|
|
|
|
finitely generated modules over a principal ideal domain
|
(Topic)
|
|
|
Let be a principal ideal domain and let be a finitely generated - module.
Lemma Let be a submodule of the -module . Then is free and finitely generated by elements.
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
. Since is surjective, 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 ![[*] [*]](http://images.planetmath.org:8080/cache/objects/4680/l2h//usr/share/latex2html/icons/crossref.png) 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 with generating set
. The homomorphism
defined by
is injective since is torsion-free. Let
be the standard basis of . Then the elements
are linearly independent in . Now let
where
. Then the cosets can be identified with the elements of , and the statement follows.
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
.
|
"finitely generated modules over a principal ideal domain" is owned by yark. [ full author list (2) | owner history (1) ]
|
|
(view preamble | get metadata)
Cross-references: sum, spanned by, standard basis, injective, implies, torsion element, coset, proof, torsion-free, finitely generated module, inverse image, multiple, component, linearly independent, basis, isomorphic, equivalent, induction hypothesis, mapping, function, image, surjective, ideal, clear, submodule, module, finitely generated, principal ideal domain
There is 1 reference to this entry.
This is version 14 of finitely generated modules over a principal ideal domain, born on 2003-09-01, modified 2008-09-11.
Object id is 4680, canonical name is FinitelyGeneratedModulesOverAPrincipalIdealDomain.
Accessed 3529 times total.
Classification:
| AMS MSC: | 13E15 (Commutative rings and algebras :: Chain conditions, finiteness conditions :: Rings and modules of finite generation or presentation; number of generators) |
|
|
|
|
|
|
Pending Errata and Addenda
|
|
|
|
|
|
|
|
|
|
|