lectures on local fields

7/26/2019 Lectures on Local Fields

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Lectures on Complete Discrete Valuation Fields

1: Discrete Valuation Fields

(1.1). Valuations. One can generalize the properties of the -adic valuation and proceed

to the concept of valuation. Let be an additively written totally ordered abelian group. Add

to a formal element + with the properties + , + + , + (+ ) = + ,

(+ ) + (+

) = +

, for each ; denote = + .

A map : with the properties

(

) = +

= 0

(

) =

(

) +

(

)

( + ) min( ( ) ( ))

is said to be a valuationon ; in this case is said to be a valuation field. The map induces

a homomorphism of to and its value group ( ) is a totally ordered subgroup of .

If ( ) = 0 , then is called thetrivial valuation. It is easy to show that ( 1) = 0, and if

(

)

(

), then

(

)

min(

(

+

)

(

))

min(

(

)

(

)) =

(

);

thus, if ( ) = ( ) then ( + ) = min( ( ) ( )).

(1.2). Basic Objects. Let

=

: (

)

0

,

=

: (

)

0

. Then

coincides with the set of non-invertible elements of . Therefore, is a local ring with the

unique maximal ideal

;

is called the ring of integers (with respect to ), and the field

= is called the residue field, or residue class field. The image of an element

in is denoted by , it is called the residueof in . The set of invertible elements of

is a multiplicative group =

, it is called thegroup of units.

Assume thatchar( ) = char(

). Then char( ) = 0 andchar(

) = 0.

Proof. Suppose that char( ) = = 0. Then = 0 in and therefore in . Hence = char(

).

(1.3).

1. A valuation on issaid tobe discrete if the totally ordered group ( ) is isomorphic

to the naturally ordered group .

For a prime and a non-zero integer let = ( ) be the maximal integer such that

divides . Extend to rational numbers putting ( ) = ( ) ( ); (0) = + .The -adic valuation is a discrete valuation with the ring of integers

=

: is relatively prime to

1


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2

The residue field

is a finite field of order

.

2. Let = ( ). For an irreducible polynomial ( ) define ( ) similarly to above.

The map (

) is a discrete valuation with the ring of integers

( )

=

(

) (

)

:

( )

( )

[

]

( )is relatively prime to ( )

and the residue field is [ ] ( ) [ ] which is a finite algebraic extension of .

The field has another discrete valuation trivial on : (

) = deg(

), its residue field

is .

(1.4). Discrete valuations. A field is said to be a discrete valuation field if it admits a

nontrivial discrete valuation (see Example 1 in (1.3)). An element is said to be

a prime element(uniformizing element) if (

) generates the group

(

). Without loss of

generality we shall often assume that the homomorphism : issurjective.

Let be a discrete valuation field, and be a prime element. Then the ring of

integers

is a principal ideal ring, and every proper ideal of

can be written as

for

some 0. In particular,

= . The intersection of all proper ideals of is the zero

ideal.

Each element can be uniquely written as for some and .

Proof. Let be a proper ideal of . Then there exists = min ( ) : and hence

for some unit . It follows that and = . If belongs to

the intersection of all proper ideals in , then ( ) = + , i.e., = 0.

Let = ( ). Then and = for . If 1 = 2, then

+

(

1) = + ( 2). As 1 2 , we deduce = , 1 = 2.

(1.5). Completion. Completion of a discrete valuation field is an object which is easier to

work with than with the original field.

Let be a field with a discrete valuation (as usual, ( ) = ). is a metric space

with respect to the norm = (1 2) (

). So one can introduce the notion of a fundamental

sequence: a sequence (

)

0 of elements of is called a fundamental sequence if for every

real there is ( ) 0 such that (

) for ( ).

If (

) is a fundamental sequence then for every integer there is such that for all

we have (

) . We can assume 1 2

. If for every there

is

such that (

) = (

+1), then (

) and (

) for

, and

hence lim (

) = + . Otherwise lim (

) is finite.

The set of all fundamental sequences forms a ring with respect to componentwise

addition and multiplication. The set of all fundamental sequences (

)

0 with

0 as

+

forms a maximal ideal

of

. The field

is a discrete valuation field with its

discrete valuation defined by ((

)) = lim (

) for a fundamental sequence (

)

0.

Proof. A sketch of the proof is as follows. It suffices to show that is a maximal ideal of

. Let (

) 0 be a fundamental sequence with

0 as

+

. Hence, there is an 0 0 such that

= 0 for 0. Put

= 0 for 0 and

= 1

for 0 .

Then (

)

0 is a fundamental sequence and (

)(

) (1) +

. Therefore, is maximal.


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3

(1.6). A discrete valuation field is called a complete discrete valuation field if every

fundamental sequence (

)

0 is convergent, i.e., there exists = lim

with respect to

. A field with a discrete valuation is called acompletionof if it is complete, = ,

and

is a dense subfield in

with respect to

.

Every discrete valuation field has a completion which is unique up to an iso-

morphism over .

Proof. We verify that the field with the valuation is a completion of . is

embedded in by the formula ( ) mod . For a fundamental sequence (

)

0

and real , let 0 0 be such that (

) for all 0. Hence, for

0

we have ((

0)

(

)

0) , which shows that is dense in . Let (( (

)

)

)

be a fundamental sequence in with respect to . Let (0), (1),

be an increasing

sequence of integers such that ( ( )

2

(

)

1)

for 1 , 2 ( ). Then ( (

)

(

))

is a

fundamental sequence in and it is the limit of (( ( )

)

)

with respect to in . Thus,

we obtain the existence of the completion , .

If there are two completions

1 , 1 and

2 , 2 , then we put

( ) = for andextend this homomorphism by continuity from , as a dense subfield in 1, to 1 . It is easy

to verify that the extension

: 1

2 is an isomorphism and 2

= 1 .

We shall denote the completion of the field with respect to by or simply .

1.7. Examples of complete valuation fields. 1. The completion of with respect to of

(1.3) is denoted by and is called thefield of -adic numbers. Certainly, the completion of

with respect to the absolute value of (1.1) is . Embeddings of in for all prime

and in is a tool to solve various problems over . An example is theMinkowskiHasse

Theorem: an equation = 0 for has a nontrivial solution in if and only

if it admits a nontrivial solution in and in for all prime

.

The ring of integers of

is denoted by

and is called the ring of

-adic integers. Theresidue field of is the finite field consisting of elements.

2. The completion of ( ) with respect to

is the formal power series field (( ))

of all formal series +

with

and

= 0 for almost all negative . The

ring of integers with respect to

is [[

]], that is, the set of all formal series

+

0

,

. Its residue field may be identified with .

(1.8). Representatives. For simplicity, we will often omit the index in notations , ,

, . We fix a prime element of .

A set is said to be a set of representatives for a valuation field if , 0

and is mapped bijectively on under the canonical map = . Denote by

rep: the inverse bijective map. For a set denote by ( )+

the set of all sequences

( )

, . Let ( )+

denote the union of increasing sets ( )+

where .

The additive group has a natural filtration

+1

The factor filtration of this filtration is easy to calculate:

+1 .


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Let be a complete field with respect to a discrete valuation . Let

for each

be an element of

with

(

) =

. Then the map

Rep: (

)

+

(

)

+

rep(

)

is a bijection. Moreover, if ( )

= (0)

then

(Rep(

)) = min

: = 0

.

Proof. The map Rep is well defined, because for almost all

0 we get rep( ) = 0 and the

series rep( ) converges in . If ( ) = ( ) and

= min

: =

then (

) = . Since (

) for

, we deduce that

(Rep(

)

Rep(

)) =

Therefore Rep is injective.

In particular, (Rep( )) = min

: = 0 . Further, let . Then = with

, . We also get =

for some . Let

be the image of in ;

then

= 0 and 1 = rep(

)

+1 . Continuing in this way for 1, we obtain a

convergent series = rep( ) . Therefore, Rep is surjective.

We oftentake

= . Therefore, by the preceding Lemma, every element

can be uniquely expanded as

=

+

and

= 0 for almost all

0

If , we write mod .

