Algebraic Number Theory (Fesenko)

8/3/2019 Algebraic Number Theory (Fesenko)

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Algebraic Number Theory 2002/2003

1. Field extensions (revision of algebraic prerequisites)

1.1. General

1.1.1.

Definition.For a field

define the ring homomorphism

by

1

. Itskernel is an ideal of such that

is isomorphic to the image of in

. The

latter is an integral domain, so is a prime ideal of , i.e. = 0 or =

for a

prime number . In the first case

is said to have characteristic 0, in the second

characteristic .

DefinitionLemma. Let

be a subfield of a field . An element

is called

algebraic over

if one of the following equivalent conditions is satisfied:

(i) " ( ) = 0 for a non-zero polynomial " ( # )

[ # ];

(ii) elements 1 $ $ 2 $ ' ' ' are linearly dependent over

;

(iii)

-vector space

[ ] = 01 4

4

: 4

8

is of finite dimension over

;

(iv)

[ ] =

( ).

Proof. (i) implies (ii): if " ( # ) = 1 @4 =0 B 4 #

4

,B

0 $B

@

D

= 0, then 1B

4

4

= 0.

(ii) implies (iii): if 1@

4 =0 B 4 4

= 0,B

@

D

= 0, then @

= F 1 @ G1

4 =0 B G1

@

B

4

4

, @

+1 =

@

= F 1 @G

14 =0

B

G

1

@

B

4

4 +1 = F 1 @G

24 =0

B

G

1

@

B

4

4 +1 +B

G

1

@

B

@G

1 1 @ G1

4 =0B

G

1

@

B

4

4

, etc.

(iii) implies (i): if @

is a linear combination 1 @G

14 =0

R

4

4

then is a root of

" ( # ) = #@

F

1

R

4#

4

.

(i) implies (iv): 1

= FB

G

10 1 @4 =1 B 4

4

G

1.

(iv) implies (i): if 1

is equal to 1 T 4 4

, then is a root of 1 T 4 #4 +1

F 1.

For an element algebraic over

denote by

"X

( # )

[ # ]

the monic polynomial of minimal degree such that " X ( ) = 0.This polynomial is irreducible: if " X = Y ` , then Y ( ) ` ( ) = 0, so Y ( ) = 0 o r

` ( ) = 0, contradiction. It is called the monic irreducible polynomial of over

.

For example, "X

(# ) is a linear polynomial iff

.


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2 Alg number theory

Lemma. Define a ring homomorphism

[ # ] , Y ( # ) Y ( ). Its kernel is the

principal ideal generated by " X ( # ) and its image is

( ) , so

[#

]

("

X

(#

))d

(

)'

Proof. The kernel consists of those polynomials Y over

which vanish at . Using

the division algorithm write Y = "X `

+ e where e = 0 or the degree of e is smaller

than that of " X . Now e ( ) = Y ( ) F " X ( ) ` ( ) = 0, so the definition of " X implies

e = 0 which means that " X divides Y .

Definition. A field is called algebraic overits subfield

if every element of is

algebraic over

. The extension

is called algebraic.

Definition. Let

be a subfield of a field . The dimension of as a vector space

over

is called the degreeh

:

h of the extension

.

Transitivity of the degree h :

h = h : i h h i :

h follows from the observation:if r

4form a basis of i over

and tu

form a basis of over i then r4 t u

form a

basis of over

.

Every extension

of finite degree is algebraic: if t

, then h

( t ) :

hx

h :

h is finite, so by (iii) above t is algebraic over

.

An algebraic extension

( 0 4

8

) of

is is the composite of extensions

( 4),

and since 4 is algebraic h

( 4 ) :

h is finite, thus every algebraic extension is the

composite of finite extensions.

1.1.2. Definition. An extension

of of finite degree is called an algebraic number

field, the degree h

: h

is called the degree of

.

Examples. 1. Every quadratic extension

of

can be written as

( T

) for asquare-free integer T . Indeed, if 1 $

ris a basis of over , then r 2 = 1 + 2 r with

rational 4, so r is a root of the polynomial # 2 F

2 # F 1 whose roots are of the

form 2 2 R

2 whereR

is the discriminant. Write

R

= " Y with integer " $ Y

and notice that (R

) = (R

Y

2) = ( "Y

). Obviously we can get rid of all square

divisors of "Y

without changing the extension ( "Y

).

2. Cyclotomic extensions = (

) of where

is a primitive th root of

unity.

1.1.3. Definition. Let two fields $

contain a field

. A homo(iso)morphism : such that h is the identity map is called a

-homo(iso)morphism of

into .

The set ofall

-homomorphisms from to is denoted by Hom ( $ ). Noticethat every

-homomorphism is injective: its kernel is an ideal of

and 1 does not

belong to it, so the ideal is the zero ideal.

The set of all

-isomorphisms from to

is denoted by Iso ( $

).


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2002/2003 3

Two elements $

are calledconjugate over

ifthereisa

-isomorphism such that ( ) = . If $ are algebraic over

and isomorphic over

, they are

called conjugate over

.

Lemma. (i) Any two roots of an irreducible polynomial over

are conjugate.

(ii) An element is conjugate to over

iff " X

= " X .

(iii) The polynomial " X ( # ) is divisible by ( # F 4 ) in [ # ] , where 4 are all

distinct conjugate to elements, is the field

( 0 4

8

) generated by 4

over

.

Proof. (i) Let " ( # ) be an irreducible polynomial over

and $

be its roots in a

field extension of

. Then we have an

-isomorphism

( ) d

[ # ] ( "X

( # )) =

[ # ] ( "( # )) d

( )$

and therefore is conjugate to over

.

(ii) 0 = "X

( ) = "X

( ) = "X

( ), hence "

X= "

X

. If "X

= "X

, use (i).

(iii) If 4

is a rootof "X

then by the division algorithm "X ( # ) is divisible by # F 4

in [ # ].

1.1.4. Definition. A field is called algebraically closed if it does not have algebraic

extensions.

Theorem (without proof). Every field

has an algebraic extension m which is

algebraically closed. The field m is called an algebraic closure of

. Every two

algebraic closures of

are isomorphic over

.

Example. The field of rational numbers is contained in algebraically closed

field n . The maximal algebraic extension X

of is obtained as the subfield

of complex numbers which contains all algebraic elements over . The field X

is algebraically closed (it is much smaller than n , since its cardinality is countable

whereas the cardinality of complex numbers is uncountable).

Everywhere below we denote by m an algebraic closure of

.

Elements of Hom (

( ) $ m ) are in one-to-one correspondence with distinct roots

of "X

( # )

[ # ]: for each such root 4, as in the proof of (i) above we have

:

( ) m , 4 ; and conversely each such Hom (

( ) $ m ) maps to one

of the roots 4.

1.2. Galois extensions

1.2.1. Definition. A polynomial " ( # )

[ # ] is called separable if all its roots in m

are distinct.

Recall that if is a multiple root of " ( # ), then "( ) = 0. So a polynomial " is

separable iff the polynomials " and"

dont have common roots.


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4 Alg number theory

Example. Separable polynomials: irreducible polynomials over fields of characteris-

tic zero, irreducible polynomials over finite fields.

Proof: if " is an irreducible polynomial over a field of characteristic zero, then

its derivative " is non-zero and has degree strictly smaller than " ; and so if " has amultiple root, than a g.c.d. of " and " would be of positive degree strictly smaller than

" which contradicts the irreducibility of " . For the case of irreducible polynomials

over finite fields see section 1.3.

Definition. Let be a field extension of

. An element

is called separable

over

if " X (# ) is separable. The extension

is called separable if every element

of is separable over

.

Example. Every algebraic extension of a field of characteristic zero or a finite field is

separable.

1.2.2. Lemma. Let i be a field extension of

and be a finite extension of i .

Then every

-homomorphism

: i m can be extended to an

-homomorphism

: m

.

Proof. Let z i

and "X

( # ) = 1B

4#

4

be the minimal polynomial of over

i . Then ( " X )( # ) = 1 B

4#

4

is irreducible over i . Let be its root. Then

"X = " . Consider the

-homomorphism } : i [ # ] m , } ( 1 4 #4

) = 1 ( 4 ) 4

.

Its image is ( i )( ) and its kernel is generated by" X

. Since i [ # ] (" X

( # )) di

( ),

} determines an extension

: i ( ) m of . Since h : i ( ) h h : i h , by

induction can be extended to an

-homomorphism : m such that h = .

1.2.3. Theorem. Let be a finite separable extension of

of degree . Then there

exist exactly distinct

-homomorphisms of into m , i.e. h Hom ( $ m

) h =h

:

h .

Proof. Use the previous Lemma.

The first part: the number of distinct

-homomorphisms of into m is x

is

valid for any extension of degree . To prove this, argue by induction on h :

h and

use the fact that every

-homomorphism :

( ) m

sends to one of roots of

"X ( # ) and that root determines completely.

To show that there are distinct

-homomorphisms for separable

consider

first the case of =

( ). From separability we deduce that the polynomial "X

( # ) has

distinct roots 4

which give distinct

-homomorphisms of into m :

4.

Now argue by induction on degree. For z

consider i =

( ). There

are = h i :

h distinct

-homomorphisms 4 of i into m . Let 4

: m

be an extension of 4 which exists according to 1.2.2. By induction there are

distinct

(

4

(

))-homomorphisms

4 u

of

4

(

) intom

. Now

4 u

4

are distinct -homomorphisms of into m .

1.2.4. Proposition. Every finite subgroup of the multiplicative group

of a field

is cyclic.


