aarjaneiro/planetmath huggingface dataset:
algebraic closure of a finite field

algebraic closure of a finite field

Fix a prime

p

in

ℤ

. Then the

Galois fields

G

⁢

F

⁢

(

p

e

)

denotes the

finite field

of order

p

e

,

e

≥

1

. This can be concretely constructed as
the

splitting field

of the

polynomials

x

p

e

-

x

over

ℤ

p

. In so doing we
have

G

⁢

F

⁢

(

p

e

)

⊆

G

⁢

F

⁢

(

p

f

)

whenever

e

|

f

. In particular, we have an

infinite chain

:

G

⁢

F

⁢

(

p

1

!

)

⊆

G

⁢

F

⁢

(

p

2

!

)

⊆

G

⁢

F

⁢

(

p

3

!

)

⊆

⋯

⊆

G

⁢

F

⁢

(

p

n

!

)

⊆

⋯

.

So we define

G

⁢

F

⁢

(

p

∞

)

=

⋃

n

=

1

∞

G

⁢

F

⁢

(

p

n

!

)

.

Theorem 1

.

G

⁢

F

⁢

(

p

∞

)

is an algebraically closed field of

characteristic

p

.
Furthermore,

G

⁢

F

⁢

(

p

e

)

is a

contained

in

G

⁢

F

⁢

(

p

∞

)

for all

e

≥

1

.
Finally,

G

⁢

F

⁢

(

p

∞

)

is the

algebraic closure

of

G

⁢

F

⁢

(

p

e

)

for any

e

≥

1

.

Proof.

Given

elements

x

,

y

∈

G

⁢

F

⁢

(

p

∞

)

then there exists some

n

such that

x

,

y

∈

G

⁢

F

⁢

(

p

n

!

)

. So

x

+

y

and

x

⁢

y

are contained in

G

⁢

F

⁢

(

p

n

!

)

and also
in

G

⁢

F

⁢

(

p

∞

)

. The properties of a field are thus inherited and we have
that

G

⁢

F

⁢

(

p

∞

)

is a field. Furthermore, for any

e

≥

1

,

G

⁢

F

⁢

(

p

e

)

is
contained in

G

⁢

F

⁢

(

p

e

!

)

as

e

|

e

!

, and so

G

⁢

F

⁢

(

p

e

)

is contained in

G

⁢

F

⁢

(

p

∞

)

.

Now given

p

⁢

(

x

)

a polynomial over

G

⁢

F

⁢

(

p

∞

)

then there exists some

n

such that

p

⁢

(

x

)

is a polynomial over

G

⁢

F

⁢

(

p

n

!

)

. As the splitting field
of

p

⁢

(

x

)

is a

finite extension

of

G

⁢

F

⁢

(

p

n

!

)

, so it is a finite field

G

⁢

F

⁢

(

p

e

)

for some

e

, and hence contained in

G

⁢

F

⁢

(

p

∞

)

. Therefore

G

⁢

F

⁢

(

p

∞

)

is

algebraically closed

.
∎

We say

G

⁢

F

⁢

(

p

∞

)

is

the

algebraic closure indicating that up to

field
isomorphisms

, there is only one algebraic closure of a field. The actual objects
and constructions may vary.

Corollary 2

.

The algebraic closure of a finite field is

countable

.

Proof.

By construction the algebraic closure is a countable union of

finite sets

so
it is countable.
∎

References

1

McDonald, Bernard R.,

Finite rings with

identity

, Pure and Applied Mathematics, Vol. 28,
Marcel Dekker Inc., New York, 1974, p. 48.

Title

algebraic closure of a finite field

Canonical name

AlgebraicClosureOfAFiniteField

Date of creation

2013-03-22 16:40:51

Last modified on

2013-03-22 16:40:51

Owner

Algeboy (12884)

Last modified by

Algeboy (12884)

Numerical id

5

Author

Algeboy (12884)

Entry type

Derivation

Classification

msc 12F05

Related topic

FiniteField

Related topic

FiniteFieldCannotBeAlgebraicallyClosed

Generated on Fri Feb 9 16:41:48 2018 by

LaTeXML
algebraic extension

algebraic extension

Definition 1

.

