On Čebotarev Sets

by Kay Wingberg at Heidelberg

The aim of this paper is to define a topology with sufficiently good properties on the set P_{K} of prime ideals of a number field K. The idea is, roughly speaking, that open sets are given by so-called Čebotarev sets, i.e. sets of the form

P_{L|K}(σ) = \{𝔭 ∈ P_{K} | 𝔭 is unramified in L, σ = (\frac{L|K}{𝔓} ), 𝔓|𝔭\},

where L|K is a finite Galois extension with Galois group G(L|K), σ ∈ G(L|K) and (\frac{L|K}{𝔓}) denotes the Frobenius automorphism with respect to 𝔓, 𝔓 an arbitrary extension of 𝔭 to L. The precise definition of the topology T_{K} of P_{K} is slightly more complicated (see §2) since we want that the natural map

φ_{K′|K} : (P_{K′}, T_{K′}) ⟶(P_{K}, T_{K}), 𝔓 ↦ 𝔓 ∩ K,

is continuous if K′|K is a finite extension. We will show that (P_{K}, T_{K}) is a strongly zero-dimensional (and so totally disconnected) Hausdorff space with countable base, and so metrizable, hence normal and completely regular (and not discrete). In particular, every point of (P_{K}, T_{K}) has a base of neighbourhoods consisting of both open and closed sets.

Furthermore we will see that the isolated points of this space have to be prime ideals whose underlying prime numbers ramify in the extension K|ℚ (and so the set of isolated points is finite), and that every open neighbourhood of a prime ideal whose underlying prime number is unramified in K|ℚ has infinitely many points.

In section 3 we consider uniform structures on P_{K} inducing the topology T_{K}. If 𝔘_{K} is the uniformity defined by finite partitions of P_{K} given by both open and closed sets, then the completion (P̂_{K}, 𝔘̂_{K}) of (P_{K}, 𝔘_{K}) is a profinite space, i.e. compact and totally disconnected. Finally we define in section 4 a metric on P_{K} (in the case K = ℚ) inducing the topology T_{K} and calculate it for small prime numbers.

The good properties of this topology are consequences of deep theorems in algebraic number theory. The Hausdorff property may illustrate this: it follows easily by considering certain number fields with suitable local behaviour. But the existence of these fields is a consequence of the theorem of Grunwald/Wang.

1 Čebotarev Sets

Let K be a number field and let P_{K} be the set of all prime ideals 𝔭 ≠ (0) of K. For a finite Galois extension L|K with Galois group G(L|K) we denote by

U(L|K) the set of prime ideals of K which are unramified in L,
D(L|K) the set of prime ideals of K which are completely decomposed in L,
R(L|K) the set of prime ideals of K ramifying in L.

For an element σ ∈ G(L|K) let

P_{L|K}(σ) = \{𝔭 ∈ U(L|K) | σ = (\frac{L|K}{𝔓} ) for a prime ideal 𝔓|𝔭 of L\},

where (\frac{L|K}{𝔓}) denotes the Frobenius automorphism with respect to 𝔓. Obviously, this set depends only on the conjugacy class ⟨⟨σ⟩⟩ = {τστ^{−1} | τ ∈ G(L|K)} of σ and

P_{L|K}(σ) ∩ P_{L|K}(τ) = ∅ if ⟨⟨σ⟩⟩ ≠ ⟨⟨τ⟩⟩,

and P_{L|K}(1) = D(L|K). By Čebotarev’s density theorem we have

δ(P_{L|K}(σ)) = \frac{#⟨⟨σ⟩⟩}{#G(L|K)} ,

where δ(S) = δ_{K}(S) denotes the Dirichlet density of a set S of primes of K. Observe that for a finite Galois extension L|K and a set S(K) of primes of K we have

δ_{L}(S(L)) = δ_{K}(S(K) ∩ D(L|K))) · [L : K],

where S(L) denotes the set of all extensions of S(K) to L. For sets S_{1} and S_{2} of primes, we use the notation

S_{1} ⊂_{∼} S_{2} :⟺ δ(S_{1}\S_{2}) = 0,

i.e. S_{1} is contained in S_{2} up to a set of primes of density zero, and

S_{1} =_{∼} S_{2} :⟺ S_{1} ⊂_{∼} S_{2} and S_{2} ⊂_{∼} S_{1}.

Proposition 1.1 Let N|L|K be finite Galois extensions, H = G(N|L) and let σ̄ ∈ G(L|K). Then

U(N|K) ∩ P_{L|K}(σ̄) = ⋃ ._{⟨⟨τ⟩⟩∩σH≠∅}P_{N|K}(τ) ,

where σ is a lifting of σ̄ to G(N|K); in particular

U(N|K) ∩ D(L|K) = ⋃ ._{⟨⟨τ⟩⟩∩H≠∅}P_{N|K}(τ) .

Proof: Let 𝔭 be a prime ideal of K which is unramified in N|K. Then 𝔭 ∈ P_{L|K}(σ̄) if and only if there exists a prime \overline{𝔓}|𝔭 of L such that σ̄ = (\frac{L|K}{\overline{𝔓}}), i.e. if

there exists a prime 𝔓|𝔭 of N such that σH = (\frac{N|K}{𝔓})H. This ist equivalent to the assertion that there exists an element in σH which is contained in the conjugacy class ⟨⟨τ⟩⟩ of τ = (\frac{N|K}{𝔓}) for some prime ideal 𝔓|𝔭 of N, i.e. if 𝔭 ∈ P_{N|K}(τ) for some τ ∈ G(N|K) with ⟨⟨τ⟩⟩ ∩ σH ≠ ∅. □

Corollary 1.2 Let L_{1}|K and L_{2}|K be finite Galois extensions, H_{i} = G(L_{1}L_{2}|L_{i}) and σ̄_{i} ∈ G(L_{i}|K), i = 1, 2. Then

P_{L_{1}|K}(σ̄_{1}) ∩ P_{L_{2}|K}(σ̄_{2}) = ⋃ ._{\begin{matrix}⟨⟨τ⟩⟩ ∩ σ_{1}H_{1} ≠ ∅ \\ ⟨⟨τ⟩⟩ ∩ σ_{2}H_{2} ≠ ∅ \end{matrix}} P_{L_{1}L_{2}|K}(τ)

Proof: Since

U(L_{1}|K) ∩ U(L_{2}|K) = U(L_{1}L_{2}|K),

the corollary is a consequence of proposition 1.1. □

If σ̄_{1} = 1 = σ̄_{2}, then the corollary above is just the assertion D(L_{1}|K) ∩ D(L_{2}|K) = D(L_{1}L_{2}|K).

Corollary 1.3 Let L_{1}|K and L_{2}|K be finite Galois extensions and σ_{i} ∈ G(L_{i}|K), i = 1, 2. Then

P_{L_{1}|K}(σ_{1}) ∩ P_{L_{2}|K}(σ_{2}) ≠ ∅

if and only if

(σ_{1})^{−1}(σ_{2})^{ρ} ∈ G(L_{1}L_{2}|L_{1} ∩ L_{2}) for some ρ ∈ G(L_{1}L_{2}|K)

(here σ_{i} denotes an arbitrary lifting of σ_{i} to G(L_{1}L_{2}|K)).

In particular, if L_{1} and L_{2} are linearly disjoint over K, then all sets P_{L_{1}|K}(σ_{1}) and P_{L_{2}|K}(σ_{2}) have a non-trivial intersection.

For an element τ of a finite group G we denote the stabilizer of τ under conjugation by G by St_{G}(τ).

Proposition 1.4 Let L_{1} and L_{2} be finite Galois extensions of K. For an element σ_{i} ∈ G(L_{i}|K) we denote its restriction to L_{1} ∩ L_{2} by \overline{σ_{i}}, i = 1, 2. Then the following assertions are equivalent:

(i) P_{L_{1}|K}(σ_{1}) ⊂_{∼} P_{L_{2}|K}(σ_{2}),

(ii) ⟨⟨\overline{σ_{1}}⟩⟩ = ⟨⟨\overline{σ_{2}}⟩⟩ and #St_{G(L_{2}|K)}(σ_{2}) = #St_{G(L_{1}∩L_{2}|K)}(\overline{σ_{2}}).

In particular, P_{L_{1}|K}(σ_{1}) =_{∼} P_{L_{2}|K}(σ_{2}) if and only if ⟨⟨\overline{σ_{1}}⟩⟩ = ⟨⟨\overline{σ_{2}}⟩⟩ and

#St_{G(L_{1}|K)}(σ_{1}) = #St_{G(L_{1}∩L_{2}|K)}(\overline{σ_{1}}) = #St_{G(L_{2}|K)}(σ_{2}).

Proof: Let N = L_{1}L_{2}, H_{i} = G(N|L_{i}), i = 1, 2, and H = G(L_{2}|L_{1} ∩ L_{2}) ≅ H_{1}. We lift σ_{i} to G(N|K) and denote it again by σ_{i}. Assume that (i) holds, i.e.

P_{L_{1}|K}(σ_{1}) =_{∼} ⋃ ._{⟨⟨τ⟩⟩∩σ_{1}H_{1}≠∅}P_{N|K}(τ) ⊂_{∼} ⋃ ._{⟨⟨τ⟩⟩∩σ_{2}H_{2}≠∅}P_{N|K}(τ) =_{∼} P_{L_{2}|K}(σ_{2}).

Since the sets P_{N|K}(τ) have positive density, it follows that for every h_{1} ∈ H_{1} there exist h_{2} ∈ H_{2} and ρ ∈ G(N|K) such that σ_{1}h_{1} = (σ_{2})^{ρ}h_{2}, and so ⟨⟨\overline{σ_{1}}⟩⟩ = ⟨⟨\overline{σ_{2}}⟩⟩.

If h ∈ H is a fixed element and h̃ a lifting of h to G(N|K), then it follows that for every h_{1} ∈ H_{1} there exist h_{2} ∈ H_{2} and ρ ∈ G(N|K) such that σ_{1}h_{1} = (σ_{2}h̃)^{ρ}h_{2}, and so

⋃ ._{⟨⟨τ⟩⟩∩σ_{1}H_{1}≠∅}P_{N|K}(τ) ⊂_{∼} ⋃ ._{⟨⟨τ⟩⟩∩(σ_{2}h̃)H_{2}≠∅}P_{N|K}(τ).

We obtain P_{L_{1}|K}(σ_{1}) ⊂_{∼} P_{L_{2}|K}(σ_{2}h) for every h ∈ H, i.e.

P_{L_{1}|K}(σ_{1}) ⊂_{∼} ⋂_{h∈H} P_{L_{2}|K}(σ_{2}h).

Since

P_{L_{2}|K}(σ_{2}) ∩ P_{L_{2}|K}(σ_{2}h) ≠ ∅ if and only if ⟨⟨σ_{2}⟩⟩ = ⟨⟨σ_{2}h⟩⟩,

it follows that

P_{L_{1}∩L_{2}|K}(σ_{2}H) =_{∼} ⋃ ._{⟨⟨σ_{2}h⟩⟩,h∈H} P_{L_{2}|K}(σ_{2}h) = P_{L_{2}|K}(σ_{2}),

and therefore

\frac{#⟨⟨σ_{2}H⟩⟩_{G(L_{1}∩L_{2}|K)}}{#G(L_{1} ∩ L_{2}|K)} = \frac{#⟨⟨σ_{2}⟩⟩_{G(L_{2}|K)}}{#G(L_{2}|K)} .

