Let $R$ be a commutative Noetherian ring and $I\subseteq R$ an ideal. Then $\operatorname{Ann}(I)=0$ if and only if $I$ contains a non zero divisor.
$\Leftarrow$ is trivial, if $I$ contains a non zero divisor then its annilhator must be zero.
What's trickier is $\Rightarrow$
$\DeclareMathOperator{\ann}{ann}$
The statement is true. Here is a sketch of the proof of the nontrivial direction using standard Noetherian ring theory.
Let $R$ be a commutative Noetherian ring. Recall than an associated prime ideal of $R$ is a prime ideal of the form $P=\ann(x)$ for some $x\in R$. It is well-known that the set of zero divisors of $R$ is the union of the associated prime ideals of $R$ and that any ideal consisting of zero divisors is contained in an associated prime ideal. It follows immediately that if $I$ is an ideal of zero divisors, then $xI=0$ for some nonzero $x\in R$.
This fails if we only assume $R$ is a commutative ring and $I$ is a finitely generated ideal. See this answer.