Let

L

/

K

be an extension of fields.

L

/

K

is said to be an

algebraic extension

of fields if every element of

L

is
algebraic over

K

. If

L

/

K

is not

algebraic

then we say that it is a

transcendental extension

of fields.

Examples:

1.

Let

L

=

ℚ

⁢

(

2

)

. The

extension

L

/

ℚ

is an
algebraic extension. Indeed, any element

α

∈

L

is of the
form

α

=

q

+

t

⁢

2

∈

L

for some

q

,

t

∈

ℚ

. Then

α

∈

L

is a root of

X

2

-

2

⁢

q

⁢

X

+

q

2

-

2

⁢

t

2

=

0

2.

The field extension

ℝ

/

ℚ

is not an algebraic
extension. For example,

π

∈

ℝ

is a

transcendental number

over

ℚ

(see pi). So

ℝ

/

ℚ

is a transcendental extension of fields.

3.

Let

K

be a field and denote by

K

¯

the

algebraic closure

of

K

. Then the extension

K

¯

/

K

is
algebraic.

4.

In general, a

finite extension

of fields is an algebraic
extension. However, the converse is not true. The extension

ℚ

¯

/

ℚ

is

far

from finite.

5.

The extension

ℚ

⁢

(

π

)

/

ℚ

is transcendental because

π

is a transcendental number, i.e.

π

is not the root of any

polynomial

p

⁢

(

x

)

∈

ℚ

⁢

[

x

]

.

Title

algebraic extension

Canonical name

AlgebraicExtension

Date of creation

2013-03-22 13:57:27

Last modified on

2013-03-22 13:57:27

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

7

Author

alozano (2414)

Entry type

Definition

Classification

msc 12F05

Synonym

algebraic field extension

Related topic

Algebraic

Related topic

FiniteExtension

Related topic

AFiniteExtensionOfFieldsIsAnAlgebraicExtension

Related topic

ProofOfTranscendentalRootTheorem

Related topic

EquivalentConditionsForNormalityOfAFieldExtension

Defines

examples of field extension

Defines

transcendental extension

Generated on Fri Feb 9 16:41:50 2018 by

LaTeXML
algebraically closed

algebraically closed

A field

K

is

algebraically closed

if every non-constant

polynomial

in

K

⁢

[

X

]

has a root in

K

.

An

extension field

L

of

K

is an

algebraic closure

of

K

if

L

is algebraically closed and every element of

L

is algebraic over

K

. Using the

axiom of choice

, one can show that any field has an algebraic closure. Moreover, any two algebraic closures of a field are

isomorphic

as fields, but not necessarily canonically isomorphic.

Title

algebraically closed

Canonical name

AlgebraicallyClosed

Date of creation

2013-03-22 12:12:06

Last modified on

2013-03-22 12:12:06

Owner

djao (24)

Last modified by

djao (24)

Numerical id

10

Author

djao (24)

Entry type

Definition

Classification

msc 12F05

Defines

algebraic closure

Generated on Fri Feb 9 16:41:52 2018 by

LaTeXML
algebraically dependent

algebraically dependent

Let

L

be a

field extension

of a field

K

. Two

elements

α

,

β

of

L

are

algebraically dependent

if there exists a non-zero

polynomial

f

⁢

(

x

,

y

)

∈

K

⁢

[

x

,

y

]

such that

f

⁢

(

α

,

β

)

=

0

. If no such polynomial exists,

α

and

β

are said to be

algebraically independent

.

More generally, elements

α

1

,

…

,

α

n

∈

L

are said to be algebraically dependent if there exists a non-zero polynomial

f

⁢

(

x

1

,

…

,

x

n

)

∈

K

⁢

[

x

1

,

…

,

x

n

]

such that

f

⁢

(

α

1

,

α

2

,

…

,

α

n

)

=

0

. If no such polynomial exists, the

collection

of

α

’s are said to be algebraically independent.