From this equation we get

#St_{G(L_{2}|K)}(σ_{2}) = #St_{G(L_{1}∩L_{2}|K)}(σ_{2}H).

Conversely, using the arguments above in the other direction, we obtain from the assertion (ii) that

P_{L_{1}∩L_{2}|K}(σ_{2}H) =_{∼} P_{L_{2}|K}(σ_{2}).

Since ⟨⟨\overline{σ_{1}}⟩⟩ = ⟨⟨\overline{σ_{2}}⟩⟩, we get

P_{L_{1}|K}(σ_{1}) ⊂_{∼} P_{L_{1}∩L_{2}|K}(σ_{2}H)

and so (i). This finishes the proof of the proposition. □

Remark: From the proposition above it follows that it is possible that

P_{L|K}(\overline{σ}) =_{∼} P_{N|K}(σ),

where L is a proper subfield of the extension N|K, σ is an element of G(N|K) and \overline{σ} = σG(N|L). As an example let G(N|K) be the non-abelian group of order p^{3}, p an odd prime number, with generators σ, τ, ρ and the defining relations

σ^{p} = τ^{p} = ρ^{p} = [σ, ρ] = [τ, ρ] = 1 and [σ, τ] = ρ,

and let G(N|L) be the normal subgroup generated by ρ.

The following corollary is a generalization of a theorem of M.Bauer (see [3], theorem (13.9)).

Corollary 1.5 Let L_{1} and L_{2} be finite Galois extensions of K and let σ_{i} be an element of G(L_{i}|K), i = 1, 2. Assume that σ_{2} lies in the center of G(L_{2}|K). Then the following assertions are equivalent:

(i) P_{L_{1}|K}(σ_{1}) ⊂_{∼} P_{L_{2}|K}(σ_{2}),

(ii) L_{2} ⊆ L_{1} and ⟨⟨σ_{2}⟩⟩ = ⟨⟨σ_{1} mod G(L_{1}|L_{2})⟩⟩.

In particular, if σ_{i} lies in the center of G(L_{i}|K), i = 1, 2, then

P_{L_{1}|K}(σ_{1}) =_{∼} P_{L_{2}|K}(σ_{2}) if and only if L_{1} = L_{2} and ⟨⟨σ_{1}⟩⟩ = ⟨⟨σ_{2}⟩⟩.

Proof: By assumption σ_{2} lies in the center of G(L_{2}|K), and so

#St_{G(L_{2}|K)}(σ_{2}) = #St_{G(L_{1}∩L_{2}|K)}(\overline{σ_{2}})

if and only if G(L_{2}|L_{1} ∩ L_{2}) = 1, i.e. L_{2} ⊆ L_{1}. Now the corollary follows from proposition 1.4. □

Application to quadratic fields

We will use the following notation: the non-trivial element of the Galois group G(ℚ(\sqrt{a})|ℚ), a a squarefree integer, is denoted by −1. If σ ∈ \{±1\}, then σ′ ∈ \{±1\} is defined by σσ′ = −1. For x, y ∈ P_{ℚ}, y odd, \binom{x}{y} is the Legendre symbol. Let ε(ℓ) = (ℓ − 1)/2 if ℓ is odd, and zero otherwise, and let

L_{p} = ℚ(\sqrt{(−1)^{ε(p)}p}, p ∈ P_{ℚ}.

For σ̃ = (σ_{p})_{p} ∈ ∏_{p∈P_{ℚ}}{±1} and ε ∈ {±1} let

δ^{ε}(σ̃) = δ({p ∈ P_{ℚ} | σ_{p} = ε})

be the density of the set of prime numbers p such that σ_{p} has a given value.

Proposition 1.6 Let σ̃ = (σ_{p})_{p} ∈ ∏_{p∈P_{ℚ}}{±1} and S ⊆ P_{ℚ} a set of density equal to 1. Then

(i) #(P_{ℚ}\ ⋃_{p∈S} P_{L_{p}|ℚ}(σ_{p})) ≤ 1,

(ii) if #(P_{ℚ}\ ⋃_{p∈S} P_{L_{p}|ℚ}(σ_{p})) = 1, then δ^{1}(σ̃) = δ^{−1}(σ̃) = 1/2.

Proof: Suppose that

q_{1}, q_{2} ∈ ⋂_{p∈S} (P_{L_{p}|ℚ}(σ′_{p}) ∪ {p}).

If q_{1} and q_{2} are odd, then

\binom{(−1)^{ε(p)}p}{q_{1}} = σ′_{p} = \binom{(−1)^{ε(p)}p}{q_{2}}, and so \binom{q_{1}}{p} = \binom{q_{2}}{p}

for all p ∈ S\{2, q_{1}, q_{2}}. From corollary 1.5 it follows that q_{1} = q_{2}. If q_{2} = 2, then

2 ∈ P_{L_{p}|ℚ}(σ′_{p}), p odd, if and only if \binom{2}{p} = σ′_{p}

((−1)^{ε(p)}p ≡ 1 mod 4 and 2 splits in L_{p} if and only if (−1)^{ε(p)}p ≡ 1 mod 8, i.e. \binom{2}{p} = 1.) Again it follows that \binom{q_{1}}{p} = \binom{2}{p} for all p ∈ S\\{2, q_{1}\}, and so q_{1} = 2. Since

P_{ℚ}\ ⋃_{p∈S} P_{L_{p}|ℚ}(σ_{p}) = ⋂_{p∈S} (P_{L_{p}|ℚ}(σ′_{p}) ∪ {p}),

we proved (i).

If q ∈ ⋂_{p∈S} (P_{L_{p}|ℚ}(σ′_{p}) ∪ {p}), and so

q ∈ ⋂_{p∈S,p≠q} P_{L_{p}|ℚ}(σ′_{p}),

then \binom{q}{p} = σ′_{p} for all p ∈ S\\{2, q\}. It follows that δ^{1}(σ̃) = δ^{−1}(σ̃) = 1/2. □

Definition 1.7 A set S of prime ideals of K is called Čebotarev set if there exist a finite Galois extension L of K and an element σ ∈ G(L|K) such that

S = P_{L|K}(σ).

We set

C_{K} = {S ⊆ P_{K} is a Čebotarev set}.

For a finite extension K′|K let

φ_{K′|K} : P_{K′} ⟶ P_{K}, 𝔓 ↦ 𝔭 = 𝔓 ∩ K,

and we also denote the corresponding map on the set of all subsets of P_{K′} by φ_{K′|K}. For a subfield E ⊆ K, a finite Galois extension F |E and an element σ ∈ G(F |E),

[figure: K; F; Galois; E]

let

P_{F |E}^{K}(σ) = φ_{K|E}^{−1}P_{F |E}(σ),
U^{K}(F |E) = φ_{K|E}^{−1}U(F |E) and R^{K}(F |E) = φ_{K|E}^{−1}R(F |E).

Definition 1.8 For a number field K let

B_{K} = ⋃_{E⊆K} φ_{K|E}^{−1}C_{E} ={P_{F |E}^{K}(σ) | E ⊆ K, F |E finite Galois , σ ∈ G(F |E)}

and

A_{K} ={ ⋂_{i=1}^{n} P_{F_{i}|E_{i}}^{K}(σ_{i}) | n ∈ ℕ, P_{F_{i}|E_{i}}^{K}(σ_{i}) ∈ B_{K}}.

Observe that C_{K} ⊆ B_{K} ⊆ A_{K}.

In the next section we will need the following two lemmas.

Lemma 1.9 Let K|ℚ be a finite Galois extension and for i = 1, …, n let E_{i} ⊆ K be subfields of K, F_{i}|E_{i} finite Galois extensions and σ_{i} ∈ G(F_{i}|E_{i}). Then

P_{K|ℚ}^{K}(1) ∩ ⋂_{i=1}^{n} P_{F_{i}|E_{i}}^{K}(σ_{i})

is empty or has positive density.

Proof: Let F |E be one of the extensions F_{i}|E_{i}. Then

φ_{K|ℚ}^{−1}P_{K|ℚ}(1) ∩ φ_{K|E}^{−1}P_{F |E}(σ) = φ_{K|ℚ}^{−1}P_{K|ℚ}(1) ∩ ⋃ ._{⟨⟨τ⟩⟩∩σH≠∅}P_{F K|K}(τ),

where H = G(F K|F). Indeed, let 𝔓_{K} ∈ φ_{K|ℚ}^{−1}P_{K|ℚ}(1) ⊆ P_{K} and consider the diagram of fields

[figure: F K; F; K; F ∩ K; E.]

Let 𝔓_{F} be an extension of 𝔭 = 𝔓_{K} ∩ E to F and 𝔓_{F K} an extension of 𝔓_{F} to F K. Then 𝔓′_{K} = 𝔓_{F K} ∩ K is conjugated to 𝔓_{K}. Since 𝔓_{F K} is unramified over K and the residue degree f(𝔓′_{K}|𝔭) = 1, we have

(\frac{F K|K}{𝔓_{F K}} )_{|F} = (\frac{F |E}{𝔓_{F}} ).

Now the equality stated above follows easily. Thus we obtain

P_{K|ℚ}^{K}(1) ∩ ⋂_{i=1}^{n} P_{F_{i}|E_{i}}^{K}(σ_{i})
= φ_{K|ℚ}^{−1}P_{K|ℚ}(1) ∩ ⋂_{i=1}^{n} ⋃ ._{⟨⟨τ_{i}⟩⟩∩σ_{i}H_{i}≠∅}P_{F_{i}K|K}(τ_{i})
= φ_{K|ℚ}^{−1}P_{K|ℚ}(1) ∩ ⋃ ._{⟨⟨τ_{1}⟩⟩∩σ_{1}H_{1}≠∅}⋯ ⋃ ._{⟨⟨τ_{n}⟩⟩∩σ_{n}H_{n}≠∅}(P_{F_{1}K|K}(τ_{1}) ∩ ⋯ ∩ P_{F_{n}K|K}(τ_{n})).

From corollary 1.2 it follows that the sets

P_{F_{1}K|K}(τ_{1}) ∩ ⋯ ∩ P_{F_{n}K|K}(τ_{n})

are empty or have positive density. Since the density of φ_{K|ℚ}^{−1}P_{K|ℚ}(1) is equal to 1, we proved the lemma. □

Lemma 1.10 Let K be a number field and for i = 1, …, n let E_{i} ⊆ K be subfields of K, F_{i}|E_{i} finite Galois extensions, E = ⋂_{i} E_{i} and σ_{i} ∈ G(F_{i}|E_{i}). Then the set

S = φ_{K|E}^{−1}U(K|E) ∩ ⋂_{i=1}^{n} P_{F_{i}|E_{i}}^{K}(σ_{i})

is empty or infinite.

