Algebraic Number Theory (Fesenko)
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Transcript of 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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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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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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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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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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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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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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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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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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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 .
2.3.4. Proposition. Let
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
.
2.3.5. Proposition. Let
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
.
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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.
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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
) ) '
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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 [
].
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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.
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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.
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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
.
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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.
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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)
+ @
=
+ @
$
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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.
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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.
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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 .
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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( # )) '
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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 .
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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
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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.
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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 '' ' $
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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.
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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) '
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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.
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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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34 Alg number theory
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
(
)