Title

algebraically dependent

Canonical name

AlgebraicallyDependent

Date of creation

2013-03-22 13:58:13

Last modified on

2013-03-22 13:58:13

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

8

Author

mathcam (2727)

Entry type

Definition

Classification

msc 12F05

Classification

msc 11J85

Related topic

DependenceRelation

Defines

algebraically independent

Defines

algebraic dependence

Defines

algebraic independence

Generated on Fri Feb 9 16:41:53 2018 by

LaTeXML
alternative definition of algebraically closed

alternative definition of algebraically closed

Proposition 1

.

If

K

is a field, the following are

equivalent

:

(1)

K

is

algebraically closed

, i.e. every nonconstant

polynomial

f

in

K

⁢

[

x

]

has a root in

K

.

(2)

Every nonconstant polynomial

f

in

K

⁢

[

x

]

splits completely over

K

.

(3)

If

L

|

K

is an

algebraic extension

then

L

=

K

.

Proof.

If (1) is true then we can prove by

induction

on degree of

f

that every nonconstant polynomial

f

splits completely over

K

.

Conversely

, (2)

⇒

(1) is

trivial

.

(2)

⇒

(3) If

L

|

K

is

algebraic

and

α

∈

L

, then

α

is a root of a polynomial

f

∈

K

⁢

[

x

]

. By (2)

f

splits over

K

, which implies that

α

∈

K

. It follows that

L

=

K

.

(3)

⇒

(1) Let

f

∈

K

⁢

[

x

]

and

α

a root of

f

(in some

extension

of

K

). Then

K

⁢

(

α

)

is an algebraic extension of

K

, hence

α

∈

K

.
∎

Examples

1) The

field of real

numbers

ℝ

is not algebraically closed. Consider the equation

x

2

+

1

=

0

. The square of a

real number

is always positive and cannot be

-

1

so the equation has no roots.

2) The

p

-adic field

ℚ

p

is not algebraically closed because the equation

x

2

-

p

=

0

has no roots. Otherwise

x

2

=

p

implies

2

⁢

v

p

⁢

x

=

1

, which is false.

Title

alternative definition of algebraically closed

Canonical name

AlternativeDefinitionOfAlgebraicallyClosed

Date of creation

2013-03-22 16:53:23

Last modified on

2013-03-22 16:53:23

Owner

polarbear (3475)

Last modified by

polarbear (3475)

Numerical id

8

Author

polarbear (3475)

Entry type

Derivation

Classification

msc 12F05

Generated on Fri Feb 9 16:41:55 2018 by

LaTeXML
conjugate fields

conjugate fields

If

ϑ

1

,

ϑ

2

,

…

,

ϑ

n

are the

algebraic conjugates

of the

algebraic number

ϑ

1

, then the

algebraic number fields

ℚ

⁢

(

ϑ

1

)

,

ℚ

⁢

(

ϑ

2

)

,

…

,

ℚ

⁢

(

ϑ

n

)

are the

conjugate fields

of

ℚ

⁢

(

ϑ

1

)

.

Notice that the conjugate fields of

ℚ

⁢

(

ϑ

1

)

are always

isomorphic

but not necessarily distinct.

All conjugate fields are equal, i.e. (

http://planetmath.org/Ie

)

ℚ

⁢

(

ϑ

1

)

=

ℚ

⁢

(

ϑ

2

)

=

…

=

ℚ

⁢

(

ϑ

n

)

, or equivalently

ϑ

1

,

…

,

ϑ

n

belong to

ℚ

⁢

(

ϑ

1

)

, if and only if the

extension

ℚ

⁢

(

ϑ

1

)

/

ℚ

is a

Galois extension

of fields. The reason for this is that if

ϑ

1

is an algebraic number and

m

⁢

(

x

)

is the

minimal polynomial

of

ϑ

1

then the roots of

m

⁢

(

x

)

are precisely the algebraic conjugates of

ϑ

1

.

For example, let

ϑ

1

=

2

. Then its only

conjugate

is

ϑ

2

=

-

2

and

ℚ

⁢

(

2

)

is Galois and contains both

ϑ

1

and

ϑ

2

. Similarly, let

p

be a prime and let

ϑ

1

=

ζ

be a

primitive

p

th

root of unity

(

http://planetmath.org/PrimitiveRootOfUnity

). Then the algebraic conjugates of

ζ

are

ζ

2

,

…

,

ζ

p

-

1

and so all conjugate fields are equal to

ℚ

⁢

(

ζ

)

and the extension

ℚ

⁢

(

ζ

)

/

ℚ

is Galois. It is a

cyclotomic extension

of

ℚ

.

