aarjaneiro/planetmath huggingface dataset:
invertible ideals in semi-local rings

invertible ideals in semi-local rings

Theorem

.

Let

R

be a

commutative ring

in which there are only finitely many

maximal ideals

. Then, a

fractional ideal

over

R

is invertible if and only if it is principal and generated by a

regular element

.

In particular, a semi-local (

http://planetmath.org/SemiLocalRing

)

Dedekind domain

is a

principal ideal domain

and every

finitely generated

ideal in a semi-local Prüfer domain is principal.

Proof.

First, if

a

is

regular

then

(

a

)

is invertible, with

inverse

(

a

-

1

)

, so only the converse needs to be shown.

Suppose that

𝔞

is invertible, and

𝔞

⁢

𝔟

=

R

.
Then let the maximal ideals of

R

be

𝔪

1

,

…

,

𝔪

n

. As

𝔞

⁢

𝔟

⊈

𝔪

k

, there exist

a

k

∈

𝔞

,

b

k

∈

𝔟

such that

a

k

⁢

b

k

∈

R

∖

𝔪

k

.

By maximality,

𝔪

j

⊈

𝔪

k

whenever

j

≠

k

, so we may choose

λ

j

⁢

k

∈

𝔪

j

∖

𝔪

k

.
Setting

λ

k

=

∏

j

≠

k

λ

j

⁢

k

gives

λ

k

∈

𝔪

j

for all

j

≠

k

and, as

𝔪

k

is prime (

http://planetmath.org/PrimeIdeal

),

λ

k

∉

𝔪

k

.
Then, writing

a

=

λ

1

⁢

a

1

+

⋯

+

λ

n

⁢

a

n

∈

𝔞

,

b

=

λ

1

⁢

b

1

+

⋯

+

λ

n

⁢

b

n

∈

𝔟

we can expand the product to get

a

⁢

b

=

∑

i

,

j

λ

i

⁢

λ

j

⁢

a

i

⁢

b

j

.

(1)

However,

a

i

⁢

b

j

∈

𝔞

⁢

𝔟

⊆

R

so

λ

i

⁢

λ

j

⁢

a

i

⁢

b

j

is in

𝔪

k

whenever either

i

or

j

is not equal to

k

. On the other hand,

λ

k

⁢

λ

k

⁢

a

k

⁢

b

k

∉

𝔪

k

and, consequently, there is exactly one term on the

right hand side

of (

1

) which is not in

𝔪

k

, so

a

⁢

b

∉

𝔪

k

.

We have shown that

a

⁢

b

is not in any maximal ideal of

R

, and must therefore be a unit. So a is regular and,

(

a

)

⊆

𝔞

=

a

⁢

b

⁢

𝔞

⊆

a

⁢

𝔟

⁢

𝔞

=

a

⁢

R

=

(

a

)

as required.
∎

Title

invertible ideals in semi-local rings

Canonical name

InvertibleIdealsInSemilocalRings

Date of creation

2013-03-22 18:36:17

Last modified on

2013-03-22 18:36:17

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

6

Author

gel (22282)

Entry type

Theorem

Classification

msc 11R04

Classification

msc 13F05

Related topic

PruferDomain

Generated on Fri Feb 9 15:41:43 2018 by

LaTeXML
localizations of Dedekind domains are Dedekind

localizations of Dedekind domains are Dedekind

If

R

is an

integral domain

with

field of fractions

k

and

S

⊆

R

∖

{

0

}

is a

multiplicative set

, then the

localization

at

S

is given by

S

-

1

⁢

R

=

{

s

-

1

⁢

x

:

x

∈

R

,

s

∈

S

}

(up to

isomorphism

). This is a

subring

of

k

, and the following theorem states that localizations of

Dedekind domains

are again Dedekind domains.

Theorem

.

Let

R

be a Dedekind domain and

S

⊆

R

∖

{

0

}

be a multiplicative set. Then

S

-

1

⁢

R

is a Dedekind domain.