(1.9). Units. The group1 + is called the group of principal units 1 and its elements are

called principal units. Introduce alsohigher groups of units:

= 1 +

for

1.

The multiplicative group has a natural filtration 1 2

. We

describe the factor filtration of the introduced filtration on .

Let be a discrete valuation field. Then

(1) The choice of a prime element ( 1 ) induces an isomorphism

.

(2) The canonical map = induces the surjective homomorphism

0: ;

0 maps 1 isomorphically onto .

(3) The map

:

1 +

for induces the isomorphism

of

+1 onto for

1.


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Proof. (2) The kernel of

0 coincides with 1 and

0 is surjective. (3) The induced map

+1 is a homomorphism, since

(1 + 1

)(1 + 2

) = 1 + ( 1+ 2)

+ 1 2 2

This homomorphism is bijective, since

(1 + rep( )

) = .

Let be not divisible by char(

). Raising to the

th power induces an auto-

morphism of

+1 for

1. If is complete, then the group

for

1 is uniquely

-divisible.

Proof. If = 1 +

with , then 1 +

mod +1 . Absence of nontrivial

-torsion in the additive group implies the first property. It also shows that has no

nontrivial -torsion.

For an element = 1 +

with we have = (1 + 1

) 1 with 1 +1 .

Applying the same argument to 1 and so on, we get an th root of in in the case of

complete .

(1.10). Raising to

th power. Let char( ) =

0. Lemma (1.2) shows that eitherchar(

) = or char( ) = 0. We shall study the operation of raising to the th power. Denote

this homomorphism by :

The first and simplest case is char( ) =

.

Let char( ) = char(

) =

0. Then the homomorphism

maps

injectively into for

1. For

1

(1 +

)

1 +

mod +1

Proof. Since (1 +

)

= 1 +

and there is no nontrivial

-torsion in and , the

assertion follows.

Let be a field of characteristic

0 and let

be perfect, i.e

=

. Then maps the quotient group

+1 isomorphically onto the quotient group +1 for

1.

We now consider the case of char( ) = 0, char( ) = 0. As = 0 in the residue field

, we conclude that and, therefore, for the surjective discrete valuation of we get

( ) = 1.

The number = ( ) = ( ) is called the absolute ramification index of .

Let be a prime element in

. Let

be a set of representatives, and let

0 be the

element of uniquely determined by the relation

0 +1

.

Let be a discrete valuation field of characteristic zero with residue field of

positive characteristic

. Then the homomorphism

maps to for

(

1), and

to +

for

( 1). Moreover, for

(1 +

)

1 +

mod

+1

if

(

1)

(1)


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(1 +

)

1 + (

+

0 )

mod

+1

if

= (

1)

(2)

(1 +

)

1 +

0 +

mod +

+1 if

( 1) (3)

The induced homomorphisms on the quotient filtration are injective in cases (1), (3) andsurjective in case(3).

If a primitive

th root of unity is contained in , then (1 ) = ( 1) and the

kernel of the induced homomorphisms in case (2)is of order .

If is complete, then

+ =

for

( 1). If is complete and ( 1) ,

then the homomorphism in(2)is injective iff there is no nontrivial -torsion in .

Proof. Let 1 + . We get

(1 + )

= 1 + +

( 1)

2

2 + +

1 +

and ( ) = +

, ( 1)

2

2 =

+ 2

, (

1) = + (

1)

, (

) =

, so

((1 +

)

1) =

(

+ )

if (

)

= (

)

((1 +

)

1)

(

+ )

otherwise

Note (

) ( ) if and only if

( 1). For a unit we obtain the first statement of

the proposition.

Further, the homomorphism

is an isomorphism in case (3) and injective in case (1).

Assume that . From the previous (1 ) = (

1) and (

1)

.

Therefore, the homomorphism

+

0 is not injective. Its kernel 1

0 in this

case is of order

.

If is complete, then due to surjectivity of the homomorphisms in case (3) for

( 1) we get = +1

= +2

= =

. Now let ( 1) be an integer.

Assume that the horizontal homomorphism in case (2) is not injective. Let 0 satisfy the

equation

0 +

0

0 = 0. Then (1 +

0

(

1)

)

for some

(

1). Therefore(1 + 0)

(

1))

=

1 for some 1

(

1)+1 . Thus, (1 + 0) (

1)) 11

(

1) is a

primitive

th root of unity.

Let be a complete discrete valuation field.

Ifchar( ) = 0, then is an open subgroup in for 1. Ifchar( ) = 0, then

is an open subgroup in

if and only if

is relatively prime to

.

contains finitely many roots of unity of order a power of .

Proof. If char( ) = 0, then we get 1 for 1. It means that is open. If

char( ) = , then 1 for ( ) = 1 and is open. In this case, if char( ) =

,

then 1 +

for (

) = 1. Then

is not open. If char( ) = 0, then we obtain

when

( 1) + ( 1) . Therefore is open for 1.

This corollary demonstrates that for complete discrete valuation fields of characteristic 0 with

residue field of characteristic the topological properties are closely related with the algebraic

ones. The case char( ) = is very different.


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(1.11). Product representation. Now we deduce a multiplicative analog of the expansion in

the corollary of (1.8).

Let be a complete discrete valuation field. Let be a set of

representatives. Then for there exist uniquely determined ,

for

0,

0 , such that can be expanded in the convergent product

=

0

1

(1 +

)

Proof. The existence and uniqueness of and

0 immediately follow. Assume that

,

then find

with (1 +

)

1

+1. Proceeding by induction, we obtain an

expansion of in a convergent product. If there are two such expansions (1 +

) =

(1 +

), then the residues

,

coincide in . Thus,

=

.

(1.12).

-Structure of The Group of Principal Units. Everywhere in this section is a

complete discrete valuation field with residue field of positive characteristic .

If 1 then

1 as + . This enables us to define

= lim

if lim

=

Let 1, . Then 1 is well defined and

+ =

,

= (

) ,

( )

=

for 1, . The multiplicative group 1 is a -module under

the operation of raising to a power. Moreover, the structure of the -module

1 is compatible

with the topologies of

and 1.

Proof. Assume that lim

= lim

; hence

0 as + and lim

= 1.

A map

1 1 ( ( ) ) is continuous with respect to the -adic topology on

and the discrete valuation topology on 1. This argument can be applied to verify the other

assertions of the lemma.

Let be of characteristic

with perfect residue field. Let be a set of

representatives, and let 0 be a subset of it such that the residues of its elements in form a

basis of as a vector space over

. Let an index-set numerate the elements of 0 . Let

be the -adic valuation.

Then every element 1 can be uniquely represented as a convergent product

=

( )=1 0

(1 +

)

0

and the sets = : ( ) are finite for all 0, (

) = 1.

Proof. We first show that the element can be written modulo

for 1 in the desired

form with . Proceeding by induction, it will suffice to consider an element

modulo

+1. Let 1 +

mod

+1 ,

. If ( ) = 1, then one can find

1

0 and 1

such that 1 +

=1(1 +

) mod

+1 for

some . If = with an integer , ( ) = 1, then one can find

1

0 and


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1

such that 1 +

=1(1 +

)

mod

+1 for some . Now due

to the continuity we get the desired expression for 1 with the above conditions on the sets

.

Assume that there is a convergent product for 1 with

, . Let (

0 ) = 1 and 0 be

such that = (

0

0 )

0

(

) for all (

) = 1, . Then the choice of 0 imply

(1 +

)

+1, which concludes the proof.

The group 1 has a free topological basis 1 +

where where

0,

(

) = 1.

If = ( ) is divisible by 1, let

: be the map

+

0 . Then raising to

the th power in case (2) of the proposition in (1.10) is described by

.

Let be of characteristic 0 with perfect residue field of characteristic

.

Let be a set of representatives and let

0 (resp. 0 ) be a subset of it suchthat the residues

of its elements in form a basis of as a vector space over (resp. are -generators of

(

) ). Let the index-set

(resp.

)numerate the elements of

0 (resp.

0 ). Let =

:

1

( 1) (

) = 1

Then every element 1 can be represented as a convergent product

=

(1 +

)

(1 +

(

1))

0 0

(the second product occurs when ( 1) is an integer)and the sets

=

:

(

)

=

: (

)

are finite for all 0,

.