Now let

ϑ

1

=

2

3

and let

ζ

be a primitive

3

rd root of unity (i.e.

ζ

is a root of

x

2

+

x

+

1

, so we can pick

ζ

=

-

1

+

-

3

2

). Then the conjugates of

ϑ

1

are

ϑ

1

,

ϑ

2

=

ζ

⁢

2

3

, and

ϑ

3

=

ζ

2

⁢

2

3

. The three conjugate fields

ℚ

⁢

(

ϑ

1

)

,

ℚ

⁢

(

ϑ

2

)

, and

ℚ

⁢

(

ϑ

3

)

are distinct in this case. The

Galois closure

of each of these fields is

ℚ

⁢

(

ζ

,

2

3

)

.

Title

conjugate fields

Canonical name

ConjugateFields

Date of creation

2013-03-22 17:10:28

Last modified on

2013-03-22 17:10:28

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Definition

Classification

msc 12F05

Classification

msc 11R04

Related topic

PropertiesOfMathbbQvarthetaConjugates

Generated on Fri Feb 9 16:42:02 2018 by

LaTeXML
every algebraically closed field is perfect

every algebraically closed field is perfect

Proposition 1

.

Every algebraically closed field is

perfect

Proof.

Let

K

be an algebraically closed field of prime

characteristic

p

. Take

a

∈

K

. Then the

polynomial

X

p

-

a

admits a zero in

K

. It follows that

a

admits a

p

th root in

K

. Since

a

is arbitrary we have proved that the field

K

is perfect.∎

Title

every algebraically closed field is perfect

Canonical name

EveryAlgebraicallyClosedFieldIsPerfect

Date of creation

2013-03-22 16:53:06

Last modified on

2013-03-22 16:53:06

Owner

polarbear (3475)

Last modified by

polarbear (3475)

Numerical id

6

Author

polarbear (3475)

Entry type

Result

Classification

msc 12F05

Generated on Fri Feb 9 16:42:04 2018 by

LaTeXML
examples of trace and norm

examples of trace and norm

Let

ω

be a

complex root

of unity different than 1. Then

ω

and

ω

2

are the

conjugate

roots of the

minimal polynomial

x

2

+

x

+

1

.
Since

ℚ

⁢

(

ω

)

is the

splitting field

of

x

2

+

x

+

1

, it is Galois over

ℚ

. Moreover the

Galois group

G

a

l

(

ℚ

(

ω

)

/

ℚ

)

)

is formed by the

identity

and the

automorphism

g

⁢

(

ω

)

=

ω

2

The elements of

ℚ

⁢

(

ω

)

have the form

a

+

b

⁢

ω

,

a

,

b

∈

ℚ

.
Then we obtain

N

ℚ

⁢

(

ω

)

ℚ

⁢

(

a

+

b

⁢

ω

)

=

(

a

+

b

⁢

ω

)

⁢

(

a

+

b

⁢

ω

2

)

=

a

2

-

a

⁢

b

+

b

2

,

T

⁢

r

ℚ

⁢

(

ω

)

ℚ

⁢

(

a

+

b

⁢

ω

)

=

(

a

+

b

⁢

ω

)

+

(

a

+

b

⁢

ω

2

)

=

2

⁢

a

-

b

Title

examples of trace and norm

Canonical name

ExamplesOfTraceAndNorm

Date of creation

2013-03-22 15:55:45

Last modified on

2013-03-22 15:55:45

Owner

polarbear (3475)

Last modified by

polarbear (3475)

Numerical id

16

Author

polarbear (3475)

Entry type

Example

Classification

msc 12F05

Generated on Fri Feb 9 16:42:09 2018 by

LaTeXML
existence of extensions of field isomorphisms to splitting fields

existence of extensions of field isomorphisms to splitting fields

The following

theorem

implies the essential uniqueness of

splitting fields

and

algebraic closures

.

