aarjaneiro/planetmath huggingface dataset:
left and right unity of ring

left and right unity of ring

If a ring

(

R

,

+

,

⋅

)

left identity

element

e

, i.e. if

e

⋅

a

=

a

∀

a

,

then

e

is called the

left unity

of

R

.

If a ring

R

right identity element

e

′

, i.e. if

a

⋅

e

′

=

a

∀

a

,

then

e

′

is called the

right unity

of

R

.

A ring may have several left or right unities (see e.g. the

Klein four-ring

).

If a ring

R

has both a left unity

e

and a right unity

e

′

, then they must coincide, since

e

′

=

e

⋅

e

′

=

e

.

This situation means that every right unity equals to

e

, likewise every left unity.  Then we speak simply of a unity of the ring.

Title

left and right unity of ring

Canonical name

LeftAndRightUnityOfRing

Date of creation

2013-03-22 15:10:54

Last modified on

2013-03-22 15:10:54

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

6

Author

rspuzio (6075)

Entry type

Definition

Classification

msc 20-00

Classification

msc 16-00

Related topic

InversesInRings

Defines

left unity

Defines

right unity

Generated on Fri Feb 9 19:15:57 2018 by

LaTeXML
Minkowski sum

Minkowski sum

Definition

Suppose

A

and

B

are sets in a

vector space

V

over a field

K

,
and suppose

λ

∈

K

. Then

A

+

B

=

{

a

+

b

∣

a

∈

A

,

b

∈

B

}

,

A

-

B

=

{

a

-

b

∣

a

∈

A

,

b

∈

B

}

,

λ

⁢

A

=

{

λ

a

∣

a

∈

A

}

,

-

A

=

(

-

1

)

⁢

A

.

The set

A

+

B

is called the

Minkowski sum

of

A

and

B

.
If either

A

or

B

is a single point (a

singleton

), say

B

=

{

x

}

, then we
write

A

+

x

instead of

A

+

{

x

}

. Similarly we define

A

-

x

,

x

-

A

and

x

+

A

.

Properties

Suppose

A

,

B

,

V

, and

λ

are as above. Then

•

A

+

B

=

B

+

A

•

λ

⁢

(

A

+

B

)

=

λ

⁢

A

+

λ

⁢

B

•

2

⁢

A

⊆

A

+

A

,

3

⁢

A

⊆

A

+

A

+

A

, etc,
but in general,

A

+

A

≠

2

⁢

A

.
(Consider

A

=

{

(

0

,

0

)

,

(

0

,

1

)

}

in

ℝ

2

.)

Title

Minkowski sum

Canonical

name

MinkowskiSum

Date of creation

2013-03-22 15:16:22

Last modified on

2013-03-22 15:16:22

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

4

Author

matte (1858)

Entry type

Definition

Classification

msc 20-00

Classification

msc 15-00

Classification

msc 13-00

Classification

msc 16-00

Related topic

VectorSpace

Related topic

Sumset

Generated on Fri Feb 9 19:15:59 2018 by

LaTeXML
modules over decomposable rings

modules over decomposable rings

Let

R

1

,

R

2

be two, nontrivial,

unital rings

and

R

=

R

1

⊕

R

2

. If

M

1

is a

R

1

-module and

M

2

is a

R

2

-module, then obviously

M

1

⊕

M

2

is a

R

-module via

(

r

,

s

)

⋅

(

m

1

,

m

2

)

=

(

r

⋅

m

1

,

s

⋅

m

2

)

. We will show that every

R

-module can be obtain in this way.

Proposition

.

If

M

is a

R

-module, then there exist

submodules

M

1

,

M

2

⊆

M

such that

M

=

M

1

⊕

M

2

and for any

r

∈

R

1

,

s

∈

R

2

,

m

1

∈

M

1

and

m

2

∈

M

2

we have

(

r

,

s

)

⋅

m

1

=

(

r

,

0

)

⋅

m

1

(

r

,

s

)

⋅

m

2

=

(

0

,

s

)

⋅

m

2

,

i.e. ring action on

M

1

(respectively

M

2

) does not depend on

R

2

(respectively

R

1

).

Proof.

