aarjaneiro/planetmath huggingface dataset:
example of a projective module which is not free

example of a projective module which is not free

Let

R

1

and

R

2

be two nontrivial,

unital rings

and let

R

=

R

1

⊕

R

2

. Furthermore let

π

i

:

R

→

R

i

be a

projection

for

i

=

1

,

2

. Note that in this case both

R

1

and

R

2

are (left) modules over

R

via

⋅

:

R

×

R

i

→

R

i

;

(

r

,

s

)

⋅

x

=

π

i

⁢

(

r

,

s

)

⁢

x

,

where on the right side we have the

multiplication

in a ring

R

i

.

Proposition

.

Both

R

1

and

R

2

are projective

R

-modules, but neither

R

1

nor

R

2

is free.

Proof.

Obviously

R

1

⊕

R

2

is

isomorphic

(as a

R

-modules) with

R

thus both

R

1

and

R

2

are projective as a

direct summands

of a

free module

.

Assume now that

R

1

is free, i.e. there exists

ℬ

=

{

e

i

}

i

∈

I

⊆

R

1

which is a basis. Take any

i

0

∈

I

. Both

R

1

and

R

2

are nontrivial and thus

1

≠

0

in both

R

1

and

R

2

. Therefore

(

1

,

0

)

≠

(

1

,

1

)

in

R

, but

(

1

,

1

)

⋅

e

i

0

=

π

1

⁢

(

1

,

1

)

⁢

e

i

0

=

1

⁢

e

i

0

=

π

1

⁢

(

1

,

0

)

⁢

e

i

0

=

(

1

,

0

)

⋅

e

i

0

.

This situation is impossible in free modules (

linear combination

is uniquely determined by scalars).

Contradiction

. Analogously we prove that

R

2

is not free.

□

Title

example of a

projective module

which is not free

Canonical name

ExampleOfAProjectiveModuleWhichIsNotFree

Date of creation

2013-03-22 18:49:55

Last modified on

2013-03-22 18:49:55

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

6

Author

joking (16130)

Entry type

Example

Classification

msc 16D40

Generated on Fri Feb 9 18:38:23 2018 by

LaTeXML
faithfully flat

faithfully flat

Let

A

be a

commutative ring

. Then

M

if

faithfully flat

if for any

A

-modules

P

,

Q

, and

R

, we have

0

→

P

→

Q

→

R

→

0

is exact if and only if the

M

-tensored

sequence

0

→

M

⊗

A

P

→

M

⊗

A

Q

→

M

⊗

A

R

→

0

is exact. (Note that the “if and only if” clause makes this stronger than the definition of flatness).

Equivalently, an

A

-module

M

is faithfully flat iff

M

is flat and the

functor

-

⊗

A

M

is a

faithful functor

(and hence the name).

Title

faithfully flat

Canonical name

FaithfullyFlat

Date of creation

2013-03-22 14:35:55

Last modified on

2013-03-22 14:35:55

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

5

Author

mathcam (2727)

Entry type

Definition

Classification

msc 16D40

Generated on Fri Feb 9 18:38:28 2018 by

LaTeXML
finitely generated projective module

finitely generated projective module

Let

R

be a

unital ring

.
A

finitely generated

projective right

R

-module is of the form

e

⁢

R

n

,

n

∈

ℕ

, where

e

is an

idempotent

in

End

R

(

R

n

)

.

Let

A

be a unital

C

*

-

algebra

and

p

be a

projection

in

End

A

(

A

n

)

,

n

∈

ℕ

.
Then,

ℰ

=

p

⁢

A

n

is a finitely generated projective right

A

-module.
Further,

ℰ

is a pre-Hilbert

A

-module with (

A

-valued)

inner product

⟨

u

,

v

⟩

=

∑

i

=

1

n

u

i

*

⁢

v

i

,

u

,

v

∈

ℰ

.

