aarjaneiro/planetmath huggingface dataset:
properties of discriminant in algebraic number field

properties of discriminant in algebraic number field

Theorem 1.

Let

α

1

,

α

2

,

…

,

α

n

and

β

1

,

β

2

,

…

,

β

n

be elements of the

algebraic number field

ℚ

⁢

(

ϑ

)

of degree (

http://

planetmath

.org/NumberField

)

n

.  If they satisfy the equations

α

i

=

∑

j

=

1

n

c

i

⁢

j

⁢

β

j

for

i

=

1

,

2

,

…

,

n

,

where all

coefficients

c

i

⁢

j

are

rational numbers

, then the

http://planetmath.org/node/12060

discriminants

are via the equation

Δ

⁢

(

α

1

,

α

2

,

…

,

α

n

)

=

det

⁡

(

c

i

⁢

j

)

2

⋅

Δ

⁢

(

β

1

,

β

2

,

…

,

β

n

)

.

As a special case one obtains the

Theorem 2.

If

α

i

=

c

i

⁢

1

+

c

i

⁢

2

⁢

ϑ

+

…

+

c

i

⁢

n

⁢

ϑ

n

-

1

for

i

=

1

,

2

,

…

,

n

(1)

are the

canonical forms

of the

elements

α

i

in

ℚ

⁢

(

ϑ

)

, then

Δ

⁢

(

α

1

,

α

2

,

…

,

α

n

)

=

det

⁡

(

c

i

⁢

j

)

2

⋅

Δ

⁢

(

1

,

ϑ

,

…

,

ϑ

n

-

1

)

,

where

Δ

⁢

(

1

,

ϑ

,

…

,

ϑ

n

-

1

)

is a

Vandermonde determinant

thus
having the

product

form

Δ

⁢

(

1

,

ϑ

,

…

,

ϑ

n

-

1

)

=

(

∏

1

≤

i

≤

j

(

ϑ

j

-

ϑ

i

)

)

2

=

∏

1

≤

i

≤

j

(

ϑ

i

-

ϑ

j

)

2

(2)

where

ϑ

1

=

ϑ

,

ϑ

2

,

…

,

ϑ

n

are the

algebraic conjugates

of

ϑ

.

Since the (2) is also the

polynomial discriminant

of the irreducible

minimal polynomial

of

ϑ

, the numbers

ϑ

i

are inequal.  It follows the

Theorem 3.

When (1) are the canonical forms of the numbers

α

i

, one has

Δ

⁢

(

α

1

,

α

2

,

…

,

α

n

)

=

0

⟺

det

⁡

(

c

i

⁢

j

)

=

0

.

The powers

1

,

ϑ

,

…

,

ϑ

n

-

1

of the

primitive element

(

http://planetmath.org/SimpleFieldExtension

) form a basis (

http://planetmath.org/Basis

) of the

field extension

ℚ

⁢

(

ϑ

)

/

ℚ

(see the canonical form of element of number field).  By the theorem 3 we may write the

Theorem 4.

The numbers

α

1

,

α

2

,

…

,

α

n

of

ℚ

⁢

(

ϑ

)

are

linearly independent

over

ℚ

if and only if

Δ

⁢

(

α

1

,

α

2

,

…

,

α

n

)

≠

0

.

Theorem 5.

ℚ

⁢

(

α

)

=

ℚ

⁢

(

ϑ

)

⟺

Δ

⁢

(

1

,

α

,

α

2

,

…

,

α

n

-

1

)

≠

0

.  Here, the the discriminant is the discriminant of the

algebraic number

(

http://planetmath.org/DiscriminantOfAlgebraicNumber

)

α

.

