HUA-CHIEH LI

In [4] 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 → N*_{L/K}*L*^{∗}

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 = N**L/K**L** ^{∗}*. 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

*r**L/K* *: G(L/K)*^{ab ∼}*−→ K*^{∗}*/N**L/K**L*^{∗}*.*

In this note, we employ Neukirch’s method [4] for the construction of the reciprocity
map ˆ*r** _{L/K}*. For the cohomology version of this construction, we recommend Serre’s
presentation [5].

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.

1

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 Q* _{p}* or F

_{p}*((t)). We will use the following notation.*

*• v**K* *is the discrete valuation normalized by v**K**(K** ^{∗}*) = Z.

*• O**K* *= {a ∈ K | v**K**(a) ≥ 0} is the valuation ring.*

*• P**K* *= {a ∈ K | v**K**(a) > 0} is the maximal ideal.*

*• U**K* *= {a ∈ K | v**K**(a) = 0} is the unit group.*

*• U*_{K}^{(n)}*= {a ∈ K | v**K**(a − 1) ≥ n}*

*• π**K* is a prime element, i.e., P*K**= π**K**O**K*.

*• eK is the maximal unramified extension of K.*

*Furthermore, for a finite extension L/K,*

*• L*0*= L ∩ eK is the maximal unramified subextension of L/K,*

*• e*_{L/K}*= v*_{L}*(π*_{K}*) = [L : L*_{0}] and

*• f**L/K* *= [O**L**/P**L**: O**K**/P**K**] = [L*0*: K].*

*We remark that K is locally compact with respect to the discrete valuation*
*topology, and the ring of integers O**K* and the maximal ideal P*K* are compact. The
*multiplicative group K*^{∗}*is also locally compact, and the unit group U**K* is compact.

1.1. The residue class field of e*K 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 ' F**q*. This isomorphism associates to the Frobenius automorphism
*x 7→ x*^{q}*in G(k/k) and the Frobenius automorphism φ**K* *in G( eK/K) which is given*
by

*a*^{φ}^{K}*≡ a** ^{q}* (mod P

_{K}_{e}

*),*

*a ∈ O*

_{K}_{e}

*.*

*The subgroup hφ**K**i = {φ*^{n}_{K}*| 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 hφ*_{K}*i 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 F*q**/F**q*.
Observe first that F*q* = S_{∞}

*n=1*F*q** ^{n}* and F

*q*

^{m}*⊆ F*

*q*

^{n}*if and only if m | n. Let φ*

*be the Frobenius automorphism in G(F*

*q*

*/F*

*q*

*). Since φ|*F

*qm*

*has order m, if a ≡ b*

*(mod m) then φ*

^{a}*|*F

_{qm}*= φ*

^{b}*|*F

_{qm}*. Therefore, if we choose a sequence {a*

*n*

*}*

*n∈N*of

*integers satisfying a**n* *≡ a**m* *(mod m) whenever m | n, then the map ψ satisfying*
*ψ|*F*qm* *= φ*^{a}^{m}*|*F*qm* is an automorphism of F*q**. In particular, if we choose {a**n**}**n∈N*

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

*G(F**q**/F**q**) ' b*Z = lim_{←−−}

*n* *Z/nZ*

*and what we did amounted to writing down the element (. . . , a**n**, . . . ) ∈ 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∈N**Z/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

b*Z/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 b*Z. 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 b*Z → 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

*H*^{0}*(G, A) = A*^{G}*/N**G**A and H*^{−1}*(G, A) =* *N**G**A/I**G**A,*
where

*A*^{G}*= {a ∈ A | a*^{σ}*= a, ∀ σ ∈ G},* *N**G**A = {N**G**a =* Y

*σ∈G*

*a*^{σ}*| a ∈ A},*

*N**G**A = {a ∈ A | N**G**a = 1}*

*and I**G**A is the subgroup of* *N**G**A which is generated by all elements a** ^{σ−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 I**G**A is simply the group A*^{σ−1}*= {a*^{σ−1}*| a ∈ A}.*

*In fact, the formal identity σ*^{m}*−1 = (1+σ+· · ·+σ*^{m−1}*)(σ−1) implies a*^{σ}^{m}^{−1}*= b*^{σ−1}*with b =*Q_{m−1}

*i=0* *a*^{σ}^{i}*.*

*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*

*H*^{0}*(G, A)* *−→* *H*^{0}*(G, B)*

*%* *&*

*H*^{−1}*(G, C)* *H*^{0}*(G, C)*

*-* *.*

*H*^{−1}*(G, B) ←− H*^{−1}*(G, A)*

*An excellent tool for studying H*^{0}*(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) =* *#H*^{0}*(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 −→ A*^{G}*−→ A*^{σ−1}*−→ I**G**A −→ 1 and 1 −→* *N**G**A −→ A−→ N*^{N}^{G}*G**A −→ 1,*
*show that #A = #A*^{G}*· #I**G**A = #**N**G**A · #N**G**A, 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*

*#H*^{0}*(G(L/K), L*^{∗}*) = [L : K] and #H*^{−1}*(G(L/K), L*^{∗}*) = 1.*

*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*

^{∗}*) = H*

^{0}

*(G, L*

^{∗}*) = [L : K], where*

*we have put G = G(L/K). The exact sequence 1 → U*

*L*

*→ L*

^{∗ v}*−→ Z → 0, in which*

^{L}*Z has to be viewed as the trivial G-module, yields, by Proposition 1.2.1,*

*h(G, L*^{∗}*) = h(G, U**L**)h(G, Z) = h(G, U**L**)[L : K].*

*Hence we have to show that h(G, U**L*) = 1.

