aarjaneiro/planetmath huggingface dataset:
natural homomorphism

natural homomorphism

The

natural homomorphism

R

→

R

/

I

where

R

is a ring and

I

an ideal of

R

is the map

α

↦

α

+

I

. Coset

operations

, e.g

(

α

+

β

)

+

I

=

(

α

+

I

)

+

(

β

+

I

)

guarantee this will be a

homomorphism

. Natural homomorphisms for groups can be defined similarly.

Title

natural homomorphism

Canonical name

NaturalHomomorphism

Date of creation

2013-03-22 14:33:32

Last modified on

2013-03-22 14:33:32

Owner

iwnbap (1760)

Last modified by

iwnbap (1760)

Numerical id

8

Author

iwnbap (1760)

Entry type

Definition

Classification

msc 13B10

Related topic

Homomorphism

Related topic

QuotientRing

Defines

Natural homomorphism

Generated on Fri Feb 9 17:25:17 2018 by

LaTeXML
ring homomorphism

ring homomorphism

Let

R

and

S

be rings. A

ring homomorphism

is a

function

f

:

R

⟶

S

such that:

•

f

⁢

(

a

+

b

)

=

f

⁢

(

a

)

+

f

⁢

(

b

)

for all

a

,

b

∈

R

•

f

⁢

(

a

⋅

b

)

=

f

⁢

(

a

)

⋅

f

⁢

(

b

)

for all

a

,

b

∈

R

A

ring isomorphism

is a ring homomorphism which is a

bijection

. A

ring monomorphism

(respectively,

ring epimorphism

) is a ring homomorphism which is an

injection

(respectively,

surjection

).

When working in a

context

in which all rings have a

multiplicative identity

, one also requires that

f

⁢

(

1

R

)

=

1

S

. Ring homomorphisms which

satisfy

this

property

are called

unital

ring homomorphisms.

Title

ring homomorphism

Canonical

name

RingHomomorphism

Date of creation

2013-03-22 11:48:50

Last modified on

2013-03-22 11:48:50

Owner

djao (24)

Last modified by

djao (24)

Numerical id

12

Author

djao (24)

Entry type

Definition

Classification

msc 13B10

Classification

msc 16B99

Classification

msc 81P05

Related topic

Ring

Defines

unital

Defines

ring isomorphism

Defines

ring epimorphism

Defines

ring monomorphism

Defines

homomorphism

Defines

isomorphism

Defines

epimorphism

Defines

monomprhism

Generated on Fri Feb 9 17:25:21 2018 by

LaTeXML
integral

integral

Let

B

be a ring with a subring

A

. We will assume that

A

is contained in the center of

B

(in particular,

A

is

commutative

). An element

x

∈

B

is

integral

over

A

if there exist elements

a

0

,

…

,

a

n

-

1

∈

A

such that

x

n

+

a

n

-

1

⁢

x

n

-

1

+

⋯

+

a

1

⁢

x

+

a

0

=

0

.

The ring

B

is

integral

over

A

if every element of

B

is integral over

A

.

Title

integral

Canonical

name

Integral

Date of creation

2013-03-22 12:07:44

Last modified on

2013-03-22 12:07:44

Owner

djao (24)

Last modified by

djao (24)

Numerical id

9

Author

djao (24)

Entry type

Definition

Classification

msc 13B21

Related topic

IntegralBasis

Generated on Fri Feb 9 17:25:24 2018 by

LaTeXML
integral closures in separable extensions are finitely generated

integral closures in separable extensions are finitely generated

The theorem below generalizes to arbitrary integral ring

extensions

(under certain conditions) the fact that the

ring of integers

of a

number field

is

finitely generated

over

ℤ

. The proof parallels the proof of the number field result.

Theorem 1

.

Let

B

be an

integrally closed

Noetherian domain with

field of fractions

K

. Let

L

be a finite

separable extension

of

K

, and let

A

be the

integral closure

of

B

in

L

. Then

A

is a finitely generated

B

-module.

Proof.

