An Explicit Construction of Abelian Extensions via Formal Groups

Danielle Li

May 25, 2004

Abstract

We introduce method of approaching local class field theory from the perspective of Lubin-Tate formal groups. Our primarily aim is to demonstrate how to construct abelian extensions using these groups.

1 Introduction

Local class field theory centers about the study of abelian extensions of local fields, most notably the p-adic numbers. Traditionally, local class field theory has been approached either as an offshoot of global, or classical class field theory, or in more recent tradition of Nakayama and Hochschild, from a cohomological perspective as in Artin [1] or Serre [5]. In the mid 1960s, Lubin and Tate [3] showed that many theorems of local class field theory could be deduced from formal groups over local fields, particularly their valuation rings. This paper seeks to provide a basic introduction to that approach.

Although all the results we present are by Lubin and Tate, we follow primarily the exposition of Iwasawa [2] and Milne [4]. Iwasawa’s work follows more closely the original Lubin-Tate paper and does so in more generality. He begins with the maximal unramified extension of K, which we, like Milne, exclude when introducing formal group constructions. The primary benefit of this approach is that it makes our notation less cumbersome.^{1} Our expository innovation is to begin with a motivational example of how formal groups can be used to generate abelian extensions, presented in Section 2. Section 3 provides an introduction to formal groups in general and specifically Lubin-Tate formal groups. Finally, Section 4 uses these formal groups to construct abelian extensions of a local field K.

1In general, I found Milne’s exposition to be more clear, although both works cover the preliminarly material in roughly the same detail. Iwasawa, however, relies on formal groups in the rest of his book, proving the main theorems of local class field theory from this perspective, while Milne’s discussion is primarily a digression on the construction of abelian groups.

2 Motive

To get a feel for Lubin and Tate’s insight about how formal groups can generate abelian extensions, we first consider an incomplete, unrigorous, but enlightening example.

Let K be a locally compact non-archimeadean field. We define the following objects that act on K.

Definition 1 The multiplicative group G_{m} is a functor from the category of fields to the category of groups which gives us the units in a field. That is,

G_{m}(L) = L^{∗}

where L^{∗} is the group of units.

We also define a subfunctor, μ_{p^{n}} that takes fields to their roots of unity. That is, μ_{p^{n}}(L) is group of p^{n}th roots of unity contained in L. When we write μ_{p^{n}}, we mean all roots of unity.

We say that μ_{p^{n}} is the p^{n}-torsion of G_{m} because μ_{p^{n}}(L) is the p^{n}th torsion of G_{m}(L). These are exactly the elements that are killed when we raise to the p^{n}th power in the image of G_{m}.

Now, consider A_{n} = Aut(μ_{p^{n}}) and let z_{n} be a primitive nth root of unity. z_{n} generates μ_{p^{n}} so that any automorphism of μ_{p^{n}} is defined by where it sends z_{n}. The map

z_{n} → z_{n}^{a}

defines an automorphism of μ_{p^{n}} as long as a is a unit in ℤ/p^{n}ℤ so that

A_{n} = (Z/p^{n})^{∗}

Further, we have the reduction map

π : A_{n} → A_{n−1}

defined by taking a, a unit mod p^{n} and mapping it to a mod p^{n−1}. Letting n vary, we choose z_{n}, z_{n+1}, etc., such that

z_{n}^{p} = z_{n−1}

In this way, we define all the A_{n}’s to be compatible. The advantage of this construction is that we can now take the inverse limit.

Definition 2 Consider a sequence of groups {G_{i}}_{i=1}^{∞} which are related by surjective homomorphisms f_{i} : G_{i} → G_{i−1} such that:

G_{1} ←^{f_{2}} G_{2} ←^{f_{3}} G_{3} …

We define the inverse limit of (G_{i}, f_{i}), denoted lim inf G_{i} to be the subset of ∏_{i=1}^{∞} G_{i} consisting of sequences

(g_{1}, g_{2}, g_{3}, …)

such that f_{i}(g_{i}) = g_{i−1}.

We define another function associated with inverse limits.

Definition 3 We define the Tate Module T_{p} which takes a group to the inverse limit of its p^{n}- torsion.

