A generalization of upper nilradical Let $R$ be a ring not necessarily commutative and not necessarily has unity.
The lower nilradical of $R$ is defined by $\bigcap \text{prime ideal}$.
The upper nilradical of $R$ is defined by $\langle \bigcup \text{nil ideal}\rangle$.
We know that 
$$\bigcap \text{prime ideal}\subseteq \langle \bigcup \text{nil ideal}\rangle\subseteq \{r\in R\mid r^n=0\text{ for some }n\in \Bbb{N}^+\}.$$
I want to generalize the notion of upper nilradical:
Let $I$ be an ideal in $R$. Consider the ideal generated by the union of all nil ideal contains $I$.
That is, $$\langle \bigcup_{I\subseteq\text{ nil ideal}L}L\rangle.$$
We already know that 
$$\bigcap_{I\subseteq \text{ prime ideal }P}P\subseteq \{r\in R\mid r^n\in I\text{ for some }n\in \Bbb{N}^+\}.$$
Do we have 
$$\bigcap_{I\subseteq \text{ prime ideal }P}P\stackrel{?}{\subseteq} \langle \bigcup_{I\subseteq\text{ nil ideal}L}L\rangle\subseteq \{r\in R\mid r^n\in I\text{ for some }n\in \Bbb{N}^+\}.$$
Remark: I already know that the second inclusion holds.

Consider $R=F\times F$ for a field $F$, and let $I=F\times \{0\}$.
The intersection of primes over $I$ is just $I$ itself, and $I$ contains an idempotent, so no ideal containing it can be nil. So the ideal generated by the union is the ideal generated by the emptyset, which is the zero ideal of $R$. So the containment in question does not hold.
I think what you actually want is to replace that set with the ideal $A$ such that $A/I=Nil^\ast(R/I)$. This pattern has been followed already for a long time: if you have some sort of radical for a ring, you can define radicals of ideals by looking at the radicals of the quotient ring.
