finitely generated modules over a principal ideal domain

finitely generated modules over a principal ideal domain

\PMlinkescapephrase

generated by

\PMlinkescapephrase

generating set

Let

R

be a

principal ideal domain

and let

M

be a

finitely generated

R

-
module.

Lemma

.

Let

M

be a

submodule

of the

R

-module

R

n

. Then

M

is
free and finitely generated by

s

≤

n

elements

.

Proof.

For

n

=

1

this is clear,
since

M

is an ideal of

R

and is generated by some element

a

∈

R

.
Now suppose that the statement is true
for all submodules of

R

m

,

1

≤

m

≤

n

-

1

.

For a submodule

M

of

R

n

we define

f

:

M

→

R

by

(

k

1

,

…

,

k

n

)

↦

k

1

.
The image of

f

is an ideal

ℑ

in

R

.
If

ℑ

=

{

0

}

, then

M

⊆

ker

⁡

(

f

)

=

(

0

)

×

R

n

-

1

.
Otherwise,

ℑ

=

(

g

)

,

g

≠

0

.
In the first case, elements of

ker

⁡

(

f

)

can be bijectively mapped to

R

n

-

1

by the

function

ker

⁡

(

f

)

→

R

n

-

1

given by

(

0

,

k

1

,

…

,

k

n

-

1

)

↦

(

k

1

,

…

,

k

n

-

1

)

;
so the image of

M

under this

mapping

is a submodule of

R

n

-

1

,
which by the

induction hypothesis

is finitely generated and free.

Now let

x

∈

M

such that

f

⁢

(

x

)

=

g

⁢

h

and

y

∈

M

with

f

⁢

(

y

)

=

g

.
Then

f

⁢

(

x

-

h

⁢

y

)

=

f

⁢

(

x

)

-

f

⁢

(

h

⁢

y

)

=

0

,
which is

equivalent

to

x

-

h

⁢

y

∈

ker

⁡

(

f

)

∩

R

n

:=

N

which is

isomorphic

to a submodule of

R

n

-

1

.
This shows that

R

⁢

x

+

N

=

M

.

Let

{

g

1

,

…

,

g

s

}

be a basis of

N

.
By

assumption

,

s

≤

n

-

1

.
We’ll show that

{

x

,

g

1

,

…

,

g

s

}

is

linearly independent

.
So let

r

⁢

x

+

∑

i

=

1

s

r

i

⁢

g

i

=

0

.
The first

component

of the

g

i

are 0,
so the first component of

r

⁢

x

must also be 0.
Since

f

⁢

(

x

)

is a

multiple

of

g

≠

0

and

0

=

r

⋅

f

⁢

(

x

)

, then

r

=

0

.
Since

{

g

1

,

…

,

g

s

}

are linearly independent,

{

x

,

g

1

,

…

,

g

s

}

is a generating set of

M

with

s

+

1

≤

n

elements.
∎

Corollary

.

If

M

is a finitely generated

R

-module
over a PID generated by

s

elements
and

N

is a submodule of

M

,
then

N

can be generated by

s

or fewer elements.

Proof.

Let

{

g

1

,

…

,

g

s

}

be a generating set of

M

and

f

:

R

s

→

M

,

(

r

1

,

…

,

r

s

)

↦

∑

i

=

1

s

r

i

⁢

g

i

.
Then the

inverse image

N

′

of

N

is a submodule of

R

s

,
and according to lemma

Lemma

.

can be generated by

s

or fewer elements.
Let

n

1

,

…

,

n

t

be a generating set of

n

′

;
then

t

≤

s

, and since

f

is

surjective

,

f

⁢

(

n

1

)

,

…

,

f

⁢

(

n

t

)

is a generating set of

N

.
∎

Theorem

.

Let

M

be a finitely generated module over a principal ideal domain

R

.

(I)

Note that

M

/

tor

⁡

(

M

)

is

torsion-free

, that is,

tor

⁡

(

M

/

tor

⁡

(

M

)

)

=

{

0

}

.
In particular, if

M

is torsion-free, then

M

is free.

(II)

Let

tor

⁡

(

M

)

be a

proper submodule

of

M

.

Then there exists a finitely generated free submodule

F

of

M

such that

M

=

F

⊕

tor

⁡

(

M

)

.

Proof of (I): Let

T

=

tor

⁡

(

M

)

.
For

m

∈

M

,

m

¯

denotes the coset modulo

T

generated by

m

.
Let

m

be a

torsion element

of

M

/

T

,
so there exists

α

∈

R

∖

{

0

}

such that

α

⋅

m

¯

=

0

,
which means

α

⋅

m

¯

⊆

T

.
But then

α

⋅

m

is a member of

T

,
and this implies that

M

/

T

has no non-zero torsion elements
(which is obvious if

M

=

tor

⁡

(

M

)

).

Now let

M

be a finitely generated torsion-free

R

-module.
Choose a

maximal

linearly independent subset

S

of

M

,
and let

F

be the submodule of

M

generated by

S

.
Let

{

m

1

,

…

,

m

n

}

be a set of

generators

of

M

.
For each

i

=

1

,

…

,

n

there is a non-zero

r

i

∈

R

such that

r

i

⋅

m

i

∈

F

.
Put

r

=

∏

i

=

1

n

r

i

.
Then

r

is non-zero,
and we have

r

⋅

m

i

∈

F

for each

i

=

1

,

…

,

n

.
As

M

is torsion-free, the

multiplication

by

r

is

injective

,
so

M

≅

r

⋅

M

⊆

F

.
So

M

is isomorphic to a submodule of a

free module

, and is therefore free.

Proof of (II): Let

π

:

M

→

M

/

T

be defined by

a

↦

a

+

T

.
Then

π

is surjective,
so

m

1

,

…

,

m

t

∈

M

can be chosen such that

π

⁢

(

m

i

)

=

n

i

,
where the

n

i

’s are a basis of

M

/

T

.
If

0

M

=

∑

i

=

1

t

a

i

⁢

m

i

, then

0

n

=

∑

i

=

1

t

a

i

⁢

n

i

.
Since

n

1

,

…

,

n

t

are linearly independent in

N

it follows

0

=

a

1

=

…

=

a

t

.
So the submodule spanned by

m

1

,

…

,

m

t

of

M

is free.

Now let

m

be some element of

M

and

π

⁢

(

m

)

=

∑

i

=

1

t

a

i

⁢

n

i

.
This is equivalent to

m

-

(

∑

i

=

1

t

a

i

⁢

n

i

)

∈

ker

⁡

(

π

)

=

T

.
Hence, any

m

∈

M

is a sum of the form

f

+

t

,
for some

f

∈

F

and

t

∈

T

.
Since

F

is torsion-free,

F

∩

T

=

{

0

}

,
and it follows that

M

=

F

⊕

T

.

Title

finitely generated modules over a principal ideal domain

Canonical name

FinitelyGeneratedModulesOverAPrincipalIdealDomain

Date of creation

2013-03-22 13:55:22

Last modified on

2013-03-22 13:55:22

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

21

Author

yark (2760)

Entry type

Topic

Classification

msc 13E15

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