In our case,

T_{p}G_{m} = lim inf μ_{p^{n}}.

It consists of infinite sequences x = (x_{1}, x_{2}, …) such that x_{i} is a p^{i}th root of unity and x_{i}^{p} = x_{i−1}.

Now let G =Gal(K̄/K) for some field K of char≠ p. We have that x_{i} ∈ K̄ for all i. Thus, any element g ∈ G acts on x_{i} by the Galois action. Using this, we can define a natural Galois action on T_{p}G_{m} by

gx = g(x_{1}, x_{2}, …) = (gx_{1}, gx_{2}, …)

Further, we have that every element of G induces an automorphism of mu_{p^{n}} since g ∈ G is a Galois automorphism of K̄ and thus respects multiplication in μ_{p^{n}}. So we get the maps:

ϕ_{n} : G → Aut(μ_{p^{n}}) = (Z/p^{n})^{∗}

This formulation gives us a bit more in fact. Let Ker(ϕ_{n}) = G_{n} be a subgroup of the Galois group G. Let K_{n} be the fixed field of G_{n}. From group theory, we have that G/Ker(ϕ_{n}) = G/G_{n} = Im(G). But since we are also dealing with Galois groups, we also have that G/G_{n} = Gal(K_{n}/K). So we have that

Gal(K_{n}/K) = Im(G)

a subgroup of (Z/p^{n})^{∗}. But subgroups of abelian groups are abelian and so we have constructed an abelian extension of K.

The inverse limit allows us to recover all this information from the map:

ϕ : G → Aut(T_{p}G_{m}) = Z_{p}^{∗}

Thus, we can get these abelian extensions back as well.

We now show that we can get these same abelian extensions from a formal group law as well. This is desireable because T_{p}G_{m} is pretty unwieldy, but the formal group we will find is very explicit. Although the abelian extensions we just found do not represent all possible abelian extensions, (although, by Kronecker-Weber, this is true in the case of ℚ), this is a good starting point for the remainder of the paper where we do construct an arbitrary extension.

Recall that we are in K a locally compact non-archimedean field. Note that this means K is complete, discrete, and has finite residue field. Further, its finite extensions L are complete with respect to the unique extension of our underlying valuation.

We will define a formal group in more generality later, but for now, consider the function:

f(X, Y) = X + Y + XY

Loosely, we call something a formal group as opposed to a group because a formal group is just the composition law, without the actual set. Thus, we can consider f(X, Y) as a way of defining a binary operation on a set by:

x +_{f} y = f(x, y) = x + y + xy

We will take our x,y as elements of 𝔪_{L}, the maximal ideal in the valuation ring O_{L} of a finite extension of K. Recall that we define O_{L} = {x ∈ L||x| ≤ 1} and 𝔪_{L} = {x ∈ L||x| < 1}.

Proposition 1 𝔪_{L} together with the composition defined by f(X, Y) is a group.

Proof :

1. Closure. |x + y + xy| ≤ max{|x|, |y|, |x||y|} < 1

2. Inverses. If |x| ≤ 1, then |1/x| = 1/|x| < 1

3. Identity. 0 works.

Thus, we can define F as the functor from the category of finite extensions, L, to the category of groups (𝔪_{L}, +_{f}) where f defines a binary operation on the set 𝔪_{L}, the maximal ideal of O_{L}. That is,

F (L) = (𝔪_{L}, +_{f})

Theorem 1 T_{p}F, the inverse limit of the p^{n}-torsion of F is isomorphic to T_{p}G_{m}.

Proof : Here is a roadmap of our proof.

1. We show that the set 1 + 𝔪_{L} with multiplication is a subgroup of L^{∗} = G_{m}(L).

2. We show that (𝔪_{L}, +_{f}) ≅ (1 + 𝔪_{L}, ∗)

3. We show that the p^{n}-torsion of F (L) = (𝔪_{L}, +_{f}) is contained in the p^{n}-torsion of G_{m}(L) = L^{∗}.

4. We show the other inclusion. This will allow us to conclude that T_{p}G_{m} is isomorphic to T_{p}F because we defined them to be the inverse images of the p^{n} torsion.

