Show that if $R$ is principal, $N $ is pure and $Ann(x+N)= Rd$ then there exists $y \in M$ such that $x+N=y+N$ and $Ann(y)=Rd$

Question

Let $R$ a integral domain and $M$ a $R$-module. A submodule $N$ of $M$ is pure if for all $x \in M$ and $a \in R$ such that $ax \in N$, there exists $y \in N$ such that $ax=ay$.

Show that if $R$ is principal, $N $ is pure and $Ann(x+N)= Rd$ then there exists $y \in M$ such that $x+N=y+N$ and $Ann(y)=Rd$

Comments: Showed that if $N$ is adding direct then $N$ is pure.

Answer 1

Hint(s). $d(x+N)=0$ implies $dx\in N$, and since $N$ is pure there is $z\in N$ such that $dx=dz$. Take $y=x-z$.