aarjaneiro/planetmath huggingface dataset:
ordinary quiver of an algebra

ordinary quiver of an algebra

Let

k

be a field and

A

an algebra over

k

.

Denote by

rad

⁢

A

the (Jacobson)

radical

of

A

and

rad

2

⁢

A

=

(

rad

⁢

A

)

2

a square of radical.

Since

A

is

finite-dimensional

, then we have a

complete set of primitive orthogonal idempotents

(

http://planetmath.org/CompleteSetOfPrimitiveOrthogonalIdempotents

)

E

=

{

e

1

,

…

,

e

n

}

.

Definition.

The

ordinary quiver

of a finite-dimensional

algebra

A

is defined as follows:

1.

The set of vertices is equal to

Q

0

=

{

1

,

…

,

n

}

which is in bijective correspondence with

E

.

2.

If

a

,

b

∈

Q

0

, then the number of arrows from

a

to

b

is equal to the dimension of the

k

-

vector space

e

a

⁢

(

rad

⁢

A

/

rad

2

⁢

A

)

⁢

e

b

.

It can be shown that the ordinary quiver is well-defined, i.e. it is

independent

on the choice of a

complete set

of primitve

orthogonal idempotents

. Also finite dimension of

A

implies, then the ordinary quiver is finite.

Title

ordinary quiver of an algebra

Canonical name

OrdinaryQuiverOfAnAlgebra

Date of creation

2013-03-22 19:17:41

Last modified on

2013-03-22 19:17:41

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

4

Author

joking (16130)

Entry type

Definition

Classification

msc 16S99

Classification

msc 20C99

Classification

msc 13B99

Generated on Fri Feb 9 18:43:14 2018 by

LaTeXML
Peirce decomposition

Peirce decomposition

Let

e

be an

idempotent

of a ring

R

, not necessarily with an

identity

.
For any subset

X

of

R

, we introduce the notations:

(

1

-

e

)

⁢

X

=

{

x

-

e

⁢

x

∣

x

∈

X

}

and

X

⁢

(

1

-

e

)

=

{

x

-

x

⁢

e

∣

x

∈

X

}

.

If it happens that

R

has an

identity element

, then

1

-

e

is a legitimate element of

R

, and this notation agrees with the usual product of an element and a set.

It is easy to see that

X

⁢

e

∩

X

⁢

(

1

-

e

)

=

0

=

e

⁢

X

∩

(

1

-

e

)

⁢

X

for any set

X

which contains

0

.

Applying this first on the right with

X

=

R

and then on the left with

X

=

R

⁢

e

and

X

=

R

⁢

(

1

-

e

)

,
we obtain:

R

=

e

⁢

R

⁢

e

⊕

e

⁢

R

⁢

(

1

-

e

)

⊕

(

1

-

e

)

⁢

R

⁢

e

⊕

(

1

-

e

)

⁢

R

⁢

(

1

-

e

)

.

This is called the

Peirce Decompostion

of

R

with respect to

e

.

Note that

e

⁢

R

⁢

e

and

(

1

-

e

)

⁢

R

⁢

(

1

-

e

)

are

subrings

,

e

⁢

R

⁢

(

1

-

e

)

is an

e

⁢

R

⁢

e

-

(

1

-

e

)

⁢

R

⁢

(

1

-

e

)

-

bimodule

,
and

(

1

-

e

)

⁢

R

⁢

e

is a

(

1

-

e

)

⁢

R

⁢

(

1

-

e

)

-

e

⁢

R

⁢

e

-bimodule.

This is an example of a

generalized matrix ring

:

R

≅

(

e

⁢

R

⁢

e

e

⁢

R

⁢

(

1

-

e

)

(

1

-

e

)

⁢

R

⁢

e

(

1

-

e

)

⁢

R

⁢

(

1

-

e

)

)

More generally, if

R

has an identity element,
and

e

1

,

e

2

,

…

,

e

n

is a complete set of orthogonal idempotents,
then

R

≅

(

e

1

⁢

R

⁢

e

1

e

1

⁢

R

⁢

e

2

…

e

1

⁢

R

⁢

e

n

e

2

⁢

R

⁢

e

1

e

2

⁢

R

⁢

e

2

…

e

2

⁢

R

⁢

e

n

⋮

⋮

⋱

⋮

e

n

⁢

R

⁢

e

1

e

n

⁢

R

⁢

e

2

…

e

n

⁢

R

⁢

e

n

)

is a generalized matrix ring.