A special case of this is the localization at a

prime ideal

𝔭

, which is defined as

R

𝔭

≡

(

R

∖

𝔭

)

-

1

⁢

R

, and is therefore a Dedekind domain. In fact, if

𝔭

is nonzero then it can be shown that

R

𝔭

is a

discrete valuation ring

.

Title

localizations of Dedekind domains are Dedekind

Canonical name

LocalizationsOfDedekindDomainsAreDedekind

Date of creation

2013-03-22 18:35:13

Last modified on

2013-03-22 18:35:13

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

4

Author

gel (22282)

Entry type

Theorem

Classification

msc 11R04

Classification

msc 13F05

Classification

msc 13H10

Related topic

Localization

Generated on Fri Feb 9 15:41:45 2018 by

LaTeXML
minimality of integral basis

minimality of integral basis

The

discriminant

Δ

:=

Δ

⁢

(

α

1

,

α

2

,

…

,

α

s

)

of the set

{

α

1

,

α

2

,

…

,

α

s

}

of

integers

of an

algebraic number field

K

is a

rational integer

.  If this set is an

integral basis

of

K

, then

|

Δ

|

has the least possible (

positive integer

) value in the field

K

, and conversely.  The value

d

=

Δ

is equal for all integral bases of

K

, and it is called the

discriminant

or

fundamental number

of the field.

Title

minimality of integral basis

Canonical name

MinimalityOfIntegralBasis

Date of creation

2013-03-22 15:20:38

Last modified on

2013-03-22 15:20:38

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

9

Author

Mathprof (13753)

Entry type

Theorem

Classification

msc 11R04

Related topic

CanonicalBasis

Related topic

PropertiesOfDiscriminantInAlgebraicNumberField

Defines

fundamental number

Defines

discriminant of field

Generated on Fri Feb 9 15:41:47 2018 by

LaTeXML
multiples of an algebraic number

multiples of an algebraic number

Theorem

.

If

α

is an

algebraic number

, then there exists a non-zero

multiple

(

http://

planetmath

.org/GeneralAssociativity

) of

α

which is an

algebraic integer

.

Proof.

Let

α

be a of the equation

x

n

+

r

1

⁢

x

n

-

1

+

r

2

⁢

x

n

-

2

+

⋯

+

r

n

=

0

,

where

r

1

,

r

2

, …,

r

n

are

rational numbers

(

n

>

0

).  Let

l

be the least common multiple of the

denominators

of the

r

j

’s.  Then we have

0

=

l

n

⁢

(

α

n

+

r

1

⁢

α

n

-

1

+

r

2

⁢

α

n

-

2

+

⋯

+

r

n

)

=

(

l

⁢

α

)

n

+

l

⁢

r

1

⁢

(

l

⁢

α

)

n

-

1

+

l

2

⁢

r

2

⁢

(

l

⁢

α

)

n

-

2

+

⋯

+

l

n

⁢

r

n

,

i.e. the the

algebraic equation

x

n

+

l

⁢

r

1

⁢

x

n

-

1

+

l

2

⁢

r

2

⁢

x

n

-

2

+

⋯

+

l

n

⁢

r

n

=

0

with

rational integer

coefficients

.

According to the theorem, any algebraic number

ξ

is a

quotient

(

http://planetmath.org/Division

) of an algebraic integer
(of the field

ℚ

⁢

(

ξ

)

) and a rational integer.

Title

multiples of an algebraic number

Canonical name

MultiplesOfAnAlgebraicNumber

Date of creation

2014-05-16 19:58:36

Last modified on

2014-05-16 19:58:36

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Theorem

Classification

msc 11R04

Related topic

TheoryOfAlgebraicNumbers

Related topic

AlgebraicSinesAndCosines

Related topic

SomethingRelatedToAlgebraicInteger

Related topic

RationalAlgebraicIntegers

Generated on Fri Feb 9 15:41:50 2018 by

LaTeXML
norm and trace of algebraic number

norm and trace of algebraic number

Theorem 1

.

