This gives a polynomial Q_{4} satisfied by the z_{i}^{2}:

Q_{4}(x) = x^{4} − 5 x^{3} + \frac{20261200695}{3175710433} x^{2} + \frac{34560}{3175710433} x^{2} γ^{2} − \frac{47690820}{3175710433} x^{2} γ

+ \frac{36679035170}{9527131299} x − \frac{28800}{3175710433} x γ^{2} + \frac{39742350}{3175710433} x γ − \frac{203476507483}{38108525196}

− \frac{72000}{3175710433} γ^{2} − \frac{56249419}{12702841732} γ.

We write its zeros as β_{1}^{2}, β_{2}^{2}, β_{3}^{2}, β_{4}^{2} say. They are real and close to −1, 1, 2, and 3. (The values for the invariants were chosen to be close to the values they would have had if z_{i}^{2}, i = 1, …, 4 had been exactly −1, 1, 2, 3.) Furthermore, its discriminant 223967999/97200 is not a square in ℚ(γ). Now, shifting x in this quartic by 5/4 to obtain a polynomial z^{4} + b_{2}z^{2} + b_{1}z + b_{0} having zero cubic term, its cubic resolvent z^{3} + 2b_{2}z^{2} + (b_{2}^{2} − 4b_{0})z − b_{1}^{2} is readily checked to be irreducible over ℚ(γ). Hence by [Gar86, Ex. 14.7, p. 117], the Galois closure of ℚ(γ, β) over ℚ(γ) has Galois group S_{4}. Then, as β_{1}^{2} < 0 < β_{2}^{2} < β_{3}^{2} < β_{4}^{2}, we have [ℚ(β_{1}, β_{2}, β_{3}, β_{4}) : ℚ] = 2^{4} · 4! = 384, on applying Corollary 11 with k = ℚ(γ).

If we now take the resultant of Q_{4}(x^{2}) and the minimal polynomial of γ, to eliminate γ, we obtain the degree 24 auxiliary polynomial

P_{24}(x) =

x^{24} − 15 x^{22} + \frac{375}{4} x^{20} − \frac{2405}{8} x^{18} + \frac{65435}{128} x^{16} − \frac{25905}{64} x^{14} − \frac{181583}{3072} x^{12} + \frac{8367137}{18432} x^{10}

− \frac{28198575}{65536} x^{8} + \frac{1338226651}{5308416} x^{6} − \frac{895964239}{8847360} x^{4} + \frac{4234139}{294912} x^{2} − \frac{24389830879}{1592524800}.

This polynomial is irreducible, with zeros \frac{1}{2}(±β_{1} ± β_{2} ± β_{3} ± β_{4}) as well as ±β_{1}, ±β_{2}, ±β_{3}, ±β_{4}. Now (1, 2, 3, 5)^{T} is not a fixed point of any g ≠ I in W (F_{4}). It follows that α = β_{1} + 2β_{2} + 3β_{3} + 5β_{4} has n(α) = 4 and degree d(α) = 1152, its conjugates being the numbers (β_{1}, β_{2}, β_{3}, β_{4})g(1, 2, 3, 5)^{T} for g ∈ W (F_{4}).

4. Conjugate dimensions over other fields

4.1. General results. The conjugate dimension can behave differently if we use ground fields other than ℚ. For a field k and a positive integer n, let D(k, n) be the maximal degree of α ∈ k^{s} of k-conjugate dimension at most n. For instance D(ℚ, n) = d_{max}(n). If the degree is unbounded, we set D(k, n) = ∞. This can happen even for Hilbertian fields of characteristic zero. For example, D(ℂ(t), 1) = ∞, because for each d ⩾ 1 a d-th root of t generates the Galois extension ℂ(t^{1/d}) of degree d, and all conjugates of t^{1/d} generate the same 1-dimensional space. Nevertheless we can generalize some of our results to various ground fields other than ℚ. We find:

Theorem 14.

(i) If k is a number field of degree m over ℚ, then d_{max}(n) ⩽ D(k, n) ⩽ d_{max}(mn) for all n ⩾ 1.

