rings whose every module is free

rings whose every module is free

Recall that if

R

is a (nontrivial) ring and

M

is a

R

-module, then (nonempty) subset

S

⊆

M

is called

linearly independent

if for any

m

1

,

…

,

m

n

∈

M

and any

r

1

,

⋯

,

r

n

∈

R

the equality

r

1

⋅

m

1

+

…

+

r

n

⋅

m

n

=

0

implies that

r

1

=

…

=

r

n

=

0

. If

S

⊆

M

is a linearly independent subset of

generators

of

M

, then

S

is called a basis of

M

. Of course not every module has a basis (it even doesn’t have to have linearly independent subsets).

R

-module is called free, if it has basis. In particular if

R

is a field, then it is well known that every

R

-module is free. What about the

converse

?

Proposition

.

Let

R

be a

unital ring

. Then

R

is a

division ring

if and only if every left

R

-module is free.

Proof.

,,

⇒

” First assume that

R

is a divison ring. Then obviously

R

has only two (left) ideals, namely

0

and

R

(because every nontrivial ideal contains invertible element and thus it contains

1

, so it contains every element of

R

). Let

M

be a

R

-module and

m

∈

M

such that

m

≠

0

. Then we have

homomorphism

of

R

-modules

f

:

R

→

M

such that

f

⁢

(

r

)

=

r

⋅

m

. Note that

ker

⁢

(

f

)

≠

R

(because

f

⁢

(

1

)

≠

0

) and thus

ker

⁢

(

f

)

=

0

(because

ker

⁢

(

f

)

is a

left ideal

). It is clear that this implies that

{

m

}

is linearly independent subset of

M

. Now let

Λ

=

{

P

⊆

M

|

P

⁢

is linearly independent

}

.

Therefore we proved that

Λ

≠

∅

. Note that

(

Λ

,

⊆

)

is a poset (where ,,

⊆

” denotes the

inclusion

) in which every chain is bounded. Thus we may apply Zorn’s lemma. Let

P

0

∈

Λ

be a

maximal element

in

Λ

. We will show that

P

0

is a basis (i.e.

P

0

generates

M

). Assume that

m

∈

M

is such that

m

∉

P

0

. Then

P

0

∪

{

m

}

is linearly dependent (because

P

0

is

maximal

) and thus there exist

m

1

,

⋯

,

m

n

∈

M

and

λ

,

λ

1

,

⋯

,

λ

n

∈

R

such that

λ

≠

0

and

λ

⋅

m

+

λ

1

⋅

m

1

+

⋯

⁢

λ

n

⋅

m

n

=

0

.

Since

λ

≠

0

, then

λ

is

invertible

in

R

(because

R

is a divison ring) and therefore

m

=

(

-

λ

-

1

⁢

λ

1

)

⋅

m

1

+

⋯

+

(

-

λ

-

1

⁢

λ

n

)

⋅

m

n

.

Thus

P

0

generates

M

, so every

R

-module is free. This

completes

this

implication

.

,,

⇐

” Assume now that every left

R

-module is free. In particular every left

R

-module is projective, thus

R

is

semisimple

and therefore

R

is

Noetherian

. This implies that

R

has

invariant basis number

. Let

I

⊆

R

be a nontrivial left ideal. Thus

I

is a

R

-module, so it is free and since all modules are projective (because they are free), then

I

is

direct summand

of

R

. If

I

is proper, then we have a

decomposition

of a

R

-module

R

≃

I

⊕

I

′

,

but rank of

R

is

1

and rank of

I

⊕

I

′

is at least

2

.

Contradiction

, because

R

has invariant basis number. Thus the only left ideals in

R

are

0

and

R

. Now let

x

∈

R

. Then

R

⁢

x

=

R

, so there exists

β

∈

R

such that

β

⁢

x

=

1

.

Thus every

element

is

left invertible

. But then every element is invertible. Indeed, if

β

⁢

x

=

1

then there exist

α

∈

R

such that

α

⁢

β

=

1

and thus

1

=

α

⁢

β

=

α

⁢

(

β

⁢

x

)

⁢

β

=

(

α

⁢

β

)

⁢

x

⁢

β

=

x

⁢

β

,

so

x

is right invertible. Thus

R

is a divison ring.

□

Remark.

Note that this proof can be dualized to the case of

right modules

and thus we obtained that a unital ring

R

is a divison ring if and only if every right

R

-module is free.

Title

rings whose every module is free

Canonical name

RingsWhoseEveryModuleIsFree

Date of creation

2013-03-22 18:50:20

Last modified on

2013-03-22 18:50:20

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

10

Author

joking (16130)

Entry type

Example

Classification

msc 16D40

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