aarjaneiro/planetmath huggingface dataset:
de Moivre identity, proof of

de Moivre identity, proof of

To prove the

de Moivre identity

, we will first prove by

induction

on

n

that the

identity

holds for all

natural numbers

.

For the case

n

=

0

, observe that

cos

⁡

(

0

⁢

θ

)

+

i

⁢

sin

⁡

(

0

⁢

θ

)

=

1

+

i

⁢

0

=

(

cos

⁡

(

θ

)

+

i

⁢

sin

⁡

(

θ

)

)

0

.

Assume that the identity holds for a certain value of

n

:

cos

⁡

(

n

⁢

θ

)

+

i

⁢

sin

⁡

(

n

⁢

θ

)

=

(

cos

⁡

(

θ

)

+

i

⁢

sin

⁡

(

θ

)

)

n

.

Multiply both sides of this identity by

cos

⁡

(

θ

)

+

i

⁢

sin

⁡

(

θ

)

and expand the left side to obtain

cos

⁡

(

θ

)

⁢

cos

⁡

(

n

⁢

θ

)

-

sin

⁡

(

θ

)

⁢

sin

⁡

(

n

⁢

θ

)

+

i

⁢

cos

⁡

(

θ

)

⁢

sin

⁡

(

n

⁢

θ

)

+

i

⁢

sin

⁡

(

θ

)

⁢

cos

⁡

(

n

⁢

θ

)

=

(

cos

⁡

(

θ

)

+

i

⁢

sin

⁡

(

θ

)

)

⁢

(

cos

⁡

(

n

⁢

θ

)

+

i

⁢

sin

⁡

(

n

⁢

θ

)

)

=

(

cos

⁡

(

θ

)

+

i

⁢

sin

⁡

(

θ

)

)

n

+

1

.

By the

angle sum identities

,

cos

⁡

(

θ

)

⁢

cos

⁡

(

n

⁢

θ

)

-

sin

⁡

(

θ

)

⁢

sin

⁡

(

n

⁢

θ

)

=

cos

⁡

(

n

⁢

θ

+

θ

)

cos

⁡

(

θ

)

⁢

sin

⁡

(

n

⁢

θ

)

+

sin

⁡

(

θ

)

⁢

cos

⁡

(

n

⁢

θ

)

=

sin

⁡

(

n

⁢

θ

+

θ

)

Therefore,

cos

⁡

(

(

n

+

1

)

⁢

θ

)

+

i

⁢

sin

⁡

(

(

n

+

1

)

⁢

θ

)

=

(

cos

⁡

(

θ

)

+

i

⁢

sin

⁡

(

θ

)

)

n

+

1

.

Hence by induction de Moivre’s identity holds for all natural

n

.

Now let

-

n

be any negative integer. Then using the fact that

cos

is an even and

sin

an odd function, we obtain that

cos

⁡

(

-

n

⁢

θ

)

+

i

⁢

sin

⁡

(

-

n

⁢

θ

)

=

cos

⁡

(

n

⁢

θ

)

-

i

⁢

sin

⁡

(

n

⁢

θ

)

=

cos

⁡

(

n

⁢

θ

)

-

i

⁢

sin

⁡

(

n

⁢

θ

)

cos

2

⁡

(

n

⁢

θ

)

+

sin

2

⁡

(

n

⁢

θ

)

=

1

cos

⁡

(

n

⁢

θ

)

+

i

⁢

sin

⁡

(

n

⁢

θ

)

⋅

cos

⁡

(

n

⁢

θ

)

-

i

⁢

sin

⁡

(

n

⁢

θ

)

cos

⁡

(

n

⁢

θ

)

-

i

⁢

sin

⁡

(

n

⁢

θ

)

=

1

cos

⁡

(

n

⁢

θ

)

+

i

⁢

sin

⁡

(

n

⁢

θ

)

,

the denominator of which is

(

cos

⁡

(

n

⁢

θ

)

+

i

⁢

sin

⁡

(

n

⁢

θ

)

)

n

. Hence

cos

⁡

(

-

n

⁢

θ

)

+

i

⁢

sin

⁡

(

-

n

⁢

θ

)

=

(

cos

⁡

(

θ

)

+

i

⁢

sin

⁡

(

θ

)

)

-

n

.