Proof. We shall show how to obtain the required form for

modulo

+1 . Let

= 1 +

mod

+1,

. There are four cases to consider:

(1) . One can find

1

0 and 1

satisfying the congruence

1 +

=1(1 +

) mod

+1 for some .

(2)

( 1), =

with . Then thereexist

1

0, 1

such that

1 +

=1

(1 +

)

mod

+1 for some

(3) (

1)

, =

(

1). The definition of 0 imply that if =

with

, then there exist

1

0, 1

0 , 1

, 1

such

that

1 +

=1

(1 +

)

=1

(1 +

)

mod

+1 for some

(4) ( 1). If = min : ( 1) and = , then

1 +

(1 +

)

mod

+1 for some


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Now applyingthe arguments of the precedingcasesto 1+

, we can write 1 +

mod

+1

in the required form.

Let

be of characteristic 0 with perfect residue field of characteristic

.(1) If

does not contain nontrivial

th roots of unity then the representation in the proposition

is unique. Therefore the elements of the proposition form a topological basis of 1 .

(2) If contains a nontrivial th root of unity, let be the maximal integer such that

contains a primitive

th root of unity. Then the numbers

of the proposition

are determined uniquely modulo

. Therefore the elements of the proposition form a

topological basis of 1

1 .

(3) If the residue field of is finite then 1 is the direct sum of a free -module of rank

and the torsion part.

Proof. (1) If then all horizontal homomorphisms of the diagrams in the second

Proposition of (1.10) are injective.

(2) Argue by induction on . Write a primitive

th root

in the form

=

(1 +

)

(1 +

(

1))

and raise the right hand side to the

th power which demonstrates the non-uniqueness.

Now if

1 =

(1 +

)

(1 +

(

1))

then we deduce that =

=

with

-adic integers . Then

(1 +

)

(1 +

(

1))

is a th root of unity, and so is equal to

(1 +

)

(1 +

(

1))

1

for some integer . Now by the induction assumption all

1

1

are

divisible by

1. Thus, all are divisible by

.

(3) If the residue field of is finite then 1 is a module of finite type over the principal

ideal domain

, so by the structure theorem on such modules it is a direct sum of a free

module and a finite torsion module. If a primitive th root of unity is in , then the kernel of

is of order . Hence :

( ) = , since is finite. The cardinality of is equal to

= [

(

1)]

[[

(

1)]

].


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2: Extensions of Complete Fields

(2.1). Henselian Property. Let be a commutative ring. For two polynomials

( ) =

+ 0, ( ) =

+ + 0 their resultant is the determinant of the matrix

1

0

1

0

. . . . . .

. . . . . .

1

0

1

0

1

0

. . . . . .

. . . . . .

1

0

This determinant (

) is zeroiff

and have a common root; in general (

) =

1 +

1

for some polynomials 1

1

[

]. If (

) =

=1(

),

(

) =

=1(

),then their resultant (

) is

( ). In particular, ( ( )) = ( ).

If

[

] then

(

)

. We shall use the following properties of the resultant:

if

mod [

] then (

)

(

1 ) mod

; if (

)

+1 then

[ ]

[ ] + [ ].

Let be a complete discrete valuation field with the ring of integers , the maximal ideal

, and the residue field . For a polynomial

( ) =

+

+

0 [ ] we will denote

the polynomial

+

+

0 by

( )

[

]. We will write

( )

( ) mod

if

( )

( )

[

].

Let

0 0

be polynomials over such thatdeg

= deg 0+ deg 0 and the

leading coefficient of

coincides with that of

0 0. Let

( 0 0)

+1

0 0 mod 2 +1

for an integer

0.

Then there exist polynomials (

)

(

) such that

=

deg = deg 0 deg = deg 0

0 mod +1

0 mod +1

Proof. We first construct polynomials ( ) ( ) [ ] with the following properties:

deg(

0) deg

0, deg( 0) deg 0

(

)

1( ) mod +

(

)

1( ) mod +

( ) ( ) ( ) mod +2 +1

Proceeding by induction, we can assume that the polynomials ( ) ( ), for

1,

have been constructed. For a prime element put

(

) =

1( ) + +

(

)

(

) =

1( ) + +

(

)


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11

with

( )

( ) [ ], deg

( ) deg 0( ), deg

( ) deg 0( ). Then

(

)

(

)

1( )

1( ) +

1( )

(

) +

1( )

(

) mod

+2 +1

Since by the induction assumption

( )

1(

)

1(

) =

+2

1(

) for a suitable

1( ) [ ] of degree smaller than that of

, we deduce that it suffices for

(

),

(

)

to satisfy the congruence

1( )

1( )

( ) +

1( )

( ) mod

+1 .

However, (

1( )

1( )) (

0( ) 0( )) 0 mod +1. Then the properties

of the resultant imply the existence of polynomials

,

satisfying the congruence. Write

=

1 +

1 with polynomial

1 of degree smaller than that of

1 . Then it is easy

to see that the degree of

=

+

1 is smaller that the degree of

1 . The polynomials

are the required ones.

Now put ( ) = lim ( ) ( ) = lim ( ) and get

( ) = ( ) ( ).

The following statement is often called Hensel Lemma; it was proved by K. Hensel for -adic

numbers and byK. Rychlikfor complete valuation fields.

Let

0 0 be monic polynomials with coefficients in . Let

=

0 0 andsuppose that 0 0 are relatively prime in [ ]. Then there exist monic polynomials

with

coefficients in , such that

( ) = (

) (

)

( ) = 0( ) ( ) = 0( )

Proof. We have (

0( )

0( )) and we can apply the previous proposition for = 0.

The polynomials ( ) and ( ) may be assumed to be monic, as it follows from the proof of

the proposition.

Valuation fields satisfying the assertion of Corollary 1 are said to be Henselian. Corollary 1

demonstrates that complete discrete valuation fields are Henselian.

Let

( ) be a monic polynomial with coefficients in . Let

( 0)

2 +1

( 0)

+1

for some 0 and integer 0. Then there exists such that 0

+1 and

( ) = 0.

Proof. Put 0( ) = 0 and write

( ) =

1( )( 0) + with . Then

2 +1. Put 0( ) =

1( ) [ ]. Hence

( ) 0( ) 0( ) mod 2 +1 and

(

0) = 0( 0) +1. Hence ( 0( ) 0( ))

+1, and the proposition implies the

existence of polynomials ( ) ( ) [ ] such that ( ) = 0 mod +1 ,

and

( ) = ( ) ( ).

If the residue field of is finite , then for every

the polynomial

(

) =

1

1 has a root

such that

=

. So one hasmultiplicative representativesof

consisting of0 and all roots of unity in of order dividing

1 (or, equivalently, all roots

of unity in of order prime to ).

For every positive integer there is such that 1 + .


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12

Proof. Put

( ) = with 1+ . Let

+1. Then

(1)

+1 .

Therefore for every 1 + 2

+1the polynomial

( ) has a root 1 mod

+1 due to

Corollary 2.

Let

( ) = +

1 1 + + 0 be an irreducible polynomial with coefficients

in . Then the condition

0 implies for 0

1.

Proof. Assume that 0 and that is the maximal such that ( ) = min0

1 ( ).

If , then put

1( ) = 1

( )

0( ) =

+ 1

1

1 + + 1

0

0( ) = 1

+ 1

We have

1( ) =

0( ) 0( ), and

0( ) 0( ) are relatively prime. Therefore, by the

proposition

1( ) and

( ) are not irreducible.

(2.2). and

. Let be a field and an extension of with a valuation : .

Then induces the valuation 0 = : on . In this context is said to be

an extension of valuation fields. The group 0( ) is a totally ordered subgroup of ( )

and the index of 0( ) in ( ) is called the ramification index ( ). The ring of

integers 0

is a subring of the ring of integers and the maximal ideal

0coincides with

0. Hence, the residue field

0can be considered as a subfield of the residue field

. Therefore, if is an element of

0, then its residue in the field

0can be identified

with the image of as an element of in the field . We shall denote this image of

by . The degree of the extension 0

is called the intertia degree

(

). An

immediate consequence is the following lemma.

