# A BRIEF INTRODUCTION ON LOCAL CLASS FIELD THEORY HUA-CHIEH LI

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HUA-CHIEH LI

In  Neukirch wrote: “The main goal of field theory is to classify all algebraic extensions of a given field K. The law governing the constitution of extensions of K is hidden in the inner structure of the base field K itself, and should therefore be expressed in terms of entities directly associated with it.” Local class field theory solves this problem as far as the abelian extensions of the local field K are concerned. It establishes a one-to-one correspondence between these extensions and certain subgroups of K. More precisely, the rule

L → NL/KL

gives a one-to-one correspondence between the finite abelian extensions of a local field K and the open subgroups of finite index in K. This is called the existence theorem, because its essential statement is that, for every open subgroup N of finite index in K, there exists an abelian extension L/K such that N = NL/KL. This is the “class field” of N .

The existence theorem can be deduced by the local reciprocity law which says that, for every finite Galois extension L/K of local fields we have a canonical isomorphism

ˆ

rL/K : G(L/K)ab ∼−→ K/NL/KL.

In this note, we employ Neukirch’s method  for the construction of the reciprocity map ˆrL/K. For the cohomology version of this construction, we recommend Serre’s presentation .

Most of the theorems in this note are followed by a “sketch of proof” rather than a complete proof. Frequently all the major steps of a proof will be stated, with the reasons or the routine calculational details left to the reader.

Prerequisite for reading this note, apart from Galois theory, is merely a standard introduction to the theory of local fields. We recommend [1, 5] for these subjects.

Lecture notes for a summer school held by NCTS in 2006.

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1. Notations and Preliminary Results

A local field is a field which is complete with respect to a discrete valuation, and which has a finite residue class field. These are precisely the finite extensions K of the field Qp or Fp((t)). We will use the following notation.

• vK is the discrete valuation normalized by vK(K) = Z.

• OK = {a ∈ K | vK(a) ≥ 0} is the valuation ring.

• PK = {a ∈ K | vK(a) > 0} is the maximal ideal.

• UK = {a ∈ K | vK(a) = 0} is the unit group.

• UK(n)= {a ∈ K | vK(a − 1) ≥ n}

• πK is a prime element, i.e., PK= πKOK.

• eK is the maximal unramified extension of K.

Furthermore, for a finite extension L/K,

• L0= L ∩ eK is the maximal unramified subextension of L/K,

• eL/K = vLK) = [L : L0] and

• fL/K = [OL/PL: OK/PK] = [L0: K].

We remark that K is locally compact with respect to the discrete valuation topology, and the ring of integers OK and the maximal ideal PK are compact. The multiplicative group Kis also locally compact, and the unit group UK is compact.

1.1. The residue class field of eK is the algebraic closure k of the residue class field k of K. We get a canonical isomorphism

G( eK/K) ' G(k/k).

Suppose that k ' Fq. This isomorphism associates to the Frobenius automorphism x 7→ xq in G(k/k) and the Frobenius automorphism φK in G( eK/K) which is given by

aφK ≡ aq (mod PKe), a ∈ OKe.

The subgroup hφKi = {φnK| n ∈ Z} has the same fixed field K as the whole group G( eK/K). But contrary to what we are used to in finite Galois theory, we find hφKi 6= G( eK/K). In fact, there are more subgroups than fixed fields for infinite extension. In order to explain this, let us consider the extension Fq/Fq. Observe first that Fq = S

n=1Fqn and Fqm ⊆ Fqn if and only if m | n. Let φ be the Frobenius automorphism in G(Fq/Fq). Since φ|Fqm has order m, if a ≡ b (mod m) then φa|Fqm = φb|Fqm. Therefore, if we choose a sequence {an}n∈N of

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integers satisfying an ≡ am (mod m) whenever m | n, then the map ψ satisfying ψ|Fqm = φam|Fqm is an automorphism of Fq. In particular, if we choose {an}n∈N

such that there is no integer a satisfying an ≡ a (mod n) for all n ∈ N, then ψ cannot belong to hφi. In fact, passing from the isomorphism G(Fqn/Fq) ' Z/nZ to the projective limit gives an isomorphism

G(Fq/Fq) ' bZ = lim←−−

n Z/nZ

and what we did amounted to writing down the element (. . . , an, . . . ) ∈ bZ. The projective limit bZ is going to occupy quite an important position in what follows.

It contains Z as a dense subgroup (by considering bZ as the subset of Q

n∈NZ/nZ which is equipped with the product topology). The groups nbZ, n ∈ N, are precisely the open subgroups of bZ, and it is easy to verify that

bZ/nbZ ' Z/nZ.