Title

de Moivre identity, proof of

Canonical name

DeMoivreIdentityProofOf

Date of creation

2013-03-22 14:34:08

Last modified on

2013-03-22 14:34:08

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

10

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 12E10

Synonym

proof of de Moivre’s

formula

Synonym

proof of de Moivre’s

theorem

Related topic

AngleSumIdentity

Generated on Fri Feb 9 16:40:46 2018 by

LaTeXML
monic

monic

A

monic polynomial

is a

polynomial

with a leading coefficient of 1. That is, if

P

n

⁢

(

x

)

is a polynomial of degree

n

in the

variable

x

, then the

coefficient

of

x

n

in

P

n

⁢

(

x

)

is 1.

For example,

x

5

+

3

⁢

x

3

-

10

⁢

x

2

+

1

is a monic 5th-degree polynomial.

3

⁢

x

2

+

2

⁢

z

-

5

is a 2nd-degree polynomial which is not monic.

Title

monic

Canonical

name

Monic1

Date of creation

2013-03-22 12:18:42

Last modified on

2013-03-22 12:18:42

Owner

akrowne (2)

Last modified by

akrowne (2)

Numerical id

9

Author

akrowne (2)

Entry type

Definition

Classification

msc 12E10

Synonym

monic polynomial

Related topic

EisensteinCriterion

Related topic

IrreduciblePolynomial2

Related topic

AlgebraicInteger

Generated on Fri Feb 9 16:40:48 2018 by

LaTeXML
second proof of Wedderburn’s theorem

second proof of Wedderburn’s theorem

We can prove Wedderburn’s theorem,without using Zsigmondy’s theorem on the

conjugacy class formula

of the first proof;
let

G

n

set of n-th

roots of unity

and

P

n

set of n-th

primitive
roots of unity

and

Φ

d

⁢

(

q

)

the d-th

cyclotomic polynomial

.

It results

•

Φ

n

⁢

(

q

)

=

∏

ξ

∈

P

n

(

q

-

ξ

)

•

p

⁢

(

q

)

=

q

n

-

1

=

∏

ξ

∈

G

n

(

q

-

ξ

)

=

∏

d

∣

n

Φ

d

⁢

(

q

)

•

Φ

n

⁢

(

q

)

∈

ℤ

⁢

[

q

]

, it has

multiplicative identity

and

Φ

n

(

q

)

∣

q

n

-

1

•

Φ

n

(

q

)

∣

q

n

-

1

q

d

-

1

with

d

∣

n

,

d

<

n

by conjugacy class formula, we have:

q

n

-

1

=

q

-

1

+

∑

x

q

n

-

1

q

n

x

-

1

by last two previous properties, it results:

Φ

n

(

q

)

∣

q

n

-

1

,

Φ

n

(

q

)

∣

q

n

-

1

q

n

x

-

1

⇒

Φ

n

(

q

)

∣

q

-

1

because

Φ

n

⁢

(

q

)

divides the left and each addend of

∑

x

q

n

-

1

q

n

x

-

1

of the right member of the conjugacy class formula.

By third property

q

>

1

,

Φ

n

(

x

)

∈

ℤ

[

x

]

⇒

Φ

n

(

q

)

∈

ℤ

⇒

|

Φ

n

(

q

)

|

∣

q

-

1

⇒

|

Φ

n

(

q

)

|

⩽

q

-

1

If, for

n

>

1

,we have

|

Φ

n

⁢

(

q

)

|

>

q

-

1

, then

n

=

1

and the theorem is proved.