Proof: We may assume that K = F_{1} = ⋯ = F_{n} and that K|E is a Galois extension. In order to see this let M be the normal closure of KF_{1} ⋯ F_{n} over E and S_{M} = φ_{M|K}^{−1}(S). Then S_{M} is non-empty and finite if and only S is. We get

S_{M} = φ_{M|E}^{−1}U(K|E) ∩ φ_{M|E_{1}}^{−1}P_{F_{1}|E_{1}}(σ_{1}) ∩ ⋯ ∩ φ_{M|E_{n}}^{−1}P_{F_{n}|E_{n}}(σ_{n})
= φ_{M|E}^{−1}U(M|E) ∩ φ_{M|E_{1}}^{−1}P_{F_{1}|E_{1}}(σ_{1}) ∩ ⋯ ∩ φ_{M|E_{n}}^{−1}P_{F_{n}|E_{n}}(σ_{n})
= φ_{M|E}^{−1}U(M|E) ∩ ⋃ ._{⟨⟨τ_{1}⟩⟩∩σ_{1}H_{1}≠∅} φ_{M|E_{1}}^{−1}P_{M|E_{1}}(τ_{1}) ∩ ⋯ ∩ ⋃ ._{⟨⟨τ_{n}⟩⟩∩σ_{n}H_{n}≠∅} φ_{M|E_{n}}^{−1}P_{M|E_{n}}(τ_{n})
= φ_{M|E}^{−1}U(M|E) ∩ ⋃ ._{\begin{matrix}⟨⟨τ_{i}⟩⟩ ∩ σ_{i}H_{i} ≠ ∅ \\ i = 1, …, n \end{matrix}}(φ_{M|E_{1}}^{−1}P_{M|E_{1}}(τ_{1}) ∩ ⋯ ∩ φ_{M|E_{n}}^{−1}P_{M|E_{n}}(τ_{n})),

where H_{i} = G(M|F_{i}), i = 1, …, n, and so it is enough to show that the sets

S(τ_{1}, …, τ_{n}) = φ_{M|E}^{−1}U(M|E) ∩ φ_{M|E_{1}}^{−1}P_{M|E_{1}}(τ_{1}) ∩ ⋯ ∩ φ_{M|E_{n}}^{−1}P_{M|E_{n}}(τ_{n})

are empty or infinite. For a set T of primes of M let (T)_{G(M|E)} be the closure under conjugation by G(M|E). Obviously, it is sufficient to show that (S(τ_{1}, …, τ_{n}))_{G(M|E)} is empty or infinite.

Suppose that 𝔓 ∈ S(τ_{1}, …, τ_{n}) and let 𝔭 = 𝔓 ∩ E. We fix an index i. Then 𝔓_{E_{i}} = 𝔓 ∩ E_{i} ∈ P_{M|E_{i}}(τ_{i}), i.e. there exists an extension 𝔓_{M} of 𝔓_{E_{i}} in M such that τ_{i} = (\frac{M|E_{i}}{𝔓_{M}}). Since 𝔓 and 𝔓_{M} are conjugated over E_{i}, it follows that there

is an element ρ_{i} ∈ G(M|E_{i}) such that τ_{i}^{ρ_{i}} = (\frac{M|E_{i}}{𝔓}) and we may assume that

τ_{i} = (\frac{M|E_{i}}{𝔓}).

Since 𝔓 ∈ φ_{M|E}^{−1}U(M|E), we have the element σ = (\frac{M|E}{𝔓}) ∈ G(M|E) and it follows that

τ_{i} = (\frac{M|E_{i}}{𝔓} ) = (\frac{M|E}{𝔓} )^{f(𝔓_{E_{i}}|𝔭)} = σ^{f(𝔓_{E_{i}}|𝔭)},

where f(𝔓_{E_{i}}|𝔭) is the inertia degree of 𝔓_{E_{i}} over E. We claim that

φ_{M|E}^{−1}P_{M|E}(σ) ⊆ (S(τ_{1}, …, τ_{n}))_{G(M|E)}.

Indeed, let 𝔓′ ∈ φ_{M|E}^{−1}P_{M|E}((\frac{M|E}{𝔓})). Then there exists a prime 𝔓′′ of M which is

conjugated to 𝔓′ over E such that (\frac{M|E}{𝔓}) = (\frac{M|E}{𝔓′′}). Let f(𝔓′′_{E_{i}}|𝔭) be the inertia degree of 𝔓′′_{E_{i}} over E. Since

G(M|E_{i}) ∩ G_{𝔓′′}(M|E) = G_{𝔓′′}(M|E_{i}),

we get

(\frac{M|E}{𝔓′′} )^{f(𝔓_{E_{i}}|𝔭)} = (\frac{M|E}{𝔓} )^{f(𝔓_{E_{i}}|𝔭)} = (\frac{M|E_{i}}{𝔓} ) ∈ G_{𝔓′′}(M|E_{i}).

Since G_{𝔓′′}(M|E_{i}) is generated by (\frac{M|E_{i}}{𝔓′′}) = (\frac{M|E}{𝔓′′})^{f(𝔓′′_{E_{i}}|𝔭)}, f(𝔓′′_{E_{i}}|𝔭) divides f(𝔓_{E_{i}}|𝔭). Analogously,

(\frac{M|E}{𝔓} )^{f(𝔓′′_{E_{i}}|𝔭)} = (\frac{M|E}{𝔓′′} )^{f(𝔓′′_{E_{i}}|𝔭)} = (\frac{M|E_{i}}{𝔓′′} ) ∈ G_{𝔓}(M|E_{i}),

and so f(𝔓_{E_{i}}|𝔭) divides f(𝔓′′_{E_{i}}|𝔭). Therefore we obtain

(\frac{M|E_{i}}{𝔓′′} ) = (\frac{M|E}{𝔓′′} )^{f(𝔓′′_{E_{i}}|𝔭)} = (\frac{M|E}{𝔓} )^{f(𝔓_{E_{i}}|𝔭)} = (\frac{M|E_{i}}{𝔓} ) = τ_{i}.

It follows that 𝔓′′ ∈ φ_{M|E_{i}}^{−1}P_{M|E_{i}}(τ_{i}) for all i = 1, …, n, i.e. 𝔓′′ ∈ S(τ_{1}, …, τ_{n}), and so 𝔓′ ∈ (S(τ_{1}, …, τ_{n}))_{G(M|E)}. This proves the claim. Since φ_{M|E}^{−1}P_{M|E}(σ) is an infinite set, we proved the lemma. □

We finish this section with a slightly more general version of the theorem of Grunwald/Wang (see also [4], theorem (9.2.2)).

Let p be a prime number, K a number field and S ⊇ T sets of primes of K, where S contains the set S_{p} ∪ S_{∞} of archimedean primes and primes above p. Let K_{S} be the maximal extension of K which is unramified outside S. By μ_{p} we denote the group of all p-th roots of unity.

Theorem 1.11 Let K be a number field and let S ⊇ T be sets of primes of K, where S ⊇ S_{p} ∪ S_{∞}, T is finite and

δ(S ∩ D(K(μ_{p})|K)) > \frac{1}{p [K(μ_{p}) : K]} .

Then the canonical homomorphism

H^{1}(K_{S}|K, ℤ/pℤ) ⟶ ⨁_{𝔭∈T}H^{1}(K_{𝔭}, ℤ/pℤ)

is surjective.

Proof: Using [4], lemma (9.2.1), it is enough to show that the canonical map

H^{1}(K_{S}|K, μ_{p}) ⟶ ∏_{𝔭∈(S\T)(K)}H^{1}(K_{𝔭}, μ_{p})

is injective. Since [K(μ_{p}) : K] is prime to p, it is sufficient to show the injectivity of the homomorphism

H^{1}(K_{S}|K(μ_{p}), μ_{p}) ⟶ ∏_{𝔓∈(S\T)(K(μ_{p}))}H^{1}(K(μ_{p})_{𝔓}, μ_{p}).

An element of the kernel corresponds to a Galois extension L|K(μ_{p}) of degree p which is unramified outside S(K(μ_{p})) and completely decomposed at (S\T)(K(μ_{p})). Since

\begin{aligned}δ_{K(μ_{p})}((S\T)(K(μ_{p}))) & = & δ_{K(μ_{p})}(S(K(μ_{p})) \\ & = & δ_{K}(S(K) ∩ D(K(μ_{p})|K)) · [K(μ_{p}) : K] > \frac{1}{p} , \end{aligned}

such an extension has to be trivial. □

2 Topology

In this section we define a topology on the set P_{K} of non-trivial prime ideals of a number field K.

Definition 2.1 For a number field K let T_{K} be the topology on P_{K} generated by

A_{K} ={ ⋂_{i=1}^{n} P_{F_{i}|E_{i}}^{K}(σ_{i}) | n ∈ ℕ, P_{F_{i}|E_{i}}^{K}(σ_{i}) ∈ B_{K}},

i.e. A_{K} is a base and

B_{K} ={P_{F |E}^{K}(σ) | E ⊆ K, F |E a finite Galois extension , σ ∈ G(F |E)}

is a subbase of T_{K}. Obviously, the topology T_{K} has a countable base.

Remark: From proposition 1.2 it follows that

C_{ℚ} ∪ {∅}={P_{F |ℚ}(σ) | F |ℚ a finite Galois extension, σ ∈ G(F |ℚ)} ∪ {∅}

is a base of T_{ℚ}.

Proposition 2.2 If K′|K is a finite extension, then the map

φ_{K′|K} : (P_{K′}, T_{K′}) ⟶(P_{K}, T_{K}), 𝔓 ↦ 𝔓 ∩ K,

is continuous.

Proof: This follows directly from the definition of the topologies T_{K} and T_{K′}.

□

An observation, which is not quite obvious, is the following:

T_{K} is not the discrete topology on P_{K}.

In order to see this, suppose that T_{K} is the discrete topology. Then for every point 𝔭 ∈ P_{K} the set {𝔭} is open and therefore there exist finite Galois extensions F_{i}|E_{i}, E_{i} ⊆ K, and σ_{i} ∈ G(F_{i}|E_{i}), i = 1, …, n, such that

{𝔭} = ⋂_{i=1}^{n} P_{F_{i}|E_{i}}^{K}(σ_{i}).

But if 𝔭 is contained in U^{K}(K|ℚ), then this equality contradicts lemma 1.10.

For a subset W of P_{K} we denote the closure of W by \overline{W}.

Proposition 2.3

(i) Let P_{F |E}^{K}(σ) ∈ B_{K}. Then

P_{K}\(P_{F |E}^{K}(σ) ∪ R^{K}(F |E)) = ⋃ ._{⟨⟨τ⟩⟩≠⟨⟨σ⟩⟩} P_{F |E}^{K}(τ),

and so P_{F |E}^{K}(σ) ∪ R^{K}(F |E) is a closed set. In particular, \overline{P_{F |E}^{K}(σ)}\P_{F |E}^{K}(σ) is a finite set.

(ii) Let F |ℚ be a Galois extension of prime degree and let σ ∈ G(F |ℚ). Then P_{F |ℚ}(σ) ∪ R(F |ℚ) is the closure of P_{F |ℚ}(σ).

