finitely generated modules over a principal ideal domain
generated by \PMlinkescapephrasegenerating set
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.
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 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|
|Date of creation||2013-03-22 13:55:22|
|Last modified on||2013-03-22 13:55:22|
|Last modified by||yark (2760)|