aarjaneiro/planetmath huggingface dataset:
norm

norm

Let

K

/

F

be a

Galois extension

, and let

x

∈

K

. The

norm

N

F

K

⁡

(

x

)

of

x

is defined to be the product of all the

elements of

the orbit of

x

under the

group action

of the

Galois group

Gal

⁡

(

K

/

F

)

on

K

; taken with multiplicities if

K

/

F

is a

finite extension

.

In the case where

K

/

F

is a finite extension, the norm of

x

can be defined to be the

determinant

of the

linear transformation

[

x

]

:

K

→

K

given by

[

x

]

⁢

(

k

)

:=

x

⁢

k

, where

K

is regarded as a

vector space

over

F

. This definition does not require that

K

/

F

be Galois, or even that

K

be a field—for instance, it remains valid when

K

is a

division ring

(although

F

does have to be a field, in order for determinant to be defined). Of course, for finite Galois extensions

K

/

F

, this definition agrees with the previous one, and moreover the formula

N

F

K

⁡

(

x

)

:=

∏

σ

∈

Gal

⁡

(

K

/

F

)

σ

⁢

(

x

)

holds.

The norm of

x

is always an element of

F

, since any element of

Gal

⁡

(

K

/

F

)

permutes the orbit of

x

and thus fixes

N

F

K

⁡

(

x

)

.

Title

norm

Canonical name

Norm

Date of creation

2013-03-22 12:18:02

Last modified on

2013-03-22 12:18:02

Owner

djao (24)

Last modified by

djao (24)

Numerical id

5

Author

djao (24)

Entry type

Definition

Classification

msc 12F05

Generated on Fri Feb 9 16:42:25 2018 by

LaTeXML
orders in a number field

orders in a number field

If

μ

1

,

…

,

μ

m

are elements of an

algebraic number field

K

, then the subset

M

=

{

n

1

μ

1

+

…

+

n

m

μ

m

∈

K

⋮

n

i

∈

ℤ

∀

i

}

of

K

is a

ℤ

-module, called a

module in

K

.  If the module contains as many over

ℤ

linearly independent

elements as is the degree (

http://planetmath.org/NumberField

) of

K

over

ℚ

, then the module is

complete

.

If a complete module in

K

the unity 1 of

K

and is a ring, it is called an

order

(in German:

Ordnung

) in the field

K

.

A number

α

of the algebraic number field

K

is called a

coefficient

of the module

M

, if

α

⁢

M

⊆

M

.

Theorem

1.

The set

ℒ

M

of all coefficients of a complete module

M

is an order in the field.  Conversely, every order

ℒ

in the

number field

K

is a

coefficient ring

of some module.

Theorem 2.

If

α

belongs to an order in the field, then the coefficients of the

characteristic equation

(

http://planetmath.org/CharacteristicEquation

) of

α

and thus the coefficients of the

minimal polynomial

of

α

are

rational integers

.

Theorem 2 means that any order is contained in the

ring of integers

of the algebraic number field

K

.  Thus this ring

𝒪

K

, being itself an order, is the greatest order;

𝒪

K

is called the

maximal order

or the

principal order

(in German:

Hauptordnung

).  The set of the orders is partially ordered by the

set inclusion

.

Example.

In the field

ℚ

⁢

(

2

)

, the coefficient ring of the module

M

generated by

2

and

2

2

is the module

ℒ

M

generated by

1

and

2

⁢

2

.  The maximal order of the field is generated by

1

and

2

.

References

1

S. Borewicz & I. Safarevic

:

Zahlentheorie

.  Birkhäuser Verlag. Basel und Stuttgart (1966).

Title

orders in a number field

Canonical

name

OrdersInANumberField

Date of creation

2013-03-22 16:52:46

Last modified on

2013-03-22 16:52:46

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

17

Author

pahio (2872)

Entry type

Topic

Classification

msc 12F05

Classification

msc 11R04

Classification

msc 06B10

Related topic

Module

Defines

module

Defines

complete

Defines

order of a number field

Defines

principal order

Defines

maximal order

Generated on Fri Feb 9 16:42:27 2018 by

LaTeXML
p-extension

p

-extension

Definition 1

.