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2002/2003 5

Proof. Denote this subgroup by , it is an abelian group of finite order. From the

standard theorem on the stucture of finitely generated abelian groups we deduce that

d 1

where 1 divides 2, etc. We need to show that = 1 (then is cyclic). If 1,

then let a prime be a divisor of 1 . The cyclic group 1 has elements

of order and similarly, 2 has elements of order , so has at least

2

elements of order . However, all elements of order in are roots of the polynomial

# F

1 which over the field

cannot have more than roots, a contradiction. Thus,

= 1.

1.2.5. Theorem. Let

be a field of characteristic zero or a finite field. Let be

a field extension of

of degree . Then there exists an element

such that

=

( ) =

[ ].

Proof. If

is of characteristic 0, then

is infinite. By 1.2.3 there are = h :

h

distinct homomorphisms 4 : m . Put 4 u = 0 : 4 ( ) = u ( )8

. Then 4 u

are proper

-vector subspaces of for D

= , and since

is infinite, 4 u

D

= .

Then there is z ( 4 u ). Since the set 0 4 ( )8

is of cardinality , the minimal

polynomial of over

has at least distinct roots. Then h

( ) :

h = h :

h

and hence =

( ).

If is finite, then

is cyclic by 1.2.4. Let be any of its generators. Then

=

( ).

1.2.6. Definition. An algebraic extension of

(inside m ) is called the splitting

field of polynomials " 4 if =

( 0

4u

8

) where 4 u are all the roots of " 4 .

An algebraic extension of

is called a Galois extension if is the splitting

field of some separable polynomials"

4

over

.Example. Let be a finite extension of

such that =

( ). Then

is a

Galois extension if the polynomial" X

( # ) of over

has deg" X

distinct roots in .

So quadratic extensions of and cyclotomic extensions of are Galois extensions.

1.2.7. Lemma. Let be the splitting field of an irreducible polynomial " (# )

[ # ].

Then ( ) = for every Hom ( $ m

).

Proof. permutes the roots of " (# ). Thus, ( ) =

( ( 1) $ ' ' ' $ (

@

)) = .

1.2.8. Theorem. A finite extension of

is a Galois extension iff ( ) = for every Hom (

$ m) and h Hom (

$ ) h =

h :

h .

The set Hom

( $

) = Iso

( $

) with respect to the composite of field isomor-

phisms is a finite group, called the Galois group Gal(

) of the extension

.

Sketch of the proof. Let be a Galois extension of

, then by 1.2.5 =

( ) for

some

. is the splitting field of some polynomials "4

over

, and hence


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6 Alg number theory

is the splitting field of their product. By 1.2.7 and induction = . Due to the

desciption of in 1.1.4 we deduce =

( 4 ) for any root 4 of " X , and elements of

Hom

( $

) correspond to

4. Thus, elements of the Galois group correspond to

substitutions 4 .

1.2.9. Theorem (without proof). Let

be a finite Galois extension and i be an

intermediate field between

and .

Then i is a Galois extension with the Galois group

Gal( i ) = 0 Gal(

) : h = id 8

'

For a subgroup of Gal(

) denote

= 0 : ( ) = for all

8

'

This set is an intermediate field between and

.

1.2.10. Main theorem of Galois theory (without proof). Let

be a finite Galoisextension with Galois group = Gal(

).

Then

is a one-to-one correspondence between subgroups of and

subfields of which contain

; the inverse map is given by i Gal( i ). We

have Gal( i ) = .

Normal subgroups of correspond to Galois extensions i

and

Gal( i

) d '

1.3. Finite fields

Every finite field

has positive characteristic, since the homomorphism

is

not injective. Let

be of prime characteristic . Then the image of in

can

be identified with the finite field

consisting of elements. If the degree of

is , then the number of elements in

is @

. By 1.2.4 the group

is cyclic of

order @

F 1, so every non-zero element of

is a root of the polynomial #

G

1F 1.

Therefore, all @

elements of

are all @

roots of the polynomial "@

(# ) = # F # .

The polynomial "@

is separable, since its derivative in characteristic is equal to

@

#

G

1F 1 = F 1. Thus,

is the splitting field of "@

over

. We conclude that

is a Galois extension of degree = h

:

h .

Lemma. The Galois group of

is cyclic of order : it is generated by an auto-

morphism}

of

called the Frobenius automorphism:

} ( ) = for all

'

Proof. }

( ) = = iff h

.


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2002/2003 7

On the other hand, for every

1 the splitting field of "@

over

is a finite field

consisiting of @

elements.

Thus,

Theorem. For every there is a unique (up to isomorphism) finite field

consisting

of @

elements; it is the splitting field of the polynomial "@

( # ) = # F # . The finite

extension

is a Galois extension with cyclic group of degree generated

by the Frobenius automorphism }@

:

.

Lemma. Let Y ( # ) be an irreducible polynomial of degree over a finite field

.

Then Y ( # ) divides "@

( # ) and therefore is a separable polynomial.

Proof. Let bea rootof Y ( # ). Then

( )

is of degree , so

( ) =

.

Since is a root of "@

( # ), Y divides "@

. The latter is separable and so is Y .


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8 Alg number theory

2. Integrality

2.1. Integrality over rings

2.1.1. Proposition Definition. Let be a ring and be its subring. An element

is called integral over if it satisfies one of the following equivalent conditions:

(i) there exist 4

such that " ( ) = 0 where " ( # ) = #@

+ @

G

1 # @ G1 + + 0 ;

(ii) the subring of generated by and is an -module of finite type;

(iii) there exists a subring m of which contains and and which is an

-module of finite type.

Proof. (i) (ii): note that the subring [ ] of generated by and coincides

with the -module i generated by 1 $ ' ' ' $ @ G 1 . Indeed,

@

+u = F 0 u

F F

@

+ uG

1

and by induction u

i

.

(ii) (iii): obvious.

(iii) (i): let m =B

1 + +B

. Then B

4 = 1u

4

u

B

u , so 1u

( 4 u F

4u )

B

u = 0.

Denote byR

the determinant of ( 4 u F 4 u ). Note thatR

= " ( ) where " ( # ) [ # ]

is a monic polynomial. Now from linear algebra we getR

B

u = 0 for all 1 x x .

ThenR

B

= 0 for allB

m , in particular,

R

1 = 0, so " ( ) =R

= 0.

Examples. 1. Every element of is integral over .

2. If $ are fields, then an element is integral over iff is algebraic

over .3. Let = , = . A rational number

with relatively prime and

is integral over iff (

)@

+ @

G

1( ) @ G1 +

+ 0 = 0 for some integer 4 .

Multiplying by @

we deduce that divides @

, hence = 1 and . Hence

integral in elements over are just all integers.

4. If is a field, then it contains the field of fractions

of . Let

Hom ( $ m ) where m is an algebraically closed field containing . If is

integral over , then ( ) ( ) is integral over .

5. If

is a root of a non-zero polynomial " ( # ) = @

#

@

+

[ # ], then

Y ( @

) = 0 for Y ( # ) = #@

+ @

G

1 # @ G1 + +

@

@

0 , so @

is integral over . Thus,

for every algebraic over element of there is a non-zero -multiple

which

is integral over .

2.1.2. Corollary. Let be a subring of an integral domain . Let be an -module

of finite type,

. Let

satisfy the property

. Then is integral over

.


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2002/2003 9

Proof. Indeed, as in the proof of ( ) ( ) we deduce thatR

B

= 0 for allB

.

Since is an integral domain, we deduce thatR

= 0, soR

= " ( ) = 0.

2.1.3. Proposition. Let

be a subring of a ring

, and let

4

be such that

4

isintegral over [ 1 $ ' ' ' $ 4G

1] for all . Then [ 1 $ ' ' ' $ @

] is an -module of finite

type.

Proof. Induction on . = 1 is the previous proposition. If m = [ 1 $ ' ' ' $ @

G

1]

is an -module of finite type, then m = 1 4 =1

B

4

. Now by the previous proposition

m [ @

] is a m -module of finite type, so m [ @

] = 1 u =1

R

um . Thus, m [

@

] = 14

u

R

u

B

4

is an -module of finite type.

2.1.4. Corollary 1. If 1 $ 2 are integral over , then 1 + 2 $ 1 F 2 $ 1 2 are

integral over .

Certainly 1 2 isnt necessarily integral over .

Corollary 2. The set of elements of which are integral over is a subring of

containing .

Definition.

is called the integral closure of in . If is an integral domain

and is its field of fractions,

is called the integral closure of .

A ring is called integrally closed if is an integral domain and coincides

with its integral closure in its field of fractions.

Let

be an algebraic number field. The integral closure of in

is called the

ring

of (algebraic) integers of

.

Examples. 1. A UFD is integrally closed. Indeed, if =

with relatively prime

$

is a root of polynomial"

(#

) =#

@

+

+

0

[#

], then

divides

@

,so is a unit of and

.

In particular, the integral closure of in is .

2.

is integrally closed (see below in 2.1.6).

2.1.5. Lemma. Let be integrally closed. Let be a field. Then an element is

integral over iff the monic irreducible polynomial "

( # )

[ # ] over the fraction

field

of has coefficients in .

Proof. Let be a finite extension of

which contains and all ( ) for all

-homomorphisms from to an algebraically closed field m . Since

is integral

over , ( ) is integral over for every . Then " ( # ) = ( # F ( )) has

coefficients in

which belong to the ring generated by and all ( ) and therefore

are integral over . Since is integrally closed, " ( # )

[ # ].

Examples. 1. Let

be an algebraic number field. Then an element

is integral

iff its monic irreducible polynomial has integer coefficients.


