aarjaneiro/planetmath huggingface dataset:
field

field

A

field

is a set

F

together with two binary operations on

F

, called

addition

and

multiplication

, and denoted

+

and

⋅

, satisfying the following

properties

, for all

a

,

b

,

c

∈

F

:

1.

a

+

(

b

+

c

)

=

(

a

+

b

)

+

c

(

associativity

of addition)

2.

a

+

b

=

b

+

a

(

commutativity

of addition)

3.

a

+

0

=

a

for some element

0

∈

F

(existence of

zero element

)

4.

a

+

(

-

a

)

=

0

for some element

-

a

∈

F

(existence of

additive inverses

)

5.

a

⋅

(

b

⋅

c

)

=

(

a

⋅

b

)

⋅

c

(

associativity of multiplication

)

6.

a

⋅

b

=

b

⋅

a

(commutativity of multiplication)

7.

a

⋅

1

=

a

for some element

1

∈

F

, with

1

≠

0

(existence of unity element)

8.

If

a

≠

0

, then

a

⋅

a

-

1

=

1

for some element

a

-

1

∈

F

(existence of

multiplicative inverses

)

9.

a

⋅

(

b

+

c

)

=

(

a

⋅

b

)

+

(

a

⋅

c

)

(

distributive property

)

Equivalently, a field is a

commutative ring

F

with

identity

such that:

•

1

≠

0

•

If

a

∈

F

, and

a

≠

0

, then there exists

b

∈

F

with

a

⋅

b

=

1

.

Title

field

Canonical

name

Field

Date of creation

2013-03-22 11:48:43

Last modified on

2013-03-22 11:48:43

Owner

djao (24)

Last modified by

djao (24)

Numerical id

9

Author

djao (24)

Entry type

Definition

Classification

msc 12E99

Classification

msc 03A05

Generated on Fri Feb 9 16:41:22 2018 by

LaTeXML
field homomorphism

field homomorphism

Let

F

and

K

be fields.

Definition

.

A

field homomorphism

is a

function

ψ

:

F

→

K

such that:

1.

ψ

⁢

(

a

+

b

)

=

ψ

⁢

(

a

)

+

ψ

⁢

(

b

)

for all

a

,

b

∈

F

2.

ψ

⁢

(

a

⋅

b

)

=

ψ

⁢

(

a

)

⋅

ψ

⁢

(

b

)

for all

a

,

b

∈

F

3.

ψ

⁢

(

1

)

=

1

,

ψ

⁢

(

0

)

=

0

If

ψ

is

injective

and

surjective

, then we say that

ψ

is a

field isomorphism

.

Lemma

.

Let

ψ

:

F

→

K

be a field homomorphism. Then

ψ

is
injective.

Proof.

Indeed, if

ψ

is a field homomorphism, in particular it is a

ring homomorphism

. Note that the kernel of a ring homomorphism is
an ideal and a field

F

only has two ideals, namely

{

0

}

,

F

.
Moreover, by the definition of field homomorphism,

ψ

⁢

(

1

)

=

1

,
hence

1

is not in the kernel of the map, so the kernel must be
equal to

{

0

}

.
∎

Remark

: For this reason the terms “field homomorphism” and
“field monomorphism” are synonymous. Also note that if

ψ

is
a field monomorphism, then

ψ

⁢

(

F

)

≅

F

,

ψ

⁢

(

F

)

⊆

K

so there is a “copy” of

F

in

K

. In other words, if

ψ

:

F

→

K

is a field homomorphism then there exist a

subfield

H

of

K

such that

H

≅

F

.

Conversely

, suppose
there exists

H

⊂

K

with

H

isomorphic

to

F

. Then there
is an

isomorphism

χ

:

F

→

H

and we also have the

inclusion

homomorphism

ι

:

H

↪

K

Thus
the composition

ι

∘

χ

:

F

→

K

is a field homomorphism.

