## 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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