Theorem

.

Let

σ

:

F

→

F

′

be an

isomorphism

of fields,

S

=

{

f

α

:

α

∈

A

}

a set of non-constant

polynomials

in

F

⁢

[

X

]

, and

S

′

=

{

σ

⁢

(

f

α

)

:

α

∈

A

}

the corresponding set of polynomials in

F

′

⁢

[

X

]

. If

K

is a splitting field of

S

over

F

and

K

′

a splitting field of

S

′

over

F

′

, then

σ

may be extended to an isomorphism of

K

and

K

′

.

Corollary

.

If

F

is a field and

S

a set of non-constant polynomials in

F

⁢

[

X

]

, then any two splitting fields of

S

over

F

are

F

-isomorphic. In particular, any two algebraic closures of

F

are

F

-isomorphic.

Title

existence of extensions of field isomorphisms to splitting fields

Canonical name

ExistenceOfExtensionsOfFieldIsomorphismsToSplittingFields

Date of creation

2013-03-22 18:37:59

Last modified on

2013-03-22 18:37:59

Owner

azdbacks4234 (14155)

Last modified by

azdbacks4234 (14155)

Numerical id

4

Author

azdbacks4234 (14155)

Entry type

Theorem

Classification

msc 12F05

Related topic

SplittingField

Generated on Fri Feb 9 16:42:11 2018 by

LaTeXML
existence of the minimal polynomial

existence of the minimal polynomial

Proposition 1

.

Let

K

/

L

be a

finite extension

of fields and let

k

∈

K

. There
exists a unique

polynomial

m

k

⁢

(

x

)

∈

L

⁢

[

x

]

such that:

1.

m

k

⁢

(

x

)

is a

monic polynomial

;

2.

m

k

⁢

(

k

)

=

0

;

3.

If

p

⁢

(

x

)

∈

L

⁢

[

x

]

is another polynomial such that

p

⁢

(

k

)

=

0

,
then

m

k

⁢

(

x

)

divides

p

⁢

(

x

)

.

Proof.

We start by defining the following map:

ψ

:

L

⁢

[

x

]

→

K

ψ

⁢

(

p

⁢

(

x

)

)

=

p

⁢

(

k

)

Note that this map is clearly a

ring homomorphism

. For all

p

⁢

(

x

)

,

q

⁢

(

x

)

∈

L

⁢

[

x

]

:

•

ψ

⁢

(

p

⁢

(

x

)

+

q

⁢

(

x

)

)

=

p

⁢

(

k

)

+

q

⁢

(

k

)

=

ψ

⁢

(

p

⁢

(

x

)

)

+

ψ

⁢

(

q

⁢

(

x

)

)

•

ψ

⁢

(

p

⁢

(

x

)

⋅

q

⁢

(

x

)

)

=

p

⁢

(

k

)

⋅

q

⁢

(

k

)

=

ψ

⁢

(

p

⁢

(

x

)

)

⋅

ψ

⁢

(

q

⁢

(

x

)

)

Thus, the kernel of

ψ

is an ideal of

L

⁢

[

x

]

:

Ker

⁡

(

ψ

)

=

{

p

⁢

(

x

)

∈

L

⁢

[

x

]

∣

p

⁢

(

k

)

=

0

}

Note that the kernel is a

non-zero

ideal. This fact relies
on the fact that

K

/

L

is a finite extension of fields, and
therefore it is an

algebraic extension

, so every element of

K

is
a root of a non-zero polynomial

p

⁢

(

x

)

with

coefficients

in

L

,
this is,

p

⁢

(

x

)

∈

Ker

⁡

(

ψ

)

.

Moreover, the ring of polynomials

L

⁢

[

x

]

is a

principal ideal
domain

(see example of PID).
Therefore, the kernel of

ψ

is a

principal ideal

, generated by
some polynomial

m

⁢

(

x

)

:

Ker

⁡

(

ψ

)

=

(

m

⁢

(

x

)

)

Note that the only units in

L

⁢

[

x

]

are the

constant polynomials

,
hence if

m

′

⁢

(

x

)

is another

generator

of

Ker

⁡

(

ψ

)

then

m

′

⁢

(

x

)

=

l

⋅

m

⁢

(

x

)

,

l

≠

0

,

l

∈

L

Let

α

be the leading coefficient of

m

⁢

(

x

)

. We define

m

k

⁢

(

x

)

=

α

-

1

⁢

m

⁢

(

x

)

, so that the leading coefficient of

m

k

is

1

. Also note that by the previous remark,

m

k

is
the unique generator of

Ker

⁡

(

ψ

)

which is monic.