Let

e

=

(

1

,

0

)

∈

R

and

f

=

(

0

,

1

)

∈

R

. Of course both

e

,

f

are

idempotents

and

(

1

,

1

)

=

e

+

f

. Moreover

e

⁢

f

=

f

⁢

e

=

0

and

e

,

f

are

central

, i.e.

e

,

f

∈

{

c

∈

R

|

∀

x

∈

R

c

⁢

x

=

x

⁢

c

}

. We will use

e

,

f

to construct submodules

M

1

,

M

2

. More precisely, let

M

1

=

e

⁢

M

and

M

2

=

f

⁢

M

. Because

e

,

f

are central, then it is clear that both

M

1

and

M

2

are submodules. We will show that

M

1

+

M

2

=

M

. Indeed, let

m

∈

M

. Then we have

m

=

(

1

,

1

)

⋅

m

=

(

e

+

f

)

⋅

m

=

e

⋅

m

+

f

⋅

m

.

Thus

M

1

+

M

2

=

M

. Furthermore, assume that

m

∈

M

1

∩

M

2

. Then there exist

m

1

,

m

2

∈

M

such that

e

⋅

m

1

=

m

=

f

⋅

m

2

and therefore

e

⋅

m

1

-

f

⋅

m

2

=

0

.

Now, after multiplying both sides by

e

we obtain that

0

=

(

e

⁢

e

)

⋅

m

1

-

(

e

⁢

f

)

⋅

m

2

=

e

⋅

m

1

-

0

⋅

m

2

=

e

⋅

m

1

=

m

,

thus

M

1

∩

M

2

=

0

. This shows that

M

=

M

1

⊕

M

2

. To finish the proof, we need to show that the ring action on

M

1

does not depend on

R

2

(the other case is analogous). But this is clear, since for any

(

r

,

s

)

∈

R

and

m

∈

M

we have

(

r

,

s

)

⋅

(

e

⋅

m

)

=

(

(

r

,

s

)

⁢

(

1

,

0

)

)

⋅

m

=

(

r

,

0

)

⋅

m

=

(

(

r

,

0

)

⁢

(

1

,

0

)

)

⋅

m

=

(

r

,

0

)

⋅

(

e

⋅

m

)

.

This

completes

the proof.

□

Title

modules over decomposable rings

Canonical name

ModulesOverDecomposableRings

Date of creation

2013-03-22 18:50:05

Last modified on

2013-03-22 18:50:05

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

4

Author

joking (16130)

Entry type

Theorem

Classification

msc 20-00

Classification

msc 16-00

Classification

msc 13-00

Generated on Fri Feb 9 19:16:01 2018 by

LaTeXML
non-commutative rings of order four

non-commutative rings of order four

Up to

isomorphism

, there are two

non-commutative

rings of order (

http://planetmath.org/OrderRing

) four. Since all

cyclic rings

are

commutative

(

http://planetmath.org/CommutativeRing

), one can immediately deduce that a ring of order four must have an

additive group

that is isomorphic to

𝔽

2

⊕

𝔽

2

.

One of the two

non-commutative rings of order four

is the

Klein 4-ring

, whose multiplication table is given by:

⋅

0

a

b

c

0

0

0

0

0

a

0

a

0

a

b

0

b

0

b

c

0

c

0

c

The other is closely related to the Klein 4-ring. In fact, it is

anti-isomorphic

to the Klein 4-ring; that is, its multiplication table is obtained by swapping the of the multiplication table for the Klein 4-ring:

⋅

0

a

b

c

0

0

0

0

0

a

0

a

b

c

b

0

0

0

0

c

0

a

b

c

Title

non-commutative rings of order four

Canonical name

NoncommutativeRingsOfOrderFour

Date of creation

2013-03-22 17:09:24

Last modified on

2013-03-22 17:09:24

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

12

Author

Wkbj79 (1863)

Entry type

Topic

Classification

msc 20-00

Classification

msc 16B99

Related topic

Klein4Ring

Related topic

OppositeRing

Related topic

ExampleOfKlein4Ring

Generated on Fri Feb 9 19:16:04 2018 by

LaTeXML
proof of general associativity

proof of general associativity

We suppose that “

⋅

” is an

associative

binary operation

of the set

S

.

Let

f

1

:

S

→

2

S

be the

mapping

f

1

⁢

(

a

1

)

:=

{

a

1

}

∀

a

1

∈

S

.