Title

finitely generated projective module

Canonical name

FinitelyGeneratedProjectiveModule

Date of creation

2013-03-22 13:29:37

Last modified on

2013-03-22 13:29:37

Owner

mhale (572)

Last modified by

mhale (572)

Numerical id

6

Author

mhale (572)

Entry type

Definition

Classification

msc 16D40

Synonym

finite projective module

Related topic

HilbertModule

Generated on Fri Feb 9 18:38:30 2018 by

LaTeXML
flat module

flat module

A

right module

M

over a ring

R

is

flat

if the

tensor product functor

M

⊗

R

(

-

)

is an

exact functor

.

Similarly, a left module

N

over

R

is

flat

if the tensor product functor

(

-

)

⊗

R

N

is an exact functor.

Title

flat module

Canonical name

FlatModule

Date of creation

2013-03-22 12:09:45

Last modified on

2013-03-22 12:09:45

Owner

antizeus (11)

Last modified by

antizeus (11)

Numerical id

6

Author

antizeus (11)

Entry type

Definition

Classification

msc 16D40

Synonym

flat

Generated on Fri Feb 9 18:38:32 2018 by

LaTeXML
free module

free module

Let

R

be a

commutative ring

with unity. A

free module

over

R

is a (unital) module

isomorphic

to a

direct sum

of copies of

R

. In particular, as every

abelian group

is a

ℤ

-module, a

free abelian group

is a direct sum of copies of

ℤ

. This is

equivalent

to saying that the module has a

free basis

,
i.e. a set of

elements

with the that every element of the module can be uniquely expressed as an

linear combination

over

R

of elements of the free basis. In the case that a free module over

R

is a sum of finitely many copies of

R

, then the number of copies is called the

rank

of the free module.

An alternative definition of a free module is via its

universal property

: Given a set

X

, the free

R

-module

F

⁢

(

X

)

on the set

X

is equipped with a

function

i

:

X

→

F

⁢

(

X

)

satisfying the property that for any other

R

-module

A

and any function

f

:

X

→

A

, there exists a unique

R

-module map

h

:

F

⁢

(

X

)

→

A

such that

(

h

∘

i

)

=

f

.

Title

free module

Canonical name

FreeModule

Date of creation

2013-03-22 12:10:10

Last modified on

2013-03-22 12:10:10

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

9

Author

mathcam (2727)

Entry type

Definition

Classification

msc 16D40

Related topic

FreeGroup

Defines

free module

Defines

free abelian group

Defines

free basis

Defines

rank of a free module

Generated on Fri Feb 9 18:38:33 2018 by

LaTeXML
free module

free module

Let

R

be a ring.
A

free module

over

R

is a

direct sum

of copies of

R

.

Similarly, as an

abelian group

is simply a module over

ℤ

,
a

free abelian group

is a direct sum of copies of

ℤ

.

This is equivalent to saying
that the module has a

free basis

,
i.e. a set of elements
with the

property

that every element of the module
can be uniquely expressed
as an

linear combination

over

R

of elements of the free basis.

Every free module is also a

projective module

,
as well as a

flat module

.

Title

free module

Canonical

name

FreeModule1

Date of creation

2013-03-22 14:03:50

Last modified on

2013-03-22 14:03:50

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

5

Author

Mathprof (13753)

Entry type

Definition

Classification

msc 16D40

Generated on Fri Feb 9 18:38:35 2018 by

LaTeXML
invertible ideals are projective

invertible ideals are projective

If

R

is a ring and

f

:

M

→

N

is a

homomorphism

of

R

-modules, then a

right inverse

of

f

is a homomorphism

g

:

N

→

M

such that

f

∘

g

is the

identity map

on

N

. For a right inverse to exist, it is clear that

f

must be an

epimorphism

. If a right inverse exists for every such epimorphism and all modules

M

, then

N

is said to be a

projective module

.

For

fractional ideals

over an

integral domain

R

, the property of being projective as an

R

-module is

equivalent

to being an invertible ideal.

Theorem

.

Let

R

be an integral domain. Then a fractional ideal over

R

is

invertible

if and only if it is projective as an

R

-module.

In particular, every fractional ideal over a

Dedekind domain

is invertible, and is therefore projective.