We know that

|

Φ

n

⁢

(

q

)

|

=

∏

ξ

∈

P

n

|

q

-

ξ

|

,

w

⁢

i

⁢

t

⁢

h

⁢

q

-

ξ

∈

ℂ

by the

triangle inequality

in

ℂ

|

q

-

ξ

|

⩾

|

|

q

|

-

|

ξ

|

|

=

|

q

-

1

|

as

ξ

is a primitive root of unity,
besides

|

q

-

ξ

|

=

|

q

-

1

|

⇔

ξ

=

1

but

n

>

1

⇒

ξ

≠

1

therefore, we have

|

q

-

ξ

|

>

|

q

-

1

|

=

q

-

1

⇒

|

Φ

n

⁢

(

q

)

|

>

q

-

1

Title

second proof of Wedderburn’s theorem

Canonical name

SecondProofOfWedderburnsTheorem

Date of creation

2013-03-22 13:34:39

Last modified on

2013-03-22 13:34:39

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

17

Author

Mathprof (13753)

Entry type

Proof

Classification

msc 12E15

Generated on Fri Feb 9 16:40:53 2018 by

LaTeXML
Wedderburn’s theorem

Wedderburn’s theorem

A finite division ring is a field.

One of the many consequences of this theorem
is that for a

finite projective plane

,
Desargues’ theorem implies Pappus’ theorem.

Title

Wedderburn’s theorem

Canonical name

WedderburnsTheorem

Date of creation

2013-03-22 13:03:56

Last modified on

2013-03-22 13:03:56

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

10

Author

yark (2760)

Entry type

Theorem

Classification

msc 12E15

Synonym

Wedderburn theorem

Related topic

NonZeroDivisorsOfFiniteRing

Related topic

JosephWedderburn

Generated on Fri Feb 9 16:40:54 2018 by

LaTeXML
Chevalley-Warning Theorem

Chevalley-Warning Theorem

Let

𝔽

q

be the

finite field

of

q

elements with

characteristic

p

. Let

f

i

⁢

(

x

1

,

…

,

x

n

)

,

i

=

1

,

2

,

…

,

r

,
be

polynomial

of

n

variables over

𝔽

q

. If

n

>

∑

i

=

1

r

deg

⁡

(

f

i

)

, then the number of

solutions

over

𝔽

q

to the system of equations

f

1

⁢

(

x

1

,

x

2

,

…

,

x

n

)

=

0

f

2

⁢

(

x

1

,

x

2

,

…

,

x

n

)

=

0

⋮

f

r

⁢

(

x

1

,

x

2

,

…

,

x

n

)

=

0

is

divisible

by

p

. In particular, if none of the polynomials

f

1

,

f

2

,

…

,

f

r

have constant term, then there are at least

p

solutions.

Title

Chevalley-Warning Theorem

Canonical name

ChevalleyWarningTheorem

Date of creation

2013-03-22 17:46:52

Last modified on

2013-03-22 17:46:52

Owner

kshum (5987)

Last modified by

kshum (5987)

Numerical id

6

Author

kshum (5987)

Entry type

Theorem

Classification

msc 12E20

Generated on Fri Feb 9 16:40:56 2018 by

LaTeXML
Frobenius map

Frobenius map

Let

K

be any field of

characteristic

p

>

0

, and suppose

K

contains the

finite field

𝔽

q

of size

q

, where

q

=

p

r

. The

q

th

power

Frobenius map

on

K

is the map

Frob

q

:

K

⟶

K

defined by

Frob

q

⁡

(

x

)

:=

x

q

.

If

K

is

perfect

, then

Frob

q

is an

automorphism

of

K

which fixes

𝔽

q

, and accordingly is a member of the

Galois group

Gal

⁡

(

K

/

𝔽

q

)

.

Title

Frobenius map

Canonical

name

FrobeniusMap

Date of creation

2013-03-22 12:34:52

Last modified on

2013-03-22 12:34:52

Owner

djao (24)

Last modified by

djao (24)

Numerical id

6

Author

djao (24)

Entry type

Definition

Classification

msc 12E20

Classification

msc 11T99

Related topic

FrobeniusAutomorphism

Generated on Fri Feb 9 16:41:01 2018 by

LaTeXML
irreducible polynomials over finite field

irreducible polynomials over finite field

Theorem.

Over a

finite field

F

, there exist

irreducible polynomials

of any degree.

Proof.