*First, we choose a normal basis {α*^{σ}*| σ ∈ G} of L/K with α ∈ O** _{L}* (see [2,

*Chapter VIII]), and consider the open G-module M =*P

*σ∈G**O**K**α** ^{σ}*. Then consider
the open sets

*V*^{n}*= 1 + π*_{K}^{n}*M, n ∈ N.*

*Since M is open, we have π*^{N}_{K}*O**L* *⊆ M for suitable N , and for n ≥ N the V** ^{n}* are

*subgroups of U*

*L*

*. Via the correspondence 1 + π*

^{n}

_{K}*β 7→ β (mod π*

*K*

*M ), we obtain*

*G-isomorphisms*

*V*^{n}*/V*^{n+1}*' M/π*_{K}*M =*M

*σ∈G*

*(O*_{K}*/P*_{K}*)α*^{σ}*.*

It is easy to check that
*H*^{0}*(G,*M

*σ∈G*

*(O**K**/P**K**)α*^{σ}*) = H*^{−1}*(G,*M

*σ∈G*

*(O**K**/P**K**)α*^{σ}*) = 0,*

and hence

*H*^{0}*(G, V*^{n}*/V*^{n+1}*) = H*^{−1}*(G, V*^{n}*/V*^{n+1}*) = 1,* *for n ≥ N .*

*This implies that H*^{0}*(G, V*^{n}*) = 1, for n ≥ N . Indeed, if a ∈ (V** ^{n}*)

*, then*

^{G}*a = (N*

*G*

*b*0

*)a*1

*, with b*0

*∈ V*

^{n}*and a*1

*∈ (V*

*)*

^{n+1}*. Continuing in this way, by*

^{G}*the completeness yields a = N*

*G*

*b, with b =*Q

_{∞}*i=0**b**i**∈ V*^{n}*, so that H*^{0}*(G, V** ^{n}*) = 1.

*Similarly, we have H*^{−1}*(G, V*^{n}*) = 1, for n ≥ N . Because U**L* *is compact, U**L**/V** ^{n}* is
finite. Therefore, by Proposition 1.2.1, we obtain

*h(G, U**L**) = h(G, U**L**/V*^{n}*)h(G, V*^{n}*) = 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*
*H*^{0}*(G(L/K), U**L**) = H*^{−1}*(G(L/K), U**L*) = 1

*and*

*H*^{0}*(G(L/K), U*_{L}^{(n)}*) = H*^{−1}*(G(L/K), U*_{L}^{(n)}*) = 1,* *n ∈ N.*

*Proof. Let G = G(L/K). As H*^{−1}*(G, L*^{∗}*) = 1, every element u ∈ U**L* such that
*N*_{L/K}*(u) = 1 is of the form u = a*^{φ}^{L/K}^{−1}*, with a ∈ L*^{∗}*and φ*_{L/K}*= φ**K**|**L*. Since
*L/K is unramified, π**K* *is also a prime element of L. So writing a = επ*^{m}* _{K}* with

*ε ∈ U*

*L*

*, we obtain u = ε*

^{φ}

^{L/K}

^{−1}*. This shows that H*

^{−1}*(G, U*

*L*) = 1. Since we have

*proved that h(G, U*

_{L}*) = 1 in Theorem 1.2.2, this shows that H*

^{0}

*(G, U*

*) = 1.*

_{L}*In order to prove H*^{0}*(G, U*_{L}^{(n)}*) = H*^{−1}*(G, U*_{L}* ^{(n)}*) = 1, we first show that

*H*

^{0}

*(G, l*

^{∗}*) = H*

^{−1}*(G, l*

^{∗}*) = 1 and H*

^{0}

*(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 H*

^{0}

*(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*

#*N**G**l = #{x ∈ l |*

*f −1*X

*i=0*

*x*^{q}^{i}*= 0} ≤ q*^{f −1}*and #I**G**l = q*^{f −1}*,*

*since k is the fixed field of the Frobenius automorphism l−→ l. Therefore*^{↑q}*H*^{−1}*(G, l) =* *N**G**l/I**G**l = 0.*