(ii) If k is a Hilbertian field of characteristic not dividing ℓ and k contains ℓ roots of

unity, then D(k, n) ⩾ ℓ^{n}n!. (iii) If k is a finitely generated transcendental extension of ℂ, then D(k, n) = ∞ for all

n ⩾ 1. (iv) If k is a finite field of q elements, then D(k, n) = q^{n} − 1.

(v) If k is a finitely generated transcendental extension of a finite field k_{0}, then D(k, 1) = q − 1 where q is the size of the largest finite subfield of k, and D(k, n) = ∞ for all n ⩾ 2.

Proof.

(i) By Proposition 2, if α ∈ k^{s} has degree d and conjugate dimension n then there exists a d-element subgroup of GL_{n}(k). If [k : ℚ] = m, then an n-dimensional vector space over k can be viewed as an mn-dimensional vector space over ℚ, so we get an injection GL_{n}(k) ↪ GL_{mn}(ℚ). Hence d ⩽ d_{max}(mn). For the lower bound, note that the specialization made in Proposition 8 can, by [Sch00, Theorem 46, p. 298], be made in such a way that the minimal polynomial of the algebraic number with conjugate dimension n remains irreducible over the field k. This gives an example of an algebraic number of degree d_{max}(n) over k and k-conjugate dimension at most n, so d_{max}(n) ⩽ D(k, n).

(ii) If k contains ℓ roots of unity then GL_{n}(k) contains the group of size ℓ^{n}n! consisting of the permutation matrices whose entries are roots of unity in k. Moreover, the invariant ring of this group is polynomial, being generated by the elementary symmetric functions of the ℓ-th powers of the coordinates. Thus the invariant field is purely transcendental over k. Therefore, by Propositions 5 and 8, there exist α ∈ k^{s} of conjugate dimension n and degree ℓ^{n}n!.

(iii) This follows from (ii), using the fact that every such field is Hilbertian ([Sch00, Theorem 49, p. 308]).

(iv) The Galois group of any k(α)/k with n(α) = n must be contained in GL_{n}(k), but must also be cyclic because k is a finite field 𝔽_{q}. Hence #G ⩽ q^{n} − 1, as may be seen using the characteristic equation of an invertible matrix in GL_{n}(k). We claim that the field of q^{q^{n}−1} elements is generated by an element α of conjugate dimension n over k. Let g be a generator of 𝔽_{q^{n}}^{∗}, and let f(x) = ∑_{i=0}^{n} c_{i}x^{i} be its minimal polynomial over 𝔽_{q}. Let α ∈ \overline{𝔽}_{q}^{∗} be a zero of ∑_{i=0}^{n} c_{i}X^{q^{i}}. Make the 𝔽_{q}-vector space \overline{𝔽}_{q} into a module over the polynomial ring 𝔽_{q}[τ] by letting τ act as the endomorphism z ↦ z^{q}. Then the ideal I of 𝔽_{q}[τ] that annihilates α contains f(τ), but I ≠ (1). Since f is irreducible, I = (f(τ)). Thus the 𝔽_{q}-span of α and its conjugates is an 𝔽_{q}[τ]-module isomorphic to 𝔽_{q}[τ]/(f(τ)). In particular, n(α) = deg f = n. Also d(α) is the smallest d such that τ^{d}(α) = α, which is the smallest d such that τ^{d} = 1 in 𝔽_{q}[τ]/(f(τ)); by choice of g, we get d = q^{n} − 1.

(v) Without loss of generality, suppose that k_{0} is the largest finite subfield of k, so #k_{0} = q. Suppose α ∈ \overline{k} has n(α) = 1. Proposition 2 bounds d(α) by the size of the largest finite subgroup of GL_{1}(k) = k^{∗}. Elements of finite order in k^{∗} are roots of unity, hence contained in k_{0}^{∗}. Thus D(k, 1) ⩽ q − 1. The opposite inequality follows from (ii) since, by [Sch00, Theorem 47, p. 301], k is Hilbertian.