Let

n

be a

positive integer

,

p

be the

characteristic

of

F

,

𝔽

p

be the

prime subfield

, and

p

r

be the order (

http://planetmath.org/FiniteField

) of the field

F

.  Since

p

r

-

1

is a

divisor

of

p

r

⁢

n

-

1

, the zeros of the

polynomial

X

p

r

-

X

form in

G

:=

𝔽

p

r

⁢

n

a

subfield

isomorphic

to

F

.  Thus, one can regard

F

as a subfield of

G

.  Because

[

G

:

F

]

=

[

G

:

𝔽

p

]

[

F

:

𝔽

p

]

=

r

⁢

n

r

=

n

,

the

minimal polynomial

of a

primitive element

of the

field extension

G

/

F

is an irreducible polynomial of degree

n

in the ring

F

⁢

[

X

]

.

Title

irreducible polynomials over finite field

Canonical name

IrreduciblePolynomialsOverFiniteField

Date of creation

2013-03-22 17:43:14

Last modified on

2013-03-22 17:43:14

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Theorem

Classification

msc 12E20

Classification

msc 11T99

Related topic

FiniteField

Generated on Fri Feb 9 16:41:03 2018 by

LaTeXML
Hilbert’s irreducibility theorem

Hilbert’s irreducibility theorem

In this entry,

K

is a field of characteristic zero and

V

is an

irreducible

algebraic variety

over

K

.

Definition 1

.

A

variety

V

satisfies the

Hilbert property

over

K

if

V

⁢

(

K

)

is not a

thin algebraic set

.

Definition 2

.

A field

K

is said to be Hilbertian if there exists an

irreducible variety

V

/

K

of

dim

⁡

V

≥

1

which has the Hilbert property.

Theorem

(Hilbert’s irreducibility theorem).

A

number field

K

is Hilbertian. In particular, for every

n

, the

affine space

A

n

⁢

(

K

)

has the Hilbert property over

K

.

However, the

field of real

numbers

ℝ

and the field of

p

-adic

rationals

ℚ

p

are not Hilbertian.

References

1

J.-P. Serre,

Topics in

Galois Theory

,
Research Notes in Mathematics, Jones and Barlett Publishers, London.

Title

Hilbert’s irreducibility theorem

Canonical name

HilbertsIrreducibilityTheorem

Date of creation

2013-03-22 15:14:16

Last modified on

2013-03-22 15:14:16

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

5

Author

alozano (2414)

Entry type

Theorem

Classification

msc 12E25

Synonym

Hilbertian

Defines

Hilbert property

Defines

Hilbertian field

Generated on Fri Feb 9 16:41:05 2018 by

LaTeXML
thin algebraic set

thin algebraic set

Definition 1

.

Let

V

be an

irreducible

algebraic variety

(we assume it to be integral and quasi-projective) over a field

K

with characteristic zero. We regard

V

as a

topological space

with the usual

Zariski topology

.

1.

A subset

A

⊂

V

⁢

(

K

)

is said to be of type

C

1

if there is a

closed subset

W

⊂

V

, with

W

≠

V

, such that

A

⊂

W

⁢

(

K

)

. In other words,

A

is not dense in

V

(with respect to the Zariski topology).

2.

A subset

A

⊂

V

⁢

(

K

)

is said to be of type

C

2

if there is an

irreducible variety

V

′

of the same

dimension

as

V

, and a (

generically

)

surjective

algebraic morphism

ϕ

:

V

′

→

V

of degree

≥

2

, with

A

⊂

ϕ

⁢

(

V

′

⁢

(

K

)

)

Example

.

Let

K

be a field and let

V

⁢

(

K

)

=

𝔸

⁢

(

K

)

=

𝔸

1

⁢

(

K

)

=

K

be the

1

-dimensional

affine space

. Then, the only Zariski-closed subsets of

V

are finite subsets of points. Thus, the only subsets of type

C

1

are subsets formed by a finite number of points.

Let

V

′

⁢

(

K

)

=

𝔸

⁢

(

K

)

be affine space and define:

ϕ

:

V

′

→

V

by

ϕ

⁢

(

k

)

=

k

2

. Then

deg

⁡

(

ϕ

)

=

2

. Thus, the subset:

A

=

{

k

2

:

k

∈

𝔸

⁢

(

K

)

}

, i.e.

A

is the subset of

perfect squares

in

K

, is a subset of type

C

2

.

Definition 2

.

A subset

A

of an irreducible variety

V

/

K

is said to be a

thin algebraic set

(or

thin set

, or “mince” set) if it is a union of a finite number of subsets of type

C

1

and type

C

2

.