*This implies that H*^{0}*(G, l) = 0 as h(G, l) = 1.*

*Applying now the exact hexagon to the exact sequence of G-modules*
*1 → U*_{L}^{(1)}*→ U**L**→ l*^{∗}*→ 1,*

we obtain

*H*^{0}*(G, U*_{L}^{(1)}*) = H*^{0}*(G, U*_{L}*) = 1 and H*^{−1}*(G, U*_{L}^{(1)}*) = H*^{−1}*(G, U*_{L}*) = 1.*

*From the exact sequence of G-modules*

*1 → U*_{L}^{(n+1)}*→ U*_{L}^{(n)}*→ l → 0,*
we now deduce by induction just as above, that

*H*^{0}*(G, U*_{L}^{(n+1)}*) = H*^{0}*(G, U*_{L}^{(n)}*) = 1 and H*^{−1}*(G, U*_{L}^{(n+1)}*) = H*^{−1}*(G, U*_{L}^{(n)}*) = 1.*

¤

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 φ**K**the Frobenius automorphism in G( eK/K).*

*Let d**K* *: G( eK/K) → bZ be the isomorphism such that d**K**(φ**K*) = 1. We pass from
*the Galois extension L/K to the extension eL/K and consider the function d**L/K* :
*G(eL/K) → bZ such that d**L/K**(σ) = d**K**(σ|*_{K}_{e}*), for σ ∈ G(eL/K). In particular, since*
*φ**L**|*_{K}_{e} *= φ*^{f}_{K}^{L/K}*, one has d*_{L/K}*(φ**L**) = f*_{L/K}*. Notice that every element in G( eK/K)*
*can be extended to an element in G(eL/K). Therefore, d**L/K* *: G(eL/K) → b*Z is
surjective.

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

Frob(e*L/K) = {σ ∈ G(eL/K) | d**L/K**(σ) ∈ N}.*

*In other words, σ ∈ Frob(eL/K) if and only if σ ∈ G(eL/K) and σ|*_{K}_{e} *= φ*^{n}* _{K}*, for

*some n ∈ N. Because d*

*L/K*

*(1) = 0 and d*

*L/K*

*(σ*1

*σ*2

*) = d*

*L/K*

*(σ*1

*) + d*

*L/K*

*(σ*2), 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 d*

*L/K*

*is surjective, d*

*L/K*

maps Frob(e*L/K) onto N. Firstly, we have the*

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

*is surjective.*

*Proof. Let ˜φ ∈ Frob(eL/K) be an element such that d**L/K*( ˜*φ) = 1. Then ˜φ|*_{K}_{e} *= φ**K*.
*Remark here that L*0 *= L ∩ eK is the maximal unramified subextension of L/K.*

*For ρ ∈ G(L/K), since ρ|**L*0 *∈ G(L*0*/K) = h ˜φ|**L*0*i, there exists n ∈ N such that*
*ρ|**L*0 = ˜*φ*^{n}*|**L*0*. Since ρ ˜φ*^{−n}*|**L**∈ G(L/L*0) and the mapping

*G(eL/ eK) → G(L/L*0*),* *τ 7→ τ |**L*

*is an isomorphism, there is τ ∈ G(eL/ eK) such that τ |**L* *= ρ ˜φ*^{−n}*|**L*. Therefore,
*σ = τ ˜φ*^{n}*∈ G(eL/K) is an element satisfying σ ∈ Frob(eL/K) (because σ|*_{K}_{e} *= φ*^{n}* _{K}*)

*such that σ|**L**= τ ˜φ*^{n}*|**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 e*Σ = e*L, f**Σ/K* *= d**L/K**(σ) and e**Σ/K* *= e**L/K**.*
*Moreover, σ = φ*Σ*.*

*Proof. We first show that e*Σ = e*L. Since Σ ⊆ eL, one has eΣ ⊆ eL. The canonical*
*surjection G(eL/Σ) → G(eΣ/Σ), τ 7→ τ |*_{Σ}_{e} *must be bijective, because G(eL/Σ) = hσi*
is a quotient of b*Z ' G(eΣ/Σ). But G(eL/Σ) = G(eΣ/Σ) implies that e*Σ = e*L. This*
*also implies that e** _{Σ/K}*= [eΣ : e

*K] = [eL : eK] = e*

*.*

_{L/K}*Suppose that d*_{L/K}*(σ) = d. In other words, σ|*_{K}_{e} *= φ*^{d}* _{K}*. Because the fixed field

*of σ|*

_{K}_{e}

*is Σ ∩ eK = Σ*0

*, we have f*

*Σ/K*= [Σ0

*: K] = d. Therefore,*

*[Σ : K] = e*_{Σ/K}*f*_{Σ/K}*= e*_{L/K}*d ≤ [L : K]d*_{L/K}*(σ)*
is finite.