Now suppose n ⩾ 2. Choose a finite Galois extension L of k with [L : k] = n − 1. (For instance, let L be the compositum of a suitable subfield of a cyclotomic extension of k with some Artin-Schreier extensions of k.) Let V be the 𝔽_{q}-span of a Gal(L/k)-stable finite subset of L that spans L as a k-vector space. Define

P_{V,ε}(X) := ∏_{x∈V}(X − x) + ε ∈ k[X, ε],

where ε is an indeterminate. Then P_{V,0}(X) is a q-linearized polynomial in X, that is, a k-linear combination of X, X^{q}, X^{q^{2}}, . . .. (See [Gos96, Corollary 1.2.2], for instance.) It has distinct roots, namely the elements of V. Therefore P_{V,ε}(X), considered as a polynomial in X, has distinct roots, which constitute a translate of V in the separable closure of k(ε). Moreover, P_{V,ε}(X) is irreducible, because it is a monic polynomial in ε of degree 1. Since k is Hilbertian, it contains c ≠ 0 such that P_{V,c} ∈ k[X] is irreducible. Let α be a zero of P_{V,c}. Then α is an element of k^{s} of degree #V. Since the set of conjugates of α is {α + v | v ∈ V }, the k-span of this set equals the span of V ∪ {α}. However α ∉ L since d(α) = #V ⩾ q^{n−1} > n − 1. So, as the k-span of V is L, n(α) = [L : k] + 1 = n. Thus D(k, n) ⩾ #V. Since V can be taken arbitrarily large, D(k, n) = ∞. □

4.2. Results for cyclotomic fields. Theorem 1 generalizes to finite cyclotomic extensions of ℚ. Let ω_{ℓ} be a primitive ℓ-th root of unity.

Theorem 15. Fix an integer n ⩾ 0 and an even integer ℓ ⩾ 4. If α ∈ \overline{ℚ} has conjugate dimension n over ℚ(ω_{ℓ}) then the degree d of α over ℚ(ω_{ℓ}) satisfies

n ⩽ d ⩽ D(ℚ(ω_{ℓ}), n),

where D(ℚ(ω_{ℓ}), n) is defined by Table 2. In particular, D(ℚ(ω_{ℓ}), n) = ℓ^{n}n! for

(n, ℓ) ∉ {(2, 4), (2, 8), (2, 10), (2, 20), (4, 4), (4, 6), (4, 10), (5, 4), (6, 6), (6, 10), (8, 4)}.

Furthermore, for each pair (n, ℓ) with n ⩾ 1 and ℓ ⩾ 4 even, there exist α ∈ \overline{ℚ} attaining the lower and upper bounds.

Table 2 is a list of groups isomorphic to maximal-order finite subgroups G of GL_{n}(ℚ(ω_{ℓ})), quoted from Feit [Fei96]. (An error in the first line of his table has been corrected.) In

n ℓ  D(ℚ(w_{ℓ}), n)/(ℓ^{n}n!) Maximal-order subgroup G D(ℚ(ω_{ℓ}), n) = #G
2 4  3 ST_{8} = ⟨GL_{2}(𝔽_{3}), ω_{4}I⟩  96
2 8  3/2 ST_{9} = ⟨GL_{2}(𝔽_{3}), ω_{8}I⟩  192
2 10  3 ST_{16} = ⟨ω_{5}I⟩ × SL_{2}(𝔽_{5})  600
2 20  3/2 ST_{17} = ⟨SL_{2}(𝔽_{5}), ω_{20}I⟩  1200
4 4  15/2 ST_{31}  46080
4 6  5 ST_{32}  155520
4 10  3 ST_{16} ≀ S_{2}  720000
5 4  3/2 ST_{31} × ⟨ω_{4}I⟩  184320
6 6  7/6 ST_{34}  39191040
6 10  9/5 ST_{16} ≀ S_{3}  1296000000
8 4  45/28 ST_{31} ≀ S_{2}  4246732800
all other (n, ℓ), ℓ ⩾ 4 even 1 ST_{2}(ℓ, 1, n) = (ℤ/ℓℤ)^{n} ⋊ S_{n}  ℓ^{n}n!

