THE IDELIC APPROACH TO NUMBER THEORY

TOM WESTON

1. Introduction

In classical algebraic number theory one embeds a number field into the cartesian product of its completions at its archimedean absolute values. This embedding is very useful in the proofs of several fundamental theorems. However, it was noticed by Chevalley and Weil that the situation was improved somewhat if the number field is embedded in the cartesian product of its completions at all of its absolute values. With a few additional restrictions, these objects are known as the adeles, and the units of this ring are called the ideles.

When considering the adeles and ideles, it is their topology as much as their algebraic structure that is of interest. Many important results in number theory translate into simple statements about the topologies of the adeles and ideles. For example, the finiteness of the ideal class group and the Dirichlet unit theorem are equivalent to a certain quotient of the ideles being compact and discrete.

We will begin by reviewing the construction of local fields, first algebraically and then topologically. We will then prove the basic global results combining all of the local data, namely the product formula and the approximation theorem. Next we will define the adeles and the ideles and prove their basic topological properties. We will then define the idele class group, and relate it to the usual ideal class group. We will conclude with proofs of the finiteness of the ideal class group and the Dirichlet unit theorem, using idelic methods.

I have tried in this paper to emphasize the topological details in these constructions, and hopefully have not ignored any important points. We will assume some familiarity with number fields, at the level of [3, Chapter 1], [4, Chapters 1-3] or [8, Chapter 1].

We fix the following notation throughout this paper: we let k be a number field (that is, a finite extension of ℚ) of degree n over ℚ. We let 𝔬_{k} be the ring of integers of k. If 𝔭 is a prime ideal of 𝔬_{k} with 𝔭 ∩ ℤ = (p), we write e_{𝔭} for the ramification degree and f_{𝔭} for the inertial degree of 𝔭 over p. That is, e_{𝔭} is the largest power of 𝔭 dividing p𝔬_{k}, and f_{𝔭} is the degree of the residue field extension 𝔬_{k}/𝔭 over ℤ/(p). If the prime 𝔭 is clear from context, then we will just write e = e_{𝔭} and f = f_{𝔭}.

Part 1. Classical Algebraic Number Theory

2. Local Fields : Algebraic Description

Recall that the local ring 𝔬_{𝔭} ⊆ k is a discrete valuation ring. Let π be a uniformizing element of 𝔬_{𝔭}; that is, π generates the unique non-zero prime ideal 𝔭𝔬_{𝔭} of 𝔬_{𝔭}. Then any α ∈ k^{∗} can be written as α = uπ^{m} for a unique integer m and some unit u of 𝔬_{𝔭}^{∗}. We say that this m is the 𝔭-adic valuation v_{𝔭}(α) of α. Setting v_{𝔭}(0) = ∞, we have defined a discrete valuation

v = v_{𝔭} : k → ℤ ∪ {∞},

which is easily checked to satisfy the usual properties of a discrete valuation:

1. v_{𝔭}(αβ) = v_{𝔭}(α) + v_{𝔭}(β) for all α, β ∈ k;

2. v_{𝔭}(α + β) ≥ min{v_{𝔭}(α), v_{𝔭}(β)} for all α, β ∈ k;

3. v_{𝔭}(α) = ∞ if and only if α = 0.

We use the standard conventions for arithmetic with ∞, including ∞ · 0 = ∞. 𝔬_{𝔭} is precisely the set of elements of k with nonnegative valuation; the units 𝔬_{𝔭}^{∗} are precisely the elements of k^{∗} with valuation 0. We also have

𝔬_{𝔭}/𝔭𝔬_{𝔭} = (𝔬_{k}/𝔭)_{𝔭𝔬_{k}/𝔭} = (𝔬_{k}/𝔭)_{(0)} = 𝔬_{k}/𝔭

since 𝔬_{k}/𝔭 is already a field.

Note that p = uπ^{e} for some unit u ∈ 𝔬_{𝔭}^{∗}, so that

v_{𝔭}(p) = e,

where e is the ramification degree of 𝔭.

We will now construct the completion of 𝔬_{𝔭}. Let A_{m} = 𝔬_{𝔭}/(π^{m}). Then we have natural maps

φ_{m} : A_{m} → A_{m−1}

given by simply considering a residue class modulo π^{m} modulo π^{m−1}. The A_{m} form an inverse system, and we define the 𝔭-adic integers 𝔬_{v} to be the inverse limit of the A_{m}. That is, 𝔬_{v} is the subset of elements (a_{1}, a_{2}, …) of the infinite product

∏_{m=1}^{∞} A_{m}

such that

φ_{m}(a_{m}) = a_{m−1}

for all m. It is easy to see that if (a_{m}) and (b_{m}) lie in 𝔬_{v}, then so do (a_{m} + b_{m}) and (a_{m}b_{m}), so 𝔬_{v} inherits a ring structure from ∏_{m=1}^{∞} A_{m}.

Note that we have natural inclusions

𝔬_{k} ↪ 𝔬_{𝔭} ↪ 𝔬_{v}

given by sending α ∈ 𝔬_{𝔭} to (ᾱ, ᾱ, …) ∈ 𝔬_{v}. These are clearly ring homomorphisms, so 𝔬_{k} and 𝔬_{𝔭}, are naturally subrings of 𝔬_{v}.

We would like to have a better way to think of elements of 𝔬_{v}. To do this, first fix representatives 0 = c_{0}, …, c_{p^{f} −1} ∈ 𝔬_{𝔭} of the residue classes in the finite field 𝔬_{𝔭}/(π) ≅ 𝔬_{k}/𝔭. (In the case where 𝔬_{𝔭}/(π) = 𝔽_{p} we have the standard representatives 0, 1, …, p − 1, but in general there are no such obvious choices.) Now, take any element a = (a_{m}) of 𝔬_{v}. There is then a unique c_{i_{0}} such that

a_{1} ≡ c_{i_{0}} (mod π).

Next consider a − c_{i_{0}} = (a_{m} − c_{i_{0}}) ∈ 𝔬_{v}, The first coordinate of a − c_{i_{0}} will be 0, and thus divisible by π. It follows from the compatibility of the (a_{m} − c_{i_{0}}) under the φ_{m} that each a_{m} − c_{i_{0}} is then divisible by π. Thus, we can write

a − c_{i_{0}} = (a_{m} − c_{i_{0}})
= (πa′_{m})
= π(a′_{m})
= πa′

for some a′ = (a′_{m}) in 𝔬_{v}. Repeating this procedure, we can write

a′ − c_{i_{1}} = πa′′

for some unique c_{i_{1}} and some a′′ ∈ 𝔬_{v}.

Continuing this and combining all of the terms, we have

a = c_{i_{0}} + (a − c_{i_{0}})
= c_{i_{0}} + πa′
= c_{i_{0}} + π (c_{i_{1}} + (a′ − c_{i_{1}}))
= c_{i_{0}} + c_{i_{1}}π + π(π(c_{i_{2}} + (a′′ − c_{i_{2}})))
= c_{i_{0}} + c_{i_{1}}π + c_{i_{2}}π^{2} + ⋯

for some uniquely determined c_{i_{0}}, c_{i_{1}}, c_{i_{2}}, …, each chosen from our fixed set of representatives. Further, it is clear that any such expression

∑_{m=0}^{∞} c_{i_{m}}π^{m}

corresponds to some a = (a_{m}) ∈ 𝔬_{v} by setting

a_{m} ≡ c_{i_{0}} + c_{i_{1}}π + ⋯ + c_{i_{m−1}}π^{m−1} (mod π^{m}).

Thus, we can think of the elements of 𝔬_{v} as some sort of infinite power series in the uniformizing element π, with coefficients chosen from representatives of the finite field 𝔬_{𝔭}/(π) = 𝔬_{k}/𝔭. We will think of the c_{i_{m}} as the digits of x ∈ 𝔬_{v}.