*Finally, since f** _{Σ/K}* = [Σ0

*: K] = d, one has φ*Σ

*|*

_{K}_{e}

*= φ*

^{d}

_{K}*= σ|*

_{K}_{e}. Hence,

*σ ∈ G(eΣ/Σ) and φ*Σ

*∈ G(eΣ/Σ) are identical in Σ eK = eΣ. Thus σ = φ*Σ. ¤

Our goal is to define a canonical homomorphism
*r**L/K* *: G(L/K) → K*^{∗}*/N**L/K**L*^{∗}

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

*Definition 2.1.3. The reciprocity map*

*r*_{L/K}_{e} : Frob(e*L/K) → K*^{∗}*/N**L/K**L** ^{∗}*
is defined by

*r*_{L/K}_{e} *(σ) = N**Σ/K**(π*Σ*) (mod N**L/K**L*^{∗}*),*
*where Σ is the fixed field of σ.*

*Observe that the definition of r*_{L/K}_{e} *(σ) does not depend on the choice of the*
*element π*Σ*. For another prime element differs from π*Σ*only by an element u ∈ U*Σ,

*and for this we have N*_{Σ/K}*(u) ∈ N*_{L/K}*L*^{∗}*. To see this, we let M = LΣ. Applying*
*Corollary 1.2.3 to the unramified extension M/Σ (because fM = e*Σ), one finds
*u = N*_{M/Σ}*(ε), for some ε ∈ U**M*, and thus

*N**Σ/K**(u) = N**Σ/K**(N**M/Σ**(ε)) = N**L/K**(N**M/L**(ε)) ∈ N**L/K**L*^{∗}*.*

*Next we want to show that the reciprocity map r*_{L/K}_{e} is multiplicative. In other
*words, we have to show that if σ*1*σ*2*= σ*3is an equation in Frob(e*L/K) and Σ**i* the
*fixed field of σ**i**, for i = 1, 2, 3, then*

*N*_{Σ}_{1}_{/K}*(π*Σ1*)N*_{Σ}_{2}_{/K}*(π*Σ2*) ≡ N*_{Σ}_{3}_{/K}*(π*Σ3*) (mod N*_{L/K}*L*^{∗}*).*

To do this, we need two Lemmas.

*Our first lemma makes N*_{Σ}_{i}_{/K}*(π*Σ*i**), 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*
*d**L/K**(φ) = 1 and d**L/K**(σ) = n. If Σ is the fixed field of σ and a ∈ Σ, then*

*N*_{Σ/K}*(a) = N*_{L/ e}_{e} * _{K}*(

*n−1*Y

*i=0*

*a*^{φ}^{i}*).*

*Proof. Since e*Σ = e*L (Proposition 2.1.2), for a ∈ Σ we have N**Σ/Σ*0*(a) = N*_{e}_{Σ/ e}_{K}*(a) =*
*N*_{L/ e}_{e} _{K}*(a). On the other hand, since [Σ*0*: K] = f*_{Σ/K}*= n and G(Σ*0*/K) is generated*
*by φ**K**|*Σ0*= φ|*Σ0*, one has N*_{Σ}_{0}_{/K}*(b) =*Q_{n−1}

*i=0* *b*^{φ}^{i}*, for b ∈ Σ*0*. For a ∈ Σ we thus get*
*N*_{Σ/K}*(a) = N*_{Σ}_{0}_{/K}*(N*_{Σ/Σ}_{0}*(a)) =*

*n−1*Y

*i=0*

*N*_{L/ e}_{e} _{K}*(a)*^{φ}^{i}*= N*_{L/ e}_{e} * _{K}*(

*n−1*Y

*i=0*

*a*^{φ}^{i}*).*

*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 N**L/K**L** ^{∗}*.

*Lemma 2.1.5. For a finite Galois extension L/K, let φ ∈ Frob(eL/K) satisfy*

*d*

_{L/K}*(φ) = 1. Suppose that u ∈ U*

_{L}_{e}

*such that u*

*=Q*

^{φ−1}

_{r}*i=1**u*^{τ}_{i}^{i}^{−1}*, for some u**i**∈ U*_{L}_{e}
*and τ**i**∈ G(eL/ eK). Then N*_{L/ e}_{e} _{K}*(u) ∈ N**L/K**L*^{∗}*.*

*Proof. Let M/K be a finite Galois subextension of eL/K such that u, u**i* *∈ U**M* and
*L ⊆ M . Let [M : K] = n, σ = φ*^{n}*and let Σ be the fixed field of σ. Since fM = e*Σ
*and f**M/K* *| n = f**Σ/K**, we have that M ⊆ Σ. Further, let Σ*^{0}*/Σ be the unramified*
*extension of degree n. By Corollary 1.2.3, we can then find elements ˜u, ˜u**i**∈ U*Σ* ^{0}*such

*that u = N*Σ

^{0}*/Σ*(˜

*u) and u*

*i*

*= N*Σ

^{0}*/Σ*(˜

*u*

*i*

*). Since G(Σ*

^{0}*/Σ) is generated by σ|*Σ

^{0}*= φ*Σ

*|*Σ

^{0}*and φG(eL/ eK) = G(eL/ eK)φ, by the assumption we have ˜u*^{φ−1}*= λ*Q_{r}

*i=1**u*˜^{τ}_{i}^{i}* ^{−1}*, for

*an element λ ∈ U*Σ^{0}*such that N*_{Σ}^{0}_{/Σ}*(λ) = 1. Hence, again by Corollary 1.2.3,*
*λ = µ*^{σ−1}*= µ*^{φ}^{n}^{−1}*, with µ ∈ U*Σ* ^{0}*. We may thus write