Title

Peirce decomposition

Canonical name

PeirceDecomposition

Date of creation

2013-03-22 14:39:17

Last modified on

2013-03-22 14:39:17

Owner

mclase (549)

Last modified by

mclase (549)

Numerical id

7

Author

mclase (549)

Entry type

Definition

Classification

msc 16S99

Generated on Fri Feb 9 18:43:17 2018 by

LaTeXML
properties of the ordinary quiver

properties of the ordinary quiver

Let

k

be a field and

A

be a

finite-dimensional

algebra over

k

. Denote by

Q

A

the ordinary quiver (

http://planetmath.org/OrdinaryQuiverOfAnAlgebra

) of

A

.

Theorem.

The following statements hold:

1.

If

A

is basic and

connected

, then

Q

A

is a connected quiver.

2.

If

Q

is a finite quiver and

I

is an

admissible

ideal (

http://planetmath.org/AdmissibleIdealsBoundQuiverAndItsAlgebra

) in

k

⁢

Q

and

A

=

k

⁢

Q

/

I

, then

Q

A

and

Q

are

isomorphic

.

3.

If

A

is basic and connected, then

A

is isomorphic to

k

⁢

Q

A

/

I

for some (not necessarily unique) admissible ideal (

http://planetmath.org/AdmissibleIdealsBoundQuiverAndItsAlgebra

)

I

.

For proofs please see

[

1

, Chapter II.3]

.

References

1

I. Assem, D. Simson, A. Skowronski,

Elements of the

Representation Theory

of

Associative Algebras

, vol 1.

, Cambridge University Press 2006, 2007

Title

properties of the ordinary quiver

Canonical name

PropertiesOfTheOrdinaryQuiver

Date of creation

2013-03-22 19:17:44

Last modified on

2013-03-22 19:17:44

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

4

Author

joking (16130)

Entry type

Theorem

Classification

msc 16S99

Classification

msc 20C99

Classification

msc 13B99

Generated on Fri Feb 9 18:43:18 2018 by

LaTeXML
the characteristic embedding of the Burnside ring

the characteristic embedding of the Burnside ring

Let

G

be a

finite group

,

H

its

subgroup

and

X

a finite

G

-set. By the

H

-

fixed point

subset

of

X

we understand the set

X

H

=

{

x

∈

X

;

∀

h

∈

H

h

x

=

x

}

.

Denote by

|

X

|

the

cardinality

of a set

X

.

It is easy to see that for any

G

-sets

X

,

Y

we have:

|

(

X

⊔

Y

)

H

|

=

|

X

H

|

+

|

Y

H

|

;

|

(

X

×

Y

)

H

|

=

|

X

H

|

⋅

|

Y

H

|

.

Denote by

Sub

(

G

)

=

{

H

⊆

G

;

H

is

a

subgroup

of

G

}

. Recall that any

H

,

K

∈

Sub

⁢

(

G

)

are said to be

conjugate

iff there exists

g

∈

G

such that

H

=

g

⁢

K

⁢

g

-

1

.

Conjugation

is an

equivalence relation

. Denote by

Con

⁢

(

G

)

the quotient set.

One can check that for any

H

,

K

∈

Sub

⁢

(

G

)

such that

H

is conjugate to

K

and for any finite

G

-set

X

we have

|

X

H

|

=

|

X

K

|

.

Thus we have a well defined

ring homomorphism

:

φ

:

Ω

⁢

(

G

)

→

⊕

(

H

)

∈

Con

⁢

(

G

)

ℤ

;

φ

⁢

(

[

X

]

-

[

Y

]

)

=

(

|

X

H

|

-

|

Y

H

|

)

(

H

)

∈

Con

⁢

(

G

)

.

This

homomorphism

is known as the

characteristic

embedding

, since it is

monomorphism

(see

[

1

]

for proof).

References

1

T. tom Dieck,

Transformation groups

and

representation theory

, Lecture Notes in Math. 766, Springer-Verlag, Berlin, 1979.