Table 2. Maximal-order subgroups of GL_{n}(ℚ(ω_{ℓ})) for ℓ ⩾ 4 even

this table ST_{j} refers to the j-th unitary reflection group in [ST54, Table VII], and wreath product G ≀ S_{n} is the semidirect product (G × ⋯ × G) ⋊ S_{n} in which S_{n} acts on the n-fold product of G by permuting the coordinates. See also [Smi95, Table 7.3.1].

Proof. The proof is a generalization of that of Theorem 1. For fixed ℓ, D(ℚ(ω_{ℓ}), n) is a strictly increasing function of n. Thus to carry over the proof, it remains to show that the invariant subfield ℚ(ω_{ℓ})(x_{1}, …, x_{n})^{G} is purely transcendental over ℚ(ω_{ℓ}) in each case of Table 2. This is immediate for all the Shephard-Todd groups in the table, by the extension of Chevalley’s Theorem to unitary reflection groups by Shephard and Todd ([ST54]; see also [Bou81, p. 115, Thm. 4] and [Hum90, p. 65]). For example, when G = (ℤ/ℓℤ)^{n} ⋊ S_{n}, the field of invariants ℚ(ω_{ℓ})(x_{1}, …, x_{n})^{G} is ℚ(ω_{ℓ})(e_{1}, …, e_{n}), where e_{j} is the j-th elementary symmetric function of x_{1}^{ℓ}, …, x_{n}^{ℓ}. The three remaining cases are handled by Lemma 17 below. □

Lemma 16. Let k be a field. Let the symmetric group S_{m} act on

K = k(x_{1}^{(1)}, …, x_{1}^{(m)}; …; x_{n}^{(1)}, …, x_{n}^{(m)})

by acting on the superscripts. Then K^{S_{m}} is purely transcendental over k.

Proof. If E/F is a Galois extension of fields with Galois group G, and V is an E-vector space equipped with a semilinear action of G, there exists an E-basis of V consisting of G-invariant vectors [Sil92, II.5.8.1].

Apply this to E = k(x_{1}^{(1)}, …, x_{1}^{(m)}), G = S_{m}, F = E^{G} (the purely transcendental extension of k generated by the symmetric functions in x_{1}^{(1)}, …, x_{1}^{(m)}), and V the E-subspace of K spanned by all the x_{i}^{(j)} with i ≥ 2. Choose an E-basis {v_{s}} of G-invariant vectors as above. Let K_{0} = k({v_{s}}). Since EK_{0} = K, we have [K : K_{0}] ≤ [E : F] = m!, On the other hand, K_{0} ⊆ K^{G} with [K : K^{G}] = m!, so K_{0} = K^{G}. Since the x_{i}^{(j)} are algebraically independent over E, the v_{s} are algebraically independent over k. □

Lemma 17. Let k be a field, and let G be a finite subgroup of GL_{n}(k) whose field of invariants k(x_{1}, …, x_{n})^{G} is purely transcendental over k. Let G ≀ S_{m} act on

L = k(x_{1}^{(1)}, …, x_{n}^{(1)}; …; x_{1}^{(m)}, …, x_{n}^{(m)})

by letting the i-th of the m copies of G act linearly on the span of x_{1}^{(i)}, …, x_{n}^{(i)} while S_{m} acts on the superscripts. Then L^{G≀S_{m}} is purely transcendental over k.

Proof. Since G ≀ S_{m} is a semidirect product of S_{m} by G^{m}, we have L^{G≀S_{m}} = (L^{G^{m}})^{S_{m}}. If k(x_{1}, …, x_{n})^{G} = k(I_{1}, …, I_{n}), then

L^{G^{m}} = k(I_{1}^{(1)}, …, I_{n}^{(1)}; …; I_{1}^{(m)}, …, I_{n}^{(m)}),

and S_{m} acts on this by acting on superscripts. Now apply Lemma 16. □

Example. Using the elimination procedure outlined in Section 3.2, we can give an example of an algebraic number α of degree 96 over ℚ(i) with ℚ(i)-conjugate dimension 2 and Galois group ST_{8}, as in Table 2. Now ST_{8} = ⟨( \begin{matrix}0 & 1 \\ 1 & i \end{matrix}) , ( \begin{matrix}0 & 1 \\ −i & 0 \end{matrix} )⟩, with invariants