˜

*u*^{φ−1}*= µ*^{φ}^{n}* ^{−1}*
Y

*r*

*i=1*

˜

*u*^{τ}_{i}^{i}* ^{−1}*= (

*n−1*Y

*j=1*

*µ*^{φ}* ^{j}*)

*Y*

^{φ−1}*r*

*i=1*

˜
*u*^{τ}_{i}^{i}^{−1}*.*

*Applying N*_{L/ e}_{e} _{K}*gives N*_{L/ e}_{e} * _{K}*(˜

*u)*

^{φ−1}*= N*

_{L/ e}_{e}

*(Q*

_{K}

_{n−1}*j=1**µ*^{φ}* ^{j}*)

*, so that*

^{φ−1}*N*

_{L/ e}_{e}

*(˜*

_{K}*u) = N*

_{L/ e}_{e}

*(*

_{K}*n−1*Y

*j=1*

*µ*^{φ}^{j}*)z,*

*for some z ∈ U**K**. Finally, applying N*_{Σ}^{0}_{/Σ}*, we obtain, observing n = [M : K] =*
[Σ* ^{0}*: Σ] and using Lemma 2.1.4, that

*N*_{L/ e}_{e} _{K}*(u) = N*_{L/ e}_{e} _{K}*(N*Σ^{0}*/Σ*(˜*u)) = N*Σ^{0}*/Σ**(N*_{L/ e}_{e} * _{K}*(˜

*u)) = N*Σ

^{0}*/Σ*

*(N*

_{L/ e}_{e}

*(*

_{K}*n−1*Y

*j=1*

*µ*^{φ}^{j}*))z*^{n}

*= N*_{L/ e}_{e} * _{K}*(

*n−1*Y

*j=1*

*N*Σ^{0}*/Σ**(µ)*^{φ}^{j}*)z*^{n}*= N**Σ/K**(N*Σ^{0}*/Σ**(µ))N**M/K**(z) ∈ N**L/K**L*^{∗}*.*

¤ 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 σ*1*σ*2 *= σ*3 be an equation in Frob(e*L/K), Σ**i* *the fixed field of σ**i* and
*π**i**= π*Σ*i**, for i = 1, 2, 3. We have to show that*

*N*Σ1*/K**(π*1*)N*Σ2*/K**(π*2*) ≡ N*Σ3*/K**(π*3*) (mod N**L/K**L*^{∗}*).*

*Suppose that d*_{L/K}*(σ**i**) = n**i**, for i = 1, 2, 3. In order to apply Lemma 2.1.5, we*
*choose a fixed φ ∈ Frob(eL/K) such that d**L/K**(φ) = 1 and put*

*τ**i**= σ*_{i}^{−1}*φ*^{n}^{i}*∈ G(eL/ eK),* *i = 1, 2, 3.*

*From σ*1*σ*2*= σ*3*and n*1*+ n*2*= n*3*, we then deduce that τ*3*= τ*2*(φ*^{−n}^{2}*σ*1*φ*^{n}^{2})^{−1}*φ*^{n}^{1}.
*Putting σ*4*= φ*^{−n}^{2}*σ*1*φ*^{n}^{2} *and n*4*= d**L/K**(σ*4*) = n*1 *and τ*4 *= σ*^{−1}_{4} *φ*^{n}^{4}, we find that
*τ*3 *= τ*2*τ*4 *and N*_{Σ}_{4}_{/K}*(π*4*) = N*_{Σ}_{1}_{/K}*(π*1), where Σ4 = Σ^{φ}_{1}* ^{n2}* is the fixed field of

*σ*4

*and π*4

*= π*

^{φ}_{1}

*is a prime element of Σ4. We may therefore pass to show the congruence*

^{n2}*N*_{Σ}_{2}_{/K}*(π*2*)N*_{Σ}_{4}_{/K}*(π*4*) ≡ N*_{Σ}_{3}_{/K}*(π*3*) (mod N*_{L/K}*L*^{∗}*).*

From Lemma 2.1.4, if we put
*u =*¡* ^{n}*Y

^{2}

^{−1}*i=0*

*π*^{φ}_{2}* ^{i}*¢¡

*Y*

^{n}^{4}

^{−1}*i=0*

*π*_{4}^{φ}* ^{i}*¢¡

*Y*

^{n}^{3}

^{−1}*i=0*

*(π*^{−1}_{3} )^{φ}* ^{i}*¢

*∈ U*_{L}_{e}*,*

*then the congruence amounts simply to the relation N*_{L/ e}_{e} _{K}*(u) ∈ N*_{L/K}*L*^{∗}*. For this,*
however, Lemma 2.1.5 gives us all that we need.

