Let $R$ be a commutative ring with identity and let $I$ be an ideal of $R$. Define $\operatorname{Rad}(I)=\{a\in R:\exists n\in\mathbb N, a^n\in I\}$. Show that $Rad(I)$ is the intersection of all prime ideals $P$ of $R$ containing $I$.
I could prove that $$\operatorname{Rad}(I)\subset \bigcap_{I\subset P \text{ prime}}P$$
However I am not able to prove the other direction. I found the same question has been asked before but I do not follow the arguments placed there. I know nothing about localization or that of maximal ideal disjoint from a subring, etc. I would be obliged if someone helps me to prove this in the simplest terms possible.
Here's the link to the "identical" question: $\operatorname{rad}(I)=\bigcap_{I\subset P,~P\text{ prime}}P$
I don't actually get any of the answers, from the first line themselves.
Let $x\notin Rad(I)$. By definition of $Rad(I)$ you know $S=\{1,x,x^2,x^3,\ldots\}$ is a multiplicatively closed set disjoint from $I$.
At this point, I highly recommend you get your head around the idea in item 1 of this post. It is a standard bit of commutative algebra that you should not avoid. It is literally Theorem 1 on page 1 of Kaplansky's Commutative Rings, so you should not be afraid of learning it. You don't have to go clear into localization, but it is well worth your time to get this bit down.
After you understand why
it is not hard to adapt the proof to prove that
After that is established and you find your prime $P$, you'll have to see why $x\notin P$. At that point, you will have established that $x\notin Rad(I)$ implies $x\notin \cap \{P\mid P\text{ prime and } I\subseteq P\}$, and that's the contrapositive of the containment that you are seeking.