Writing elements of 𝔬_{v} in this form, it is now clear how to extend our discrete valuation v_{𝔭} to 𝔬_{v}: given any element

a = c_{i_{0}} + c_{i_{1}}π + c_{i_{2}}π^{2} + ⋯

of 𝔬_{v}, we define v_{𝔭}(a) = m, where π^{m} is the first power of π with a non-zero coefficient. It is easy to see that v_{𝔭} agrees with our original discrete valuation on the image of 𝔬_{𝔭} in 𝔬_{v}, and that it still satisfies the axioms of a discrete valuation. Therefore, 𝔬_{v} is a discrete valuation ring. The maximal ideal 𝔭_{v} of 𝔬_{v} is generated by any element of valuation 1; for example, our original uniformizer π. Thus, 𝔭_{v} = 𝔭𝔬_{v}. Further, it is clear from our expansion of elements of 𝔬_{v} in terms of π that

𝔬_{v}/𝔭_{v} ≅ 𝔬_{𝔭}/𝔭𝔬_{𝔭} ≅ 𝔬_{k}/𝔭,

so that all of the residue fields coincide.

The units 𝔬_{v}^{∗} of 𝔬_{v} are precisely those elements of 𝔬_{v} of valuation 0, and the non-units 𝔭 are precisely those elements of 𝔬_{v} divisible by π. Since any element of 𝔬_{v} can be written as uπ^{m} for some u ∈ 𝔬_{v}^{∗} and some nonnegative integer n, the field of fractions of 𝔬_{v} is obtained simply by inverting π. We write k_{v} for this field; then we have

k_{v} = 𝔬_{v}[π^{−1}].

Therefore, elements x of k_{v} have the form

x = ∑_{m=m_{0}}^{∞} c_{i_{m}}π^{m}

for some integer m_{0}. If we require that c_{i_{m_{0}}} ≠ 0, then m_{0} is just v_{𝔭}(x). Since k is the field of fractions of 𝔬_{𝔭}, and 𝔬_{𝔭} embeds in 𝔬_{v}, k must embed in k_{v}.

3. Local Fields : Topological Description

We remain in the situation of the preceding section; that is, we have a number field k and a prime 𝔭 of the ring of integers 𝔬_{k} lying over a prime p of ℤ, and we let v = v_{𝔭} be the corresponding discrete valuation on k. In this section we will give a topological description of the completion k_{v}, in terms of the valuation v_{𝔭}.

First, observe that the valuation v induces a norm ∥·∥_{v} on k by

∥α∥_{v}= p^{−fv(α)}.

(We could have set ∥α∥_{v}= c^{v(α)} for any real constant c between 0 and 1; it does not matter for the purposes of this section. However, the choice c = p^{−f} is necessary for the product formula of Section 4. For the moment, note that the residue class ring 𝔬_{𝔭}/(α) has size 1/ ∥ α ∥_{v}, so this choice of c has at least something recommending it.) It follows immediately from the axioms of a discrete valuation that ∥·∥_{v} satisfies the axioms of a norm :

1. ∥αβ ∥_{v}=∥α∥_{v}∥β ∥_{v} for all α, β ∈ k;

2. ∥α + β ∥_{v}≤∥α∥_{v} + ∥β ∥_{v} for all α, β ∈ k. In fact, ∥·∥_{v} satisfies the stronger axiom

3. ∥α + β ∥_{v}≤ min{∥α∥_{v}, ∥β ∥_{v}} for all α, β ∈ k. For this reason we call ∥·∥_{v} a non-archimedean norm.

We can now in the usual way use ∥·∥_{v} to give us a metric on k, defined by

d_{v}(α, β) =∥α − β ∥_{v} .

This then gives us a topology on k. One can now define the completion of k with respect to v to be the usual completion of a metric space, in terms of equivalence classes of Cauchy sequences. This in fact gives rise to exactly the field k_{v} we constructed in Section 2. We will not go through the construction here; for a detailed exposition of it, see [2, Chapter 1, Section 4]. We will instead show that our algebraically constructed field of Section 2 has the necessary topological properties.

Note that we can use the discrete valuation on our field k_{v} of Section 2 to define a metric on it, which will agree with the metric defined above on the subfield k. We will show that under this topology k_{v} is a complete, locally compact topological ring, with compact subsets 𝔬_{v} and 𝔬_{v}^{∗}.

Let us describe the basic open sets on k_{v}; these are simply the open balls

B(x, r) = {y ∈ k_{v}; ∥x − y ∥_{v}< r}.

First, note that since our metric takes on only a discrete set of values, for any r there is some small ε so that

B(x, r) = \overline{B(x, r − ε)},

if ε is chosen small enough so that the metric does not take on any values between r − ε and r. Thus every basic open set is also closed. (Of course, it does not follow that every open set is also closed.)

Next, let us consider in more detail the open balls around 0 ∈ k_{v}. It is enough to consider the balls

B(0, p^{−fs})

for integral s, since these are the only values the metric takes. An element y of k_{v} satisfies ∥ y − 0 ∥_{v}< p^{−fs} if and only if v_{𝔭}(y) > s; that is, if and only if y has the form

∑_{m=s+1}^{∞} c_{i_{m}}π^{m}.

Thus, the ball B(0, p^{−fs}) consists precisely of those y which have coefficients 0 up to (and including) the coefficient of π^{s}. In the same way, for any x ∈ k_{v}, the ball B(x, p^{−fs}) consists of those y ∈ k_{v} agreeing with x up to (and including) the coefficient of π^{s}.

In particular, we see that

𝔬_{v} = B(0, p^{f})

and

𝔭_{v} = B(0, 1)

are open and closed subsets of k_{v}. Then

𝔬_{v}^{∗} = 𝔬_{v} − 𝔭_{v}

is open and closed as well.

We can actually describe the induced topology on 𝔬_{v} in a different way. Recall that we defined 𝔬_{v} to be a subset of an infinite product

∏_{m=1}^{∞} A_{m}.

Each set A_{m} is finite, and we give it the discrete topology. We then give ∏_{m=1}^{∞} A_{m} the product topology, and 𝔬_{v} the topology it inherits as a subspace.

Let us now try to describe the basic open sets in this topology. They have the form

∏_{m=1}^{∞} B_{m} ∩ 𝔬_{v}

where B_{m} is any subset of A_{m}, and B_{m} = A_{m} for all but finitely many m. Equivalently, they are the sets of the form

( ∏_{m=1}^{m_{0}} B_{m} × ∏_{m>m_{0}} A_{m}) ∩ 𝔬_{v} (1)

for positive integers m_{0}, where the B_{m} are any subsets of A_{m}. Now, consider any set U of the form (1), and pick any point x in it. It is then clear that

x ∈ B(x, p^{−f(m_{0}−1)}) ⊆ U,

so that U is open in the metric topology on 𝔬_{v}.

Conversely, consider any open ball B(x, p^{−fs}) in the metric topology. Such a set is actually of the form (1), taking m_{0} = s + 1 and the first s + 1 B_{m} to be a single point. We therefore have established:

Proposition 3.1. The two topologies we have defined on 𝔬_{v} are the same.

We are now ready to give the fundamental topological description of k_{v}.

Theorem 3.2. k_{v} is a locally compact, complete topological field, with compact open and closed subsets 𝔬_{v} and 𝔬_{v}^{∗}.

Proof. To say that k_{v} is a topological field is to say that addition, negation, multiplication and inversion are all continuous maps. The proofs of these facts are nearly identical to the proofs of the corresponding facts for the real numbers ℝ; in fact, the non-archimedean nature of our norm makes the proofs even easier.

Next we will show that 𝔬_{v} is compact. This is easy, using our alternate description of the topology on 𝔬_{v}: each finite set A_{m} is compact, so by the Tychonoff theorem, ∏_{m=1}^{∞} A_{m} is compact. Thus it is enough to show that 𝔬_{v} is closed as a subspace of ∏_{m=1}^{∞} A_{m}. So take (a_{m}) ∉ 𝔬_{v}. Then there is an m_{0} with φ_{m_{0}}(a_{m_{0}}) ≠ a_{m_{0}−1}. Consider the set

{a_{1}} × {a_{2}} × ⋯ × {a_{m_{0}}} × ∏_{m>m_{0}} A_{m}.

This is open in the product topology, and it is disjoint from 𝔬_{v} since every point in it has the same incompatible beginning. Thus, the complement of 𝔬_{v} is open, so 𝔬_{v} is closed, and thus compact.