Remark

: Let

ψ

:

F

→

K

be a field homomorphism. We claim that the

characteristic

of

F

and

K

must be the same. Indeed, since

ψ

⁢

(

1

F

)

=

1

K

and

ψ

⁢

(

0

F

)

=

0

K

then

ψ

⁢

(

n

⋅

1

F

)

=

n

⋅

1

K

for all

natural numbers

n

. If the characteristic of

F

is

p

>

0

then

0

=

ψ

⁢

(

p

⋅

1

)

=

p

⋅

1

in

K

, and so the characteristic of

K

is also

p

. If the characteristic of

F

is

0

, then the characteristic of

K

must be

0

as well. For if

p

⋅

1

=

0

in

K

then

ψ

⁢

(

p

⋅

1

)

=

0

, and since

ψ

is injective by the lemma, we would have

p

⋅

1

=

0

in

F

as well.

Title

field homomorphism

Canonical name

FieldHomomorphism

Date of creation

2013-03-22 13:54:54

Last modified on

2013-03-22 13:54:54

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

9

Author

alozano (2414)

Entry type

Definition

Classification

msc 12E99

Synonym

field monomorphism

Related topic

RingHomomorphism

Defines

field homomorphism

Defines

field isomorphism

Generated on Fri Feb 9 16:41:24 2018 by

LaTeXML
field homomorphisms fix prime subfields

field homomorphisms fix prime subfields

Theorem

.

Let

F

and

K

be fields having the same

prime subfield

L

and

φ

:

F

→

K

be a

field homomorphism

. Then

φ

fixes

L

.

Proof.

Without loss of generality, it will be assumed that

L

is either

ℚ

or

ℤ

/

c

⁢

ℤ

.

Since

φ

is a field homomorphism,

φ

⁢

(

0

)

=

0

,

φ

⁢

(

1

)

=

1

, and, for every

x

∈

F

,

φ

⁢

(

-

x

)

=

-

φ

⁢

(

x

)

.

Let

n

∈

ℤ

and

c

be the

characteristic

of

F

. Then

φ

⁢

(

n

)

≡

φ

⁢

(

sign

⁡

(

n

)

⁢

|

n

|

)

⁢

mod

⁡

c

, where

sign

denotes the

signum function

≡

sign

⁡

(

n

)

⁢

φ

⁢

(

|

n

|

)

⁢

mod

⁡

c

≡

sign

⁡

(

n

)

⁢

φ

⁢

(

∑

j

=

1

|

n

|

1

)

⁢

mod

⁡

c

≡

sign

⁡

(

n

)

⁢

∑

j

=

1

|

n

|

φ

⁢

(

1

)

⁢

mod

⁡

c

≡

sign

⁡

(

n

)

⁢

∑

j

=

1

|

n

|

1

⁢

mod

⁡

c

≡

sign

⁡

(

n

)

⁢

|

n

|

⁢

mod

⁡

c

≡

n

⁢

mod

⁡

c

.

This the proof in the case that

c

is prime.

Now consider

c

=

0

. Let

x

∈

ℚ

. Then there exist

a

,

b

∈

ℤ

with

b

>

0

such that

x

=

a

b

. Thus,

b

⁢

φ

⁢

(

x

)

=

∑

j

=

1

b

φ

⁢

(

a

b

)

=

φ

⁢

(

∑

j

=

1

b

a

b

)

=

φ

⁢

(

a

)

=

a

. Therefore,

φ

⁢

(

x

)

=

a

b

=

x

. Hence,

φ

fixes

ℚ

.
∎

Title

field homomorphisms fix prime subfields

Canonical name

FieldHomomorphismsFixPrimeSubfields

Date of creation

2013-03-22 16:19:54

Last modified on

2013-03-22 16:19:54

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

10

Author

Wkbj79 (1863)

Entry type

Theorem

Classification

msc 12E99

Generated on Fri Feb 9 16:41:26 2018 by

LaTeXML
Frobenius homomorphism

Frobenius homomorphism

Let

F

be a field of

characteristic

p

>

0

. Then for any

a

,

b

∈

F

,

(

a

+

b

)

p

=

a

p

+

b

p

,

(

a

⁢

b

)

p

=

a

p

⁢

b

p

.

Thus the map

ϕ

:

F

→

F

a

↦

a

p

is a

field homomorphism

, called the

Frobenius homomorphism

, or simply the

Frobenius map

on

F

.
If it is surjective then it is an

automorphism

, and is called the

Frobenius automorphism

.

Note: This

morphism

is sometimes also called the “small Frobenius” to distinguish it from the map

a

↦

a

q

, with

q

=

p

n

. This map is then also referred to as the “big Frobenius” or the “power Frobenius map”.