The Galois group G(Ω/K) of any Galois extension Ω/K carries a canonical topology. This topology is called the Krull topology and is obtained as follows. for every σ ∈ G(Ω/K), we take the cosets σG(Ω/L) as a basis of neighborhoods of σ, with L/K ranging over finite Galois subextensions of Ω/K. The main theorem of Galois theory for infinite extensions can now be formulated as follows.

Theorem 1.1.1. Let Ω/K be a Galois extension. Then the assignment M 7→

G(Ω/M ) is a 1-1-correspondence between the subextensions M/K of Ω/K and the closed subgroups of G(Ω/K).

In particular, if σ ∈ G(Ω/K) and Σ is the fixed field of σ, then G(Ω/Σ) is the closure hσi of the subgroup hσi. G(Ω/Σ) is a quotient of the group bZ. In fact, we have for every n the projective homomorphism

Z/nZ → G(Ω/Σ)/G(Ω/Σ)n, 1 (mod nZ) 7→ σ (mod G(Ω/Σ)n), and passing to the projective limit yields a continuous surjection bZ → G(Ω/Σ).

1.2. Let G be a finite group and A a (multiplicative) G-module. At the center of class field theory, there are two groups

H0(G, A) = AG/NGA and H−1(G, A) = NGA/IGA, where

AG = {a ∈ A | aσ= a, ∀ σ ∈ G}, NGA = {NGa = Y

σ∈G

aσ| a ∈ A},

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NGA = {a ∈ A | NGa = 1}

and IGA is the subgroup of NGA which is generated by all elements aσ−1, with a ∈ A and σ ∈ G.

We will be mainly interested in the case where G is a finite cyclic group. If G is cyclic and σ is a generator, then IGA is simply the group Aσ−1= {aσ−1| a ∈ A}.

In fact, the formal identity σm−1 = (1+σ+· · ·+σm−1)(σ−1) implies aσm−1= bσ−1 with b =Qm−1

i=0 aσi.

Suppose that G is a finite cyclic group. If 1 → A → B → C → 1 is an exact sequence of G-modules, then we obtain an exact hexagon

H0(G, A) −→ H0(G, B)

% &

H−1(G, C) H0(G, C)

- .

H−1(G, B) ←− H−1(G, A)

An excellent tool for studying H0(G, A) and H−1(G, A) is the Herbrand quotient.

The Herbrand quotient of the G-module A is defined to be h(G, A) = #H0(G, A)

#H−1(G, A),

provided that both orders are finite. In particular, if G = hσi and A is a finite G-module, then the exact sequences

1 −→ AG−→ Aσ−1−→ IGA −→ 1 and 1 −→ NGA −→ A−→ NNG GA −→ 1, show that #A = #AG· #IGA = #NGA · #NGA, and hence h(G, A) = 1.

Using the exact hexagon, we can deduce the multiplicativity of the Herbrand quotient.

Proposition 1.2.1. Let G be a finite cyclic group. If 1 → A → B → C → 1 is an exact sequence of G-modules, then one has

h(G, B) = h(G, A) · h(G, C)

in the sense that, whenever two of these quotients are defined, so is the third and the identity holds.

In local class field theory, the crucial point is to verify for the multiplicative group of a local field the class field axiom :

Theorem 1.2.2. For a cyclic extension L/K of local fields, one has

#H0(G(L/K), L) = [L : K] and #H−1(G(L/K), L) = 1.

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Proof. For #H−1(G(L/K), L) = 1, this is the famous “Hilbert 90”. So all we have to show is that the Herbrand quotient is h(G, L) = H0(G, L) = [L : K], where we have put G = G(L/K). The exact sequence 1 → UL→ L∗ v−→ Z → 0, in whichL Z has to be viewed as the trivial G-module, yields, by Proposition 1.2.1,

h(G, L) = h(G, UL)h(G, Z) = h(G, UL)[L : K].

Hence we have to show that h(G, UL) = 1.

First, we choose a normal basis {ασ | σ ∈ G} of L/K with α ∈ OL (see [2, Chapter VIII]), and consider the open G-module M =P

σ∈GOKασ. Then consider the open sets

Vn= 1 + πKnM, n ∈ N.

Since M is open, we have πNKOL ⊆ M for suitable N , and for n ≥ N the Vn are subgroups of UL. Via the correspondence 1 + πnKβ 7→ β (mod πKM ), we obtain G-isomorphisms

Vn/Vn+1' M/πKM =M

σ∈G

(OK/PKσ.