Proof: Assertion (i) follows from the equation

P_{K} = ⋃ ._{⟨⟨τ⟩⟩} φ_{K|E}^{−1}P_{F |E}(τ) ∪. φ_{K|E}^{−1}R(F |E).

In order to prove (ii), suppose the contrary is true. Then there exists a prime number p ∈ R(F |ℚ) and an open neighbourhood U = P_{L|ℚ}(τ) of p, L|ℚ a finite Galois extension, such that U does not meet P_{F |ℚ}(σ). From corollary 1.3 it follows that F and L are not linearly disjoint over ℚ, and so F ⊆ L. But p is unramified in L and ramifies in F. This contradiction shows assertion (ii). □

Remark: In general the set P_{F |E}^{K}(σ) ∪ R^{K}(F |E) is not necessarily the closure of P_{F |E}^{K}(σ), since there may be isolated points in the set R^{K}(F |E), see proposition 2.9, or there may exist subextensions of F |E in which elements of R^{K}(F |E) are unramified.

Proposition 2.4

(i) For every two different points 𝔭_{1} and 𝔭_{2} of (P_{K}, T_{K}) there exists a both open and closed neighbourhood W ∈ A_{K} of 𝔭_{1} such that 𝔭_{2} ∉ W.

(ii) Let 𝔭_{1}, …, 𝔭_{n} be pairwise different points of (P_{K}, T_{K}). Then there exist both open and closed neighbourhoods U(𝔭_{i}) of 𝔭_{i} such that

U(𝔭_{i}) ∩ U(𝔭_{j}) = ∅ for i ≠ j.

Proof: In order to prove (i) let L|K be a cyclic extension of degree m > 2 such that 𝔭_{1} is unramified in L|K and let σ ∈ G(L|K) with 𝔭_{1} ∈ P_{L|K}(σ). We denote the open neighbourhood P_{L|K}(σ) of 𝔭_{1} by U.

Let N|K be a quadratic extension of K which is unramified at all primes of U, completely decomposed at R(L|K) ∪ {𝔭_{2}} and inert at 𝔭_{1}; if V = P_{N|K}(τ), where τ is the non-trivial element of G(N|K), then 𝔭_{1} ∈ V and 𝔭_{2} ∉ V. Such an extension exists. Indeed, let

T = S_{2} ∪ S_{∞} ∪ R(L|K) ∪ {𝔭_{1}, 𝔭_{2}} and S = (P_{K}\U) ∪ T,

then

δ_{K}(S) = 1 − \frac{1}{m} > \frac{1}{2},

and so we can apply theorem 1.11: there exists an element φ ∈ H^{1}(K_{S}|K, ℤ/2ℤ) such that

res_{𝔭}(φ) = 0 ∈ H^{1}(K_{𝔭}, ℤ/2ℤ)) for 𝔭 ∈ T \{𝔭_{1}}

and

0 ≠ res_{𝔭_{1}}(φ) ∈ H_{nr}^{1}(K_{𝔭_{1}}, ℤ/2ℤ) ⊂ H^{1}(K_{𝔭_{1}}, ℤ/2ℤ).

If ker φ = G(K_{S}|N), then N is a quadratic extension of K with the desired properties.

Now W = U ∩ V is an open neighbourhood of 𝔭_{1} and 𝔭_{2} ∉ W. It remains to show that W is closed. Let \overline{W} be the closure of W. Using proposition 2.3(i), we get

\overline{U ∩ V} ⊆ \overline{U} ∩ \overline{V} ⊆ (U ∪ R(L|K)) ∩ (V ∪ R(N|K)) = U ∩ V,

and so \overline{W} = W. This finishes the proof of (i).

In order to prove (ii) we use induction with respect to n. Assume that we have found open and closed neighbourhoods W (𝔭_{i}) of 𝔭_{i}, i = 1, …, n − 1, which are pairwise disjoint. By (i) it follows that for every i ∈ {1, …, n−1} there exists an open and closed neighbourhood W_{i}(𝔭_{n}) of 𝔭_{n} such that 𝔭_{i} ∉ W_{i}(𝔭_{n}). Then

U(𝔭_{n}) = ⋂_{i=1}^{n−1} W_{i}(𝔭_{n})

is an open and closed neighbourhood of 𝔭_{n} such that 𝔭_{i} ∉ U(𝔭_{n}) for all i = 1, …, n − 1. Now the open and closed neighbourhoods U(𝔭_{i}) = W (𝔭_{i})\U(𝔭_{n}), i = 1, …, n − 1, and U(𝔭_{n}) have the desired property. □

Recall that a Hausdorff space X is called zero-dimensional if every point of X has a fundamental system of neighbourhoods which are both open and closed, and X is called strongly zero-dimensional if for every closed subset A of X and each neighbourhood U of A there is an open and closed neighbourhood of A contained in U.

Proposition 2.5 The space (P_{K}, T_{K}) has the following properties: it is

(i) a Hausdorff space, (ii) strongly zero-dimensional (and so totally disconnected), (iii) metrizable (and so normal and completely regular), (iv) every point of (P_{K}, T_{K}) has a base of neighbourhoods consisting of both open

and closed sets.

Proof: By proposition 2.4(i) there exists for every two different points x and y of P_{K} an open and closed neighbourhood W of x such that y ∉ W. It follows that P_{K}\W is an open neighbourhood of y being disjoint to W. Therefore P_{K} is a Hausdorff space.

Now we proved (iv). Let 𝔭 ∈ (P_{K}, T_{K}) and let

U = ⋂_{i=1}^{n} P_{F_{i}|E_{i}}^{K}(σ_{i}) ∈ A_{K}

be an open neighbourhood of 𝔭. We have to find and open and closed neighbourhood of 𝔭 being contained in U. Obviously we may assume that U = P_{F |E}^{K}(σ). By proposition 2.4(ii) there exist open und closed, pairwise disjoint neighbourhoods U(𝔭_{i}) of 𝔭_{i} i = 0, …, n, where {𝔭_{1} …, 𝔭_{n}} = R^{K}(F |E) and 𝔭_{0} = 𝔭. Then

U_{R} = ⋃ ._{i=1}^{n} U(𝔭_{i})

is an open and closed neighbourhood of R^{K}(F |E) not containing 𝔭. Let

V = P_{F |E}^{K}(σ)\U_{R},

then V is open and contains 𝔭. But V is also closed, since we get for the closure \overline{V} of V, using proposition 2.3(i),

\overline{P_{F |E}^{K}(σ)\U_{R}} = \overline{P_{F |E}^{K}(σ) ∩ (P_{K}\U_{R})}
⊆ \overline{P_{F |E}^{K}(σ)} ∩ (P_{K}\U_{R})
⊆ (P_{F |E}^{K}(σ) ∪ R^{K}(F |E)) ∩ (P_{K}\U_{R})
= P_{F |E}^{K}(σ)\U_{R}.

This finishes the proof of (iv). The other assertions follow from [2] IX.6 exercise 2(b) since the considered space has a countable base. □

Remarks:

1. The space (P_{K}, T_{K}) is not compact. Otherwise (P_{ℚ}, T_{ℚ}) would be compact. But the following is true:

Every compact subset A of P_{ℚ} has no interior point.

Suppose the contrary, i.e. there is an open subset U of P_{ℚ} contained in A. Then, by the remark following definition 2.1, we may assume that U is a Čebotarev set: U = P_{F |ℚ}(σ). Let L_{p_{1}}, …, L_{p_{n}} be the finitely many quadratic fields with prime number discriminant contained in F; here we use the notation of proposition 1.6. Using this proposition, we get

⋃_{p≠p_{1},...,p_{n}}(P_{L_{p}|ℚ}(1) ∩ A) = A,

and so there finitely many prime numbers q_{1}, …, q_{m} ∉ {p_{1}, …, p_{n}} such that

⋃_{i=1}^{m}(P_{L_{q_{i}}|ℚ}(1) ∩ A) = A, i.e. A ⊆ ⋃_{i=1}^{m} P_{L_{q_{i}}|ℚ}(1).

Since

(P_{L_{q_{1}}|ℚ}(1) ∪ ⋯ ∪ P_{L_{qm}|ℚ}(1)) ∩ (P_{L_{q_{1}}|ℚ}(−1) ∩ ⋯ ∩ P_{L_{qm}|ℚ}(−1)) = ∅,

and, by the propositions 1.2 and 1.3,

A ∩ ⋂_{i=1}^{m} P_{L_{q_{i}}|ℚ}(−1) ⊇ P_{F |ℚ}(σ) ∩ ⋂_{i=1}^{m} P_{L_{q_{i}}|ℚ}(−1) ≠ ∅,

we get a contradiction.

2. If the set P′_{K} = P_{K} ∪ {(0)} of all prime ideals of the ring of integers O_{K} of K is equipped with the sum of the topologies T_{K} of P_{K} and the discrete topology on {(0)}, then the identity map

(P′_{K}, T′_{K}) ⟶^{id} Spec_{Zar} O_{K}

is continuous, since the non-trivial closed subsets of Spec O_{K} with respect to the Zariski topology are the finite sets not containing (0) and (P_{K}, T_{K}) is a Hausdorff space.

Lemma 2.6 If (X, T ) is a strongly zero-dimensional topological space, then for every finite open covering ⋃_{i=1}^{n} U_{i} = X of X there exist pairwise disjoint, open and closed sets V_{i} ⊆ U_{i}, i = 1, …, n, such that

⋃ ._{i=1}^{n} V_{i} = X.

Proof: Assume that we have found open and closed set W_{i} ⊆ U_{i}, i = 1, …, m, with m ≤ n such that

⋃_{i=1}^{m} W_{i} ∪ ⋃_{i=m+1}^{n} U_{i} = X.

If A = X\(⋃_{i=1}^{m} W_{i} ∪ ⋃_{i=m+2}^{n} U_{i}), then A is closed and A ⊆ U_{m+1}. Since X is strongly zero-dimensional, there exists an open and closed set W_{m+1} with A ⊆ W_{m+1} ⊆ U_{m+1}. It follows that

⋃_{i=1}^{m+1} W_{i} ∪ ⋃_{i=m+2}^{n} U_{i} = X,

and finally ⋃_{i=1}^{n} W_{i} = X. Now the sets

V_{i} = W_{i}\ ⋃_{j<i} W_{j} ⊆ U_{i}

are open and closed, pairwise disjoint and they form a covering of X. □

Proposition 2.7 If ⋃_{i=1}^{n} U_{i} = P_{K} is a finite open covering of the space (P_{K}, T_{K}), then there exist pairwise disjoint, open and closed sets V_{i} ⊆ U_{i}, i = 1, …, n, such that

⋃ ._{i=1}^{n} V_{i} = P_{K}.

Proof: This follows from lemma 2.6 and proposition 2.5. □

Proposition 2.8

(i) Let 𝔭 ∈ (P_{K}, T_{K}) be a prime ideal of K such that p = 𝔭 ∩ ℚ is completely decomposed in K. Then every open neighbourhood of 𝔭 has positive density.

(ii) Let 𝔭 ∈ (P_{K}, T_{K}) be a prime ideal of K such that p = 𝔭 ∩ ℚ is unramified in K. Then every open neighbourhood of 𝔭 has infinitely many points.