Title

Frobenius homomorphism

Canonical name

FrobeniusHomomorphism

Date of creation

2013-03-22 12:22:50

Last modified on

2013-03-22 12:22:50

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

11

Author

mathcam (2727)

Entry type

Definition

Classification

msc 12E99

Synonym

Frobenius endomorphism

Synonym

Frobenius map

Related topic

FrobeniusMorphism

Related topic

FrobeniusMap

Defines

Frobenius automorphism

Generated on Fri Feb 9 16:41:28 2018 by

LaTeXML
homomorphisms from fields are either injective or trivial

homomorphisms from fields are either injective or trivial

Suppose

F

is a field,

R

is a ring, and

ϕ

:

F

→

R

is a

homomorphism

of rings. Then

ϕ

is either

trivial

or

injective

.

Proof.

We use the fact that kernels of

ring homomorphism

are ideals. Since

F

is a field, by the above result, we have that the kernel of

ϕ

is an ideal of the field

F

and hence either empty or all of

F

. If the kernel is empty, then since a ring homomorphism is injective iff the kernel is trivial, we get that

ϕ

is injective. If the kernel is all of

F

, then

ϕ

is the

zero map

from

F

to

R

.
∎

Finally, it is clear that both of these possibilities are in fact achieved:

•

The map

ϕ

:

ℚ

→

ℚ

given by

ϕ

⁢

(

n

)

=

0

is trivial (has all of

ℚ

as a kernel)

•

The

inclusion

ℚ

→

ℚ

⁢

[

x

]

is injective (i.e. the kernel is trivial).

Title

homomorphisms from fields are either injective or trivial

Canonical name

HomomorphismsFromFieldsAreEitherInjectiveOrTrivial

Date of creation

2013-03-22 14:39:07

Last modified on

2013-03-22 14:39:07

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

4

Author

mathcam (2727)

Entry type

Corollary

Classification

msc 12E99

Generated on Fri Feb 9 16:41:30 2018 by

LaTeXML
integral element

integral element

An element

a

of a field

K

is an

integral element

of the field

K

, iff

|

a

|

≤

1

for every

non-archimedean valuation

|

⋅

|

of this field.

The set

𝒪

of all integral elements of

K

is a

subring

(in fact, an

integral domain

) of

K

, because it is the

intersection

of all

valuation rings

in

K

.

Examples

1.

K

=

ℚ

.  The only non-archimedean valuations of

ℚ

are the

p

-adic

valuations

|

⋅

|

p

(where

p

is a

rational prime

) and the

trivial valuation

(all values are 1 except the value of 0).  The valuation ring

𝒪

p

of

|

⋅

|

p

consists of all so-called

p-integral

rational numbers

whose

denominators

are not

divisible

by

p

.  The valuation ring of the trivial valuation is, generally, the whole field.  Thus,

𝒪

is, by definition, the intersection of the

𝒪

p

’s for all

p

;  this is the set of rationals whose denominators are not divisible by any prime, which is exactly the set

ℤ

of ordinary

integers

.

2.

If

K

is a

finite field

, it has only the trivial valuation.  In fact, if

|

⋅

|

is a valuation and

a

any non-zero

element of

K

, then there is a

positive integer

m

such that

a

m

=

1

,  and we have

|

a

|

m

=

|

a

m

|

=

|

1

|

=

1

,  and therefore

|

a

|

=

1

.  Thus,

|

⋅

|

is trivial and

𝒪

=

K

.  This means that all elements of the field are integral elements.

3.

If

K

is the field

ℚ

p

of the

p

-adic numbers (

http://planetmath.org/NonIsomorphicCompletionsOfMathbbQ

), it has only one non-trivial valuation, the

p

-adic valuation, and now the ring

𝒪

is its valuation ring, which is the ring of

p

-adic integers

(

http://planetmath.org/PAdicIntegers

);  this is visualized in the 2-adic (

dyadic

) case in the article “

p

-adic

canonical form

”.

Title

integral element

Canonical name

IntegralElement

Date of creation

2013-03-22 14:15:56

Last modified on

2013-03-22 14:15:56

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

31

Author

pahio (2872)

Entry type

Definition

Classification

msc 12E99

Related topic

PAdicCanonicalForm

Related topic

PAdicValuation

Related topic

KummersCongruence

Generated on Fri Feb 9 16:41:31 2018 by

LaTeXML
integrity characterized by places

integrity characterized by places

Theorem

.