We first show that the trace (

http://planetmath.org/Trace2

)

T

⁢

r

K

L

maps

A

to

B

. Choose

u

∈

A

and let

f

=

I

⁢

r

⁢

r

⁢

(

u

,

K

)

∈

K

⁢

[

x

]

be the

minimal polynomial

for

u

over

K

; assume

f

is of degree

d

. Let the

conjugates

of

u

in some

splitting field

be

u

=

a

1

,

…

,

a

d

. Then the

a

i

are all integral over

B

since they satisfy

u

’s

monic polynomial

in

B

⁢

[

x

]

. Since the

coefficients

of

F

are

polynomials

in the

a

i

, they too are integral over

B

. But the coefficients are in

K

, and

B

is integrally closed (in

K

), so the coefficients are in

B

. But

T

⁢

r

K

L

⁢

(

u

)

is just the coefficient of

x

d

-

1

in

f

, and thus

T

⁢

r

K

L

⁢

(

u

)

∈

B

. This proves the claim.

Now, choose a basis

ω

1

,

…

,

ω

d

of

L

/

K

. We may assume

ω

i

∈

A

by multiplying each by an appropriate element of

B

. (To see this, let

I

⁢

r

⁢

r

⁢

(

ω

i

,

K

)

∈

K

⁢

[

x

]

=

x

d

+

k

1

⁢

x

d

-

1

+

…

+

k

d

. Choose

b

∈

B

such that

b

⁢

k

i

∈

B

⁢

∀

i

. Then

(

b

⁢

ω

)

d

+

b

⁢

k

1

⁢

(

b

⁢

ω

)

d

-

1

+

…

+

b

d

⁢

k

d

=

0

and thus

b

⁢

ω

∈

A

). Define a linear map

φ

:

L

→

K

d

:

a

↦

(

T

⁢

r

K

L

⁢

(

a

⁢

ω

1

)

,

…

,

T

⁢

r

K

L

⁢

(

a

⁢

ω

d

)

)

.

φ

is 1-1, since if

u

∈

ker

⁡

φ

,

u

≠

0

, then

T

⁢

r

⁢

(

u

⁢

L

)

=

0

. But

u

⁢

L

=

L

, so

T

⁢

r

K

L

is identically zero, which cannot be since

L

is

separable

over

K

(it is a standard result that separability is

equivalent

to nonvanishing of the trace map; see for example

[

1

]

, Chapter 8).

But

T

⁢

r

K

L

:

A

→

B

by the above, so

φ

:

A

↪

B

d

. Since

B

is

Noetherian

, any

submodule

of a finitely generated module is also finitely generated, so

A

is finitely generated as a

B

-module.
∎

References

1

P. Morandi,

Field and

Galois Theory

, Springer, 2006.

Title

integral closures in separable extensions are finitely generated

Canonical name

IntegralClosuresInSeparableExtensionsAreFinitelyGenerated

Date of creation

2013-03-22 17:02:12

Last modified on

2013-03-22 17:02:12

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

5

Author

rm50 (10146)

Entry type

Theorem

Classification

msc 13B21

Classification

msc 12F05

Related topic

IntegralClosureIsRing

Generated on Fri Feb 9 17:25:29 2018 by

LaTeXML
integrality is transitive

integrality is transitive

Let

C

⊂

B

⊂

A

be rings. If

B

is integral over

C

and

A

is integral over

B

, then

A

is integral over

C

.

Proof.

Choose

u

∈

A

. Then

u

n

+

b

1

⁢

u

n

-

1

+

⋯

+

b

n

=

0

,

b

i

∈

B

. Thus

C

⁢

[

b

1

,

…

,

b

n

,

u

]

is

integral

and thus

module-finite

over

C

⁢

[

b

1

,

…

,

b

n

]

. Each

b

i

is integral over

C

, so

C

⁢

[

b

1

,

…

,

b

n

]

is integral hence module-finite over

C

. Thus

C

⁢

[

b

1

,

…

,

b

n

,

u

]

is module-finite, hence integral, over

C

, so

u

is integral over

C

.

Title

integrality is transitive

Canonical name

IntegralityIsTransitive

Date of creation

2013-03-22 17:01:25

Last modified on

2013-03-22 17:01:25

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

6

Author

rm50 (10146)

Entry type

Theorem

Classification

msc 13B21

Generated on Fri Feb 9 17:25:33 2018 by

LaTeXML
separability is required for integral closures to be finitely generated

separability is required for integral closures to be finitely generated

The parent

theorem

assumed that

L

was a

separable extension

of

K

. Here is an example that shows that separability is in fact a necessary condition for the result to hold.