Since 𝔬_{v}^{∗} is a closed subset of the compact set 𝔬_{v}, it is also compact. In fact, every basic open subset of k_{v} is compact, since we have a homeomorphism

ψ : B(x, p^{−fs}) → B(0, p^{f}) = 𝔬_{v}

given by ψ(y) = π^{−s−1}(y − x). (ψ is continuous since k_{v} is a topological field, and it is easy to see that it is bijective.) This shows that k_{v} is locally compact. Finally, since k_{v} is locally compact, it is complete.

Proposition 3.3. 𝔬_{v}^{∗}, 𝔬_{v} and k_{v} are the topological closures of 𝔬_{k}^{∗}, 𝔬_{k} and k respectively.

Proof. Since we clearly have 𝔬_{k}^{∗} ⊆ 𝔬_{v}^{∗}, 𝔬_{k} ⊆ 𝔬_{v} and k ⊆ k_{v} and the larger sets are all closed, it is enough to show that any element of the larger set can be arbitrarily closely approximated by an element of the smaller set. But this is clear; for example, if we pick our representatives c_{i} to all lie in 𝔬_{k} (which we can do since 𝔬_{𝔭}/𝔭𝔬_{𝔭} = 𝔬_{k}/𝔭), as well as π, then given x = ∑_{m=0}^{∞} c_{i_{m}}π^{m} ∈ 𝔬_{v}, the sequence

(c_{i_{0}}, c_{i_{0}} + c_{i_{1}}π, c_{i_{0}} + c_{i_{1}}π + c_{i_{2}}π^{2}, …)

of elements of 𝔬_{k} converges to x.

Note that since 𝔬_{k}^{∗} ⊆ 𝔬_{𝔭}^{∗} ⊆ 𝔬_{v}^{∗} and 𝔬_{k} ⊆ 𝔬_{𝔭} ⊆ 𝔬_{v}, we also have \overline{𝔬_{𝔭}^{∗}} = 𝔬_{v}^{∗} and \overline{𝔬_{𝔭}} = 𝔬_{v}.

4. Global Fields

We will now consider the description of k in terms of local data. Each prime 𝔭 gives rise to a valuation v_{𝔭} on k, and thus to a non-archimedean absolute value ∥·∥_{v_{𝔭}} on k. We also have archimedean absolute values arising from the embeddings of k in ℂ. Precisely, if

σ : k ↪ ℂ

is an embedding, then we have an absolute value ∥·∥_{σ} on k given by

∥α∥_{σ}= |σ(α)|,

where | · | is the usual absolute value on ℂ.

Lemma 4.1. If σ_{1} and σ_{2} are two different embeddings of k into ℂ and ∥α∥_{σ_{1}}=∥ α∥_{σ_{2}} for all α ∈ k, then σ_{1} = \overline{σ_{2}}.

Proof. Pick α ∈ k such that k = ℚ(α). Set σ_{i}(α) = x_{i} + y_{i}i. Then, since |σ_{1}(α)| = |σ_{2}(α)|, we have

x_{1}^{2} + y_{1}^{2} = x_{2}^{2} + y_{2}^{2}.

Further,

|σ_{1}(α + 1)| = |σ_{2}(α + 1)|
|σ_{1}(α) + 1| = |σ_{2}(α) + 1|
|(x_{1} + 1) + y_{1}i| = |(x_{2} + 1) + y_{2}i|
x_{1}^{2} + 2x_{1} + 1 + y_{1}^{2} = x_{2}^{2} + 2x_{2} + 1 + y_{2}^{2}.

Combining these two equations, we have

2x_{1} + 1 = 2x_{2} + 1,

so that x_{1} = x_{2}. Therefore y_{1} = ±y_{2}, and since σ_{1} ≠ σ_{2} we must have y_{1} = −y_{2}. Since α generates k over ℚ, this implies that σ_{1} = \overline{σ_{2}}, as desired.

Now, if σ is a real embedding of k (meaning that the image of σ lies in ℝ), then σ = \overline{σ}, so we get exactly one absolute value from this embedding. If σ is complex (meaning that the image of σ is strictly larger than ℝ), then σ and \overline{σ} are distinct embeddings of k into ℂ, giving rise to the same absolute value. Lemma 4.1 guarantees that this is the only situation in which this can happen. Thus, if k has r_{1} real embeddings and r_{2} pairs of complex conjugate embeddings, then we have r_{1} + 2r_{2} = n, but we get only r_{1} + r_{2} distinct absolute values.

We define the canonical set M_{k} of k to be the set of all of these absolute values : one non-archimedean absolute value for each prime 𝔭 (normalized as in Section 3), one archimedean absolute value for each real embedding and one archimedean absolute value for each pair of complex conjugate embeddings. We denote by S_{∞} the subset of M_{k} of archimedean valuations. These absolute values are all independent, in the sense that they induce different topologies on k. We will prove this fact later, while proving the approximation theorem.

To simplify notation, we will use v for both embeddings and valuations, so that we can write ∥·∥_{v} for any of the absolute values of M_{k}, not just the non-archimedean ones. We will also say valuation even when we more properly mean absolute value; in particular, we will often use v ∈ M_{k} to mean ∥·∥_{v}∈ M_{k}.

Note that if v is a real embedding, then the (topological) completion k_{v} of k with respect to ∥ · ∥_{v} is ℝ, since ℚ ⊆ k ⊂ ℝ, and the completion of ℚ is ℝ. Similarly, if v is a complex embedding, then the completion k_{v} of k with respect to ∥ · ∥_{v} is ℂ. If we let v_{1}, …, v_{r_{1}} be the real embeddings and v_{r_{1}+1}, \overline{v_{r_{1}+1}}, …, v_{r_{1}+r_{2}}, \overline{v_{r_{1}+r_{2}}} be the complex embeddings of k, then, thinking of each ℂ as ℝ^{2}, we have

∏_{i=1}^{r_{1}+r_{2}} k_{v} = ∏_{i=1}^{r_{1}} ℝ × ∏_{i=r_{1}+1}^{r_{1}+r_{2}} ℝ^{2} = ℝ^{n}.

Since k embeds in each k_{v}, k has a natural embedding in this ℝ^{n}. It is this embedding which is used classically.

We will now prove the product formula, which is an important relation between the absolute values in M_{k}. We will need a lemma.

Lemma 4.2. Let 𝔭_{1}, …, 𝔭_{m} be the primes of 𝔬_{k} lying over p ∈ ℤ. Let v_{1}, …, v_{m} be the corresponding valuations on k, and let v be the p-adic valuation on ℚ. Then for any α ∈ k^{∗},

∥N(α)∥_{v}= ∏_{i=1}^{m} ∥α∥_{v_{i}},

where N is the norm from k to ℚ.

Proof. By unique factorization of fractional ideals we can write

(α) = ∏_{i=1}^{m} 𝔭_{i}^{n_{i}} × I′

for some fractional ideal I′ prime to 𝔭_{1}, …, 𝔭_{m}. Then n_{i} = v_{i}(α), so ∥α∥_{v_{i}}= p^{−f_{i}n_{i}}, where f_{i} is the inertial degree of 𝔭_{i} over p. Therefore,

∏_{i=1}^{m} ∥α∥_{v_{i}}= p^{− ∑_{i=1}^{m} f_{i}n_{i}}.

Now, since the norm is multiplicative and N(𝔭_{i}) = p^{f_{i}}, we have

N(α) = p^{∑_{i=1}^{m} f_{i}n_{i}} · N(I′).

N(I′) is prime to p, so this has p-adic absolute value

p^{− ∑_{i=1}^{m} f_{i}n_{i}},

which proves the lemma.

Note that if α ∈ k, then

N(α) = ∏_{σ:k↪ℂ} |σ(α)| = ∏_{v∈S_{∞},v real} ∥α∥_{v} · ∏_{v∈S_{∞},v complex} ∥α∥_{v}^{2} .

If we define n_{v} = 1 for v real (and also, for later use, for v non-archimedean) and n_{v} = 2 for v complex, then this takes the form

N(α) = ∏_{v∈S_{∞}} ∥α∥_{v}^{n_{v}} .

