The set of zero divisors is the union of radicals of annihilators

Question

I am trying to figure out why the statement $$\text{the set of zero divisors }=\bigcup_{0\ne x\in R} \sqrt{\text{Ann}(x)}$$ is true. Here $R$ is a commutative ring, $\text{Ann}(x)=\{r\in R\mid rx=0\}$ the annihilator of $x\in R$ and $\sqrt I=\{r\in R\mid \exists n:\ r^n\in I\}$ the radical of some ideal $I\lhd R$.

I can certainly see that the set of zero divisors is $\bigcup_{0\ne x\in R}\text{Ann}(x)$, but is it still true if we take the radical of the annihilator?

Answer 1

If $r^n \in \mathrm{Ann}(x)$ and $r$ is not a zero divisor, then show by descending induction that $r^k x =0$ for $k \leq n$, in particular $x=0$, contradiction.