Let

k

=

(

ℤ

/

p

⁢

ℤ

)

⁢

(

b

1

,

b

2

,

…

)

where the

b

i

are

indeterminates

.

Let

B

be the

subring

of

k

⁢

[

[

t

]

]

(power series in

t

over

k

) consisting of

{

∑

i

=

0

∞

c

i

⁢

t

i

}

such that

{

c

0

,

c

1

,

c

2

,

…

}

generates

a

finite extension

of

(

ℤ

/

p

⁢

ℤ

)

⁢

(

b

1

p

,

b

2

p

,

…

)

. Then every element of

B

is a power of

t

times a unit (first factor out the largest power of

t

. What’s left is

a

⁢

(

1

+

c

1

⁢

t

+

c

2

⁢

t

2

+

⋯

)

; its

inverse

is

a

-

1

⁢

(

1

-

c

1

⁢

t

-

⋯

)

). Hence the ideals of

B

are powers of

t

, so

B

is a PID (in fact, it is a DVR).

Now, let

u

=

b

0

+

b

1

⁢

t

+

b

2

⁢

t

2

+

⋯

. Now,

u

∉

B

because it uses all of the

b

i

and thus the

coefficients

do not define a finite extension of

(

ℤ

/

p

⁢

ℤ

)

⁢

(

b

1

p

,

b

2

p

,

…

)

. However,

u

is integral over

B

:

b

i

p

∈

B

⇒

u

p

∈

B

which implies that the degree of

u

over the

field of fractions

is

p

. Hence a basis for

K

⁢

(

u

)

/

K

is

{

1

,

u

,

u

2

,

…

,

u

p

-

1

}

. There are other elements integral over

B

:

b

1

+

b

2

⁢

t

+

b

3

⁢

t

2

+

⋯

=

u

-

b

0

t

=

-

b

0

t

+

1

t

⁢

u

b

2

+

b

3

⁢

t

+

b

4

⁢

t

2

+

⋯

=

u

-

b

0

-

b

1

⁢

t

t

2

=

-

b

0

-

b

1

⁢

t

t

2

+

1

t

2

⁢

u

⋮

Clearly the denominators are getting bigger, so the

integral closure

of

B

cannot be

finitely generated

as a

B

-module.

Title

separability is required for integral closures to be finitely generated

Canonical

name

SeparabilityIsRequiredForIntegralClosuresToBeFinitelyGenerated

Date of creation

2013-03-22 17:02:15

Last modified on

2013-03-22 17:02:15

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

6

Author

rm50 (10146)

Entry type

Example

Classification

msc 13B21

Classification

msc 12F05

Generated on Fri Feb 9 17:25:37 2018 by

LaTeXML
congruence in algebraic number field

congruence in algebraic number field

Definition.

Let

α

,

β

and

κ

be

integers

(

http://planetmath.org/AlgebraicInteger

) of an

algebraic number field

K

and

κ

≠

0

.  One defines

α

≡

β

(

mod

κ

)

(1)

if and only if

κ

∣

α

-

β

,  i.e. iff there is an integer

λ

of

K

with

α

-

β

=

λ

⁢

κ

.

Theorem

.

The

congruence

“

≡

” modulo

κ

defined above is an

equivalence relation

in the

maximal order

of

K

.  There are only a finite amount of the

equivalence classes

, the

residue classes

modulo

κ

.

Proof.

For justifying the

transitivity

of “

≡

”, suppose (1) and

β

≡

γ

(

mod

κ

)

; then there are the integers

λ

and

μ

of

K

such that

α

-

β

=

λ

⁢

κ

,

β

-

γ

=

μ

⁢

κ

.  Adding these equations we see that

α

-

γ

=

(

λ

+

μ

)

⁢

κ

with the integer

λ

+

μ

of

K

.  Accordingly,

α

≡

γ

(

mod

κ

)

.