Let

p

be a

prime number

. A

Galois extension

of fields

E

/

F

, with

G

=

Gal

⁡

(

E

/

F

)

, is said to be a

p

-

extension

if

G

is a

p

-group.

Example 1

.

Let

d

be a square-free integer. Then the field extension

Q

⁢

(

d

)

/

Q

is a

2

-extension.

Example 2

.

Let

p

>

2

be a prime and, for any

n

, let

ζ

p

n

be a

primitive

p

n

th

root of unity

. The

cyclotomic extension

:

ℚ

⁢

(

ζ

p

n

)

/

ℚ

⁢

(

ζ

p

)

is a

p

-extension. Indeed:

G

n

=

Gal

⁡

(

ℚ

⁢

(

ζ

p

n

)

/

ℚ

)

≅

(

ℤ

/

p

n

⁢

ℤ

)

×

Thus,

|

G

n

|

=

φ

⁢

(

p

n

)

=

p

(

n

-

1

)

⁢

(

p

-

1

)

and

|

G

1

|

=

φ

⁢

(

p

)

=

p

-

1

, where

φ

is the

Euler phi function

. Therefore the extension above is of degree

p

(

n

-

1

)

.

Title

p

-extension

Canonical

name

Pextension

Date of creation

2013-03-22 15:02:56

Last modified on

2013-03-22 15:02:56

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

5

Author

alozano (2414)

Entry type

Definition

Classification

msc 12F05

Synonym

p-extension

Related topic

PGroup4

Related topic

UnramifiedExtensionsAndClassNumberDivisibility

Related topic

PushDownTheoremOnClassNumbers

Related topic

ClassNumberDivisibilityInPExtensions

Related topic

QuadraticExtension

Generated on Fri Feb 9 16:42:29 2018 by

LaTeXML
primitive element theorem

primitive element theorem

Theorem 1

.

Let

F

and

K

be arbitrary fields, and let

K

be an

extension

of

F

of finite degree. Then there exists an element

α

∈

K

such that

K

=

F

⁢

(

α

)

if and only if there are finitely many fields

L

with

F

⊆

L

⊆

K

.

Note that this implies that every finite

separable extension

is not only

finitely generated

, it is generated by a single element.

Let

X

be an

indeterminate

. Then

ℚ

⁢

(

X

,

i

)

is not generated over

ℚ

by a single element (and there are infinitely many intermediate fields

ℚ

⁢

(

X

,

i

)

/

L

/

ℚ

). To see this, suppose it is generated by an element

α

. Then clearly

α

must be

transcendental

, or it would generate an extension of finite degree. But if

α

is transcendental, we know it is

isomorphic

to

ℚ

⁢

(

X

)

, and this field is not isomorphic to

ℚ

⁢

(

X

,

i

)

: for example, the

polynomial

Y

2

+

1

has no roots in the first but it has two roots in the second. It is also clear that it is not sufficient for every element of

K

to be algebraic over

F

: we know that the

algebraic closure

of

ℚ

has

infinite

degree over

ℚ

, but if

α

is algebraic over

ℚ

then

[

ℚ

⁢

(

α

)

:

ℚ

]

will be finite.

This

theorem

has the corollary:

Corollary 1

.

Let

F

be a field, and let

[

F

⁢

(

β

,

γ

)

:

F

]

be finite and separable. Then there exists

α

∈

F

⁢

(

β

,

γ

)

such that

F

⁢

(

β

,

γ

)

=

F

⁢

(

α

)

. In fact, we can always take

α

to be an

F

-

linear combination

(

http://planetmath.org/LinearCombination

) of

β

and

γ

.