Let

be an extension of

and let

be a valuation on

. Let

andlet 0 be the induced valuation on . Then

( ) = ( ) ( 0)

(

) =

(

)

( 0)

(2.3). Extension of Discrete Valuation. Assume that is a finite extension and 0 is

a discrete valuation. Let elements 1

for natural ( ) be such that

( 1) + ( )

(

) + (

) are distinct in ( ) ( ). If =1 = 0 holds

with , then, as ( ) are all distinct, we get =1

= min1

(

) and

so = 0 for 1

. This shows that 1

are linearly independent over and

hence ( ) is finite. Let be a prime element with respect to 0 . Then we deduce that

there are only a finite number of positive elements in ( ) which are ( ). Consider the

smallest positive element in ( ). It generates the group ( ), and we conclude that is

a discrete valuation. Thus, we have proved the following result.

Let be a finite extension and 0 discrete for a valuation on . Then is

discrete.


13/57

13

(2.4).

for Complete Fields. Let and be fields with discrete valuations and

respectively and . The valuation is said to be an extension of the valuation , if

0 = for a positive . We shall write and use the notations ( )

( ) instead of

(

)

(

). If then ( ) = ( ) ( ).

Let be a finite extension of of degree ; then

(

)

( )

Proof. Let = ( ) and let

be a positive integer such that

( ). Let

1

be elements of such that their residues in are linearly independent over

. It suffices

to show that

are linearly independent over for 1

0 1. Assume

that

= 0

for and not all = 0.

Multiplying the coefficients by a suitable power of , we may assume that

and not all . Note that if

, then

= 0 and so .Therefore, there exists an index such that

. Let 0 be the minimal such index.

Then 0 = (

), which is impossible. We conclude that all = 0. Hence,

and ( )

( )

.

Let be the completion of with the discrete valuation . Then ( ) =

1

( ) = 1. Note that if is not complete, then : = ( )

( ). On the contrary, in

the case of complete discrete valuation fields we have

Let be an extension of and let be complete with respect to discrete

valuations

. Let

=

( ) and

=

(

)

. Let

be a prime element

with respect to and

1

elements of such that their residues form a basis of

over

. Then

is a basis of the

-space

and of the

-module

, with1

0 1. If

, then is a finite extension of degree =

.

Proof. Let be a set of representatives for . Then the set

=

=1

:

and almost all = 0

is the set of representatives for . For a prime element with respect to put

=

,

where = + 0 . Using Corollary (1.8) we obtain that an element can be

expressed as a convergent series

=

with

Writing

=

=1

with


14/57

14

we get

=

+

Thus, can be expressed as

with

=

+

1

0

1

By the proof of the previous lemma this expression for is unique. We conclude that

form a basis of over and of over .

(2.5). Uniqueness of Extension of Discrete Valuation From a Complete Field. Further we

shall assume that ( ) = for a discrete valuation . Then ( ) = : ( ) for an

extension of .

Let be a complete field with respect to a discrete valuation and a finite

extension of . Then there is precisely one extension on of the valuation

and =

1

with

=

( ). The field is complete with respect to .

Proof. Let =

. First we verify that is a valuation on . It is clear that

(

) = +

if and only if = 0 and ( ) = ( ) + ( ). Assume that ( ) ( )

for , then

( +

) = (

) +

1 +

and it suffices to show that if ( ) 0, then (1 + ) 0. Let

( ) =

+

1 1 +

+

0

be the monic irreducible polynomial of over . Then we get ( 1) 0 = ( )

( ) and if

=

: ( )

, then (( 1) 0

)

=

( ). We deduce that ( 0

) 0, and making use of

(2.1), we get ( ) 0 for 0

1. Now

( 1)

( )

(1 + ) =

( 1) = (

1) +

1( 1) 1 +

+

0

hence

( )

(1 + ) 0 and

(1 + ) 0

i.e., (1 + ) 0. Thus, we have shown that is a valuation on .

Let = : ; then ( ) = ( ) for . Hence, the valuation (1 ) is an

extension of to (note that (1 ) ( ) = in general). Let = ( (1 ) ); is

finite. Put = ( ) : , hence ( ) = ( ) = with a prime element

with respect to . Therefore, = ( )

is at once a discrete valuation on and an

extension of .

Let 1

be a basis of the -vector space . By induction on , 1 , we

shall show that

=1

(

)

0

(

)

0

for

= 1


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15

where (

)

.

The left arrow and the case = 1 are clear. For the induction step we can assume that

(

)

0 for each

= 1

. Therefore we can assume that ( (

)

) is bounded for

= 1 . Hence

1+

=2

(

)

=

(

)1

1

=1

(

)

0

where (

)

= (

(

)1

1

(

)

. Then

=2

( (

)

(

+1)

) =

=2

(

)

=2

(

+1)

0

and the induction hypothesis shows that (

)

(

+1)

0 for

= 2

. Thus, each

(

(

)

)

converges to, say,

. Finally, the sequence 1+

=2 (

)

converges both to

0 and to 1+

=2 , so 0 = 1+

=2 which contradicts the choice of .

Similarly one shows that a sequence

=1

(

)

is fundamental if and only if

(

)

is

fundamental for each

= 1

.

Thus, the completeness of implies the completeness of its finite extension with respect

to any extension of . We also have the uniqueness of the extension.

Let be a finite Galoisextension of . Then

= for the discrete valuation

on and

Gal( ). If is a prime element in , then

is a prime element and

=

,

=

.

Let be a finite extension. Let

and let

( ) be the monic irreducible

polynomial of over

. Then

( )

[

]. Conversely, let

( ) be a monic polynomial

with coefficients in . If is a root of

( ), then .

Proof. It is well known that =

is separable over for some 0. Let be a

finite Galois extension of with . Then, in fact, and the monic irreducible

polynomial ( ) of over can be written as

( ) =

=1

(

)

Gal( ) 1 = 1

By the previous corollary we get

. Hence we obtain ( )

[ ] and

( ) =

[ ]. If is a root of the polynomial

( ) = +

1 1 + + 0

[

] and , then1 =

1 1

0 , contradiction. Thus, .

(2.6). Types of Ramified Extensions. From now on is a complete discrete valuation field.Let be an algebraic extension. If

is the unique discrete valuation on which

extends the valuation = on , then we shall write ( )

( ) instead of (

),

(

). We shall write or or

or or for the ring of integers


16/57

16

, the maximal ideal

, the group of units

, a prime element with respect to , and

the residue field , respectively.

A finite extension of a complete discrete valuation field is called unramifiedif

is a separable extension of the same degree as . A finite extension is called totally

ramifiedif

( ) = 1. A finite extension is called tamely ramifiedif is a separable

extension and

( ) where = char( ) 0. A finite extension is calledtotally

ramifiedif = . A finite extension is calledwildly totally ramifiedif = and the

degree of is a power of =char( ).

(2.7). Unramified Extensions. If is unramified then we deduce From Lemma (2.4) that

(

) = 1,

( ) =

:

.

(1) Let be an unramified extension, and = (

) for some

. Let

be such

that =

. Then = (

), and is separable over

,

=

[ ];

is a simple

root of the polynomial

( ) irreducible over

, where

( ) is the monic irreducible

polynomial of over

.

(2) Let

( ) be a monic polynomial over , such that its residue is a monic separablepolynomial over

. Let

be a root of

( ) in

alg, and let = (

). Then the

extension is unramified and =

(

) for

= .

Proof. (1) By the preceding lemma

( )

[

]. We have

( ) = 0 and

( ) = 0,

deg

( ) = deg

( ). Furthermore,

:

(

) :

= deg

( ) = deg

( )

(

) :

=

:

It follows that = ( ) and

is a simple root of the irreducible polynomial

( ). Therefore,

(

) = 0 and

( ) = 0, i.e., is separable over . Thus,

= [ ].

