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

  1. May 24, 2010 #1

    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 [tex]\mathbb{Z}[/tex]-module [tex]\mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}[/tex], and want to consider the submodule generated by all elements [tex]\left(x_{1},x_{2},x_{3}\right)[/tex] satisfying certain relations [tex]\sum_{1}^{3}{a_{i,j}x_{j}} = 0[/tex] for [tex]i=1,2[/tex]. I have heard that a good way of understanding this submodule would be to consider the matrix [tex](a_{i,j})_{2,3}[/tex], 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.

    Thanks in advance!
  2. jcsd
  3. May 24, 2010 #2


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    I am not sure about this but this is my thought.

    The kernel of the linear map,


    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?
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