Proof of 1. The set 1 + 𝔪_{L} with multiplication is a subgroup of L^{∗} = G_{m}(L). It is clearly a subset so we just need to show that it’s a group.

1. Closure. Consider x, y ∈ 𝔪_{L}. So 1 + x, 1 + y are in 1 + 𝔪_{L}. Then (1 + x)(1 + y) = 1 + x + y + xy but x + y + xy ∈ 𝔪_{L} by the composition law on 𝔪_{L}. Thus, the product is in 1 + 𝔪_{L}.

2. Inverses. \frac{1}{1+x} = 1 − x + x^{2} − x^{3} + … but −x + x^{2} − x^{3} + … converges in 𝔪_{L} because |x| < 1 and we know that L is complete.

3. Identity. 1=1+0 and 0 ∈ 𝔪_{L}.

Proof of 2. (𝔪_{L}, f) ≅ (1 + 𝔪_{L}, ∗) by the map h : x → 1 + x. The map is clearly bijective and h(x +_{f} y) = h(x + y + xy) = 1 + x + y + xy = (1 + x)(1 + y) = h(x) ∗ h(y).

Proof of 3. From 1, and 2, we have that F (L) = (𝔪_{L}, +_{f}) ≅ (1 + 𝔪_{L}, ∗) ⊂ L^{∗}. Thus the p^{n} torsion of F (L) must be contained in the p^{n} torsion of G_{m}(L) = L^{∗}, which we already said is μ_{p^{n}}(L).

Proof of 4. We need to show the other inclusion, that the p^{n} torsion of G_{m}(L) is contained in the p^{n} torsion of F (L). To show this, consider the map

ϕ : O_{L} → O_{L}/𝔪_{L}

the reduction to the residue field. This induces another map from ϕ′ : O_{L}^{∗} → (O_{L}/𝔪_{L})^{∗} because we know that a ring homomorphism maps units to units. Here, 1 + 𝔪_{L} is exactly the kernel of ϕ′. Since K is a local field, we know that both O_{K}/m_{K} and O_{L}/𝔪_{L} are finite fields, and further, O_{L}/𝔪_{L} is a field extension of O_{K}/m_{K}. Since O_{K}/m_{K} is finite, it has character p, as does O_{L}/𝔪_{L}, its extension. But then it has no nontrivial p^{n}th roots of unity. To see this, notice that x^{p} − 1 = (x − 1)^{p} mod p. Thus, in going from O_{L}^{∗} to (O_{L}/𝔪_{L})^{∗}, we’ve lost all the p^{n} torsions of G_{m}(L) = L^{∗}. They must be in the kernel then. This shows that the p^{n} torsion of G_{m}(L) = L^{∗} is contained in 1 + 𝔪_{L}. This concludes the proof.

Thus, we have that T_{p}G_{m} ≅ T_{p}F. This is a remarkable result because we can use T_{p}G_{m} to recover the the specific class of abelian extension of our local field K that we constructed in the beginning. But the whole point of local class field theory is to study the abelian extensions of K. Using our formal group, we can recover the exact abelian extension we looked at earlier and get a far more explicit description!

So now we try to do this in more generality.

3 Formal Groups

In general, a formal group is a law of composition that satisfies the group axioms, but without the set. We can use them to create groups from sets as we did with 𝔪_{L} and the operation defined by f(X) = X + Y + XY in section 2.

Definition 4 Let R be a commutative ring with unit. A power series F (X, Y) in R[[X, Y]] is called a formal group over R if it satisfies:

1. F (X, Y) ≡ X + Y mod deg 2

2. F (F (X, Y), Z) = F (X, F (Y, Z))

3. F (X, Y) = F (Y, X)

Note that these definitions tell us that the operation is associative and commutative. We will be especially interested in the case where A = O_{K} where K is a nonarchimedean local field (that is, K is locally compact with respect to a nontrivial valuation - this also means that K is complete, | · | is discrete, and has finite residue field).

