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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.

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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.

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  • $\begingroup$ Oh,that is very useful ! Thanks! $\endgroup$ – user144600 Nov 21 '15 at 17:21
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If you are familiar with localizations, they provide a very nice alternate way to frame this argument. By considering the ring $S=R/B$, it suffices to show the following: if $x\in S$ is in every prime ideal of $S$, then $x^n=0$ for some $n\in\mathbb{N}$. To prove this, consider the localization $S[x^{-1}]$ (i.e., the localization with respect to the multiplicatively closed subset $\{1,x,x^2,x^3,\dots,\}$). By the explicit description of localizations as fractions under an equivalence relation, we see that $1/1=0/1$ in the ring $S[x^{-1}]$ iff $x^n=0$ for some $n$. Thus if $x^n\neq 0$ for all $n$, the ring $S[x^{-1}]$ is not the zero ring. There thus exists a maximal ideal $M\subset S[x^{-1}]$. If $\varphi:S\to S[x^{-1}]$ is the canonical map, $\varphi^{-1}(M)$ is then a prime ideal (because the inverse image of a prime ideal is always prime) such that $x\not\in\varphi^{-1}(M)$ (since $\varphi(x)$ is a unit).

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