Understanding the Proof that Nilpotent Elements Equals Intersection of all Prime Ideals.

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 $$n$$?

• 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 $a^n \in I$. Dec 26, 2017 at 22:55

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.