I_{8}(x_{1}, x_{2}) = x_{1}^{8} + 4(1 + i) x_{1}^{7} x_{2} + 14 i x_{1}^{6} x_{2}^{2} − 14(1 − i) x_{1}^{5} x_{2}^{3} − 21 x_{1}^{4} x_{2}^{4} − 14(1 + i) x_{1}^{3} x_{2}^{5} − 14 i x_{1}^{2} x_{2}^{6} + 4(1 − i) x_{1} x_{2}^{7} + x_{2}^{8}
I_{12}(x_{1}, x_{2}) = 2 x_{1}^{12} + 12(1 + i) x_{1}^{11} x_{2} + 66 i x_{1}^{10} x_{2}^{2} − 110(1 − i) x_{1}^{9} x_{2}^{3} − 231 x_{1}^{8} x_{2}^{4}

− 132(1 + i) x_{1}^{7} x_{2}^{5} − 132(1 − i) x_{1}^{5} x_{2}^{7} − 231 x_{1}^{4} x_{2}^{8} − 110(1 + i) x_{1}^{3} x_{2}^{9} − 66 i x_{1}^{2} x_{2}^{10} + 12(1 − i) x_{1} x_{2}^{11} + 2 x_{2}^{12}.

The x_{2}-resultant of I_{8} − 1 − i and I_{12} − 1 is P_{24}(x_{1})^{4}, where the auxiliary polynomial P_{24} is

P_{24}(x) = 27 x^{24} − 270(1 + i) x^{16} + 270 x^{12} − 810 i x^{8} + 54(1 + i) x^{4} − 9 + 8 i.

Two zeros β and β′ of P_{24} can be chosen so that the conjugates of β are

ωβ, ωβ′, ω(β + β′), ω(β − iβ′), ω(β + (1 − i)β′), ω((1 + i)β + β′),

where ω ∈ {±1, ±i}. Then α = β + 2β′ has degree 96 over ℚ(i), with conjugates (β, β′)g(1, 2)^{T} for g ∈ ST_{8}. The minimal polynomial of α can be computed directly as the x_{2}-resultant of I_{8}(y − 2x_{2}, x_{2}) − 1 − i and I_{12}(y − 2x_{2}, x_{2}) − 1.

4.3. D(k, n) depends on more than ℓ and n. Let k be a number field, and let ℓ be the number of roots of unity in k. It seems reasonable to guess, as in the case of cyclotomic fields ℚ(ω_{ℓ}), that D(k, n) = ℓ^{n}n! for all but finitely many n. However, it is possible that two number fields k and k′ contain the same number of roots of unity, but D(k, n) ≠ D(k′, n) for some n. For example, we can take k = ℚ(cos(2π/m), sin(2π/m)), where m > 6, and k′ = ℚ. In both cases ℓ = 2, but D(k, 2) > D(ℚ, 2) = 12. Indeed, there exist a, b ∈ k such that α = \sqrt[m]{a}(1 + bω_{m}) is of degree 2m > 12 over k. Its conjugate dimension over k is 2; its conjugates are spanned by \sqrt[m]{a} and i \sqrt[m]{a}. This example also shows that the number of exceptional cases can be arbitrarily large, since we may simply take m with 2m > 2^{n}n!.

Another example is D(ℚ(\sqrt{5} ), 3) ⩾ 120, obtained from the icosahedral subgroup of GL_{3}(ℚ(\sqrt{5} ) (reflection group ST_{23}) via Propositions 5 and 8.