To see this (in the case of

characteristic

0

), we need only show that there are finitely many intermediate fields. But any intermediate field is contained in the

splitting field

of the

minimal polynomials

of

β

and

γ

, which is Galois with finite

Galois group

. The explicit form of

α

comes from the proof of the theorem.

For more detail on this theorem and its proof see, for example,

Field and

Galois Theory

, by Patrick Morandi (Springer Graduate Texts in Mathematics 167, 1996).

Title

primitive element theorem

Canonical

name

PrimitiveElementTheorem

Date of creation

2013-03-22 11:45:48

Last modified on

2013-03-22 11:45:48

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

18

Author

alozano (2414)

Entry type

Theorem

Classification

msc 12F05

Classification

msc 65-01

Related topic

SimpleFieldExtension

Related topic

PrimitiveElementOfBiquadraticField2

Related topic

PrimitiveElementOfBiquadraticField

Generated on Fri Feb 9 16:42:31 2018 by

LaTeXML
proof of Kummer theory

proof of Kummer theory

Proof.

Let

ζ

∈

K

be a

primitive

n

th

root of unity, and denote by

𝝁

n

the

subgroup

of

K

⋆

generated by

ζ

.

(1) Let

L

=

K

⁢

(

a

n

)

; then

L

/

K

is Galois since

K

contains all

n

th

roots of unity and thus is a

splitting field

for

x

n

-

a

, which is

separable

since

n

≠

0

in

K

. Thus the elements of

Gal

⁡

(

L

/

K

)

permute the roots of

x

n

-

a

, which are

a

n

,

ζ

⁢

a

n

,

ζ

2

⁢

a

n

,

…

,

ζ

n

-

1

⁢

a

n

and thus for

σ

∈

Gal

⁡

(

L

/

K

)

, we have

σ

⁢

(

a

n

)

=

ζ

σ

⁢

a

n

for some

ζ

σ

∈

𝝁

n

.
Define a map

p

:

Gal

⁡

(

L

/

K

)

→

𝝁

n

:

σ

↦

ζ

σ

We will show that

p

is an

injective

homomorphism

, which proves the result.

Since

𝝁

n

⊂

K

, each

n

th

root of unity is fixed by

Gal

⁡

(

L

/

K

)

. Then for

σ

,

τ

∈

Gal

⁡

(

L

/

K

)

,

ζ

σ

⁢

τ

⁢

a

n

=

σ

⁢

τ

⁢

(

a

n

)

=

σ

⁢

(

ζ

τ

⁢

a

n

)

=

ζ

τ

⁢

(

σ

⁢

(

a

n

)

)

=

ζ

σ

⁢

ζ

τ

⁢

a

n

so that

ζ

σ

⁢

τ

=

ζ

σ

⁢

ζ

τ

and

p

is a homomorphism. The kernel of the map consists of all elements of

Gal

⁡

(

L

/

K

)

which fix

a

n

, so that

p

is injective and we are done.

(2) Note that

N

L

/

K

⁡

(

ζ

)

=

1

since

ζ

is a root of

x

n

-

1

, so that by Hilbert’s

Theorem

90,

ζ

=

σ

⁢

(

u

)

/

u

,

for some

⁢

u

∈

L

But then

σ

⁢

(

u

)

=

ζ

⁢

u

so that

σ

⁢

(

u

n

)

=

σ

⁢

(

u

)

n

=

ζ

n

⁢

u

n

=

u

n

and

a

=

u

n

∈

K

since it is fixed by a

generator

of

Gal

⁡

(

L

/

K

)

. Then clearly

K

⁢

(

u

)

is a splitting field of

x

n

-

a

, and the elements of

Gal

⁡

(

L

/

K

)

send

u

into distinct elements of

K

⁢

(

u

)

. Thus

K

⁢

(

u

)

admits at least

n

automorphisms

over

K

, so that

[

K

(

u

)

:

K

]

≥

n

=

[

L

:

K

]

. But

K

⁢

(

u

)

⊂

L

, so

K

⁢

(

a

n

)

=

K

⁢

(

u

)

=

L

.
∎

References

1

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

Abstract Algebra

, Third Edition

, Wiley, 2004.