It is easy to check that H0(G,M

σ∈G

(OK/PKσ) = H−1(G,M

σ∈G

(OK/PKσ) = 0,

and hence

H0(G, Vn/Vn+1) = H−1(G, Vn/Vn+1) = 1, for n ≥ N .

This implies that H0(G, Vn) = 1, for n ≥ N . Indeed, if a ∈ (Vn)G, then a = (NGb0)a1, with b0 ∈ Vn and a1 ∈ (Vn+1)G. Continuing in this way, by the completeness yields a = NGb, with b =Q

i=0bi∈ Vn, so that H0(G, Vn) = 1.

Similarly, we have H−1(G, Vn) = 1, for n ≥ N . Because UL is compact, UL/Vn is finite. Therefore, by Proposition 1.2.1, we obtain

h(G, UL) = h(G, UL/Vn)h(G, Vn) = 1.

¤ Among the cyclic extensions there are in particular the unramified extensions, so that one has

Corollary 1.2.3. For a finite unramified extension L/K, one has H0(G(L/K), UL) = H−1(G(L/K), UL) = 1

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and

H0(G(L/K), UL(n)) = H−1(G(L/K), UL(n)) = 1, n ∈ N.

Proof. Let G = G(L/K). As H−1(G, L) = 1, every element u ∈ UL such that NL/K(u) = 1 is of the form u = aφL/K−1, with a ∈ L and φL/K = φK|L. Since L/K is unramified, πK is also a prime element of L. So writing a = επmK with ε ∈ UL, we obtain u = εφL/K−1. This shows that H−1(G, UL) = 1. Since we have proved that h(G, UL) = 1 in Theorem 1.2.2, this shows that H0(G, UL) = 1.

In order to prove H0(G, UL(n)) = H−1(G, UL(n)) = 1, we first show that H0(G, l) = H−1(G, l) = 1 and H0(G, l) = H−1(G, l) = 0,

for the residue class field l of L. We have H−1(G, l) = 1 by Hilbert 90. This implies that H0(G, l) = 1, as l is finite and so h(G, l) = 1. Let f = [l : k] be the degree of l over the residue class field k of K, and let q = #k. Then we have

#NGl = #{x ∈ l |

f −1X

i=0

xqi= 0} ≤ qf −1 and #IGl = qf −1,

since k is the fixed field of the Frobenius automorphism l−→ l. Therefore↑q H−1(G, l) = NGl/IGl = 0.

This implies that H0(G, l) = 0 as h(G, l) = 1.

Applying now the exact hexagon to the exact sequence of G-modules 1 → UL(1)→ UL→ l→ 1,

we obtain

H0(G, UL(1)) = H0(G, UL) = 1 and H−1(G, UL(1)) = H−1(G, UL) = 1.

From the exact sequence of G-modules

1 → UL(n+1)→ UL(n)→ l → 0, we now deduce by induction just as above, that

H0(G, UL(n+1)) = H0(G, UL(n)) = 1 and H−1(G, UL(n+1)) = H−1(G, UL(n)) = 1.

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2. The Local Reciprocity Law

The Frobenius automorphism governs the entire class field theory like a king.

It is therefore most remarkable that in the case of a finite Galois extension L/K, every σ ∈ G(L/K) becomes a Frobenius automorphism once it is maneuvered into the right position. This point of view helps us to construct the reciprocity map which expresses the fundamental principle of class field theory to the effect that Frobenius automorphisms correspond to prime elements.

2.1. For a local field K, we denote by φKthe Frobenius automorphism in G( eK/K).

Let dK : G( eK/K) → bZ be the isomorphism such that dKK) = 1. We pass from the Galois extension L/K to the extension eL/K and consider the function dL/K : G(eL/K) → bZ such that dL/K(σ) = dK(σ|Ke), for σ ∈ G(eL/K). In particular, since φL|Ke = φfKL/K, one has dL/KL) = fL/K. Notice that every element in G( eK/K) can be extended to an element in G(eL/K). Therefore, dL/K : G(eL/K) → bZ is surjective.

Consider in the Galois group G(eL/K) the semigroup

Frob(eL/K) = {σ ∈ G(eL/K) | dL/K(σ) ∈ N}.