5. Multiplicative conjugate rank

Instead of the dimension n(α) of the ℚ-vector space spanned by the d conjugates α_{i} of an algebraic number α, we may consider the rank r(α) of the multiplicative subgroup of \overline{ℚ}^{∗} they generate. We call this the (multiplicative) conjugate rank of α. As before, we have the trivial inequality r(α) ⩽ d(α), which is sharp in the case of maximal Galois group (again by [Smy86, Lemma 1]). Unlike in the additive case, we can have no nontrivial lower bound without some further hypothesis, because if α is a root of unity then r(α) = 0 while d(α) is unbounded. However, also unlike the additive case, we have the following result over a very general field. The main difficulty in the proof below is to show that this bound is sharp for Hilbertian fields.

Theorem 18. Suppose that α is separable and algebraic of degree d(α) over a field k, and the multiplicative subgroup of (k^{s})^{∗} generated by the conjugates α_{1}, …, α_{d} of α is torsionfree. Then the rank r(α) of this subgroup satisfies r(α) ⩽ d(α) ⩽ d_{max}(r(α)), with d_{max}(·) defined by Table 1 as before. If k is Hilbertian, then for each integer r ⩾ 1 there are α ∈ k^{s} of conjugate rank r attaining the lower and upper bounds.

The upper bound is given by the same function d_{max}(·) that we found for the conjugate dimension over ℚ, and this bound is independent of the ground field k, although it need not always be sharp.

Proof. For any α ∈ k^{s}, let Γ = Γ(α) be the multiplicative group generated by the α_{i}. We observed already that the lower bound d(α) ⩾ r(α) is immediate. For the upper bound, we argue as we did for n(α). The Galois group G acts faithfully on Γ. By hypothesis, Γ ≅ ℤ^{r(α)}, so G acts faithfully also on Γ ⊗_{ℤ} ℚ, which is a ℚ-vector space of dimension r(α). Hence #G is bounded above by d_{max}(r(α)), the size of the largest finite subgroup of GL_{r(α)}(ℚ). Hence d(α) ⩽ #G ⩽ d_{max}(r(α)).

The proof that there are examples attaining equality when k is Hilbertian uses two corollaries of the following technical result.

Proposition 19. Let L/k be a finite Galois extension of fields with Galois group G, and suppose that k is not algebraic over a finite field. Then the ℤG-module L^{∗} contains a free ℤG-module of rank 1.

Proof. For each g ∈ G−{1}, choose a_{g} ∈ L that is not fixed by g. Choose b ∈ L that is not algebraic over a finite field. Let S be the union of the G-orbits of the a_{g} and of b. Then S is finite. Let L_{0} be the minimal subfield of L containing S. Let k_{0} be the subfield (L_{0})^{G} fixed by G. The action of G on S is faithful, so G acts faithfully on L_{0}, and L_{0}/k_{0} is Galois with group G. In this way we reduce to the case where k and L are finitely generated fields (finitely generated over their minimal subfield).

Choose finitely generated ℤ-algebras A ⊆ B with fraction fields k and L, respectively. Without loss of generality we may assume, by localization, that B is a finite étale Galois algebra over A. Since L is not algebraic over a finite field, dim A = dim B ⩾ 1. By [Poo01, Theorem 4], there is a maximal ideal 𝔪_{1} of B lying over a maximal ideal 𝔪 of A such that the residue field extension B/𝔪_{1} over A/𝔪 is trivial. Thus 𝔪 splits completely: if n = #G, there are n distinct maximal ideals 𝔪_{1}, …, 𝔪_{n} of B lying over 𝔪, and they are are permuted transitively by G. By [AM69, Proposition 1.11], there exists a nonzero β ∈ 𝔪_{1} lying outside all of 𝔪_{2}, …, 𝔪_{n}. We can label the conjugates β_{i} of β so that β_{i} ∈ 𝔪_{j} if and only if i = j. Any nontrivial relation ∏_{i=1}^{n} β_{i}^{b_{i}} = 1 with b_{i} ∈ ℤ, would, after moving the factors with negative exponent to the other side, give an equality between an element in 𝔪_{i} and an element outside 𝔪_{i}, for some i. Hence the ℤG-module generated by β in L^{∗} is free of rank 1. □

Corollary 20. Let k be a field that is not algebraic over a finite field. If k has a Galois extension with Galois group S_{r}, then there exists α ∈ (k^{s})^{∗} with r(α) = d(α) = r.