Definition 5 For each prime element, π, of K, let F_{π} denote the family of power series f(X) in A[[X]] such that :

f(X) ≡ πXmod deg 2 f(X) ≡ X^{q}modπ

For example, the polynomial πX + X^{q} is an element of F_{π}.

Proposition 2 Let f, g be power series in F_{π} and let ϕ_{1}(X_{1}, …, X_{n}) be a linear form with coefficients in A. Then there is a unique ϕ ∈ A[[X_{1}, …, X_{n}]] such that:

ϕ(X_{1}, …, X_{n}) = ϕ_{1}mod deg 2
f(ϕ(X_{1}, …, X_{n})) = ϕ(g(X_{1}), …, g(X_{n})).

Proof : See [4], [2], or [3].

Proposition 3 For each f ∈ F_{π}, there exists a unique formal group F_{f}(X, Y) over A admitting f as an endomorphism.

Proof : We apply Propostion 2 for f = g, and L(X, Y) = X + Y to get that there exists a unique power series F_{f}(X, Y) in A[[X, Y]] such that:

F_{f}(X, Y) ≡ X + Y mod deg 2
f(F_{f}(X, Y)) = F_{f}(f(X), f(Y)).

If F_{f}(X, Y) defines a formal group, we are done because the above means that f acts as an endomorphism on F_{f}(X, Y). We claim that this F_{f}(X, Y) is a formal group law.

1. Commutativity. We let G = F_{f}(Y, X) then we have that:

G(X, Y) ≡ X + Y mod deg 2

f(G(X, Y)) = f(F_{f}(Y, X)) = F_{f}(f(Y), f(X)) = G(f(X), f(Y)).

But by Proposition 2, F_{f}(X, Y) is the only power series with this property, so we must have that G(X, Y) = F_{f}(Y, X) = F_{f}(X, Y).

2. Associativity. Now let

G_{1}(X, Y, Z) = F_{f}(X, F_{f}(Y, Z)), G_{2}(X, Y, Z) = F_{f}(F_{f}(X, Y), Z)

Then, for i = 1, 2,

G_{i}(X, Y, Z) ≡ X + Y + Zmod deg 2
G_{i}(f(X), f(Y), f(Z)) = f(G_{i}(X, Y, Z)).

Proposition 2 says that for ϕ(X, Y, Z) = X + Y + Z, there is a unique power series G(X, Y, Z) that satisfies these criterion. Thus, G_{1} = G_{2}.

Thus, F_{f}(X, Y) satisfies all the axioms of a formal group over A. This concludes the proof of the theorem.

These F_{f}(X, Y) are the Lubin-Tate formal group laws.

Example 1 Let K_{v} = ℚ_{p} and π = p. We define f = (1 + X)^{p} − 1. Notice that f ∈ F_{p}. We also define F_{f}(X, Y) = X + Y + XY. As we showed in section 1, F_{f}(X, Y) admits f as an endomorphism. By the above proposition, we now know that F_{f}(X, Y) is unique.

4 Constructing Extensions with Formal Groups

Now, let a ∈ A and let f, g again be a power series in F_{π}. Applying Proposition 2 for ϕ(X) = aX, we get that there exists a unique power series, call it [a]_{g,f} in A[[X]] such that

[a]_{g,f}(X) ≡ aXmod deg 2
g ∘ [a]_{g,f} = [a]_{g,f} ∘ f.

Proposition 4 [a]_{g,f} is a homomorphism from F_{f}(X, Y) → F_{g}(X, Y).

Proof : Since I am lazy, let’s denote [a]_{g,f} by h. We want to show that

h(F_{f}(X, Y)) = F_{g}(h(X), h(Y)).

It follows from the definition of h that h(F_{f}(X, Y)) ≡ aX + aY mod deg 2 ≡ F_{g}(h(X), h(Y)). We also have that:

h(F_{f}(f(X), f(Y))) = (h ∘ f)(F_{f}(X, Y)) = g(h(F_{f}(X, Y)))
F_{g}(f(f(X)), h(f(Y))) = F_{g}(g(h(X), g(h(Y))) = g(F_{g}(h(X), h(Y))).

Again, we can use Proposition 2 to get uniqueness.