Title

the characteristic embedding of the

Burnside ring

Canonical name

TheCharacteristicEmbeddingOfTheBurnsideRing

Date of creation

2013-03-22 18:08:09

Last modified on

2013-03-22 18:08:09

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

7

Author

joking (16130)

Entry type

Derivation

Classification

msc 16S99

Generated on Fri Feb 9 18:43:23 2018 by

LaTeXML
(partial) tilting module

(partial) tilting module

Let

A

be an

associative

,

finite-dimensional

algebra over a field

k

. Throughout all modules are finite-dimensional.

A right

A

-module

T

is called a

partial tilting module

if the

projective dimension

of

T

is at most

1

(

pd

⁢

T

⩽

1

) and

Ext

A

1

⁢

(

T

,

T

)

=

0

.

Recall that if

M

is an

A

-module, then by

add

⁢

M

we denote the class of all

A

-modules which are

direct sums

of

direct summands

of

M

. Since

Krull-Schmidt Theorem

holds in the

category

of finite-dimensional

A

-modules, then this means, that if

M

=

E

1

⊕

⋯

⊕

E

n

for some

indecomposable modules

E

i

, then

add

⁢

M

consists of all modules which are

isomorphic

to

E

1

a

1

⊕

⋯

⊕

E

n

a

n

for some nonnegative integers

a

1

,

…

,

a

n

.

A partial tiliting module

T

is called a

tilting module

if there exists a

short exact sequence

0

→

A

→

T

′

→

T

′′

→

0

such that both

T

′

,

T

′′

∈

add

⁢

T

. Here we treat the

algebra

A

is a

right module

via multiplication.

Note that every

projective module

is partial tilting. Also a projective module

P

is tilting if and only if every

indecomposable

direct summand of

A

is a direct summand of

P

.

Title

(partial) tilting module

Canonical name

partialTiltingModule

Date of creation

2013-12-11 16:01:05

Last modified on

2013-12-11 16:01:05

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

8

Author

joking (16130)

Entry type

Definition

Classification

msc 16S99

Classification

msc 20C99

Classification

msc 13B99

Generated on Fri Feb 9 18:43:25 2018 by

LaTeXML
Prüfer domain

Prüfer domain

A

commutative

integral domain

R

is a

Prüfer domain

if every

finitely generated

nonzero ideal

I

of

R

is

invertible

.

Let

R

I

denote the

localization

of

R

at

R

\

I

. Then the following statements are equivalent:

•

i)

R

is a Prüfer domain.

•

ii) For every

prime ideal

P

in

R

,

R

P

is a

valuation domain

.

•

iii) For every

maximal ideal

M

in

R

,

R

M

is a valuation domain.

A Prüfer domain is a Dedekind domain if and only if it is

Noetherian

.

If

R

is a Prüfer domain with

quotient field

K

, then any domain

S

such that

R

⊂

S

⊂

K

is Prüfer.

References

1

Thomas W. Hungerford.

Algebra

. Springer-Verlag, 1974. New York, NY.

Title

Prüfer domain

Canonical name

PruferDomain

Date of creation

2013-03-22 13:47:34

Last modified on

2013-03-22 13:47:34

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

8

Author

mathcam (2727)

Entry type

Definition

Classification

msc 16U10

Related topic

ValuationDomain

Related topic

DedekindDomain

Related topic

PruferRing

Related topic

InvertibleIdealsInSemiLocalRings

Generated on Fri Feb 9 18:43:27 2018 by

LaTeXML
valuation domain

valuation domain

An

integral domain

R

is a

valuation domain

if for all

a

,

b

∈

R

, either

a

|

b

or

b

|

a

. Equivalently, an integral domain is a valuation domain if for any

x

in the

field of fractions

of

R

,

x

∉

R

⟹

x

-

1

∈

R

.

Some properties:

•

A valuation domain is a

discrete valuation ring

(DVR) if and only if it is a

principal ideal domain

(PID) if and only if it is

Noetherian

.

•

Every valuation domain is a

Bezout domain

, though the converse is not true. For a partial converse, any local Bezout domain is a valuation domain.

•

Valuation domains are

integrally closed

.

