Units in number fields

For no good reason, I find it a surprising fact that there are algebraic integers of modulus 1 that are not roots of unity. Clearly it is my intuition that needs adjusting, because as this paper by R. Daileda shows, such elements exist in any algebraic number field \mathbf{Q}(\alpha) such that \mathbf{Q}(\alpha) \cap \mathbf{R} admits a complex field embedding.

Here is a version of Daileda’s argument that gives a stronger group theoretic result. Given an algebraic number field K, let U(K) denote the set of units of the ring of algebraic integers \mathcal{O}_K, and let V(K)= \{\beta \in U(K) : |\beta| = 1\}. If \beta \in U(K) then we have

\beta^2 = \frac{\beta}{\beta^\star} |\beta|^2

where \beta^\star is the complex conjugate of \beta. It follows that

U(K)^2 \subseteq V(K) \times U(K \cap \mathbf{R}).

Either every unit of \mathcal{O}_K is a square, or U(K)^2 has index 2 in U(K). In the first case U(K) = V(K) \times U(K \cap \mathbf{R}), and in the second U(K) has index 2 in V(K) \times U(K \cap \mathbf{R}).

Let K \cap \mathbf{R} = \mathbf{Q}(\gamma). We apply Dirichlet’s Unit Theorem to K = \mathbf{Q}(\alpha) and Q(\gamma). By Dirichlet, the torsion-free part of U(K) has rank r(\alpha) + s(\alpha) - 1, where r(\alpha) is the number of real conjugates of \alpha and s(\alpha) is the number of pairs of complex conjugates; similarly, the torsion-free part of U(K \cap \mathbf{R}) has rank r(\gamma) + s(\gamma)-1. Hence

r(\alpha) + s(\alpha) \ge r(\gamma) + s(\gamma)

with equality if and only if V(K) is a torsion group, i.e. if and only if every algebraic integer in \mathcal{O}_K of modulus 1 is a root of unity.

By applying the Tower Law to the chain of extensions K > K \cap \mathbf{R} > \mathbf{Q} we obtain

r(\alpha) + 2s(\alpha) = [K : K \cap \mathbf{R}] (r(\gamma) + 2s(\gamma)).

It follows that

r(\alpha) \ge (2-[K : K \cap \mathbf{R}]) r(\gamma) + 2(1-[K : K \cap \mathbf{R}]) s(\gamma).

The left-hand side is non-negative, and the right-hand side is non-positive. Hence every algebraic integer in \mathcal{O}_K of modulus 1 is a root of unity if and only if r(\alpha) = 0, [K : K \cap \mathbf{R}] = 2 and s(\gamma) = 0. In particular, if s(\gamma) \ge 1 then, as claimed, there are units in \mathcal{O}_K of modulus 1 that are not roots of unity.

For example, let \gamma = 2^{1/3}, let \omega = \exp(2\pi \mathrm{i}/3) and let K = Q(\gamma, \omega). Equivalently, since K is generated by \gamma \omega, we can think of K as the splitting field of X^6 + 108 over \mathbf{Q}. A laborious calculation with discriminants shows that \mathcal{O}_K = \mathbf{Z}[\gamma,\omega]. Since s(\gamma) = 1, there should be algebraic integers in \mathcal{O}_K of modulus 1 that are not roots of unity. A brute force search shows that

u = -2 + 2\gamma - \gamma^2 + \omega(1 + \gamma -2\gamma^2)

is one such algebraic integer. Its minimum polynomial is

X^6 + 15X^5 + 78X^4 + 101X^3 + 78X^2 + 15X + 1.

(A simpler real unit is -1+\gamma.)

The answer by KcD to this question on Math.Stackexchange gives a similar example with a polynomial of degree 4, and explains the characteristic palindromic pattern of the polynomial above.

It is worth noting that the condition that \mathbf{Q}(\alpha) \cap \mathbf{R} has a complex embedding implies that complex conjugation is not in the centre of the Galois group of \mathbf{Q}(\alpha) : \mathbf{Q}. Therefore S_3 is the smallest possible Galois group for a field extension containing algebraic integer units that do not have modulus 1.


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