Let

K

be an

algebraic number field

and

α

an element of

K

.  The norm

N

⁢

(

α

)

and the trace

S

⁢

(

α

)

of

α

in the

field extension

K

/

ℚ

both are

rational numbers

and especially

rational integers

in the case

α

is an

algebraic integer

.  If

β

is another element of

K

, then

N

⁢

(

α

⁢

β

)

=

N

⁢

(

α

)

⁢

N

⁢

(

β

)

,

S

⁢

(

α

+

β

)

=

S

⁢

(

α

)

+

S

⁢

(

β

)

,

(1)

i.e. the norm is

multiplicative

and the trace

additive

.  If

[

K

:

ℚ

]

=

n

and

a

∈

ℚ

, then

N

⁢

(

a

)

=

a

n

,

S

⁢

(

a

)

=

n

⁢

a

.

Remarks

1.  The notions norm and trace were originally introduced in German as “die Norm” and “die Spur”.  Therefore in German and many other literature the symbol of trace is S, Sp or sp.  Nowadays the symbols T and Tr are common.

2.  The norm and trace of an

algebraic number

α

in the field extension

ℚ

⁢

(

α

)

/

ℚ

,  i.e. the

product

and sum of all

algebraic conjugates

of

α

, are called the

absolute norm

and the

absolute trace

of

α

.  Formulae like (1) concerning the absolute norms and traces are not sensible.

Theorem 2

.

An algebraic integer

ε

is a unit if and only if

N

⁢

(

ε

)

=

±

1

,

i.e. iff the absolute norm of

ε

is a rational unit.  Thus in the

minimal polynomial

of an algebraic unit is always

±

1

.

Example.

The minimal polynomial of the number

2

+

3

, which is the

fundamental unit

of the

quadratic field

ℚ

⁢

(

3

)

, is

x

2

-

4

⁢

x

+

1

.

Title

norm and trace of algebraic number

Canonical name

NormAndTraceOfAlgebraicNumber

Date of creation

2013-03-22 15:19:08

Last modified on

2013-03-22 15:19:08

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

15

Author

pahio (2872)

Entry type

Theorem

Classification

msc 11R04

Related topic

TheoryOfAlgebraicNumbers

Related topic

AlgebraicNumberTheory

Related topic

IdealNorm

Related topic

UnitsOfRealCubicFieldsWithExactlyOneRealEmbedding

Related topic

IndependenceOfCharacteristicPolynomialOnPrimitiveElement

Defines

absolute norm

Defines

absolute trace

Generated on Fri Feb 9 15:41:52 2018 by

LaTeXML
number of prime ideals in a number field

number of prime ideals in a number field

Theorem.

The

ring of integers

of an

algebraic number field

contains infinitely many

prime ideals

.

Proof.

Let

𝒪

be the ring of integers of a

number field

.  If

p

is a

rational prime

number, then the

principal ideal

(

p

)

of

𝒪

does not coincide with

(

1

)

=

𝒪

and thus

(

p

)

has a set of prime ideals of

𝒪

as factors.  Two different (

positive

) rational primes

p

and

q

satisfy

gcd

⁡

(

(

p

)

,

(

q

)

)

=

(

p

,

q

)

=

(

1

)

,

since there exist

integers

x

and

y

such that

x

⁢

p

+

y

⁢

q

=

1

and consequently

1

∈

(

p

,

q

)

.  Therefore, the principal ideals

(

p

)

and

(

q

)

of

𝒪

have no common prime ideal factors.  Because there are

http://planetmath.org/node/3036

infinitely many rational prime numbers, also the corresponding principal ideals have infinitely many different prime ideal factors.

Title

number of prime ideals in a number field

Canonical name

NumberOfPrimeIdealsInANumberField

Date of creation

2013-03-22 19:12:51

Last modified on

2013-03-22 19:12:51

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

4

Author

pahio (2872)

Entry type

Theorem

Classification

msc 11R04

Generated on Fri Feb 9 15:41:53 2018 by

LaTeXML
polynomial equation with algebraic coefficients

polynomial equation with algebraic coefficients

If

α

1

,

…

,

α

k

are

algebraic numbers

[resp.

algebraic integers

] and

f

1

⁢

(

α

1

,

…

,

α

k

)

,

…

,

f

n

⁢

(

α

1

,

…

,

α

k

)

polynomials

in

α

1

,

…

,

α

k

with

rational

[resp.

integer

]

coefficients

, then all

complex roots

of the equation

x

n

+

f

1

⁢

(

α

1

,

…

,

α

k

)

⁢

x

n

-

1

+

…

+

f

n

⁢

(

α

1

,

…

,

α

k

)

=

0

(1)

are algebraic numbers [resp. algebraic integers].