Title

valuation domain

Canonical name

ValuationDomain

Date of creation

2013-03-22 13:47:31

Last modified on

2013-03-22 13:47:31

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

7

Author

mathcam (2727)

Entry type

Definition

Classification

msc 16U10

Classification

msc 13G05

Classification

msc 13F30

Related topic

PruferDomain

Generated on Fri Feb 9 18:43:29 2018 by

LaTeXML
classical ring of quotients

classical ring of quotients

Let

R

be a ring. An element of

R

is called

regular

if it is not a

right zero divisor

or a left zero divisor in

R

.

A ring

Q

⊃

R

is a

left classical ring of quotients

for

R

(resp.

right classical ring of quotients

for

R

) if it satisifies:

•

every regular element of

R

is

invertible

in

Q

•

every element of

Q

can be written in the form

x

-

1

⁢

y

(resp.

y

⁢

x

-

1

) with

x

,

y

∈

R

and

x

regular.

If a ring

R

has a left or right classical ring of quotients, then it is unique up to

isomorphism

.

If

R

is a

commutative

integral domain

, then the left and right classical rings of quotients always exist – they are the

field of fractions

of

R

.

For non-commutative rings, necessary and

sufficient

conditions are given by Ore’s

Theorem

.

Note that the goal here is to construct a ring which is not too different from

R

, but in which more

elements

are invertible. The first condition says which elements we want to be invertible. The second condition says that

Q

should just enough extra elements to make the regular elements invertible.

Such rings are called classical rings of quotients, because there are other

rings of quotients

. These all attempt to enlarge

R

somehow to make more elements invertible (or sometimes to make ideals invertible).

Finally, note that a ring of quotients is not the same as a

quotient ring

.

Title

classical ring of quotients

Canonical name

ClassicalRingOfQuotients

Date of creation

2013-03-22 14:03:01

Last modified on

2013-03-22 14:03:01

Owner

mclase (549)

Last modified by

mclase (549)

Numerical id

5

Author

mclase (549)

Entry type

Definition

Classification

msc 16U20

Classification

msc 16S90

Synonym

left classical ring of quotients

Synonym

right classical ring of quotients

Related topic

OreCondition

Related topic

ExtensionByLocalization

Related topic

FiniteRingHasNoProperOverrings

Defines

regular

Generated on Fri Feb 9 18:43:31 2018 by

LaTeXML
Goldie’s theorem

Goldie’s theorem

Let

R

be a ring with an

identity

. Then

R

has a

right classical ring of quotients

Q

which is

semisimple

Artinian

if and only if

R

is a

semiprime

right Goldie ring

. If this is the case, then the

composition

length of

Q

is equal to the

uniform dimension

of

R

.

An immediate corollary of this is that a semiprime

right Noetherian ring

always has a right classical ring of quotients.

This result was discovered by Alfred in the late 1950’s.

Title

Goldie’s

theorem

Canonical name

GoldiesTheorem

Date of creation

2013-03-22 14:04:17

Last modified on

2013-03-22 14:04:17

Owner

mclase (549)

Last modified by

mclase (549)

Numerical id

7

Author

mclase (549)

Entry type

Theorem

Classification

msc 16U20

Classification

msc 16P60

Related topic

OresTheorem2

Generated on Fri Feb 9 18:43:32 2018 by

LaTeXML
m-system

m

-system

Let

R

be a ring. A subset

S

of

R

is called an

m

-

system

if

•

S

≠

∅

, and

•

for every two

elements

x

,

y

∈

S

, there is an element

r

∈

R

such that

x

⁢

r

⁢

y

∈

S

.

m

-Systems are a

generalization

of multiplicatively closet subsets in a ring. Indeed, every

multiplicatively closed

subset of

R

is an

m

-system: any

x

,

y

∈

S

, then

x

⁢

y

∈

S

, hence

x

⁢

y

⁢

y

∈

S

. However, the

converse

is not true. For example, the set

{

r

n

∣

r

∈

R

⁢

and

⁢

n

⁢

is an odd positive integer

}

is an

m

-system, but not multiplicatively closed in general (unless, for example, if

r

=

1

).

Remarks

.

m

-Systems and

prime ideals

of a ring are intimately related. Two basic relationships between the two notions are

1.