(2) Since the residue of

is separable, it is separable too. Let

( ) =

=1

(

) be the

decomposition of

( ) into irreducible monic factors in [ ]. Then every root of

( ) is in

the ring of integers of any finite Galois extension of which contains all of them; and therefore

(

)

[

]. Suppose that is a rootof

1( ). Then

1( ) =

1( ) is a monic separable

polynomial over . The Henselian property of implies that

1( ) is irreducible over .

We get

. Since

=

, we obtain (

) and

deg

1( ) = : : (

) :

= deg 1( ) = deg

1( )

Thus, = (

), and is unramified.

(1) If are unramified, then is unramified.

(2) If is unramified, is an algebraic extension of and is the discrete valuation

field with respect to the extension of the valuation of , then

is unramified.

(3) If

1

2 are unramified, then

1

2 is unramified.

Proof. (1) follows from the multiplicativity of the ramification index. To verify (2) let

= (

) with

,

( )

[

] as in the first part of the proposition. Then

because = ( ). Observing that = ( ), we denote the irreducible monic polynomial


17/57

17

of over by

1( ). By the Henselian property of we obtain that

1( ) is a power of

an irreducible polynomial over . However,

1( ) divides

( ), hence

1( ) is irreducible

separable over . Applying the second part of the proposition, we conclude that is

unramified.

(3) follows from (1) and (2).

An algebraic extension of is calledunramifiedif are separable extensions

and ( ) = 1, where is the discrete valuation on , and is the unique extension of

on .

The third assertion of the Corollary shows that the compositum of all finite unramified

extensions of in a fixed algebraic closure alg of is unramified. This extension is not

a complete field in general, but a Henselian discrete valuation field. It is called the maximal

unramified extension ur of . Its maximality implies

ur = ur for any automorphism of

the separable closure sep over . Thus, ur is Galois.

(1) Let be an unramified extension and let be a Galois extension. Then is

Galois.

(2) Let be an unramified Galois extension. Then is Galois. For an automorphism

Gal(

) let

be the automorphism in Gal( ) satisfying the relation

=

for every

. Then the map

induces an isomorphism of Gal( ) onto

Gal( ).

Proof. (1) It suffices to verify the first assertion for a finite unramified extension . Let

= (

) and let ( ) be the irreducible monic polynomial of

over . Then

( ) =

=1

(

)

with

1 =

. Let

( ) be a monic polynomial over of the same degree as ( )

and

(

) =

(

). The Henselian property implies

( ) =

=1

(

)

with

=

. The first proposition above shows that = ( 1), and we deduce that

is Galois.

(2) Note that the automorphism

is well defined. Indeed, if

with = , then

(

)

and

=

. It suffices to verify the second assertion for a finite unramified

Galois extension . Let

( ) be as in the first part of the first proposition. Since all

roots of

( )belong to , we obtain that allroots of

( ) belong to and is Galois. The

homomorphism Gal( ) Gal( ) defined by

is surjective because the condition

=

implies

=

for the root of

( ) with =

. Since Gal( ) Gal( )

are of the same order, we conclude that Gal(

) is isomorphic to Gal(

).

The residue field of ur coincides with the separable closure sep

of and

Gal(

ur

) Gal(

sep

).


18/57

18

Proof. Let

sep, let ( ) be the monic irreducible polynomial of

over , and

( ) as

in the second part of the first proposition. Let be all the roots of

( ) and = ( ).

Then ur and

=

ur for a suitable

. Hence, ur = sep

.

(2.8). Maximal Unramified Field. Let be an algebraic extension of , and let be a

discrete valuation field with perfect residue field. We will assume that alg = alg.

Let be an algebraic extension of and let be a discrete valuation field.

Then ur =

ur, 0 = ur is the maximal unramified subextension of

which is

contained in , and 0 is totally ramified.

Proof. We have ur ur. Since the residue field of ur coincides with the residue field

of ur, we deduce ur = ur. An unramified subextension of in is contained in 0 ,

and 0 is unramified.

Let be a finite Galoisextension of and let 0 be the maximal unramified

subextension in

. Then

0

and

0

are Galois, and the map

defined in thesecond proposition of(2.7)induces the surjective homomorphism

Gal( )

Gal( 0 ) Gal( 0 ) = Gal( )

The extension ur

is Galois and Gal( ur 0) Gal(

ur

) Gal( ur

ur),

Gal( ur

ur) Gal( 0), Gal(

ur

) Gal(

ur 0).

Proof. Let

Gal(

). Then

0 is unramified over , hence 0 = 0 and 0

is Galois. The surjectivity of the homomorphism Gal( ) Gal( 0 ) follows from the

second proposition of (2.7). Since and

ur

are Galois extensions, we obtain that

ur is a Galois extension. Then ur = ur by the previous proposition. The remaining

assertions are easily deduced by Galois theory.

(2.9). Tamely Ramified Extensions.

(1) Let be a finite tamely ramified extension of , and let 0 be the maximal unramified

subextension in . Then = 0( ) and = 0 [ ] with a prime element in

satisfying the equation 0 = 0 for some prime element 0 in 0, where = ( ).

(2) Let 0 be a finite unramified extension, = 0( ) with = 0. Let

if

= char( )

0. Then

is separable tamely ramified.

Proof. (1) (2.8) shows that 0 is totally ramified. Let 1 be a prime element in 0 , then

1 =

for a prime element

in and

. Since = 0 , there exists 0 such

that = . Hence 1 1 =

for the principal unit = 1

. For the polynomial

( ) = we have

(1)

,

(1) = . Now Corollary 2 of (2.1) shows the existence

of an element

with

=

,

= 1. Therefore,

= 1

1

, 0 =

are the elementsdesired for the first part of the Proposition.

(2) Let = 1 for a prime element 1 in 0 and a unit 0 . The polynomial

( ) =

is separablein 0[ ] and we canapply the Henselianproperty to

( ) =


19/57

19

and a root

sep of

( ). We deduce that 0( ) 0 is unramified and hence it suffices

to verify that 0 for = ( ) 0 = 0( ), is tamely ramified. We get = 0( 1)

with 1 = 1 , 1 = 1 . Put = g.c.d.( ). Then 0( 2 ) with

2 =

1

and a primitive th root of unity. Since the extension 0

( )

0 is unramified (this can

be verified by the same arguments as above), 1 is a prime element in 0( ). Let be the

discrete valuation on 0( 2 ). Then ( ) ( 1) ( ) and ( 1) ( ) , because

and are relatively prime. This shows that 0( 2 ) 0( )

. However,

0( 2) : 0( ) , and we conclude that 0( 2) 0( ) is tamely and totally

ramified. Thus, 0( 2) 0 and 0 are tamely ramified extensions.

(1) If are tamely ramified, then is separable tamely ramified.

(2) If is tamely ramified, is an algebraic extension, and is discrete, then

is tamely ramified.

(3) If 1 2 are tamely ramified, then 1 2 is tamely ramified.

Proof. It is carried out similarly to (2.8). To verify (2) one can find the maximal unramified

subextension 0 in . Then it remains to show that 0 is tamely ramified. Put

= 0( ) with = 0. Then we get = 0( ), and the second part of the proposition

yields the required assertion.

(2.10). Totally Ramified Extensions. A polynomial

( ) = +

1 1 + + 0 over

is called an Eisenstein polynomialif

0

1 0 2

(1) The Eisenstein polynomial

( ) is irreducible over . If is a root of

( ), then

(

)

is a totally ramified extension of degree

,

is a prime element in

(

),

( )

=

[ ].

(2) Let be a separable totally ramified extension of degree , and let be a prime

element in . Then is a root of an Eisenstein polynomial over of degree .

Proof. (1) Let be a root of

( ), = ( ), = ( ). Then

( ) =

1

=0

min

0

1(

(

) +

( ))

where and

are the discrete valuations on and . It follows that

( )

0. Since

(

0) ( ) +

( ) for

0, one has

( ) =

(

0) = .Then ( ) = 1 =

= 1, and

=

[ ].

(2) Let be a prime element in . Then =

(

). Let

( ) =

+

1 1 +

+

0

be the irreducible polynomial of over . Then = and

( ) = min0

1 ( ) +

, hence ( ) 0, and = ( 0), ( 0) = 1.