Proof: Let 𝔭 ∈ (P_{K}, T_{K}) such that p = 𝔭 ∩ ℚ is completely decomposed in K and let U be an open neighbourhood of 𝔭. The prime number p is also completely decomposed in the normal closure N of K|ℚ. If 𝔓 is an extension of 𝔭 to N, then V = φ_{N|K}^{−1}(U) is an open neighbourhood of 𝔓. Since every open neighbourhood of a point of (P_{N}, T_{N}) contains a set which is a finite intersection of sets of B_{N}, it follows from lemma 1.9 that V has positive density, and so U has. This proves assertion (i) and (ii) follows from lemma 1.10. □

Recall that a point x of a topological space X is called isolated if {x} is an open set in X.

If G(F |E) is the Galois group of a finite Galois extension F |E and 𝔓 a prime of F, then we denote the decomposition group and the inertia subgroup of G(F |E) with respect to 𝔓 by G_{𝔓} = G_{𝔓}(F |E) and T_{𝔓} = T_{𝔓}(F |E), respectively. If ℓ is a prime number, then G(ℓ) is a ℓ-Sylow group of a group G.

Proposition 2.9 Let K|ℚ be a finite extension and let 𝔭 ∈ φ_{K|ℚ}^{−1}(R(K|ℚ)).

(i) Assume that K|ℚ is normal and that G_{𝔭}(K|ℚ) has the following property: there exists a prime number ℓ such that G_{𝔭}(ℓ) is not cyclic and the quotient G_{𝔭}(ℓ)/T_{𝔭}(ℓ) is non-trivial. Then 𝔭 is an isolated point of (P_{K}, T_{K}).

(ii) For every prime ideal 𝔓|𝔭 of the normal closure N of K|ℚ there exists a finite Galois extension L|N such that 𝔓 and all G(N|ℚ)-conjugates of 𝔓 are inert in L|N and their unique extensions to L are isolated in (P_{L}, T_{L}).

Proof: Let K_{0} ⊆ K be the fixed field of [G_{𝔭}(ℓ), G_{𝔭}(ℓ)]. From our assumptions it follows that K_{0} has subfields E_{i}, i = 0, 1, 2, such that K_{0} = E_{1}E_{2}, E_{0} = E_{1} ∩ E_{2} and

G(K_{0}|E_{0}) ≅ ℤ/ℓℤ × ℤ/ℓℤ,

and 𝔭 ∩ E_{1} is inert and 𝔭 ∩ E_{2} is ramified in K_{0}. Let E_{3} be any extension of E_{0} in K_{0} of degree ℓ different to E_{1} and E_{2}:

[figure: K_{0}; i; i; r; E_{1}; r; E_{3}; r; E_{2}; i; E_{0}.]

The letters i and r indicate whether 𝔭 ∩ E_{0} resp. its unique extensions to the fields E_{i}, i = 1, 2, 3, are inert or ramify in the considered extensions. Now we consider the open set

U = P_{K_{0}|E_{1}}^{K_{0}}(σ) ∩ P_{K_{0}|E_{3}}^{K_{0}}(τ) = φ_{K_{0}|E_{1}}^{−1}P_{K_{0}|E_{1}}(σ) ∩ φ_{K_{0}|E_{3}}^{−1}P_{K_{0}|E_{3}}(τ)

of (P_{K_{0}}, T_{K_{0}}), where

σ = (\frac{K_{0}|E_{1}}{𝔭 ∩ K_{0}} ), τ = (\frac{K_{0}|E_{3}}{𝔭 ∩ K_{0}} ).

Observe that σ ≠ 1 ≠ τ and 𝔭_{0} = 𝔭 ∩ K_{0} ∈ U.

Let 𝔭′ be a prime ideal contained in U. Since K_{0}|E_{0} is not cyclic and 𝔭′ ∩E_{1} is inert in K_{0}|E_{1}, 𝔭′ ∩E_{0} is completely decomposed or ramifies in E_{1}|E_{0}. In the first case its extensions to E_{3} would also be completely decomposed in K_{0}|E_{3}, and so 𝔭′ can not be contained in φ_{K_{0}|E_{3}}^{−1}P_{K_{0}|E_{3}}(τ). It follows that U ⊆ φ_{K_{0}|E_{0}}^{−1}R(K_{0}|E_{0}), and so U is finite. Therefore {𝔭_{0}} ⊆ U is also open (the finite set U\{𝔭_{0}} is closed as (P_{K_{0}}, T_{K_{0}}) is a Hausdorff space). Therefore 𝔭_{0} is an isolated point of (P_{K_{0}}, T_{K_{0}}), and so 𝔭 = φ_{K|K_{0}}^{−1}(𝔭_{0}) is an isolated point of (P_{K}, T_{K}). This proves assertion (i).

In order to prove (ii) let 𝔓 be a prime ideal contained in φ_{N|ℚ}^{−1}(R(N|ℚ)) and let ℓ be any prime number dividing the order of the inertia subgroup T_{𝔓} of G_{𝔓} = G_{𝔓}(N|ℚ). Let L_{0}|ℚ be a cyclic extension of ℓ-power degree such that 𝔓 ∩ ℚ is inert in L_{0}|ℚ and L_{0} ⊊ N. Let L = NL_{0}. Then all G(N|ℚ)-conjugates of 𝔓 are inert in L|N and G_{𝔓_{L}}(L|ℚ) fulfills the condition of (i), where 𝔓_{L} denotes the unique extension of 𝔓 to L. It follows that 𝔓_{L} is isolated in (P_{L}, T_{L}). □

Definition 2.10 Let K be a number field and N the normal closure of K|ℚ. A point 𝔭 ∈ (P_{K}, T_{K}) is called potentiell isolated if for every 𝔓|𝔭 of N there exists a finite Galois extension L|N such that

(i) all G(N|ℚ)-conjugates of 𝔓 are unramified in L|N, (ii) all points of φ_{L|N}^{−1}(𝔓) are isolated in (P_{L}, T_{L}).

We denote the set of all isolated points and the set of all potentiell isolated points of (P_{K}, T_{K}) by (P_{K})_{iso} and (P_{K})_{p.iso}, respectively.

Without condition (i) in the definition above, i.e. φ_{N|ℚ}^{−1}(𝔓 ∩ ℚ) ⊆ U(L|N), all points of P_{K} would be potentiell isolated, since for every 𝔭 ∈ P_{K} there exists a finite Galois extension K′|K in which 𝔭 ramifies, and we can apply proposition 2.9(ii) to the field K′. Furthermore we would like to mention (although it is completely trivial) that

P_{ℚ} has no isolated points,

since every open set of P_{ℚ} has positive density. The following proposition considers the general case.

Theorem 2.11 Let K be a number field. Then the following is true:

(i) (P_{K})_{iso} ⊆ φ_{K|ℚ}^{−1}(R(K|ℚ)) = (P_{K})_{p.iso} ,

(ii)

φ_{K|ℚ}^{−1}(U(K|ℚ)) ⊆ \{𝔭 ∈ P_{K} | \begin{matrix}every open neighbourhood of 𝔭 \\ has infinitely many points \end{matrix} \},

(iii)

φ_{K|ℚ}^{−1}(D(K|ℚ)) ⊆ \{𝔭 ∈ P_{K} | \begin{matrix}every open neighbourhood of 𝔭 \\ has positive density \end{matrix} \}.

If K|ℚ is a Galois extension, then we have equality in (iii).

Proof: Let N be the normal closure over K over ℚ. The inclusion

φ_{K|ℚ}^{−1}(R(K|ℚ)) ⊆ (P_{K})_{p.iso}

is just proposition 2.9(ii). In order to prove the other inclusion suppose that 𝔭 ∈ (P_{K})_{p.iso} is not contained in φ_{K|ℚ}^{−1}(R(K|ℚ)). Then the extensions 𝔓 of 𝔭 to N are contained in φ_{N|ℚ}^{−1}(U(N|ℚ)). Let 𝔓_{0} be one of these extensions and let L|N be a finite Galois extension such that all G(N|ℚ)-conjugates of 𝔓_{0} are unramified in L|N and all points 𝔓_{0L} ∈ φ_{L|N}^{−1}(𝔓_{0}) are isolated in (P_{L}, T_{L}). Then 𝔓_{0L} ∈ φ_{L|ℚ}^{−1}(U(L|ℚ)). This contradicts proposition 2.8(ii) and therefore we proved the equality stated in (i). Assertions (ii) (and so the inclusion in (i)) and the inclusion (iii) follow from proposition 2.8(ii) and (i), respectively.

Now we show that for every point 𝔓 ∈ φ_{N|ℚ}^{−1}(U(N|ℚ))\φ_{N|ℚ}^{−1}(D(N|ℚ)) there exists an open neighbourhood of density equal to 0. Indeed, let N_{0} ⊂ N be its decomposition field and observe that by assumption N ≠ N_{0}. Therefore τ = (\frac{N|N_{0}}{𝔓}) ∈ G(N|N_{0}) is not equal to 1. Obviously, 𝔓 ∈ φ_{N|N_{0}}^{−1}P_{N|N_{0}}(τ) and this open set has density equal to 0 since every prime ideal of P_{N|N_{0}}(τ) is inert in the extension N|N_{0}. So we get

φ_{N|ℚ}^{−1}(D(N|ℚ)) = \{𝔓 ∈ P_{N} | \begin{matrix}every open neighbourhood of 𝔓 \\ has positive density \end{matrix} \}

showing also the last assertion of the theorem. □

Remark 1: The inclusion in (ii) may be strict (even if K|ℚ is a Galois extension), i.e. there may exist ramified primes having only infinite open neighbourhoods, or with other words, it is possible that there are ramified points which are not isolated. An example of this situation is the following:

Let K|ℚ be a cyclic extension of prime degree ramified at p. Suppose that the unique extension 𝔭 of p to K is an isolated point. Then there exist finitely many Galois extensions L|K and E|ℚ and elements σ ∈ G(L|K), τ ∈ G(E|ℚ), such that

{𝔭} = ⋂_{L|K} P_{L|K}(σ) ∩ ⋂_{E|ℚ}φ_{K|ℚ}^{−1}P_{E|ℚ}(τ).

Using corollary 1.2, it follows that

{𝔭} = P_{L|K}(σ) ∩ φ_{K|ℚ}^{−1}P_{E|ℚ}(τ)

for one extension L|K and one extension E|ℚ. Furthermore, we may assume that E ⊆ L. This follows from proposition 1.1 with N = LE, since the extensions of 𝔭 to L are unramified in LE (as 𝔭 ∈ φ_{K|ℚ}^{−1}P_{E|ℚ}(τ) ⊆ φ_{K|ℚ}^{−1}U(E|ℚ)). So we have the following diagram of fields:

[figure: L; 𝔓_{L}; EK; 𝔭; K; E; 𝔓_{E}; p; ℚ.]

We claim that without lost of generality σ_{|E} = τ.

