Prove that if R is an integral domain then Torsion of M is a submodule of R-module M I can't quite get this one.  I can show that two non-zero elements m,n of M have a non-zero product but they belong to the R-module, not the integral domain so I don't know if it's necessarily true that their product be nonzero.
 A: Your prove should go like this: If $m_1,m_2$ are torsion elements, then there exist two elements $a_1,a_2$ of $R$, $a_1\neq 0, a_2 \neq 0$ such that
$$
a_1m_1=0 \qquad a_2 m_2=0
$$
have to show: for any $r_1,r_2 \in R$ the Element $m:=r_1m_1+r_2m_2 \in M$ is a torsion element.
first note that $a_1a_2 \neq 0$ because $R$ is a domain
and this element anihilates the linear combination (R commutative)
$$
a_1a_2 (r_1m_1+r_2m_2) = a_1a_2r_1m_1+a_1a_2r_2m_2 = a_2r_1(a_1m_1)+a_1r_2(a_2m_2)=0+0=0
$$
So you have found a non-zero Element ($a:=a_1a_2$) such that $am=0$. 
A: Let $M$ be an left $R$-module such that $r\cdot m=rm$ is well defined. Then if $m_{1},m_{2}$ are torsion elements, $r_{1}m_{1}+r_{2}m_{2}$ must also be a torsion element for all $r_{1},r_{2}\in R$. We need to show that $M_{1}$ generated by $m$ of the kind
$$
m=\sum^{n}_{i=1}r_{i}m_{i}, \exists R_{i}m_{i}=0,r_{i},R_{i}\in R
$$
is a well defined submodule. But this just follows from definition. Note that we used "integral domain" implicitly as $R_{i}\in R$ are not zero divisors. 