Theorem 4.3 (The Product Formula). For all α ∈ k^{∗},

∏_{v∈M_{k}} ∥α∥_{v}^{n_{v}}= 1.

(Note that this infinite product makes sense, since v(α) = 0 (and therefore ∥α∥_{v}= 1) for all but finitely many v ∈ M_{k}.)

Proof. We first prove this formula in the case k = ℚ. In this case it is enough to check it for α = p a prime number, since both sides are multiplicative. But then ∥ α ∥_{v_{q}}= 1 for q ≠ p, so the only two terms which contribute to the product are ∥ α ∥_{v_{p}}= 1/p and the standard absolute value |α| = p, which multiply to 1. This proves the formula for k = ℚ.

Now let k be any number field. Then

∏_{v∈M_{k}} ∥α∥_{v}^{n_{v}} = ∏_{v∈M_{k}−S_{∞}} ∥α∥_{v} · ∏_{v∈S_{∞}} ∥α∥_{v}^{n_{v}}

= ∏_{𝔭} ∥α∥_{v_{𝔭}} ·N(α)

= ∏_{p} ∏_{𝔭|p} ∥α∥_{v_{𝔭}} ·N(α)

= ∏_{p} ∥N(α)∥_{v_{p}} ·N(α)

= ∏_{v∈M_{ℚ}} ∥N(α)∥_{v}

= 1

since we know the product formula for ℚ.

We chose the normalization we did on the absolute values ∥ · ∥_{v} in order to get the product formula to work out.

Next we will need some way to find elements of k satisfying certain local conditions. The standard result along these lines is the approximation theorem. We will prove it through a sequence of lemmas.

Lemma 4.4. For any distinct v_{1}, v_{2} ∈ M_{k}, there exists α ∈ k with ∥ α ∥_{v_{1}}> 1 and ∥α∥_{v_{2}}< 1.

Proof. If both v_{1} and v_{2} are non-archimedean, say v_{1} = v_{𝔭_{1}} and v_{2} = v_{𝔭_{2}}, then pick α ∈ 𝔭_{2} −𝔭_{1} and β ∈ 𝔭_{1} −𝔭_{2}. (These must exist since 𝔭_{1} and 𝔭_{2} are maximal ideals.) We must have ∥ α ∥_{v_{1}}≥ 1, ∥ α ∥_{v_{2}}< 1, ∥ β ∥_{v_{1}}< 1 and ∥ β ∥_{v_{2}}≥ 1. If ∥ α ∥_{v_{1}}≠ 1, then α will work; if ∥β ∥_{v_{2}}≠ 1, then β^{−1} will work. If ∥α∥_{v_{1}}=∥β ∥_{v_{2}}= 1, then αβ^{−1} will work.

If v_{1} is archimedean and v_{2} is non-archimedean, say v_{2} = v_{𝔭}, then any non-zero element of 𝔭∩ℤ will work. In the opposite case, the inverse of any non-zero element of 𝔭 ∩ ℤ works.

This leaves the case where both v_{1} and v_{2} are archimedean. By Lemma 4.1 we know that there is α ∈ k with ∥ α ∥_{v_{1}}≠∥ α ∥_{v_{2}} . Suppose that ∥ α ∥_{v_{1}}>∥ α ∥_{v_{2}} . Pick some rational number r in between. Then α/r will satisfy the conditions of the lemma. If we have the opposite inequality, then r/α will work.

Corollary 4.5. Any two distinct v_{1}, v_{2} ∈ M_{k} induce different topologies on k.

Proof. Note that for any absolute value ∥·∥_{v}, the ball B(0, 1) is precisely the set of α ∈ k with lim_{m→∞} α^{m} = 0. Thus B(0, 1) depends only on the topology, and not on the metric. But by Lemma 4.4, B_{v_{1}}(0, 1) and B_{v_{2}}(0, 1) are different, so v_{1} and v_{2} must induce different topologies.

Lemma 4.6. Let v_{1}, v_{2} be distinct elements of M_{k}. Suppose α ∈ k satisfies ∥ α∥_{v_{1}}> 1 and ∥α∥_{v_{2}}< 1. Then the sequence

z_{m} = \frac{α^{m}}{1 + α^{m}}

converges to 1 with respect to v_{1} and converges to 0 with respect to v_{2}.

Proof. This is intuitively quite clear, since the denominator is dominated by α^{m} with respect to v_{1} and by 1 with respect to v_{2}. We will leave the details to the reader.

Lemma 4.7. For any distinct v_{1}, v_{2}, …, v_{s} ∈ M_{k}, there exists α ∈ k with ∥α∥_{v_{1}}> 1 and ∥α∥_{v_{i}}< 1 for i = 2, …, s.

Proof. We prove this by induction on s. The case s = 2 is Lemma 4.4. Now, suppose we have α ∈ k with ∥ α ∥_{v_{1}}> 1 and ∥ α ∥_{v_{i}}< 1 for i = 2, …, s − 1. If ∥ α ∥_{v_{s}}< 1, then we are done. If ∥ α ∥_{v_{s}}= 1, then pick y ∈ k with ∥ y ∥_{v_{1}}> 1 and ∥ y ∥_{v_{s}}< 1. Then for large enough m, α^{m}y will satisfy the necessary conditions, since ∥α^{m} ∥_{v_{i}} can be made arbitrarily small for i = 2, …, s − 1.

This leaves the case where ∥α∥_{v_{s}}> 1. By Lemma 4.6 the sequence

z_{m} = \frac{α^{m}}{1 + α^{m}}

will converge to 1 with respect to v_{1} and v_{s}, and will converge to 0 with respect to v_{2}, …, v_{s−1}. Using Lemma 4.4 to pick y ∈ k with ∥ y ∥_{v_{1}}> 1 and ∥ y ∥_{v_{s}}< 1, we then see that for sufficiently large m the element z_{m}y will satisfy the conditions of the lemma.

Theorem 4.8 (The Approximation Theorem). For any distinct v_{1}, v_{2}, …, v_{s} ∈ M_{k}, α_{1}, …, α_{s} ∈ k and ε > 0 there exists α ∈ k satisfying

∥α − α_{i} ∥_{v_{i}}< ε

for all i.

Proof. By Lemma 4.7 we can find y_{i} ∈ k with ∥y_{i} ∥_{v_{i}}> 1 and ∥y_{i} ∥_{v_{j}}< 1 for j ≠ i. Then using Lemma 4.6 we can find z_{1}, …, z_{s} ∈ k with z_{i} arbitrarily close to 1 with respect to v_{i} and arbitrarily close to 0 with respect to v_{j}, j ≠ i. The element α = z_{1}α_{1} + ⋯ + z_{s}α_{s} will then satisfy the conditions of the theorem, since

∥α − α_{i} ∥_{v_{i}}≤∥z_{1} ∥_{v_{i}}∥α_{1} ∥_{v_{i}} + ⋯ + ∥z_{i} − 1∥_{v_{i}}∥α_{i} ∥_{v_{i}} + ⋯ + ∥z_{s} ∥_{v_{i}}∥α_{s} ∥_{v_{i}}

which we can make arbitrarily small.

Part 2. Adeles and Ideles

5. Adeles

We continue to let k be a number field of degree n over ℚ. Let M_{k} be its canonical set of absolute values. In analogy with our embedding of k into ℝ^{n} in the previous section, we would like to embed k into the product of its completions k_{v} at all v ∈ M_{k}. However, we need to restrict this product somewhat, in order to respect the important property that any α ∈ k is a 𝔭-adic integer for all but finitely many primes 𝔭 of 𝔬_{k}.

We will do this as follows: First, consider the usual direct product

∏_{v∈M_{k}} k_{v}.