2

Kaplansky, I.,

Fields and Rings

, University of Chicago Press, 1969.

Title

proof of Kummer theory

Canonical name

ProofOfKummerTheory

Date of creation

2013-03-22 18:42:07

Last modified on

2013-03-22 18:42:07

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

5

Author

rm50 (10146)

Entry type

Proof

Classification

msc 12F05

Generated on Fri Feb 9 16:42:36 2018 by

LaTeXML
proof of primitive element theorem

proof of primitive element theorem

Theorem

.

Let

F

and

K

be arbitrary fields, and let

K

be an

extension

of

F

of finite degree. Then there exists an

element

α

∈

K

such that

K

=

F

⁢

(

α

)

if and only if there are finitely many fields

L

with

F

⊆

L

⊆

K

.

Proof.

Let

F

and

K

be fields, and let

[

K

:

F

]

=

n

be finite.

Suppose first that

K

=

F

⁢

(

α

)

. Since

K

/

F

is finite,

α

is algebraic over

F

. Let

m

⁢

(

x

)

be the

minimal polynomial

of

α

over

F

. Now, let

L

be an intermediary field with

F

⊆

L

⊆

K

and let

m

′

⁢

(

x

)

be the minimal polynomial of

α

over

L

. Also, let

L

′

be the field generated by the

coefficients

of the

polynomial

m

′

⁢

(

x

)

. Thus, the minimal polynomial of

α

over

L

′

is still

m

′

⁢

(

x

)

and

L

′

⊆

L

. By the properties of the minimal polynomial, and since

m

⁢

(

α

)

=

0

, we have a

divisibility

m

′

(

x

)

|

m

(

x

)

, and so:

[

K

:

L

]

=

deg

(

m

′

(

x

)

)

=

[

K

:

L

′

]

.

Since we know that

L

′

⊆

L

, this implies that

L

′

=

L

. Thus, this shows that each intermediary

subfield

F

⊆

L

⊆

K

corresponds with the field of definition of a (monic) factor of

m

⁢

(

x

)

. Since the polynomial

m

⁢

(

x

)

has only finitely many monic factors, we conclude that there can be only finitely many subfields of

K

containing

F

.

Now suppose

conversely

that there are only finitely many such intermediary fields

L

. If

F

is a

finite field

, then so is

K

, and we have an explicit description of all such possibilities; all such extensions are generated by a single element. So assume

F

(and therefore

K

) are

infinite

. Let

α

1

,

α

2

,

…

,

α

n

be a basis for

K

over

F

. Then

K

=

F

⁢

(

α

1

,

…

,

α

n

)

. So if we can show that any field extension generated by two elements is also generated by one element, we will be done: simply apply the result to the last two elements

α

j

-

1

and

α

j

repeatedly until only one is left.

So assume

K

=

F

⁢

(

β

,

γ

)

. Consider the set of elements

{

β

+

a

⁢

γ

}

for

a

∈

F

×

. By

assumption

, this set is infinite, but there are only finitely many fields intermediate between

K

and

F

; so two values must generate the same extension

L

of

F

, say

β

+

a

⁢

γ

and

β

+

b

⁢

γ

. This field

L

contains

(

β

+

a

⁢

γ

)

-

(

β

+

b

⁢

γ

)

a

-

b

=

γ

and

(

β

+

a

⁢

γ

)

/

a

-

(

β

+

b

⁢

γ

)

/

b

1

/

a

-

1

/

b

=

β

and so letting

α

=

β

+

a

⁢

γ

, we see that

F

⁢

(

α

)

=

L

=

F

⁢

(

β

,

γ

)

=

K

.

