 Solution

The idele group is the multiplicative restricted productwith respect to ${\mathcal{O}}_v^{\times }$ at the finite places of a number field; $K_v$ is the completion of a valued field at the place $v$. Thus each tuple has nonzero components, and all but finitely many finite components are units. Its restricted product topology on the idele group has basic open sets ${\prod }_{v\in S}{W}_v\times {\prod }_{v\notin S}{\mathcal{O}}_v^{\times }$, where $S$ is finite and contains the infinite places, and each $W_v$ is open in $K_v^{\times }$. In particular $U_K$ is an open subgroup.

Embed $K^{\times }$ diagonally. Take a neighbourhood of one whose finite components all lie in ${\mathcal{O}}_v^{\times }$ and whose infinite components satisfy $|x_v-1|<1/2$ in the usual real or complex modulus. A diagonal element there is an algebraic unit $a$. If $a\ne 1$, then $a-1$ is a nonzero algebraic integer, so its field norm is a nonzero integer. Buta contradiction. Thus $K^{\times }$ is discrete. It is also closed: in a topological group, a subgroup with an isolated identity cannot have an external accumulation point, since quotients of two nearby subgroup elements would approach the identity.

Send an idele to its associated fractional ideal byOnly finitely many exponents are nonzero. This homomorphism is onto, by choosing powers of local uniformizers, and its kernel is $U_K$. Diagonal elements map to principal fractional ideals. The resulting quotient givesIt is a topological isomorphism when the ideal class group is given the discrete topology, since $U_K$ is open.

Use normalized local moduli: real modulus, squared complex modulus, and $|{\pi }_v{|}_v=(N{\mathfrak{p}}_v{)}^{-1}$ at a finite place. The idelic modulus $\Vert x\Vert ={\prod }_v|x_v{|}_v$ defines $J_K^1=\ker \Vert \cdot \Vert$, the norm-one idele group. The product formula puts $K^{\times }$ inside this kernel. Every ideal class has a representative in $J_K^1$, because an infinite component can be rescaled to correct the modulus without altering its fractional ideal. The compact space $J_K^1/K^{\times }$ therefore maps continuously onto the discrete ideal class group. Its image must be finite, proving $\mathrm{Cl}(K)$ is finite.

For the Dirichlet unit theorem, put $U^1=U_K\cap J_K^1$ and $E=K^{\times }\cap U^1={\mathcal{O}}_K^{\times }$. Let $r_1$ count real embeddings and $r_2$ count conjugate complex pairs. Infinite logarithms define a continuous surjectionusing $\log |x_v|$ at real places and $2\log |x_v|$ at complex places. Its kernel is compact: it consists of real signs, complex unit circles, and the product of compact finite-place unit groups. More generally the inverse image of a bounded closed subset of $H$ is compact. Since $E$ is closed and discrete, its intersection with each such inverse image is finite. Hence $\Lambda =\ell (E)$ is discrete in $H$, and the kernel $E\cap \ker \ell$ is a finite group. It is exactly the roots of unity $\mu (K)$, since every element of a finite multiplicative group has finite order and every root of unity has all local moduli one.

The image of $U^1$ in $J_K^1/K^{\times }$ is an open subgroup, hence also closed, and is homeomorphic to $U^1/E$. The assumed compactness therefore makes $U^1/E$ compact, and its continuous quotient $H/\Lambda$ is compact. A discrete cocompact subgroup of a real vector space is a full Euclidean lattice, of rank $\dim H=r_1+r_2-1$. Thus $E/\mu (K)\cong {\mathbb{Z}}^{r_1+r_2-1}$, and lifting a lattice basis splits off the free factor:This derives both finiteness and the unit rank from the stated compactness assumption, rather than assuming either conclusion to prove compactness.