An ideal

P

in a ring

R

is a prime ideal iff

R

-

P

is an

m

-system.

Proof.

P

is prime iff

x

⁢

R

⁢

y

⊆

P

implies

x

or

y

∈

P

, iff

x

,

y

∈

R

-

P

implies that there is

r

∈

R

with

x

⁢

r

⁢

y

∉

P

iff

R

-

P

is an

m

-system.
∎

2.

Given an

m

-system

S

of

R

and an ideal

I

with

I

∩

S

=

∅

. Then there exists a prime ideal

P

⊆

R

with the

property

that

P

contains

I

and

P

∩

S

=

∅

, and

P

is the largest among all ideals with this property.

Proof.

Let

𝒞

be the

collection

of all ideals containing

I

and

disjoint

from

S

. First,

I

∈

𝒞

. Second, any chain

K

of ideals in

𝒞

, its union

⋃

K

is also in

𝒞

. So Zorn’s lemma applies. Let

P

be a

maximal element

in

𝒞

. We want to show that

P

is prime. Suppose otherwise. In other words,

a

⁢

R

⁢

b

⊆

P

with

a

,

b

∉

P

. Then

⟨

P

,

a

⟩

and

⟨

P

,

b

⟩

both have non-empty

intersections

with

S

. Let

c

=

p

+

f

⁢

a

⁢

g

∈

⟨

P

,

a

⟩

∩

S

and

d

=

q

+

h

⁢

b

⁢

k

∈

⟨

P

,

b

⟩

∩

S

,

where

p

,

q

∈

P

and

f

,

g

,

h

,

k

∈

R

. Then there is

r

∈

R

such that

c

⁢

r

⁢

d

∈

S

. But this implies that

c

⁢

r

⁢

d

=

(

p

+

f

⁢

a

⁢

g

)

⁢

r

⁢

(

q

+

h

⁢

b

⁢

k

)

=

p

⁢

(

r

⁢

q

+

r

⁢

h

⁢

b

⁢

k

)

+

(

f

⁢

a

⁢

g

⁢

r

)

⁢

q

+

f

⁢

(

a

⁢

(

g

⁢

r

⁢

h

)

⁢

b

)

⁢

k

∈

P

as well, contradicting

P

∩

S

=

∅

. Therefore,

P

is prime.
∎

m

-Systems are also used to define the non-commutative version of the

radical

of an ideal of a ring.

Title

m

-system

Canonical

name

Msystem

Date of creation

2013-03-22 17:29:09

Last modified on

2013-03-22 17:29:09

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

11

Author

CWoo (3771)

Entry type

Definition

Classification

msc 16U20

Classification

msc 13B30

Synonym

m-system

Related topic

MultiplicativelyClosed

Related topic

NSystem

Related topic

PrimeIdeal

Generated on Fri Feb 9 18:43:42 2018 by

LaTeXML
multiplicative sets in rings and prime ideals

multiplicative sets in rings and prime ideals

Proposition

.

Let

R

be a

commutative ring

,

S

⊆

R

a mutliplicative subset of

R

such that

0

∉

S

. Then there exists

prime ideal

P

⊆

R

such that

P

∩

S

=

∅

.

Proof.

Consider the family

𝒜

=

{

I

⊆

R

|

I

⁢

is an ideal and

⁢

I

∩

S

=

∅

}

. Of course

𝒜

≠

∅

, because the

zero ideal

0

∈

𝒜

. We will show, that

𝒜

is inductive (i.e.

satisfies

Zorn’s Lemma’s

assumptions

) with respect to

inclusion

.

Let

{

I

k

}

k

∈

K

be a chain in

𝒜

(i.e. for any

a

,

b

∈

K

either

I

a

⊆

I

b

or

I

b

⊆

I

a

). Consider

I

=

⋃

k

∈

K

I

k

. Obviously

I

is an ideal. Furthermore, if

x

∈

I

∩

S

, then there is

k

∈

K

such that

x

∈

I

k

∩

S

=

∅

. Thus

I

∩

S

=

∅

, so

I

∈

𝒜

. Lastely each

I

k

⊆

I

, which

completes

this part of proof.

