Adèles and the Finiteness of the Class Number

Andrei Jorza

February 18, 2005

Abstract This note is trying to be slick, so all the proofs are most efficient and

neat.

1 Adèles

Let K be a number field, i.e., an extension K/ℚ of degree n.

For each prime ideal 𝔭 of O_{K}, the ring of integers of K, we have a valuation v_{𝔭} : K → ℤ given by v_{𝔭}(x) is the exponent of 𝔭 in the prime decomposition of the ideal xO_{K}. Each v_{𝔭} has an associated metric |x|′_{𝔭} = 2^{−v_{𝔭}(x)}. The field K is not complete with respect to the metric v_{𝔭} so we can take K_{v_{𝔭}} to be the completion of K.

There exist embeddings i_{1}, …, i_{r_{1}} : K ↪ ℝ and j_{1}, …, j_{r_{2}} : K ↪ ℂ (then r_{1} + 2r_{2} = n = [K : ℚ]) and natural inclusions i_{𝔭} : K ↪ K_{v_{𝔭}}. We will call each i_{k}, j_{k}, v_{𝔭} a place and we will denote a general place by v. If v is of the form i_{k} we call it a real place and if it is of the form j_{k} we will call it a complex place. We will call these infinite places; each place v of the form v_{𝔭} is called a finite place and the valuation v_{𝔭} will be denoted simply by v; it has the property that v(xy) = v(x) + v(y) and v(x + y) ≥ min(v(x), v(y)).

Every finite place has, by definition, an associated prime ideal 𝔭_{v}. We will write K_{v} for the completion of K. Consider the ring of integers O_{v} = {x ∈ K_{v}|v(x) ≥ 0} which is a local ring with maximal ideal ℘_{v} = {x ∈ K_{v}|v(x) > 0}. It is a principal ideal domain and a generator π_{v} of ℘_{v} is called a uniformizer of K_{v}. Every fractional ideal of O_{v} is generated by π_{v}^{m} for some m ∈ ℤ. Let k_{v} = O_{v}/℘_{v} be the (finite) residue field at v and let q_{v} = #k_{v}.

For each real place v we write K_{v} = ℝ and for each complex place we write K_{v} = ℂ. Each field K_{v} has a canonical norm on it. For v real it is |x|_{v} = |i_{v}(x)| where i_{v} is the real embedding. For v complex it is |x|_{v} = |i_{v}(x)|^{2}, where i_{v} is the complex embedding. For v finite it is |x|_{v} = q_{v}^{−v(x)}.

Let S be a finite set of places that includes all the infinite places. Define

𝔸_{K,S} = ∏_{v∈S} K_{v} ∏_{v∉S} O_{v}.

Endow 𝔸_{K,S} with the product topology.

Define the ring of adèles over K to be

𝔸_{K} = ⋃_{S} 𝔸_{K,S},

together with the topology consisting of sets U such that U ∩ 𝔸_{K,S} is open in 𝔸_{K,S} for all finite sets S that contains the infinite places.

Lemma 1.1. 1. The ring 𝔸_{K} is Hausdorff.

2. The ring 𝔸_{K} is locally compact.

Proof. 1. Points in 𝔸_{K} are sequences (x_{v}) such that x_{v} ∈ O_{v} for almost all v. Let (x_{v}) ≠ (y_{v}) be two such points and assume that x_{μ} ≠ y_{μ} for some finite place μ. Then consider U_{μ} ∋ x_{μ} and V_{μ} ∋ y_{μ} be disjoint neighborhoods (The rings O_{v} are metric spaces and so Hausdorff). Then the preimages of U_{μ} and V_{μ} under the projection map to the μ component will separate (x_{v}) and (y_{v}).

2. Around each point (x_{v}) the neighborhood ∏_{v=∞}{x_{v}} ∏_{v<∞} O_{v} is compact by Tychonov’s theorem.

An annoying to prove, but true, fact is that 𝔸_{K} is a topological group under component-wise addition and multiplication. Define 𝔸_{K}^{×} to be the multiplicative subgroup of 𝔸_{K}, consisting of all sequences (x_{v}) such that v(x_{v}) = 0 for almost all finite v. The topology on 𝔸_{K}^{×} is the direct limit product topology on the multiplicative groups K_{v}^{×} and O_{v}^{×}.

For every x ∈ K there are only finitely many prime ideals that divide xO_{K} so K ↪ 𝔸_{K} but also K^{×} ↪ 𝔸_{K}^{×}.

2 Topology

We would like to understand the topological properties of K ⊂ 𝔸_{K} and K^{×} ⊂ 𝔸_{K}^{×}.

Proposition 2.1. K is discrete in 𝔸_{K} and 𝔸_{K}/K is compact.

Proof. We will interpret K and O_{K} as embedded in 𝔸_{K}. Let 𝔸_{∞} = ∏_{v=∞} K_{v}. By the Chinese Remainder Theorem we essentially get that 𝔸_{K} = K + 𝔸_{K,∅}. Clearly K ∩ 𝔸_{K,∅} = O_{K}. But O_{K} is discrete in 𝔸_{∞} so it is discrete in 𝔸_{K,∅} so K is discrete in 𝔸_{K}.

Let C = ∏_{v=ℝ}[−1/2, 1/2] ∏_{v=ℂ}{|z| ≤ 1/2} ∏_{v<∞} O_{v} which is compact by Tychonov. Then note that 𝔸_{K,∅} = O_{K} + C, again by the Chinese Remainder Theorem and so 𝔸_{K} = K +C. This means that 𝔸_{K}/K = (K +C)/K is compact being a closed subset of a compact set.