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20

(2.11). Ramification Groups. Let be a finite Galois extension of ,

= Gal( ). Put

=

:

+1

for all

1

Then

1 =

by Lemma (2.11) and

+1 is a subset of

.

Let

be the discrete valuation of . For a real number define

=

:

(

) + 1 for all

Certainly each of

is equal to

with the least integer

.

are normal subgroups of

.

Proof. Let

. Then

+1

. Hence

1( )

1(

+1

) =

+1

by Lemma (2.11), i.e.,

1

. Let

. Then

( )

=

( ( )

) +

( )

+1

i.e.,

. Furthermore, let

. Then ( )

for

and

( )

+1

, 1

( )

+1

, 1

.

The groups

are called (lower)ramification groupsof

= Gal( ).

Let be a finite Galois extension of , and let be a separable extension of

. Then

0 = Gal( 0) and the

th ramification groups of

0 and

coincide for

0.

Moreover,

=

0 : +1

for a prime element in , and

=

1

for sufficiently large

.

Proof. Note that

0 if and only if Gal( ) is trivial. Then

0 coincides with

the kernel of the homomorphism Gal( ) Gal( ). The first proposition of (2.11) and

the second proposition of (2.7) imply that this kernel is equal to Gal( 0). Since

is a

subgroup of

0 for

0, we deduce the assertion about the

th ramification group of

0 . We

get = 0 [ ]. Let =

=0

be an expansion of with coefficients in 0 .As

= for

0 it follows that

=

=0

(

)

Now we deducethe description of

, since

(

)

. If

max

(

) :

,

then

=

1

.

The group

0 is called theinertia group of

, and the field 0 is called the inertia subfield

of .

Let be a finite Galois extension of , a separable extension of

, and

a prime element in . Introduce the maps

0:

0

:

(

0)

by the formulas

(

) =

(

). Then

is a homomorphism with the kernel

+1 for

0.


21/57

21

Proof. The proof follows from the congruence

( )

=

mod

+1

for

. The kernel of

consists of those automorphisms

, for which

1 +

+1

, i.e.,

+2

.

Let be a finite Galois extension of , and a separable extension of .

If char( ) = 0, then

1 = 1 and

0 is cyclic. If char( ) =

0, then the group

0

1

is cyclic of order relatively prime to

,

+1 are abelian

-groups, and

1 is the maximal -subgroup of

0.

Proof. The previous proposition permits us to transform the assertions of this corollary into

the following: a finite subgroup in is cyclic (of order relatively prime to char( ) when

char( )

= 0 ); there are no nontrivial finite subgroups in the additive group of ifchar( ) = 0;

ifchar( ) = 0 then a finite subgroup in is a -group.

Let be a finite Galois extension of and a separable extension of

.

Then the group

1 coincides with Gal( 1), where 1 is the maximal tamely ramified

subextension in .

Proof. The extension 1 0 is totally ramified by the first proposition of (2.11) and is the

maximal subextension in 0 of degree relatively prime with char( ). Now Corollary 1

implies

1 = Gal( 1).

Let be a finite Galois extension of and a separable extension of

.

Then

0 is a solvable group. If, in addition, is a solvable extension, then is solvable.

Proof. It follows from Corollary 1.

(2.12). The Norm Map for Cyclic Extensions. Let be a local field and its Galois

extension of prime degree . Then there are four possible cases:

is unramified;

is tamely and totally ramified;

is totally ramified of degree = char( ) 0;

Let be a separable extension of prime degree ,

. Then

(1 + ) = 1 +

( ) + Tr

( ) + Tr

( )

with some

such that

( ) 2

( ) (

and Tr

are the norm and the trace

maps, respectively).

Proof. Recall that for distinct embeddings

of over into the algebraic closure of ,

1

, one has

=

=1

( )

Tr

=

=1

( )


22/57

22

Hence

(1 + ) =

=1

(1 +

( )) = 1 +

=1

( ) +

=1

1

( ) + +

=1

( )

For = 1

( ) +

we get

( ) 2

( ).

Our nearest purpose is to describe the action of the norm map

with respect to the

filtration on and .

Let be an unramified extension of degree . Then a prime element

in is a prime element in . Let

= 1 +

,

= 1 +

. Then the following

diagrams are commutative:

0

0

Tr

Proof. The commutativity of the first two diagrams is easy. The preceding Lemma shows that

(1 +

) = 1 + (Tr

)

+ (

)

+ Tr

( )

with

( ) 2

and, consequently, Tr

( ) 2

. Thus, we get

(1 +

) 1 + (Tr

)

mod +1

and the commutativity of the third diagram.

In the case under consideration

1 = 1 .

Let be a totally and tamely ramified cyclic extension of degree . Thenfor some prime element

in , the element =

is prime in and = . Let

= 1 +

,

= 1 +

. Then the following diagrams

id

0

0

=

are commutative, where id is the identity map,

takes an element to its th power, is

the multiplication by ,

1. Moreover,

=

+1 if

.

Proof. Since

=

and is Galois, then Gal( ) is cyclic of order and

(

) =

for a generator

ofGal( ), where is a primitive th root of unity, .It is easy to see that the first diagram is commutative.

Corollary in (2.5) shows that

( ) =

for

Gal(

),

, and we get the

commutativity of the second diagram.


23/57

23

If =

, then 1 +

for

, and

(1 +

) = (1 +

) 1 +

mod

+1

We deduce

=

=

Finally, 1 = 1

=0 (

), therefore for

and for

one has

(1 +

) =

1

=0

(1 +

) = 1 (

)

Thus

=

+1 .

In the case under consideration

1 = 1 . If is algebraically closed

then

= .

Now we treat the most complicated case where

is a totally ramified Galois extensionof degree = char( ) 0. In this case

=

[

], = (

) for a prime element

in , and = . Let

be a generator of Gal( ), then

(

)

. One can write

(

)

=

with

1 +

. Then

2(

)

=

(

)

=

2

( )

and

1 =

(

)

=

( )

1( )

This shows that

1 +

and

1 +

, because raising to the th power is an injective

homomorphism of . Thus, we obtain

(

)

1 +

. Put

(

)

= 1 +

with

= (

)

1 ( )

Note that does not depend on the choice of the prime element

and of the generator

of

= Gal( ). Indeed, we have

(

)

1 +

mod

+1

and

( )

1 mod

+1

for an element

. We also deduce that

( )

for every element . This means that

=

,

+1 = 1 .

Let

( ) =

+

1

1 + +

0 be the irreducible polynomial of over .

Then

Tr

(

) =

0 if 0 2

1 if

= 1


24/57

24

Proof. Since

(

) for 0

1 are all the roots of the polynomial

( ), we get

1

( )

=

1

=0

1

(

)

(

)

Putting = 1 and performing the calculations in the field (( )), we consequently deduce

( ) =

(1 +

1 + + 0

)

1

( )

=

1 +

1 + + 0

mod +1

1

(

) =

1

(

) =

0

(

) +1

(because 1 (1 ) =

0

in (( )) ). Hence

0

1

=0

(

) +1

( )

mod +1

or

Tr

(

) =

1

=0

(

)

(

) =

0 if 0

2

1 if = 1

as desired.

Let [ ] denote the maximal integer

. For an integer

0 put

(

) =

+ 1 + (

1

)

. Then

Tr

(

) =

( )

Proof. One has

(

) =

1 =1

(

) and

(

)

1 +

mod

+1

.

Then

(

) = ( 1)!( )

1

(

1)( +1)

with some 1 + (

1)( +1)+1

. Since = , for a prime element in one has the

representation =

with

. The previous lemma implies

Tr

+ +1

+ +1

=

0 if 0 1

+1

if = 1

for + +1 = ( )

+1

( 1)!(

)

1

. Since Tr

(

) =

Tr

( ) we can choose

the units

+

+1 , for every integer

, such that Tr

(

+ +1

+

+1) = 0 if

(

+ 1) and=

+( +1)

if ( + 1). Thus, since the -module

is generated by

,

, we

conclude that Tr

(

) = ( )

.


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25

Let be a totally ramified Galois extension of degree = char( ) 0.