Let

ω

be an arbitrary integer of

K

and

{

ω

1

,

ω

2

,

…

,

ω

n

}

a

minimal basis

of the field.  Then we can write

ω

=

a

1

⁢

ω

1

+

a

2

⁢

ω

2

+

…

+

a

n

⁢

ω

n

,

where the

a

i

’s are

rational integers

.  For

i

=

1

,

2

,

…

,

n

, the

division algorithm

determines the rational integers

q

i

and

r

i

with

a

i

=

N

⁢

(

κ

)

⁢

q

i

+

r

i

,

0

≦

r

i

<

|

N

⁢

(

κ

)

|

,

whence

ω

=

N

⁢

(

κ

)

⁢

(

q

1

⁢

ω

1

+

q

2

⁢

ω

2

+

…

+

q

n

⁢

ω

n

⏟

=

π

)

+

(

r

1

⁢

ω

1

+

r

2

⁢

ω

2

+

…

+

r

n

⁢

ω

n

⏟

=

ϱ

)

.

So we have

ω

=

N

⁢

(

κ

)

⁢

π

+

ϱ

,

(2)

where

π

and

ϱ

are some integers of the field.  If

κ

(

1

)

,

κ

(

2

)

,

…

,

κ

(

n

)

are the

algebraic conjugates

of

κ

=

κ

(

1

)

,  then

N

⁢

(

κ

)

=

κ

(

1

)

⏟

integer

⁢

κ

(

2

)

⁢

⋯

⁢

κ

(

n

)

⏟

integer

=

κ

⁢

κ

′

∈

ℤ

.

Hence,

κ

divides

N

⁢

(

κ

)

in the

ring of integers

of

K

, and (2) implies

ω

≡

ϱ

(

mod

κ

)

.

Since any number

r

i

has

|

N

⁢

(

κ

)

|

different possible values

0

,

1

,

…

,

|

N

⁢

(

κ

)

|

-

1

, there exist

|

N

⁢

(

κ

)

|

n

different

ordered tuplets

(

r

1

,

r

2

,

…

,

r

n

)

.  Therefore there exist at most

|

N

⁢

(

κ

)

|

n

different

residues

and residue classes in the ring.

Title

congruence in algebraic number field

Canonical name

CongruenceInAlgebraicNumberField

Date of creation

2013-03-22 18:17:11

Last modified on

2013-03-22 18:17:11

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Theorem

Classification

msc 13B22

Synonym

congruence in number field

Related topic

CongruenceRelationOnAnAlgebraicSystem

Related topic

ChineseRemainderTheoremInTermsOfDivisorTheory

Related topic

Congruences

Defines

residue class

Generated on Fri Feb 9 17:25:43 2018 by

LaTeXML
examples of ring of integers of a number field

examples of ring of integers of a number field

Definition 1

.

Let

K

be a

number field

. The

ring of integers

of

K

, usually denoted by

O

K

, is the set of all

elements

α

∈

K

which are roots of some

monic polynomial

with

coefficients

in

Z

, i.e. those

α

∈

K

which are integral over

Z

. In other words,

O

K

is the integral closure of

Z

in

K

.

Example 1

.

Notice that the only

rational numbers

which are roots of monic polynomials with

integer

coefficients are the integers themselves. Thus, the ring of integers of

ℚ

is

ℤ

.

Example 2

.

Let

𝒪

K

denote the ring of integers of

K

=

ℚ

⁢

(

d

)

, where

d

is a square-free integer. Then:

𝒪

K

≅

{

ℤ

⊕

1

+

d

2

⁢

ℤ

,

if

⁢

d

≡

1

⁢

mod

⁡

4

,

ℤ

⊕

d

⁢

ℤ

,

if

⁢

d

≡

2

,

3

⁢

mod

⁡

4

.

In other words, if we let

α

=

{

1

+

d

2

,

if

⁢

d

≡

1

⁢

mod

⁡

4

,

d

,

if

⁢

d

≡

2

,

3

⁢

mod

⁡

4

.

then

𝒪

K

=

{

n

+

m

⁢

α

:

n

,

m

∈

ℤ

}

.

Example 3

.