We define the adeles A to be the subset of this direct product consisting of a = (a_{v})_{v∈M_{k}} with a_{v} a 𝔭-adic integer for all but finitely many of the non-archimedean valuations v_{𝔭}. (We do not impose any restriction on the components of a_{v} at archimedean valuations.) However, we do not give A the subspace topology; we take as our basis of open sets sets of the form

∏_{v∈M_{k}} U_{v}

where U_{v} ⊆ k_{v} is open in k_{v} for all v ∈ M_{k}, and U_{v} = 𝔬_{v} for all but finitely many v ∈ M_{k}. (We will always assume that in any such condition the archimedean absolute values are included in the finite excluded set.) We can write such a set as

∏_{v∈S} U_{v} × ∏_{v∉S} 𝔬_{v} (2)

where S is some finite subset of M_{k}, which we take to include S_{∞}, and each U_{v} is an open subset of k_{v}.

A is said to be the restricted direct product of the k_{v} with respect to the 𝔬_{v}. A inherits a ring structure from the direct product, so that operations are done componentwise.

Note that we can actually take our basis of open sets to be sets of the form (2) with each U_{v} a basic open set of k_{v}. The simplest such sets are the sets

A_{S} = ∏_{v∈S} k_{v} × ∏_{v∉S} 𝔬_{v},

where S is some finite subset of M_{k} containing S_{∞}. The product ∏_{v∉S} 𝔬_{v} is a product of compact sets, and thus compact by the Tychonoff theorem. A_{S} is therefore a finite product of locally compact sets, and thus is locally compact. Since any x ∈ A certainly lies in some A_{S} (take S to be the finite set of valuations v at which x does not lie in 𝔬_{v}), it follows that the space A is locally compact. This would not be the case if we had given A the subspace topology.

The sets A_{S} are actually closed as well; given any a ∉ A_{S}, we must either have a_{v_{0}} ∉ U_{v_{0}} for some v_{0} ∈ S, or else a_{v_{0}} ∉ 𝔬_{v_{0}} for v_{0} ∉ S. In the first case take a neighborhood of a with v_{0}-component disjoint from U_{v_{0}}; in the second, take a neighborhood of a with v_{0}-component disjoint from 𝔬_{v_{0}}. This gives us a neighborhood of a disjoint from A_{S}, so A_{S} is closed.

Proposition 5.1. The adeles A are a topological ring.

Proof. We will show that addition is continuous; negation and multiplication are similar. So, let f : A × A → A be defined by f(a, b) = a + b. Let U be some basic open set of the form

∏_{v∈S} B(c_{v}, r_{v}) × ∏_{v∉S} 𝔬_{v}

where S is some finite subset of M_{k} containing S_{∞}. We must show that f^{−1}(U) is open in A × A.

Pick (a, b) ∈ f^{−1}(U). Then

∥a_{v} + b_{v} − c_{v} ∥_{v}< r_{v}

for all v ∈ S. Now, define sets U_{1}, U_{2} ∈ A by

U_{1} = ∏_{v∈S} B (a_{v}, \frac{r_{v}}{4} ) × ∏_{v∉S} 𝔬_{v}

and

U_{2} = ∏_{v∈S} B (b_{v}, \frac{r_{v}}{4} ) × ∏_{v∉S} 𝔬_{v}.

Then U_{1} and U_{2} are both open in A, so U_{1} ×U_{2} is open in A×A. The v-component of f(U_{1} ×U_{2}) is just 𝔬_{v} for v ∉ S, and is contained in B(a_{v} +b_{v}, r_{v}/2), which in turn is contained in B(c_{v}, r_{v}), for v ∈ S. Thus (a, b) ∈ U_{1} × U_{2} ⊆ f^{−1}(U). Therefore f^{−1}(U) is open, and f is continuous.

Note that we have a natural embedding of k into A, sending each α ∈ k to the vector (α, α, …) ∈ A. This is well-defined, since any α ∈ k is a 𝔭-adic integer for all but finitely many primes 𝔭. We give k the subspace topology.

Proposition 5.2. k is embedded as a discrete subring of the adeles A.

Proof. k is a topological subring of A since we have given it the subspace topology. It remains to show that it is discrete. We will show that 0 has a neighborhood disjoint from k − {0}, and the general case follows by translation.

Define a set U = ∏_{v∈M_{k}} U_{v} by taking U_{v} = 𝔬_{v} for all non-archimedean v ∈ M_{k}, and U_{v} = B(0, 1/2) for all v ∈ S_{∞}. Then U is a basic open neighborhood of 0. Suppose there is a non-zero α ∈ U ∩ k. Then ∥ α ∥_{v}≤ 1 for all non-archimedean v, and ∥α∥_{v}< 1/2 for all archimedean v. Thus,

∏_{v∈M_{k}} ∥α∥_{v}^{n_{v}}< \frac{1}{2}.

But this contradicts the product formula, since this product is always 1 for nonzero α. Thus, U ∩ k = 0, which shows that {0} is open in k, and thus that k is discrete.

We also have embeddings of k_{v} into A for each v ∈ M_{k} given by sending x ∈ k_{v} to (0, …, x, …) ∈ A, with the x in the v-component.

Proposition 5.3. This embeds k_{v} as a closed subring of A, and the topology k_{v} inherits as a subspace of A is just the usual topology on k_{v}.

Proof. It is clear that k_{v} inherits its usual topology, since U_{v} × ∏_{v′∈M_{k},v′≠v} 𝔬_{v′} is open in A, and for any basic open set U of A (and thus for any open set of A), U ∩ k_{v} will be an open set of k_{v}.

It remains to show that k_{v} is closed in A. So take a ∈ A − k_{v}. Then there is some v′ ≠ v with a_{v′} ≠ 0. The open set

B (a_{v}, \frac{∥a_{v} ∥_{v}}{2} ) × ∏_{v′′∈M_{k}−S_{∞},v′′≠v′} 𝔬_{v′′} × ∏_{v′′∈S_{∞}} k_{v′′}

will then be an open neighborhood of a disjoint from k_{v}. Thus k_{v} is closed in A.

Note that the embedding k_{v} ↪ A induces an embedding k ↪ A, inducing the v-adic topology on k. Thus, in A we have copies of k with each of our topologies from M_{k}, together with a copy of k with the discrete topology.

Of course, A is not an integral domain. The units A^{∗} are the elements a = (a_{v}) with a_{v} ∈ k_{v}^{∗} for all v ∈ M_{k}, and with a_{v} ∈ 𝔬_{v}^{∗} for all but finitely many v ∈ M_{k}. (If there are infinitely many a_{v} ∉ 𝔬_{v}^{∗}, then the inverse a^{−1} = (a_{v}^{−1}) is not an adele.) Such an element is called an idele. Unfortunately, the ideles are not a topological (multiplicative) subgroup of A, since inversion is not continuous. We will leave the demonstration of this to the reader.

Since k is an additive subgroup of A, it makes sense to consider the quotient A/k as a topological group in the quotient topology. We will show that A/k is compact. We will first need a lemma.

Lemma 5.4.

k + A_{S_{∞}} = A,

in the sense that any adele can be written as a sum of an element of k and an element in the basic open set A_{S_{∞}}.

Proof. This means precisely that for any adele a, there is some α ∈ k such that a − α ∈ 𝔬_{v} for all non-archimedean v ∈ M_{k}, since A_{S_{∞}} is just the set

∏_{v∈S_{∞}} k_{v} × ∏_{v∉S_{∞}} 𝔬_{v}.

We find such an α as follows: First, find an integer c ∈ ℤ such that ca ∈ 𝔬_{v} for all v ∉ S_{∞}. We can do this since a_{v} ∈ 𝔬_{v} for all but finitely many v ∉ S_{∞}, so we can just take some c highly divisible by the finitely many primes of ℤ lying under the 𝔭 with a_{v} ∉ 𝔬_{v_{p}}.

Let S be the set of primes of 𝔬_{k} dividing c𝔬_{k}; this is finite, but may be larger than the set of primes with a_{v_{𝔭}} ∉ 𝔬_{v_{p}}, since more than one prime of 𝔬_{k} can divide p ∈ ℤ. Now, by the approximation theorem we can find α ∈ 𝔬_{k} with

α ≡ ca_{v} (mod 𝔭^{m})

for all 𝔭 ∈ S and some large m.

I claim that a − α/c satisfies the conditions of the lemma. First, if 𝔭 ∉ S, then c is in 𝔬_{v}^{∗}, so α/c ∈ 𝔬_{v} and a_{v} − α/c will still be in 𝔬_{v}. If 𝔭 ∈ S, then so long as we took m larger than the power of 𝔭 dividing c, a_{v} − α/c will still be in 𝔬_{v}. This completes the proof.

