Proposition: An ideal $I$ of the ring $R$ is a semiprime ideal if and only if the quotient ring $R/I$ has no nonzero nilpotent elements.
Proof?: Let $I$ be an ideal of $R$ s.t. $I=\sqrt{I}$; that is, $\forall r\in R$ such that $r^n \in I$, for some $n \in \Bbb{Z}^+$, r must be in $I$, as well. Thus, $\forall \pi \in R$ such that $\pi + I \neq I$, $\forall k \in \Bbb{Z}^+$, $(\pi + I)^k = \pi^k + I \neq I$, because $\pi^k$ cannot be in $I$. Conversely, if $R/I$ has no nonzero nilpotent elements, then $\forall k \in \Bbb{Z}^+, \forall r \in R$, $(r+I)^k=r^k + I\neq I$. So $r^k \notin I$. Note that $r+I \neq I \Leftrightarrow r \notin I$. Thus, if $r \notin I$, then $r^k \notin I$, $\forall k \in \Bbb{Z}^+$. In other words, $\forall r \in R$, $r^k \in I \Leftrightarrow r \in I$. Hence, $I$ is a semiprime ideal of $R$.