Came across this question while reviewing for an exam and I'm pretty stumped. We let $\displaystyle R$ be a Principal Ideal Domain (P.I.D.) and $\displaystyle M$ be a free module of rank 2 with some basis set $\displaystyle \{b_1,b_2\}$. Then $\displaystyle \forall m \in M$ define $\displaystyle Rm$ to be the submodule of $\displaystyle M$ generated by $\displaystyle m$ (this is also a free module since $\displaystyle R$ is a P.I.D.). Let $\displaystyle m=xb_1+yb_2 \in M$ for some $\displaystyle x,y \in R$. For $\displaystyle x=y$, show that $\displaystyle Rm$ is a direct summand of $\displaystyle M$ (i.e there exists some submodule $\displaystyle N$ of $\displaystyle M$ such that $\displaystyle M = Rm \bigoplus N$) iff $\displaystyle x$ is invertible in $\displaystyle R$. Furthermore for any $\displaystyle x,y$, $\displaystyle Rm$ is a direct summand of $\displaystyle M$ iff the ideal $\displaystyle (a,b)=R$. If $\displaystyle R=\mathbb{Z}$, $\displaystyle x=3$, and $\displaystyle y=5$, find some $\displaystyle v \in M$ such that $\displaystyle \mathbb{Z}m \bigoplus \mathbb{Z}v = M$. Any help here would be appreciated thanks.