In other words, σ ∈ Frob(eL/K) if and only if σ ∈ G(eL/K) and σ|Ke = φnK, for some n ∈ N. Because dL/K(1) = 0 and dL/K1σ2) = dL/K1) + dL/K2), one knows that 1 6∈ Frob(eL/K) and Frob(eL/K) is closed with respect to multiplication (but not closed with respect to inversion). Moreover, since dL/K is surjective, dL/K

maps Frob(eL/K) onto N. Firstly, we have the

Proposition 2.1.1. For a finite Galois extension L/K, the mapping Frob(eL/K) → G(L/K), σ 7→ σ|L,

is surjective.

Proof. Let ˜φ ∈ Frob(eL/K) be an element such that dL/K( ˜φ) = 1. Then ˜φ|Ke = φK. Remark here that L0 = L ∩ eK is the maximal unramified subextension of L/K.

For ρ ∈ G(L/K), since ρ|L0 ∈ G(L0/K) = h ˜φ|L0i, there exists n ∈ N such that ρ|L0 = ˜φn|L0. Since ρ ˜φ−n|L∈ G(L/L0) and the mapping

G(eL/ eK) → G(L/L0), τ 7→ τ |L

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is an isomorphism, there is τ ∈ G(eL/ eK) such that τ |L = ρ ˜φ−n|L. Therefore, σ = τ ˜φn ∈ G(eL/K) is an element satisfying σ ∈ Frob(eL/K) (because σ|Ke = φnK)

such that σ|L= τ ˜φn|L= ρ. ¤

Thus every element σ ∈ G(L/K) may be lifted to an element in Frob(eL/K).

The following proposition shows that this lifting, considered over its fixed field, is actually the Frobenius automorphism.

Proposition 2.1.2. Let σ ∈ Frob(eL/K) and let Σ be the fixed field of σ. Then Σ/K is a finite extension such that eΣ = eL, fΣ/K = dL/K(σ) and eΣ/K = eL/K. Moreover, σ = φΣ.

Proof. We first show that eΣ = eL. Since Σ ⊆ eL, one has eΣ ⊆ eL. The canonical surjection G(eL/Σ) → G(eΣ/Σ), τ 7→ τ |Σe must be bijective, because G(eL/Σ) = hσi is a quotient of bZ ' G(eΣ/Σ). But G(eL/Σ) = G(eΣ/Σ) implies that eΣ = eL. This also implies that eΣ/K= [eΣ : eK] = [eL : eK] = eL/K.

Suppose that dL/K(σ) = d. In other words, σ|Ke = φdK. Because the fixed field of σ|Ke is Σ ∩ eK = Σ0, we have fΣ/K = [Σ0: K] = d. Therefore,

[Σ : K] = eΣ/KfΣ/K= eL/Kd ≤ [L : K]dL/K(σ) is finite.

Finally, since fΣ/K = [Σ0 : K] = d, one has φΣ|Ke = φdK = σ|Ke. Hence, σ ∈ G(eΣ/Σ) and φΣ∈ G(eΣ/Σ) are identical in Σ eK = eΣ. Thus σ = φΣ. ¤

Our goal is to define a canonical homomorphism rL/K : G(L/K) → K/NL/KL

for every finite Galois extension L/K. To this end, we define first a mapping on Frob(eL/K).

Definition 2.1.3. The reciprocity map

rL/Ke : Frob(eL/K) → K/NL/KL is defined by

rL/Ke (σ) = NΣ/KΣ) (mod NL/KL), where Σ is the fixed field of σ.

Observe that the definition of rL/Ke (σ) does not depend on the choice of the element πΣ. For another prime element differs from πΣonly by an element u ∈ UΣ,

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and for this we have NΣ/K(u) ∈ NL/KL. To see this, we let M = LΣ. Applying Corollary 1.2.3 to the unramified extension M/Σ (because fM = eΣ), one finds u = NM/Σ(ε), for some ε ∈ UM, and thus

NΣ/K(u) = NΣ/K(NM/Σ(ε)) = NL/K(NM/L(ε)) ∈ NL/KL.

Next we want to show that the reciprocity map rL/Ke is multiplicative. In other words, we have to show that if σ1σ2= σ3is an equation in Frob(eL/K) and Σi the fixed field of σi, for i = 1, 2, 3, then

NΣ1/KΣ1)NΣ2/KΣ2) ≡ NΣ3/KΣ3) (mod NL/KL).

To do this, we need two Lemmas.

Our first lemma makes NΣi/KΣi), for i = 1, 2, 3 as a norm over the same fields.

Lemma 2.1.4. For a finite Galois extension L/K, let φ, σ ∈ Frob(eL/K) with dL/K(φ) = 1 and dL/K(σ) = n. If Σ is the fixed field of σ and a ∈ Σ, then

NΣ/K(a) = NL/ ee K(

n−1Y

i=0

aφi).

