finitely generated modules over a principal ideal domain
Let be a principal ideal domain and let be a finitely generated
-
module.
Proof.
For this is clear, since is an ideal of and is generated by some element . Now suppose that the statement is true for all submodules of .
For a submodule of we define by . The image of is an ideal in . If , then . Otherwise, . In the first case, elements of can be bijectively mapped to by the function given by ; so the image of under this mapping is a submodule of , which by the induction hypothesis is finitely generated and free.
Now let such that and with .
Then ,
which is equivalent to
which is isomorphic
to a submodule of .
This shows that .
Let be a basis of .
By assumption, .
We’ll show that is linearly independent
.
So let .
The first component
of the are 0,
so the first component of must also be 0.
Since is a multiple of and , then .
Since are linearly independent,
is a generating set of with elements.
∎
Corollary.
If is a finitely generated -module over a PID generated by elements and is a submodule of , then can be generated by or fewer elements.
Proof.
Let be a generating set of
and , .
Then the inverse image of is a submodule of ,
and according to lemma Lemma. can be generated by or fewer elements.
Let be a generating set of ;
then , and since is surjective
,
is a generating set of .
∎
Theorem.
Let be a finitely generated module over a principal ideal domain .
- (I)
-
Note that is torsion-free, that is, . In particular, if is torsion-free, then is free.
- (II)
-
Let be a proper submodule of . Then there exists a finitely generated free submodule of such that .
Proof of (I): Let . For , denotes the coset modulo generated by . Let be a torsion element of , so there exists such that , which means . But then is a member of , and this implies that has no non-zero torsion elements (which is obvious if ).
Now let be a finitely generated torsion-free -module.
Choose a maximal linearly independent subset of ,
and let be the submodule of generated by .
Let be a set of generators of .
For each there is a non-zero
such that .
Put .
Then is non-zero,
and we have for each .
As is torsion-free, the multiplication by is injective
,
so .
So is isomorphic to a submodule of a free module
, and is therefore free.
Proof of (II): Let be defined by . Then is surjective, so can be chosen such that , where the ’s are a basis of . If , then . Since are linearly independent in it follows . So the submodule spanned by of is free.
Now let be some element of and . This is equivalent to . Hence, any is a sum of the form , for some and . Since is torsion-free, , and it follows that .
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 |