Title

invertible ideals are projective

Canonical name

InvertibleIdealsAreProjective

Date of creation

2013-03-22 18:35:47

Last modified on

2013-03-22 18:35:47

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

5

Author

gel (22282)

Entry type

Theorem

Classification

msc 16D40

Classification

msc 13A15

Related topic

ProjectiveModule

Related topic

FractionalIdeal

Generated on Fri Feb 9 18:38:37 2018 by

LaTeXML
minimal projective presentation

minimal projective presentation

Let

R

be a ring and

M

a (right) module over

R

. A

short exact sequence

of modules

\xymatrix

⁢

P

1

⁢

\ar

⁢

[

r

]

p

1

⁢

&

⁢

P

0

⁢

\ar

⁢

[

r

]

p

0

⁢

&

⁢

M

⁢

\ar

⁢

[

r

]

⁢

&

⁢

0

is called a

minimal projective presentation

of

M

if both

p

0

:

P

0

→

M

and

p

1

:

P

1

→

ker

⁢

p

0

are

projective covers

.

Minimal

projective presenetations are unique in the following sense: if

\xymatrix

⁢

P

1

⁢

\ar

⁢

[

r

]

p

1

⁢

&

⁢

P

0

⁢

\ar

⁢

[

r

]

p

0

⁢

&

⁢

M

⁢

\ar

⁢

[

r

]

⁢

&

⁢

0

⁢

P

1

′

⁢

\ar

⁢

[

r

]

p

1

′

⁢

&

⁢

P

0

′

⁢

\ar

⁢

[

r

]

p

0

′

⁢

&

⁢

M

⁢

\ar

⁢

[

r

]

⁢

&

⁢

0

are both minimal projective presentations of

M

, then this

diagram

can be completed to the following

commutative

one:

\xymatrix

⁢

P

1

⁢

\ar

⁢

[

r

]

p

1

⁢

\ar

⁢

[

d

]

a

⁢

&

⁢

P

0

⁢

\ar

⁢

[

r

]

p

0

⁢

\ar

⁢

[

d

]

b

⁢

&

⁢

M

⁢

\ar

⁢

[

r

]

⁢

\ar

⁢

[

d

]

=

⁢

&

⁢

0

⁢

P

1

′

⁢

\ar

⁢

[

r

]

p

1

′

⁢

&

⁢

P

0

′

⁢

\ar

⁢

[

r

]

p

0

′

⁢

&

⁢

M

⁢

\ar

⁢

[

r

]

⁢

&

⁢

0

were both

a

,

b

are

isomorphisms

.

It can be shown, that if

R

is a finite-dimensional algebra over a field

k

, then every

finitely generated

R

-module

M

admits minimal projective presentation (indeed,

R

is semiperfect (

http://planetmath.org/PerfectAndSemiperfectRings

) in this case).

Title

minimal projective presentation

Canonical name

MinimalProjectivePresentation

Date of creation

2013-03-22 19:18:00

Last modified on

2013-03-22 19:18:00

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

4

Author

joking (16130)

Entry type

Definition

Classification

msc 16D40

Generated on Fri Feb 9 18:38:39 2018 by

LaTeXML
perfect and semiperfect rings

perfect and semiperfect rings

A ring

R

is called

left/right perfect

if for any left/right

R

-module

M

there exists a

projective cover

p

:

P

→

M

.

A ring

R

is called

left/right semiperfect

if for any left/right finitely-generated

R

-module

M

there exists a projective cover

p

:

P

→

M

.

It can be shown that there are rings which are left perfect, but not right perfect. However being semiperfect is left-right

symmetric

property

.

Some examples of semiperfect rings include:

1.

perfect rings;

2.

left/

right Artinian

rings;

3.

finite-dimensional

algebras over a field

k

.