Proof. Since eΣ = eL (Proposition 2.1.2), for a ∈ Σ we have NΣ/Σ0(a) = NeΣ/ eK(a) = NL/ ee K(a). On the other hand, since [Σ0: K] = fΣ/K= n and G(Σ0/K) is generated by φK|Σ0= φ|Σ0, one has NΣ0/K(b) =Qn−1

i=0 bφi, for b ∈ Σ0. For a ∈ Σ we thus get NΣ/K(a) = NΣ0/K(NΣ/Σ0(a)) =

n−1Y

i=0

NL/ ee K(a)φi= NL/ ee K(

n−1Y

i=0

aφi).

The last equation follows from φG(eL/ eK) = G(eL/ eK)φ. ¤ Next lemma provide us a method to identify an element which is in NL/KL. Lemma 2.1.5. For a finite Galois extension L/K, let φ ∈ Frob(eL/K) satisfy dL/K(φ) = 1. Suppose that u ∈ ULe such that uφ−1=Qr

i=1uτii−1, for some ui∈ ULe and τi∈ G(eL/ eK). Then NL/ ee K(u) ∈ NL/KL.

Proof. Let M/K be a finite Galois subextension of eL/K such that u, ui ∈ UM and L ⊆ M . Let [M : K] = n, σ = φn and let Σ be the fixed field of σ. Since fM = eΣ and fM/K | n = fΣ/K, we have that M ⊆ Σ. Further, let Σ0/Σ be the unramified extension of degree n. By Corollary 1.2.3, we can then find elements ˜u, ˜ui∈ UΣ0such that u = NΣ0u) and ui= NΣ0ui). Since G(Σ0/Σ) is generated by σ|Σ0 = φΣ|Σ0

and φG(eL/ eK) = G(eL/ eK)φ, by the assumption we have ˜uφ−1 = λQr

i=1u˜τii−1, for

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an element λ ∈ UΣ0 such that NΣ0(λ) = 1. Hence, again by Corollary 1.2.3, λ = µσ−1= µφn−1, with µ ∈ UΣ0. We may thus write

˜

uφ−1 = µφn−1 Yr i=1

˜

uτii−1= (

n−1Y

j=1

µφj)φ−1 Yr i=1

˜ uτii−1.

Applying NL/ ee K gives NL/ ee Ku)φ−1 = NL/ ee K(Qn−1

j=1µφj)φ−1, so that NL/ ee Ku) = NL/ ee K(

n−1Y

j=1

µφj)z,

for some z ∈ UK. Finally, applying NΣ0, we obtain, observing n = [M : K] =0: Σ] and using Lemma 2.1.4, that

NL/ ee K(u) = NL/ ee K(NΣ0u)) = NΣ0(NL/ ee Ku)) = NΣ0(NL/ ee K(

n−1Y

j=1

µφj))zn

= NL/ ee K(

n−1Y

j=1

NΣ0(µ)φj)zn= NΣ/K(NΣ0(µ))NM/K(z) ∈ NL/KL.

¤ Now we are ready to show the

Proposition 2.1.6. For a finite Galois extension L/K, the reciprocity map is multiplicative.

Proof. Let σ1σ2 = σ3 be an equation in Frob(eL/K), Σi the fixed field of σi and πi= πΣi, for i = 1, 2, 3. We have to show that

NΣ1/K1)NΣ2/K2) ≡ NΣ3/K3) (mod NL/KL).

Suppose that dL/Ki) = ni, for i = 1, 2, 3. In order to apply Lemma 2.1.5, we choose a fixed φ ∈ Frob(eL/K) such that dL/K(φ) = 1 and put

τi= σi−1φni ∈ G(eL/ eK), i = 1, 2, 3.

From σ1σ2= σ3and n1+ n2= n3, we then deduce that τ3= τ2−n2σ1φn2)−1φn1. Putting σ4= φ−n2σ1φn2 and n4= dL/K4) = n1 and τ4 = σ−14 φn4, we find that τ3 = τ2τ4 and NΣ4/K4) = NΣ1/K1), where Σ4 = Σφ1n2 is the fixed field of σ4 and π4 = πφ1n2 is a prime element of Σ4. We may therefore pass to show the congruence

NΣ2/K2)NΣ4/K4) ≡ NΣ3/K3) (mod NL/KL).

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From Lemma 2.1.4, if we put u =¡nY2−1

i=0

πφ2i¢¡nY4−1

i=0

π4φi¢¡nY3−1

i=0

−13 )φi¢

∈ ULe,

then the congruence amounts simply to the relation NL/ ee K(u) ∈ NL/KL. For this, however, Lemma 2.1.5 gives us all that we need.

