Recall \(\mathcal{O}_{K} \subseteq K_{\mathbb{R}}\) with respect to the Minkowski’s embedding.
Let \(K_{\mathbb{R}}^{*} \subseteq K_{\mathbb{R}}\) be the set \(\{ x \in K_{\mathbb{R}} \ | \ \mathop{\mathrm{N}}(x) \neq 0 \}\). From this, we can define the following logarithm mapping \[\begin{align} \mathop{\mathrm{\mathbf{log}}}: K_{\mathbb{R}}^{*} \rightarrow \ \ \ \mathbb{R}^{r_1 + r_2} \end{align}\] \[\begin{equation} x \mapsto \left( \sigma \mapsto \begin{cases} \log | \sigma(x)| & \text{ real } \\ 2\log | \sigma(x)| & \sigma \text{ complex } \\ \end{cases} \right) \end{equation}\]
This is a homomorphism of abelian groups. We know that \(\mathcal{O}_{K}^{*} \subset K_{\mathbb{R}}^{*}\) maps under this to a discrete subgroup of rank \(r_1+r_2-1\) and has a kernel of exactly the torsional units. The image of the \(\mathop{\mathrm{\mathbf{log}}}\) embedding is the vector subspace of \(\mathbb{R}^{r_1+r_2}\) perpendicular to the vector \((1,1,1,\cdots)\)
Dirichlet’s unit theorem says that the group \(\mathcal{O}_{K}^{*}\) has finite covolume inside the abelian group \(K_{\mathbb{R}}^{(1)} = \{ x \in K_{\mathbb{R}} \ | \ |\mathop{\mathrm{N}}(x)|=1\}\). And the kernel of \(\mathcal{O}_K^{*} \rightarrow \mathop{\mathrm{\mathbf{log}}}\left( \mathcal{O}_K^{*} \right)\) is a cyclic group.
The covolume of \(\mathop{\mathrm{\mathbf{log}}}(\mathcal{O}_{K}^{*}) \subseteq (1,1,\cdots)^{\perp}\) is finite. It is equal to the regulator of the field \(K\), up to a constant. The regulator is denoted as \(R_K\).
The kernel of \(\mathcal{O}_K^{*} \rightarrow \mathop{\mathrm{\mathbf{log}}}\left( \mathcal{O}_K^{*} \right)\) is a cyclic group whose size is denoted as \(w_K\).
(DUT) \[\begin{align} \mathcal{O}_{K}^{*} \simeq \left( \frac{\mathbb{Z}}{ w_{K} \mathbb{Z}} \right) \times \mathbb{Z}^{r_1+r_2-1}. \end{align}\]
If we have a finite index subring \(\mathcal{O} \subseteq \mathcal{O}_{K}\), then one can still conclude Dirichlet’s unit theorem. This is because \(\mathcal{O}_{K}^{\times}/ \mathcal{O} \cap \mathcal{O}_{K}^{\times} \subseteq \mathcal{O}_{K}/\mathcal{O}\) is a finite set and therefore \(\mathcal{O}^{\times} \subseteq \mathcal{O}_{K}\) is a finite index sublattice.
Finite index unital subgrings of \(\mathcal{O}_{K}\) are called orders.
DUT basically asserts that \(K_{\mathbb{R}}^{(1)}/\mathcal{O}_{K}^{*}\) is compact. This is basically a Lie group modulo an arithmetic subgroup. The fact that this space is compact also follows from a special case of Borel-Harish-Chandra theorem
To do this, one has to consider the group \(K^{\times}/\mathbb{Q}^{\times}\) as an arithmetic group over \(\mathbb{Q}\).
This page was updated on January 21, 2026.
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