*Since σ**i* *fixes π**i**, we have π*_{i}^{φ}^{ni}^{−1}*= π*_{i}^{σ}^{−1}^{i}^{φ}^{ni}^{−1}*= π*_{i}^{τ}^{i}^{−1}*, and hence u** ^{φ−1}* =

*π*

^{τ}_{2}

^{2}

^{−1}*π*

_{4}

^{τ}^{4}

^{−1}*π*

_{3}

^{1−τ}^{3}. Because eΣ2 = eΣ3 = eΣ4

*, we have π*2

*= u*2

*π*4

*, π*3

*= u*

^{−1}_{3}

*π*4 and

*π*

^{τ}_{4}

^{2}

*= u*

^{−1}_{4}

*π*4

*, for u*2

*, u*3

*, u*4

*∈ U*

_{L}_{e}

*. We obtain u*

^{φ−1}*= u*

^{τ}_{2}

^{2}

^{−1}*u*

^{τ}_{3}

^{3}

^{−1}*u*

^{τ}_{4}

^{4}

^{−1}*. By Lemma*

*2.1.5, we do get N*_{L/ e}_{e} _{K}*(u) ∈ N*_{L/K}*L*^{∗}*.* ¤

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

*Proposition 2.2.1. For every finite Galois extension L/K, there is a canonical*
*homomorphism r**L/K* *: G(L/K) → K*^{∗}*/N**L/K**L*^{∗}*given by*

*r*_{L/K}*(σ) = N*_{Σ/K}*(π*Σ*) (mod N*_{L/K}*L*^{∗}*),*

*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 r** _{L/K}* is independent of the choice of
the extension ˜

*σ ∈ Frob(eL/K) of σ ∈ G(L/K). For this, let ˜σ*

^{0}*∈ Frob(eL/K) be*another extension and Σ

^{0}*its fixed field. If d*

*L/K*(˜

*σ) = d*

*L/K*(˜

*σ*

*), then ˜*

^{0}*σ|*

_{K}_{e}= ˜

*σ*

^{0}*|*

_{K}_{e}and ˜

*σ|*

*L*= ˜

*σ*

^{0}*|*

*L*, so that ˜

*σ = ˜σ*

*, and there is nothing to show. However, if we have,*

^{0}*say, d*

*L/K*(˜

*σ) < d*

*L/K*(˜

*σ*

^{0}*), then denote τ = ˜σ*

^{−1}*σ*˜

^{0}*. The automorphism τ ∈ G(eL/L)*

*and d*

_{L/K}*(τ ) = d*

*(˜*

_{L/K}*σ*

^{0}*) − d*

*(˜*

_{L/K}*σ) ∈ N. Hence τ ∈ Frob(eL/K) and the fixed field*Σ

^{00}*of τ contains L. Therefore, Proposition 2.1.6 shows that*

*N*Σ^{0}*/K**(π*Σ^{0}*) ≡ N*Σ^{00}*/K**(π*Σ^{00}*)N**Σ/K**(π*Σ*) ≡ N**Σ/K**(π*Σ*) (mod N**L/K**L*^{∗}*).*

*This means that r** _{L/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*
*r**L/K* *: G(L/K) → K*^{∗}*/N**L/K**L** ^{∗}*
is defined by

*r**L/K**(σ) = N**Σ/K**(π*Σ*) (mod N**L/K**L*^{∗}*),*

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

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 r*_{L/K}*: G(L/K) → K*^{∗}*/N*_{L/K}*L*^{∗}*is given by*

*r*_{L/K}*(φ**K**|**L**) ≡ π**K* *(mod N*_{L/K}*L** ^{∗}*)

*and is an isomorphism.*

*Proof. In this case eL = eK, the Frobenius automorphism φ**K* *∈ Frob(eL/K) is an ex-*
*tension of φ**K**|**L**. The fixed filed of φ**K* *is K, and hence by definition, r**L/K**(φ**K**|**L**) ≡*
*π**K* *(mod N**L/K**L** ^{∗}*).

*Consider the valuation map v**K**: K*^{∗}*→ Z. It induces an isomorphism*
*K*^{∗}*/N*_{L/K}*L*^{∗}*' Z/nZ,*

*with n = [L : K]. Indeed, if v**K**(a) ≡ 0 (mod nZ), then a = uπ*_{K}* ^{nr}*, and since

*u = N*

*L/K*

*(ε) for some ε ∈ U*

*L*

*(Corollary 1.2.3), we find a = N*

*L/K*

*(επ*

^{r}

_{K}*) ≡ 1*

*(mod N*

*L/K*

*L*

^{∗}*). This shows that π*

*K*

*(mod N*

*L/K*

*L*

*) generates the cyclic group*

^{∗}*K*

^{∗}*/N*

_{L/K}*L*

^{∗}*of order [L : K]. Since φ*

*K*

*|*

*L*

*also generates the cyclic group G(L/K),*

*we have r**L/K* is an isomorphism. ¤

*The homomorphism r** _{L/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 r*_{L/K}*: G(L/K) → K*^{∗}*/N*_{L/K}*L*^{∗}*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|**L*is a generator
*of G(L/K). Let σ*1= ˜*σφ**L**∈ G(eL/K). Since*

*d*_{L/K}*(σ*1*) = d**K**(σ*1*|*_{K}_{e}*) = d**K**(φ**L**|*_{K}_{e}*) = f** _{L/K}* = 1

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

*L/K containing Σ and L and let M*e 0 *= M ∩ eK. Since fM = e*Σ = e*L, we have that*
*G(M/M*0*) ' G(Σ/K) ' G(L/K) and N**M/M*0*|*Σ*= N**Σ/K**, N**M/M*0*|**L**= N**L/K*.