By Zorn’s Lemma there is a

maximal element

P

∈

𝒜

. We will show that this ideal is prime. Let

x

,

y

∈

R

be such that

x

⁢

y

∈

P

. Assume that neither

x

∉

P

nor

y

∉

P

. Then

P

⊂

P

+

(

x

)

and

P

⊂

P

+

(

y

)

and these inclusions are proper. Therefore both

P

+

(

x

)

and

P

+

(

y

)

do not belong to

𝒜

(because

P

is maximal). This implies that there exist

a

∈

(

P

+

(

x

)

)

∩

S

and

b

∈

(

P

+

(

y

)

)

∩

S

. Thus

a

=

m

1

+

r

1

⁢

x

∈

S

;

b

=

m

2

+

r

2

⁢

y

∈

S

;

where

m

1

,

m

2

∈

P

and

r

1

,

r

2

∈

R

. Note that

a

⁢

b

∈

S

. We calculate

a

⁢

b

=

(

m

1

+

r

1

⁢

x

)

⁢

(

m

2

+

r

2

⁢

y

)

=

m

1

⁢

m

2

+

m

2

⁢

r

1

⁢

x

+

m

1

⁢

r

2

⁢

y

+

x

⁢

y

⁢

r

1

⁢

r

2

.

Of course

m

1

⁢

m

2

,

m

2

⁢

r

1

⁢

x

,

m

1

⁢

r

2

⁢

y

∈

P

, because

m

1

,

m

2

∈

P

and

x

⁢

y

⁢

r

1

⁢

r

2

∈

P

by our assumption that

x

⁢

y

∈

P

. This shows, that

a

⁢

b

∈

P

. But

a

⁢

b

∈

S

and

P

∈

𝒜

.

Contradiction

.

□

Corollary.

Let

R

be a commutative ring,

I

an ideal in

R

and

S

⊆

R

a

multiplicative subset

such that

I

∩

S

=

∅

. Then there exists prime ideal

P

in

R

such that

I

⊆

P

and

P

∩

S

=

∅

.

Proof.

Let

π

:

R

→

R

/

I

be the

projection

. Then

π

⁢

(

S

)

⊆

R

/

I

is a multiplicative subset in

R

/

I

such that

0

+

I

∉

π

⁢

(

S

)

(because

I

∩

S

=

∅

). Thus, by proposition, there exists a prime ideal

P

in

R

/

I

such that

P

∩

π

⁢

(

S

)

=

∅

. Of course the

preimage

of a prime ideal is again a prime ideal. Furthermore

I

⊆

π

-

1

⁢

(

P

)

. Finaly

π

-

1

⁢

(

P

)

∩

S

=

∅

, because

P

∩

π

⁢

(

S

)

=

∅

. This completes the proof.

Title

multiplicative sets in rings and prime ideals

Canonical name

MultiplicativeSetsInRingsAndPrimeIdeals

Date of creation

2013-03-22 19:03:58

Last modified on

2013-03-22 19:03:58

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

5

Author

joking (16130)

Entry type

Theorem

Classification

msc 16U20

Classification

msc 13B30

Generated on Fri Feb 9 18:43:44 2018 by

LaTeXML
Ore condition

Ore condition

A ring

R

satisfies the

left Ore condition

(resp.

right Ore condition

) if and only if for all elements

x

and

y

with

x

regular

, there exist elements

u

and

v

with

v

regular such that

u

⁢

x

=

v

⁢

y

(resp.

⁢

x

⁢

u

=

y

⁢

v

⁢

).

A ring which satisfies the (left, right) Ore condition is called a (

left

,

right

)

Ore ring

.

Title

Ore condition

Canonical name

OreCondition

Date of creation

2013-03-22 14:03:04

Last modified on

2013-03-22 14:03:04

Owner

mclase (549)

Last modified by

mclase (549)

Numerical id

6

Author

mclase (549)

Entry type

Definition

Classification

msc 16U20

Related topic

ClassicalRingOfQuotients

Related topic

OresTheorem2

Defines

Ore ring

Defines

left Ore condition

Defines

right Ore condition

Defines

left Ore ring

Defines

right Ore ring

Generated on Fri Feb 9 18:43:46 2018 by

LaTeXML
Ore’s theorem

Ore’s theorem

A ring has a (left, right)

classical ring of quotients

if and only if it satisfies the (left, right)

Ore condition

.