Let

R

be a

subring

of the field

K

,

1

∈

R

.  An

element

α

of the field is integral over

R

if and only if all places (

http://planetmath.org/PlaceOfField

)

φ

of

K

satisfy

the

implication

φ

⁢

is

⁢

finite

⁢

in

⁢

R

⇒

φ

⁢

(

α

)

⁢

is

⁢

finite

.

1.

Let

R

be a subring of the field

K

,

1

∈

R

.  The

integral closure

of

R

in

K

is the

intersection

of all

valuation domains

in

K

which contain the ring

R

.  The integral closure is

integrally closed

in the field

K

.

2.

Every valuation domain is integrally closed in its

field of fractions

.

Title

integrity characterized by places

Canonical name

IntegrityCharacterizedByPlaces

Date of creation

2013-03-22 14:56:57

Last modified on

2013-03-22 14:56:57

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

11

Author

pahio (2872)

Entry type

Theorem

Classification

msc 12E99

Classification

msc 13B21

Related topic

Integral

Related topic

PlaceOfField

Generated on Fri Feb 9 16:41:33 2018 by

LaTeXML
inverse number

inverse number

The

inverse number

or

reciprocal number

of a non-zero real or

complex number

a

may be denoted by

a

-

1

, and it the

quotient

1

a

(so, it is really the

-

1

th

power of

a

).

•

Two numbers are inverse numbers of each other if and only if their product is equal to 1 (cf. opposite

inverses

).

•

If

a

(

≠

0

) is given in a quotient form

b

c

, then its inverse number is simply

(

b

c

)

-

1

=

c

b

.

•

Forming the inverse number is also a

multiplicative function

, i.e.

(

b

⁢

c

)

-

1

=

b

-

1

⁢

c

-

1

(to be more precise, it is an

automorphism

of the

multiplicative group

of

ℝ

resp.

ℂ

).

Title

inverse number

Canonical name

InverseNumber

Date of creation

2013-03-22 14:53:46

Last modified on

2013-03-22 14:53:46

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

12

Author

pahio (2872)

Entry type

Definition

Classification

msc 12E99

Classification

msc 00A05

Synonym

inverse

Synonym

reciprocal

Related topic

ConditionOfOrthogonality

Related topic

InverseFormingInProportionToGroupOperation

Defines

reciprocal number

Generated on Fri Feb 9 16:41:35 2018 by

LaTeXML
long division

long division

In this entry we treat two cases of

long division

.

1

Integers

Theorem 1

(Integer Long Division).

For every pair of

integers

a

,

b

≠

0

there exist unique integers

q

and

r

such that:

1.

a

=

b

⋅

q

+

r

,

2.

0

≤

r

<

|

b

|

.

Example 1

.

Let

a

=

10

and

b

=

-

3

. Then

q

=

-

3

and

r

=

1

correspond to the long division:

10

=

(

-

3

)

⋅

(

-

3

)

+

1

.

Definition 1

.

The number

r

as in the theorem is called the remainder of the

division

of

a

by

b

. The numbers

a

,

b

and

q

are called the dividend,

divisor

and

quotient

respectively.

2

Polynomials

Theorem 2

(Polynomial Long Division).

Let

R

be a

commutative ring

with

non-zero unity

and let

a

⁢

(

x

)

and

b

⁢

(

x

)

be two

polynomials

in

R

⁢

[

x

]

, where the leading coefficient of

b

⁢

(

x

)

is a unit of

R

. Then there exist unique polynomials

q

⁢

(

x

)

and

r

⁢

(

x

)

in

R

⁢

[

x

]

such that:

1.

a

⁢

(

x

)

=

b

⁢

(

x

)

⋅

q

⁢

(

x

)

+

r

⁢

(

x

)

,

2.

0

≤

deg

⁡

(

r

⁢

(

x

)

)

<

deg

⁡

b

⁢

(

x

)

or

r

⁢

(

x

)

=

0

.

Example 2

.

Let

R

=

ℤ

and let

a

⁢

(

x

)

=

x

3

+

3

,

b

⁢

(

x

)

=

x

2

+

1

. Then

q

⁢

(

x

)

=

x

and

r

⁢

(

x

)

=

-

x

+

3

, so that:

x

3

+

3

=

x

⁢

(

x

2

+

1

)

-

x

+

3

.