We define recursively the mapping

f

n

:

S

×

S

×

…

×

S

⏟

n

→

2

S

such that

f

n

⁢

(

a

1

,

…

,

a

n

)

:=

⋃

r

=

1

n

-

1

f

r

⁢

(

a

1

,

…

,

a

r

)

⋅

f

n

-

r

⁢

(

a

r

+

1

,

…

,

a

n

)

(1)

for

n

=

2

,

3

,

4

,

…

For example,

f

2

⁢

(

a

1

,

a

2

)

=

{

a

1

}

⋅

{

a

2

}

=

{

a

1

⋅

a

2

}

,

f

3

⁢

(

a

1

,

a

2

,

a

3

)

=

{

a

1

⋅

(

a

2

⋅

a

3

)

}

∪

{

(

a

1

⋅

a

2

)

⋅

a

3

}

=

{

a

1

⋅

(

a

2

⋅

a

3

)

,

(

a

1

⋅

a

2

)

⋅

a

3

}

=

{

(

a

1

⋅

a

2

)

⋅

a

3

}

;

the last equality due to the associativity.  It’s clear that always

|

f

1

⁢

(

a

1

)

|

=

1

,

|

f

2

⁢

(

a

1

,

a

2

)

|

=

1

,

|

f

3

⁢

(

a

1

,

a

2

,

a

3

)

|

=

1

.

We shall show by

induction

that

|

f

n

⁢

(

a

1

,

a

2

,

…

,

a

n

)

|

=

1

(2)

for each

positive integer

n

.  This means that all groupings of the

n

fixed elements using parentheses in forming the

products

with “

⋅

” yield one single

element

.

We make the

induction hypothesis

, that (2) is true for all

n

<

k

.

Now let

z

and

z

′

be arbitrary elements of

f

k

⁢

(

a

1

,

…

,

a

k

)

.
Then there exist the elements

x

,

y

,

x

′

,

y

′

of

S

and the

integers

r

,

s

∈

{

1

,

…

,

k

-

1

}

such that

z

=

x

⋅

y

,

x

∈

f

r

⁢

(

a

1

,

…

,

a

r

)

,

y

∈

f

k

-

r

⁢

(

a

r

+

1

,

…

,

a

k

)

,

z

′

=

x

′

⋅

y

′

,

x

′

∈

f

s

⁢

(

a

1

,

…

,

a

s

)

,

y

′

∈

f

k

-

s

⁢

(

a

s

+

1

,

…

,

a

k

)

.

If specially

r

=

s

,  then, by the induction hypothesis,

x

=

x

′

and

y

=

y

′

,  whence

z

=

x

⋅

y

=

x

′

⋅

y

′

=

z

′

.  If on the contrary,

r

≠

s

,  e.g.

r

<

s

,  then the induction hypothesis guarantees the

existence

of an element

v

of

S

such that

f

s

-

r

⁢

(

a

r

+

1

,

…

,

a

s

)

=

{

v

}

and

x

′

=

x

⋅

v

,

y

=

v

⋅

y

′

.

Since “

⋅

” is associative, we have

z

=

x

⋅

y

=

x

⋅

(

v

⋅

y

′

)

=

(

x

⋅

v

)

⋅

y

′

=

x

′

⋅

y

′

=

z

′

.

Thus the

equation

(2) is in force for

n

=

k

.

Title

proof of general associativity

Canonical name

ProofOfGeneralAssociativity

Date of creation

2014-05-11 15:12:33

Last modified on

2014-05-11 15:12:33

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Proof

Classification

msc 20-00

Related topic

PowerSet

Related topic

Cardinality

Generated on Fri Feb 9 19:16:07 2018 by

LaTeXML
proof of second isomorphism theorem for rings

proof of second isomorphism theorem for rings

In the

context

of rings, the

Second Isomorphism Theorem

can be phrased as follows:

If

A

is an ideal in a ring

R

and

S

is a

subring

of

R

, then

•

S

+

A

is a subring of

R

,

•

A

is an ideal in

S

+

A

,

•

S

∩

A

is an ideal in

S

,

•

There is an

isomorphism

S

/

(

S

∩

A

)

≅

(

S

+

A

)

/

A

.

It should be emphasized that the proof of this is exactly the same as the proof of the corresponding statement for groups. Again, the main idea is to use the

First Isomorphism Theorem

. (It is quite routine to verify the above statements concerning what’s a subring of what and what’s an ideal in what. The heart of the matter is the isomorphism, and that is what we are up to here.)

