# A submodule of a free module is torsion-free?

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!

• 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 $t$? – Daniel Fischer Aug 16 '13 at 20:00
• @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. – RghtHndSd Aug 16 '13 at 20:07
• @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. – Leo Spencer Aug 16 '13 at 20:13
• @LeoSpencer Thanks for clarifying. If they are all domains then things are simplified considerably :) – rschwieb Aug 16 '13 at 20:14
• @DanielFischer good hint so I think the answer should true for #1 with proof – 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.

• 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. – rschwieb Aug 16 '13 at 20:30
• yes i meant that I was going to have to still put it here not figure it out – Leo Spencer Aug 16 '13 at 20:33