I'm trying to prove that $\bigcap\limits_{P\in \mathcal{P}}P\subseteq Nil(R)$, where $R$ is a commutative unital ring and $\mathcal{P}$ is a collection of prime ideals of $R$.
Suppose $a\not\in Nil(R)$ and let $\mathcal{S}=\{I: I \text{ is an ideal of $R$, such that } I\ne R, a^n\not\in I \text{ for all $n>0$} \}$. Then $\mathcal{S}$ is non-empty because $0\in Nil(R)$, so $0\in I$ for some $I\in \mathcal{S}$, thus $\langle 0\rangle\in \mathcal{S}$. Also, by Zorn's Lemma, since $R$ is an upper bound for any chain in the poset $(\mathcal{S}, \subseteq)$, there exists a maximal set $M\in \mathcal{S}$ (which, I think, must be an ideal, since $\mathcal{S}$ is a collection of ideals).
Now the part I don't get. I need to prove in this part that $x,y\not\in M$ implies that $\exists n,m>0$ such that $a^n\in M\cup\{x\}$ and $a^m\in M\cup\{y\}$. We know that, since $x,y\not\in M$ and $\mathcal{S}$ contains only sets with nilpotent elements, and $M$ is maximal (and it is related to all other sets in $\mathcal{S}$ by inclusion), then $a^n$ must be equal to $x$ and $a^m$ must be equal to $y$, and that $x,y$ must be non-nilpotent. But here's the part I don't understand:
- How can it possibly be the case that for two arbitrary $x$ and $y$, there must necessarily exist some element $a$ such that $a^n=x$ and $a^m=y$? What is the mystery here?
You can prove $\bigcap_{P\in\mathcal{P}}P\subseteq\operatorname{Nil}(R)$ if $\mathcal{P}$ is the collection of all prime ideals of $R$.
The union of a chain of ideals is an ideal: given two elements in the union, they belong to one and the same element of the chain, so their sum too; the other condition is easy. Thus Zorn’s lemma applies to the set $\mathcal{S}$ consisting of the ideals that contain no power of $a$, where $a\notin\operatorname{Nil}(R)$. This set is not empty, because $(0)\in\mathcal{S}$.
It's clear that the union of a chain in $\mathcal{S}$ is an element of $\mathcal{S}$, so the family contains a maximal element $M$.
We want to show that $M$ is prime, that is,
Suppose $x\notin M$ and $y\notin M$.
Note: we have to consider $M+(x)$ and $M+(y)$, not $M\cup\{x\}$ and $M\cup\{y\}$ (which are not ideals).
Since $x\notin M$, we have that $M+(x)$ properly contains $M$. By maximality of $M$ in $\mathcal{S}$, we conclude $M+(x)\notin\mathcal{S}$, so $M+(x)$ must contain $a^m$, for some $m$. Similarly, $a^n\in M+(y)$ for some $n$. Thus $a^m=r+xs$ and $a^n=t+yu$, for $r,t\in M$ and $s,u\in R$. Consequently $$ a^{m+n}=rt+xst+yru+xysu $$ If $xy\in M$, then we conclude that $a^{m+n}\in M$, because $rt,xst,yru\in M$. Contradiction. Therefore $xy\notin M$.
A simpler approach. We want to show that if $a$ is not nilpotent, then there exists a prime ideal that contains no power of $a$. Consider the multiplicative set $S=\{a^n:n\in\mathbb{N}\}$. Then the ring $S^{-1}R$ is nonzero, so it contains a maximal ideal $M$. The inverse image of $M$ under the canonical ring homomorphism $R\to S^{-1}R$ is the required prime ideal.