Title

properties of discriminant in algebraic number field

Canonical name

PropertiesOfDiscriminantInAlgebraicNumberField

Date of creation

2013-03-22 19:09:28

Last modified on

2013-03-22 19:09:28

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

12

Author

pahio (2872)

Entry type

Result

Classification

msc 11R29

Related topic

MinimalityOfIntegralBasis

Related topic

ConditionForPowerBasis

Generated on Fri Feb 9 15:44:15 2018 by

LaTeXML
ray class group

ray class group

Let

𝔪

be a

modulus

for a

number field

K

. The

ray class group

of

K

mod

𝔪

is the group

𝕀

𝔪

/

K

𝔪

,

1

, where

•

𝕀

𝔪

is the subgroup of the

ideal group

of

K

generated by all

prime ideals

which do not occur in the factorization of

𝔪

.

•

K

𝔪

,

1

is the subgroup of

𝕀

𝔪

consisting of all

principal ideals

in the

ring of integers

of

K

having the form

(

α

)

where

α

is

multiplicatively congruent

to

1

mod

𝔪

.

Title

ray class group

Canonical name

RayClassGroup

Date of creation

2013-03-22 12:50:19

Last modified on

2013-03-22 12:50:19

Owner

djao (24)

Last modified by

djao (24)

Numerical id

4

Author

djao (24)

Entry type

Definition

Classification

msc 11R29

Generated on Fri Feb 9 15:44:17 2018 by

LaTeXML
root-discriminant

root-discriminant

Definition 1

.

Let

K

be a

number field

, let

d

K

be its

discriminant

and let

n

=

[

K

:

Q

]

be the degree over

Q

. The quantity:

|

d

K

n

|

is called the

root-discriminant

of

K

and it is usually denoted by

rd

K

.

The following lemma is one of the motivations for the previous definition:

Lemma 1

.

Let

E

/

F

be an

extension

of number fields which is

unramified

at all

finite primes

. Then

rd

E

=

rd

F

. In particular, the

Hilbert class field

of a number field has the same root-discriminant as the number field.

Proof.

Notice that the relative

discriminant ideal

(or different) for

E

/

F

is the

ring of integers

in

F

. Therefore we have:

|

d

E

|

=

|

d

F

|

[

E

:

F

]

The results follows by taking

[

E

:

ℚ

]

-th roots on both sides of the previous equation.
∎

Title

root-discriminant

Canonical name

Rootdiscriminant

Date of creation

2013-03-22 15:05:44

Last modified on

2013-03-22 15:05:44

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

5

Author

alozano (2414)

Entry type

Definition

Classification

msc 11R29

Synonym

root discriminant

Related topic

ExistenceOfHilbertClassField

Generated on Fri Feb 9 15:44:19 2018 by

LaTeXML
Stickelberger’s theorem

Stickelberger’s theorem

Theorem 1

(Stickelberger).

Let

L

=

Q

⁢

(

ζ

m

)

be a

cyclotomic field

extension

of

Q

with

Galois group

G

=

{

σ

a

}

a

∈

(

Z

/

m

⁢

Z

)

×

, and consider the

group ring

Q

⁢

[

G

]

. Define the

Stickelberger element

θ

∈

Q

⁢

[

G

]

by

θ

=

1

m

⁢

∑

1

≤

a

≤

m

,

(

a

,

m

)

=

1

a

⁢

σ

a

-

1

,

and take

β

∈

Z

⁢

[

G

]

such that

β

⁢

θ

∈

Z

⁢

[

G

]

as well. Then

β

⁢

θ

is an

annihilator

for the

ideal class group

of

Q

⁢

(

ζ

m

)

.

Note that

θ

itself need not be an annihilator, just that any

multiple

of it in

ℤ

⁢

[

G

]

is.

This result allows for the most basic between the (otherwise hard to determine) of a cyclotomic ideal class group and its of annihilators. For an application of Stickelberger’s

theorem

, see Herbrand’s theorem.

