Let $K/\mathbb{Q}$ be a finite extension of degree $d > 1$. Suppose that $\omega_1, \cdots, \omega_d$ is a basis for $K$ over $\mathbb{Q}$. Further, we assume that $\omega_1, \cdots, \omega_d \in \mathcal{O}_K$, the ring of integers of $K$, and that it is an integral basis for $\mathcal{O}_K$.

Recall that a unit in $\mathcal{O}_K$ is an invertible element in $\mathcal{O}_K$, or equivalently, an element of norm $\pm 1$ in $\mathcal{O}_K$. By the norm we mean the function

$$\displaystyle N_{K/\mathbb{Q}}(u) = \prod_{j=1}^d \sigma_j(u),$$

where $\sigma_1, \cdots, \sigma_d$ are the distinct embeddings of $K$ into $\mathbb{C}$.

It is then easy to see that if $\alpha \in \mathcal{O}_K^\ast$, then for all $u \in \mathcal{O}_K$ we have

$$\displaystyle N_{K/\mathbb{Q}}(\alpha u ) = N_{K/\mathbb{Q}}(u).$$

If we consider the polynomial

$$\displaystyle N(x_1, \cdots, x_n) = N_{K/\mathbb{Q}}(\omega_1 x_1 + \cdots + \omega_n x_n)$$

for $n \leq d$, then it is clear that on writing $\mathbf{x} = (x_1, \cdots, x_n)$ and $A$ for the linear operator induced by multiplication by $\alpha \in \mathcal{O}_K^\ast$ that

$$\displaystyle N(A \mathbf{x}) = N(\mathbf{x}).$$

This equation only makes sense in the vector space $K/\mathbb{Q}$, because any unit in $\mathcal{O}_K^\ast$ can be realized as a linear endomorphism of $K \cong \mathbb{Q}^d$. However, we want to make sure that $A$ is defined over the linear span of $\{\omega_1, \cdots, \omega_n\}$, say $S$. Further, if $n < d$, we do not want $\{\omega_1, \cdots, \omega_n\}$ to all lie in a proper subfield of $K$. Indeed, we want $K = \mathbb{Q}(\omega_1, \cdots, \omega_n)$.

Hence, our goal is to find a subgroup of positive rank of the unit group $\mathcal{O}_K^\ast$ which can be realized as a subgroup of $\operatorname{GL}(S)$. Does anyone know how to go about this problem?

a(any old) representation, or do you intend to stick to the one induced by the multiplication? $\endgroup$ – GNiklasch Jul 15 '15 at 11:52