Proof.

Let the

minimal polynomial

x

m

+

a

1

⁢

x

m

-

1

+

…

+

a

m

of

α

1

over

ℤ

have the zeros (

http://planetmath.org/ZeroOfAFunction

)

α

1

(

1

)

=

α

1

,

α

1

(

2

)

,

…

,

α

1

(

m

)

and denote by

F

⁢

(

x

;

α

1

,

α

2

,

…

,

α

k

)

the left hand side of the equation (1).  Consider the equation

G

⁢

(

x

;

α

2

,

…

,

α

k

)

:=

∏

i

=

1

m

F

⁢

(

x

;

α

1

(

i

)

,

α

2

,

…

,

α

k

)

=

0

.

(2)

Here, the coefficients of the polynomial

G

are polynomials in the numbers

α

1

(

1

)

,

α

1

(

2

)

,

…

,

α

1

(

m

)

,

α

2

,

…

,

α

k

with rational [resp. integer] coefficients.  Thus the coefficients of

G

are

symmetric polynomials

g

j

in the numbers

α

1

(

i

)

:

G

=

∑

j

g

j

⁢

x

j

By the

fundamental theorem of symmetric polynomials

, the coefficients

g

j

of

G

are polynomials in

α

2

,

…

,

α

k

with rational [resp. integer] coefficients.  Consequently,

G

has the form

G

=

x

h

+

a

1

′

⁢

(

α

2

,

…

,

α

k

)

⁢

x

h

-

1

+

…

+

a

h

′

⁢

(

α

2

,

…

,

α

k

)

where the coefficients

a

i

′

⁢

(

α

2

,

…

,

α

k

)

are polynoms in the numbers

α

j

with rational [resp. integer] coefficients.  As one continues similarly, removing one by one also

α

2

,

…

,

α

k

which go back to the rational [resp. integer] coefficients of the corresponding minimal polynomials, one shall finally arrive at an equation

x

s

+

A

1

⁢

x

s

-

1

+

…

+

A

s

=

0

,

(3)

among the roots of which there are the roots of (1); the coefficients

A

ν

do no more explicitely depend on the algebraic numbers

α

1

,

…

,

α

k

but are rational numbers [resp. integers].

Accordingly, the roots of (1) are algebraic numbers [resp. algebraic integers], Q.E.D.

Title

polynomial equation with algebraic coefficients

Canonical name

PolynomialEquationWithAlgebraicCoefficients

Date of creation

2013-03-22 19:07:37

Last modified on

2013-03-22 19:07:37

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Theorem

Classification

msc 11R04

Synonym

monic equation with algebraic coefficients

Generated on Fri Feb 9 15:41:56 2018 by

LaTeXML
power basis over ℤ

power basis over

ℤ

Let

K

be a

number field

with

[

K

:

ℚ

]

=

n

and

𝒪

K

denote the

ring of integers

of

K

. Then

𝒪

K

has a

power basis

over

ℤ

(sometimes shortened simply to

power basis

) if there exists

α

∈

K

such that the set

{

1

,

α

,

…

,

α

n

-

1

}

is an

integral basis

for

𝒪

K

. An

equivalent

(

http://planetmath.org/Equivalent3

) condition is that

𝒪

K

=

ℤ

⁢

[

α

]

. Note that if such an

α

exists, then

α

∈

𝒪

K

and

K

=

ℚ

⁢

(

α

)

.