References

1

J.-P. Serre,

Topics in

Galois Theory

,
Research Notes in Mathematics, Jones and Barlett Publishers, London.

Title

thin algebraic set

Canonical name

ThinAlgebraicSet

Date of creation

2013-03-22 15:14:13

Last modified on

2013-03-22 15:14:13

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

5

Author

alozano (2414)

Entry type

Definition

Classification

msc 12E25

Synonym

thin set

Synonym

mince set

Generated on Fri Feb 9 16:41:07 2018 by

LaTeXML
characteristic

characteristic

Let

(

F

,

+

,

⋅

)

be a field. The

characteristic

Char

⁡

(

F

)

of

F

is commonly given by one of three

equivalent

definitions

:

•

if there is some positive integer

n

for which the result of adding any

element

to itself

n

times yields

0

, then the characteristic of the field is the least such

n

. Otherwise,

Char

⁡

(

F

)

is defined to be

0

.

•

if

f

:

ℤ

→

F

is defined by

f

⁢

(

n

)

=

n

⋅

1

then

Char

⁡

(

F

)

is the least

strictly

positive

generator of

ker

⁡

(

f

)

if

ker

⁡

(

f

)

≠

{

0

}

; otherwise it is

0

.

•

if

K

is the

prime subfield

of

F

, then

Char

⁡

(

F

)

is the size of

K

if this is finite, and

0

otherwise.

Note that the first definition also applies to arbitrary rings, and not just to fields.

The characteristic of a field (or more generally an integral domain) is always prime. For if the characteristic of

F

were composite, say

m

⁢

n

for

m

,

n

>

1

, then in particular

m

⁢

n

would equal zero. Then either

m

would be zero or

n

would be zero, so the characteristic of

F

would actually be smaller than

m

⁢

n

, contradicting the minimality condition.

Title

characteristic

Canonical

name

Characteristic

Date of creation

2013-03-22 12:05:01

Last modified on

2013-03-22 12:05:01

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

16

Author

Mathprof (13753)

Entry type

Definition

Classification

msc 12E99

Generated on Fri Feb 9 16:41:09 2018 by

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characterization of field

characterization of field

Proposition 1

.

Let

R

≠

0

be a

commutative ring

with

identity

. The ring

R

(as above) is a field if and only if

R

has exactly two ideals:

(

0

)

,

R

.

Proof.

(

⇒

)

Suppose

ℛ

is a field and let

𝒜

be a non-zero ideal of

ℛ

. Then there
exists

r

∈

𝒜

⊆

ℛ

with

r

≠

0

.
Since

ℛ

is a field and

r

is a non-zero element,
there exists

s

∈

ℛ

such that

s

⋅

r

=

1

∈

ℛ

Moreover,

𝒜

is an ideal,

r

∈

𝒜

,

s

∈

𝒮

, so

s

⋅

r

=

1

∈

𝒜

. Hence

𝒜

=

ℛ

. We have proved that the only ideals of

ℛ

are

(

0

)

and

ℛ

as desired.

(

⇐

)

Suppose the ring

ℛ

has only two
ideals, namely

(

0

)

,

ℛ

. Let

a

∈

ℛ

be a
non-zero element; we would like to prove the existence of a

multiplicative inverse

for

a

in

ℛ

. Define the
following set:

𝒜

=

(

a

)

=

{

r

∈

ℛ

∣

r

=

s

⋅

a

,

for
some

⁢

s

∈

ℛ

}

This is clearly an ideal, the

ideal
generated by

the element

a

. Moreover, this ideal is not the

zero
ideal

because

a

∈

𝒜

and

a

was assumed to be
non-zero. Thus, since there are only two ideals, we conclude

𝒜

=

ℛ

. Therefore

1

∈

𝒜

=

ℛ

so there exists an element

s

∈

ℛ

such that

s

⋅

a

=

1

∈

ℛ

Hence for all non-zero

a

∈

ℛ

,

a

has a multiplicative inverse in

ℛ

, so

ℛ

is, in fact, a field.
∎

Title

characterization of field

Canonical name

CharacterizationOfField

Date of creation

2013-03-22 13:57:03

Last modified on

2013-03-22 13:57:03

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

7

Author

alozano (2414)

Entry type

Theorem

Classification

msc 12E99

Synonym

a field only has two ideals

Related topic

Field

Related topic

Ring

Related topic

Ideal

Generated on Fri Feb 9 16:41:10 2018 by

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characterizations of integral

characterizations of integral

Theorem

.