Let

K

=

ℚ

⁢

(

ζ

n

)

be a

cyclotomic extension

of

ℚ

, where

ζ

n

is a primitive

n

th

root of unity

. Then the ring of integers of

K

is

𝒪

K

=

ℤ

⁢

[

ζ

n

]

, i.e.

𝒪

K

=

{

a

0

+

a

1

⁢

ζ

n

+

a

2

⁢

ζ

n

2

+

…

+

a

n

-

1

⁢

ζ

n

n

-

1

:

a

i

∈

ℤ

}

.

Example 4

.

Let

α

be an

algebraic integer

and let

K

=

ℚ

⁢

(

α

)

. It is

not true in general

that

𝒪

K

=

ℤ

⁢

[

α

]

(as we saw in Example

2

, for

d

≡

1

mod

4

).

Example 5

.

Let

p

be a

prime number

and let

F

=

ℚ

⁢

(

ζ

p

)

be a cyclotomic extension of

ℚ

, where

ζ

p

is a primitive

p

th root of unity. Let

F

+

be the maximal real

subfield

of

F

. It can be shown that:

F

+

=

ℚ

⁢

(

ζ

p

+

ζ

p

-

1

)

.

Moreover, it can also be shown that the ring of integers of

F

+

is

𝒪

F

+

=

ℤ

⁢

[

ζ

p

+

ζ

p

-

1

]

.

Title

examples of ring

of integers of a number field

Canonical name

ExamplesOfRingOfIntegersOfANumberField

Date of creation

2013-03-22 15:08:09

Last modified on

2013-03-22 15:08:09

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

7

Author

alozano (2414)

Entry type

Example

Classification

msc 13B22

Related topic

NumberField

Related topic

AlgebraicNumberTheory

Related topic

CanonicalBasis

Related topic

IntegralBasisOfQuadraticField

Generated on Fri Feb 9 17:25:55 2018 by

LaTeXML
integral closure

integral closure

Let

B

be a ring with a

subring

A

. The

integral closure

of

A

in

B

is the set

A

′

⊂

B

consisting of all elements of

B

which are integral over

A

.

It is a theorem that the integral closure of

A

in

B

is itself a ring. In the special case where

A

=

ℤ

, the integral closure

A

′

of

ℤ

is often called the

ring of integers

in

B

.

Title

integral closure

Canonical name

IntegralClosure

Date of creation

2013-03-22 12:07:53

Last modified on

2013-03-22 12:07:53

Owner

djao (24)

Last modified by

djao (24)

Numerical id

8

Author

djao (24)

Entry type

Definition

Classification

msc 13B22

Related topic

IntegrallyClosed

Defines

ring of integers

Generated on Fri Feb 9 17:25:58 2018 by

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integral closure is ring

integral closure is ring

Theorem.

Let

A

be a

subring

of a

commutative ring

B

having

nonzero unity

.  Then the

integral closure

of

A

in

B

is a subring of

B

containing

A

.

Proof.

Let

x

be an arbitrary element of the integral closure

A

′

of

A

in

B

.  Then there are the elements

a

0

,

a

1

,

…

,

a

n

-

1

of

A

such that

a

0

+

a

1

⁢

x

+

…

+

a

n

-

1

⁢

x

n

-

1

+

x

n

=

0

where

n

>

0

.  If

f

⁢

(

X

)

=

c

0

+

c

1

⁢

X

+

…

+

c

m

⁢

X

m

is a

polynomial

in

A

⁢

[

X

]

with degree

m

>

n

,  we have

f

⁢

(

x

)

=

c

0

+

c

1

⁢

x

+

…

+

c

m

-

1

⁢

x

m

-

1

+

c

m

⁢

x

m

-

n

⁢

(

-

a

0

-

a

1

⁢

x

-

…

-

a

n

-

1

⁢

x

n

-

1

)

=

c

0

′

+

c

1

′

⁢

x

+

…

+

c

m

-

1

′

⁢

x

m

-

1

where  the elements

c

i

′

belong to

A

.  This procedure may be repeated until we see that

f

⁢

(

x

)

is an element of the

A

-module generated by

1

,

x

,

…

,

x

n

.  Accordingly,

A

⁢

[

x

]

=

A

+

A

⁢

x

+

…

+

A

⁢

x

n

is a

finitely generated

A

-module.

