Recall that a submodule $N$ of a module $M$ is called small if $N+K=M$ implies that $K=M$ for any submodule $K$ of $M$.
Recall that for any module $M$ over a ring $R$, $ann(M)=\{r\in R|rM=0\}$.
The following result has been used in a reasearch paper (title: small compressible and small retractable modules)
Result: Let $M=M_1\oplus M_2$ be a module over a ring $R$ with identity $1$ such that $ann(M_1)+ann(M_2)=R$. If $N$ is a small submodule of $M$, then there exists small submodules $K_1$ in $M_1$ and $K_2$ in $M_2$ such that $N=K_1\oplus K_2$.
I have no idea how to prove this result. Suggest any idea if possible. It will be helpful to me.
Best Answer
Fix $r\in \operatorname{ann}(M_1)$ and $s\in\operatorname{ann}(M_2)$ such that $r+s=1$, and let $K_1=sN$ and $K_2=rN$. Since $r+s=1$, $K_1+K_2=N$, and $K_1\subseteq M_1$ since $s$ annihilates $M_2$ and $K_2\subseteq M_2$ since $r$ annihilates $M_1$.
It remains to show that $K_i$ is small in $M_i$; by symmetry it suffices to show this for $i=1$. Suppose $K\subseteq M_1$ and $M_1=K+K_1$. Then $M=K+K_1+M_2$. Since $K_1\subseteq N$ and $N$ is small in $M$, this implies $K+M_2=M$. Since $K\subseteq M_1$ and $M$ is the direct sum of $M_1$ and $M_2$, this implies $K$ is all of $M_1$.