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10 Alg number theory

For example, R

for integerR

is integral.

IfR

1 mod 4 then the monic irreducible polynomial of (1 + R

) 2 over is

#

2F

# + (1 FR

) 4 [ # ], so (1 + R

) 2 is integral.

The integral closure of in (R

) contains the subring [R

].

An element +

R

with integer

and

D

= 0 is integral if # 2 F 2 #

+ ( 2 FR

2)

[ # ]. If = (2 e + 1) 2 for an integer e , then 2 FR

2

iff = (2 + 1) 2 with

integer and (2 e + 1)2 FR

(2 + 1)2 is divisible by 4. The latter is equivalent toR

being a quadratic residue mod 4, i.e.R

1 mod 4. Now, ifR

1 mod 4 then every

element (2 e + 1) 2 + (2 + 1)R

is integral.

Thus, integral elements of (R

) are equal to

[R

] ifR

D

1 mod 4

[(1 + R

) 2] if R

1 mod 4

2.

is equal to [

] (see section 2.4).

2.1.6. Definition. is said to be integral over if every element of is integral

over . If is of characteristic zero, its elements integral over are called integral

elements of .

Lemma. If is integral over and m is integral over , then m is integral over

.

Proof. LetB

m be a root of the polynomial " ( # ) = #

@

+ @

G

1 # @ G1 + + 0 with

4

. ThenB

is integral over [ 0 $ ' ' ' $ @

G

1]. Since 4 are integral over ,

proposition 2.1.3 implies that [ 0 $ ' ' ' $ @

G

1 $B

] is an -module of finite type. From

2.1.1 we conclude that B is integral over

.

Corollary. is integrally closed

Proof. An element of

integral over is integral over due to the previous

lemma.

2.1.7. Proposition. Let be an integral domain and be its subring such that is

integral over . Then is a field iff is a field.

Proof. If is a field, then [ ] for z 0 is a vector space of finite dimension over

, and the -linear map " : [ ] [ ] $ " (B

) = B

is injective, therefore surjective,

so is invertible in .

If isa fieldand z

0, then the inverse G

1

satisfies

G@

+ @

G

1 G @+1 +

+ 0 = 0 with some 4 . Then G1 = F

@G

1 F F 0 @ G1 , so

G

1

.


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2002/2003 11

2.2. Norms and traces

2.2.1. Definition. Let

be a subring of a ring

such that

is a free

-moduleof finite rank . For

its trace Tr

( ), norm

( ) and characteristic

polynomial Y

( # ) are the trace, the norm and the characteristic polynomial of the

endomorphism :

, (B

) = B

. In other words, Y ( # ) = det( # F i

),

Tr ( ) = Tr i , = det i .

2.2.2. First properties.

Tr( + ) = Tr( ) + Tr( ) $ Tr(

) = Tr( ) $ Tr( ) = $

( ) = ( ) ( ) $ ( ) = @

( ) $ ( ) = @

for

.

2.2.3. Everywhere below in this section

is either a finite field of a field of charac-

teristic zero. Then every finite extension of

is separable.

Proposition. Let be an algebraic extension of

of degree . Let

and

1 $ ' ' ' $ @

be roots of the minimal polynomial of over

each one repeated h :

( ) h times. Then the characteristic polynomial Y ( # ) of with respect to

is

( F 1)

:

( #

F 4) , and Tr

( ) = 1 4 $

( ) =

4.

Proof. If =

( ), then use the basis 1 $ $ ' ' ' $ @

G

1 to calculate Y . Let " ( # ) =

#

@

+B

@G

1 # @ G1 +

+

B

0 be the monic irreducible polynomial of over

, then the

matrix of

is

i

=

0 1 0 ' ' ' 0

0 0 1'

' '

0......

......

...

F

B

0 FB

1 FB

2 ' ' ' FB

@G

1

'

Hence det( # F i ) = " ( # ) and det i =

4 , Tr i = 1 4 .

In the general case when h

( ) :

h =

choose a basis 1 $ ' ' ' $ @

of

over

( ) and take 1 $ ' ' ' $ 1 G1

$ 2 $ ' ' ' $ 2 G

1$

' ' 'as a basis of over

. The matrix i

is a block matrix with the same block repeated

times on the

diagonal and everything else being zero. Therefore,Y

( # ) = ( F 1)

: "

( # )

: ( )

where " ( # ) is the monic irreducible polynomial of over

.

Example. Let

= , = (R

) with square-free integerR

. Then

Y

X + ( # ) = ( # F F

R

)(# F + R

) = # 2 F 2 # + ( 2 FR

2) $

so

Tr ( ) ( +

R

) = 2 $

( ) ( +

R

) = 2 FR

2'


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In particular, an integer numberB

isa sum of two squares iffB

(G

1) (G

1) .

More generally,B

is in the form 2 + T 2 with integer $

and square-free T iff

B

( )

[

T

]

2.2.4. Corollary 1. Let 4 be distinct

-homomorphisms of into m . Then

Tr

( ) = 1 4 ,

( ) = 4 ( ).

Proof. In the previous proposition 4

= 4 ( ).

Corollary 2. Let be an integral domain,

be its field of fractions. Let be an

extension of

of finite degree. Let be the integral closure of in

. Then for

an integral element

over Y

( # )

[ # ] and Tr

( ) $

( ) belong to

.

Proof. All 4

are integral over .

Corollary 3. If, in addition, is integrally closed, then Tr

( ) $

( ) .

Proof. Since is integrally closed,

= .

2.2.5. Lemma. Let

be a finite field of a field of characteristic zero. If is a finite

extension of

and i

is a subextension of

, then Tr

= Tr

Tr

and

=

.

Proof. Let 1 $ ' ' ' $

be all distinct

-homomorphisms of i into m ( = h i :

h ). Let 1 $ ' ' ' $ @

be all distinct i -homomorphisms of into m (

=h

:

ih ). The field u ( ) is a finite extension of

, and by 1.2.5 there is an element u m

such that

u

(

) =

(

u

). Let

be the minimal subfield ofm

containingi

and all

u. Using 1.2.3 extend 4 to

4

: m

. Then the composition 4

u

: m

is

defined. Note that 4

u =

4

1

u

1implies 4 =

4

u h =

4

1

u

1h

= 41

, so

= 1, and then = 1. Hence

4

u

for 1 x x $

1 x x

are all distinct

-homomorphisms of into m . By Corollary 3 in 2.2.4

(

( )) =

(

u ( )) = 4

(

u ( )) = ( 4

u )( ) =

( ) '

Similar arguments work for the trace.

2.3. Integral basis

2.3.1. Definition. Let be a subring of a ring such that is a free -module

of rank . Let 1 $ ' ' ' $ @

. Then the discriminant ( 1 $ ' ' ' $

@

) is defined as

det(Tr

( 4 u

)).


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2002/2003 13

2.3.2. Proposition. IfB

4 and

B

4 = 1 4 u u , 4 u , then (

B

1 $ ' ' ' $B

@

) =

(det( 4 u ))2 ( 1 $ ' ' ' $ @

).

Proof. (Tr(B

B

)) = (

4

)(Tr(

4 u

))(

u

)

.2.3.3. Definition. The discriminant of over is the principal ideal of

generated by the discriminant of any basis of over .


contain a non zero-divisor. Then a set 1 $ ' ' ' $ @

is a

basis of over iff ( 1 $ ' ' ' $ @

) = .

Proof. Let ( 1 $ ' ' ' $ @

) =

. If

contains a non zero-divisor, then

R

= ( 1 $ ' ' ' $ @

) is not a divisor of zero. LetB

1 $ ' ' ' $B

@

be a basis of over and

let 4

= 1u

4 u

B

u . ThenR

= det( 4 u

)2 (B

1 $ ' ' ' $B

@

).

If ( 1 $ ' ' ' $ @

) = (B

1 $ ' ' ' $B

@

) , we get R

= (B

1 $ ' ' ' $B

@

) for some

.

ThenR

(1 F det( 4 u )2) = 0 and det( 4 u ) is invertible in , so the matrix ( 4 u ) is

invertible and 1 $ ' ' ' $

@

is a basis of

over

.


be a finite field or a field of characteristic zero. Let be

an extension of

of degree and let 1 $ ' ' ' $

@

be distinct

-homomorphisms of

into m . Let 1 $ ' ' ' $ @

be a basis of over

. Then

( 1 $ ' ' ' $ @

) = det( 4 ( u ))2D

= 0 '

Proof. det(Tr( 4 u

)) = det( 1

( 4)

( u

)) = det((

( 4)) (

( u

))) = det( 4 ( u

))2 .

If det( 4 ( u

)) = 0, then there exist 4

not all zero such that 14

4

4 ( u

) = 0 for

all . Then 14

4

4 ( ) = 0 for every . Let 1 4

4 ( ) = 0 for all with the

minimal number of non-zero

4

. Then 1

4

4 ( B

) = 1

4

4 ( ) 4 (B

) = 0 for every

B

and 1(B

)( 1

4

4 ( )) F 1

4

4 ( ) 4 (B

) = 14

1 4 (

1(B

) F 4 (B

)) 4 ( ) = 0, a

contradiction.

Corollary. Underthe assumptions of the proposition the linear map

Hom ( $

):

(B

Tr

( B

)) between -dimnesional

-vector spaces is injective, and hence

bijective. Therefore for a basis 1 $ ' ' ' $ @

of

there is a dual basisB

1 $ ' ' ' $B

@

of

such that Tr

( 4B

u ) = 4 u .