Now we have evidently

A

⊆

A

′

.  Let

y

be another element of

A

′

.  Then

A

⁢

[

x

,

y

]

=

A

⁢

[

x

]

⁢

[

y

]

is a finitely generated

A

⁢

[

x

]

-module, whence

A

⁢

[

x

,

y

]

is a finitely generated

A

-module.  Because the elements

x

-

y

and

x

⁢

y

belong to

A

⁢

[

x

,

y

]

,  they are integral over

A

and thus belong to

A

′

.  Consequently,

A

′

is a subring of

B

(see the

http://planetmath.org/node/2738

subring condition).

References

1

M. Larsen & P. McCarthy

:

Multiplicative theory of ideals

.  Academic Press, New York (1971).

Title

integral closure is ring

Canonical name

IntegralClosureIsRing

Date of creation

2013-03-22 19:15:40

Last modified on

2013-03-22 19:15:40

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Theorem

Classification

msc 13B22

Related topic

PolynomialRing

Related topic

RingAdjunction

Related topic

IntegralClosuresInSeparableExtensionsAreFinitelyGenerated

Generated on Fri Feb 9 17:26:03 2018 by

LaTeXML
integrally closed

integrally closed

A

subring

R

of a

commutative ring

S

is said to be

integrally closed

in

S

if whenever

θ

∈

S

and

θ

is integral over

R

, then

θ

∈

R

.

The

integral closure

of

R

in

S

is integrally closed in

S

.

An

integral domain

R

is said to be

integrally closed

(or ) if it is integrally closed in its

fraction field

.

Title

integrally closed

Canonical name

IntegrallyClosed

Date of creation

2013-03-22 12:36:34

Last modified on

2013-03-22 12:36:34

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

15

Author

rmilson (146)

Entry type

Definition

Classification

msc 13B22

Classification

msc 11R04

Synonym

normal ring

Related topic

IntegralClosure

Related topic

AlgebraicClosure

Related topic

AlgebraicallyClosed

Generated on Fri Feb 9 17:26:06 2018 by

LaTeXML
proof of the ring of integers of a number field is finitely generated over ℤ

proof of the ring of integers of a number field is finitely generated over

ℤ

Proof:

Choose any basis

α

1

,

…

,

α

n

of

K

over

ℚ

. Using the theorem in the entry

multiples

of an

algebraic number

, we can multiply each element of the basis by an

integer

to get a new basis

α

1

,

…

,

α

n

with each

α

i

∈

𝒪

K

.

Consider the

group homomorphism

φ

:

K

→

ℚ

n

:

γ

↦

(

Tr

ℚ

K

⁡

(

γ

⁢

α

1

)

,

…

,

Tr

ℚ

K

⁡

(

γ

⁢

α

n

)

)

where

Tr

ℚ

K

is the trace (

http://planetmath.org/trace2

) from

K

to

ℚ

. Note that

φ

is

1

-

1

, since if

γ

≠

0

and

φ

⁢

(

γ

)

=

0

, then

n

=

Tr

ℚ

K

⁡

(

1

)

=

Tr

ℚ

K

⁡

(

γ

⁢

γ

-

1

)

=

Tr

ℚ

K

⁡

(

γ

⁢

∑

r

i

⁢

α

i

)

=

∑

r

i

⁢

Tr

ℚ

K

⁡

(

γ

⁢

α

i

)

=

0

where the last equality holds since

γ

∈

ker

⁡

φ

.

Hence

φ

:

𝒪

K

↪

ℤ

n

, so

𝒪

K

is

finitely generated

and

torsion-free

. It has rank

≥

n

since the

α

i

are

linearly independent

, and rank

≤

n

since it injects into

ℤ

n

, so it has rank

n

.

Title

proof of the

ring of integers

of a

number field

is finitely generated over

ℤ

Canonical name

ProofOfTheRingOfIntegersOfANumberFieldIsFinitelyGeneratedOvermathbbZ

Date of creation

2013-03-22 16:03:07

Last modified on

2013-03-22 16:03:07

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

9

Author

rm50 (10146)

Entry type

Proof

Classification

msc 13B22

Generated on Fri Feb 9 17:26:10 2018 by

LaTeXML
ring of S-integers

ring of

S

-integers

Definition

.