Consider the

mapping

f

:

S

→

(

S

+

A

)

/

A

where

f

⁢

(

s

)

=

s

+

A

. Note that this is a

ring homomorphism

. Furthermore, it is

surjective

(or “onto”): If

(

s

+

a

)

+

A

is an arbitrary element of

(

S

+

A

)

/

A

with

s

∈

S

and

a

∈

A

, then

f

⁢

(

s

)

=

s

+

A

=

s

+

(

a

+

A

)

=

(

s

+

a

)

+

A

.

Consequently, the First Isomorphism Theorem tells us that

S

/

k

⁢

e

⁢

r

⁢

(

f

)

≅

(

S

+

A

)

/

A

,

where

⁢

k

⁢

e

⁢

r

⁢

(

f

)

⁢

denotes the kernel of

⁢

f

.

So it all comes down to showing that

k

⁢

e

⁢

r

⁢

(

f

)

=

S

∩

A

.

Let

x

∈

k

⁢

e

⁢

r

⁢

(

f

)

. Then

A

=

f

⁢

(

x

)

=

x

+

A

. So

x

∈

A

. Moreover, since

k

⁢

e

⁢

r

⁢

(

f

)

⊆

S

, we have

x

∈

S

. Thus

x

∈

S

∩

A

and so

k

⁢

e

⁢

r

⁢

(

f

)

⊆

S

∩

A

.

Conversely

, suppose

x

∈

S

∩

A

. Then

x

∈

A

. So

f

⁢

(

x

)

=

x

+

A

=

A

. That is,

x

∈

k

⁢

e

⁢

r

⁢

(

f

)

and so

S

∩

A

⊆

k

⁢

e

⁢

r

⁢

(

f

)

.

Therefore,

k

⁢

e

⁢

r

⁢

(

f

)

=

S

∩

A

.

Title

proof of second isomorphism theorem for rings

Canonical name

ProofOfSecondIsomorphismTheoremForRings

Date of creation

2013-03-22 15:22:39

Last modified on

2013-03-22 15:22:39

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

14

Author

CWoo (3771)

Entry type

Proof

Classification

msc 20-00

Classification

msc 16-00

Classification

msc 13-00

Related topic

ProofOfSecondIsomorphismTheorem

Generated on Fri Feb 9 19:16:09 2018 by

LaTeXML
subgroup of a group defines an equivalence relation on the group, proof that a

subgroup of a group defines an equivalence relation on the group, proof that a

Let

H

be a

subgroup

of

G

. Then

a

∼

b

⟺

a

⁢

b

-

1

∈

H

defines an

equivalence relation

in

G

.

Proof.

We need to show that the

relation

is

reflexive

,

symmetric

and

transitive

.

1.

Reflexive:

a

⁢

a

-

1

=

e

∈

H

therefore

a

∼

a

.

2.

Symmetric: We have

a

∼

b

⇒

a

⁢

b

-

1

∈

H

⇒

(

a

⁢

b

-

1

)

-

1

∈

H

⇒

b

⁢

a

-

1

∈

H

⇒

b

∼

a

3.

Transitive: If

a

∼

b

and

b

∼

c

then we have that

a

⁢

b

-

1

∈

H

,

and

b

⁢

c

-

1

∈

H

but then

(

a

⁢

b

-

1

)

⁢

(

b

⁢

c

-

1

)

∈

H

which gives

a

⁢

c

-

1

∈

H

that is,

a

∼

c

.

∎

Title

subgroup of a group defines an equivalence relation on the group, proof that a

Canonical name

SubgroupOfAGroupDefinesAnEquivalenceRelationOnTheGroupProofThatA

Date of creation

2013-03-22 15:32:46

Last modified on

2013-03-22 15:32:46

Owner

Dr_Absentius (537)

Last modified by

Dr_Absentius (537)

Numerical id

5

Author

Dr_Absentius (537)

Entry type

Proof

Classification

msc 20-00

Generated on Fri Feb 9 19:16:16 2018 by

LaTeXML
submodule

submodule

Given a ring

R

and a left

R

-module

T

, a subset

A

of

T

is called a (

left

)

submodule

of

T

, if

(

A

,

+

)

is a

subgroup

of

(

M

,

+

)

and

r

⁢

a

∈

A

for all

elements

r

of

R

and

a

of

A

.