Since σi fixes πi, we have πiφni−1 = πiσ−1i φni−1 = πiτi−1, and hence uφ−1 = πτ22−1π4τ4−1π31−τ3. Because eΣ2 = eΣ3 = eΣ4, we have π2 = u2π4, π3 = u−13 π4 and πτ42 = u−14 π4, for u2, u3, u4∈ ULe. We obtain uφ−1= uτ22−1uτ33−1uτ44−1. By Lemma

2.1.5, we do get NL/ ee K(u) ∈ NL/KL. ¤

2.2. From the surjectivity of the mapping Frob(eL/K) → G(L/K), we now have the

Proposition 2.2.1. For every finite Galois extension L/K, there is a canonical homomorphism rL/K : G(L/K) → K/NL/KL given by

rL/K(σ) = NΣ/KΣ) (mod NL/KL),

where Σ is the fixed field of an extension ˜σ ∈ Frob(eL/K) of σ ∈ G(L/K).

Proof. We first show that the definition of rL/K is independent of the choice of the extension ˜σ ∈ Frob(eL/K) of σ ∈ G(L/K). For this, let ˜σ0 ∈ Frob(eL/K) be another extension and Σ0 its fixed field. If dL/Kσ) = dL/Kσ0), then ˜σ|Ke = ˜σ0|Ke and ˜σ|L= ˜σ0|L, so that ˜σ = ˜σ0, and there is nothing to show. However, if we have, say, dL/Kσ) < dL/Kσ0), then denote τ = ˜σ−1σ˜0. The automorphism τ ∈ G(eL/L) and dL/K(τ ) = dL/Kσ0) − dL/Kσ) ∈ N. Hence τ ∈ Frob(eL/K) and the fixed field Σ00of τ contains L. Therefore, Proposition 2.1.6 shows that

NΣ0/KΣ0) ≡ NΣ00/KΣ00)NΣ/KΣ) ≡ NΣ/KΣ) (mod NL/KL).

This means that rL/K is well defined.

The fact that the mapping is a homomorphism follows directly from Proposition

2.1.6. ¤

Definition 2.2.2. The reciprocity homomorphism rL/K : G(L/K) → K/NL/KL is defined by

rL/K(σ) = NΣ/KΣ) (mod NL/KL),

where Σ is the fixed field of an extension ˜σ ∈ Frob(eL/K) of σ ∈ G(L/K).

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The definition of the reciprocity map expresses the fundamental principle of class field theory to the effect that Frobenius automorphisms correspond to prime elements. This principle appears at its purest in the

Proposition 2.2.3. If L/K is a finite unramified extension, then the reciprocity homomorphism rL/K : G(L/K) → K/NL/KL is given by

rL/KK|L) ≡ πK (mod NL/KL) and is an isomorphism.

Proof. In this case eL = eK, the Frobenius automorphism φK ∈ Frob(eL/K) is an ex- tension of φK|L. The fixed filed of φK is K, and hence by definition, rL/KK|L) ≡ πK (mod NL/KL).

Consider the valuation map vK: K→ Z. It induces an isomorphism K/NL/KL' Z/nZ,

with n = [L : K]. Indeed, if vK(a) ≡ 0 (mod nZ), then a = uπKnr, and since u = NL/K(ε) for some ε ∈ UL (Corollary 1.2.3), we find a = NL/K(επrK) ≡ 1 (mod NL/KL). This shows that πK (mod NL/KL) generates the cyclic group K/NL/KL of order [L : K]. Since φK|L also generates the cyclic group G(L/K),

we have rL/K is an isomorphism. ¤

The homomorphism rL/K in general is not an isomorphism. This can be clearly seen when G(L/K) is not abelian. Finite unramified extension is always a cyclic extension. Next, we treat the other extreme case.

Proposition 2.2.4. If L/K is a finite extension which is cyclic and totally ram- ified, then the reciprocity homomorphism rL/K : G(L/K) → K/NL/KL is an isomorphism.