Let

R

be a subring of a field

K

,

1

∈

R

and let

α

be a non-zero element of

K

.  The following conditions are

equivalent

:

1.

α

is integral over

R

.

2.

α

belongs to

R

⁢

[

α

-

1

]

.

3.

α

is unit of

R

⁢

[

α

-

1

]

.

4.

α

-

1

⁢

R

⁢

[

α

-

1

]

=

R

⁢

[

α

-

1

]

.

Proof.

Supposing the first condition that an

equation

α

n

+

a

1

⁢

α

n

-

1

+

…

+

a

n

-

1

⁢

α

+

a

n

=

0

,

with

a

j

’s belonging to

R

, holds.  Dividing both by

α

n

-

1

gives

α

=

-

a

1

-

a

2

⁢

α

-

1

-

…

-

a

n

⁢

α

-

n

+

1

.

One sees that

α

belongs to the ring

R

⁢

[

α

-

1

]

even being a unit of this (of course

α

-

1

∈

R

⁢

[

α

-

1

]

).  Therefore also the

principal ideal

α

-

1

⁢

R

⁢

[

α

-

1

]

of the ring

R

⁢

[

α

-

1

]

coincides with this ring.

Conversely

, the last circumstance implies that

α

is integral over

R

.

References

1

Emil Artin: .  Lecture notes.  Mathematisches Institut, Göttingen (1959).

Title

characterizations of integral

Canonical name

CharacterizationsOfIntegral

Date of creation

2013-03-22 14:56:54

Last modified on

2013-03-22 14:56:54

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

11

Author

pahio (2872)

Entry type

Theorem

Classification

msc 12E99

Classification

msc 13B21

Synonym

characterisations of integral

Generated on Fri Feb 9 16:41:12 2018 by

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division by zero

division by zero

Suppose that we construct a

number system

where we can divide by zero. What are the

consequences

?

Dividing by

x

means there exists an

inverse

1

x

so that

x

⋅

1

x

=

1

. So if we could divide by

0

then we would have a number we would call

1

0

.
We know

0

⋅

x

=

0

for any

x

, so

0

⋅

1

0

=

0

. However

0

⋅

1

0

=

1

because

x

⋅

1

x

=

1

for all

x

. Therefore

0

=

1

.

The problem is worse than

forcing

0

=

1

. Indeed,

1

⋅

x

=

x

for all

x

, so

x

=

1

⋅

x

=

0

⋅

x

=

0

.

So indeed every number is then 0.

Therefore, in a number system where we can add, subtract, multiply and divide (a field), we do not allow

division

by 0 because it would force the number system to only contain 0.

Hence, most

calculators

, from the the very basic to the full-fledged

computer algebra system

, generate some kind of an error message when asked to divide by 0. In fact, such an

operation

on a basic calculator causes the device to become inoperable until either the error clear key is pressed or the device is reset. A computer algebra system like

Mathematica

will complain about encountering “the infinite expression

1

0

” but deliver the result “

ComplexInfinity

” and remain operable.

But of course the limitations of machinery alone are not enough to forbid something from reality. So there are various thought experiments that can also be used to explain the omission of

division by zero

, for

instance

:

How could you slice a pizza into slices that have 0 width?

Since this would correspond to division by zero, it seems intellectually impossible. There is however an unfortunate side effect from such thinking.
For example:

How could you slice a pizza into slices of width -7, since width is always a

positive

quantity?
Furthermore, how could you actually slice a pizza into slices of width

7

or

e

π

, and so on? Yet we can actually divide by each of these
numbers.

This is an example of how mathematical abstraction benefits the understanding
of a problem.