By construction,

m

k

⁢

(

k

)

=

0

, since

m

k

belongs to the kernel
of

ψ

, so it satisfies

(

2

)

.

Finally, if

p

⁢

(

x

)

is any polynomial such that

p

⁢

(

k

)

=

0

, then

p

⁢

(

x

)

∈

Ker

⁡

(

ψ

)

. Since

m

k

generates

this
ideal, we know that

m

k

must divide

p

⁢

(

x

)

(this is property

(

3

)

).

For the uniqueness, note that any polynomial satisfying

(

2

)

and

(

3

)

must be a generator of

Ker

⁡

(

ψ

)

, and, as
we pointed out, there is a unique monic generator, namely

m

k

⁢

(

x

)

.

∎

Title

existence of the

minimal polynomial

Canonical name

ExistenceOfTheMinimalPolynomial

Date of creation

2013-03-22 13:57:24

Last modified on

2013-03-22 13:57:24

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

7

Author

alozano (2414)

Entry type

Theorem

Classification

msc 12F05

Related topic

FiniteExtension

Related topic

Algebraic

Generated on Fri Feb 9 16:42:14 2018 by

LaTeXML
expressible

expressible

Let

F

be a field and

α

be algebraic (

http://planetmath.org/AlgebraicElement

) over

F

. Then

α

is

expressible

over

F

if

F

⁢

(

α

)

/

F

is a

radical extension

. On the other hand,

α

is

inexpressible

over

F

if

F

⁢

(

α

)

/

F

is not a radical extension.

Title

expressible

Canonical

name

Expressible

Date of creation

2013-03-22 16:55:41

Last modified on

2013-03-22 16:55:41

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

7

Author

Wkbj79 (1863)

Entry type

Definition

Classification

msc 12F05

Classification

msc 12F10

Defines

inexpressible

Generated on Fri Feb 9 16:42:16 2018 by

LaTeXML
finite extension

finite extension

Let

K

an

extension field

of

F

. We say that

K

is a

finite extension

if

[

K

:

F

]

is finite. That is,

K

is a

finite dimensional

space over

F

.

An important result on finite extensions establishes that any finite extension is also an

algebraic extension

.

Title

finite extension

Canonical name

FiniteExtension

Date of creation

2013-03-22 11:56:55

Last modified on

2013-03-22 11:56:55

Owner

drini (3)

Last modified by

drini (3)

Numerical id

9

Author

drini (3)

Entry type

Definition

Classification

msc 12F05

Synonym

finite field extension

Related topic

ExtensionField

Related topic

AlgebraicElement

Related topic

ExistenceOfTheMinimalPolynomial

Related topic

AlgebraicExtension

Related topic

ExtensionMathbbRmathbbQIsNotFinite

Related topic

AlgebraicSumAndProduct

Related topic

FiniteExtensionsOfDedekindDomainsAreDedekind

Generated on Fri Feb 9 16:42:17 2018 by

LaTeXML
finite field cannot be algebraically closed

finite field cannot be algebraically closed

Theorem

.

A

finite field cannot be algebraically closed

.

Proof.