Not all rings of integers have power bases. (See the entry

biquadratic field

for more details.) On the other hand, any ring of integers of a

quadratic field

has a power basis over

ℤ

, as does any ring of integers of a

cyclotomic field

. (See the entry

examples of ring

of

integers

of a number field for more details.)

Title

power basis over

ℤ

Canonical name

PowerBasisOvermathbbZ

Date of creation

2013-03-22 15:56:55

Last modified on

2013-03-22 15:56:55

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

17

Author

Wkbj79 (1863)

Entry type

Definition

Classification

msc 11R04

Synonym

power basis

Synonym

power bases

Related topic

ConditionForPowerBasis

Generated on Fri Feb 9 15:41:58 2018 by

LaTeXML
proof of localizations of Dedekind domains are Dedekind

proof of localizations of Dedekind domains are Dedekind

Let

R

be a

Dedekind domain

with

field of fractions

k

and

S

⊆

R

∖

{

0

}

be a

multiplicative set

. We show that the

localization

at

S

,

S

-

1

⁢

R

≡

{

s

-

1

⁢

x

:

x

∈

R

,

s

∈

S

}

,

is again a Dedekind domain.

We use the characterization of Dedekind domains as

integral domains

in which every nonzero ideal is

invertible

(

http://planetmath.org/FractionalIdeal

) (see proof that a domain is Dedekind if its ideals are invertible).

Let

𝔞

be a nonzero integral ideal of

S

-

1

⁢

R

. Then

𝔞

∩

R

is a nonzero ideal of the Dedekind domain

R

, so it has an inverse

(

𝔞

∩

R

)

⁢

𝔟

=

R

.

Here,

𝔟

is a

fractional ideal

of

R

. Also let

S

-

1

⁢

𝔟

be the fractional ideal of

S

-

1

⁢

R

generated by

𝔟

,

S

-

1

⁢

𝔟

=

{

s

-

1

⁢

x

:

x

∈

𝔟

,

s

∈

S

}

.

The equalities

𝔞

⁢

(

S

-

1

⁢

𝔟

)

=

S

-

1

⁢

(

(

𝔞

∩

R

)

⁢

𝔟

)

=

S

-

1

⁢

R

show that

𝔞

is invertible, so

S

-

1

⁢

R

is a Dedekind domain.

Title

proof of localizations of Dedekind domains are Dedekind

Canonical name

ProofOfLocalizationsOfDedekindDomainsAreDedekind

Date of creation

2013-03-22 18:35:16

Last modified on

2013-03-22 18:35:16

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

4

Author

gel (22282)

Entry type

Proof

Classification

msc 11R04

Classification

msc 13F05

Classification

msc 13H10

Generated on Fri Feb 9 15:42:00 2018 by

LaTeXML
proof of transcendental root theorem

proof of transcendental root theorem

Proposition 1

.

Let

F

⊂

K

be a

field extension

with

K

an algebraically closed field. Let

x

∈

K

be

transcendental

over

F

. Then for any

natural number

n

≥

1

, the

element

x

1

/

n

∈

K

is also transcendental over

F

.

Proof.

Suppose

x

is transcendental over a field

F

, and assume for a

contradiction

that

x

1

/

n

is algebraic over

F

. Thus, there is a

polynomial

P

⁢

(

y

)

∈

F

⁢

[

y

]

such that

P

⁢

(

x

1

/

n

)

=

0

(note that the polynomial

y

n

-

x

is

not

a polynomial with

coefficients

in

F

, so

P

⁢

(

y

)

might be more involved). Then the field

H

=

F

⁢

(

x

1

/

n

)

⊆

K

is a finite

algebraic extension

of

F

, and every element of

H

is algebraic over

K

. However

x

∈

H

, so

x

is algebraic over

F

which is a contradiction.
∎

Title

proof of transcendental root theorem

Canonical name

ProofOfTranscendentalRootTheorem

Date of creation

2013-03-22 14:11:41

Last modified on

2013-03-22 14:11:41

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

6

Author

alozano (2414)

Entry type

Proof

Classification

msc 11R04

Related topic

AlgebraicElement

Related topic

AlgebraicClosure

Related topic

Algebraic

Related topic

AlgebraicExtension

Related topic

AFiniteExtensionOfFieldsIsAnAlgebraicExtension

Generated on Fri Feb 9 15:42:07 2018 by

LaTeXML
properties of ℚ⁢(ϑ)-conjugates

properties of

ℚ

⁢

(

ϑ

)

-conjugates

Lemma.

