I am studying for a comprehensive exam and looking at a large bank of problems. One problem has six statements about module and asks for a a proof or a counter-example of the statements. I am able to solve all except two related to torsion . Assume that R is an integral domain and the modules below are R-modules. So an element $m \in M$ is torsion if there is an $r \in R-0$ s.t. $rm=0$.

  1. A submodule of a free module is torsion-free.
  2. A submodule of a torsion module is a torsion module.

For #1, I know that a submodule of a free module is not necessarily free and I know that a free module is torsion free but I can't put these to use to find a counterexample. For #2, this seems logical but again I am unable to provide a proof. By the way, to be clear a torsion module has only torsion elements. Thanks!

  • 1
    $\begingroup$ Let $F$ the free module, and $S$ the submodule. Suppose $t \in S$ is a torsion element. Considering $t \in F$, what does that say about $t$? $\endgroup$ – Daniel Fischer Aug 16 '13 at 20:00
  • $\begingroup$ @Leo: rschwieb gave you two options and you answered "Yes". Also, it would be helpful to state this in the question itself rather than to only clarify in the comments. $\endgroup$ – RghtHndSd Aug 16 '13 at 20:07
  • $\begingroup$ @rghthndsd Assume that R is an integral domain and the modules below are R-modules. When he first posted the question there wasn't two options or at least it didn't show up on my screen. Although it turns out my response was wrong. I didn't see in my question that R was an integral domain. Also, I have now fixed the question so let us delete out comments so we don't confuse others. $\endgroup$ – Leo Spencer Aug 16 '13 at 20:13
  • $\begingroup$ @LeoSpencer Thanks for clarifying. If they are all domains then things are simplified considerably :) $\endgroup$ – rschwieb Aug 16 '13 at 20:14
  • $\begingroup$ @DanielFischer good hint so I think the answer should true for #1 with proof $\endgroup$ – Leo Spencer Aug 16 '13 at 20:18

Number 1 is true.

Consider $r \in R-0$ and $t \in S \subset M$. Thus, since $M$ is free. $t=r_1m_1+ \ldots + r_nm_n$. Assume $rt=0$. Then $rr_1m_1+ \ldots + rr_nm_n$, then $rr_i=0$ and since $R$ is an integral domain, then $r_i=0$ implying $t=0$ and $S$ is torsion-free. QED

Thanks to @danielfischer for a good hint.

Number 2 is true too.

Consider $t \in S \subset M$, since $t\in M$, $\exists r \in R-0$ s.t. $rt=0$. Thus, S is a torsion module. QED

I'm not sure how I missed this one to begin with, had a brain freeze I guess.

  • 1
    $\begingroup$ But #2 is also clear, right? If $mr=0$ for some $m\neq 0$ in a submodule and $r\neq 0$ in the ring, then of course the same $mr=0$ in the big module. $\endgroup$ – rschwieb Aug 16 '13 at 20:30
  • $\begingroup$ yes i meant that I was going to have to still put it here not figure it out $\endgroup$ – Leo Spencer Aug 16 '13 at 20:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.