Examples

1.

The subsets

{

0

}

and

T

are always submodules of the module

T

.

2.

The set

{

t

∈

T

:

r

⁢

t

=

t

⁢

∀

r

∈

R

}

of all

invariant

elements of

T

is a submodule of

T

.

3.

If

X

⊆

T

and

𝔞

is a

left ideal

of

R

, then the set

𝔞

⁢

X

:=

{

finite

⁢

∑

ν

a

ν

⁢

x

ν

:

a

ν

∈

𝔞

,

x

ν

∈

X

⁢

∀

ν

}

is a submodule of

T

.  Especially,

R

⁢

X

is called the submodule

generated

by the subset

X

; then the elements of

X

are

generators

of this submodule.

There are some

operations

on submodules.  Given the submodules

A

and

B

of

T

, the

sum

A

+

B

:=

{

a

+

b

∈

T

:

a

∈

A

∧

b

∈

B

}

and the

intersection

A

∩

B

are submodules of

T

.

The notion of sum may be extended for any family

{

A

j

:

j

∈

J

}

of submodules:  the sum

∑

j

∈

J

A

j

of submodules consists of all finite sums

∑

j

a

j

where every

a

j

belongs to one

A

j

of those submodules.  The sum of submodules as well as the intersection

⋂

j

∈

J

A

j

are submodules of

T

.  The submodule

R

⁢

X

is the intersection of all submodules containing the subset

X

.

If

T

is a ring and

R

is a

subring

of

T

, then

T

is an

R

-module; then one can consider the

product

and the

quotient

of the left

R

-submodules

A

and

B

of

T

:

•

A

⁢

B

:=

{

finite

⁢

∑

ν

a

ν

⁢

b

ν

:

a

ν

∈

A

,

b

ν

∈

B

⁢

∀

ν

}

•

[

A

:

B

]

:=

{

t

∈

T

:

t

B

⊆

A

}

Also these are left

R

-submodules of

T

.

Title

submodule

Canonical name

Submodule

Date of creation

2013-03-22 15:15:26

Last modified on

2013-03-22 15:15:26

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

19

Author

PrimeFan (13766)

Entry type

Definition

Classification

msc 20-00

Classification

msc 16-00

Classification

msc 13-00

Related topic

SumOfIdeals

Related topic

QuotientOfIdeals

Defines

R-submodule

Defines

generated submodule

Defines

generator

Defines

sum of submodules

Defines

product submodule

Defines

quotient of submodules

Generated on Fri Feb 9 19:16:18 2018 by

LaTeXML
subring

subring

Let

(

R

,

+

,

*

)

a ring. A

subring

is a subset

S

of

R

with the

operations

+

and

*

of

R

restricted to

S

and such that

S

is a ring by itself.

Notice that the restricted operations inherit the

associative

and

distributive properties

of

+

and

*

, as well as

commutativity

of

+

.
So for

(

S

,

+

,

*

)

to be a ring by itself, we need that

(

S

,

+

)

be a

subgroup

of

(

R

,

+

)

and that

(

S

,

*

)

be closed.
The subgroup condition is

equivalent

to

S

being non-empty and having the

property

that

x

-

y

∈

S

for all

x

,

y

∈

S

.

A subring

S

is called a

left ideal

if for all

s

∈

S

and all

r

∈

R

we have

r

*

s

∈

S

.

Right ideals

are defined similarly, with

s

*

r

instead of

r

*

s

.
If

S

is both a left ideal and a right ideal, then it is called a

two-sided ideal

. If

R

is

commutative

, then all three

definitions

coincide. In ring theory, ideals are far more important than subrings, as they play a role analogous to

normal subgroups

in

group theory

.

Example:

Consider the ring

(

ℤ

,

+

,

⋅

). Then

(

2

⁢

ℤ

,

+

,

⋅

)

is a subring, since the

difference

and

product

of two

even numbers

is again an even number.