Let

K

be a

number field

and let

S

be a finite set of

absolute values

of

K

, containing all

archimedean valuations

. The ring of

S

-

integers

of

K

, usually denoted by

R

S

, is the ring:

R

S

=

{

k

∈

K

:

ν

⁢

(

k

)

≥

0

⁢

for all valuations

⁢

ν

∉

S

}

.

Notice that, for any set

S

as above, the

ring of integers

of

K

,

𝒪

K

, is always contained in

R

S

.

Example

.

Let

K

=

ℚ

and let

S

=

{

ν

p

,

|

⋅

|

}

where

p

is a prime and

ν

p

is the usual

p

-adic

valuation

, and

|

⋅

|

is the usual absolute value. Then

R

S

=

ℤ

⁢

[

1

p

]

, i.e.

R

S

is the result of adjoining (as a new ring element)

1

/

p

to

ℤ

(i.e. we allow to invert

p

).

Title

ring of

S

-integers

Canonical name

RingOfSintegers

Date of creation

2013-03-22 15:57:27

Last modified on

2013-03-22 15:57:27

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

4

Author

alozano (2414)

Entry type

Definition

Classification

msc 13B22

Synonym

ring of S-integers

Generated on Fri Feb 9 17:26:14 2018 by

LaTeXML
the ring of integers of a number field is finitely generated over ℤ

the ring of integers of a number field is finitely generated over

ℤ

Theorem

.

Let

K

be a

number field

of degree

n

over

Q

and let

O

K

be the

ring of integers

of

K

. The ring

O

K

is a

free abelian group

of rank

n

. In other words, there exists a finite

integral basis

(with

n

elements) for

K

, i.e. there exist

algebraic integers

α

1

,

…

,

α

n

such that every element of

O

K

can be expressed uniquely as a

Z

-

linear combination

of the

α

i

.

Corollary

.

Every ideal of

O

K

is

finitely generated

.

Proof of the corollary.

By the theorem,

𝒪

K

is a free abelian group of rank

n

, and therefore it is finitely generated. Notice that an ideal is an

additive

subgroup

. Finally a subgroup of a finitely generated free abelian group is also finitely generated.
∎

This is the first step to prove that

𝒪

K

is a

Dedekind domain

. Notice that the

field of fractions

of

𝒪

K

is the field

K

itself. Therefore, by definition,

𝒪

K

is

integrally closed

in

K

.

Title

the ring of integers of a number field is finitely generated over

ℤ

Canonical name

TheRingOfIntegersOfANumberFieldIsFinitelyGeneratedOvermathbbZ

Date of creation

2013-03-22 15:08:22

Last modified on

2013-03-22 15:08:22

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

7

Author

alozano (2414)

Entry type

Theorem

Classification

msc 13B22

Generated on Fri Feb 9 17:26:18 2018 by

LaTeXML
total integral closure

total integral closure

A

commutative

unitary ring

R

is said to be

totally

integrally closed

if it does not have an

overring

which is both an

integral

and an

essential extension

of

R

.

All totally integrally closed rings are reduced (

http://planetmath.org/ReducedRing

).

Suppose that

R

is any

commutative ring

and that

R

¯

is an integral and essential extension of

R

. If

R

¯

is a totally integrally closed ring, then

R

¯

is called a

total integral closure

of

R

.

For fields the concept totally integrally closed, integrally closed and

algebraically closed

coincide.

Let

A

be an

integral domain

, then its total integral closure is the

integral closure

of

A

in the

algebraic closure

of

Quot

⁢

(

A

)

.

Enochs has first proven that all commutative

reduced rings

have total integral closure and this is unique up to

ring isomorphism

.

Title

total integral closure

Canonical name

TotalIntegralClosure

Date of creation

2013-03-22 18:51:44

Last modified on

2013-03-22 18:51:44

Owner

jocaps (12118)

Last modified by

jocaps (12118)

Numerical id

14

Author

jocaps (12118)

Entry type

Definition

Classification

msc 13B22

Defines

total integral closure

Generated on Fri Feb 9 17:26:22 2018 by

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