Locally compact groups, such as 𝔸_{K} have something called a Haar measure, which is a dμ(x), which is unique up to multiplication. As such, for every y ∈ 𝔸_{K} if we look at dμ(yx) we get another measure, so by uniqueness there exists a scalar |y| ∈ ℝ_{>0} such that dμ(yx) = |y|dμ(x). Then we have the property that

|x| = ∏_{v} |x_{v}|_{v},

where x = (x_{v}). Note that |x| is convergent since almost all terms in the product at ≤ 1.

Lemma 2.2. Let 𝔸_{K}^{1} ⊂ 𝔸_{K}^{×} be the set {x ∈ 𝔸_{K}^{×}||x| = 1} whose topology is inherited from 𝔸_{K} and 𝔸_{K}^{×} (the inherited topologies are the same). Then K^{×} ↪ 𝔸_{K}^{1}.

Proof. Conceptually this is a simple problem. But we will use the neatest method.

Since 𝔸_{K}/K is compact, it will have a finite volume relative to dμ(x). But for every y ∈ K we have α_{y} : x ↦ yx is an automorphism of 𝔸_{K}/K. (If x ∈ K then xy ∈ K and vice-versa so it is well-defined.) Therefore,

∫_{𝔸_{K}/K} dμ(x) = ∫_{𝔸_{K}/K} dμ(α_{y}(x)),

since all we are doing is a change of variables. But then ∫_{𝔸_{K}/K} dμ(x) = |y| ∫_{𝔸_{K}/K} dμ(x) so |y| = 1.

We also have a discrete embedding K^{×} ↪ 𝔸_{K}^{1} by the above Lemma.

Lemma 2.3. Let a ∈ 𝔸_{K} such that

|a| > \frac{vol(𝔸_{K}/K)}{vol(C)}.

Prove that there exists an x_{a} ∈ K such that |x_{a}|_{v} ≤ |a_{v}|_{v}.

Proof. Let A_{a} = aC. Then vol(A_{a}) > vol(A_{K}/K). Therefore the map aA → 𝔸_{K}/K is not an injection so there exist u, v ∈ A_{a} such that u − v ∈ K. But then by construction of C we have

|(u − v)|_{v} ≤ |a_{v}|_{v},

for every place v so u − v ∈ aA ∩ K.

Proposition 2.4. 𝔸_{K}^{1}/K^{×} is a compact topological group.

Proof. This is no longer as simple as 𝔸_{K}/K compact, but it is of a similar flavor. Instead of taking the surjection K + C → 𝔸_{K}/K to imply that 𝔸_{K}/K is compact, we will look for a compact set W and a surjection W → 𝔸_{K}^{1}/K^{×}.

Take W = {x ∈ 𝔸_{K}^{×}||x_{v}|_{v} ≤ |a_{v}|_{v}} where a ∈ 𝔸_{K}^{×} such that

|a| > \frac{vol(𝔸_{K}/K)}{vol(C)}.

By the previous Lemma the map W → 𝔸_{K}^{1}/K^{×} must be surjective because if t ∈ 𝔸_{K}^{1} then there exists x_{a} such that |x_{a}|_{v} ≤ |a_{v}/t_{v}|_{v} so x_{a}t ∈ W.

3 The Finiteness of the Class Number of K

Let Cl(K) be the class group of K and let I(K) be the group of fractional ideals of K.

Consider the map id : 𝔸_{K}^{×} → I(K) defined by

id : (x_{v}) ↦ ∏_{v<∞} 𝔭_{v}^{v(x_{v})}.

This map is well-defined because the product of prime ideals is a finite one. Moreover, if we restrict to id : 𝔸_{K}^{1} → I(K) we get a surjection because the preimage of ∏ 𝔭_{v}^{e_{v}} can be taken to be a ∈ 𝔸_{K} such that a_{v} = 1 for all infinite places v ≠ i_{1}, a_{v} = q_{v}^{e_{v}} for all finite places v and a_{i_{1}} ∈ ℝ to be whatever is needed to make a ∈ 𝔸_{K}^{1}.

The kernel of this map is clearly A_{∞} ∏_{v<∞} O_{v}^{×} so we get a bijection

id : 𝔸_{K}^{1}/(A_{∞} ∏_{v<∞} O_{v}^{×}) → I(K).

We get a projection id : 𝔸_{K}^{1}/(A_{∞} ∏_{v<∞} O_{v}^{×}) → Cl(K) = I(K)/K^{×}. What is its kernel? We need all a ∈ 𝔸_{K}^{1} such that a = (a_{v}) maps to a principal ideal.

But principal ideals correspond to a factorization xO_{K} = ∏_{v<∞} 𝔭_{v}^{v(x)} and the (unique) preimage via the injective map id in 𝔸_{K}^{1}/(A_{∞} ∏_{v<∞} O_{v}^{×}) is x ∈ K^{×}. Therefore we get a bijection

(𝔸_{K}^{1}/K^{×})/(A_{∞} ∏_{v<∞} O_{v}^{×}) → Cl(K).

Note that A_{∞} ∏_{v<∞} O_{v}^{×} is open in 𝔸_{K}^{1}/K^{×} which is compact so Cl(K) must be finite.