Proof. If = 1 4 4

, 4

and Tr

( B

) = 0 for allB

, then we get equations

1

4 Tr

( 4 u ) = 0 this is a system of linear equations in 4 with nondegenerate

matrix Tr

( 4 u ), so the only solution is 4 = 0.

2.3.6. Theorem. Let be an integrally closed ring and

be its field of fractions. Let

be an extension of

of degree and be the integral closure of in . Let

be of characteristic 0. Then

is an -submodule of a free -module of rank .

Proof. Let T 1 $ ' ' ' $ T@

be a basis of

-vector space . Then due to Example 5 in

2.1.1 there is

such that 4

= T 4

form a basis of

.


14/38


LetB

1 $ ' ' ' $B

@

be the dual basis for 1 $ ' ' ' $ @

. Claim: 1B

4 . Indeed, let

B

= 1 4B

4 . Then 4 = Tr

(B

4 ) by 2.2.5. Now 1

B

4 =

B

4 , since

0

B

4

8

is a basis of

.

2.3.7. Theorem (on integral basis). Let be a principal ideal ring and

be its

field of fractions of characteristic 0. Let be an extension of

of degree . Then

the integral closure of in is a free -module of rank .

In particular, the ring of integers of a number field

is a free -module of

rank equal to the degree of

.

Proof. The description of modules of finite type over PID and the previous theorem

imply that is a free -module of rank x . On the other hand, by the first part

of the proof of the previous theorem

contains

-linear independent elements

over . Thus, = .

Definition. The discriminantR

of anyintegral basis of

is called the discriminant

of

. Since every two integral bases are related via an invertible matrix with integer

coefficients (whose determinant is therefore 1 ), 2.3.2 implies thatR

is uniquely

determined.

2.3.8. Examples. 1. LetR

be a square-free integer. By 2.1.5 the ring of integers of

(R

) has an integral basis 1 $r

where r = R

if D

1 mod 4 and r = (1 + R

) 2

ifR

1 mod 4.

The discriminant of (R

) is therefore equal to

4R

ifR

D

1 mod 4 , andR

ifR

1 mod 4 '

2. Let

be an algebraic number field of degree and

be an integral

element over . If (1 $ $ ' ' ' $ @

G

1) is a square free integer, then proposition 2.3.2

implies that 1 $ $ ' ' ' $ @

G

1 is a basis of over , so = [ ].

2.3.9. Example. Let

be of characteristic zero and =

( ) be an extension of

degree over

. Let " ( # ) be the minimal polynomial of over

whose roots are

4 . Then (1 $

$ ' ' ' $

@G

1) = det( u

4

)2 = 4

= u ( 4 F u )2 = ( F 1)

@

(@

G

1) 2

4=u ( 4 F u ) =

( F 1)@

(@

G

1) 2

( " ( )).

Let " ( # ) = #@

+ #

+B

. Then T ="

( ) = ( F F

B

G

1) + , so

= F B

( T + ( F 1) )G

1 . The minimal polynomial Y ( ) of T over

is the numerator

ofB

G

1" ( F

B

( +( F 1) )G

1), i.e. ( +( F 1) )@

F ( +( F 1) )

@G

1 +( F 1)@

@

B

@G

1.

Hence

( " ( )) = the free term of Y ( ) times ( F 1)@

G

1 = @

B

@G

1+( F 1)@

G

1( F 1)@

G

1

@

$

so

(1 $ $ ' ' ' $ @

G

1) = ( F 1)@

(@

G

1) 2( @

B

@G

1 + ( F 1)@

G

1( F 1)@

G

1

@

) '

For = 2 one has 2 F 4B

, for = 3 one has F 27B

2F 4 3.


15/38

2002/2003 15

For example, let " ( # ) = # 3 + # + 1. It is irreducible over . Its discriminant is

equal to ( F 31), so according to example 2.5.3 = [ ] where is a root of " ( # )

and

= [ ].

2.4. Cyclotomic fields

2.4.1. Definition. Let @

be a primitive th root of unity. The field ( @

) is called

the ( th) cyclotomic field.

2.4.2. Theorem. Let be a prime number and be a primitive th root of unity. The

cyclotomic field (

) is of degree F 1 over . Its ring of integers coincides with

[

].

Proof. Denote =

. Let " ( # ) = ( # F 1) ( #F

1) = # G

1 +

+ 1. Recall that

F 1 is a root of the polynomial Y ( ) = " (1 + ) =

G

1 + + is a -Eisenstein

polynomial, so " ( # ) is irreducible over ,h

( ) : h

= F

1 and 1 $ $ ' ' ' $

G

2 is

a basis of the -vector space ( ).

Let be the ring of integers of ( ). Since the monic irreducible polynomial

of over has integer coefficients,

. Since G

1 is a primitive root of unity,

G

1

. Thus, is a unit of .

Then 4

for all

( G

1 = G

1 ). We have 1 F 4

= (1 F )(1+ + 4

G

1)

(1 F

) .

Denote by Tr and the trace and norm for ( )

. Note that Tr( ) = F 1 and

since 4

for 1 x x F 1 are primitive th roots of unity, Tr(4

) = F 1; Tr(1) = F 1.

Hence

Tr(1 F 4

) = for 1 x x F 1 .

Furthermore, ( F

1) is equal to the free term of Y ( ) times ( F 1)G

1 , so ( F

1) =

( F 1)G

1 and

(1 F ) =

1

4

G

1

(1 F 4

) = $

since 1 F

4

are conjugate to 1 F

over . Therefore

is contained in the ideal

= (1 F

)

.

If = , then 1 F would be a unit of and so would be its conjugates 1 F 4

,

which then implies that as their product would be a unit of . Then G

1

= ,

a contradiction. Thus,

= (1 F ) = '

Now we prove another auxiliary result:

Tr((1 F

) ) '


16/38


Indeed, every conjugate of (1 F ) for is of the type 4 (1 F 4

) with appropriate

4 , so Tr( (1 F )) = 1 4 (1 F

4

) = .

Now let = 1 0

4

G

2 4 4

with 4

. We aim to show that all 4

belong

to . From the calculation of the traces of 4

it follows that Tr((1 F ) ) = 0 Tr(1 F )+1 0

4

G

2 4 Tr(4

F

4 +1) = 0 and so 0 Tr((1 F ) ) ; therefore, 0 .

Since isaunitof , we deduce that 1 = G1(

F 0) = 1+ 2 + +

G

2 G3

.

By the same arguments 1 . Looking at 4 = G1( 4

G

1 F 4G

1) we conclude

4 for all . Thus = [ ].

2.4.3. In general, the extension (

) is a Galois extension and elements of the

Galois group Gal( (

)

) are determined by their action on the primitive th root

of unity:

: (

) = 4

$ ( $ ) = 1 '

This map induces a group isomorphism

Gal( (

) ) ( )

'

One can prove that the ring of integers of (

) coincides with [

].


17/38

2002/2003 17

3. Dedekind rings

3.1. Noetherian rings

3.1.1. Recall that a module i over a ring is called a Noetherian module if one of the

following equivalent properties is satisfied:

(i) every submodule of i is of finite type;

(ii) every increasing sequence of submodules stabilizes;

(iii) every nonempty family of submodules contains a maximal element with respect

to inclusion.

A ring is called Noetherian if it is a Noetherian -module.

Example. A PID is a Noetherian ring, since every ideal of it is generated by one

element.

Lemma. Let i be an -module and is a submodule of i . Then i is a Noethe-

rian -module iff and i are.

Corollary 1. If 4 are Noetherian -modules, so is @4 =1 4 .

Corollary 2. Let be a Noetherian ring and let i be an -module of finite type.

Then i is a Noetherian -module.

3.1.2. Proposition. Let be a Noetherian integrally closed ring. Let be its field of

fractions and let be a finite extension of . Let be the integral closure of in

. Suppose that is of characteristic 0. Then is an -module of finite type andis a Noetherian ring.

Proof. According to 2.3.8 is a submodule of a free -module of finite rank. Hence

is a Noetherian -module. The ideals of are -submodules of , so every

increasing sequence of them stabilizes.

3.1.3. Example. The ring of integers of a number field

is a Noetherian ring.

It is a -module of rank where is the degree of

.

Every nonzero element of

z0 0

8

factorizes into a product of prime elements

(not uniquely in general).

Indeed, assume the family of principal ideals ( ) which are generated by elements

which are not products of prime elements is nonempty and then choose a maximalelement ( ) in this family. The element isnt prime, i.e., there is a factorization

= B

with both $

B

D

. Then ( ) $ (B

) are strictly larger than ( ), so andB

are

products of prime elements. Then is, a contradiction.


18/38


3.2. Dedekind rings

3.2.1. Definition. An integral domain is called a Dedekind ring if(i) is a Noetherian ring;

(ii) is integrally closed;

(iii) every non-zero (proper) prime ideal of is maximal.

Example. Every principal ideal domain is a Dedekind ring: for (i) see 3.1.1 and

for (ii) see 2.1.4. If ( ) is a prime ideal and ( ) ( )D

= , then isnt a unit of and

divides . Write = B

. Since ( ) is prime, either orB

belongs to ( ). Since

doesnt,B

must belong to ( ), soB

= R

for someR

. Therefore = B

= R

which means that is a unit of , a contradiction. Thus, property (iii) is satisfied as

well.

3.2.2. Lemma. Let be an integral domain. Let be its field of fractions and let

be a finite extension of . Let be the integral closure of in . Let be a

non-zero prime ideal of . Then is a non-zero prime ideal of .

Proof. Let be a non-zero prime ideal of . It is clear thatB

$

R

andB

R

implies that eitherB

orR

. Hence

is a prime ideal of .

