Let $R$ be an infinite commutative ring with unity (not an integral domain). Does there exist an infinite number of distinct zero-divisors $x$ and $y$ in $R$ such that $\mathrm{ann}(x)=\mathrm{ann}(y)$? If not true any counterexample?
Here, $\mathrm{ann}(x)=\{ r\in R: rx=0\}$ is an ideal of $R$.
I have a result: if $R$ is infinite, then there are infinitely many zero-divisors in $R$. Will this imply there are infinite number of elements with $\mathrm{ann}(x)=\mathrm{ann}(y)$?
No. For instance, let $R=\mathbb{F}_2^X$ for an infinite set $X$. Then all elements of $R$ have distinct annihilators, since if $f,g\in R$ are such that $f(x)=0$ and $g(x)=1$ for some $x\in X$, then $f$ is annihilated by the function that is $1$ at $x$ and $0$ everywhere else, but $g$ is not.