Title

Stickelberger’s theorem

Canonical name

StickelbergersTheorem

Date of creation

2013-03-22 14:12:31

Last modified on

2013-03-22 14:12:31

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

6

Author

mathcam (2727)

Entry type

Theorem

Classification

msc 11R29

Defines

Stickelberger element

Generated on Fri Feb 9 15:44:21 2018 by

LaTeXML
Thaine’s theorem

Thaine’s theorem

Let

F

/

ℚ

be a totally real

abelian number field

. By the

Kronecker-Weber theorem

, there exists an

m

such that

F

⊂

ℚ

⁢

(

ζ

m

)

. Let

G

be the

Galois group

of the

extension

F

/

Q

. Let

𝒪

F

×

denote the

group of units

in the

ring of integers

of

F

, let

C

be the subgroup of

𝒪

F

×

consisting of units

η

of the form

η

=

±

N

ℚ

⁢

(

ζ

m

)

/

F

⁢

(

∏

a

∈

(

ℤ

/

m

⁢

ℤ

)

×

(

ζ

m

a

-

1

)

b

a

)

for some

collection

of

b

a

∈

ℤ

. (Here,

N

denotes the norm

operator

and

ζ

m

is a primitive

m

-th

root of unity

.) Finally, let

A

denote the

ideal class group

of

F

.

Theorem 1

(Thaine).

Suppose

p

is a

rational prime

not dividing the degree

[

F

:

Q

]

and suppose

θ

∈

Z

⁢

[

G

]

annihilates the Sylow

p

-subgroup of

E

/

C

′

. Then

2

⁢

θ

annihilates the Sylow

p

-subgroup of

A

.

This is one of the most sophisticated results concerning the

annihilators

of an ideal class group. It is a direct, but more complicated, version of Stickelberger’s

theorem

, applied to totally real fields (for which Stickelberger’s theorem gives no information).

Title

Thaine’s theorem

Canonical name

ThainesTheorem

Date of creation

2013-03-22 14:12:34

Last modified on

2013-03-22 14:12:34

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

6

Author

mathcam (2727)

Entry type

Theorem

Classification

msc 11R29

Generated on Fri Feb 9 15:44:24 2018 by

LaTeXML
Vandiver’s conjecture

Vandiver’s conjecture

Let

K

=

ℚ

⁢

(

ζ

p

)

+

, the maximal real subfield of the

p

-th

cyclotomic field

. Vandiver’s conjecture states that

p

does not divide

h

K

, the

class number

of

K

.

For comparison, see the entries on

regular primes

and

irregular primes

.

A proof of Vandiver’s conjecture would be a landmark in

algebraic number theory

, as many theorems hinge on the assumption that this conjecture is true. For example, it is known that if Vandiver’s conjecture holds, that the

p

-rank of the ideal class group of

ℚ

⁢

(

ζ

p

)

equals the number of

Bernoulli numbers

divisible

by

p

(a remarkable strengthening of Herbrand’s theorem).

Title

Vandiver’s conjecture

Canonical name

VandiversConjecture

Date of creation

2013-03-22 15:01:11

Last modified on

2013-03-22 15:01:11

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

5

Author

mathcam (2727)

Entry type

Conjecture

Classification

msc 11R29

Related topic

ClassNumbersAndDiscriminantsTopicsOnClassGroups

Generated on Fri Feb 9 15:44:28 2018 by

LaTeXML
additive form of Hilbert’s theorem 90

additive form of Hilbert’s theorem 90

Let

L

/

K

be a finite cyclic

Galois extension

and

σ

a

generator

of

Gal

⁡

(

L

/

K

)

. An element

y

∈

L

satisfies

Tr

⁡

(

y

)

=

0

if and only if there exists

x

∈

L

satisfying

y

=

x

-

σ

⁢

(

x

)

. Here

Tr

denotes the trace in

L

/

K

.