Let

, D

= 0. Then satisfies a polynomial relation @

+ @

G

1 @ G1+

+ 0 = 0

with 4

. We can assume that 0D

= 0. Then 0 , so is a non-zero

prime ideal of .

3.2.3. Theorem. Let be a Dedekind ring. Let be its field of fractions and let

be a finite extension of . Let be the integral closure of in . Suppose that

is of characteristic 0. Then is a Dedekind ring.

Proof. is Noetherian by 3.1.2. It is integrally closed due to 2.1.6. By 3.2.2 if is a

non-zero proper prime ideal of , then is a maximal ideal of . The quotient

ring

is integral over ( ), and by 2.1.7 is a field. Thus, is a maximal

ideal of .

3.2.4. Example. The ring of integers of a number field

is a Dedekind ring.

3.3. Factorization in Dedekind rings

3.3.1. Lemma. Every non-zero ideal in a Dedekind ring contains a product of

maximal ideals.


19/38

2002/2003 19

Proof. If not, then the set of non-zero ideals which do not contains products of maximal

ideals is non-empty. Let be a maximal element with this property. The ideal isnt

maximal, since it doesnt contain a product of maximal ideals. Therefore there are

$

such that

and

$

D

. Since

+

and

+

are strictlygreater than , there are maximal ideals 4 and u such that 4 + and

u

+

. Then 4 u

( +

)( +

)

, a contradiction.

3.3.2. Lemma. Let a prime non-zero ideal contain a product 1 ' ' '

of ideals.

Then contains one of them.

Proof. If

D

for all 1 xe x

, then take

z

and consider the product

1 ' ' '

. It belongs to , therefore one of 4

belongs to , a contradiction.

3.3.3. The next proposition shows that for every non-zero ideal of a Dedekind ring

there is an ideal such that

is a principal non-zero ideal of . Moreover, the

proposition gives an explicit description of .

Proposition. Let be a non-zero ideal of a Dedekind ring and be a non-zero

element of . Let be the field of fractions of . Define

= 0

:

8

'

Then is an ideal of and = .

Proof. We have . For we have .

Use Noetherian and integrality property of Dedekind rings: Since is an -module

of finite type, by Remark in 2.1.1 is integral over . Since is integrally closed,

.

The set is closed with respect to addition and multiplication by elements of ,

is an ideal of . It is clear that . Assume that D

= and get acontradiction.

Since G

1

is a proper ideal of , and hence it is contained in a maximal ideal

. Note that

. So 2

and

G

1

,

G

1

. By 3.3.1 there are

non-zero prime ideals 4

such that 1 ' ' '

. Let be the minimal number

with this property.

We have

1 ' ' '

G

1

'

By 3.3.2 contains one of 4. Without loss of generality we can assume that 1 .

Since 1 is maximal, 1 = .

If = 1, then

G

1

, so =

. Since

we get

,

so = or = . The definition of implies in the first case = and =

and in the second case = and again = .

Let

1. Note that 2 ' ' '

D

. Therefore there isR

2 ' ' '

such

thatR

D

. ThenR

G

1

R

2 ' ' '

. ThereforeR

G

1

.


20/38


Since is an -module of finite type, by 2.1.1R

G

1 belongs to , i.e.R

, a

contradiction.

3.3.4. Corollary 1 (Cancellation property). Let

$ $

be non-zero ideals of

,then = implies = .

Proof. Let be an ideal such that = is a principal ideal. Then =

and = .

3.3.5. Corollary 2 (Factorization property). Let and be ideals of . Then

if and only if =

for an ideal .

Proof. If and is non-zero, then let be an ideal of such that =

is a principal ideal. Then

, so = G

1

is an ideal of . Now

=

G

1

= G

1

= G

1

=

'

3.3.6. Theorem. Every proper ideal in a Dedekind domain is uniquely representable

as a product of maximal ideals.

Proof. Let be a non-zero ideal of . There is a maximal ideal 1 which contains

. Then by factorization property 3.3.5 = 1 1 for some ideal 1. Note that

1 is a proper inclusion, since otherwise 1 = 1 = 1 1 and by cancellation

property 3.3.4 1 = , a contradiction. If 1D

= , then there is a maximal ideal 2such that 1 = 2 2. Continue this way: eventually we have = 1 ' ' '

@

@

and

1

@

are all proper inclusions. Since is Noetherian,

= for

some and then = 1 ' ' '

.

If 1 ' ' '

= 1 ' ' ' @

, then 1 1 ' ' ' @

and by 3.3.2 1 being a prime

ideal contains one of

4

, so

1 =

4

. Using 3.3.4 cancel

1 on both sides and useinduction.

3.3.7. Remark. A maximal ideal of is involved in the factorization of iff

.

Indeed, if

, then =

by 3.3.5.

3.3.8. Example. Let = [ F 5]. This is a Dedekind ring, since F 5D

1 mod 4.

We have the norm map ( + F 5) = 2 + 5 2. If an element is a unit of

then

= 1 for some

, and the product of two integers ( ) and ( ) is 1,

thus ( ) = 1. Conversely, if ( ) = 1 then times its conjugate is one, and so

is a unit of .

The norms of 2 $ 3 $ 1 F 5 are 4 $ 9 $ 6. It is easy to see that 2 $ 3 are not in the

image ( ).If, say, 2 were not a prime element in , then 2 = 1 2 and 4 = ( 1) ( 2)

with both norms being proper divisors of 4, a contradiction. Hence 2 is a prime element

of , and similarly 3 $ 1 F 5 are.


21/38

2002/2003 21

Now 2 $ 3 $ 1 F 5 are not associated with each other prime elements of and

6 = 2 3 = (1 + F 5)(1 F F 5) $

so

isnt a UFD.The ideals

(2 $ 1 + F 5) $ (3 $ 1 + F 5) $ (3 $ 1 F F 5)

are maximal.

For instance, h (2) h = 4, and it is easy to show that D

= (2 $ 1 + F 5)D

= (2), so

h (2 $ 1 + F 5) h = 2, therefore (2 $ 1 + F 5) is isomorphic to 2 , i.e. is a field.

Now we get factorization of ideals

(2) = (2 $ 1 + F 5)2 $

(3) = (3 $ 1 + F 5)(3 $ 1 F F 5) $

(1 + F 5) = (2 $ 1 + F 5)(3 $ 1 + F 5) $

(1 F F 5) = (2 $ 1 + F 5)(3 $ 1 F F 5) '

So

(2) (3) = (2 $ 1 + F 5)2(3 $ 1 + F 5)(3 $ 1 F F 5)

= (2 $ 1 + F 5)(3 $ 1 + F 5)(2 $ 1 + F 5)(3 $ 1 F F 5)

= (1 + F 5)(1 F F 5) '

3.3.9. Lemma. Let + = . Then @

+ = for every $ Y T 1.

Proof. We can assume = 1. Now = ( + ) '' '

( + ) = ( '' '

) +

+

,

so + = >

Proposition. Let be a maximal ideal of . Then there is an element suchthat

=

+ @

for every

2.

Hence the ideal @

is a principal ideal of the factor ring @

. Moreover, it

is the only maximal ideal of that ring.

Every ideal of the ring

@

is principal of the form

@

= (

+ @

)

@

for some x .

Proof. If = 2, then = by cancellation property, a contradiction. Let

z

2 . Since

+ @

, factorization property implies that

+ @

=

for

an ideal

.Note that D

, since otherwise 2 , a contradiction.

Therefore, + = . The Lemma implies @

G

1 + = . Then

= ( + @

G

1)

+ @

=

+ @

$


22/38


so = + @

.

We have + @

, so = +

@

.

Let be a proper ideal of containing @

. Then by factorization property

@ = with some ideal . Thus, factorization of involves powers of only, = , 0 x .

Remark.

@

is a local ring. Indeed, the ideal

@

is a maximal ideal of

@

,

since the quotient ring is isomorphic to

which is a field. Every proper ideal of

@

is of the form

@

@

.

3.3.10. Corollary. Every ideal in a Dedekind ring is generated by 2 elements.

Proof. Let be a non-zero ideal, and let be a non-zero element of . Then

= @

11 ' ' ' @

with distinct maximal ideals 4.

By Lemma 3.3.9 we have @ 11 + @

= if D

= e , so we can apply the Chinese

remainder theorem which gives

d @

11 @

'

For the ideal of we get

d ( + @ 11 ) @

11 ( + @

) @

'

Each of ideals ( + @

4

) @

4

is of the form (

4

+ @

4

) @

4

by 3.3.9. Hence

is isomorphic to (

4

+ @

4

) @

4

. Using the Chinese remainder theorem find

such that F

4

belongs to @

4

for all . Then = ( + ) and

= + .

3.3.11. Theorem. A Dedekind ring is a UFD if and only if is a PID.

Proof. Let

be not a PID. Since every proper ideal is a product of maximal ideals,there is a maximal ideal which isnt principal. Consider the family of non-zero ideals

such that

is principal. It is nonempty by 3.3.5. Let be a maximal element of

this family and

=

, D

= 0.

Note that isnt principal, because otherwise =

and

=

=

, so

is divisible by . Put = G

1 , then ( ) = ( )( ) and by 3.3.4 = ( ), a

contradiction.

Claim: is a prime element of . Proof: If = B

, then B

, so either

orB

. By 3.3.5 then either = or

B

= for an appropriate ideal of

. Since , we get = = and . Due to maximality

of we deduce that = , and hence either

orB

is equal to

. Then one of

$

B

is asociated to , so is a prime element.

D

, since otherwise = , so = , a contradiction.

