Characterizing the Multiplicative Group of a Real Closed Field in terms of its Divisible Maximal Subgroup

M. Mahdavi-Hezavehi

Abstract

Let F be a field and M be a maximal subgroup of the multiplicative group F^{∗} = F \ {0}. It is proved that if M is divisible, then F is Euclidean. Furthermore, it is shown that F^{∗} contains a divisible maximal subgroup if and only if F^{∗} is isomorphic to the multiplicative group of a real closed field.

1 Introduction

Given the field of real numbers ℝ, denote by ℝ^{∗} and ℝ^{+} the multiplicative group of real numbers and the multiplicative group of positive real numbers, respectively. We recall that a nontrivial multiplicative abelian group G is divisible if and only if G has no maximal subgroup if and only if G = G^{p} for each prime p. It is easily seen that ℝ^{+} is a divisible maximal subgroup of ℝ^{∗} and ℝ is Euclidean. The object of this note is to show that this property on the multiplicative group of a field F gives rise to F being Euclidean. Furthermore, if R is a real closed field, then it is easily seen that (cf. Theorem A below) R^{∗} contains a unique maximal subgroup which is divisible. Here, we also characterize the multiplicative group of a real closed field in terms of its divisible maximal subgroup. To be more precise, it is proved that F^{∗} contains a divisible maximal subgroup if and only if F^{∗} is isomorphic to the multiplicative group of a real closed field. We begin our investigation with the following easy

Lemma 1. Let G be a multiplicative abelian group and M be a maximal subgroup of G. If M is divisible, then M is the unique maximal subgroup of G.

Proof. Assume that M_{1} ≠ M is another maximal subgroup of G. Then, we have G = MM_{1} and hence G/M_{1} ≅ M/M ∩ M_{1}, i.e., M ∩ M_{1} is a maximal subgroup of M. Since M is divisible we conclude that M ∩M_{1} = 1. Therefore, G/M_{1} ≅ M ≅ C_{q} for some prime number q, where C_{q} is the cyclic group of q elements. This last relation also leads to a contradiction since a finite group cannot be divisible, and so M = M_{1} as required. □

We shall also need the following theorem to prove our main result:

Theorem A [1, p. 107]. If F is a real closed field, then F^{∗} ≅ ℤ_{2}×ℚ^{|F |}. Conversely, for any infinite cardinal λ, the group ℤ_{2}×ℚ^{λ} is isomorphic to the multiplicative group of a suitable real closed field.

Theorem 1. Let M be a maximal subgroup of F^{∗}. Then, we have

(1) If M is divisible, then the Brauer group of F is non-trivial and F is Euclidean.

(2) F^{∗} contains a divisible maximal subgroup if and only if F^{∗} is isomorphic to the multiplicative group of a real closed field.

