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I am trying to show that the set of all nilpotent elements is equal to the intersection of all prime ideals. This is a quote of a quote from this post:

"To show the converse, it suffices to show that for any non-nilpotent element $a$, there is some prime ideal that does not contain $a$.

So suppose that $a$ is an element of $A$ that is not nilpotent. Let $S$ be the set of ideals of $A$ that do not contain any element of the form $a^n$. Since $(0) \in S$, $S$ is not empty; then by Zorn's Lemma, $S$ has a maximal element $\mathfrak{m}$.

It suffices to show that $\mathfrak{m}$ is a prime ideal. Indeed, suppose otherwise; then there exist elements $x,y \notin \mathfrak{m}$ for which $xy \in \mathfrak{m}$. Then the set of elements $z$ for which $xz \in \mathfrak{m}$ is evidently an ideal of $A$ that properly contains $\mathfrak{m}$; it therefore contains $a^n$, for some integer $n$. By similar reasoning, the set of elements $z$ for which $a^n$ $z \in \mathfrak{m}$ is an ideal that properly contains $\mathfrak{m}$, so this set contains $a^m$, for some integer $m$. Then $a^{n+m} \in \mathfrak{m}$, a contradiction.

Therefore $\mathfrak{m}$ is a prime ideal that does not contain $a$."

Let $I$ be the ideal consisting of all $z$ such that $xz \in \frak{m}$. Why can't it be the case that $I = \frak{m}$? Also, even if there exists an $r \in I - \frak{m}$, why must $r=a^n$ for some $n$?

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  • $\begingroup$ Because $y \in I \setminus \mathfrak{m}$. And such an $r$ need not be of the form $a^n$, but by maximality, since the inclusion is strict, there is an $n$ with $a^n \in I$. $\endgroup$ Dec 26, 2017 at 22:55

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By assumption, we have $xy \in \mathfrak{m}$, whence $y\in I$. And also by assumption, $y \notin \mathfrak{m}$, so $y \in I \setminus \mathfrak{m}$. There is no reason to believe that every $r \in I \setminus \mathfrak{m}$ has the form $a^n$, but since $\mathfrak{m}$ is maximal among the ideals not containing any $a^n$, it follows that $I$ contains an element of that form.

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