Representations of the unit group in a ring of integers

Question

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?

Answer 1

Notation: I'll use $S$ to denote the $\mathbb{Z}$-linear span of $\lbrace \omega_1, \ldots, \omega_n\rbrace$. (In the question, the $\mathbb{Q}$-span may have been meant.)

When you allow yourself to change the representation, and look for arbitrary homomorphisms from the abstract finitely generated abelian group $\mathcal{O}_K^\ast$ to $\mathrm{GL}(S)$, not much can be said, because too much can happen. For example, pick one non-torsion generator and map it to any invertible matrix, and map the other generators to the identity. (Here, $\mathrm{GL}(S)$ would just mean the group of invertible endomorphisms of the free abelian group $S$, an additive group which isn't a ring in general - so we can't even exploit the multiplication in $K$ to constrain such homomorphisms.)

So let us limit ourselves to multiplication by units. For simplicity, let me also assume $\omega_1=1$, and let us let a subgroup $G \le \mathcal{O}_K^\ast$ act.

In order for this to work, $S$ must contain at least the image of $\omega_1$, which is just $G$ again, and thus it must contain the additive group spanned by $G$, which is a ring - an image of the group ring $\mathbb{Z}[G]$. Its quotient field $L$ is just the $\mathbb{Q}$-linear span of $G$ inside the additive group of $K$ (because denominators can be replaced by their norms to make them rational integers). If $S$ is of smaller rank than $\mathcal{O}_K$, then $L$, being contained in the $\mathbb{Q}$-span of $S$, must be a proper subfield of $K$, and then $G$ must be contained in the units $\mathcal{O}_L^\ast$ of $L$. This will now give you a lot of examples, by taking $S$ no larger than necessary, but they may not be very interesting.

When $G$ is all of $\mathcal{O}_K^\ast$, the above means that all the units in $K$ come from a proper subfield $L$, and by Dirichlet's unit theorem this can only happen when $L$ is totally real and $K$ is a quadratic totally complex extension (a CM-extension) of $L$ with no roots of unity other than $\pm1$.