∎

Title

proof of primitive element theorem

Canonical name

ProofOfPrimitiveElementTheorem

Date of creation

2013-03-22 14:16:27

Last modified on

2013-03-22 14:16:27

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

8

Author

alozano (2414)

Entry type

Proof

Classification

msc 12F05

Generated on Fri Feb 9 16:42:38 2018 by

LaTeXML
quadratic closure

quadratic closure

A field

K

is said to be

quadratically closed

if it has no

quadratic extensions

. In other words, every element of

K

is a square. Two obvious examples are

ℂ

and

𝔽

2

.

A field

K

is said to be a

quadratic closure

of another field

k

if

1.

K

is quadratically closed, and

2.

among all quadratically closed

subfields

of the

algebraic closure

k

¯

of

k

,

K

is the smallest one.

By the second condition, a quadratic closure of a field is unique up to

field isomorphism

. So we say

the

quadratic closure of a field

k

, and we denote it by

k

~

Alternatively, the second condition on

K

can be replaced by the following:

K

is the smallest

field extension

over

k

such that, if

L

is any field extension over

k

obtained by a finite number of quadratic extensions starting with

k

, then

L

is a subfield of

K

.

Examples.

•

ℂ

=

ℝ

~

.

•

If

𝔼

is the

Euclidean field

in

ℝ

, then the quadratic extension

𝔼

⁢

(

-

1

)

is the quadratic closure

ℚ

~

of the

rational numbers

ℚ

.

•

If

k

=

𝔽

5

, consider the chain of fields

k

≤

k

⁢

(

2

)

≤

k

⁢

(

2

4

)

≤

⋯

≤

k

⁢

(

2

2

n

)

≤

⋯

Take the union of all these fields to obtain a field

K

. Then it can be shown that

K

=

k

~

.

Title

quadratic closure

Canonical name

QuadraticClosure

Date of creation

2013-03-22 15:42:43

Last modified on

2013-03-22 15:42:43

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

8

Author

CWoo (3771)

Entry type

Definition

Classification

msc 12F05

Defines

quadratically closed

Generated on Fri Feb 9 16:42:40 2018 by

LaTeXML
radical

radical

Let

F

be a field and

α

be

algebraic

(

http://planetmath.org/Algebraic

) over

F

. Then

α

is a

radical

over

F

if there exists a

positive integer

n

with

α

n

∈

F

.

Note that, if

K

/

F

is a

field extension

and

α

is a radical over

F

, then

α

is automatically a radical over

K

.

Following are some examples of radicals:

1.

All numbers of the form

a

b

n

, where

n

is a positive integer and

a

and

b

are

integers

with

b

≠

0

are radicals over

ℚ

.

2.

The number

2

4

is a radical over

ℚ

⁢

(

2

)

since

(

2

4

)

2

=

2

∈

ℚ

⁢

(

2

)

.

Title

radical

Canonical name

Radical1

Date of creation

2013-03-22 16:55:36

Last modified on

2013-03-22 16:55:36

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

9

Author

Wkbj79 (1863)

Entry type

Definition

Classification

msc 12F05

Classification

msc 12F10

Related topic

RadicalExtension

Related topic

NthRoot

Related topic

SolvableByRadicals

Related topic

Radical6

Generated on Fri Feb 9 16:42:44 2018 by

LaTeXML
rational integers in ideals

rational integers in ideals

Any non-zero ideal of an

algebraic number field

K

, i.e. of the

maximal order

𝒪

K

of

K

, contains

positive

rational integers

.

Proof.

Let

𝔞

≠

(

0

)

be any ideal of

𝒪

K

.  Take a nonzero element

α

of

𝔞

.  The norm (

http://planetmath.org/NormInNumberField

) of

α

is the product

N

⁢

(

α

)

=

α

(

1

)

⁢

α

(

2

)

⁢

⋯

⁢

α

(

n

)

⏟

γ

where

n

is the degree of the

number field

and

α

(

1

)

,

α

(

2

)

,

⋯

,

α

(

n

)

is the set of the

http://planetmath.org/node/12046

K

-

conjugates

of

α

=

α

(

1

)

.  The number

γ

=

N

⁢

(

α

)

α

belongs to the field

K

and it is an

algebraic integer

, since

α

(

2

)

,

⋯

,

α

(

n

)

are, as algebraic conjugates of

α

, also algebraic integers.  Thus

γ

∈

𝒪

K

.  Consequently, the non-zero

integer

N

⁢

(

α

)

=

α

⁢

γ

belongs to the ideal

𝔞

, and similarly its opposite number.  So,

𝔞

contains

positive integers

, in fact infinitely many.

