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I have been trying to work out a proof of structure theorem for finitely generated modules over PID's using localization. This is what my plan is:

1.Prove that every finitely generated torsion free module over a PID is free.

2.Consider the canonical map $f:M\rightarrow M/M_{tor}$. It is easy to see that $M/M_{tor}$ is torsion free, so using (1) it is free(call it $F$). It turns out that $M$ is isomorphic to $kerf\bigoplus F$ and since $kerf$ is $M_{tor}$, $M$ is $M_{tor} \bigoplus F$.

So if (1) is done the target is almost fulfilled except for the structure of $M_{tor}$.

I am able to prove that if $E$ is a finitely generated torsion free module over a PID(say $R$) and $S^{-1}E$ is isomorphic to $K$(the quotient field of $R$) then $E$ is isomorphic to $R$. Where $S = R-{0}$

Now, if we can find a sub-module $L$ of $E$ such that $S^{-1}(E/L)$ is isomorphic to $K$ then I can proceed by induction to prove (1).

So any ideas how to find this $L$?

Thanks

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  • $\begingroup$ Why don't you take a sub-module $E^\prime$ of $E$ generated by a nonzero element of $E$? Since $E$ is torsion-free, such an $E^\prime$ is isomorphic to $R$, so that $S^{-1}E^\prime\cong K$, and you can use induction, using $E^\prime$ and $E/E^\prime$. $\endgroup$
    – Mahdi Majidi-Zolbanin
    Commented Feb 28, 2012 at 16:14
  • $\begingroup$ @Mahdi Majidi-Zolbanin So if we can prove $E/E'$ is torsion free with lesser generators than the minimal no.of generators of $E$ then we are done. Right? $\endgroup$
    – Dinesh
    Commented Feb 28, 2012 at 16:45
  • $\begingroup$ @Dinesh: I think it suffices to show $E/E^\prime$ is torsion-free. You can choose the generator of $E^\prime$ to be one of the elements of a minimal set of generators of $E$. That way $E/E^\prime$ can be generated with one less than number of generators of $E$. $\endgroup$
    – Mahdi Majidi-Zolbanin
    Commented Feb 28, 2012 at 17:02
  • $\begingroup$ I think this approach is too naive, and you don't get some torsion-free quotient for free ... $\endgroup$
    – Martin Brandenburg
    Commented Feb 28, 2012 at 19:13
  • $\begingroup$ I agree with Martin, it wont work like this. $\endgroup$
    – Mahdi Majidi-Zolbanin
    Commented Feb 28, 2012 at 20:30

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