Further, we have that:

[a + b]_{g,f} = [a]_{g,f} +_{F_{g}} [b]_{g,f}, [ab]_{g,f} = [a]_{h,g} ∘ [b]_{g,f}

and that for u ∈ A^{∗} a unit, [u]_{f,g} and [u^{−1}]_{g,f}] are inverse isomorphisms. The proofs of these assertions follow by writing out the properties of the associated power series and using uniqueness as we did in the last few proofs.

This means that for f, g ∈ F_{π}, F_{f}(X, Y) ≅ F_{g}(X, Y). If we take f = g, then we get that there exists a unique endomorphim [a]_{f} : F_{f} → F_{f} such that

[a]_{f}(X) ≡ aXmod deg 2

and

f ∘ [a]_{f} = [a]_{f} ∘ f.

Thus we get an injection A → End(F_{f}) given by a → [a]_{f}.

Now consider 𝔪K̄ where K̄, the algebriac closure of K and 𝔪K̄ is the maximal ideal of the valuation ring. As in section 2, we use 𝔪K̄ as the set on which we define operations derived from our formal group.

Proposition 5 𝔪K̄ is an A-module with the operations:

1. For x, y ∈ 𝔪K̄, let x + F_{f}y = F_{f}(x, y).

2. For a ∈ A and x ∈ 𝔪K̄, let a ∗ x = [a]_{f}x. When we refer to 𝔪K̄ as an A-module, we call it W_{f}.

Proof : To show this, we need to show that 𝔪K̄ under addition forms an abelian group. The proof is identical to the one given in section 2, except that we are working with a more general power series. The only thing we need to add is that we can be assured that F_{f}(x, y) converges because |x|, |y| < 1. To show that it satisfies the rest of the module axioms is straightforward if we keep the fact of convergence in mind.

Associated to W_{f}, we define:

Definition 6 Denote by W_{f}^{n} the submodule of W_{f} defined as the set:

W_{f}^{n} = {x ∈ W_{f}|[π]_{f}^{n} ∗_{F_{f}} x = 0}

where [π]_{f} is defined to be unique power series in A[[X]] such that

[π]_{f}(X) ≡ πXmod deg 2
f ∘ [π]_{f} = [π]_{f} ∘ f.

However, note that this means [π]_{f}(X) = f(X) because we defined f(X) ∈ F_{π} to satisfy these requirements. We have uniqueness as a consequence of Proposition 4.

W_{f}^{n} is, in fact, exactly the [π]_{f}^{n}-torsion of W_{f}! We can also notice that W_{f}^{n} is equivalently the set of roots of [π]_{f}^{n} = f^{n} in 𝔪K̄.

Example 2 Continuing our previous example, we now take

W_{f}^{n} = {x ∈ ℤ̄_{p}|(x − 1)^{p^{n}} = 1} = {ζ|ζ^{p^{n}} − 1}

this is exactly equal to μ_{p^{n}}.

Lemma 1 [π]_{f}^{n} = f^{n}(X) is a monic, separable polynomial of degree q^{n} in A[X] and W_{f}^{n} is the set of all roots of f^{n}(X) in K̄. So the order of W_{f}^{n} is q^{n}.

Proof : We know that W_{f}^{n} consists of the roots of f^{n} in W_{f}, equal to 𝔪K̄ as a set. We claim that W_{f}^{n} is actually the set of all roots of f^{n} in K̄. First, recall that

f^{n} = f ∘ … ∘ f

For illustrative purposes,footnotewe don’t need this form to prove the proposition - all we require is that f(X) ∈ F_{π}, but it is easier to the calculation this way. suppose that f(X) takes the form:

f(X) = πX + … + X^{q}

Then we have

(f ∘ f)(X) = f(f(X)) = π(πX + … + X^{q}) + … + (πX + … + X^{q})^{q} = πX^{2} + … + X^{q^{2}}

By induction, we get that:

f^{n} = π^{n}X + … + X^{q^{n}}

Now suppose that x ∈ K̄ were a root. We would have |π^{n}x + … + x^{q^{n}}| = 0 which implies that |x|^{q^{n}} ≤ max{|x|^{lesserpowers}} and so |x| ≤ 1 so x ∈ 𝔪K̄.