D

, since otherwise

, and

= , is principal, a contradiction.

Thus, there areR

andT

not divisible by . We also have TR

=

is divisible by . This can never happen in UFD. Thus, isnt a UFD.


23/38

2002/2003 23

3.4. The norm of an ideal

In this subsection

is a number field of degree , is the ring of integers of

.

3.4.1. Proposition. For a non-zero element

h

: h

=h

( ) h

'

Proof. We know that

isa free -module of rank . The ideal

is a submodule

of

of rank , since if 1 $ ' ' ' $

are generators of

, then G

1 1 $ ' ' ' $ G

1

are generators of

, so

. By the theorem on the structure of modules over

principal ideal domains, there is a basis 1 $ ' ' ' $ @

of

such that T 1 1 $ ' ' ' $ T@

@

is a basis of

with appropriate T 1 h ' ' ' h T@

. Then

is isomorphic to

T4

, so h

:

H h=

h

T4

h. By the definition

( ) is equal to the

determinant of the matrix of the linear operator"

:

,

. Note that

has another basis: 1 $ ' ' ' $ @

, so ( 1 $ ' ' ' $ @

) = ( T 1 1 $ ' ' ' $ T@

@

) i with

an invertible matrix i with integer entries. Thus, the determinant of i is 1 and

( ) is equal to T 4 .

3.4.2. Corollary. h : h = h h

@

for a non-zero .

Proof.

( ) = @

.

3.4.3. Definition. The norm ( ) of a non-zero ideal of

is h

: h.

Note that if D

= 0 then ( ) is a finite number.

Indeed, by 3.4.1 (

) =h

( ) h for a non-zero which belongs to . Then

and ( ) x ( ) = h ( ) h .

3.4.4. Proposition. If $ are non-zero ideal of , then ( ) = ( ) ( ).

Proof. Since every ideal factors into a product of maximal ideals by 3.3.6, it is sufficient

to show that ( ) = ( ) ( ) for a maximal ideal of .

The LHS = h

: h

= h

: h h

: h

. Recall that is a maximal ideal of

, so

is a field.

The quotient can be viewed as a vector space over . Its subspaces

correspond to ideals between and according to the description of ideals of the

factor ring. If

, then by 3.3.5 =

for an ideal of

.

By 3.3.3 there is a non-zero ideal

such that

is a principal non-zero ideal

.

Then

implies

implies

. Therefore either

=

andthen =

or = and then = . Thus, the only subspaces of the vector space

are itself and the zero subspace

. Hence

is of dimension one over

and therefore

h :

h= h

:

h.


24/38


3.4.5. Corollary. If is a non-zero ideal of and ( ) is prime, then is a

maximal ideal.

Proof. If

=

, then

(

)

(

) is prime, so, say,

(

) = 1 and

=

. So

has not proper prime divisors, and therefore is a maximal ideal.

3.5. Splitting of prime ideals in field extensions

In this subsection

is a number field and is a finite extension of

. Let

and

be their rings of integers.

3.5.1. Proposition-Definition. Let be a maximal ideal of and a maximal

ideal of

. Then is said to lie over and is said to lie under if one of the

following equivalent conditions is satisfied:

(i)

;

(ii)

;

(iii)

= .

Proof. (i) is equivalent to (ii), since 1

. (ii) implies

contains , so

either = or = , the latter is impossible since 1D

. (iii)

implies (ii).

3.5.2. Proposition. Every maximal ideal of

lies over a unique maximal ideal of

. For a maximal ideal of the ideal

is a proper non-zero ideal of

.

Let

= 4 be the factorization into a product of prime ideals of

. Then 4

are exactly those maximal ideals of

which lie over .

Proof. The first assertion follows from 3.2.2.Note that by 3.3.3 for

z

2 there is an ideal of

such that = .

Then D

, since otherwise

2, a contradiction. Take an elementB

z

.

ThenB

.

If

=

, thenB

=B

, soB

G

1

=

and

B

, a contradiction. Thus,

is a proper ideal of

.

According to 3.5.1 a prime ideal of

lies over iff

which is

equivalent by 3.3.7 to the fact that is involved in the factorization of

.

3.5.3. Lemma. Let be a maximal ideal of which lie under a maximal ideal

of

. Then the finite field is a subfield of the finite field

.

Proof.

is finite by 3.4.3. The kernel of the homomorphism

isequal to = , so can be identified with a subfield of

.

3.5.4. Corollary. Let be a maximal ideal of

. Then

=

for a prime

number and ( ) is a positive power of .


25/38

2002/2003 25

Proof. = for a prime number by 3.2.2. Then is a vector space

over of finite positive dimension, therefore h : h is a power of .

3.5.5. Definition.Let a maximal ideal

of

lie under a maximal ideal

of

. The degree of

over

is called the inertia degree " ( h

). If

=

4

is the factorization of

with distinct prime ideals 4

of

, then

T 4 is called the ramification index T ( 4 h ).

3.5.6. Lemma. Let i be a finite extension of and be maximal

ideals of

,

and

correspondingly. Then " ( h

) = " ( h

) " ( h

) and

T ( h ) = T ( h ) T ( h ).

Proof. The first assertion follows from 1.1.1. Since

= (

)'

' ', we get

= (

)

= ( ) (

)

= ( ( )) (

)'

' ', so the second

assertion follows.

3.5.7. Theorem. Let 1 $ ' ' '

be different maximal ideals of

which lie over amaximal ideal of

. Let =

h :

h . Then

4 =1

T ( 4 h ) " ( 4 h ) = '

Proof. We consider only the case

= . Apply the norm to the equality

=

4

. Then by 3.4.2, 3.4.4

@

= (

) = ( 4 )

= (

) (

)'

3.5.8. Example. One can describe in certain situations how a prime ideal ( ) factorizes

in finite extensions of

, provided the factorization of the monic irreducible polynomialof an integral generator (if it exists) modulo is known.

Let the ring of integers

of an algebraic number field

be generated by one

element r : = [ r ], and " ( # ) [ # ] be the monic irreducible polynomial of r

over .

Let "4( # )

[ # ] be monic polynomials such that

" ( # ) =

4 =1

"4 ( # )

[ # ]

is the factorization of " ( # ) where "4( # ) is an irreducible polynomial over

. Since

d [ # ] ( " (# )), we have

( ) d [# ] ( $ " ( # )) d

[ # ] ( " ( # )) $

and

(

$ " 4( r )) d

[ # ] (

$ "( # ) $

" 4( # )) d

[ # ] ( "4( # )) '


26/38


Putting 4 = ( $ " 4 ( r )) we see that 4 is isomorphic to the field

[ # ] ( " 4 ( # )),

hence 4 is a prime ideal of dividing ( ). We also deduce that

(

4

) =h

[#

]

("

4

(#

)) :

h

=

deg

'

Now

4

= ( $ " 4 ( r ))

( ), since " 4 ( r )

= " ( r ) = 0. On the other

hand, if ( ) =

4

u

is the factorization of ( ) in

then, since already

( 4 )

= 1

deg

= @

, from 3.5.7 we conclude that ( ) = 4 =1

4

is the

factorization of ( ).

So we have proved

Theorem. Let the ring of integers

of an algebraic number field

be generated by

one element r : = [ r ] , and " ( # ) [# ] be the monic irreducible polynomial

of r over . Let " 4 ( # )

[ # ] be monic polynomials such that

"

(#

) =

4 =1

"4

(#

)

[#

]

is the factorization of " ( # ) where " 4 ( # ) is an irreducible polynomial over

.

Then in

( ) =

4 =1

4

where 4

= ($ " 4

( r )) is a maximal ideal of

with norm deg

.

Thus, remains prime in iff " is irreducible over

; splits ( =

1 ' ' '

) iff " is separable, and ramifies (

= ) iff " is a power of an

irreducible polynomial over

.

3.5.9. In particular, if

= (R

) then one can take R

forR

D

1 mod 4 and (1 +

R

) 2 forR

1 mod 4 as r . Then " ( # ) = # 2 FR

and " ( # ) = # 2 F#

+ (1 FR

) 4

resp.

We have # 2 F# + (1 F

R

) 4 = 1 4( 2 FR

) where = 2 # F 1, so if is

odd (so the image of 2 is invertible in

), the factorization of " ( # ) corresponds to

the factorization of # 2 FR

independently of whatR

is. The factorization of # 2 FR

certainly depends on whetherR

is a quadratic residue modulo , or not.

For = 2 " ( # ) = ( #F

R

)2 2 [ # ] and " ( # ) = #

2 + # + (1 FR

) 4 2 [ # ]

resp.

Thus, we get

Theorem. If is odd prime, then splits in = (

R

) iffR

is a quadratic residue mod .

ramifies in iffR

is divisible by .

remains prime in iffR

is a quadratic non-residue mod .


27/38

2002/2003 27

If = 2 then

ifR

1 mod 8, then 2 splits in (R

),

ifR

D

1 mod 4 then 2 ramifies in (R

);

ifR

1 mod 4 $R

D

1 mod 8 then 2 remains prime in (R

).

3.5.10. Let be an odd prime. Recall from 2.4.2 that the ring of integers of the

th cyclotomic field (

) is generated by

. Its irreducible monic polynomial is

" ( # ) = # G

1 + + 1 = ( # F 1) ( # F 1). Since # F 1

( # F 1) mod we

deduce that ( " ( # ) $ ) = (( # F 1)G

1$

). Therefore by 3.5.8 = (

F 1)G

1 ramifies

in (

)

. For any other prime the polynomial " ( # ) remains irreducible modulo

, since with its root r it contains r

. Thus, no other prime ramifies in (

)

.