Since 𝔭 ∈ P_{L|K}(σ) ∩ φ_{K|ℚ}^{−1}P_{E|ℚ}(τ), there exist primes 𝔓_{L}|𝔭 of L and 𝔓_{E}|p of E such that

σ = (\frac{L|K}{𝔓_{L}} ) and τ = (\frac{E|ℚ}{𝔓_{E}}).

Since the residue degree f(𝔭|p) = 1, we have

(\frac{L|ℚ}{𝔓_{L}}) = (\frac{L|ℚ}{𝔓_{L}})^{f(𝔭|p)} = (\frac{L|K}{𝔓_{L}} ) = σ

and so

σ_{|E} = (\frac{L|ℚ}{𝔓_{L}})_{|E} = (\frac{E|ℚ}{𝔓_{L} ∩ E} ) = (\frac{E|ℚ}{ρ𝔓_{E}} ) = τ^{ρ}

for some ρ ∈ G(E|ℚ). This shows the claim.

Since p ramifies in K|ℚ, the subset

V = P_{L|K}(σ) ∩ φ_{K|ℚ}^{−1}(P_{E|ℚ}(τ) ∩ P_{K|ℚ}(1))

of P_{L|K}(σ) ∩ φ_{K|ℚ}^{−1}P_{E|ℚ}(τ) = {𝔭} does not contain 𝔭, and so V = ∅. But this is a contradiction, since for a prime 𝔭′ in P_{L|K}(σ) ∩ φ_{K|ℚ}^{−1}P_{K|ℚ}(1) (this set has positive

density) it follows that there is a prime 𝔓′_{L}|𝔭′ of L such that (\frac{L|K}{𝔓′_{L}}) = σ, and so

(\frac{EK|K}{𝔓′_{L} ∩ EK} ) = (\frac{L|K}{𝔓′_{L}} )_{|EK} = σ_{|EK}

from which follows that

(\frac{E|ℚ}{𝔓′_{L} ∩ E} ) = (\frac{E|ℚ}{𝔓′_{L} ∩ E} )^{f(𝔭′|p)} = (\frac{EK|K}{𝔓′_{L} ∩ EK} )_{|E} = σ_{|E} = τ.

Thus 𝔭′ ∈ φ_{K|ℚ}^{−1}(P_{E|ℚ}(τ)), and so 𝔭′ ∈ V. It follows that 𝔭 is not an isolated point.

Remark 2: We will show that

P_{K} has infinite compact subsets.

Indeed, let x ∈ P_{K} be a point not contained in (P_{K})_{p.iso} and let

V = {V_{i} | i ∈ ℕ, V_{i} both open and closed, x ∈ V_{i}}

be a base of open and closed neighbourhoods of x such that

V_{i+1} ⊆ V_{i}, V_{i+1} ≠ V_{i}, i ∈ ℕ.

Such a base exists, since the topology T_{K} is countable and every open neighbourhood of x contains infinitely many points by theorem 2.11 and we can separate two different points by both open and closed sets. Let x_{i} ∈ V_{i}\V_{i+1} and

A = {x} ∪ {x_{1}, x_{2}, … }.

Then A is infinite and compact.

Proposition 2.12 Let K|ℚ be a number field. Then there exists a finite Galois extension L|K such that

φ_{L|ℚ}^{−1}(R(L|ℚ)) = (P_{L})_{iso} .

Proof: Without lost of generality we may assume that K|ℚ is normal. Let 𝔭 ∈ φ_{K|ℚ}^{−1}(R(K|ℚ)), ℓ a prime number dividing the order of the inertia group T_{𝔭}(K|ℚ) and K′ = K(μ_{2ℓ}). Furthermore let q_{1}, …, q_{t}, t > [K′ : ℚ]/2 + 1, be auxiliary pairwise different prime numbers and different to ℓ,

S = {𝔮 | 𝔮 = 𝔭 or 𝔮|ℓ · q_{1} ⋯ q_{t}} ∪ S_{∞}

and K′_{S}(ℓ) the maximal ℓ-extension of K′ unramified outside S. From [4] (10.7.8), (10.7.9) and (10.6.4) it easily follows that the local groups G(K′_{𝔮}(ℓ)|K′_{𝔮}) of the maximal ℓ-extension K′_{𝔮}(ℓ) of K′_{𝔮}, 𝔮 ∈ S\S_{∞}, are contained in G(K′_{S}(ℓ)|K′) (if the group G(K′_{S}(ℓ)|K′) would be degenerated, then its ℤ_{ℓ}-rank has to be equal to the number of complex places of K′ plus 1). Since these groups are not (pro-)cyclic, there exists a finite Galois extension L^{𝔭} of K′ such that the decomposition groups of all primes of S(L^{𝔭}) fulfill the assumptions of proposition 2.9(i). Now the finite Galois extension

L = ∏_{𝔭∈R^{K}(K|ℚ)}L^{𝔭}

of K has the property that all points of R^{L}(L|ℚ) are isolated in (P_{L}, T_{L}). □

Now we consider the inverse limit topology TK̄ on the space

PK̄ = lim_{⟵}_{K′|K} P_{K′}

where the inverse limit is taken over all spaces (P_{K′}, T_{K′}), where K′|K runs through the finite subextensions contained in a fixed algebraic closure K̄ of K and the transition maps are the continuous maps φ_{K′|K}.

Proposition 2.13 The space (PK̄, TK̄) is a locally compact, totally disconnected Hausdorff space which is countable at infinity. The Alexandroff-compactification PK̄ ∪ {ω} of (PK̄, TK̄) is a profinite space, i.e. it is compact and totally disconnected.

Proof: Since the spaces (P_{K′}, T_{K′}), K′|K finite, are totally disconnected Hausdorff spaces, the same is true for their inverse limit.

Let x_{K′} be a point of (P_{K′}, T_{K′}), K′|K finite. Then there exists a finite extension of K′ such that x_{K′} ramifies in this extension, and so, by proposition 2.12, there exists a finite extension K′′|K′ such that the set φ_{K′′|K′}^{−1}(x_{K′}) consists of isolated points of (P_{K′′}, T_{K′′}). If we denote the canonical (continuous) map from the inverse limit to P_{K′} by

φ_{K′} : (PK̄, TK̄) ⟶(P_{K′}, T_{K′}),

then it follows that the set

φ_{K′}^{−1}(x_{K′}) = φ_{K′′}^{−1}φ_{K′′|K′}^{−1}(x_{K′})

is open and closed in (PK̄, TK̄). Since φ_{K′′′|K′}^{−1}(x_{K′}) is finite for every finite extension K′′′|K′′, it follows that φ_{K′}^{−1}(x_{K′}) is compact and φ_{K′} is surjective, see [2] I.9.6, cor.1. It follows that PK̄ is the union of the open compact sets φ_{K}^{−1}(x_{K}), x_{K} ∈ P_{K}. Therefore (PK̄, TK̄) is locally compact and countable at infinity because P_{K} is countable.

Finally we show that the Alexandroff-compactification PK̄ ∪ {ω} is totally disconnected. Indeed, the sets

(PK̄\φ_{K}^{−1}(x)) ∪ {ω}, x ∈ P_{K},

are both open and closed neighbourhoods of ω, and

⋂_{x∈P_{K}} ((PK̄\φ_{K}^{−1}(x)) ∪ {ω}) = {ω}.

Therefore the connected component of ω is equal to {ω}. □

3 Uniformity

In this section we consider uniformities on P_{K} which induce the topology T_{K}. First we recall some facts concerning uniform structures on a normal topological space (X, T ).

1. The uniformity of finite partitions by open and closed subsets of X.

A partition of X is a family (U_{i})_{i∈I} of pairwise disjoint subsets such that their union is the whole space. Let

Part^{oc} = {(V_{1}, …, V_{n}) | n ∈ ℕ, V_{i} ⊆ (X, T ) open and closed, ⋃ ._{i=1}^{n} V_{i} = X}

be the set of finite partitions of X by open and closed subsets of (X, T ) and let

𝔙^{oc} = { V_{Q} = ⋃ ._{i=1}^{n}(V_{i} × V_{i}) ⊆ X × X | n ∈ ℕ, Q = (V_{1}, …, V_{n}) ∈ Part^{oc}}.

Obviously, 𝔙^{oc} is a base for the uniform structure

𝔘^{oc} = {U ⊆ X × X | U contains a subset V ∈ 𝔙^{oc}}

on X. We denote the completion of (X, 𝔘^{oc}) by (X̂, 𝔘̂^{oc}).

2. The uniformity of finite open coverings on X.

Let

Cov^{o} = {(U_{1}, …, U_{n}) | n ∈ ℕ, U_{i} ∈ T , ⋃_{i=1}^{n} U_{i} = X}

be the set of finite coverings of X by open subsets of (X, T ) and let

𝔙^{o} = { V_{Q} = ⋃_{i=1}^{n}(U_{i} × U_{i}) ⊆ X × X | n ∈ ℕ, Q = (U_{1}, …, U_{n}) ∈ Cov^{o}}.

This is a base for the uniform structure

𝔘^{o} = {U ⊆ X × X | U contains a subset V ∈ 𝔙^{o}}

on X, the so-called uniformity of finite open coverings, see [2] IX.4 exercise 17.

3. The Stone-Čech compactification of X.

Let βX be the Stone-Čech compactification of (X, T ). Then βX is the completion of X with respect to the coarsest uniformity 𝔘^{Sč} on X for which all continuous mappings of X into [0, 1] are uniformly continuous (see [2] IX.1 exercise 7).

Concerning these three uniformities on X we have the following

Proposition 3.1 Let (X, T ) be a strongly zero-dimensional Hausdorff space. Then the uniform structures 𝔘^{oc}, 𝔘^{o} and 𝔘^{Sč} on X are equal, now denoted by 𝔘. The topology induced by 𝔘 on X is equal to T .

Proof: Since the space (X, T ) is normal, the uniformity 𝔘^{o} is equal to the uniformity 𝔘_{K}^{Sč} and 𝔘^{o} = 𝔘^{Sč} induces the topology T on X, see [2] IX.4 exercise 17(b).

By definition 𝔘^{oc} is coarser than 𝔘^{o} and, using proposition 2.7, there exists for every V_{(U_{1},...,U_{n})} ∈ 𝔙^{o} an element V_{(V_{1},...,V_{n})} ∈ 𝔙^{oc} with V_{i} ⊆ U_{i}, i = 1, …, n, i.e. V_{(V_{1},...,V_{n})} ⊆ V_{(U_{1},...,U_{n})}. Thus 𝔘^{oc} is finer than 𝔘^{o}, and so they are equal. □

Proposition 3.2 Let X be a strongly zero-dimensional Hausdorff space. Then the completion (X̂, 𝔘̂) of X equipped with the uniformity 𝔘 = 𝔘^{oc} is a profinite space, i.e. it is compact and totally disconnected.

Proof: From proposition 3.1 it follows that (X̂, 𝔘̂) = β(X, T ) and the compact space β(X, T ) is totally disconnected, see [2] IX.6 exercise 1(b). □

Proposition 3.3 Let X be a strongly zero-dimensional Hausdorff space and let

i : (X, 𝔘) ⟶(X̂, 𝔘̂)

be the canonical mapping (𝔘 = 𝔘^{oc}) and we identify X with i(X). Let OC_{X} and OCX̂ be the set of both open and closed subsets of X and X̂, respectively.