Title

additive form of Hilbert’s theorem 90

Canonical name

AdditiveFormOfHilbertsTheorem90

Date of creation

2013-03-22 15:20:35

Last modified on

2013-03-22 15:20:35

Owner

mathwizard (128)

Last modified by

mathwizard (128)

Numerical id

7

Author

mathwizard (128)

Entry type

Theorem

Classification

msc 11R32

Classification

msc 12F10

Related topic

HilbertTheorem90

Generated on Fri Feb 9 15:44:30 2018 by

LaTeXML
algebraic sum and product

algebraic sum and product

Let

α

,

β

be two elements of an

extension field

of a given field

K

.  Both these elements are algebraic over

K

if and only if both

α

+

β

and

α

⁢

β

are algebraic over

K

.

Proof.

Assume first that

α

and

β

are

algebraic

.  Because

[

K

(

α

,

β

)

:

K

]

=

[

K

(

α

,

β

)

:

K

(

α

)

]

[

K

(

α

)

:

K

]

and both here are finite (

http://planetmath.org/ExtendedRealNumbers

), then

[

K

⁢

(

α

,

β

)

:

K

]

is finite.  So we have a

finite field extension

K

⁢

(

α

,

β

)

/

K

which thus is also algebraic, and therefore the elements

α

+

β

and

α

⁢

β

of

K

⁢

(

α

,

β

)

are algebraic over

K

.  Secondly suppose that

α

+

β

and

α

⁢

β

are algebraic over

K

.  The elements

α

and

β

are the roots of the

quadratic equation

x

2

-

(

α

+

β

)

⁢

x

+

α

⁢

β

=

0

(cf.

properties of quadratic equation

) with the

coefficients

in

K

⁢

(

α

+

β

,

α

⁢

β

)

.  Thus

[

K

(

α

,

β

)

:

K

]

=

[

K

(

α

,

β

)

:

K

(

α

+

β

,

α

β

)

]

[

K

(

α

+

β

,

α

β

)

:

K

]

≦

2

[

K

(

α

+

β

,

α

β

)

:

K

]

.

Since

[

K

⁢

(

α

+

β

,

α

⁢

β

)

:

K

]

is finite,  then also

[

K

⁢

(

α

,

β

)

:

K

]

is, and in the

finite extension

(

http://planetmath.org/FiniteExtension

)

K

⁢

(

α

,

β

)

/

K

the elements

α

and

β

must be algebraic over

K

.

Title

algebraic sum and product

Canonical name

AlgebraicSumAndProduct

Date of creation

2013-03-22 15:28:03

Last modified on

2013-03-22 15:28:03

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Theorem

Classification

msc 11R32

Classification

msc 11R04

Classification

msc 13B05

Synonym

sum and product algebraic

Related topic

FiniteExtension

Related topic

TheoryOfAlgebraicNumbers

Related topic

FieldOfAlgebraicNumbers

Generated on Fri Feb 9 15:44:32 2018 by

LaTeXML
existence of Hilbert class field

existence of Hilbert class field

Let

K

be a

number field

. There exists a

finite extension

E

of

K

with the following properties:

1.

[

E

:

K

]

=

h

K

, where

h

K

is the

class number

of

K

.

2.

E

is Galois over

K

.

3.

The ideal class group of

K

is

isomorphic

to the

Galois group

of

E

over

K

.

4.

Every ideal of

𝒪

K

is a

principal ideal

of the ring extension

𝒪

E

.

5.

Every

prime ideal

𝒫

of

𝒪

K

decomposes into the product of

h

K

f

prime ideals in

𝒪

E

, where

f

is the order (

http://planetmath.org/Order

)
of

[

𝒫

]

in the ideal class group of

𝒪

E

.

There is a unique field

E

satisfying the above five properties, and it is known as the

Hilbert class field

of

K

.

The field

E

may also be characterized as the

maximal

abelian

unramified

(

http://planetmath.org/AbelianExtension

)

extension

of

K

. Note that in this context, the term ‘unramified’ is meant not only for the

finite places

(the classical ideal theoretic ) but also for the infinite places. That is, every

real embedding

of

K

extends to a real embedding of

E

. As an example of why this is necessary, consider some

real quadratic field

.