Theorem 5.5. The group A/k is compact.

Proof. First, recall that we had an embedding

k ↪ ∏_{v∈S_{∞}} k_{v} = ℝ^{n}.

Under this embedding the integers 𝔬_{k} form a lattice. If α_{1}, …, α_{n} is a ℤ-basis for 𝔬_{k}, then the images of the α_{i} in ℝ^{n} will still be linearly independent. (This is because the computation to check this is essentially just the computation of the discriminant, which is non-zero; for details, see [9, Chapter 8].) This shows that this lattice has rank n.

Let P ∈ ℝ^{n} be a fundamental parallelotope for the lattice 𝔬_{k}. Then, since the lattice 𝔬_{k} has the same rank as the dimension of the space ℝ^{n}, P is bounded, and therefore \overline{P} is compact. Also, note that

A_{S_{∞}} = ∏_{v∉S_{∞}} 𝔬_{v} × ℝ^{n}.

Now, given any a ∈ A, we first take α ∈ k such that a − α ∈ A_{S_{∞}}. Next we take some β ∈ 𝔬_{k} such that

a − α − β ∈ ∏_{v∉S_{∞}} 𝔬_{v} × \overline{P} .

We can do this since P is a fundamental domain for 𝔬_{k}. Call this set Q. It is a product of compact sets, and thus is compact. Therefore translation by the element −(α + β) ∈ k takes a into a compact set. Therefore A/k = Q/(k ∩ Q), which is a quotient of a compact space and thus is compact.

6. Ideles

Recall that we defined the ideles to be the units A^{∗} of the adele ring A. However, as we observed, inversion is not continuous on A^{∗} with this topology. So A^{∗} is not a topological group, at least when viewed as a subspace of A. To remedy this we will give A^{∗} a different topology, which we can describe in two different ways.

The first is in analogy with the restricted direct product topology on A. We define the ideles J to be the subset of the direct product

∏_{v∈M_{k}} k_{v}^{∗}

consisting of those elements for which all but finitely many components lie in 𝔬_{v}^{∗}. That is, a = (a_{v}) ∈ J if and only if a_{v} ∈ 𝔬_{v}^{∗} for all but finitely many nonarchimedean valuations v. (Again, we impose no restriction on the values a takes on at archimedean v.) Then, as sets, J is just the units A^{∗} of A. However, we take for a basis of open sets of J the sets of the form

∏_{v∈M_{k}} U_{v}

where U_{v} ⊆ k_{v}^{∗} is open in k_{v}^{∗}, and U_{v} = 𝔬_{v}^{∗} for all but finitely many non-archimedean valuations v. (We give k_{v}^{∗} the topology it inherits as a subspace of k_{v}.) As with the adeles, we can write such a set as

∏_{v∈S} U_{v} × ∏_{v∉S} 𝔬_{v}^{∗} (3)

where S is some finite subset of M_{k} containing S_{∞}, and U_{v} is open in k_{v}^{∗} for each v ∈ S. As with the adeles, we can actually take our basis to consist of sets of the form (3) with each U_{v} a basic open set of k_{v}^{∗}.

There is another way to define this topology. First, consider the product set A × A, with the product topology. We then have a map

φ : J → A × A

given by φ(a) = (a, a^{−1}). We claim that this is a topological embedding. It is clear that φ is injective, so we must only show that it is bicontinuous.

First, note that if

U = (∏_{v∈S} U_{v} × ∏_{v∉S} 𝔬_{v}, ∏_{v∈S} U′_{v} × ∏_{v∉S} 𝔬_{v})

is a basic open set in A × A (we can take the same set S in each factor simply by taking some of the U_{v} and U′_{v} to be all of 𝔬_{v}), then

U ∩ φ(J) = (∏_{v∈S} (U_{v} ∩ (U′_{v})^{−1}) × ∏_{v∉S} 𝔬_{v}^{∗}, ∏_{v∈S} (U_{v}^{−1} ∩ (U′_{v})^{−1}) × ∏_{v∉S} 𝔬_{v}^{∗}) .

This is the image of

∏_{v∈S} (U_{v} ∩ (U′_{v})^{−1}) × ∏_{v∉S} 𝔬_{v}^{∗},

which is open in J since inversion is continuous in each k_{v}^{∗}. This implies immediately that φ is bicontinuous.

With both of these descriptions of the topology on J in hand, it is now easy to prove the basic facts about its topology. First, as with the adeles, we define sets

J_{S} = ∏_{v∈S} k_{v}^{∗} × ∏_{v∉S} 𝔬_{v}^{∗}

where S is some finite subset of M_{k} containing S_{∞}. Then J_{S} is open and locally compact, so J is locally compact. As in the adelic case, J_{S} is also closed, as is easy to see.

Proposition 6.1. J is a topological group.

Proof. Viewing J as embedded in A × A, it is clear that inversion is continuous, since it is just the natural map reversing the order of the direct product. It is also clear that multiplication is continuous, since the map from A × A × A × A given by

(a_{1}, a_{2}, a_{3}, a_{4}) ↦ (a_{1}a_{3}, a_{2}a_{4})

is continuous, and this restricts to give precisely multiplication in J.

We have natural embeddings k^{∗} ↪ k_{v}^{∗} ↪ J, along the v-component, and k^{∗} ↪ J along the diagonal. An element of k^{∗} viewed as an idele in this second way is called a principal idele.

Proposition 6.2. k^{∗} is embedded along the diagonal in J as a discrete subgroup.

Proof. This is essentially the same as the proof for the adeles, taking neighborhoods B(1, 1/2) for the archimedean valuations and using the product formula to derive a contradiction about α − 1 for α ∈ k.

Proposition 6.3. k_{v}^{∗} is embedded as a closed subgroup of J.

Proof. This again is essentially the same as the adelic case.

Summarizing, we have shown that the ideles J are a locally compact topological group, containing k^{∗} as a discrete subgroup along the diagonal, and in the v-adic topology for each v inside of k_{v}^{∗} ↪ J.

7. The Idele Class Group

The ideles possess two important natural homomorphisms into other multiplicative groups. The first is a map

∥·∥ : J → ℝ^{+}

defined by

∥a∥ = ∏_{v∈M_{k}} ∥a_{v} ∥_{v}^{n_{v}} .

This makes sense, since for any idele a all but finitely many of the terms in the product are 1. This map is also continuous; to show this, we must show that the preimage of an open interval (b_{0}, b_{1}) ⊆ ℝ^{+} is open in J. So take any a ∈ J with ∥ a ∥ ∈ (b_{0}, b_{1}). Pick one archimedean valuation v_{0}, let S be the set of valuations at which a is not in 𝔬_{v}^{∗}, and consider the open sets

U_{r} = ∏_{v∈S,v≠v_{0}} B(a_{v}, 1) × ∏_{v∉S} 𝔬_{v}^{∗} × B(a_{v_{0}}, r)

where we let r vary. Then these open sets contain a, and we can make r sufficiently small to force ∥U_{r} ∥ to lie in (b_{0}, b_{1}). This shows that ∥· ∥ is continuous.

Since {1} is closed in ℝ^{+}, the kernel of ∥ · ∥ (which is just the preimage of {1}) is closed in J. We denote this by J^{0}. Note that the product formula shows that k^{∗} ⊆ J^{0}, and is a discrete subgroup.

The second natural homomorphism is given as follows: Let I be the multiplicative group of fractional ideals of k, and let P be the subgroup of principal ideals. We define a map

J → I

by associating to a ∈ J the fractional ideal

∏_{v_{𝔭}∈M_{k}−S_{∞}} 𝔭^{v_{𝔭}(a_{v})},

which we will denote (a). This is a fractional ideal, since only finitely many of the v_{𝔭}(a_{v}) are non-zero. This is clearly surjective, with kernel J_{S_{∞}}. Further, if α ∈ k^{∗} ⊆ J, then (α) is just the usual principal ideal generated by α. Thus (k^{∗}) ⊆ P, so that we obtain an induced surjective homomorphism from J/k^{∗} to the ideal class group I/P. We call J/k^{∗} the idele class group, and write it as C. It contains the closed subgroup C^{0} = J^{0}/k^{∗}.