Let

be a prime element in . Then =

is a prime element in . Let

=

1 +

= 1 +

. Then the following diagrams are commutative:

id

0

0

=

=

1

+

+

=

(

1)

+

+

where 1

in the third diagram and

0 is the last diagram.

Moreover,

(

+

) =

(

+ +1 ) for

0

.

Proof. The commutativity of the first and the second diagrams is easy.

To treat the remaining diagrams, put = 1 +

with

. Then, by the first lemma,

we get

= 1 +

(

)

+ Tr

(

) + Tr

(

)

with

( ) 2

. The previous proposition implies that

Tr

(

) + 1 +

1

Tr

( ) + 1 +

2

1

and for

Tr

(

)

+ 1 Tr

( )

+ 1

Therefore, the third diagram is commutative.

Further, using ( ), write

1 =

(

)

1 +

( )

+ Tr

(

) mod

+1

We deduce that

Tr

(

)

( )

mod

+1

Since

( )

mod

in view of

1 , we conclude that

(1 +

) 1

(

)

+1

for

. This implies the commutativity of the fourth (putting

) and the fifth (when

) diagrams.Finally, if

, then

(1 +

)

1 +

1 +

+

mod

+ +1


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26

This means that

(1 +

+

)

+ +1 and

( + ) =

( + +1 ).

Compare the behaviour of the norm map with the behaviour of raising to the

thpower in Proposition (1.10).

+1 =

+1 .

If is algebraically closed then

= .

Proof. It follows immediately from the last diagram of the proposition, since the multiplication

by ( )

1 is an isomorphism of the additive group .

3: Local Class Field Theory

This section focuses on complete discrete valuation fields with finite residue field.

3.1. Useful Results on Local Fields with Finite Residue Field

(3.1.1). Structures. Let be a local field with finite residue field = ,

=

elements.

Since char( ) = , is of characteristic 0 or of characteristic .

In the first case ( ) 0 for the discrete valuation in , hence the restriction of on

is equivalent to the -adic valuation (by Ostrowskis Theorem). Then we can view the field

of -adic numbers as a subfield of . Let = ( ) = ( ) be the absolute ramification index

of . Then is a finite extension of of degree =

. Such a field was called a local

number field.

In the second case is isomorphic (with respect to the field structure and the discrete

valuation topology) to the field of formal power series (( )) with prime element , since

the multiplicative representatives of the residue field form a finite subfield of . Such a field

was called a local functional field.

is a locally compact topological space with respect to the discrete valuation

topology. The ring of integers and the maximal ideal

are compact. The multiplicative

group is locally compact, and the group of units is compact.

Proof. Assume that is not compact. Let ( ) be a covering by open subsets in , i.e.,

= , such that isnt covered by a finite union of . Let be a prime element of .

Since is finite, there exists an element

0 such that the set

0+ is not contained

in the union of a finite number of . Similarly, there exist elements

1

such that

0+

1 + +

+ +1 is not contained in the union of a finite number of . However,the element

= lim

+

=0

belongs to some

, a contradiction. Hence,

is

compact and , as the union of

+ with

= 0, is compact.


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27

(3.1.2). Galois Extensions.

The Galois group of every finite extension of is solvable.

Proof. Follows from (2.11).

For every

1 there exists a unique unramified extension of of degree

: = (

1). The extension is cyclic and the maximal unramified extension ur of

is a Galois extension. Gal(

ur

) is isomorphic to and topologically generated by an

automorphism such that

( )

mod

ur for ur

The automorphism is called the Frobenius automorphism of

.

Proof. First we note that, by (2.1) contains the group

1 of ( 1) th roots of unity

which coincides with the set of nonzero multiplicative representatives of in . Moreover,

the unit group is isomorphic to

1 1 .

The field

has the unique extension

of degree

, whichis cyclic over

. (2.7) showsthat there is a unique unramified extension of degree over and hence = (

1).

Now let be an unramified extension of and . Then ( ) is of finite degree.

Therefore, ur is contained in the union of all finite unramified extensions of . We have

Gal(

ur

) lim

Gal(

)

It is well known that Gal( sep

) is topologically generated by the automorphism

such

that

( ) =

for sep

. Hence, Gal( ur ) is topologically generated by the Frobenius

automorphism .

If

1 , then

(

)

mod

and (

)

1. The uniqueness of the multiplicative representative for

implies

now that (

) =

.

(3.1.3).

Let

be a primitive th root of unity. Put

(

)

=

(

). Then

(

)

(

) = 0

and

belongs to the ring of integers of ( )

. Let

( ) =

1

1 1

=

(

1) 1

+

(

2) 1

+ + 1

Then

is a root of

( ), and hence ( )

:

( 1) 1. The elements

,

0

, are roots of

( ). Hence

( ) =

0

(

) and =

(1) =

0

(1

)


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28

However,

(1

)(1

) 1 = 1 +

+ +

1

belongs to the ring of integers of ( )

. For the same reason, (1

)(1

)

1 belongs to

the ring of integers of ( )

. Thus, (1

)(1

)

1 is a unit and = (1

)

1(

1)

for some unit . Therefore, ( ( )

)

(

1)

1, and ( )

is a cyclic totally ramified

extension with the prime element 1

, and of degree ( 1) 1 over . In particular,

(

)

=

[1

] =

[

]

(3.1.4). The Group of Principal Units. If char( ) = , then Proposition (6.2) Ch. I shows

that every element 1 can be uniquely expressed as the convergent product

=

0

(1 +

)

where the index-set numerates

elements in , such that their residues form a basis of

over , and the elements

belong to this set of

elements; are elements of with

(

) =

, and . Thus, 1 has the infinite topological basis 1 +

.

Now let char( ) = 0. Every element 1 can be expressed as a convergent product

=

(1 +

)

where = 1

(

1),

, = ( ); the index-set numerates

elements in

, such that their residues form a basis of over , and the elements

belong to this set

of

elements; are elements of with ( ) =

, and .

If a primitive th root of unity does not belong to , then = 1 = 0 and the above

expression for is unique; 1 is a free -module of rank =

= :

.

If a primitive th root of unity belongs to , then = 1 +

(

1) is a principal unit

such that

, and

. In this case the above expression for

is not unique.

1 isisomorphic to the product of copies of and the -torsion group

, where 1 is the

maximal integer such that

.

If char( ) = 0, then is an open subgroup of finite index in for 1. If

char( ) = , then is an open subgroup of finite index in for . If char( ) =

and

, then is not open and is not of finite index in .

Proof. It follows (1.9) and (1.10) and the previous considerations.

(3.1.5). The Norm Groups. Now we have a look at the norm group

( ) for a finite

extension of . Recall that the norm map

:

is surjective when

.Let be a finite unramified extension. Then (2.12) implies that

= in the

case of an unramified extension and

= , where is a prime element

in , = : . This means, in particular, that

is a cyclic group of order


29/57

29

in this case. Conversely, every subgroup of finite index in

that contains

coincides with

the norm group

for a suitable unramified extension .

Let be a totally and tamely ramified Galois of degree . (2.12) shows that

1

=

1

for a suitable prime element in (e.g. such that = (

), and

for

if and only if

. Since is Galois, we get

and (

1).

Hence, the subgroup

is of index in

, and the quotient group

is cyclic. We

conclude that

=

1

with an element

, such that its residue

generates

. So

is

cyclic of order . Conversely, every subgroup of index relatively prime to char( ) coincides

with the norm group

for a suitable cyclic extension .

Let be a totally ramified Galois extension of degree

. The right vertical

homomorphism of the fourth diagram in the last proposition of (2.12)

1

has

kernel of order

; therefore its cokernel is also of order

. Let

be such that

does not belong to the image of this homomorphism. Since is perfect, we deduce that

1 +

1 . The other diagrams imply that

is a cyclic group of order

and generated by 1 +

mod

. If char( ) = 0, then

(

1), where

=

(

), and if then = ( 1) and a primitive th root of unity belongs to ,

and = (

) for a suitable prime element in . In this case

is generated

by mod

.

(3.1.6). Completion of ur . The field ur is a Henselian discrete valuation field with

algebraically closed residue field and its completion is a local field with algebraically closed

residue field sep

.

