In my algebra class we are discussing Noetherian rings, and the definition we were given is that a ring is Noetherian if every chain of ideals satisfies the ascending chain condition; that is, if $I_1\subseteq I_2\subseteq\cdots$, then there exists $n\in\mathbb{N}$ such that $I_n=I_{n+1}=\cdots$. But this to me seems to discount the possibility that we might have an uncountable chain of ideals.
Is it not possible to have an uncountable totally-ordered set $\mathscr{A}$ and an ascending chain of ideals $\{I_\alpha:\alpha\in\mathscr{A}\}$? If it is possible, is the ascending chain condition actually that for every ascending chain of ideals $\{I_\alpha:\alpha\in\mathscr{A}\}$, there is an $\alpha_0\in\mathscr{A}$ such that $I_{\alpha}=I_{\alpha_0}$ for all $\alpha\geq\alpha_0$? Or is an ascending chain of ideals of a ring $R$ necessarily countable?
Best Answer
I think the added generality you ask about gives you back the same notion, nothing new.
Let's temporarily call a ring "Noetherian" if every countable ascending chain of ideals stabilizes, and a ring "strongly Noetherian" if every ascending chain (of possible uncountable number of) of ideals stabilizes, where stabilizes means exactly what you suggest: for an ascending chain of ideals $\{I_\alpha:\alpha\in\mathscr{A}\}$, there is an $\alpha_0\in\mathscr{A}$ such that $I_{\alpha}=I_{\alpha_0}$ for all $\alpha\geq\alpha_0$
I think noetherian and strongly noetherian are equivalent notions.
I think it's clear that every strongly noetherian ring is noetherian.
Now assume we have a noetherian ring and let's show it is strongly noetherian. Let $\{I_\alpha\}_{\alpha \in A}$ be a possibly uncountable ascending chain of ideals. Assume for contradiction that it did not stabilize, meaning, for every $\alpha \in A$, there is an $\beta \in A$ such that $I_{\alpha} \subsetneq I_{\beta}$. So pick some $\alpha_0 \in A$ (if $A$ is empty the chain stabilizes), then we get an $\alpha_1 \in A$, such that $I_{\alpha_0} \subsetneq I_{\alpha_1}$. Apply this reasoning again now to $\alpha=\alpha_1$ so there is $\alpha_2 \in A$ such that $I_{\alpha_1} \subsetneq I_{\alpha_2}$. And then by axiom of dependent choice we can continue and find a sequence $\alpha_3, \alpha_4, \ldots$ such that $$I_{\alpha_0} \subsetneq I_{\alpha_1} \subsetneq I_{\alpha_3} \subsetneq I_{\alpha_4} \cdots$$ i.e. a countable ascending chain that does not stabilize, contradicting the hypothesis that our ring was noetherian.