The following is a structure theorem for certain modules that we will take for granted:

Proposition 6 Let A be a principal ideal domain with only one prime up to conjugation (notice that A = O_{K} satisfies these conditions). Then every finitely generated torsion A-module, M, decomposes into a direct sum of cyclic modules:

M ≅ A/(π^{n_{1}}) ⊕ … ⊕ A/(π^{n_{r}}), n_{1} ≤ … ≤ n_{r}

and the sequence n_{1}, …, n_{r} is uniquely determined.

Lemma 2 Let M be an A-module, and let M_{n} = Ker(π^{n} : M → M). If we have that:

1. If M_{1} has q := (A : (π)) elements, and

2. π : M → M is surjective, then M_{n} ≅ A/(π^{n}); and thus it has q^{n} elements.

It is worth clarifying that by the map π^{n} : M → M, we mean the map that takes x ∈ M to π ∗ x = [π]_{f}^{n}(x). Proof : This is done by induction on n. Base case: n = 1. We want to show that M_{1} ≅ A/(π). From 1, we have that M_{1} has q elements. Thus, applying the structure theorem for modules, we can write M_{1} as a product of cyclic modules. The only way to do this to write M_{1} ≅ A/(π) because it has q elements (since (π) is a prime and thus maximal ideal so A/(π) is the residue field which we assumed has order q). More generally, note that A/(π^{n}) has q^{n} elements. Induction case: We assume that M_{n−1} ≅ A/(π^{n−1}). Now consider the sequence:

0 → M_{1} → M_{n} →_{π} M_{n−1} → 0

Condition 2. implies that this sequence is exact. By the first isomorphism theorem, then, we have:

M_{n}/M_{1} ≅ M_{n−1}

Using our hypotheses, this gives:

M_{n}/(A/(π)) ≅ (A/π^{n−1})

and so M_{n} has q^{n} elements. Further, M_{n} must be cyclic because M_{1} and M_{n−1} are (we need the above equation to hold). We know from the structure theorem that A/(π^{n}) is the unique module that satisfies these conditions.

Proposition 7 The A-module, W_{f}^{n} is isomorphic to A/(π^{n}). Thus we have that End_{A}(W_{f}^{n}) = A/(π^{n}) and Aut_{A}(W_{f}^{n}) = (A/(π^{n}))^{×}.

Proof : We will prove this by showing that W_{f} satisfies the hypothesis for M in the previous lemma. We already showed that W_{f} is an A-module. We need to show that:

1. W_{f_{1}} = Ker(π : W_{f} → W_{f}) has q elements. Note that W_{f_{1}} is just W_{f}^{1} by definition. The kernel of π is precisely the roots of [π]_{f}(X) = f(X) for f(X) ∈ F_{π}. It does not matter which f(X) we pick because an A-isomorphism h : F_{f(X,Y)} → F_{g(X,Y)} of formal groups induces an isomorphism of A-modules W_{f} → W_{g}. Thus, we can take f(X) to be of the form πX + … + X^{q}. This is an Eisenstein polynomial so it has q distinct roots. we have already shown that these roots all have valuation less than 1 and so they lie in W_{f}. This means that the kernel of the map π has exactly q elements, all in W_{f}.

2. π : W_{f} → W_{f} is surjective. This follows because [π]_{f} = f(X) which we showed is an automorphism.

Thus, we apply the lemma on W_{f} to get that W_{f}^{n} ≅ A/(π^{n}). It follows immediately that the action of A on W_{f}^{n} induces an isomorphism A/(π^{n}) → End_{A}(W_{f}^{n}). The automorphisms, then, are the subset which are isomorphisms, that is, (A/(π^{n}))^{×}.