Example 3

.

The theorem is not true in general if the leading coefficient of

b

⁢

(

x

)

is not a unit. For example, if

a

⁢

(

x

)

=

x

3

+

3

and

b

⁢

(

x

)

=

3

⁢

x

2

+

1

then there are no

q

⁢

(

x

)

and

r

⁢

(

x

)

with coefficients in

ℤ

with the required properties.

Title

long division

Canonical name

LongDivision

Date of creation

2013-03-22 15:09:07

Last modified on

2013-03-22 15:09:07

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

7

Author

alozano (2414)

Entry type

Theorem

Classification

msc 12E99

Classification

msc 00A05

Classification

msc 11A05

Synonym

division algorithm

Related topic

Polynomial

Related topic

PolynomialLongDivision

Related topic

MixedFraction

Defines

dividend

Defines

remainder

Generated on Fri Feb 9 16:41:37 2018 by

LaTeXML
prime subfield

prime subfield

The

prime subfield

of a field

F

is the

intersection

of all

subfields

of

F

, or equivalently the smallest subfield of

F

. It can also be constructed by taking the

quotient field

of the

additive

subgroup

of

F

generated by the

multiplicative identity

1

.

If

F

has

characteristic

p

where

p

>

0

is a prime, then the prime subfield of

F

is

isomorphic

to the field

ℤ

/

p

⁢

ℤ

of

integers

mod

p

. When

F

has characteristic zero, the prime subfield of

F

is isomorphic to the field

ℚ

of

rational numbers

.

Title

prime subfield

Canonical name

PrimeSubfield

Date of creation

2013-03-22 12:37:47

Last modified on

2013-03-22 12:37:47

Owner

djao (24)

Last modified by

djao (24)

Numerical id

4

Author

djao (24)

Entry type

Definition

Classification

msc 12E99

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subfield

subfield

Let

F

be a field and

S

a subset of

F

such that

S

with the inherited operations from

F

is also a field.  Then we say that

S

is a

subfield

of

F

.

The name “subfield” is used also in the case that

F

is only a skew field and its subset

S

forms a skew field.

Title

subfield

Canonical

name

Subfield

Date of creation

2013-03-22 14:01:45

Last modified on

2013-03-22 14:01:45

Owner

drini (3)

Last modified by

drini (3)

Numerical id

7

Author

drini (3)

Entry type

Definition

Classification

msc 12E99

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

LaTeXML
subfield criterion

subfield criterion

Let

K

be a skew field and

S

its subset.  For

S

to be a

subfield

of

K

, it’s necessary and

sufficient

that the following three conditions are fulfilled:

1.

S

a non-zero element of

K

.

2.

a

-

b

∈

S

always when

a

,

b

∈

S

.

3.

a

⁢

b

-

1

∈

S

always when

a

,

b

∈

S

and

b

≠

0

.

Proof.

Because the conditions are fulfilled in every skew field, they are necessary.  For proving the sufficience, suppose now that the subset

S

these conditions.  The condition 1 guarantees that

S

is not empty and the condition 2 that

(

S

,

+

)

is an subgroup of

(

K

,

+

)

;  thus all the required

properties

of

addition

for a skew field hold in

S

.  If

b

is a non-zero element of

S

, then, according to the condition 3, we have

0

≠

1

=

b

⁢

b

-

1

∈

S

.  Moreover,

a

⋅

1

=

1

⋅

a

=

a

∈

S

for all

a

∈

S

⊆

K

.  The laws of

multiplication

(associativity and left and distributivity over addition) hold in

S

since they hold in whole

K

.  So

S

fulfils all the

postulates

for a skew field.

Title

subfield criterion

Canonical

name

SubfieldCriterion

Date of creation

2013-03-22 16:26:34

Last modified on

2013-03-22 16:26:34

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Theorem

Classification

msc 12E99

Classification

msc 12E15

Related topic

FieldOfAlgebraicNumbers

Generated on Fri Feb 9 16:41:42 2018 by

LaTeXML
a condition of algebraic extension

a condition of algebraic extension

Theorem.

