Dirichlet's Unit Theorem

We prove $\mathcal{O}_K^\times \cong \mu_K \times \mathbb{Z}^r$ in four steps.

Step 1 — The kernel of $\ell$ is $\mu_K$. Let $u \in \mathcal{O}_K^\times$ with $\ell(u) = 0$. Then $|\sigma(u)| = 1$ for every archimedean embedding $\sigma$. Hence $u$ and all its conjugates lie on the unit circle, so the set $\{u, u^2, u^3, \dots\}$ has all conjugates bounded by $1$. The minimal polynomials of $u^m$ are monic integer polynomials whose roots are bounded, so their coefficients are bounded; only finitely many such polynomials exist, forcing the powers $u^m$ to take finitely many values. Therefore $u^a = u^b$ for some $a > b$, so $u^{a-b} = 1$ and $u$ is a root of unity. Conversely roots of unity have $|\sigma(\zeta)| = 1$ everywhere, so $\ell(\zeta) = 0$. Thus

Derivation hint: Kronecker's theorem — an algebraic integer all of whose conjugates have absolute value $1$ is a root of unity; finiteness comes from boundedness of conjugates plus integrality.

Step 2 — $\Gamma := \ell(\mathcal{O}_K^\times)$ is a discrete subgroup (a lattice) of $H$. For any bound $B$, the units $u$ with $|\sigma(u)| \le e^B$ for all $\sigma$ form a finite set: such $u$ are algebraic integers with bounded conjugates, hence have minimal polynomials with bounded integer coefficients. Intersecting $\Gamma$ with any bounded ball thus gives finitely many points, so $\Gamma$ is discrete. A discrete subgroup of a finite-dimensional real vector space is a free abelian group whose rank equals the dimension of the $\mathbb{R}$-span. Hence $\Gamma$ is a lattice of some rank $s \le \dim H = r$. Derivation hint: Discreteness ⇒ free of rank = dimension of $\mathbb{R}$-span; finiteness of bounded units gives discreteness.

Step 3 — Structure of $\mathcal{O}_K^\times$. The short exact sequence $1 \to \mu_K \to \mathcal{O}_K^\times \xrightarrow{\ell} \Gamma \to 0$ has $\Gamma \cong \mathbb{Z}^s$ free, so it splits:

Derivation hint: An extension of a free abelian group by any group splits, because $\mathbb{Z}^s$ is projective. It remains to show $s = r$, i.e. $\Gamma$ has full rank in $H$.

Step 4 — Full rank via Minkowski's lattice-point theorem. This is the crux. We must produce enough units so that $\ell(\mathcal{O}_K^\times)$ spans $H$. Equivalently: for each coordinate index $k$, there is a unit whose $\ell$-image is not nonpositive in coordinate $k$ alone — concretely, it suffices to show that for every nonzero linear functional vanishing on... we instead argue directly. Embed $\mathcal{O}_K$ as a full lattice $\Lambda$ in the Minkowski space $\mathbb{R}^{r_1} \times \mathbb{C}^{r_2} \cong \mathbb{R}^n$ via $x \mapsto (\sigma_i(x); \tau_j(x))$; its covolume is $2^{-r_2}\sqrt{|d_K|}$ where $d_K$ is the discriminant. Fix a target index $m$. Choose positive reals $c_1, \dots, c_{r_1+r_2}$ with $\prod_k c_k^{e_k} > (2/\pi)^{r_2}\sqrt{|d_K|}$ (here $e_k = 1$ for real, $2$ for complex places). By Minkowski's theorem, the symmetric convex region $\{|\sigma_i(x)| \le c_i,\ |\tau_j(x)| \le c_{r_1+j}\}$ contains a nonzero $x \in \mathcal{O}_K$. Its norm satisfies $1 \le |N(x)| \le \prod_k c_k^{e_k} =: C$ (bounded independently of which $c_k$ we shrink/grow). By varying the $c_k$ so that one chosen coordinate is forced small while their product stays $\ge$ the Minkowski bound, one generates infinitely many algebraic integers $x_i$ with $|N(x_i)| \le C$. Since only finitely many ideals have norm $\le C$, infinitely many of the principal ideals $(x_i)$ coincide; then ratios $x_i/x_j$ are units, and the construction can be arranged so that these units' logarithmic images point "in a chosen direction." Doing this for each direction shows $\Gamma$ is not contained in any proper subspace of $H$, hence

Derivation hint: Minkowski guarantees lattice points in large symmetric convex bodies; pigeonholing by ideal norm produces units, and steering the convex body's shape steers their log-images to span $H$. Combining Steps 3 and 4 gives $\mathcal{O}_K^\times \cong \mu_K \times \mathbb{Z}^{r}$. $\blacksquare$