Title

subring

Canonical

name

Subring

Date of creation

2013-03-22 12:30:19

Last modified on

2013-03-22 12:30:19

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

17

Author

yark (2760)

Entry type

Definition

Classification

msc 20-00

Classification

msc 16-00

Classification

msc 13-00

Related topic

Ideal

Related topic

Ring

Related topic

Group

Related topic

Subgroup

Defines

ideal

Generated on Fri Feb 9 19:16:19 2018 by

LaTeXML
uniqueness of additive inverse in a ring

uniqueness of additive inverse in a ring

Lemma

.

Let

R

be a ring, and let

a

be any element of

R

. There exists a unique element

b

of

R

such that

a

+

b

=

0

, i.e. there is a unique

additive inverse

(

http://planetmath.org/Ring

) for

a

.

Proof.

Let

a

be an element of

R

. By

definition

of ring, there exists at least one additive inverse (

http://planetmath.org/Ring

) of

a

, call it

b

1

, so that

a

+

b

1

=

0

. Now, suppose

b

2

is another additive inverse of

a

, i.e. another element of

R

such that

a

+

b

2

=

0

where

0

is the

zero element

(

http://planetmath.org/Ring

) of

R

. Let us show that

b

1

=

b

2

. Using

properties

for a ring and the above

equations

for

b

1

and

b

2

yields

b

1

=

b

1

+

0

(definition of zero)

=

b

1

+

(

a

+

b

2

)

(

b

2

⁢

is an additive inverse of

⁢

a

)

=

(

b

1

+

a

)

+

b

2

(

associativity in

⁢

R

)

=

0

+

b

2

(

b

1

⁢

is an additive inverse of

⁢

a

)

=

b

2

(definition of zero)

.

Therefore, there is a unique additive inverse for

a

.
∎

Title

uniqueness of additive inverse in a ring

Canonical

name

UniquenessOfAdditiveInverseInARing

Date of creation

2013-03-22 14:13:54

Last modified on

2013-03-22 14:13:54

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

7

Author

alozano (2414)

Entry type

Theorem

Classification

msc 20-00

Classification

msc 16-00

Classification

msc 13-00

Related topic

UniquenessOfInverseForGroups

Generated on Fri Feb 9 19:16:21 2018 by

LaTeXML
uniqueness of inverse (for groups)

uniqueness of inverse (for groups)

Lemma

Suppose

(

G

,

∗

)

is a group. Then every element in

G

has a
unique

inverse

.

Proof.

Suppose

g

∈

G

. By the group axioms we know that there
is an

h

∈

G

such that

g

∗

h

=

h

∗

g

=

e

,

where

e

is the

identity element

in

G

. If there is also a

h

′

∈

G

satisfying

g

∗

h

′

=

h

′

∗

g

=

e

,

then

h

=

h

∗

e

=

h

∗

(

g

∗

h

′

)

=

(

h

∗

g

)

∗

h

′

=

e

∗

h

′

=

h

′

,

so

h

=

h

′

, and

g

has a unique inverse.

□

Title

uniqueness of inverse (for groups)

Canonical name

UniquenessOfInverseforGroups

Date of creation

2013-03-22 14:14:33

Last modified on

2013-03-22 14:14:33

Owner

waj (4416)

Last modified by

waj (4416)

Numerical id

5

Author

waj (4416)

Entry type

Result

Classification

msc 20-00

Classification

msc 20A05

Related topic

UniquenessOfAdditiveIdentityInARing

Related topic

IdentityElementIsUnique

Generated on Fri Feb 9 19:16:23 2018 by

LaTeXML
unity

unity

The

unity

of a ring

(

R

,

+

,

⋅

)

is the

multiplicative identity

of the ring, if it has such.  The unity is often denoted by

e

,

u

or 1.  Thus, the unity satisfies

e

⋅

a

=

a

⋅

e

=

a

∀

a

∈

R

.

If

R

consists of the mere 0, then

0

is its unity, since in every ring,

0

⋅

a

=

a

⋅

0

=

0

.  Conversely, if 0 is the unity in some ring

R

, then

R

=

{

0

}

(because

a

=

0

⋅

a

=

0

⁢

∀

a

∈

R

).

Note.

When considering a ring

R

it is often mentioned that “…having

1

≠

0

” or that “…with

non-zero unity

”, sometimes only “…with unity” or “…with ”; all these exclude the case

R

=

{

0

}

.

Theorem

.

An element

u

of a ring

R

is the unity iff

u

is an

idempotent

and

regular element

.

Proof.