Title

existence of Hilbert class field

Canonical name

ExistenceOfHilbertClassField

Date of creation

2013-03-22 12:36:45

Last modified on

2013-03-22 12:36:45

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

16

Author

mathcam (2727)

Entry type

Theorem

Classification

msc 11R32

Classification

msc 11R29

Classification

msc 11R37

Related topic

IdealClass

Related topic

Group

Related topic

NumberField

Related topic

ClassNumberDivisibilityInExtensions

Related topic

RootDiscriminant

Related topic

ExtensionsWithoutUnramifiedSubextensionsAndClassNumberDivisibility

Related topic

ClassNumbersAndDiscriminantsTopicsOnClassGroups

Defines

Hilbert class field

Generated on Fri Feb 9 15:44:34 2018 by

LaTeXML
Galois criterion for solvability of a polynomial by radicals

Galois criterion for solvability of a polynomial by radicals

Let

f

∈

F

⁢

[

x

]

be a

polynomial

over a field

F

, and let

K

be its

splitting field

. If

K

is a

radical extension

of

F

, then the

Galois group

Gal

⁡

(

K

/

F

)

is a

solvable group

.

Conversely, if the Galois group

Gal

⁡

(

K

/

F

)

is a solvable group, then

K

is a radical extension of

F

provided that the

characteristic

of

K

is either

0

or greater than

deg

⁡

(

f

)

.

Title

Galois criterion for solvability of a polynomial by

radicals

Canonical name

GaloisCriterionForSolvabilityOfAPolynomialByRadicals

Date of creation

2013-03-22 12:08:58

Last modified on

2013-03-22 12:08:58

Owner

djao (24)

Last modified by

djao (24)

Numerical id

7

Author

djao (24)

Entry type

Theorem

Classification

msc 11R32

Generated on Fri Feb 9 15:44:36 2018 by

LaTeXML
Galois groups of finite abelian extensions of ℚ

Galois groups of finite abelian extensions of

ℚ

Theorem

.

Let

G

be a finite abelian group with

|

G

|

>

1

. Then there exist infinitely many

number fields

K

with

K

/

Q

Galois and

Gal

⁡

(

K

/

Q

)

≅

G

.

Proof.

This will first be proven for

G

cyclic.

Let

|

G

|

=

n

. By Dirichlet’s theorem on primes in arithmetic progressions, there exists a prime

p

with

p

≡

1

⁢

mod

⁡

n

. Let

ζ

p

denote a

p

th

root of unity

. Let

L

=

ℚ

⁢

(

ζ

p

)

. Then

L

/

ℚ

is Galois with

Gal

⁡

(

L

/

ℚ

)

cyclic of order (

http://planetmath.org/OrderGroup

)

p

-

1

. Since

n

divides

p

-

1

, there exists a

subgroup

H

of

Gal

⁡

(

L

/

ℚ

)

such that

|

H

|

=

p

-

1

n

. Since

Gal

⁡

(

L

/

ℚ

)

is cyclic, it is

abelian

, and

H

is a

normal subgroup

of

Gal

⁡

(

L

/

ℚ

)

. Let

K

=

L

H

, the

subfield

of

L

fixed (

http://planetmath.org/FixedField

) by

H

. Then

K

/

ℚ

is Galois with

Gal

⁡

(

K

/

ℚ

)

cyclic of order

n

. Thus,

Gal

⁡

(

K

/

ℚ

)

≅

G

.

