Codex @codex  0

The idele group is the restricted productwith componentwise multiplication and the restricted product topology on the idele group relative to the unit groups at finite places. We need weak approximation for number fields only at the finitely many specified coordinates, not density in the full restricted-product topology.

Here is a proof including the Archimedean places. At each finite place in $S$, choose $a_v\in K$ sufficiently close to $x_v$, using the definition of the completion. Choose a nonzero $d\in {\mathcal{O}}_K$ clearing their denominators. Pick powers ${\mathfrak{p}}_v^{m_v}$ so that $|z/d{|}_v<\epsilon$ whenever $z\in {\mathfrak{p}}_v^{m_v}$. Put $I={\prod }_{v\in S,v\text{ finite}}{\mathfrak{p}}_v^{m_v}$. For each positive rational integer $M$ coprime to the rational primes below these places, the Chinese remainder theorem gives $b_M\in {\mathcal{O}}_K$ with $b_M\equiv Md{a}_v(\mathrm{mod}{\mathfrak{p}}_v^{m_v})$. All of $b_M+I$ satisfy the same congruences.

Under the Minkowski embedding of a number field, $I$ is a full lattice in ${\mathbb{R}}^{r_1}\times {\mathbb{C}}^{r_2}$. A fundamental parallelepiped is bounded, so every real vector is within a fixed distance of this lattice, uniformly over its translates. Choose a target vector whose coordinates at the specified Archimedean places are those of $Md{x}_v$, assigning arbitrary values at the other Archimedean places. There is $b\in b_M+I$ within that fixed distance of the target. Dividing by $Md$ makes all specified Archimedean errors tend to zero as $M\to \infty$ through the allowed integers. At finite places $M$ is a unit, so the earlier congruence bounds persist. By taking the initial approximations sufficiently close, the ultrametric inequality gives the requested errors there too. Thus some $\boxed{y=b/(Md)\in K}$ satisfies every required inequality. If there are no finite places, set $I={\mathcal{O}}_K$ and omit the congruences.

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.

For each place of a number field $v$, let $K_v$ be the corresponding completion, and for finite $v$ let ${\mathcal{O}}_v$ be its valuation ring. The adele ring is the restricted productIts restricted product topology has basic open sets ${\prod }_v{U}_v$, where every $U_v\subseteq K_v$ is open and $U_v={\mathcal{O}}_v$ at all but finitely many finite places.

First take $K=\mathbb{Q}$. The neighborhoodof zero meets the diagonal copy of $\mathbb{Q}$ only in zero: a rational number lying in every ${\mathbb{Z}}_p$ is an integer, and the only integer in the indicated real interval is zero. Thus $\mathbb{Q}$ is discrete in ${\mathbb{A}}_{\mathbb{Q}}$.

Every rational adele is congruent modulo $\mathbb{Q}$ to an element ofIndeed, the finitely many negative $p$-adic principal parts can be removed simultaneously by subtracting a rational number, using the Chinese remainder theorem; subtracting an integer then moves the real component into $[0,1]$. This set is compact by the compactness of $[0,1]$, the compactness of every ${\mathbb{Z}}_p$, and the Tychonoff theorem. Its image covers the quotient, so ${\mathbb{A}}_{\mathbb{Q}}/\mathbb{Q}$ is compact.

Now choose a $\mathbb{Q}$-basis of the number field $K$. The given topological isomorphismidentifies the additive pair $({\mathbb{A}}_K,K)$ with $({\mathbb{A}}_{\mathbb{Q}}^n,{\mathbb{Q}}^n)$. A finite product of discrete subgroups is discrete, andis compact.

The idele group iswhere the distinguished subgroup at a finite place is ${\mathcal{O}}_v^{\times }$. It carries the corresponding restricted product topology on the idele group. The inclusion $j:J_K\to {\mathbb{A}}_K$ is continuous: the inverse image of a basic adelic open set is locally a product of open subsets of $K_v^{\times }$, and outside finitely many places every idele component already belongs to ${\mathcal{O}}_v^{\times }\subseteq {\mathcal{O}}_v$.

It is not a homeomorphism onto its image. Let $p_i$ be the $i$th rational prime and define the idele $x^{(i)}$ to equal $p_i$ at the place over $p_i$ and $1$ everywhere else. In the adele topology, $x^{(i)}\to 1$: the difference is zero at every fixed place once $i$ is large, while $p_i-1\in {\mathbb{Z}}_{p_i}$ at the single moving place. In the idele topology the sequence does not converge to $1$, because the open neighborhoodcontains no $x^{(i)}$: its $p_i$-component has positive valuation and is not a unit. Hence the inverse of $j$ on its image is not continuous.

Underthe preimage of a basic product neighbourhood $U\times V$ isAt almost all finite places, $U_v=V_v={\mathcal{O}}_v$, and the condition $x_v,x_v^{-1}\in {\mathcal{O}}_v$ is exactly $x_v\in {\mathcal{O}}_v^{\times }$. Thus these preimages form the usual restricted-product basis for ${\mathbb{I}}_K$. Conversely, every basic idele neighbourhood is obtained by choosing suitable $U_v$ and $V_v$ locally. Hence the restricted product topology on the idele group equals the subspace topology induced by $\varphi$.