3.6. Finiteness of the ideal class group

In this subsection is the ring of integers of a number field

.

3.6.1. Definition. For two non-zero ideals and of

define the equivalence

relation

if there are non-zero $

such that

=

. Classes of equiv-

alence are called ideal classes. Define the product of two classes with representatives

and as the class containing

. By 3.3.3 every ideal class is invertible, thus ideal

classes form an abelian group which is called the ideal class group m of the number

field

.

The ideal class group shows how far from PID the ring is. Note that m

consists of one element iff

is a PID iff

is a UFD.

3.6.2. Proposition. There is a positive real numberB

such that every non-zero ideal

of contains a non-zero element with

h

( ) h xB

( ) '

Proof. Let = h

: h . According to 2.5.9 there is a basis 1 $ ' ' ' $ @

of the

-module

which is also a basis of the -vector space

. Let 1 $ ' ' ' $

@

be all

distinct -homomorphisms of

into n . Put

B

=@

4 =1

@

u =1

h

4

uh

'

Then B 0.For a non-zero ideal let be the positive integer satisfying the inequality

@

x

( ) ( + 1)@

. In particular, h

: h

( + 1)@

. Consider ( + 1)@

elements1 @

u =1 u u with 0 x u x , u . There are two of them which


28/38


have the same image in . Their difference 0D

= = 1 @u =1 u u belongs to and

satisfies h u h x .

Now

h

( ) h =

@

4 =1

h

4 h =

@

4 =1

h

@

u =1

u

4 u h x

@

4 =1

@

u =1

h u h h

4 u h

x

@

B

x

B

( ) '

3.6.3. Corollary. Every ideal class of contains an ideal with ( ) xB

.

Proof. Given ideal class, consider an ideal of the inverse ideal class. Let be as

in the theorem. By 3.3.3 there is an ideal such that

=

, so ( )( ) = (

) = 1

in m

. Then belongs to the given ideal class. Using 3.4.1 and 3.4.4 we deduce that

( ) ( ) = (

) = (

) =h

( ) hx

B

( ). Thus, ( ) xB

.

3.6.4. Theorem. The ideal class group m

is finite. The number hm h

is called the

class number of

.

Proof. By 3.5.4 and 3.5.2 for each prime there are finitely many maximal ideals

lying over ( ), and ( ) =

for

1. Hence there are finitely many ideals

4

satisfying (

4

) xB

.

Example. The class number of ( F 19) is 1, i.e. every ideal of the ring of integers

of ( F 19) is principal.

Indeed, by 2.3.8 we can take 1 = 1, 2 = (1 + F 19) 2 as an integral basis of the

ring of integers of ( F 19). Then

B

=

1 + h (1 + F 19) 2 h

1 + h (1 F F 19) 2 h = 13 ' 5 ' ' ' '

So every ideal class of (

G

19) contains an ideal with ( ) x 13. Let =

4

be the factorization of

, then

(

4

)x

13 for every

.By Corollary 3.5.4 we know that (

4) is a positive power of a prime integer, say

4. From 3.5.2 we know that

4is a prime divisor of the ideal (

4) of

(G

19).

So we need to look at prime integer numbers not greater than 13 and their prime ideal

divisors as potential candidates for non-principal ideals. Now odd prime numbers 3 and

13 have the property that -19 is a quadratic non-residue modulo them, so by Theorem

3.5.9 they remain prime in (

G

19) .

Odd prime numbers 5, 7, 11 have the property that -19 is a quadratic residue module

them, so by Theorem 3.5.9 they split in (

G

19). It is easy to check that

5 =

(1 + F 19) 2

(1 F F 19) 2 $

7 =

(3 + F 19) 2

(3 F F 19) 2 $

11 =

(5 + F 19) 2

(5 F F 19) 2 '

Each of ideals generated by a factor on the right hand side is prime by 3.4.5, since its

norm is a prime number. So prime ideal factors of (5) $ (7) $ (11) are principal ideals.


29/38

2002/2003 29

Finally, 2 remains prime in (

G

19) , as follows from 3.5.9.

Thus, every prime divisor of a prime integer not greater than 13 is a principal ideal,

and therefore (

G

19) is a principal ideal domain.

Remark. The bound given byB

is not good in practical applications. A more refined

estimation is given by Minkowskis Theorem 3.6.6.

3.6.5. Definition. Let 1 $ ' ' ' $

@

be -homomorphisms of

into n . Let

: n n

be the complex conjugation. Then

4 is a -homomorphisms of

into n , so it

is equal to certain u . Note that 4 = 4 iff 4 (

) . Let 1 be the number

of -homomorphisms of this type, say, after renumeration, 1 $ ' ' ' $

1. For every

1 we have

u

D

= u , so we can form couples ( u $ u ). Then F 1 is an

even number 2 2, and 1 + 2 2 = .

Renumerate the

u s so that

4 + 2 =

4 for 1 + 1 x x 1 + 2. For

define the canonical embedding of

by

( ) = ( 1( ) $ ' ' ' $

1+ 2 ( ))

1

n

2'

The field

is isomorphic to its image (

)

1

n

2 . The image (

) is called

the geometric image of

and it can be partially studied by geometric tools.

3.6.6. Minkowskis Bound Theorem. Let

be an algebraic number field of degree

with parameters 1 $ 2 . Then every class of m contains an ideal such that its

norm ( ) satisfies the inequality

( ) x (4

)

2 !

h

R

h

@

whereR

is the discriminant of

.

Proof. Use the geometric image of

and some geometric combinatorial considera-

tions. In particular, one uses Minkowskis Lattice Point Theorem:

Let be a free -module of rank in an -dimensional Euclidean vector space

over (then is called a complete lattice in ). Denote by Vol ( ) the volume

of the set

0 1 T 1 + +

@

T

@

: 0 x 4 x

18

$

where T 1 $ ' ' ' $ T@

is a basis of . Notice that Vol ( ) does not depend on the choice

of basis.

Let # be a centrally symmetric convex subset of . Suppose that Vol ( # )

2@

Vol (

). Then#

contains at least one nonzero point of

.

3.6.7. Examples. 1. Let

= ( 5). Then 1 = 2, 2 = 0, = 2, hR

h = 5.

(4

)

2 !

h

R

h

@

= 2! 5 22 = 1 ' 1 '' ' $


30/38


so ( ) = 1 and therefore = . Thus, every ideal of is principal and

m = 0 1

8

.

2. Let

= ( F 5). Then 1 = 0, 2 = 1, = 2, hR

h = 20, (2 )

h 20 h 3.

Hence, similar to Example in 3.6.4 we only need to look at prime number 2 ( 3) and

prime ideal divisors of the ideal (2) as potential candidates for non-principal ideals.

From 3.3.8 we know that (2) = (2 $ 1 + F 5)2 and 2 = (2 $ 1 + F 5). The ideal

(2 $ 1 + F 5) is prime by 3.4.5.

The ideal (2 $ 1 + F 5) is not principal: Indeed, if (2 $ 1 + F 5) =

then

2 = (2 $ 1 + F 5) = (

) =h

( ) h . If =B

+R

F 5 withB

$

R

we

deduce thatB

2 + 5R

2 = 2, a contradiction.

The second power of (2 $ 1 + F 5) is a principal ideal, and so we conclude that

m

( G

5)is a cyclic group of order 2.

3. Let

= ( 14). Then 1 = 2, 2 = 0, = 2, hR

h = 56 and (1 2) 56 =

3'

7'

' '

4. So we only need to inspect prime ideal divisors of (2) and of (3).Now 2 = (4+ 14)(4 F 14), so (2) = (4+ 14)(4 F 14). Since (4 14) = 2,

3.4.5 implies that the principal ideals (4 + 14), (4 F 14) are prime.

14 is quadratic non-residue modulo 3, so by Theorem 3.5.9 we deduce that 3

remains prime in

. Thus, every ideal of the ring of integers of ( 14) is principal,

m

( 14) = 0 18

.

4. It is known that for negative square-freeR

the only quadratic fields (R

) with

class number 1 are the following:

( F 1)$

( F 2)$

( F 3)$

( F 7)$

( F 11) $

( F 19) $ ( F 43) $ ( F 67) $ ( F 163) '

ForR

0 there are many more quadratic fields with class number 1. Gaussconjectured that there are infinitely many such fields, but this is still unproved.

3.6.8. Now we can state one of the greatest achivements of Kummer.

Kummers Theorem. Let be an odd prime. Let

= (

) be the th cyclotomic

field.

If doesnt divide h m h ,

or, equivalently, does not divide numerators of (rational) Bernoulli numbers

2 $ 4 $ ' ' ' $

G

3 given by

T

F 1=

4 =0

4

!

4

$

then the Fermat equation

#

+ =

does not have positive integer solutions.


31/38

2002/2003 31

Among primes 100 only 37, 59 and 67 dont satisfy the condition that does

not divide h m h , so Kummers theorem implies that for any other prime number smaller

100 the Fermat equation does not have positive integer solutions.

3.7. Units of rings of algebraic numbers

3.7.1. Definition. A subgroup of @

is called discrete if for every bounded closed

subset of @

the intersection

is finite.

Example: points of @

with integer coordinates form a discrete subgroup.

3.7.2. Proposition. Let be a discrete subgroup of @

. Then there are linearly

independent over vectors 1 $ ' ' '

such that 1 $ ' ' ' $

is a basis of the

-module

.