Let

p

and

q

be distinct primes with

p

≡

1

⁢

mod

⁡

n

and

q

≡

1

⁢

mod

⁡

n

. Then there exist subfields

K

1

and

K

2

of

ℚ

⁢

(

ζ

p

)

and

ℚ

⁢

(

ζ

q

)

, respectively, such that

Gal

⁡

(

K

1

/

ℚ

)

≅

G

and

Gal

⁡

(

K

2

/

ℚ

)

≅

G

. Note that

K

1

∩

K

2

=

ℚ

since

ℚ

⊆

K

1

∩

K

2

⊆

ℚ

⁢

(

ζ

p

)

∩

ℚ

⁢

(

ζ

q

)

=

ℚ

. Thus,

K

1

≠

K

2

. Therefore, for every prime

p

with

p

≡

1

⁢

mod

⁡

n

, there exists a distinct number field

K

such that

K

/

ℚ

is Galois and

Gal

⁡

(

K

/

ℚ

)

≅

G

. The theorem in the cyclic case follows from using the full of Dirichlet’s theorem on primes in arithmetic progressions: There exist infinitely many primes

p

with

p

≡

1

⁢

mod

⁡

n

.

The general case follows immediately from the above , the fundamental theorem of finite abelian groups (

http://planetmath.org/FundamentalTheoremOfFinitelyGeneratedAbelianGroups

), and a theorem regarding the

Galois group

of the

compositum

of two

Galois extensions

.
∎

Title

Galois groups of finite

abelian extensions

of

ℚ

Canonical name

GaloisGroupsOfFiniteAbelianExtensionsOfmathbbQ

Date of creation

2013-03-22 16:18:40

Last modified on

2013-03-22 16:18:40

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

11

Author

Wkbj79 (1863)

Entry type

Theorem

Classification

msc 11R32

Classification

msc 11N13

Classification

msc 11R20

Classification

msc 12F10

Related topic

AbelianNumberField

Generated on Fri Feb 9 15:44:39 2018 by

LaTeXML
Hilbert Theorem 90

Hilbert Theorem 90

Let

L

/

K

be a finite

Galois extension

with

Galois group

G

=

Gal

⁡

(

L

/

K

)

. The modern formulation of Hilbert’s Theorem 90 states that the first

Galois cohomology

group

H

1

⁢

(

G

,

L

*

)

is 0.

The original statement of Hilbert’s Theorem 90 differs somewhat from the modern formulation given above, and is nowadays regarded as a corollary of the above fact. In its original form, Hilbert’s Theorem 90 says that if

G

is cyclic with

generator

σ

, then an element

x

∈

L

has norm 1 if and only if

x

=

y

/

σ

⁢

(

y

)

for some

y

∈

L

. Note that elements of the form

y

/

σ

⁢

(

y

)

are obviously contained within the kernel of the norm map; it is the converse that forms the content of the theorem.

Title

Hilbert Theorem 90

Canonical name

HilbertTheorem90

Date of creation

2013-03-22 12:10:41

Last modified on

2013-03-22 12:10:41

Owner

djao (24)

Last modified by

djao (24)

Numerical id

7

Author

djao (24)

Entry type

Theorem

Classification

msc 11R32

Classification

msc 11S25

Classification

msc 11R34

Synonym

Hilbert’s Theorem 90

Synonym

Satz 90

Generated on Fri Feb 9 15:44:45 2018 by

LaTeXML
proof of Galois group of the compositum of two Galois extensions

proof of Galois group of the compositum of two Galois extensions

Proof.

Consider the diagram

\xymatrix

⁢

@

⁢

R

⁢

1

⁢

p

⁢

c

⁢

@

⁢

C

⁢

1

⁢

p

⁢

c

⁢

&

⁢

\ar

⁢

@

-

[

l

⁢

d

]

⁢

\ar

⁢

@

-

[

r

⁢

d

]

⁢

E

⁢

F

⁢

\ar

⁢

@

-

[

r

⁢

d

]

⁢

E

⁢

&

⁢

&

⁢

\ar

⁢

@

-

[

l

⁢

d

]

⁢

F

⁢

&

⁢

\ar

⁢

@

-

[

d

]

⁢

E

∩

F

⁢

&

⁢

K

(1): Let

p

⁢

(

x

)

∈

K

⁢

[

x

]

with a root

α

∈

E

∩

F

. Then since

E

(resp.

