Let R be a PID. Then, every nonempty set of ideals of R has a maximal element

Question

I came across the following proof for this proposition but there are some things I don't understand.

Proof

Let $S$ be the set of all proper ideals of $R$. It follows that $S$ is non-empty and it is partially ordered by inclusion. Let $I_1 \subseteq I_2 \subseteq $ ... be an arbitrary increasing chain of ideals in $S$. Let $I=\bigcup_{n} I_n$.

Since the chain of $I_n$’s are nonempty, it follows that $I$ is nonempty. $I$ is an ideal. Since $R$ is a PID, $I = (a)$. We find that $a \in I=\bigcup_{n} I_n$ so $a \in I_n$ for some $n$. We get $I_n = I_{n+1} =$ ....

Each chain of ideals has an upper bound. By Zorn’s lemma, the nonempty set of $I_n$'s of $R$ has a maximal element,the maximal ideal containing $I$.

Questions

Why can we use the set of all proper ideals of $R$ for our proof when it talks about any set?

Why is it true that $I_n = I_{n+1}$? I'm hoping it's a typo because what it really uses is that the previous is containned in the next.

I'm also not sure why the maximal would contain $I$, I understand it's not relevant to the proof since it's only asking us to prove it exists but I got curious.

Answer 1

This is a mess; what you quote (up to but not including the last paragraph) is a perfectly fine proof that $R$ has maximal ideals; and it can easily be modified to prove the statement given. But as written, it is not a proof of the proposition that every nonempty collection of ideals of $R$ has a maximal element.

And the last part of the last paragraph is completely garbled. "The maximal ideal containing $I$" is problematic in several ways, not least the implicit assumption that there is a unique maximal ideal that contains $I$.

Of your first two questions, the first is "because they are doing things incorrectly" (or "because they are trying to prove that $R$ has maximal ideals).

For the second: since $a\in I_n$ and $I_n$ is an ideal, it contains $(a)$. But $(a)=I$. So we have $$I = (a) \subseteq I_n \subseteq I_{n+1}\subseteq I,$$ and this shows that we have equality throughout. In particular, $I_n=I_{n+1}$.

Now, how to fix this argument to prove the desired result? Let $S$ be a nonempty collection of ideals of $I$. We want to show that $S$, ordered by inclusion, satisfies the hypotheses of Zorn's Lemma: every chain in $S$ has an upper bound in $S$. Let $\{I_j\}_{j\in J}$ be a chain in $S$. If $J=\varnothing$, then any element of $S$ (which exist, because $S$ is nonempty) is an upper bound.

(What follows is essentially their argument)

If $J$ is nonempty, let $I = \cup_{j\in J}I_j$. Then $I$ is an ideal of $R$, hence principal, $I=(a)$. Since $a\in\cup_{j\in J}$, there exists $n\in J$ such that $a\in I_n$. I claim that $I_n=I$ is an upper bound for $\{I_j\}_{j\in J}$ in $S$. Note that because $a\in I_n$ and $I_n$ is an ideal, then $(a)\subseteq I_n$. But $I_n\subseteq I=(a)$, so $I=I_n$.

Indeed, take any $j\in S$. We need to prove that $I_j\subseteq I_n$.

Since we have a chain, either $I_j\subseteq I_n$, or $I_n\subseteq I_j$. If $I_j\subseteq I_n$, there is nothing to do. If $I_n\subseteq I_j$, then $$I = (a) = I_n \subseteq I_j\subseteq I = I_n,$$ so in fact $I_j=I_n$. Thus, $I_n$ is an upper bound for the chain. Finally, because $I_n\in S$, it is an upper bound in $S$.

This completes the verification that $S$ satisfies the hypotheses of Zorn's Lemma. Thus, $S$ has maximal elements, as desired.

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The proof easily generalizes to rings in which every ideal is finitely generated. These are called Noetherian rings. In fact, the following are equivalent:

1. Every ideal of $R$ is finitely generated.
2. (ACC on ideals) Every ascending chain of ideals of $R$, $$I_1\subseteq I_2\subseteq\cdots\subseteq I_n\subseteq\cdots$$ stabilizes: there exists $n$ such that for all $k\geq n$, $I_n=I_k$.
3. Every nonempty collection of ideals of $R$ has maximal elements.