(i) The maps

OC_{X} ⟶ OCX̂, S ↦ \overline{S}, and OCX̂ ⟶ OC_{X}, S ↦ S ∩ X

are bijections, where \overline{S} is the closure of S in X̂. (ii) For the set of isolated points of X and X̂ we have i(X_{iso}) = X̂_{iso}.

Proof: It is clear that the second map is well-defined. Let S ∈ OC_{X}. Since S and X\S are closed sets of X, we get from S ∪. (X\S) = X the partition

\overline{S} ∪. \overline{X\S} = X̂,

see [2] IX.4 exercise 17(c), and so X̂\\overline{S} = \overline{X\S}. Thus the closed set \overline{S} is also open in X̂, and so also the first map is well-defined.

If S ∈ OC_{X}, then S ⊆ \overline{S}∩X. Let x ∈ \overline{S}∩X and suppose that x ∈ X\S. Then x ∈ \overline{X\S} = X̂\\overline{S} which is a contradiction, and it follows that x ∈ S. Therefore S = \overline{S} ∩ X.

If S ∈ OCX̂, then \overline{S ∩ X} ⊆ S, since S is closed. Since S is also open, S ∩ X is dense in S, and so \overline{S ∩ X} = S. This proves that the considered maps are bijections.

In order to prove (ii) let x̂ ∈ X̂_{iso}. Then {x̂} is open in X̂. Since i(X) is dense in X̂, the set {x̂} ∩ i(X) is not empty and so x̂ ∈ i(X). Thus {x̂} is an open subset of i(X).

Conversely, let x ∈ X_{iso}. Since the set {x} is open and closed in X, the same is true, by (i), for its closure \overline{\{x\}} in X̂. Consider the open set

U = \overline{\{x\}}\\{i(x)\} ⊆ X̂

(observe that {i(x)} is closed in the Hausdorff space X̂). Suppose that U is not empty. Then, using (i), we get the contradiction

∅ ≠ U ∩ i(X) = (\overline{\{x\}} ∩ i(X))\\{i(x)\} = \{i(x)\}\\{i(x)\}.

Therefore U is empty, i.e. \overline{\{x\}} = \{i(x)\}, and so \{i(x)\} is open in X̂. □

Now let (X, T ) = (P_{K}, T_{K}). This space is a strongly zero-dimensional Hausdorff space by proposition 2.5(ii). If 𝔘_{K} = 𝔘_{K}^{oc} denotes the uniformity of finite partitions of P_{K} by both open and closed subsets of (P_{K}, T_{K}), then we obtain Theorem 3.4 The Hausdorff uniform space (P_{K}, 𝔘_{K}) is pre-compact and strongly zero-dimensional, and its completion (P̂_{K}, 𝔘̂_{K}) is a profinite space. The canonical map

i : (P_{K}, 𝔘_{K}) ⟶(P̂_{K}, 𝔘̂_{K})

induces an isomorphism of (P_{K}, 𝔘_{K}) onto a dense subspace of (P̂_{K}, 𝔘̂_{K}).

Furthermore the sets (P_{K})_{iso} and (P̂_{K})_{iso} of isolated points are isomorphic and finite.

4 A Metric for P_{ℚ}

In this section we will define a metric on P_{ℚ} which induces the topology T_{ℚ}. The idea is that two points x, y ∈ P_{ℚ} are near, if they induce in many fields with large discriminant the same Frobenius automorphism. We start by defining another uniformity on P_{ℚ}:

the uniformity of finite open coverings of (P_{ℚ}, T_{ℚ}) defined by the discriminant of finite Galois extensions F |ℚ.

Let d ∈ ℕ and let

S_{d} = {F |ℚ a finite Galois extension, |D(F |ℚ)| ≤ d},

where D(F |ℚ) denotes the discriminant of F. The set S_{d} is finite by Hermite’s theorem, see [3] III. (2.16). For x ∈ P_{ℚ} let

S_{d,x} = {F |ℚ finite Galois , x ∈ U(F |ℚ), |D(F |ℚ)| ≤ d}

and

V_{d}(x) = ⋂_{F |ℚ∈S_{d,x}} P_{F |ℚ}((\frac{F |ℚ}{x_{F}})),

where x_{F} is an extension of x to F. Furthermore let

R_{d} = ⋃_{F |ℚ∈S_{d}} R(F |ℚ),

G_{d} = ∏_{F |ℚ∈S_{d}} G(F |ℚ).

For σ̃ = (σ_{F |ℚ})_{F |ℚ} ∈ G_{d} we put

V_{d}(σ̃) = ⋂_{F |ℚ∈S_{d}}P_{F |ℚ}(σ_{F |ℚ}).

Observe that V_{d}(σ̃) = V_{d}(x) for all x ∈ V_{d}(σ̃). We obtain a finite open covering

Cov^{o}(d) : P_{ℚ} = ⋃_{σ̃∈G_{d}} V_{d}(σ̃) ∪ ⋃_{α∈R_{d}} V_{d}(α)

of P_{ℚ}. Finally we define

V_{d} = ⋃_{σ̃∈G_{d}} (V_{d}(σ̃) × V_{d}(σ̃)) ∪ ⋃_{α∈R_{d}} (V_{d}(α) × V_{d}(α)).

Obviously, we have V_{d′} ⊆ V_{d} for d ≤ d′.

Proposition 4.1 The set 𝔙_{ℚ}^{D} = { V_{d} , d ∈ ℕ} is a base for a uniform structure 𝔘_{ℚ}^{D} on P_{ℚ} inducing the topology T_{ℚ}.

Proof: Let

V_{d} = ⋃_{σ̃∈G_{d}} (V_{d}(σ̃) × V_{d}(σ̃)) ∪ ⋃_{α∈R_{d}} (V_{d}(α) × V_{d}(α)) ∈ 𝔙_{ℚ}^{D}.

By proposition 2.4(ii) we find open (and closed) neighbourhoods U(α) of α ∈ R_{d} which are pairwise disjoint and defined by finitely many extensions of ℚ (see the proof of 2.4(ii)). Let Ṽ_{d}(α) = V_{d}(α) ∩ U(α) and Ṽ_{d}(σ̃) = V_{d}(σ̃)\V (R_{d}), where V (R_{d}) = ⋃ ._{α∈R_{d}} Ṽ_{d}(α) is an open and closed neighbourhood of the set R_{d}. Then

C̃ov(d) : P_{ℚ} = ⋃ ._{σ̃∈G_{d}} Ṽ_{d}(σ̃) ∪. ⋃ ._{α∈R_{d}} Ṽ_{d}(α)

is a partition of P_{ℚ} by open sets which is a refinement of Cov^{o}(d). Furthermore these open sets are defined by finitely many extensions of ℚ. Let

d′ = max_{F} {|D(F |ℚ)|},

where F runs through all extensions of ℚ which appear in a definition of the open sets in C̃ov(d). Then Cov^{o}(d′) is a refinement of C̃ov(d). Indeed, let U ∈ Cov^{o}(d′), i.e. U = V_{d′}(σ̃) for some σ̃ ∈ G_{d′} or U = V_{d′}(α) for some α ∈ R_{d′}. Then U = V_{d′}(x) for some x ∈ P_{ℚ}. Let V ∈ C̃ov(d) be the unique open set containing x. It follows that U ⊆ V, since V is the (finite) union of open sets of the form ⋂_{i=1}^{r} P_{F_{i}|ℚ}(σ_{i}), where |D(F_{i}|ℚ)| ≤ d′ for all i, and at least one of these sets contains x.

Now we prove that the set 𝔙_{ℚ}^{D} = { V_{d} , d ∈ ℕ} is a base for a uniform structure on P_{ℚ}. The only axiom, which is not obvious, is the following: for V_{d} ∈ 𝔙_{ℚ}^{D} there

exists a V_{d′} ∈ 𝔙_{ℚ}^{D} such that V_{d′}^{2} ⊆ V_{d}. But, taking d′ as above, and let

W_{d} = ⋃ ._{σ̃∈G_{d}} (Ṽ_{d}(σ̃) × Ṽ_{d}(σ̃)) ∪. ⋃ ._{α∈R_{d}} (Ṽ_{d}(α) × Ṽ_{d}(α)),

then it follows by the consideration above that

V_{d′}^{2} ⊆ W_{d}^{2} = W_{d} ⊆ V_{d}.

Finally, the topology induced by 𝔘_{ℚ}^{D} is obviously coarser than T_{ℚ}. On the other hand, let x ∈ P_{ℚ} and let U(x) ∈ T_{ℚ} be an open neighbourhood of x. We may assume that U(x) = ⋂_{i=1}^{r}P_{F_{i}|ℚ}(σ_{i}) for some finite Galois extensions F_{i}|ℚ and some σ_{i} ∈ G(F_{i}|ℚ). Let d = max{|D(F_{i}|ℚ)|, i = 1, …, r} and let V ∈ C̃ov(d) be the unique open set containing x, and so V ⊆ U(x). The neighbourhood of x induced by the entourage V_{d′} with d′ as above is

U_{d′}(x) = \{ \begin{matrix}V_{d′}(σ̃) & ∪ & ⋃_{α∈R_{d′}(x)} V_{d′}(α), & if x is unramified in all F ∈ S_{d′}, \\ & & ⋃_{α∈R_{d′}(x)} V_{d′}(α), & otherwise. \end{matrix}

where R_{d′}(x) = {α ∈ R_{d} | x ∈ V_{d′}(α)} and σ̃ is given by the condition that x ∈ V_{d′}(σ̃). Since Cov^{o}(d′) is a refinement of C̃ov(d), we obtain that

U_{d′}(x) ⊆ V ⊆ U(x).

Thus the topology induced by 𝔘_{ℚ}^{D} is finer than T_{ℚ}. This proves the proposition. □

Obviously, 𝔘_{ℚ}^{D} is coarser than 𝔘_{ℚ} = 𝔘_{ℚ}^{o} (and it seems unlikely that they are equal), but this uniformity defines a nice metric on P_{ℚ}.

Theorem 4.2 The map

δ : P_{ℚ} × P_{ℚ} ⟶ [0, 1] , (x, y) ↦ δ(x, y) = \frac{1}{n} ,

where

n = sup{ d | (x, y) ∈ V_{d}},

defines an ultra-metric on P_{ℚ} which induces the uniformity 𝔘_{ℚ}^{D}.

Proof: Obviously, δ is symmetric, δ(x, x) = 0 for every x ∈ P_{ℚ},

δ(x, y) ≤ max(δ(x, z), δ(z, y)) for all x, y, z ∈ P_{ℚ}

and the (quasi-)metric δ induces the uniformity 𝔘_{ℚ}^{D}. Since T_{ℚ} is a Hausdorff topology, δ is an (ultra-)metric. □

Remark: Since #R_{d} ≤ 1 for d ≤ 3, it follows that δ(P_{ℚ} × P_{ℚ}) ⊆ [0, \frac{1}{3}].

