# Show that if $M/M'$ is a torsion module over an integral domain then $M$ is a torsion module

Let's define $M$ a left $K$-module and $M'$ a submodule of $M$. We suppose $M'$ and $M/M'$ are torsion modules. I want to show that if $K$ is an integral domain then $M$ is a torsion module.

According to the definition if $M'$ is a torsion, then for each $x \in M'$ there exists a $k \in K$ such that $km=0$.

We also know that there is a natural map $\pi:M \rightarrow M/M'$ defined as $\pi(m)=m+M'$, which satisfied $\pi(m+t)=\pi(m)+\pi(t)$, and $\pi(\lambda m) = \lambda \pi(m)$.

But i'm not sure how to combine this to show that for each $m\in M$ there exists an element $r$ in $K$ such that $rm=0$. Can anyone provide a hint or an answer?

• You don't need a single $r \in K$, you need to prove that, for each $m \in M$, there exists an $0 \ne r \in K$, which can depend on $m$, such that $rm=0$. Note that there exists $0 \ne s$ such that $sm \in M'$ and then $0 \ne t$ such that $tsm=0$, so you can choose $r=ts$. – Derek Holt Mar 29 '17 at 11:33
• thank you, i modified the question – Xaving Mar 29 '17 at 12:17

Since $M/M'$ is torsion, for every $m\in M$, there exists $k\in K$ such that $k\pi(m)=0$, you deduce that $km\in M'$, since $M'$ is torsion, $k'km=0, k'\in K$, since $K$ is integral, $k'k\neq 0$.