*For the injectivity of r*_{L/K}*, we claim that: if r*_{L/K}*(σ*^{m}*) ≡ 1 (mod N*_{L/K}*L** ^{∗}*),

*where 0 ≤ m < n = [L : K], then m = 0.*

*Let π**L* *∈ O**L**and π*Σ*∈ O*Σ*be prime elements. Since Σ, L ⊆ M and fM = e*Σ = e*L,*
*π**L* *and π*Σ *are both prime elements of M . Putting π*_{Σ}^{m}*= uπ*^{m}_{L}*, with u ∈ U**M*, we
obtain

*r*_{L/K}*(σ*^{m}*) ≡ N*_{Σ/K}*(π*^{m}_{Σ}*) ≡ N*_{M/M}_{0}*(u)N*_{L/K}*(π*_{L}^{m}*) ≡ N*_{M/M}_{0}*(u) (mod N*_{L/K}*L*^{∗}*).*

*From r*_{L/K}*(σ*^{m}*) ≡ 1 (mod N*_{L/K}*L*^{∗}*), it thus follows that N*_{M/M}_{0}*(u) = N*_{L/K}*(ε) for*
*some ε ∈ U*_{L}*. Since G(M/M*_{0}) is cyclic, from Theorem 1.2.2 (Hilbert 90), we may
*write u*^{−1}*ε = a*^{˜}^{σ−1}*for some a ∈ M** ^{∗}*and have

*(π*_{L}^{m}*ε)*^{σ−1}^{˜} *= (π*_{L}^{m}*ε)*^{σ}^{1}* ^{−1}* (because ˜

*σ|*

*L*

*= σ*1

*|*

*L*)

*= (π*_{Σ}^{m}*· a*^{˜}* ^{σ−1}*)

^{σ}^{1}

^{−1}*= (a*^{σ}^{1}* ^{−1}*)

^{σ−1}^{˜}

*Hence we have b = π*^{m}_{L}*εa*^{1−σ}^{1} *∈ M*_{0}^{∗}*with v**M**(b) = m.* *However, v**M**(b) =*
*e*_{M/M}_{0}*v**M*0*(b) = nv**M*0*(b) implies that one has m = 0, and so r** _{L/K}* is injective.

The surjectivity the follows from Theorem 1.2.2

*#K*^{∗}*/N*_{L/K}*L*^{∗}*= [L : K] = #G(L/K).*

¤
*The reciprocity homomorphism r**L/K* exhibits the following functorial behavior.

*Proposition 2.2.5. Let L/K and L*1*/K*1 *be finite Galois extensions such that*
*K*1*/K and L*1*/L are finite separable extensions. Then we have the commutative*
*diagram*

*G(L*1*/K*1) *−−−−−−→ K*^{r}* ^{L1/K1}* 1

*∗*

*/N*

*L*

_{1}

*/K*

_{1}

*L*

^{∗}_{1}

y^{|}^{L}

y^{N}*K1/K*

*G(L/K)* *−−−−−−→*^{r}^{L/K}*K*^{∗}*/N**L/K**L*^{∗}

*where the left vertical homomorphism are given by the restriction σ*1*|**L* *of σ*1 *∈*
*G(L*1*/K*1*) and the right vertical homomorphism is induced by the norm map N*_{K}_{1}_{/K}*.*
*Proof. Let σ*^{0}*∈ G(L*1*/K*1*) and σ = σ*^{0}*|**L* *∈ G(L/K). If ˜σ*^{0}*∈ Frob(eL*1*/K*1) is
*an extension of σ** ^{0}*, then ˜

*σ = ˜σ*

^{0}*|*

_{L}_{e}

*∈ Frob(eL/K) is an extension of σ, because*

*d*

*(˜*

_{L/K}*σ) = f*

_{K}_{1}

_{/K}*d*

_{L}_{1}

_{/K}_{1}(˜

*σ*

^{0}*) ∈ N. Let Σ*

*be the fixed field of ˜*

^{0}*σ*

*. Then the fixed*

^{0}field of ˜*σ is Σ = Σ*^{0}*∩ eL = Σ*^{0}*∩ e*Σ = Σ^{0}_{0} *and hence f*_{Σ}^{0}* _{/Σ}* = [Σ

^{0}_{0}: Σ] = 1. If now

*π*Σ

*is a prime element of Σ*

^{0}

^{0}*, then π*Σ

*= N*Σ

^{0}*/Σ*

*(π*Σ

*) is a prime element of Σ. The commutativity of the diagram follows from the equality of norms*