Let

α

1

,

α

2

,

…

,

α

s

be

algebraic numbers

belonging to the

number field

ℚ

⁢

(

ϑ

)

of degree (

http://planetmath.org/NumberField

)

n

and

α

i

(

j

)

their

http://planetmath.org/node/12046

ℚ

⁢

(

ϑ

)

-

conjugates

.  If

P

⁢

(

x

1

,

x

2

,

…

,

x

s

)

is a

polynomial

with

rational

coefficients and if

P

⁢

(

α

1

,

α

2

,

…

,

α

s

)

=

0

,

then also

P

⁢

(

α

1

(

j

)

,

α

2

(

j

)

,

…

,

α

s

(

j

)

)

=

0

for each

j

=

1

,

2

,

…

,

n

.  In the special case of two elements

α

and

β

of

ℚ

⁢

(

ϑ

)

one may infer the formulae

(

α

⁢

β

)

(

j

)

=

α

(

j

)

⁢

β

(

j

)

,

(

α

+

β

)

(

j

)

=

α

(

j

)

+

β

(

j

)

.

(1)

The lemma implies easily the following theorems.

Theorem 1.

All

conjugate fields

of

ℚ

⁢

(

ϑ

)

are

isomorphic

.

Theorem 2.

The norm and the trace in the field

ℚ

⁢

(

ϑ

)

satisfy

N

⁢

(

α

⁢

β

)

=

N

⁢

(

α

)

⁢

N

⁢

(

β

)

,

S

⁢

(

α

+

β

)

=

S

⁢

(

α

)

+

S

⁢

(

β

)

.

Cf. the entry

norm and trace of algebraic number

.

Title

properties of

ℚ

⁢

(

ϑ

)

-conjugates

Canonical name

PropertiesOfmathbbQvarthetaconjugates

Date of creation

2013-03-22 19:09:14

Last modified on

2013-03-22 19:09:14

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Topic

Classification

msc 11R04

Classification

msc 11C08

Classification

msc 12E05

Classification

msc 12F05

Related topic

ConjugateFields

Related topic

IndependenceOfCharacteristicPolynomialOnPrimitiveElement

Generated on Fri Feb 9 15:42:09 2018 by

LaTeXML
rational algebraic integers

rational algebraic integers

Theorem

.

A

rational number

is an

algebraic integer

iff it is a

rational integer

.

Proof.

1

∘

.  Any rational integer

m

has the

minimal polynomial

x

-

m

, whence it is an algebraic integer.

2

∘

.  Let the rational number

α

=

m

n

be an algebraic integer where

m

,

n

are coprime integers and

n

>

0

.  Then there is a

polynomial

f

⁢

(

x

)

=

x

k

+

a

1

⁢

x

k

-

1

+

…

+

a

k

with

a

1

,

…

,

a

k

∈

ℤ

such that

f

⁢

(

α

)

=

(

m

n

)

k

+

a

1

⁢

(

m

n

)

k

-

1

+

…

+

a

k

=

0

.

Multiplying this equation termwise by

n

k

implies

m

k

=

-

a

1

⁢

m

k

-

1

⁢

n

-

…

-

a

k

⁢

n

k

,

which says that

n

∣

m

k

(see divisibility in rings).  Since

m

and

n

are

coprime

and

n

positive

, it follows that

n

=

1

.  Therefore,

α

=

m

∈

ℤ

.