Corollary 4.3 The completion (P̂_{ℚ}, 𝔘̂_{ℚ}^{D}) of (P_{ℚ}, 𝔘_{ℚ}^{D}) is a profinite space.

Proof: Since 𝔘_{ℚ}^{D} is coarser than 𝔘_{ℚ}, we get a surjection

\xymatrix{(P̂_{ℚ}, 𝔘̂_{ℚ}) \ar@{->>}[r] & (P̂_{ℚ}, 𝔘̂_{ℚ}^{D}),}

and so (P̂_{ℚ}, 𝔘̂_{ℚ}^{D}) is compact. Furthermore, the extension of δ to (P̂_{ℚ}, 𝔘̂_{ℚ}^{D}) is also an ultra-metric and so the completion is totally disconnected. □

Remark: Analogously one can define a uniformity 𝔘_{K}^{D} on (P_{K}, T_{K}) and a corresponding metric having the properties stated in 4.1, 4.2 and 4.3.

It is obvious that the metric δ is not easily to calculate (at least if d > 21 when not only quadratic fields are involved). But we do some calculations for d ≤ 5. We have three non-trivial extensions F |ℚ with absolute discriminant |D(F |ℚ)| ≤ 5:

ℚ(\sqrt{−3}), |D| = 3, ℚ(\sqrt{−1}), |D| = 4, and ℚ(\sqrt{5}), |D| = 5.

From now on we use the notation (a|γ) for P_{ℚ(\sqrt{a})|ℚ}(γ), γ ∈ G(ℚ(\sqrt{a})|ℚ), and we denote the non-trivial element of G(ℚ(\sqrt{a})|ℚ) by −1. Then

V_{3} = P_{ℚ} × P_{ℚ},

for V_{4} we have to use the covering

P_{ℚ} = ((−3|−1) ∩ (−1| 1)) ∪ (−3| 1) ∪ (−1|−1),

and for V_{5} the covering

\begin{aligned}P_{ℚ} & = & & ((−3| 1) ∩ (−1| 1) ∩ (5|−1)) ∪ ((−3|−1) ∩ (−1|−1) ∩ (5| 1)) \\ & & ∪ & ((−3| 1) ∩ (5| 1)) ∪ ((−1|−1) ∩ (5|−1)) ∪ ((−3|−1) ∩ (−1| 1)). \end{aligned}

It follows that for prime numbers x < y ≤ 19

δ(x, y) = \frac{1}{4} , if (x, y) = (2, 7), (2, 13), (3, 11), (3, 19), (7, 11), (7, 13), (7, 19),
(11, 19), (13, 19),
δ(x, y) ≤ \frac{1}{5} , if (x, y) = (2, 19), (3, 7), (5, 17),

and for all other pairs we have δ(x, y) = \frac{1}{3}.

5 Concluding Remark

For m ∈ ℕ, m ≥ 2, we define

B_{K}^{m} ={P_{F |E}^{K}(σ) | E ⊆ K, F |E finite Galois, σ ∈ G(F |E), [F : E] ≤ m}.

Let T_{K}^{m} be the topology on P_{K} with subbase B_{K}^{m}. Then proposition 2.5 remains true. Crucial is proposition 2.4(i) and its proof is not infected if m > 2. For m = 2 we need a different argument.

Proposition 5.1 For every two different points 𝔭_{1} and 𝔭_{2} of (P_{K}, T_{K}^{2}) there exist both open and closed neighbourhoods W_{1} ∈ T_{K}^{2} and W_{2} ∈ T_{K}^{2} of 𝔭_{1} and 𝔭_{2}, respectively, such that

W_{1} ∩ W_{2} = ∅.

Proof: First we reduce to the case K = ℚ. By the theorem of Grunwald/Wang there exist quadratic extensions L_{i}|K, i = 1, 2, such that 𝔭_{1} and all its conjugates are inert and 𝔭_{2} is completely decomposed in L_{1}|K and the same holds if we interchange the indices 1 and 2. If σ_{i} denotes the non-trivial element of G(L_{i}|K), then it follows that

𝔭_{1} ∈ P_{L_{1}|K}(σ_{1}), 𝔭_{2} ∉ P_{L_{1}|K}(σ_{1}) and 𝔭_{2} ∈ P_{L_{2}|K}(σ_{2}), 𝔭_{1} ∉ P_{L_{2}|K}(σ_{2}),

and

R(L_{i}|K) ∩ {𝔭′_{i} | 𝔭′_{i} conjugated to 𝔭_{i}} = ∅, i = 1, 2.

Let U_{i} = P_{L_{i}|K}(σ_{i}), i = 1, 2. These open neighbourhoods need not to be disjoint or closed. Let

φ_{K|ℚ}(R(L_{1}|K)) = {q_{1}, …, q_{s}}

and let p_{1} be the underlying prime number of 𝔭_{1} which is by construction different to the prime numbers q_{1}, …, q_{s}. Assume that we have proved the proposition in the case K = ℚ. Then, using the argument of the proof of proposition 2.4(ii), there are pairwise disjoint both open and closed neighbourhoods U(p_{1}), U(q_{1}), …, U(q_{s}) ∈ T_{ℚ}^{2} of the considered prime numbers. It follows that

𝔭_{1} ∈ U_{1} ∩ φ_{K|ℚ}^{−1}(U(p_{1})) and R(L_{1}|K) ⊆ ⋃_{i=1}^{s} φ_{K|ℚ}^{−1}(U(q_{i})).

The open set V_{1} = U_{1} ∩ φ_{K|ℚ}^{−1}(U(p_{1})) is also closed since

\overline{U_{1} ∩ φ_{K|ℚ}^{−1}(U(p_{1}))} ⊆ \overline{U_{1}} ∩ φ_{K|ℚ}^{−1}(U(p_{1}))
⊆ (U_{1} ∪ R(L_{1}|K)) ∩ φ_{K|ℚ}^{−1}(U(p_{1}))
= U_{1} ∩ φ_{K|ℚ}^{−1}(U(p_{1}))

(observe that the map φ_{K|ℚ} is continuous and that the sets φ_{K|ℚ}^{−1}(U(p_{1})) and φ_{K|ℚ}^{−1}(U(q_{i})), i = 1, …, s, are also pairwise disjoint). Constructing an open and closed neighbourhood V_{2} of 𝔭_{2} in the same way and observing that

𝔭_{1} ∈ V_{1}, 𝔭_{2} ∉ V_{1} and 𝔭_{2} ∈ V_{2}, 𝔭_{1} ∉ V_{2},

then the neighbourhoods W_{1} = V_{1}\V_{2} of 𝔭_{1} and W_{2} = V_{2}\V_{1} of 𝔭_{2} have the desired properties.

Now let K = ℚ and let p_{1} and p_{2} be two different prime numbers. Assume that p_{1} is odd and let

σ = \binom{p_{2}}{p_{1}} ∈ \{±1\}.

Let q be an odd prime number, different to p_{1} and p_{2}, such that

\binom{p_{1}}{q} = −σ, \binom{p_{2}}{q} = σ, \binom{−1}{q} = 1.

Such a prime number exists since the set

P_{ℚ(\sqrt{p_{1}})|ℚ}(−σ) ∩ P_{ℚ(\sqrt{p_{2}})|ℚ}(σ) ∩ P_{ℚ(\sqrt{−1})|ℚ}(1),

has positive density, and so it is not empty. Let

L_{q} = ℚ(\sqrt{(−1)^{ε(q)}q}) and L_{p_{1}} = ℚ(\sqrt{(−1)^{ε(p_{1})}p_{1}} ),

where ε(ℓ) = (ℓ − 1)/2. Then p_{1} ∉ P_{L_{q}|ℚ}(σ) and q ∉ P_{L_{p_{1}}|ℚ}(σ) since

\binom{(−1)^{ε(q)}q}{p_{1}} = \binom{p_{1}}{q} = −σ and \binom{(−1)^{ε(p_{1})}p_{1}}{q} = \binom{p_{1}}{q}\binom{−1}{q}^{ε(p_{1})} = −σ.

Furthermore, p_{2} ∈ P_{L_{q}|ℚ}(σ) ∩ P_{L_{p_{1}}|ℚ}(σ): if p_{2} is odd, then

\binom{(−1)^{ε(q)}q}{p_{2}} = \binom{p_{2}}{q} = σ and \binom{(−1)^{ε(p_{1})}p_{1}}{p_{2}} = \binom{p_{2}}{p_{1}} = σ.

If p_{2} = 2, then p_{2} splits in L_{q}|ℚ if and only if (−1)^{ε(q)}q ≡ 1 mod 8, i.e. if and only if \binom{2}{q} = 1. Since σ = \binom{2}{q} = \binom{2}{p_{1}}, we have

2 ∈ P_{L_{q}|ℚ}(σ),

and also 2 ∈ P_{L_{p_{1}}|ℚ}(σ).

It follows that V_{2} = P_{L_{p_{1}}|ℚ}(σ) ∩ P_{L_{q}|ℚ}(σ) is a neighbourhood of p_{2} which is both open and closed (since R(L_{q}|ℚ) = {q} and R(L_{p_{1}}|ℚ) = {p_{1}}), and does not contain p_{1}.

If p_{2} is odd, then we construct an open and closed neighbourhood V_{1} of p_{1} in the same way, and the neighbourhoods W_{1} = V_{1}\V_{2} of p_{1} and W_{2} = V_{2}\V_{1} of p_{2} have the desired properties.

If p_{2} = 2, then we have also to construct an open and closed neighbourhood V_{1} of p_{1} with the analogous properties. Again let

σ = \binom{2}{p_{1}},

and let q be an odd prime number such that

\binom{p_{1}}{q} = σ and \binom{2}{q} = −σ.

If

L_{q} = ℚ(\sqrt{(−1)^{ε(q)}q}) and L_{2} = ℚ(\sqrt{2} ),

then p_{1} ∈ P_{L_{q}|ℚ}(σ) ∩ P_{L_{2}|ℚ}(σ):

\binom{(−1)^{ε(q)}q}{p_{1}} = \binom{p_{1}}{q} = σ, \binom{2}{p_{1}} = σ.

Furthermore, q ∉ P_{L_{2}|ℚ}(σ) because

\binom{2}{q} = −σ,

and 2 ∉ P_{L_{q}|ℚ}(σ) because 2 ∈ P_{L_{q}|ℚ}(\binom{2}{q}) = P_{L_{q}|ℚ}(−σ). It follows that V_{1} = P_{L_{2}|ℚ}(σ) ∩ P_{L_{q}|ℚ}(σ) is a neighbourhood of p_{1} which is both open and closed and does not contain p_{2}. This finishes the proof of the proposition.

References

[1] Arhangel’skii, A.V. General Topology III. Encyclopaedia of Math. Sciences Vol. 51, Springer 1995

[2] Bourbaki, N. General Topology. Hermann 1966

[3] Neukirch, J. Algebraic Number Theory. Springer 1999

[4] Neukirch, J., Schmidt, A., Wingberg, K. Cohomology of Number Fields. Springer 2000 Mathematisches Institut der Universität Heidelberg Im Neuenheimer Feld 288 69120 Heidelberg Germany

e-mail: wingberg@mathi.uni-heidelberg.de