Proof. Since L/K is totally ramified, we have an isomorphism G(eL/ eK) → G(L/K) given by restriction. Let ˜σ ∈ G(eL/ eK) be a generator. Then σ = ˜σ|Lis a generator of G(L/K). Let σ1= ˜σφL∈ G(eL/K). Since

dL/K1) = dK1|Ke) = dKL|Ke) = fL/K = 1

and σ1|L = ˜σ|L = σ, we have that σ1 ∈ Frob(eL/K) is an extension of σ. We thus find for the fixed field Σ/K of σ1 that fΣ/K = dL/K1) = 1 (Proposition 2.1.2), and so Σ0 = Σ ∩ eK = K. Let M/K be a finite Galois subextension of

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L/K containing Σ and L and let Me 0 = M ∩ eK. Since fM = eΣ = eL, we have that G(M/M0) ' G(Σ/K) ' G(L/K) and NM/M0|Σ= NΣ/K, NM/M0|L= NL/K.

For the injectivity of rL/K, we claim that: if rL/Km) ≡ 1 (mod NL/KL), where 0 ≤ m < n = [L : K], then m = 0.

Let πL ∈ OLand πΣ∈ OΣbe prime elements. Since Σ, L ⊆ M and fM = eΣ = eL, πL and πΣ are both prime elements of M . Putting πΣm= uπmL, with u ∈ UM, we obtain

rL/Km) ≡ NΣ/KmΣ) ≡ NM/M0(u)NL/KLm) ≡ NM/M0(u) (mod NL/KL).

From rL/Km) ≡ 1 (mod NL/KL), it thus follows that NM/M0(u) = NL/K(ε) for some ε ∈ UL. Since G(M/M0) is cyclic, from Theorem 1.2.2 (Hilbert 90), we may write u−1ε = a˜σ−1for some a ∈ Mand have

Lmε)σ−1˜ = (πLmε)σ1−1 (because ˜σ|L= σ1|L)

= (πΣm· a˜σ−1)σ1−1

= (aσ1−1)σ−1˜

Hence we have b = πmLεa1−σ1 ∈ M0 with vM(b) = m. However, vM(b) = eM/M0vM0(b) = nvM0(b) implies that one has m = 0, and so rL/K is injective.

The surjectivity the follows from Theorem 1.2.2

#K/NL/KL= [L : K] = #G(L/K).

¤ The reciprocity homomorphism rL/K exhibits the following functorial behavior.

Proposition 2.2.5. Let L/K and L1/K1 be finite Galois extensions such that K1/K and L1/L are finite separable extensions. Then we have the commutative diagram

G(L1/K1) −−−−−−→ KrL1/K1 1/NL1/K1L1

y|L

 yNK1/K

G(L/K) −−−−−−→rL/K K/NL/KL

where the left vertical homomorphism are given by the restriction σ1|L of σ1 G(L1/K1) and the right vertical homomorphism is induced by the norm map NK1/K. Proof. Let σ0 ∈ G(L1/K1) and σ = σ0|L ∈ G(L/K). If ˜σ0 ∈ Frob(eL1/K1) is an extension of σ0, then ˜σ = ˜σ0|Le ∈ Frob(eL/K) is an extension of σ, because dL/Kσ) = fK1/KdL1/K1σ0) ∈ N. Let Σ0 be the fixed field of ˜σ0. Then the fixed

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field of ˜σ is Σ = Σ0∩ eL = Σ0∩ eΣ = Σ00 and hence fΣ0 = [Σ00 : Σ] = 1. If now πΣ0 is a prime element of Σ0, then πΣ= NΣ0Σ0) is a prime element of Σ. The commutativity of the diagram follows from the equality of norms

NΣ/KΣ) = NΣ/K(NΣ0Σ0) = NK1/K(NΣ0/K1Σ0)).

¤ As an easy consequence of the preceding proposition, we have the

Corollary 2.2.6. Let M/K be a Galois subextension of a finite Galois extension L/K. Then we have the commutative exact diagram

(2.1)

1 → G(L/M ) G(L/K) G(M/K) → 1

yrL/M

 yrL/K

 yrM/K

M/NL/ML NM/K−→ K/NL/KL → K/NM/KM → 1 where the central homomorphism of the lower sequence is induced by the identity map of K.

It is clear that when G(L/K) is not abelian the homomorphism rL/K is not an isomorphism. For an arbitrary group G, let G0 denote the commutator subgroup and write Gab= G/G0 for the maximal abelian quotient group. Since K/NL/KL is an abelian group, the homomorphism rL/K naturally induces the homomorphism

ˆ

rL/K : G(L/K)ab→ K/NL/KL

which represents the main theorem of class field theory, and which we will call the Local Reciprocity Law:

Theorem 2.2.7. For every finite Galois extension L/K of local fields, we have a canonical isomorphism

ˆ

rL/K : G(L/K)ab−→ K /NL/KL.

Proof. If M/K is a Galois subextension of L/K, we get from Corollary 2.2.6 the commutative exact diagram (2.1). Using this diagram, we will prove this theorem in three steps.