Let

u

be an idempotent and regular element.  For any element

x

of

R

we have

u

⁢

x

=

u

2

⁢

x

=

u

⁢

(

u

⁢

x

)

,

and because

u

is no left zero divisor, it may be cancelled from the

equation

; thus we get

x

=

u

⁢

x

.  Similarly,

x

=

x

⁢

u

.  So

u

is the unity of the ring.  The other half of the is apparent.

Title

unity

Canonical name

Unity

Date of creation

2013-03-22 14:47:17

Last modified on

2013-03-22 14:47:17

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

15

Author

pahio (2872)

Entry type

Definition

Classification

msc 20-00

Classification

msc 16-00

Classification

msc 13-00

Synonym

multiplicative identity

Synonym

characterization of unity

Related topic

ZeroDivisor

Related topic

RootOfUnity

Related topic

ZeroRing

Related topic

NonZeroDivisorsOfFiniteRing

Related topic

OppositePolynomial

Defines

non-zero unity

Defines

nonzero unity

Generated on Fri Feb 9 19:16:25 2018 by

LaTeXML
unity of subring

unity of subring

Theorem

.

Let

S

be a proper

subring

of the ring

R

.  If

S

has a

non-zero unity

u

which is not unity of

R

, then

u

is a

zero divisor

of

R

.

Proof.

Because

u

is not unity of

R

, there exists an element

r

of

R

such that

r

⁢

u

≠

r

.  Then we have

(

r

⁢

u

)

⁢

u

=

r

⁢

(

u

⁢

u

)

=

r

⁢

u

, which implies that

0

=

(

r

⁢

u

)

⁢

u

-

r

⁢

u

=

(

r

⁢

u

-

r

)

⋅

u

.  Since neither

r

⁢

u

-

r

nor

u

is 0, the element

u

is a zero divisor in

R

.

Title

unity of subring

Canonical name

UnityOfSubring

Date of creation

2013-03-22 14:49:40

Last modified on

2013-03-22 14:49:40

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Theorem

Classification

msc 20-00

Classification

msc 16-00

Classification

msc 13-00

Related topic

UnitiesOfRingAndSubring

Related topic

CornerOfARing

Generated on Fri Feb 9 19:16:26 2018 by

LaTeXML
zero times an element is zero in a ring

zero times an element is zero in a ring

Lemma 1

.

Let

R

be a ring with

zero element

0

(i.e.

0

is the

additive identity

of

R

). Then for any element

a

∈

R

we have

0

⋅

a

=

a

⋅

0

=

0

.

Proof.

0

⋅

a

=

(

0

+

0

)

⋅

a

,

by definition of zero

=

0

⋅

a

+

0

⋅

a

,

by the distributive law

Thus

0

⋅

a

=

0

⋅

a

+

0

⋅

a

. Let

b

be the

additive inverse

of

0

⋅

a

∈

R

. Hence:

b

+

0

⋅

a

=

b

+

(

0

⋅

a

+

0

⋅

a

)

(

b

+

0

⋅

a

)

=

(

b

+

0

⋅

a

)

+

0

⋅

a

0

=

0

+

0

⋅

a

0

=

0

⋅

a

as claimed. The proof of

a

⋅

0

=

0

is done analogously.
∎

Title

zero times an element is zero in a ring

Canonical name

ZeroTimesAnElementIsZeroInARing

Date of creation

2013-03-22 14:13:57

Last modified on

2013-03-22 14:13:57

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

8

Author

alozano (2414)

Entry type

Theorem

Classification

msc 20-00

Classification

msc 16-00

Classification

msc 13-00

Synonym

0

⋅

a

=

0

Related topic

1cdotAA

Related topic

AbsorbingElement

Generated on Fri Feb 9 19:16:28 2018 by

LaTeXML
generalisation of Euler–Fermat theorem

generalisation of Euler–Fermat theorem

Title

generalisation of Euler–

Fermat theorem

Canonical name

GeneralisationOfEulerFermatTheorem

Date of creation

2013-03-22 19:34:35

Last modified on

2013-03-22 19:34:35

Owner

teero (27189)

Last modified by

teero (27189)

Numerical id

6

Author

teero (27189)

Entry type

Theorem

Classification

msc 20-01

Classification

msc 11-00

Classification

msc 20A05

Generated on Fri Feb 9 19:16:30 2018 by

LaTeXML