Let $M$ be a (left) module over an integral domain $R$ and $N_{1}$, $N_{2}$ be submodules of $M$ such that $M = N_{1}\oplus N_{2}$. Let $\operatorname{Tor}(M)$ denote the torsion module of $M$. Then show that $\operatorname{Tor}(M) = \operatorname{Tor}(N_{1})\oplus\operatorname{Tor}(N_{2})$
My Attempt:
Let $n_{1}\in N_{1}$ and $n_{2}\in N_{2}$ be the torsion
elements,
$\implies \exists r,s \in R \setminus \{0\}$ such that
$rn_{1}=0$ and $sn_{1}=0$.
Now, every element of $M$ can be
written as $(n_{1},n_{2})$, therefore $rs(n_{1},n_{2})=0$ because $R$
is an integral domain.
$\implies$ $\operatorname{Tor}(M)\supset\operatorname{Tor}(N_{1})\oplus\operatorname{Tor}(N_{2})$
Let $m\in M$ be a torsion element of $M$.
$\implies \exists r_{1}\in R$ such that $r_{1}m=0$.
Now, $m$ can be written as $m=(n_{1}',n_{2}')$ where $n_{1}'\in N_{1}$ and $n_{2}'\in N_{2}$.
$\implies$ $n_{1}'$ and $n_{2}'$ are torsion elements of $N_{1}$ and $N_{2}$ respectively.
$\implies$ $\operatorname{Tor}(M)\subset\operatorname{Tor}(N_{1})\oplus\operatorname{Tor}(N_{2})$
Combining both we get $\operatorname{Tor}(M)=\operatorname{Tor}(N_{1}) \oplus\operatorname{Tor}(N_{2})$.
Can anyone spot my mistake and please give me hints to correct it and solve the question?
The idea is good, but it can be expressed better.
Let $(x,y)\in\operatorname{Tor}(M)$. Then there exist $r\in R\setminus\{0\}$ such that $r(x,y)=0$, which is the same as $rx=0$ and $ry=0$. Therefore $x\in\operatorname{Tor}(N_1)$ and $y\in\operatorname{Tor}(N_2)$. Hence $(x,y)\in\operatorname{Tor}(N_1)\oplus\operatorname{Tor}(N_2)$.
Conversely, if $x\in\operatorname{Tor}(N_1)$ and $y\in\operatorname{Tor}(N_2)$, then $rx=0$ and $sy=0$, for some $r,s\in R\setminus\{0\}$. Since $R$ is an integral domain, we have $rs\ne0$; moreover $$ rs(x,y)=(s(rx),r(sy))=(s0,r0)=(0,0) $$ so $(x,y)\in\operatorname{Tor}(M)$.