Title

Ore’s theorem

Canonical name

OresTheorem

Date of creation

2013-03-22 14:03:07

Last modified on

2013-03-22 14:03:07

Owner

mclase (549)

Last modified by

mclase (549)

Numerical id

7

Author

mclase (549)

Entry type

Theorem

Classification

msc 16U20

Related topic

OreCondition

Related topic

GoldiesTheorem

Generated on Fri Feb 9 18:43:47 2018 by

LaTeXML
saturated

saturated

Let

S

be

multiplicative subset

of

A

. We say that

S

is a

saturated

if

a

⁢

b

∈

S

⇒

a

,

b

∈

S

.

When

A

is an

integral domain

, then

S

is saturated if and only if its

complement

A

\

S

is union of

prime ideals

.

Title

saturated

Canonical name

Saturated

Date of creation

2013-03-22 12:30:17

Last modified on

2013-03-22 12:30:17

Owner

drini (3)

Last modified by

drini (3)

Numerical id

4

Author

drini (3)

Entry type

Definition

Classification

msc 16U20

Related topic

MultiplicativeSet

Related topic

Ideal

Related topic

PrimeIdeal

Related topic

IntegralDomain

Related topic

Ring

Generated on Fri Feb 9 18:43:49 2018 by

LaTeXML
group of units

group of units

Theorem

.

The set

E

of units of a ring

R

forms a

group with

respect to

ring multiplication

.

Proof.

If

u

and

v

are two units, then there are the elements

r

and

s

of

R

such that

r

⁢

u

=

u

⁢

r

=

1

and

s

⁢

v

=

v

⁢

s

=

1

.  Then we get that

(

s

⁢

r

)

⁢

(

u

⁢

v

)

=

s

⁢

(

r

⁢

(

u

⁢

v

)

)

=

s

⁢

(

(

r

⁢

u

)

⁢

v

)

=

s

⁢

(

1

⁢

v

)

=

s

⁢

v

=

1

,  similarly

(

u

⁢

v

)

⁢

(

s

⁢

r

)

=

1

.  Thus also

u

⁢

v

is a unit, which means that

E

is closed under

multiplication

.  Because

1

∈

E

and along with

u

also its

inverse

r

belongs to

E

, the set

E

is a group.

Corollary.

In a

commutative ring

, a ring product is a unit iff all are units.

The group

E

of the units of the ring

R

is called the

group of units

of the ring

.  If

R

is a field,

E

is said to be the

multiplicative group

of the field

.

Examples

1.

When

R

=

ℤ

, then

E

=

{

1

,

-

1

}

.

2.

When

R

=

ℤ

⁢

[

i

]

,  the ring of

Gaussian integers

, then

E

=

{

1

,

i

,

-

1

,

-

i

}

.

3.

When

R

=

ℤ

⁢

[

3

]

, then (

http://planetmath.org/UnitsOfQuadraticFields

)

E

=

{

±

(

2

+

3

)

n

⋮

n

∈

ℤ

}

.

4.

When

R

=

K

⁢

[

X

]

where

K

is a field, then

E

=

K

∖

{

0

}

.

5.

When

R

=

{

0

+

ℤ

,

1

+

ℤ

,

…

,

m

-

1

+

ℤ

}

is the

residue class ring

modulo

m

, then

E

consists of the

prime classes

modulo

m

, i.e. the

residue classes

l

+

ℤ

satisfying

gcd

⁡

(

l

,

m

)

=

1

.

Title

group of units

Canonical name

GroupOfUnits

Date of creation

2013-03-22 14:41:32

Last modified on

2013-03-22 14:41:32

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

24

Author

pahio (2872)

Entry type

Theorem

Classification

msc 16U60

Classification

msc 13A05

Synonym

unit group

Related topic

CommutativeRing

Related topic

DivisibilityInRings

Related topic

NonZeroDivisorsOfFiniteRing

Related topic

PrimeResidueClass

Defines

group of units of ring

Defines

multiplicative group of field

Generated on Fri Feb 9 18:43:51 2018 by

LaTeXML