Justify the use of Zorn's Lemma in a proof

Question

I was reading this proof of how the nilradical of a ring is the intersection of all prime ideals of the ring.

https://artofproblemsolving.com/wiki/index.php?title=Nilradical

In the proof, it says "Let $S$ be the set of ideals in $A$ 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 $\mathfrak{m}$."

Here is the question. Why is Zorn's Lemma valid here? More specifically, why is this set inductively ordered? It is certainly not the entire ring since $a\in A$. If the ring is Noetherian, then it is natural that all collection of ideals have a maximal element, but how does this work in general?

Thank you!

Edit1: I suppose that we could take the unions of all such ideals, but how do we prove that it is in $S$?

Edit2: We take a nonempty chain of ideals indexed by I in $S$. Since it is totally ordered, we have if $m,n\in I$, then $J_m\subset J_n$ or $J_n\subset J_m$. We take the union of all ideals in this collection $J=\bigcup_{i\in I}J_i$ and we prove that J itself is in $S$. To prove that $J$ is an ideal, take $j_1,j_2\in J$. Then we have $j_1\in J_n$ and $j_2\in J_m$ for some $m,n\in I$. Then WLOG, assume $J_n\subset J_m$. Then $j_1,j_2\in J_m$. Thus, $j_1-j_2\in J_m\subset J$. Also notice that $0\in J$ since $0\in J_i$, $\forall i\in I$. Also, if we take an element $j\in J$ and an element $a\in A$, then $j\in J_m$ for some $m\in I$ and $J_m$ is an ideal. Therefore, $aj,ja\in J_m\subset J$. Thus, $J$ is indeed an ideal and since $a^n\not\in J_i$, $\forall i\in I$, $a^n\not\in J$, $\forall n\in\mathbb{Z}^+$. Therefore, any nonempty collection of ideals in $S$ have an upper bound in $S$, so we can apply Zorn's Lemma for $S$.

Thanks to @ArturoMagidin for clarifications.

Answer 1

Zorn's Lemma does not require the partially ordered set to be bounded above.

Zorn's Lemma. Let $P$ be a partially ordered set such that every totally ordered subset of $P$ has an upper bound in $P$ (equivalently, a non-empty partially ordered set such that every non-empty totally ordered subset of $P$ has an upper bound in $P$). Then $P$ has maximal elements.

Here, the poset is the set of all ideals that do not contain any $a^n$, partially ordered by inclusion. They've noted it is nonempty. Let $\{J_i\}_{i\in I}$ be a nonempty chain in $P$. That means that each $J_i$ is an ideal that does not contain any $a^n$, and if $r,s\in I$ are any two indices, then either $J_r\subseteq J_s$ or $J_s\subseteq J_r$.

You should prove that in this situation, $$J =\bigcup_{i\in I} J_i$$ is also an ideal that does not contain any $a^n$. Thus, $J$ is an element of $P$ that is an upper bound for the chain.

This shows $P$ satisfies the hypotheses of Zorn's Lemma.