Since the kernel of the map J → I/P is k^{∗}J_{S_{∞}}, we have an isomorphism

J/k^{∗}J_{S_{∞}} ≅ I/P.

Thus we have expressed the ideal class group as a quotient of the ideles.

We can generalize this construction somewhat. For any finite subset S of M_{k} containing S_{∞}, we will call

k_{S} = J_{S} ∩ k^{∗}

the S-units of k. It is clear that this coincides with the usual notion of S-units, which are those elements of k^{∗} which are units in each 𝔬_{v} for v ∉ S. k_{S_{∞}} is just the global units 𝔬_{k}^{∗}. Since k^{∗} is a discrete subgroup of J, k_{S} is a discrete subgroup of J_{S}. We call the quotient J_{S}/k_{S} the group of S-idele classes, and denote it by C_{S}. We set J_{S}^{0} = J_{S} ∩ J^{0}, and C_{S}^{0} = J_{S}^{0}/k_{S}.

For each such S, we have a natural inclusion

C_{S} ↪ C

embedding C_{S} as an open and closed subgroup of C, since J_{S} is open and closed in J. Similarly, the natural inclusion

C_{S}^{0} ↪ C^{0}

embeds C_{S}^{0} as an open and closed subgroup of C^{0}.

We return to the ideal map defined above. We had a surjective map

C → I/P

with kernel C_{S_{∞}}. Even if we restrict this map to C^{0} it is still surjective, since if I is a fractional ideal and a an idele with (a) = I, we can modify a at one of the archimedean valuations to get some a′ ∈ J^{0} with (a′) = I. The kernel of this map is just C_{S_{∞}}^{0}, so we have an isomorphism

C^{0}/C_{S_{∞}}^{0} ≅ I/P.

We must now show that C^{0} is compact and discrete. This will turn out to be equivalent to the finiteness of the ideal class group of k and the Dirichlet unit theorem. We will first need a stronger approximation theorem, which we will prove in the next section.

8. Another Approximation Theorem

In order to prove the compactness of the idele class group we will need to exhibit elements of k satisfying conditions at all v ∈ M_{k}. The approximation theorem is not strong enough to deal with this situation, so we will need to prove a new result.

Given any idele a, we define the set L(a) ⊆ k to be the set of all α ∈ k satisfying

∥α∥_{v} ≤ ∥a_{v} ∥_{v}

for all v ∈ M_{k}. We write λ(a) for the order of L(a). We wish to show if ∥ a ∥ is sufficiently large, then L(a) is non-zero.

First, note that for any α ∈ k^{∗} L(a) and L(αa) are in canonical bijection, by

x ↦ αx.

Thus, λ(a) = λ(αa).

Theorem 8.1. Let k be a number field. Then there is a constant c_{0}, depending only on k, such that for any idele a,

λ(a) ≥ c_{0} ∥a∥ .

Proof. Let n be the degree of k over ℚ, and pick an integral basis ω_{1}, …, ω_{n} for 𝔬_{k}. Set

c_{1} = n sup_{v∈S_{∞},i}{∥ω_{i} ∥_{v}}.

Now, by the approximation theorem we can find α ∈ k^{∗} satisfying

\frac{c_{1}}{∥a∥_{v}} ≤ α ≤ \frac{2c_{1}}{∥a∥_{v}}

for all v ∈ S_{∞}. We can also pick an integer m ∈ ℤ such that

∥mαa∥_{v}≤ 1

for all v ∈ M_{k} −S_{∞}, by taking m highly divisible by the prime numbers corresponding to the finitely many valuations v for which ∥ αa ∥_{v} > 1. Now, for all v ∈ S_{∞}, we have

mc_{1} ≤ ∥mαa∥_{v} ≤ 2mc_{1}.

Since λ(mαa) = λ(a) and ∥mαa∥=∥a∥, we may replace a by mαa. This allows us to assume that

∥a∥_{v} ≤ 1

for all v ∈ M_{k} − S_{∞}, and that there is some rational integer m for which

mc_{1} ≤∥a∥_{v}≤ 2mc_{1}

for all v ∈ S_{∞}.

Now, let Λ be the set of elements of 𝔬_{k} of the form

b_{1}ω_{1} + ⋯ + b_{n}ω_{n}

with b_{i} ∈ ℤ and 0 ≤ b_{i} ≤ m. Then Λ contains (m + 1)^{n} > m^{n} elements. Now, by our normalization above, the fractional ideal (a) associated to the idele a is actually an ideal, so it makes sense to consider the quotient map

𝔬_{k} → 𝔬_{k}/(a).

The image has size N(a), so there must be a subset Λ′ ⊆ Λ containing at least

\frac{m^{n}}{N(a)}

elements of Λ, all mapping to the same class in 𝔬_{k}/(a).

Fix some x ∈ Λ′, and pick any other y ∈ Λ′. Then

x ≡ y (mod (a)),

so that

∥x − y ∥_{v}≤∥a_{v} ∥_{v}

for all non-archimedean valuations v. If v is an archimedean valuation, then

∥x − y ∥_{v} = ∥b_{x1}ω_{1} + ⋯ + b_{xn}ω_{n} − b_{y1}ω_{1} − ⋯ − b_{yn}ω_{n} ∥_{v}
= ∥(b_{x1} − b_{y1})ω_{1} + ⋯ + (b_{xn} − b_{yn})ω_{n} ∥_{v}
≤ m ∥ω_{1} ∥_{v} + ⋯ + m ∥ω_{n} ∥_{v}
≤ mc_{1}
≤ ∥a_{v} ∥_{v} .

Thus x − y ∈ L(a). Therefore,

λ(a) ≥ \frac{m^{n}}{N(a)}.

However, we also have

m^{n} ≥ 2^{−n}c_{1}^{−n} ∏_{v∈S_{∞}} ∥a_{v} ∥_{v}^{n_{v}} .

Finally, since

∥a∥ = ∏_{v∈M_{k}−S_{∞}} ∥a_{v} ∥_{v} · ∏_{v∈S_{∞}} ∥a_{v} ∥_{v}^{n_{v}}= N(a)^{−1} ∏_{v∈S_{∞}} ∥a_{v} ∥_{v}^{n_{v}},

we have

λ(a) ≥ 2^{−n}c_{1}^{−n} ∥a∥ .

This proves the theorem, with c_{0} = 2^{−n}c_{1}^{−n}.

In fact, it is possible to get a much more precise statement. If we let r_{1} be the number of real embeddings of k, r_{2} the number of complex embeddings of k and D_{k} the absolute value of the discriminant of k, then

λ(a) ≈ \frac{2^{r_{1}}(2π)^{r_{2}}}{\sqrt{D_{k}}} ∥a∥ .

See [3, Chapter 5, Section 2, Theorem 1].

9. The Compactness of the Idele Class Group

We are now in a position to prove the compactness of the idele class group. We will need a lemma. Throughout this section we let c_{0} be the constant of Theorem 8.1.

Lemma 9.1. Let a be an idele with ∥ a ∥≥ 2/c_{0}. Then there exists α ∈ k^{∗} such that

1 ≤ ∥αa_{v} ∥_{v} ≤ ∥a∥

for all v ∈ M_{k}.

Proof. By Theorem 8.1 there is some non-zero β in L(a). This β satisfies ∥β ∥_{v} ≤ ∥ a_{v} ∥_{v} for all v, so if we set α = β^{−1}, then

1 ≤ ∥αa_{v} ∥_{v}

for all v ∈ M_{k}. Also, we have

∥αa_{v} ∥_{v}= \frac{∏_{v′∈M_{k}} ∥αa_{v} ∥_{v}}{∏_{v′∈M_{k},v′≠v} ∥αa_{v} ∥_{v}} ≤ \frac{∥a∥}{1} = ∥a∥,

which completes the proof.

Theorem 9.2. C^{0} is compact.