A

field extension

L

/

K

is

algebraic

(

http://planetmath.org/AlgebraicExtension

) if and only if any subring of the

extension field

L

containing the

base field

K

is a field.

Proof.

Assume first that

L

/

K

is algebraic.  Let

R

be a subring of

L

containing

K

.  For any non-zero element

r

of

R

, naturally

K

⁢

[

r

]

⊆

R

,  and since

r

is an

algebraic element

over

K

, the ring

K

⁢

[

r

]

coincides with the field

K

⁢

(

r

)

.  Therefore we have

r

-

1

∈

K

⁢

[

r

]

⊆

R

,  and

R

must be a field.

Assume then that each subring of

L

which contains

K

is a field.  Let

a

be any non-zero element of

L

.  Accordingly, the subring

K

⁢

[

a

]

of

L

contains

K

and is a field.  So we have

a

-

1

∈

K

⁢

[

a

]

.  This means that there is a

polynomial

f

⁢

(

x

)

in the

polynomial ring

K

⁢

[

x

]

such that

a

-

1

=

f

⁢

(

a

)

.  Because

a

⁢

f

⁢

(

a

)

-

1

=

0

,  the element

a

is a zero of the polynomial

x

⁢

f

⁢

(

x

)

-

1

of

K

⁢

[

x

]

, i.e. is algebraic over

K

.  Thus every element of

L

is algebraic over

K

.

References

1

David M. Burton

:

A first course in rings and ideals

. Addison-Wesley Publishing Company. Reading, Menlo Park, London, Don Mills (1970).

Title

a

condition of algebraic extension

Canonical name

AConditionOfAlgebraicExtension

Date of creation

2013-03-22 17:53:34

Last modified on

2013-03-22 17:53:34

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Theorem

Classification

msc 12F05

Related topic

RingAdjunction

Related topic

FieldAdjunction

Related topic

Overring

Related topic

AConditionOfSimpleExtension

Related topic

SteinitzTheoremOnFiniteFieldExtension

Generated on Fri Feb 9 16:41:44 2018 by

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a finite extension of fields is an algebraic extension

a finite extension of fields is an algebraic extension

Theorem 1

.

Let

L

/

K

be a

finite field extension

. Then

L

/

K

is an

algebraic
extension

.

Proof.

In order to prove that

L

/

K

is an algebraic extension, we need to show that any element

α

∈

L

is algebraic, i.e., there exists a non-zero

polynomial

p

⁢

(

x

)

∈

K

⁢

[

x

]

such that

p

⁢

(

α

)

=

0

.

Recall that

L

/

K

is a

finite extension

of fields, by

definition

,
it means that

L

is a finite dimensional

vector space

over

K

.
Let the

dimension

be

[

L

:

K

]

=

n

for some

n

∈

ℕ

.

Consider the following set of “vectors” in

L

:

𝒮

=

{

1

,

α

,

α

2

,

α

3

,

…

,

α

n

}

Note that the cardinality of

S

is

n

+

1

, one more than the
dimension of the vector space. Therefore, the elements of

S

must
be

linearly dependent

over

K

, otherwise the dimension of

S

would be greater than

n

. Hence, there exist

k

i

∈

K

,

0

≤

i

≤

n

, not all zero, such that

k

0

+

k

1

⁢

α

+

k

2

⁢

α

2

+

k

3

⁢

α

3

+

…

+

k

n

⁢

α

n

=

0

Thus, if we define

p

⁢

(

X

)

=

k

0

+

k

1

⁢

X

+

k

2

⁢

X

2

+

k

3

⁢

X

3

+

…

+

k

n

⁢

X

n

then

p

⁢

(

X

)

∈

K

⁢

[

X

]

and

p

⁢

(

α

)

=

0

, as desired.

∎

NOTE:

The converse is not true. See the entry “algebraic
extension” for details.

Title

a finite extension of fields is an algebraic extension

Canonical

name

AFiniteExtensionOfFieldsIsAnAlgebraicExtension

Date of creation

2013-03-22 13:57:30

Last modified on

2013-03-22 13:57:30

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

6

Author

alozano (2414)

Entry type

Theorem

Classification

msc 12F05

Related topic

Algebraic

Related topic

AlgebraicExtension

Related topic

ProofOfTranscendentalRootTheorem

Generated on Fri Feb 9 16:41:46 2018 by

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