Title

perfect and semiperfect rings

Canonical

name

PerfectAndSemiperfectRings

Date of creation

2013-03-22 19:17:56

Last modified on

2013-03-22 19:17:56

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

4

Author

joking (16130)

Entry type

Definition

Classification

msc 16D40

Generated on Fri Feb 9 18:38:41 2018 by

LaTeXML
projective cover

projective cover

Let

X

and

P

be modules.
We say that

P

is a

projective cover

of

X

if

P

is a

projective module

and there exists an

epimorphism

p

:

P

→

X

such that

ker

⁡

p

is a

superfluous submodule

of

P

.

Equivalently,

P

is an projective cover of

X

if

P

is projective,
and there is an epimorphism

p

:

P

→

X

,
and if

g

:

P

′

→

X

is an epimorphism
from a projective module

P

′

to

X

,
then there exists an epimorphism

h

:

P

′

→

P

such that

p

⁢

h

=

g

.

\xymatrix

&

P

′

\ar

[

d

]

g

\ar

@

-

-

>

[

d

l

]

h

P

\ar

[

r

]

p

&

X

\ar

[

r

]

\ar

[

d

]

&

0

&

0

Title

projective cover

Canonical name

ProjectiveCover

Date of creation

2013-03-22 12:10:08

Last modified on

2013-03-22 12:10:08

Owner

antizeus (11)

Last modified by

antizeus (11)

Numerical id

6

Author

antizeus (11)

Entry type

Definition

Classification

msc 16D40

Generated on Fri Feb 9 18:38:42 2018 by

LaTeXML
projective module

projective module

A module

P

is

projective

if it

satisfies

the following

equivalent

conditions:

(a) Every

short exact sequence

of the form

0

→

A

→

B

→

P

→

0

is split (

http://planetmath.org/SplitShortExactSequence

);

(b) The

functor

Hom

⁢

(

P

,

-

)

is exact (

http://planetmath.org/ExactFunctor

);

(c) If

f

:

X

→

Y

is an

epimorphism

and there exists a

homomorphism

g

:

P

→

Y

,
then there exists a homomorphism

h

:

P

→

X

such that

f

⁢

h

=

g

.

\xymatrix

&

P

\ar

@

-

-

>

[

d

l

]

h

\ar

[

d

]

g

X

\ar

[

r

]

f

&

Y

\ar

[

r

]

&

0

(d) The module

P

is a

direct summand

of a

free module

.

Title

projective module

Canonical name

ProjectiveModule

Date of creation

2013-03-22 12:09:42

Last modified on

2013-03-22 12:09:42

Owner

antizeus (11)

Last modified by

antizeus (11)

Numerical id

7

Author

antizeus (11)

Entry type

Definition

Classification

msc 16D40

Related topic

InvertibleIdealsAreProjective

Generated on Fri Feb 9 18:38:44 2018 by

LaTeXML
proof of invertible ideals are projective

proof of invertible ideals are projective

We show that a nonzero

fractional ideal

𝔞

of an

integral domain

R

is invertible if and only if it is projective (

http://planetmath.org/ProjectiveModule

) as an

R

-module.

Let

𝔞

be an invertible fractional ideal and

f

:

M

→

𝔞

be an

epimorphism

of

R

-modules. We need to show that

f

has a

right inverse

.
Letting

𝔞

-

1

be the inverse ideal of

𝔞

, there exists

a

1

,

…

,

a

n

∈

𝔞

and

b

1

,

…

,

b

n

∈

𝔞

-

1

such that

a

1

⁢

b

1

+

⋯

+

a

n

⁢

b

n

=

1

and, as

f

is onto, there exist

e

k

∈

M

such that

f

⁢

(

e

k

)

=

a

k

. For any

x

∈

𝔞

,

x

⁢

b

k

∈

𝔞

⁢

𝔞

-

1

=

R

, so we can define

g

:

𝔞

→

M

by

g

⁢

(

x

)

≡

(

x

⁢

b

1

)

⁢

e

1

+

⋯

+

(

x

⁢

b

n

)

⁢

e

n

.

Then

f

∘

g

⁢

(

x

)

=

(

x

⁢

b

1

)

⁢

f

⁢

(

e

1

)

+

⋯

+

(

x

⁢

b

n

)

⁢

f

⁢

(

e

n

)

=

x

⁢

(

b

1

⁢

a

1

+

⋯

⁢

b

n

⁢

a

n

)

=

x

,

so

g

is indeed a right inverse of

f

, and

𝔞

is projective.