F

) is Galois over

K

, all the roots of

p

lie in

E

(resp.

F

) and thus in

E

∩

F

. The result follows.

(2): We first show that

E

⁢

F

is Galois over

K

. Choose

separable polynomials

p

⁢

(

x

)

,

q

⁢

(

x

)

∈

K

⁢

[

x

]

so that

E

(resp.

F

) is a

splitting field

for

p

(resp.

q

). Then

E

⁢

F

is a splitting field for the

squarefree part

of

p

⁢

q

, which is separable since it is

squarefree

and since

p

⁢

(

x

)

,

q

⁢

(

x

)

are separable.

Now, define

θ

:

Gal

⁡

(

E

⁢

F

/

K

)

→

Gal

⁡

(

E

/

K

)

×

Gal

⁡

(

F

/

K

)

:

σ

↦

(

σ

|

E

,

σ

|

F

)

This map is a

group homomorphism

; its kernel is precisely those elements that leave both

E

and

F

fixed. Any such element must thus leave

E

⁢

F

fixed, so that

θ

is injective. The image obviously lies in

H

=

{

(

σ

,

τ

)

:

σ

|

E

∩

F

=

τ

|

E

∩

F

}

by construction:

(

σ

|

E

)

|

E

∩

F

=

σ

|

E

∩

F

=

(

σ

|

F

)

|

E

∩

F

. We will show that

H

is precisely the image of

θ

by showing that the order of

H

is the same as the index of the

field extension

[

E

⁢

F

:

K

]

.

For each

σ

∈

Gal

⁡

(

E

/

K

)

, there are precisely

|

Gal

⁡

(

F

/

E

∩

F

)

|

elements of

Gal

⁡

(

F

/

K

)

whose restrictions to

E

∩

F

are

σ

|

E

∩

F

. Thus directly from the

definition

of

H

,

|

H

|

=

|

Gal

⁡

(

E

/

K

)

|

⋅

|

Gal

⁡

(

F

/

E

∩

F

)

|

=

|

Gal

⁡

(

E

/

K

)

|

⋅

|

Gal

⁡

(

F

/

K

)

|

|

Gal

⁡

(

(

E

∩

F

)

/

K

)

|

By the

corollary

to the theorem regarding the

compositum

of a

Galois extension

and another

extension

(

http://planetmath.org/CorollaryToTheCompositumOfAGaloisExtensionAndAnotherExtensionIsGalois

), we have

[

E

F

:

K

]

=

[

E

F

:

F

]

[

F

:

K

]

=

[

E

:

E

∩

F

]

[

F

:

K

]

=

[

E

:

K

]

[

F

:

K

]

[

E

∩

F

:

K

]

so that

|

H

|

=

[

E

F

:

K

]

∎

References

1

Dummit, D., Foote, R.M.,

Abstract Algebra, Third Edition

, Wiley, 2004.

Title

proof of

Galois group

of the compositum of two Galois extensions

Canonical

name

ProofOfGaloisGroupOfTheCompositumOfTwoGaloisExtensions

Date of creation

2013-03-22 18:42:01

Last modified on

2013-03-22 18:42:01

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

6

Author

rm50 (10146)

Entry type

Proof

Classification

msc 11R32

Classification

msc 12F99

Generated on Fri Feb 9 15:44:48 2018 by

LaTeXML
proof of Hilbert Theorem 90

proof of Hilbert Theorem 90

Remember that two

cocycles

a

,

a

′

:

G

→

L

*

are called cohomologous, denoted by

a

∼

a

′

, if there exists

b

∈

L

*

, such that

a

′

⁢

(

τ

)

=

b

⁢

a

⁢

(

τ

)

⁢

τ

⁢

(

b

-

1

)

for all

τ

∈

G

. Then

H

1

(

G

,

L

*

)

=

{

a

:

G

→

L

*

|

a

is a cocycle

}

/

∼

.