First, we show that ˆrL/K = rL/K is an isomorphism for every finite cyclic exten- sion L/K. Let M = L∩ eK in diagram (2.1) be the maximal unramified subextension of L/K. Then L/M is a cyclic totally ramified extension and M/K is a unrami- fied extension. Hence, rM/K and rL/M are isomorphisms by Propositions 2.2.3 and

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2.2.4. In the bottom sequence of diagram (2.1)

M/NL/MLNM/K−→ K/NL/KL→ K/NM/KM→ 1,

the map NM/K is injective because the groups in this sequence have the respective orders [L : M ], [L : K] and [M : K] by Theorem 1.2.2. Therefore, ˆrL/K is an isomorphism.

Next, we show that ˆrL/K = rL/K is an isomorphism for every finite abelian extension L/K. We prove this by induction on the degree. Write G(L/K) as a direct sum of cyclic subgroups Hi and let Mi be the fixed field of Hi. One has Hi = G(L/Mi) and Mi/K is an abelian subextension of L/K of smaller degree.

For every Mi, consider M = Mi in the diagram (2.1). The induction hypothesis says that rMi/K is injective. Therefore, if σ ∈ ker(rL/K), then by the commutative diagram (2.1), one has σ is in the kernel of the map G(L/K) → G(Mi/K), which is equal to G(L/Mi) = Hi. In other words, the kernel of rL/K is contained in the intersection of those Hi. Since G(L/K) is a direct sum of these Hi, the kernel of rL/K is the identity and hence ˆrL/K is injective. Surjectivity also follows by induction on the degree. Indeed, since rM/K and rL/M are surjective, so is rL/K.

Finally, we note that G(L/K) is solvable ([1, Chapter II]). If L/K is not abelian, then the commutator subgroup G(L/K)0 is neither the identity nor G(L/K). Let M be the fixed field of G(L/K)0. One has M/K is an abelian extension and G(L/M ) = G(L/K)0 ( G(L/K). Since rM/K is injective, the kernel of rL/K is contained in G(L/M ). Because K/NL/KL is abelian, G(L/M ) = G(L/K)0 is also contained in the kernel of rL/K, and hence ker(rL/K) = G(L/M ). This proves the injectivity of ˆrL/K. The surjectivity follows by induction on the degree as in the abelian case. Indeed, since [L : M ] < [L : K], by the induction hypothesis, one has rL/M and rM/K are surjective, then so is rL/K. Hence ˆrL/K is surjective. ¤ Putting Lab the maximal abelian subextension of L/K, we find G(L/K)ab = G(Lab/K). As an easy consequence of Theorem 2.2.7, we have the

Corollary 2.2.8. Let L/K is a finite Galois extension and let Lab/K be the max- imal abelian subextension in L/K. Then NL/KL= NLab/KLab∗.

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3. The Existence Theorem

The reciprocity law gives us a very simple classification of the abelian extensions of a local field K. We first formulate the existence theorem by considering the norm topology. Then we use Lubin-Tate extension to show the existence theorem for the valuation topology.

3.1. The inverse of the mapping ˆrL/K : G(L/K)ab→ K/NL/KLgives, for every finite Galois extension L/K a surjective homomorphism

( , L/K) : K→ G(L/K)ab

with kernel NL/KL. This map is called the local norm residue symbol. From Proposition 2.2.5, we have the

Proposition 3.1.1. Let L/K and L1/K1 be finite Galois extensions such that K1/K and L1/L are finite separable extension. Then we have the commutative diagram

K1 ( ,L1/K1)

−−−−−−→ G(Lab1 /K1)

 yNK1/K

 y K −−−−−−→( ,L/K) G(Lab/K)

where the left vertical homomorphism are given by the norm map NK1/K and the right vertical homomorphism is induced by the restriction σ|Lab of σ ∈ G(Lab1 /K1).

For every field K, we equip the group K with a topology by declaring the cosets aNL/KL to be a basis of neighborhoods of a ∈ K, where L/K varies over all finite Galois extensions of K. We call this topology the norm topology of K. Notice that by Corollary 2.2.8, we may just consider L/K varies over all finite abelian extensions of K. We will show latter that the norm topology is closely related to the valuation topology.

Lemma 3.1.2. For every local field, we equip with the norm topology.

(1) The open subgroups of K are precisely the closed subgroups of finite index.

(2) The valuation vK: K→ Z is continuous.

(3) If L/K is a finite extension, then NL/K : L→ K is continuous.

Proof.

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