RINGS CONTAINING A FIELD OF CHARACTERISTIC ZERO

DAVE BENSON

This note was prompted by recent email exchanges with various colleagues. We use the term “ring” to mean a ring with identity, not necessarily commutative. For a ring R, denote by R^{×} the multiplicative group of invertible elements of R.

Theorem. Let K be a field of characteristic zero, and let R be a ring containing K. Then either R^{×} = K^{×}, or K^{×} is a subgroup of infinite index in R^{×}.

Proof. Suppose, to the contrary, that |R^{×}: K^{×}| = m > 1. Letting d = m!, there is a proper normal subgroup of index dividing d in R^{×}, and contained in K^{×}. Choose s ∈ R^{×}, s ∉ K^{×}. Then s^{d} is some element of K^{×}. Let n ∈ ℤ ⊆ K with n^{d} ≠ s^{d}. Then

(s − n)(s^{d−1} + ns^{d−2} + ⋯ + n^{d−1}) = s^{d} − n^{d}

is in K^{×}, and hence s − n has a multiplicative inverse. So s − n ∈ R^{×}, and hence (s − n)^{d} ∈ K^{×}.

Choose d + 1 different values of n with n^{d} ≠ s^{d}, say n_{0}, …, n_{d}. Then we have (s−n_{i})^{d} ∈ K^{×} for i = 0, …, d. Expanding out these equations, and using the fact that we are in characteristic zero, the non-vanishing of the Vandermonde determinant then implies that 1, s, …, s^{d} are all in K. But s was chosen not to be in K. This contradiction proves the theorem. □

Institute of Mathematics, University of Aberdeen, Fraser Noble Building, Aberdeen AB24 3UE

Date: August 29, 2017.