Now let

a

:

G

→

L

*

be a cocycle. Then consider the map

α

:

L

→

L

,

c

↦

∑

τ

∈

G

a

⁢

(

τ

)

⁢

τ

⁢

(

c

)

.

Since elements of the

Galois group

are

linearly independent

,

α

is not

0

. So we can choose

c

∈

L

, such that

b

=

α

⁢

(

c

)

≠

0

. Then for

σ

∈

G

we have

σ

⁢

(

b

)

=

∑

τ

∈

G

σ

⁢

(

a

⁢

(

τ

)

⁢

τ

⁢

(

c

)

)

=

∑

τ

∈

G

σ

⁢

(

a

⁢

(

τ

)

)

⁢

(

σ

⁢

τ

)

⁢

(

c

)

=

∑

τ

∈

G

a

⁢

(

σ

)

-

1

⁢

a

⁢

(

σ

⁢

τ

)

⁢

(

σ

⁢

τ

)

⁢

(

c

)

,

since

a

is a cocycle, i.e.

a

⁢

(

σ

⁢

τ

)

=

a

⁢

(

σ

)

⁢

σ

⁢

(

a

⁢

(

τ

)

)

. Then we get

σ

⁢

(

b

)

=

a

⁢

(

σ

)

-

1

⁢

∑

τ

∈

G

a

⁢

(

σ

⁢

τ

)

⁢

(

σ

⁢

τ

)

⁢

(

c

)

=

a

⁢

(

σ

)

-

1

⁢

b

.

Thus we have that

a

⁢

(

σ

)

=

b

⁢

σ

⁢

(

b

)

-

1

is a 1-coboundary.

Now we prove the corollary. Denote the norm by

N

. Now if

x

=

y

σ

⁢

(

y

)

, we have

N

⁢

(

x

)

=

N

⁢

(

y

σ

⁢

(

y

)

)

=

∏

τ

∈

G

τ

⁢

(

y

)

τ

⁢

(

σ

⁢

(

y

)

)

=

1

.

Now let

N

⁢

(

x

)

=

1

,

n

=

|

G

|

. Since

G

is assumed cyclic, let

σ

be a

generator

of

G

.

G

is

isomorphic

to

ℤ

/

n

⁢

ℤ

. We define the map

x

~

:

ℤ

/

n

⁢

ℤ

→

L

*

by

x

~

⁢

(

[

i

]

)

=

∏

0

≤

j

≤

i

-

1

σ

j

⁢

(

x

)

,

where

[

i

]

denotes the class of

i

∈

ℤ

in

ℤ

/

n

⁢

ℤ

. Since

N

⁢

(

x

)

=

1

,

x

~

is well defined. We have

x

~

⁢

(

[

i

+

k

]

)

=

∏

0

≤

j

≤

i

+

k

-

1

σ

j

⁢

(

x

)

=

(

∏

0

≤

j

≤

i

-

1

σ

j

⁢

(

x

)

)

⁢

σ

i

⁢

(

∏

0

≤

j

≤

k

-

1

σ

j

⁢

(

x

)

)

=

x

~

⁢

(

[

i

]

)

⁢

σ

i

⁢

(

x

~

⁢

(

[

j

]

)

)

.

Therefore

x

~

is a cocycle. Because of

Hilberts Theorem 90

, there exists

y

∈

L

*

, such that

x

=

x

~

⁢

(

[

1

]

)

=

y

⁢

σ

⁢

(

y

)

-

1

.

Title

proof of Hilbert Theorem 90

Canonical name

ProofOfHilbertTheorem90

Date of creation

2013-03-22 15:19:27

Last modified on

2013-03-22 15:19:27

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

8

Author

mathcam (2727)

Entry type

Proof

Classification

msc 11R32

Classification

msc 11S25

Classification

msc 11R34

Generated on Fri Feb 9 15:44:51 2018 by

LaTeXML