Conversely, suppose that

𝔞

is projective and let

(

a

i

)

i

∈

I

generate

𝔞

(this always exists, as we can let

a

i

include every element of

𝔞

). Then let

M

be a module with

free basis

(

e

i

)

i

∈

I

and define

f

:

M

→

𝔞

by

f

⁢

(

e

i

)

=

a

i

. As

𝔞

is projective,

f

has a right inverse

g

:

𝔞

→

M

. As

e

i

freely generate

M

, we can uniquely define

g

i

:

𝔞

→

R

by

g

⁢

(

x

)

=

∑

i

∈

I

g

i

⁢

(

x

)

⁢

e

i

,

noting that all but finitely many

g

i

⁢

(

x

)

must be zero for any given

x

. Choosing any fixed nonzero

a

∈

𝔞

, we can set

b

i

=

a

-

1

⁢

g

i

⁢

(

a

)

so that

g

i

⁢

(

x

)

=

a

-

1

⁢

g

i

⁢

(

a

⁢

x

)

=

a

-

1

⁢

x

⁢

g

i

⁢

(

a

)

=

b

i

⁢

x

for all

x

∈

𝔞

, and

b

i

must equal zero for all but finitely many

i

. So, we can let

𝔟

be the fractional ideal generated by the

b

i

and, noting that

x

⁢

b

i

=

g

i

⁢

(

x

)

∈

R

we get

𝔞

⁢

𝔟

⊆

R

. Furthermore, for any

x

∈

R

,

x

=

a

-

1

⁢

f

∘

g

⁢

(

a

⁢

x

)

=

∑

i

a

-

1

⁢

g

i

⁢

(

a

⁢

x

)

⁢

f

⁢

(

e

i

)

=

∑

i

x

⁢

b

i

⁢

f

⁢

(

e

i

)

∈

𝔟

⁢

𝔞

so that

R

⊆

𝔞

⁢

𝔟

, and

𝔟

is the

inverse

of

𝔞

as required.

Title

proof of invertible ideals are projective

Canonical name

ProofOfInvertibleIdealsAreProjective

Date of creation

2013-03-22 18:35:51

Last modified on

2013-03-22 18:35:51

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

5

Author

gel (22282)

Entry type

Proof

Classification

msc 16D40

Classification

msc 13A15

Generated on Fri Feb 9 18:38:47 2018 by

LaTeXML
example of injective module

example of injective module

In the

category

of

unitary

ℤ

-modules (which is the category of Abelian groups), every

divisible Group

is

injective

, i.e. every Group

G

such that for any

g

∈

G

and

n

∈

ℕ

, there is a

h

∈

G

such that

n

⁢

h

=

g

. For example,

ℚ

and

ℚ

/

ℤ

are divisible, and therefore injective.

Proof.

We have to show that, if

G

is a divisible Group,

φ

:

U

→

G

is any

homomorphism

, and

U

is a

subgroup

of a Group

H

, there is a homomorphism

ψ

:

H

→

G

such that the

restriction

ψ

|

U

=

φ

. In other words, we want to extend

φ

to a homomorphism

H

→

G

.

Let

𝒟

be the set of pairs

(

K

,

ψ

)

such that

K

is a subgroup of

G

containing

U

and

ψ

:

K

→

G

is a homomorphism with

ψ

|

U

=

φ

. Then

𝒟

ist non-empty since it contains

(

U

,

φ

)

, and it is partially ordered by

(

K

,

ψ

)

≤

(

K

′

,

ψ

′

)

:

⟺

K

⊆

K

′

and

ψ

′

|

K

=

ψ

.

For any ascending chain

(

K

1

,

ψ

1

)

≤

(

K

2

,

ψ

2

)

≤

…

,

in

𝒟

, the pair

(

⋃

i

∈

ℕ

K

i

,

⋃

i

∈

ℕ

ψ

i

)

is in

𝒟

, and it is an upper bound for this chain. Therefore, by Zorn’s Lemma,

𝒟

contains a

maximal element

(

M

,

χ

)

.