Title

rational integers in ideals

Canonical name

RationalIntegersInIdeals

Date of creation

2013-03-22 19:08:47

Last modified on

2013-03-22 19:08:47

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Theorem

Classification

msc 12F05

Classification

msc 06B10

Classification

msc 11R04

Related topic

CharacteristicPolynomialOfAlgebraicNumber

Related topic

IdealNorm

Generated on Fri Feb 9 16:42:46 2018 by

LaTeXML
relative algebraic closure

relative algebraic closure

Let

L

be a

field extension

of

K

. We denote by

A

¯

the set of all elements of

L

that are algebraic over

K

. We will call

A

¯

the (relative)

algebraic closure

of

K

in

L

. We have that

A

¯

is a field between

K

and

L

.

Title

relative algebraic closure

Canonical name

RelativeAlgebraicClosure

Date of creation

2013-03-22 15:55:51

Last modified on

2013-03-22 15:55:51

Owner

polarbear (3475)

Last modified by

polarbear (3475)

Numerical id

5

Author

polarbear (3475)

Entry type

Definition

Classification

msc 12F05

Generated on Fri Feb 9 16:42:48 2018 by

LaTeXML
separably algebraically closed field

separably algebraically closed field

A field

K

is called

separably algebraically closed

if every

separable

element of the

algebraic closure

of

K

belongs to

K

.

In the case when

K

has

characteristic

0, it is separably algebraically closed if and only if it is

algebraically closed

.

If

K

has positive characteristic

p

,

K

is separably algebraically closed if and only if its algebraic closure is a

purely inseparable extension

of

K

.

Title

separably algebraically closed field

Canonical name

SeparablyAlgebraicallyClosedField

Date of creation

2013-03-22 15:58:30

Last modified on

2013-03-22 15:58:30

Owner

polarbear (3475)

Last modified by

polarbear (3475)

Numerical id

6

Author

polarbear (3475)

Entry type

Definition

Classification

msc 12F05

Defines

separably algebraically closed

Generated on Fri Feb 9 16:42:49 2018 by

LaTeXML
splitting field

splitting field

Let

f

∈

F

⁢

[

x

]

be a

polynomial

over a field

F

. A

splitting field

for

f

is a

field extension

K

of

F

such that

1.

f

splits (factors into a

product

of linear factors) in

K

⁢

[

x

]

,

2.

K

is the smallest field with this property (any sub-extension field of

K

which satisfies the first property is equal to

K

).

Theorem:

Any polynomial over any field has a splitting field, and any two such splitting fields are isomorphic. A splitting field is always a

normal extension

of the

ground field

.

Title

splitting field

Canonical name

SplittingField

Date of creation

2013-03-22 12:08:01

Last modified on

2013-03-22 12:08:01

Owner

djao (24)

Last modified by

djao (24)

Numerical id

7

Author

djao (24)

Entry type

Definition

Classification

msc 12F05

Related topic

NormalExtension

Generated on Fri Feb 9 16:42:51 2018 by

LaTeXML
splitting field of a finite set of polynomials

splitting field of a finite set of polynomials

Lemma 1

.

(Cauchy,Kronecker) Let

K

be a field. For any

irreducible polynomial

f

in

K

⁢

[

X

]

there is an

extension field

of

K

in which

f

has a root.

Proof.

If

I

is the

ideal generated by

f

in

K

⁢

[

X

]

, since

f

is

irreducible

,

I

is a

maximal ideal

of

K

⁢

[

X

]

, and consequently

K

⁢

[

X

]

/

I

is a field.

