example of free module with bases of diffrent cardinality

example of free module with bases of diffrent cardinality

Let

k

be a field and

V

be an infinite dimensional

vector space

over

k

. Let

{

e

i

}

i

∈

I

be its basis. Denote by

R

=

End

⁢

(

V

)

the

ring of endomorphisms

of

V

with standard

addition

and

composition

as a

multiplication

.

Let

J

be any set such that

|

J

|

≤

|

I

|

.

Proposition

.

R

and

∏

j

∈

J

R

are

isomorphic

as a

R

-modules.

Proof.

Let

α

:

I

→

J

×

I

be a

bijection

(it exists since

|

I

|

≥

|

J

|

and

I

is

infinite

) and denote by

π

1

:

J

×

I

→

J

and

π

2

:

J

×

I

→

I

the

projections

. Moreover let

δ

1

=

π

1

∘

α

and

δ

2

=

π

2

∘

α

.

Recall that

∏

j

∈

J

R

=

{

f

:

J

→

R

}

(with obvious

R

-module

structure

) and define a map

ϕ

:

∏

j

∈

J

R

→

R

by defining the

endomorphism

ϕ

⁢

(

f

)

∈

R

for

f

∈

∏

j

∈

J

R

as follows:

ϕ

⁢

(

f

)

⁢

(

e

i

)

=

f

⁢

(

δ

1

⁢

(

i

)

)

⁢

(

e

δ

2

⁢

(

i

)

)

.

We will show that

ϕ

is an

isomorphism

. It is easy to see that

ϕ

is a

R

-

module homomorphism

. Therefore it is enough to show that

ϕ

is

injective

and

surjective

.

1

)

Recall that

ϕ

is injective if and only if

ker

⁢

(

ϕ

)

=

0

. So assume that

ϕ

⁢

(

f

)

=

0

for

f

∈

∏

j

∈

J

R

. Note that

f

=

0

if and only if

f

⁢

(

j

)

=

0

for all

j

∈

J

and this is if and only if

f

⁢

(

j

)

⁢

(

e

i

)

=

0

for all

j

∈

J

and

i

∈

I

. So take any

(

j

,

i

)

∈

J

×

I

. Then (since

α

is

bijective

) there exists

i

0

∈

I

such that

α

⁢

(

i

0

)

=

(

j

,

i

)

. It follows that

δ

1

⁢

(

i

0

)

=

j

and

δ

2

⁢

(

i

0

)

=

i

. Thus we have

0

=

ϕ

⁢

(

f

)

⁢

(

e

i

0

)

=

f

⁢

(

δ

1

⁢

(

i

0

)

)

⁢

(

e

δ

2

⁢

(

i

0

)

)

=

f

⁢

(

j

)

⁢

(

e

i

)

.

Since

j

and

i

were arbitrary, then

f

=

0

which

completes

this part.

2

)

We wish to show that

ϕ

is onto, so take any

h

∈

R

. Define

f

∈

∏

j

∈

J

R

by the following

formula

:

f

⁢

(

j

)

⁢

(

e

i

)

=

h

⁢

(

e

α

-

1

⁢

(

j

,

i

)

)

.

It is easy to see that

ϕ

⁢

(

f

)

=

h

.

□

Corollary.

For any two numbers

n

,

m

∈

ℕ

there exists a ring

R

and a

free module

M

such that

M

has two bases with

cardinality

n

,

m

respectively.

Proof.

It follows from the proposition, that for

R

=

End

⁢

(

V

)

we have

R

n

≃

R

≃

R

m

.

For

finite set

J

module

∏

j

∈

J

R

is free with basis consisting

|

J

|

elements

(

product

is the same as

direct sum

). Therefore (due to

existence

of previous isomorphisms)

R

-module

R

has two bases, one of cardinality

n

and second of cardinality

m

.

□

Title

example of free module with bases of diffrent cardinality

Canonical name

ExampleOfFreeModuleWithBasesOfDiffrentCardinality

Date of creation

2013-03-22 18:07:18

Last modified on

2013-03-22 18:07:18

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

13

Author

joking (16130)

Entry type

Example

Classification

msc 16D40

Related topic

IBN

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