Nilradical equals intersection of all prime ideals Suppose we have a commutative ring R with identity 1, and A is an ideal of R. Also, we have $B=\{{x\in R\mid x^n\in A }$ for some natural number n$\}$.
Show that B is the intersection of all prime ideals that contain A.
I know that we can use Zorn's lemma to prove it, here is basic idea to prove it:
Suppose x is not in B, then we set $C=\{{1,x,x^2,x^3,...,x^n...}\}$. Next apply Zorn's lemma to the partially ordered set S if ideals I of R such that $A \subset I$ and $I \cap C=\emptyset$. Then we can find a maximal element of S, but how can I show that the maximal element of S (by inclusion) is a prime ideal of R? I feel very confused about it! Can someone tell me why?
More details: check here.

Let $\mathfrak{p}$ be a maximal element (maybe the symbol is a bit presumptuous). You should first tell me why $\mathfrak{p}$ isn't the whole ring. Now take two elements $a,b$ not contained in $\mathfrak{p}$. Then, for instance, the ideal $\mathfrak{p} + Ra$ properly contains $\mathfrak{p}$ and hence can't be one of the ideals in $S$. What sort of element has to then lie in $\mathfrak{p} + Ra$? Go through the same reasoning for $b$. Play around with the resulting elements.