Title

division by zero

Canonical

name

DivisionByZero

Date of creation

2013-03-22 16:35:53

Last modified on

2013-03-22 16:35:53

Owner

Algeboy (12884)

Last modified by

Algeboy (12884)

Numerical id

10

Author

Algeboy (12884)

Entry type

Example

Classification

msc 12E99

Classification

msc 00A05

Related topic

Field

Generated on Fri Feb 9 16:41:14 2018 by

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every ordered field with the least upper bound property is isomorphic to ℝ, proof that

every ordered field with the least upper bound property is isomorphic to

ℝ

, proof that

Let

F

be an

ordered field

with the least

upper bound

property. By the
order properties of

F

,

0

<

1

F

and by an

induction

argument

0

<

n

⋅

1

F

for any positive integer

n

. Hence the

characteristic

of the field

F

is
zero, implying that there is an

order-preserving

embedding

j

:

ℚ

→

F

.

We would like to extend this map to an embedding of

ℝ

into

F

. Let

r

∈

ℝ

and let

D

r

=

{

q

∈

ℚ

:

q

<

r

}

be the associated

Dedekind cut

. Since

D

r

is
nonempty and bounded above in

ℚ

, it follows that the set

j

⁢

(

D

r

)

is nonempty and
bounded above in

F

. Applying the least upper bound property of

F

, define a

function

ȷ

~

:

ℝ

→

F

by

ȷ

~

⁢

(

r

)

=

sup

⁡

(

j

⁢

(

D

r

)

)

.

One can check that

ȷ

~

is an order-preserving

field homomorphism

.
By replacing

F

with the

isomorphic

field

F

∖

ȷ

~

⁢

(

ℝ

)

∪

ℝ

,
we may assume that

ℝ

⊂

F

.

We claim that in fact

ℝ

=

F

. To see this, first recall that
since

F

is a

partially ordered group

with the least upper
bound property,

F

has the

Archimedean property

(

http://planetmath.org/DistributivityInPoGroups

).
So for any

f

∈

F

, there exists some positive integer

n

such that

-

n

<

f

<

n

.
Hence the set

D

f

′

=

{

r

∈

ℝ

:

r

<

f

}

⊂

ℝ

is nonempty and
bounded above, implying that

f

′

=

sup

ℝ

⁡

D

f

′

lies in

ℝ

.
Now observe that applying the least upper bound axiom in

F

gives us that

f

=

sup

F

⁡

D

f

′

. Since

f

′

is an upper bound of

D

f

′

in

F

, it follows
that

f

≤

f

′

.

Seeking a

contradiction

, suppose

f

<

f

′

. By the Archimedean property,
there is some positive integer

n

such that

f

<

f

′

-

n

-

1

<

f

′

. Because

f

′

=

sup

ℝ

⁡

D

f

′

, we obtain

f

′

-

n

-

1

<

f

, which implies the
contradiction

f

<

f

. Therefore

f

=

f

′

, and so

f

∈

ℝ

. This

completes

the proof.

Title

every ordered field with the least upper bound property is isomorphic to

ℝ

, proof that

Canonical name

EveryOrderedFieldWithTheLeastUpperBoundPropertyIsIsomorphicTomathbbRProofThat

Date of creation

2013-03-22 14:10:51

Last modified on

2013-03-22 14:10:51

Owner

mps (409)

Last modified by

mps (409)

Numerical id

9

Author

mps (409)

Entry type

Proof

Classification

msc 12E99

Classification

msc 54C30

Classification

msc 26-00

Classification

msc 12D99

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

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example of an infinite field of finite characteristic

example of an infinite field of finite characteristic

Let

K

be a field of finite

characteristic

, such as

F

p

. Then the ring of polynomials,

K

⁢

[

X

]

, is an

integral domain

. We may therefore construct its

quotient field

, namely the

field of rational

polynomials

. This is an example of an

infinite

field with finite characteristic.

Title

example of an infinite field of finite characteristic

Canonical name

ExampleOfAnInfiniteFieldOfFiniteCharacteristic

Date of creation

2013-03-22 13:11:12

Last modified on

2013-03-22 13:11:12

Owner

vitriol (148)

Last modified by

vitriol (148)

Numerical id

4

Author

vitriol (148)

Entry type

Example

Classification

msc 12E99

Generated on Fri Feb 9 16:41:18 2018 by

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