We can

construct

a canonical

monomorphism

v

from

K

to

K

⁢

[

X

]

. By tracking back the field operation on

K

⁢

[

X

]

/

I

,

v

can be extended to an

isomorphism

w

from an extension field

L

of

K

to

K

⁢

[

X

]

/

I

.

We show that

α

=

w

-

1

⁢

(

X

+

I

)

is a root of

f

.

If we write

f

=

∑

i

=

1

n

f

i

⁢

X

i

then

f

+

I

=

0

implies:

w

⁢

(

f

⁢

(

α

)

)

=

w

⁢

(

∑

i

=

1

n

f

i

⁢

α

i

)

=

∑

i

=

1

n

w

⁢

(

f

i

)

⁢

w

⁢

(

α

)

i

=

∑

i

=

1

n

v

⁢

(

f

i

)

⁢

w

⁢

(

α

)

i

=

∑

i

=

1

n

(

f

i

+

I

)

⁢

(

X

+

I

)

i

=

(

∑

i

=

1

n

f

i

⁢

X

i

)

+

I

=

f

+

I

=

0

,

which means that

f

⁢

(

α

)

=

0

.∎

Theorem 1

.

Let

K

be a field and let

M

be a

finite set

of nonconstant

polynomials

in

K

⁢

[

X

]

. Then there exists an extension field

L

of

K

such that every polynomial in

M

splits in

L

⁢

[

X

]

Proof.

If

L

is a field extension of

K

then the nonconstant polynomials

f

1

,

f

2

,

…

,

f

n

split in

L

⁢

[

X

]

iff the polynomial

∏

i

=

1

n

f

i

splits in

L

⁢

[

X

]

. Now the proof easily follows from the above lemma.
∎

Title

splitting field

of a finite set of polynomials

Canonical name

SplittingFieldOfAFiniteSetOfPolynomials

Date of creation

2013-03-22 16:53:09

Last modified on

2013-03-22 16:53:09

Owner

polarbear (3475)

Last modified by

polarbear (3475)

Numerical id

16

Author

polarbear (3475)

Entry type

Theorem

Classification

msc 12F05

Generated on Fri Feb 9 16:42:53 2018 by

LaTeXML
Steinitz theorem

Steinitz theorem

Ernst Steinitz (1871-1928) proved in

Algebraische Theorie der Körper

the existence of

algebraic closure

of a field.

Title

Steinitz theorem

Canonical name

SteinitzTheorem

Date of creation

2013-03-22 14:27:53

Last modified on

2013-03-22 14:27:53

Owner

kshum (5987)

Last modified by

kshum (5987)

Numerical id

4

Author

kshum (5987)

Entry type

Theorem

Classification

msc 12F05

Generated on Fri Feb 9 16:42:55 2018 by

LaTeXML
the field extension ℝ/ℚ is not finite

the field extension

ℝ

/

ℚ

is not finite

Theorem

.

Let

L

/

K

be a

finite field extension

. Then

L

/

K

is an

algebraic
extension

.

Corollary

.

The extension of fields

R

/

Q

is not finite.

Proof of the Corollary.

If the

extension

was finite, it would be an algebraic extension. However, the
extension

ℝ

/

ℚ

is clearly not algebraic. For example,

e

∈

ℝ

is

transcendental

over

ℚ

(see e is transcendental).
∎

Title

the field extension

ℝ

/

ℚ

is not finite

Canonical name

TheFieldExtensionmathbbRmathbbQIsNotFinite

Date of creation

2013-03-22 13:57:32

Last modified on

2013-03-22 13:57:32

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

6

Author

alozano (2414)

Entry type

Corollary

Classification

msc 12F05

Synonym

the reals is not a

finite extension

of the rationals

Related topic

Pi

Related topic

Algebraic

Related topic

FiniteExtension

Generated on Fri Feb 9 16:42:56 2018 by

LaTeXML