Lemma 3 Let L be a finite Galois extension of K, a local field with Galois group G. The, for any F ∈ O_{K}[[X_{1}, …, X_{n}]] and x_{1}, …, x_{n} ∈ 𝔪_{L}, we have that:

F (τα_{1}, …, τα_{n}) = τF (α_{1}, …, α_{n})

for all τ ∈ G

Proof : We know that the valuation on K extends uniquely to L. We also know that | · | makes L a topological field (addition, multiplication, and inverses are continuous). Thus τ is a field isomorphism that fixes O_{K}. For F a polynomial (not infinite), this is enough to show that F behaves properly. Now if F is a power series, we use the fact that τ is continuous. So it preserves limits:

lim_{m→∞} x_{m} = L → lim_{m→∞} τx_{m} = τL.

We approximate F by the polynomials F_{m} such that F = F_{m} + deg ≥ m + 1. Then we have:

τ(F (x_{1}, …)) = τ( lim_{m→∞} F_{m}(x_{1}, …)) = lim_{m→∞} τF (x_{1}, …) = lim_{m→∞} F (τx_{1}, …) = F (τx_{1}, …).

Theorem 2 Consider K[W_{f}^{n}] a field extension of K a non-archimedean local field with residue field F_{q}. The following holds:

1. For each n, K[W_{f}^{n}]/K is of degree (q − 1)q^{n−1}

2. The action of O_{K} on W_{f}^{n} defines an isomorphism:

(O_{K}/𝔪_{K})^{×} → Gal(K[W_{f}^{n}]/K).

In particular, K(W_{f}^{n})/K is an abelian extension.

Proof : We assume that f is a polynomial of the form πX +…+X^{q} (this is okay because any other f ∈ F_{π} would result in an isomorphic result). Choose a nonzero root, π_{1} of f(X). The choose a nonzero root π_{2} of f(X) − π_{1}. π_{2} is an element of W_{f} as well (by Newton’s polygon, which is an application of Hensel’s Lemma). Inductively, we choose π_{n} a nonzero root of f(X) − π_{n−1}. We get the chain of extensions:

K ⊂ K[π_{1}] ⊂ … ⊂ K[π_{n}] ⊂ K[W_{f}^{n}]

where the first extension is of degree q − 1 and the rest are of degree q. By the Tower Law, [K[W_{f}^{n}] : K] = (q − 1)q^{n−1}.

To show 2, remember that W_{f}^{n} is the set of roots of f^{n} in K̄. Thus, K[W_{f}^{n}] is the splitting field of f^{n}. Thus, K[W_{f}^{n}]/K is a Galois extension and we can identify Gal(K[W_{f}^{n}]/K) with the group of permutations of W_{f}^{n} (because these are exactly the roots of f). Now we can apply Lemma 3. This allows us to think of each element of Gal(K[W_{f}^{n}]/K) as O_{K}-module isomorphism acting on W_{f}^{n}. Because of this, the image of Gal(K[W_{f}^{n}]/K) in Sym(W_{f}^{n}) must be contained in

End_{O_{K}}(W_{f}^{n}) = (O_{K}/(π^{n}))^{×}.

Thus,

(q − 1)q^{n−1} ≥ #Gal(K[W_{f}^{n}]/K) = [K(W_{f}^{n}) : K]

But we also have:

[K[W_{f}^{n} : K] ≥ [K[W_{f}^{n}] : K] = (q − 1)q^{n−1}

So #Gal(K[W_{f}^{n}]/K) = [K(W_{f}^{n}) : K] = (q − 1)q^{n−1}

But if the order of Gal(K[W_{f}^{n})]K) is exactly (q − 1)q^{n−1} and it fits inside End_{O_{K}}(W_{f}^{n}) = (O_{K}/(π^{n}))^{×}, then they must be isomorphic. Thus, K[W_{f}^{n}]/K is an abelian extension.

References

[1] Artin, Emil. Algebriac Numbers and Algebraic Functions, Gordon and Breach: New York, 1967.

[2] Iwasawa, Kenkichi. Local Class Field Theory, Oxford University Press: Oxford, 1986.

[3] Lubin, Jonathan and John Tate. “Formal Complex Multiplication in Local Fields,” Annals of Math, 81, 1965.

[4] Milne, J.S. Class Field Theory, Online coursenotes, available: http://www.jmilne.org/ math/, 1997.

[5] Serre, Jean-Paul. Corps Locaux, Hermann: Paris, 1962.