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

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

However, we want to make sure that $A$ `makes sense' as a linear operator of $\mathbb{Q}^n$, say. Therefore it is prudent to find a representation of the unit group $\mathbb{O}_K^\ast$, which can be realized as a subgroup of $\operatorname{GL}_d(\mathbb{Z})$, in the general linear group $\operatorname{GL}_n(\mathbb{Z})$. Does anyone know how to go about this problem?