Proof. Let 1 $ ' ' ' $

be a set of linearly independent elements in over with

the maximal . Denote

= 0

@

: =

4 =1

B

4

4 : 0 xB

4x 1

8

'

The set is bounded and closed, so

is finite. For

write = 1 4 =1 4 4

with 4

. Define

= F

[ 4 ] 4 =

( 4 F [ 4 ]) 4 '

Hence the group

is generated by the finite set

and0

4

8

, and

is finitelygenerated as a -module.

Since the torsion of is trivial, the main theorem on the structure of finitely

generated modules over principal ideal domains implies the assertion of the proposition.

3.7.3. Dirichlets Unit Theorem. Let

be a number field of degree , 1 + 2 2 = .

Let be its ring of integers and ! be the group of units of . Then ! is the

direct product of a finite cyclic group " consisting of all roots of unity in

and a free

abelian group ! 1 of rank 1 + 2 F 1 :

!d "

! 1 d "

1+ 2G

1'

A basis of the free abelian group ! 1 is called a fundamental system of units in .

Proof. Consider the canonical embedding of

into

1

n

2 . Define

" : z 0

08

1 + 2$

" ( ) =

log h 1( ) h $ ' ' ' $ log h

1( ) h $ log( h

1+1( ) h2) $ ' ' ' $ log( h

1+ 2 ( ) h2) '


32/38


The map " induces a homomorphism Y : !

1 + 2 .

Note that the preimage YG

1( ) of a bounded set is finite. Indeed, if Y ( ) ,

then there isB

such that h 4 ( ) hx

B

for all . The coefficients of the characteristic

polynomial Y ' (# ) = @4 =1( # F

4 ( )) of over

being functions of

4 ( ) areintegers bounded by max(

B

@

$

B

@G

1$

' ' ' ), so the number of different characteristic

polynomials of YG

1( ) is finite, and so is YG

1( ).

Every finite subgroup of the multiplicative group of a field is cyclic by 1.2.4. Hence

the kernel of Y , being the preimage of 0, is a cyclic finite group. On the other hand,

every root of unity belongs to the kernel of Y , since Y

( ) = Y (

) = Y (1) = 0 implies

Y ( ) = 0 for the vector Y ( ). We conclude that the kernel of Y consists of all roots of

unity " in

.

Since for !

the norm

( ) =

4 ( ), as the product of units, is a unit

in , it is equal to 1. Then h 4 ( ) h = 1 and log h 1( ) h + + log h

1( ) h +

log( h

1+1( ) h2) + + log( h

1+ 2 ( ) h2) = 0. We deduce that the image Y ( ! ) is

contained in the hyperplane

1 + 2

defined by the equation

1 +

+

1+

2 = 0.Since YG

1( ) is finite for every bounded set , the intersection Y ( ! ) is finite.

Hence by 3.7.2 Y ( ! ) has a -basis 0 48

consisting of x 1 + 2 F 1 linearly

independent vectors over . Denote by ! 1 the subgroup of ! generated by 4 such

that Y ( 4 ) = 4 ; it is a free abelian group, since there are no nontrivial relations among

4. From the main theorem on group homomorphisms we deduce that !

" d Y( ! )

and hence ! = "! 1. Since ! 1 has no nontrivial torsion, " ! 1 = 0 1

8

. Then ! as a

-module is the direct product of the free abelian group ! 1 of rank and the cyclic

group " of roots of unity.

It remains to show that = 1 + 2 F 1, i.e. Y ( ! ) contains 1 + 2 F 1 linearly

independent vectors. Put = 1 + 2. As an application of Minkowskis geometric

method one can show thatfor every integer e between 1 and there is

B

0 such that for every non-zero

with Y ( ) = ( r 1 $ ' ' ' $ r

) there is a non-zero = `

( ) z0

08

such that

h

( ) h

x

B

and " ( ) = ( t 1 $ ' ' ' $ t

) with t4 r 4

for D

= e .

Fix e . Start with 1 = and construct the sequence u = `

( u

G

1) for

2. Since ( u ) = h

( u ) h x

B

, in the same way as in the proof of 3.6.4 we

deduce that there are only finitely many distinct ideals u . So u =

for

some

. Then

=

G

1u

is a unit and satisfies the property: the th coordinate of

Y (

) = " (

) F"

( u ) = ( r(

)1 $ ' ' ' $ r

(

)

) is negative for D

= e . Then r(

)

is positive,

since 1 4 r (

)4 = 0.

Havea look at units 1 $ ' ' ' $

. We claim that there are F

1 linearly independent

vectors among the images Y ( 4). To verify the claim it suffices to check that the first

F

1 columns of the matrix ( r(

)4

) are linearly independent.


33/38

2002/2003 33

If there were not, then there would be a non-zero vector (

1 $ ' ' ' $

G

1) such that

1 G

1

=1

r

(

)4

= 0 for all 1 x x . Without loss of generality one can assume that for

between 1 and F

1

4 = 1 and

ux

1 for D

= , 1 x x F

1. Then

4r

( 4 )4

= r( 4 )

4

and for D

=

ur

(u

)4

r

(u

)4

since

ux

1 and r (u

)4

0. Then

0 =

G

1

=1

r

(

)4

G

1

=1

r

(

)4

=1

r

(

)4

= 0 $

a contradiction.

Thus, = 1 + 2 F 1.

3.7.4. Example. Let

= ( R

) with a square free non-zero integerR

.

IfR

0, then the group of roots of 1 in

is 0

18

, since

and there are

only two roots of unity in .

Let

be the ring of integers of

. We have = 2 and 1 = 2 $ 2 = 0 ifR

0;

1 = 0 $ 2 = 1 ifR

0. IfR

0, then

! ( ) = "

is a finite cyclic group consisting of all roots of unity in

. It has order 4 forR

= F 1, 6

forR

= F 3, and one can show it has order 2 for all other negative square free integers.

IfR

0, ! (

) is the direct product of 3

1 4 and the infinite group generated by

a unit (fundamental unit of

):

! ( ) d 3 1 4

3 4 = 0 : e

8

'

Here is an algorithm how to find a fundamental unit ifR

D

1 mod 4 (there is a

similar algorithm for an arbitrary square free positiveR

):

Let be the minimal positive integer such that eitherR

2F 1 or

R

2 + 1 is a square

of a positive integer, say, . Then

( + R

) = 2 FR

2 = 1, so + R

1

is a unit of

.

Let 0 = T + " R

be a fundamental unit. Changing the sign of T $"

if necessary,

we can assume that T $ " are positive. Due to the definition of 0 there is an integer e

such that + R

= 0 . The sign is +, since the left hand side is positive; e 0,

since 0 1 and the left hand side is 1. From + R

= ( T + " R

) we deduce

that if e

1 then = " + some positive integer "

, a contradiction. Thus, e = 1

and + R

1 is a fundamental unit of

.

For example, 1 + 2 is a fundamental unit of ( 2) and 2 + 3 is a fundamental

unit of ( 3).

3.7.5. Now suppose that R

0, and for simplicity, RD

1 mod 4. We already knowthat if an element = +

R

of is a unit, then its norm

( ) = 2 F

R

2 is

1. On the other hand, if 2 FR

2 = 1, then

G

1 = F

R

is in

, so is a

unit. Thus, = + R

is a unit iff 2 FR

2 = 1.


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Let 0 = T + " R

be a fundamental unit.

From the previous we deduce that all integer solutions ( $ ) of the equation

#

2F

R

2

=

1satisfy +

R

= ( T + " R

) for some integer , which gives formulas for and

as functions of T $" $

.

3.8. On class field theory

3.8.1. Projective limits of groups. Let @

,

1 be a set of groups with group

homomorphisms 7@

: @

for all

such that

7

@@ = id

,7

@

= 7

7

@

for all .

The projective limit lim9F

@

of ( @

$7

@

) is the set

0 ( @

) : @

@

$ 7

@

( @

) =

for all

8

with the group operation ( @

) + ( @

) = ( @

+ @

).

For every one has a group homomorphism 7@

:lim9

F

@

$ ( @

)

.

Examples.

1. If @

= for all and 7@

= id then lim9F

@

= .

2. If @

= @

and 7@

( + @

) = + then ( @

) lim9F

@

means

min(@

)h

(

@

F

) for all

$

.The sequence (

@

) is a fundamental sequence with respect to the -adic norm, and

thus determines a -adic number = lim @

. We have a group homomorphism

" :lim9F

@

$ ( @

)

= lim @

'

It is surjective, due to the definition of -adic numbers, and its kernel is trivial, since

= 0 implies that for every e

divides @

for all sufficiently large , and so

divides

. Thus,

lim9F

@

d

'

In particular, we a surjective homomorphism

@

whose kernel equals to

@

.

3. If @

= ( @

)

and 7@

( + @

) = + , ( $ ) = 1, then similarly to

2 we have a homomorphism

" :lim9F

(

@

)

$ ( @

) lim


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2002/2003 35

(note that (

$

) = 1 and hence lim

D

). Thus, there is an isomorphism

lim9F

( @

)

A

'

4. One can extend the definition of the projective limit to the case when the maps7

@

are defined for some specific pairs ($

) and not necessarily all

. Let

@

= and let 7@

: @

be defined only if h and then 7@

( + ) =

+

.

By the Chinese remainder theorem

= 11

B

where = 11 ' ' '

B

is the factorization of . The maps 7@

induce the maps

already defined in 2 on

, and we deduce

lim9F

= lim9

F

2

lim9F

3

d 2

3

=

'

The group lim9F

is denoted D and is called the procyclic group (topologically

it is generated by its unity 1). This group is uncountable. We have a surjective

homomorphism D

whose kernel is D

.

Similarly we have

D

= lim9F

(

)

Documents

Algebraic Number Theory (Fesenko)