Let $D$ be a principal ideal domain (PID) and let $M$ be a $D$-module. A submodule $N$ of $M$ is said to be pure in $M$ if $N\cap rM=rN$ for all $r\in D$.
If $N$ is a pure submodule of $M$ and $\text{ann}(x+N)=(d)$, prove that $x$ can be chosen in its coset $x+N$ so that $\text{ann} x=(d)$.
By the definition of annihilators, we have $d(x+N)=dx+dN=0$ for $d\in D$. I am blanking right now: how does $(d)$ take part in the previous equation? Also how does the purity comes to play?
(1) For any $x' \in x + N$, we have that ${\rm ann}(x') \subset (d)$. Indeed, if $r$ annihilates $x'$ in $M$, then $r$ annihilates $x' + N = x + N$ in $M/N$, so $r$ is contained in ${\rm ann}(x + N)$, which is equal to $(d)$ by assumption.
(2) There exists a particular $x' \in x + N$ such that $(d) \subset {\rm ann}(x')$. Why? Well, we know that $d$ annihilates $x + N$ in $M/ N$. So $dx + N$ is the zero coset in $M / N$, which is to say that $dx$ is contained in $N$. Since $dx$ is obviously also in $dM$, it follows that $dx$ is contained in $N \cap dM$, which is the same as $dN$, by the purity of $N$. So there exists an $n' \in N$ such that $dx = dn'$, i.e. $d(x - n') = 0$ in $M$. Thus the element $x' := x - n'$ is an element in the coset $x + N$ that is annihilated by $d$, and hence, is annihilated by all elements of $(d)$.
Putting (1) and (2) together: if we take $x'$ as defined in $(2)$, then ${\rm ann}(x')$ must be precisely equal to $(d)$.