The proof proceeds by the method of

contradiction

. Assume that a field

F

is both finite and

algebraically closed

. Consider the

polynomial

p

⁢

(

x

)

=

x

2

-

x

as a

function

from

F

to

F

. There are two

elements

which any field (in particular,

F

) must have — the

additive identity

0

and the

multiplicative identity

1

. The polynomial

p

maps both of
these elements to

0

. Since

F

is finite and the function

p

:

F

→

F

is not

one-to-one

, the function cannot map onto

F

either, so
there must exist an element

a

of

F

such that

x

2

-

x

≠

a

for
all

x

∈

F

. In other words, the polynomial

x

2

-

x

-

a

has no root
in

F

, so

F

could not be algebraically closed.
∎

Title

finite field cannot be algebraically closed

Canonical name

FiniteFieldCannotBeAlgebraicallyClosed

Date of creation

2013-03-22 16:29:09

Last modified on

2013-03-22 16:29:09

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

6

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 12F05

Related topic

AlgebraicClosureOfAFiniteField

Generated on Fri Feb 9 16:42:19 2018 by

LaTeXML
irreducible polynomials obtained from biquadratic fields

irreducible polynomials obtained from biquadratic fields

Corollary

.

Let

m

and

n

be distinct

squarefree

integers

, neither of which is equal to

1

. Then the

polynomial

x

4

-

2

⁢

(

m

+

n

)

⁢

x

2

+

(

m

-

n

)

2

is irreducible (

http://planetmath.org/IrreduciblePolynomial2

) (over

Q

).

Proof.

By the theorem stated in the parent entry (

http://planetmath.org/PrimitiveElementOfBiquadraticField

),

m

+

n

is an

algebraic number

of degree (

http://planetmath.org/DegreeOfAnAlgebraicNumber

)

4

. Thus, a polynomial of degree

4

that has

m

+

n

as a root must be over

ℚ

. We set out to

construct

such a polynomial.

x

=

m

+

n

x

-

m

=

n

(

x

-

m

)

2

=

n

x

2

-

2

⁢

m

⁢

x

+

m

=

n

x

2

+

m

-

n

=

2

⁢

m

⁢

x

(

x

2

+

m

-

n

)

2

=

4

⁢

m

⁢

x

2

x

4

+

(

2

⁢

m

-

2

⁢

n

)

⁢

x

2

+

(

m

-

n

)

2

=

4

⁢

m

⁢

x

2

x

4

+

(

2

⁢

m

-

2

⁢

n

-

4

⁢

m

)

⁢

x

2

+

(

m

-

n

)

2

=

0

x

4

-

2

⁢

(

m

+

n

)

⁢

x

2

+

(

m

-

n

)

2

=

0

∎

Title

irreducible polynomials obtained from biquadratic fields

Canonical name

IrreduciblePolynomialsObtainedFromBiquadraticFields

Date of creation

2013-03-22 17:54:22

Last modified on

2013-03-22 17:54:22

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

5

Author

Wkbj79 (1863)

Entry type

Corollary

Classification

msc 12F05

Classification

msc 12E05

Classification

msc 11R16

Related topic

ExamplesOfMinimalPolynomials

Related topic

BiquadraticEquation2

Generated on Fri Feb 9 16:42:22 2018 by

LaTeXML
Kummer theory

Kummer theory

The following theorem is usually referred to as

Kummer theory

.

Theorem 1

(Kummer Theory).

Let

n

be a

positive integer

and let

K

be a field of

characteristic

not dividing

n

which contains the

n

-th

roots of unity

. Then:

1.

The

extension

K

⁢

(

a

n

)

for

a

∈

K

is a

cyclic extension

over

K

of degree dividing

n

.

2.

Any cyclic extension of degree

n

over

K

is of the form

K

⁢

(

a

n

)

for some

a

∈

K

.

Definition 1

.

Let

n

be a positive integer and let

K

be a field of characteristic not dividing

n

which contains the

n

-th roots of unity. An extension of

K

of the form:

K

⁢

(

a

1

n

,

a

2

n

,

…

,

a

k

n

)

with

a

i

∈

K

×

is called a

Kummer extension

of

K

. Notice that the

Galois group

of the extension is of

exponent

(

http://planetmath.org/Exponent

)

n

.

Title

Kummer theory

Canonical name

KummerTheory

Date of creation

2013-03-22 15:04:20

Last modified on

2013-03-22 15:04:20

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

5

Author

alozano (2414)

Entry type

Theorem

Classification

msc 12F05

Related topic

AbelianExtension

Related topic

CyclicExtension

Related topic

Exponent

Defines

Kummer extension

Generated on Fri Feb 9 16:42:23 2018 by

LaTeXML