Title

rational algebraic integers

Canonical name

RationalAlgebraicIntegers

Date of creation

2013-03-22 19:07:34

Last modified on

2013-03-22 19:07:34

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

5

Author

pahio (2872)

Entry type

Theorem

Classification

msc 11R04

Related topic

MultiplesOfAnAlgebraicNumber

Generated on Fri Feb 9 15:42:11 2018 by

LaTeXML
table of some fundamental units

table of some fundamental units

Below, we tabulate the

fundamental units

η

of first

real quadratic fields

ℚ

⁢

(

d

)

; the number

ω

is

1

+

d

2

for

d

≡

1

(

mod

4

)

and

d

for

d

≢

1

(

mod

4

)

.

d

η

d

η

2

1

+

ω

47

48

+

7

⁢

ω

3

2

+

ω

51

50

+

7

⁢

ω

5

ω

53

3

+

ω

6

5

+

2

⁢

ω

55

89

+

12

⁢

ω

7

8

+

3

⁢

ω

57

131

+

40

⁢

ω

10

3

+

ω

58

99

+

13

⁢

ω

11

10

+

3

⁢

ω

59

530

+

69

⁢

ω

13

1

+

ω

61

17

+

5

⁢

ω

14

15

+

4

⁢

ω

62

63

+

8

⁢

ω

15

4

+

ω

65

7

+

2

⁢

ω

17

3

+

2

⁢

ω

66

65

+

8

⁢

ω

19

170

+

39

⁢

ω

67

48842

+

5967

⁢

ω

21

2

+

ω

69

11

+

3

⁢

ω

22

197

+

42

⁢

ω

70

251

+

30

⁢

ω

23

24

+

5

⁢

ω

71

3480

+

413

⁢

ω

26

5

+

ω

73

943

+

250

⁢

ω

29

2

+

ω

74

43

+

5

⁢

ω

30

11

+

2

⁢

ω

77

4

+

ω

31

1520

+

273

⁢

ω

78

53

+

6

⁢

ω

33

19

+

8

⁢

ω

79

80

+

9

⁢

ω

34

35

+

6

⁢

ω

82

9

+

ω

35

6

+

ω

83

82

+

9

⁢

ω

37

5

+

2

⁢

ω

85

4

+

ω

38

37

+

6

⁢

ω

86

10405

+

1122

⁢

ω

39

25

+

4

⁢

ω

87

28

+

3

⁢

ω

41

27

+

10

⁢

ω

89

447

+

106

⁢

ω

42

13

+

2

⁢

ω

91

1574

+

165

⁢

ω

43

3482

+

531

⁢

ω

93

13

+

3

⁢

ω

46

24335

+

3588

⁢

ω

94

2143295

+

221064

⁢

ω

References

1

S. Borewicz & I. Safarevic

:

Zahlentheorie

.  Birkhäuser Verlag. Basel und Stuttgart (1966).

Title

table of some fundamental units

Canonical name

TableOfSomeFundamentalUnits

Date of creation

2013-03-22 17:58:39

Last modified on

2013-03-22 17:58:39

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

9

Author

pahio (2872)

Entry type

Result

Classification

msc 11R04

Classification

msc 11R11

Classification

msc 11R27

Related topic

UnitsOfQuadraticFields

Related topic

QuadraticField

Related topic

IntegralBasisOfQuadraticField

Related topic

AlgebraicNumberTheory

Generated on Fri Feb 9 15:42:14 2018 by

LaTeXML
transcendental root theorem

transcendental root theorem

Suppose a

constant

x

is transcendental over some field

F

. Then

x

n

is also transcendental over

F

for any

n

≥

1

.

Proof.

Let

F

¯

denote an

algebraic closure

of

F

. Assume for the sake of

contradiction

that

x

n

∈

F

¯

. Then since

algebraic numbers

are

closed under

multiplication

(and thus exponentiation by

positive integers

), we have

(

x

n

)

n

=

x

∈

F

¯

, so that

x

is algebraic over

F

, creating a contradiction.
∎

Title

transcendental root theorem

Canonical

name

TranscendentalRootTheorem

Date of creation

2013-03-22 14:04:23

Last modified on

2013-03-22 14:04:23

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

8

Author

mathcam (2727)

Entry type

Theorem

Classification

msc 11R04

Generated on Fri Feb 9 15:42:21 2018 by

LaTeXML