Let $R$ be a commutative ring with identity. If $\{0_R\}$ is primary ideal of $R$, then $Nil(R)=D(R)$, where $Nil(R)$ denote the nilradical of $R$ and $D(R)$ denote the set of the all divisor zero of $R$.
I was able to prove that $Nil(R) \subset D(R)$. But I could not prove the contrary.
Suppose $x \in D(R)$. By definition, there exists $y\in R$ such that $yx = 0_R$ and $y \neq 0_R$. Since $\{0_R\}$ is a primary ideal of $R$, $yx = 0_R \in \{0_R\}$ implies $y \in \{0_R\}$ or $x^n \in \{0_R\}$ for some $n \in \mathbb{Z}_{> 0}$. Since $y \neq 0_R$, we may conclude that $x^n = 0_R$ for some $n \in \mathbb{Z}_{> 0}$. Therefore, by definition, $x \in Nil(R)$.