# Does the ascending chain condition only apply to countable chains of ideals?

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?

• Any infinite totally ordered set $\mathscr{A}$ has an initial segment isomorphic to $\mathbb{N}$. The ascending chain condition is that the ideals become the same already after finitely many indices in it. So it makes no difference what the cardinality of $\mathscr{A}$ is. Commented Oct 2, 2021 at 5:33
• @Conifold that’s not quite true as stated. The collection of rationals $1/n$ for positive integers $n$ has no such initial segment when ordered in the usual way. It doesn’t even contain a copy of $\mathbb N$ as an ordered subset. I think what you meant to qualify with is a well-ordered set. Commented Oct 2, 2021 at 15:40
• Or alternatively stipulating "an infinite totally ordered set with no maximal element contains a totally ordered subset order isomorphic to $\mathbb N$" Commented Oct 4, 2021 at 19:37

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.

I've been thinking about how to answer this for quite some time. Mulling over the original question and the existing answer has crystallized some things that I had previously not appreciated.

I'm sure every student of ring theory (including myself) has had that these questions at some point: "Why is this stated in terms of a sequence? Is it the same as if you had used arbitrary chains? Is it an oversight that it isn't defined for arbitrary chains?"

After discovering the same equivalence outlined in usr0192's solution (as a student, many years ago), I felt I had settled the second and third question, but I did not have a good answer for the first one. In this solution I'm trying to address this topic to complement the existing solution.

## The index set

Firstly, notice that the linear containment order of the ideals immediately gives you a linear order on your index set. Everyone here, I think, observed this that we may as well assume the index set is linearly ordered to reflect the order of the ideals. That is one step closer to $$\mathbb N$$.

One person commented "Any infinite totally ordered set 𝒜 has an initial segment isomorphic to ℕ." and so aimed to suggest the same inductive argument given by usr0192's solution. Unfortunately, this is not true: take the set $$T=\{\frac1n\mid n\in\mathbb N^+\}\cup\{0\}$$ with its usual order in the reals. It is linearly ordered, but quite clearly does not have an initial segment isomorphic to $$\mathbb N$$, nor does it even contain a copy as a subsequence. This is not an improbable totally ordered set either: it is what the ideals of the ring of formal power series ring $$F[[X]]$$ look like, for a field $$F$$.

So an infinite total order is not quite the context to think about growing sequences.

## The meaning of it all

What the Noetherian condition means is, if you have any totally ordered set 𝐴 and any chain of ideals $$\{𝐼_𝑎\}_{𝑎∈𝐴}$$ where $$𝐼_𝑎⊆𝐼_𝑏$$ iff $$𝑎≤𝑏$$, then if you start at any fixed ideal $$𝐼_𝛼$$ in the chain and go upwards, then the set $$\{𝐼|𝐼=𝐼_𝛽 \text{ for some } 𝐼_𝛽⊇𝐼_𝛼\}$$ is actually finite.

Here was another clue that tells us that "stepping up" is not indispensable. The commenter opted to cast Noetherianness as a condition on arbitrarily totally ordered chains of ideals. Was anything lost?

Apply it to our total order $$T$$ above, indexing the ideals of $$F[[x]]$$ with $$T$$, and pick $$0$$ as the ideal you start at. We find that the set of ideals above $$I_0=\{0\}$$ is not finite at all. But $$F[[x]]$$ is a Noetherian ring...

Part of the problem with this recasting is that "going up" does not make as much sense with $$T$$ as it does with a sequence indexed by $$\mathbb N$$.

I think what these two problems are trying to tell us is that it is important to 1) start somewhere and 2) be able to step-up to successors in a well-defined way. It is not enough to only talk about "an infinite totally ordered chain," not even countable such chains.

## What does the sequence definition get us

The halting-sequence definition provides us with these things:

1. It gives you a place to start;
2. It makes it clear where the steps are; and
3. it says you arrive at a maximal element in the chain only finitely many steps from the bottom.

What usr0192's proof shows is that a chain of any cardinality with no maximal element necessarily contains a subchain order isomorphic to $$\mathbb N$$ (hence it is also well-ordered). This is what the claim in the "index set" paragraph should have said, rather than "initial segment."

## Conclusion

The definition as it is so frequently stated is sufficient to preclude strictly increasing chains of any cardinality, and you only have to check a smaller subfamily of chains, not all of them. That makes your life easier, right? Furthermore, it guarantees a theoretical $$n$$ where it stops. This is much more interesting than, say knowing a chain stabilizes at $$\omega +5$$.

Yes, general rings can have ascending chains of ideals of arbitrarily large cardinality.

But by definition, a Noetherian ring is one in which every ascending chain of ideals stabilizes after a finite number of steps.

• I'm still a bit confused on this. If we have an ascending chain of ideals which is indexed by an uncountable set, what notion of "finite number of steps" is there? The indexing set might not even have a minimal element, right? Commented Oct 2, 2021 at 0:50
• @csch2: That would be a problem even for countable index set: just index it by $\mathbb{Z}$. The point with Noetherian is that if you have a totally ordered set $I$, and ideals $\alpha_i$ with $\alpha_i\subseteq\alpha_j$ whenever $i\leq j$, then for any $\alpha_0\in I$, the chain of ideals indexed by $\{i\in I\mid i\geq \alpha_0\}$ stabilizes after a finite number of steps. Commented Oct 2, 2021 at 1:04
• @csch2: For instance, in $\mathbb{Z}$, the ideals $(2^{-n)$ ($n\leq 0$) are indexed by the negative integers; they form an infinite descending chain; because $\mathbb{Z}$ is not Artinian (but it is Noetherian). Commented Oct 2, 2021 at 1:07
• this answer given I don't think addresses the question the OP asked. The OP did not ask if there are non-noetherian rings, but rather if the definition of noetherian is changed to include uncountable chains, then do we get another notion? Commented Oct 2, 2021 at 5:34
• @csch2 It doesn't matter if the indexing set for your ascending chain of ideals has a minimal element or a maximal element, or whether it's countable or uncountable. What the Noetherian condition means is, if you have any totally ordered set $A$ and any chain of ideals $\{ I_a \}_{a \in A}$ where $I_a \subseteq I_b$ iff $a \leq b$, then if you start at any fixed ideal $I_\alpha$ in the chain and go upwards, then the set $\{ I | I = I_\beta \text{ for some } I_\beta \supseteq I_{\alpha} \}$ is actually finite. Commented Oct 2, 2021 at 18:37