Let be the set of multiplicative representatives of the residue field of if its characteristic

is . is the union of all sets

1 1 (which coincides with the set of all roots of unity

of order relatively prime to

) and of 0.Let be a finite separable extension of . Since the residue field of is algebraically

closed, is totally ramified.

The norm maps

:

:

are surjective.

Proof. Since the Galois group of is solvable, it suffices to consider the case of a Galois

extension of prime degree . Certainly, the norm of a prime element of is a prime element of

. The surjectivity of the norm maps follows from (2.12).

(3.1.7). Augmentation Group (

) .

For a finite Galois extension denote by ( ) the subgroup of 1

generated by 1 where runs through all elements of 1 and runs through all elements

ofGal( ).


30/57

30

Every unit in

can be factorized as

with

, 1 Since

1 = 1

we deduce that ( ) coincides with the subgroup of

generated by 1 ,

,

Gal(

).

Let be a finite Galois extension of . For a prime element of define

: Gal( )

(

)

(

) =

1 mod (

)

The map

is a homomorphism which does not depend on the choice of . It induces a

monomorphism

: Gal( )ab

(

) where for a group

the notation ab stands

for the maximal abelian quotient of

.

The sequence

1 Gal( )ab

( )

1

is exact.

Proof. Since 1 belongs to

, we deduce that ( 1) 1 ( ) and

1

1

1

mod

(

)

Thus, the map

is a homomorphism. It does not depend on the choice of , since ( ) 1

1 mod ( ).

Surjectivity of the norm map has already been proved.

Suppose first that Gal( ) is cyclic with generator

. The kernel of

coincides

with 1 . Since

is a homomorphism, we have

1

(

1)

mod (

). So

we deduce that 1 is equal to the product of ( ) and the image of

. This shows the

exactness in the middle term.

Note that

1 = ( 1+ + +

1) 1, so ( ) = 1

. If

1

( ), then

(

1)

=

1 for some

. Hence 1 belongs to and therefore : divides

and

= 1. This shows the injectivity of

.

Now in the general case we use the solvability of Gal( ) and argue by induction. Let

be a Galois cyclic subextension of such that = = . Put =

.Since

:

is surjective, we deduce that

(

) =

(

).

Let

= 1 for

. Then by the induction hypothesis there is Gal( )

such that

=

1

with ( ). Write =

with

(

). Then

1

1

belongs to the kernel of

and therefore by the induction hypothesis can be

written as 1 with

Gal( ), ( ). Altogether, 1 mod ( )

which shows the exactness in the middle term.

To show the injectivity of

assume that 1 ( ). Then 1

(

) and

by the previous considerations of the cyclic case

acts trivially on . So

belongs to

Gal( ). Now the maximal abelian extension of in is the compositum of all cyclic

extensions of in . Since

acts trivially on each , we conclude that

is injective.

(3.1.8). 1 Acting on . For every every element can be uniquely expandedas

=


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31

where is a prime element in .

Since : ur ur is continuous, it has exactly one extension : which acts as

.

We shall study the action of 1 on the multiplicative group.

The kernel of the homomorphism 1 is equal to and the

image is

; 1

=

for every

1.

Proof. For =

with

the condition ( ) = implies that (

) =

for

. Hence,

belongs to the residue field of and . Similarly one shows the

exactness of the sequence in the central term

+

.

Let

. We shall show the existence of a sequence

such that

1

mod

+1

and

+1 1

+1

.

Let =

0 with

, 0 1

. Let be such that

1 =

. Then 1 =

;

put 0 = .

Now assume that the elements 0 1

have already been constructed. Definethe element

+1 from the congruence

1

1

1 +

+1 +1 mod +2

There is an element

+1 such that

(

+1)

+1+

+1 =

+1

+1+

+1 0 mod

Now put

+1 =

(1 +

+1 +1). Then 1

1

+1

+2

and

+1 1

+1

.

There exists = lim

, and 1 = . When

the element can be

chosen in

as well.

Let be a finite Galois totally ramified extension. The extension ur ur is Galois with

the group isomorphic to that of . We may assume that the completion of ur is a subfield

of the completion of ur.

The extension is totally ramified of the same degree as . From (2.7) and (2.8) we

deduce that the extension is Galois with the group isomorphic to that of .

Let be such that 1 ( ). Then

belongs to the group

.

Proof. We have 1 = 1

for some

and Gal( ). By the pre-

vious proposition we have = 1

for some

. So ( 1 1

) 1 = 1 and

1 1

= with . Then

=

1

.


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32

3.2. The Neukirch Map

(3.2.1). Let be a finite Galois extension of . Then ur = ur. Recall that Gal( ur )

consists of -powers of .

Put Frob( ) =

Gal( ur

) :

ur is a positive integer power of .

The setFrob( ) is closed with respect to multiplication;it is not closed with

respect to inversion and 1 Frob( ).

The fixed field of

Frob(

) is of finite degree over

, ur = ur, and

is the

Frobenius automorphism of .

Thus, the set Frob( ) consists of the Frobenius automorphisms of finite extensions

of in ur with Gal( ur

)

.

The map Frob( )

Gal(

)

is surjective.

Proof. The first assertion is obvious.

Since

ur

we deduce that

ur

ur

ur

. The Galois group of

ur

istopologically generated by

and isomorphic to , therefore it does not have nontrivial closed

subgroups of finite order. So the group Gal( ur ur) being a subgroup of the finite group

Gal( ur

ur) should be trivial. So ur = ur.

Put 0 = ur . This field is the fixed field of

ur =

, therefore 0 : = is

finite. We deduce that

: 0 =

ur :

ur

=

ur :

ur

=

: 0

is finite. Thus, is a finite extension.

Now

is a power of and ur = 0:

ur =

ur =

ur . Therefore,

=

. Certainly, the Frobenius automorphism of a finite extension of in ur with

Gal( ur ) belongs to Frob( ).

Denote by an extension in Gal( ur ) of . Let

Gal(

), then

0 is equal

to for some positive integer . Hence

1

acts trivially on 0, and so = belongs to Gal( 0). Let Gal(

ur

ur) be such that

= . Then for

= we

deduce that

ur =

and

=

=

. Then the element

Frob(

) is mapped

to

Gal(

).

(3.2.2).

Let be a finite Galois extension. Introduce

: Frob( )

mod

where is the fixed field of

Frob(

) and is any prime element of .

The map

is well defined. If

= id

then

(

) = 1.

Proof. Let

1

2 be prime elements in . Then

1 =

2

for a unit

. Let

be thecompositum of and . Since ur , the extension is unramified. From (3.1.5)

we know that =

for some . Hence

1 =

( 2 ) =

2

(

) =

2

(

)


33/57

33

We obtain that

1

2 mod

.

If

= id

then and therefore

.

(3.2.3). The definition of the Neukirch map is very natural from the point of view of thewell known principle that a prime element in an unramified extension should correspond to

the Frobenius automorphism (see Theorem (3.2.4) below) and the functorial property of the

reciprocity map (see (3.2.5) and (3.3.4)) which forces the reciprocity map

to be defined

as it is.

Already at this stage one can prove that the map

: Frob( )

induces the Neukirch homomorphism

: Gal( )

In other words,

(

) does not depend on the choice of

Frob( ) which extends

Gal( ), and moreover,

( 1)

( 2) =

( 1 2).

We will choose a different route, which is a little longer but perhaps is more satisfying.

The plan is the following: first we easily show the existence of

for unramifiedextensions and even prove that it is an isomorphism. Then we deduce some functorial properties

of

. To treat the case of totally ramified extensions in the next section, we introduce the

Hazewinkel homomorphism

which acts in the opposite direction to

. Calculating

composites of the latter with

we shall deduce the existence of

which is expressed

by the commutative diagram

Frob( )

id

Gal( )

Then using

we prove that

is a homomorphism and that its abelian part

ab

: Gal( )ab

is an isomorphism.

Then we treat the general case of abelian extensions and then Galois extensions reducing it

to the two cases described above and using functorial properties of

. This route not only

establishes the existence of

, but also implies its isomorphism properties.

(3.2.4).

lectures on local fields

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