Proof. Let ψ : J → ℝ^{+} be defined by ψ(a) =∥ a ∥ . For any α ∈ k^{∗} we have ψ(α) = 1, so we get a well defined map

ψ : C → R^{+}.

The kernel of this map is C^{0}. Now, note that for any ρ ∈ ℝ^{+}, ψ^{−1}(ρ) is homeomorphic to C^{0}. To see this, let a_{ρ} be any idele with ∥a_{ρ} ∥= ρ. (We can always find such an a_{ρ} by adjusting the archimedean valuations.) Then we have ψ^{−1}(ρ) = a_{ρ}C^{0}, and this is a homeomorphism since J is a topological group. Thus, it will suffice to show that ψ^{−1}(ρ) is compact for some ρ ∈ ℝ^{+}.

Fix some ρ with ρ > 2/c_{0}, and pick some a ∈ ψ^{−1}(ρ). Then by Lemma 9.1 there is some α_{a} ∈ k^{∗} with

1 ≤ ∥α_{a}a_{v} ∥_{v} ≤ ρ

for all v ∈ M_{k}. However, since a non-archimedean valuation v_{𝔭} takes on no values between 1 and N𝔭, and only finitely many primes 𝔭 have N𝔭 ≤ ρ, we must have

∥α_{a}a_{v} ∥_{v_{𝔭}}= 1

for all but finitely many primes 𝔭, independent of a. Thus there is some finite set S of valuations (including S_{∞}) such that

1 ≤∥α_{a}a_{v} ∥_{v} ≤ ρ,  v ∈ S
∥α_{a}a_{v} ∥_{v} = 1,  v ∉ S.

Define

T = ∏_{v∈S}(\overline{B(0, ρ)} − B(0, 1)) × ∏_{v∉S} 𝔬_{v}^{∗}.

Then by the Tychonoff theorem T is compact, since each factor is. Further, by what we have shown above, T maps onto ψ^{−1}(ρ) under the quotient map

J → C.

(This is because we can always modify an element of ψ^{−1}(ρ) by an element of k^{∗} to get it in T.) The image of T is compact, so ψ^{−1}(ρ) is a closed subset of a compact space, and thus compact. This completes the proof.

Corollary 9.3. C_{S}^{0} is compact for any finite set S containing S_{∞}.

Proof. C_{S}^{0} is a closed subset of the compact space C^{0}, and thus is compact.

10. Applications to Algebraic Number Theory

The finiteness of the ideal class group is now an immediate corollary of everything we have done.

Theorem 10.1. For any number field k, the ideal class group I/P is finite.

Proof. Recall that we had a group isomorphism

C^{0}/C_{S_{∞}}^{0} ≅ I/P.

The space C^{0}/C_{S_{∞}}^{0} is a quotient of a compact space, and thus compact. Also, C_{S_{∞}}^{0} is an open subgroup of C^{0}, so this quotient is also discrete. (This is because each coset of C_{S_{∞}}^{0}, which become points in C^{0}/C_{S_{∞}}^{0}, is open.) Thus, the space C^{0}/C_{S_{∞}}^{0} is compact and discrete, and thus finite. Therefore I/P is finite as well.

The Dirichlet unit theorem will require a preliminary result.

Lemma 10.2. Any discrete subgroup Λ of ℝ^{s} is free abelian, of rank dim ℝΛ. (Here ℝΛ is the ℝ-vector space spanned by Λ.)

Proof. We prove this by induction on the dimension of dim ℝΛ. If dim ℝΛ = 1, then, since Λ is discrete, there must be some λ ∈ Λ closest to 0. It is then clear that Λ = ℤλ, since otherwise we could construct an element of Λ closer to 0 than λ.

Now, suppose dim Λ = m. Let λ_{1}, …, λ_{m} be a ℝ-basis for ℝΛ. If Λ_{0} is the subgroup of Λ spanned by λ_{1}, …, λ_{m−1}, then by the induction hypothesis

Λ_{0} = ℤλ_{1} ⊕ ⋯ ⊕ ℤλ_{m−1}.

Now, consider the set Λ′ of λ ∈ Λ of the form

λ = a_{1}λ_{1} + ⋯ + a_{m}λ_{m},

with 0 ≤ a_{i} < 1 for i = 1, …, m − 1, and 0 ≤ a_{m} ≤ 1. (We do not require that the a_{i} be integers.) This is a bounded subset of a discrete set, so it is finite. Pick λ′ ∈ Λ′ with minimal non-zero coefficient of λ_{m}, say

λ′ = a′_{1}λ_{1} + ⋯ + a′_{m}λ_{m}.

Now, if λ is any element of Λ, then we can find some integer t such that the coefficient a_{m} of λ_{m} in λ − tλ′ satisfies 0 ≤ a_{m} < a′_{m}. We can then further modify λ − tλ′ by some λ_{0} ∈ Λ_{0} to get λ − tλ′ − λ_{0} ∈ Λ′. But since a′_{m} was the minimal non-zero coefficient of λ_{m} in Λ′, we must have a_{m} = 0. This implies that

λ − tλ′ − λ_{0} = 0,

which, together with linear independence, shows that

Λ = ℤλ_{1} ⊕ ⋯ ⊕ ℤλ_{m},

as desired.

Theorem 10.3 (Dirichlet Unit Theorem). For any set finite set S ∈ M_{k} of size s, containing S_{∞}, the S-units k_{S} have rank s − 1.

Proof. Let v_{1}, …, v_{s} be the elements of S, ordered so that v_{s} is archimedean. We define a group homomorphism

log : J_{S} → R^{s}

by

log(a) = (log ∥a_{v_{1}} ∥_{v_{1}}^{n_{v_{1}}} , …, log ∥a_{v_{s}} ∥_{v_{s}}^{n_{vs}}).

This is continuous since it is continuous in each coordinate. Since any a ∈ J_{S}^{0} has ∥ a ∥= 1, and ∥ a_{v} ∥_{v}= 1 for v ∉ S, the image of J_{S}^{0} under this mapping lies in the hyperplane

x_{1} + ⋯ + x_{s} = 0,

where x_{1}, …, x_{s} are the usual coordinates on ℝ^{s}. Call this hyperplane H. Note that J_{S}^{0} generates H over ℝ, since log(J_{S}^{0}) contains the s − 1 linearly independent vectors

(c_{1}, 0, 0, …, 0, −c_{1})
(0, c_{2}, 0, …, 0, −c_{2})
⋮
(0, 0, 0, …, c_{s−1}, −c_{s−1}),

where the c_{i} are some non-zero constants depending on v_{i}.

Now, consider the S-units k_{S} = k^{∗} ∩ J_{S}^{0}. The set log(k_{S}) is in fact discrete. To see this, note that the elements of log(k_{S}) in any bounded region of ℝ^{s} have bounded archimedean absolute values, which in turn bounds the coefficients of the polynomials of these elements over ℤ. Since the degree is bounded by [k : ℚ], there are only finitely many such polynomials having bounded coefficients. Therefore only finitely many elements of k can map into any bounded region of ℝ^{s}, so log(k_{S}) is discrete. Thus, by Lemma 10.2, log(k_{S}) is a free abelian group. Note that the kernel of log |_{k_{S}} is just the roots of unity of k (since any algebraic integer which always has archimedean absolute value 1 is a root of unity, by an argument similar to the one just given), so it will be enough to show that log(k_{S}) has rank s − 1.

Let W be the subspace of H generated by log(k_{S}). Then we have an induced continuous homomorphism

log : J_{S}^{0}/k_{S} = C_{S}^{0} → H/W.

The image of this map generates H/W as an ℝ-vector space, since J_{S}^{0} generates H. The image is also the image of the compact set C_{S}^{0}, and thus compact. But if H/W is non-trivial it has no non-trivial compact subgroups (it is just ℝ^{n} for some n), so we must have H/W = 0, and H = W. Thus log(k_{S}) generates all of H, so it has rank s − 1.

Corollary 10.4. The group of global units of a number field k is isomorphic to

W × ℤ^{r_{1}+r_{2}−1},

where W is the subgroup of roots of unity, r_{1} is the number of real embeddings of k and r_{2} is the number of complex embeddings of k.

Proof. We simply take S = S_{∞} in the preceding theorem.

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