Finitely generated modules over a PID, and applications on abelian groups

1. May 24, 2010

Jösus

Hello!

I'm currently taking a course in group- and ring theory, and we are now dealing with a chapter on finitely generated modules over PIDs. I have stumbled across some problems that I can't really get my head around. It is one in particular that I would very much like to understand, and I would greatly appreciate some help.

The situation is the following. Assume we are given a free module, for example the $$\mathbb{Z}$$-module $$\mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}$$, and want to consider the submodule generated by all elements $$\left(x_{1},x_{2},x_{3}\right)$$ satisfying certain relations $$\sum_{1}^{3}{a_{i,j}x_{j}} = 0$$ for $$i=1,2$$. I have heard that a good way of understanding this submodule would be to consider the matrix $$(a_{i,j})_{2,3}$$, put it in smith normal form and then conclude that the entries on the generalized diagonal will be the torsion coefficients in the decomposition of the submodule as a direct sum of cyclic ones (as in the structure theorem). However, I have not managed to draw that conclusion. If someone could explain why this holds true, or if it is false tell me that (and perhaps give me a hint on how to understand said submodule), I would be extremely thankful.

2. May 24, 2010

lavinia

The kernel of the linear map,

$$(a_{i,j})_{2,3}$$

is a full free submodule of the original free module,M, and thus is generated by a linearlly independent set of vectors that can be extended to a basis for M.
The Smith form will tell you which basis vectors are in the kernel - after a suitable change of basis.

- Why is the submodule free? - because any submodule of a free module over a PID is free

- why is the submodule full? because if A.(rV) = 0 for any r in the PID the A.V = 0 by linearity.

does this seem right?