Proof. (1) Assume that M is a maximal subgroup of F^{∗} with F^{∗}/M ≅ C_{p} for some prime p. We first claim that there exists a cyclic field extension K/F of degree p such that N(K^{∗}) = F^{∗p} = M, where N is the norm of K to F. To see this, we know, by Lemma 1, that M is the unique maximal subgroup of F^{∗} such that F^{∗}/M ≅ C_{p}. Since M is divisible and maximal in F^{∗}, by 4.1.4 of [3], we have F^{∗} ≅ M × C_{p}. This means that F contains a primitive p-th root of unity. Now, it is easily seen that there is an element a ∈ F such that the equation x^{p} − a = 0 has no solutions in F. Since F has a primitive p-th root of unity we obtain a cyclic extension K = F (b) of degree p over F with b^{p} − a = 0. Now, if the norm N of K to F is surjective, i.e., N(K^{∗}) = F^{∗}, then K^{∗}/K^{1} ≅ F^{∗}, where K^{1} is the group of norm 1 elements. Hence K^{∗}/K^{1} contains a maximal subgroup M_{K}, say, containing K^{1} of index p which is divisible. By Lemma 1, M_{K} is unique. Now, M_{K} is also a maximal subgroup of K^{∗} and since it is divisible, by 4.1.4 of [3], we obtain K^{∗} ≅ M_{K} × C_{p}. This means that there exists an element 1 ≠ c ∈ C_{p} ⊂ K with c^{p} = 1. Since F contains a primitive p-th root of unity we conclude that c ∈ F. Therefore, N(c) = c^{p} = 1, i.e., c ∈ K^{1} ⊂ M_{K}. But this contradicts the fact that M_{K} ∩ C_{p} ≠ 1. Therefore, N(K^{∗}) ≠ F^{∗}. Since F^{∗}/N(K^{∗}) is torsion of bounded exponent p and M is unique, by Prũfer-Baer Theorem (cf. [3, p. 105]), we conclude that N(K^{∗}) = F^{∗p} = M, as claimed. Now, assume that the Galois group of K/F is generated by the automorphism σ of order p = [K : F]. Fix an element λ ∈ F^{∗} \ M and a symbol y. We set D = K1 ⊕ Ky ⊕ ⋯ ⊕ Ky^{p−1}, and multiply elements of D by using distributive law, and the rules y^{p} = λ, yk = σ(k)y for all k ∈ K. In this way, we obtain the cyclic algebra (K/F, σ, λ). Now, since M = N(K^{∗}) we conclude that λ ∉ N(K^{∗}). Thus, by Corollary 14.8 of [2], D is a division algebra and hence Br(F) ≠ 0. Finally, since K = F (b) with b^{p} = a ∈ F we obtain N(b) = (−1)^{p+1}a. Because N(K^{∗}) = F^{∗p} = M there is λ ∈ F such that N(b) = λ^{p}. Thus, (−1)^{p+1}b^{p} = λ^{p}. If p is odd, then in the presence of the primitive p-th root of unity in F one concludes that b ∈ F which is a contradiction and so p = 2. Thus, we have F^{∗} ≅ M × C_{2}, which shows that −1 ∉ M. The equation x^{2} + 1 = 0 over F has no root in F since a^{2} = −1 with a ∈ F implies that −1 ∈ M which is false. Now, consider the extension L = F (i) with i^{2} = −1. The above proof shows that N_{L/F} (L^{∗}) = M. We claim that M defines a positive cone for F. It is clear that M ∩ −M = ∅, MM ⊆ M, and M ∪ −M ∪ {0} = F. To show that M + M ⊆ M, take α, β ∈ M. Since M = F^{∗2}, there exist λ, μ ∈ F^{∗} such that α = λ^{2}, β = μ^{2}. Now, consider the element x = λ + μi ∈ L. We have N_{L/F} (x) = λ^{2} + μ^{2} = α + β ∈ M since N(L^{∗}) = M. Therefore, F is formally real and since M = F^{∗2} we conclude that F is Euclidean.

(2) One way is clear from Theorem A. If M is the divisible maximal subgroup of F^{∗}, then from the proof of (1) we have F^{∗} ≅ M × C_{2}. Since M is divisible from the theory of divisible abelian groups we know that M is a direct product of quasi-cyclic and full rational groups (cf. [1, p. 96] ). We claim that M contains no primitive p-th root of unity. Since −1 is not in M it suffices to consider p > 2. If ω is a primitive pth root of unity, then for p ≠ 2 we have 1^{2}+ω^{2}+⋯+ω^{2(p−1)} = (ω^{2p}−1)/(ω^{2}−1) = 0, which is not possible in a formally real field. Thus, we cannot have any copy of a quasi-cyclic group in our decomposition of M and hence M ≅ ℚ^{λ} for some cardinal λ. Since ℚ is of torsion-free rank 1, λ is the torsion-free rank of F^{∗}. Now, because CharF = 0 we have ℚ^{∗} ⊂ F^{∗}, and hence λ is infinite by Lemma 4.1.16 of [1] which asserts that Q^{∗} ≅ ℤ_{2} × ℤ^{ℵ_{0}}. Therefore, we have M ≅ ℚ^{λ} for some infinite cardinal λ. Now, by Theorem A, we obtain the result. □

We observe that in the conclusion of the theorem F need not necessarily be real closed. In fact, if F is obtained from the rationals ℚ by iteratively adjoining roots of positive real algebraic numbers, the positive cone of the resulting field F is such a maximal subgroup. But F is not real closed.

Acknowledgments. The author thanks the Research Council of Sharif University of Technology for support. Also, he wishes to thank M. G. Mahmudi for a useful discussion on the proof of the main theorem.

References

[1] G. Karpilosky, Unit Groups of Classical Rings, Clarendon press, Oxford, 1988.

[2] T. Y. Lam, A first course in non-commutative rings, GTM, No. 131, Springer-Verlag, 2nd ed. (2001).

[3] D. J. S. Robinson, A Course in the Theory of Groups, in: Grad. Text in Math., Vol. 80, Springer-Verlag, 1982.

mahdavih@sharif.edu Department of Mathematical Sciences Sharif University of Technology P. O. Box 11365-9415 Tehran, Iran