It remains to show that

M

=

H

. Suppose the opposite, and let

h

∈

H

∖

M

. Let

⟨

h

⟩

denote the subgroup of

H

generated by

h

. If

⟨

h

⟩

∩

M

=

{

0

}

, the sum

M

+

⟨

h

⟩

is in fact a

direct sum

, and we can extend

χ

to

M

+

⟨

h

⟩

by choosing an arbitrary image of

h

in

G

and extending linearly. This contradicts the maximality of

(

M

,

χ

)

.

Let us therefore suppose

⟨

h

⟩

∩

M

contains an

element

n

⁢

h

, with

n

∈

ℕ

minimal

. Since

n

⁢

h

∈

M

, and

χ

is defined on

M

,

χ

⁢

(

n

⁢

h

)

exists, and furthermore, since

G

is divisible, there is a

g

∈

G

such that

n

⁢

g

=

χ

⁢

(

n

⁢

h

)

. It is now easy to see that we can extend

χ

to

M

+

⟨

h

⟩

by defining

χ

⁢

(

h

)

:=

g

, in

contradiction

to the maximality of

(

M

,

χ

)

.

Therefore,

M

=

H

. This proves the statement.
∎

Title

example of

injective module

Canonical name

ExampleOfInjectiveModule

Date of creation

2013-03-22 17:43:40

Last modified on

2013-03-22 17:43:40

Owner

Glotzfrosch (19314)

Last modified by

Glotzfrosch (19314)

Numerical id

5

Author

Glotzfrosch (19314)

Entry type

Example

Classification

msc 16D50

Generated on Fri Feb 9 18:39:01 2018 by

LaTeXML
injective hull

injective hull

Let

X

and

Q

be modules.
We say that

Q

is an

injective hull

or

injective envelope

of

X

if

Q

is both an

injective module

and an

essential extension

of

X

.

Equivalently,

Q

is an injective hull of

X

if

Q

is injective,
and

X

is a

submodule

of

Q

,
and if

g

:

X

→

Q

′

is a

monomorphism

from

X

to an injective module

Q

′

,
then there exists a monomorphism

h

:

Q

→

Q

′

such that

h

⁢

(

x

)

=

g

⁢

(

x

)

for all

x

∈

X

.

\xymatrix

&

0

\ar

[

d

]

0

\ar

[

r

]

&

X

\ar

[

r

]

i

\ar

[

d

]

g

&

Q

\ar

@

-

-

>

[

d

l

]

h

&

Q

′

Every module

X

has an injective hull, which is unique up to

isomorphism

. The injective hull of

X

is sometimes denoted

E

⁢

(

X

)

.

Title

injective hull

Canonical name

InjectiveHull

Date of creation

2013-03-22 12:10:05

Last modified on

2013-03-22 12:10:05

Owner

mclase (549)

Last modified by

mclase (549)

Numerical id

7

Author

mclase (549)

Entry type

Definition

Classification

msc 16D50

Synonym

injective envelope

Generated on Fri Feb 9 18:39:03 2018 by

LaTeXML
injective module

injective module

A module

Q

is an

injective module

if it

satisfies

the following

equivalent

conditions:

(a) Every

short exact sequence

of the form

0

→

Q

→

B

→

C

→

0

is split (

http://planetmath.org/SplitShortExactSequence

);

(b) The

functor

Hom

⁢

(

-

,

Q

)

is exact (

http://planetmath.org/ExactFunctor

);

(c) If

f

:

X

→

Y

is a

monomorphism

and there exists a

homomorphism

g

:

X

→

Q

,
then there exists a homomorphism

h

:

Y

→

Q

such that

h

⁢

f

=

g

.