Proof. Let L be the S_{r}-extension of k. By Proposition 19, the ℤS_{r}-module L^{∗} contains a copy of ℤS_{r}, which contains a copy of the ℤS_{r}-module ℤ^{r} on which S_{r} acts by permuting coordinates. The element (1, 0, …, 0) ∈ ℤ^{r} corresponds to α ∈ L^{∗} with the desired properties. □

Corollary 21. Let k be a field that is not algebraic over a finite field, and let G be a finite group. Suppose that G = Gal(K/k) for some Galois extension K of k, and that there is a faithful r-dimensional subrepresentation V of the regular representation of G over k. Then there exists α ∈ K^{∗} with r(α) = r and d(α) = [K : k] = #G.

Proof. Apply Proposition 19 and then Lemma 3 with k = ℚ. This gives α ∈ K^{∗} ⊗_{ℤ}ℚ with the desired properties, and we replace α by a power so that it is represented by an element of K^{∗}. □

We now prove the final statement of Theorem 18. Since k is Hilbertian, k has S_{r}- extensions for all r. In particular, k is not algebraic over a finite field. Applying Corollary 20 yields α with r(α) = d(α) = r. Combining Corollaries 9 and 21 gives a different α with r(α) = r and d(α) = d_{max}(r), for any r ⩾ 1. □

We end by giving an explicit algebraic number of conjugate rank n and degree 2^{n}n! over ℚ.

Proposition 22. Let \sqrt{r_{1}}, …, \sqrt{r_{n}} be as in Proposition 12. Let s_{i} = (1 + \sqrt{r_{i}})/(1 − \sqrt{r_{i}}) and α = s_{1}s_{2}^{2} ⋯ s_{n}^{n}. Then r(α) = n and d(α) = 2^{n}n! over ℚ.

Proof. The proof of Proposition 12 showed that [ℚ(\sqrt{r_{1}}, …, \sqrt{r_{n}}) : ℚ] = 2^{n}n!, so its Galois group G is the signed permutation group W (B_{n}). The elements of G act on α by permuting the exponents 1, 2, …, n and changing their signs independently. In particular, the group generated by the conjugates of α is of finite index in the subgroup generated by the s_{i}. On the other hand, the s_{i} are multiplicatively independent since they are not roots of unity and since there is an automorphism inverting any one of them while fixing all the others. Thus α has 2^{n}n! distinct conjugates, and they generate a subgroup of rank n. □

Acknowledgments

We thank Hendrik Lenstra for suggesting the proof of Lemma 16, and Walter Feit for some helpful correspondence. N.D.E. also thanks David Moulton for communicating the topic to him, as a problem on the list compiled by Gerry Myerson at the 2001 Asilomar meeting. N.B. was supported by a UK EPSRC postgraduate award. A.D. was supported in part by the Lithuanian State Studies and Science Foundation, and the London Mathematical Society. N.D.E. was supported in part by NSF grant DMS-0200687. B.P. was supported in part by NSF grant DMS-0301280 and a Packard Fellowship.

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School of Mathematics, University of Edinburgh, James Clerk Maxwell Building, King’s Buildings, Mayfield Road, Edinburgh EH9 3JZ, Scotland, U.K.

E-mail address: neilb@maths.ed.ac.uk

Department of Mathematics and Informatics, Vilnius University, Naugarduko 24, Vilnius 2006, Lithuania

E-mail address: arturas.dubickas@maf.vu.lt

Department of Mathematics, Harvard University, Cambridge, MA 02138, USA

E-mail address: elkies@math.harvard.edu

Department of Mathematics, University of California, Berkeley, CA 94720-3840, USA

E-mail address: poonen@math.berkeley.edu

School of Mathematics, University of Edinburgh, James Clerk Maxwell Building, King’s Buildings, Mayfield Road, Edinburgh EH9 3JZ, Scotland, U.K.

E-mail address: c.smyth@ed.ac.uk