^{0}*N**Σ/K**(π*Σ*) = N**Σ/K**(N*Σ^{0}*/Σ**(π*Σ^{0}*) = N**K*1*/K**(N*Σ^{0}*/K*1*(π*Σ^{0}*)).*

¤ 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*

y^{r}^{L/M}

y^{r}^{L/K}

y^{r}^{M/K}

*M*^{∗}*/N*_{L/M}*L*^{∗}^{NM/K}*−→* *K*^{∗}*/N*_{L/K}*L*^{∗}*→ K*^{∗}*/N*_{M/K}*M*^{∗}*→ 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 r**L/K* is not an
*isomorphism. For an arbitrary group G, let G** ^{0}* denote the commutator subgroup

*and write G*

^{ab}

*= G/G*

^{0}*for the maximal abelian quotient group. Since K*

^{∗}*/N*

*L/K*

*L*

^{∗}*is an abelian group, the homomorphism r*

*naturally induces the homomorphism*

_{L/K}ˆ

*r**L/K* *: G(L/K)*^{ab}*→ K*^{∗}*/N**L/K**L*^{∗}

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

*r**L/K* *: G(L/K)*^{ab}*−→ K*^{∼}^{∗}*/N**L/K**L*^{∗}*.*

*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 ˆ*r**L/K* *= r**L/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, r*_{M/K}*and r** _{L/M}* are isomorphisms by Propositions 2.2.3 and

2.2.4. In the bottom sequence of diagram (2.1)

*M*^{∗}*/N*_{L/M}*L*^{∗}^{NM/K}*−→ K*^{∗}*/N*_{L/K}*L*^{∗}*→ K*^{∗}*/N*_{M/K}*M*^{∗}*→ 1,*

*the map N**M/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, ˆr** _{L/K}* is an
isomorphism.

Next, we show that ˆ*r**L/K* *= r**L/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 H**i* *and let M**i* *be the fixed field of H**i*. One has
*H**i* *= G(L/M**i**) and M**i**/K is an abelian subextension of L/K of smaller degree.*

*For every M**i**, consider M = M**i* in the diagram (2.1). The induction hypothesis
*says that r**M**i**/K* *is injective. Therefore, if σ ∈ ker(r**L/K*), then by the commutative
*diagram (2.1), one has σ is in the kernel of the map G(L/K) → G(M**i**/K), which*
*is equal to G(L/M**i**) = H**i**. In other words, the kernel of r**L/K* is contained in the
*intersection of those H**i**. Since G(L/K) is a direct sum of these H**i*, the kernel
*of r**L/K* is the identity and hence ˆ*r**L/K* is injective. Surjectivity also follows by
*induction on the degree. Indeed, since r**M/K* *and r**L/M* *are surjective, so is r**L/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 r**M/K* *is injective, the kernel of r**L/K* is
*contained in G(L/M ). Because K*^{∗}*/N**L/K**L*^{∗}*is abelian, G(L/M ) = G(L/K)** ^{0}* is

*also contained in the kernel of r*

_{L/K}*, and hence ker(r*

_{L/K}*) = G(L/M ). This proves*the injectivity of ˆ

*r*

*L/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 r*

*L/M*

*and r*

*M/K*

*are surjective, then so is r*

*L/K*. Hence ˆ

*r*

*L/K*is surjective. ¤

*Putting L*

^{ab}

*the maximal abelian subextension of L/K, we find G(L/K)*

^{ab}=

*G(L*

^{ab}

*/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 L*^{ab}*/K be the max-*
*imal abelian subextension in L/K. Then N**L/K**L*^{∗}*= N*_{L}^{ab}_{/K}*L*^{ab∗}*.*

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 ˆ*r*_{L/K}*: G(L/K)*^{ab}*→ K*^{∗}*/N*_{L/K}*L** ^{∗}*gives, for every

*finite Galois extension L/K a surjective homomorphism*

*( , L/K) : K*^{∗}*→ G(L/K)*^{ab}

*with kernel N*_{L/K}*L*^{∗}*. 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 L*1*/K*1 *be finite Galois extensions such that*
*K*1*/K and L*1*/L are finite separable extension. Then we have the commutative*
*diagram*

*K*1*∗* *( ,L*1*/K*1)

*−−−−−−→ G(L*^{ab}_{1} */K*1)

y^{N}*K1/K*

y
*K*^{∗}*−−−−−−→*^{( ,L/K)}*G(L*^{ab}*/K)*

*where the left vertical homomorphism are given by the norm map N*_{K}_{1}_{/K}*and the*
*right vertical homomorphism is induced by the restriction σ|*_{L}^{ab} *of σ ∈ G(L*^{ab}_{1} */K*1*).*

*For every field K, we equip the group K** ^{∗}* with a topology by declaring the

*cosets aN*

_{L/K}*L*

^{∗}*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 v**K**: K*^{∗}*→ Z is continuous.*

*(3) If L/K is a finite extension, then N*_{L/K}*: L*^{∗}*→ K*^{∗}*is continuous.*

*Proof.*