\xymatrix

0

\ar

[

r

]

&

X

\ar

[

d

]

g

\ar

[

r

]

f

&

Y

\ar

@

-

-

>

[

d

l

]

h

&

Q

Title

injective module

Canonical name

InjectiveModule

Date of creation

2013-03-22 12:02:26

Last modified on

2013-03-22 12:02:26

Owner

antizeus (11)

Last modified by

antizeus (11)

Numerical id

8

Author

antizeus (11)

Entry type

Definition

Classification

msc 16D50

Generated on Fri Feb 9 18:39:05 2018 by

LaTeXML
product of injective modules is injective

product of injective modules is injective

Proposition

.

Let

R

be a ring and

{

Q

i

}

i

∈

I

a family of

injective

R

-modules. Then the

product

Q

=

∏

i

∈

I

Q

i

is injective.

Proof.

Let

B

be an arbitrary

R

-module,

A

⊆

B

a

submodule

and

f

:

A

→

Q

a

homomorphism

. It is enough to show that

f

can be extended to

B

. For

i

∈

I

denote by

π

i

:

Q

→

Q

i

the

projection

. Since

Q

i

is injective for any

i

, then the homomorphism

π

i

∘

f

:

A

→

Q

i

can be extended to

f

i

′

:

B

→

Q

i

. Then we have

f

′

:

B

→

Q

;

f

′

⁢

(

b

)

=

(

f

i

′

⁢

(

b

)

)

i

∈

I

.

It is easy to check, that if

a

∈

A

, then

f

′

⁢

(

a

)

=

f

⁢

(

a

)

, so

f

′

is an

extension

of

f

. Thus

Q

is injective.

□

Remark.

Unfortunetly

direct sum

of

injective modules

need not be injective. Indeed, there is a

theorem

which states that direct sums of injective modules are injective if and only if ring

R

is

Noetherian

. Note that the proof presented above cannot be used for direct sums, because

f

′

⁢

(

b

)

need not be an element of the direct sum, more precisely, it is possible that

f

i

′

⁢

(

b

)

≠

0

for infinetly many

i

∈

I

. Nevertheless products are always injective.

Title

product of injective modules is injective

Canonical name

ProductOfInjectiveModulesIsInjective

Date of creation

2013-03-22 18:50:17

Last modified on

2013-03-22 18:50:17

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

5

Author

joking (16130)

Entry type

Theorem

Classification

msc 16D50

Generated on Fri Feb 9 18:39:07 2018 by

LaTeXML
central simple algebra

central simple algebra

Let

K

be a field. A

central simple algebra

A

(over

K

) is an

algebra

A

over

K

, which is finite dimensional as a

vector space

over

K

, such that

•

A

has an

identity element

, as a ring

•

A

is central: the center of

A

equals

K

(for all

z

∈

A

, we have

z

⋅

a

=

a

⋅

z

for all

a

∈

A

if and only if

z

∈

K

)

•

A

is simple: for any two sided ideal

I

of

A

, either

I

=

{

0

}

or

I

=

A

By a

theorem

of Brauer, for every central simple algebra

A

over

K

, there exists a unique (up to

isomorphism

)

division ring

D

containing

K

and a unique

natural number

n

such that

A

is isomorphic to the ring of

n

×

n

matrices with coefficients in

D

.

Title

central simple algebra

Canonical

name

CentralSimpleAlgebra

Date of creation

2013-03-22 11:49:08

Last modified on

2013-03-22 11:49:08

Owner

djao (24)

Last modified by

djao (24)

Numerical id

7

Author

djao (24)

Entry type

Definition

Classification

msc 16D60

Classification

msc 70K75

Generated on Fri Feb 9 18:39:09 2018 by

LaTeXML
completely reducible

completely reducible

A module is called

completely reducible

(or

semisimple

) if it is a

direct sum

of

irreducible

(or simple) modules.

Title

completely reducible

Canonical name

CompletelyReducible

Date of creation

2013-03-22 13:21:13

Last modified on

2013-03-22 13:21:13

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

7

Author

Mathprof (13753)

Entry type

Definition

Classification

msc 16D60

Synonym

semisimple

Defines

semisimple

Generated on Fri Feb 9 18:39:10 2018 by

LaTeXML