If every ascending chain of ideals leading up to an ideal stabilises, is the ideal finitely generated?

Question

I'm a fourth-year undergraduate currently studying a master's degree. In the last couple of weeks, I have been wondering about the interaction of the Noetherian condition with the prime ideals of a commutative ring with $1$.

While working on a result I found, I came across the following condition for an ideal $\mathfrak{b}$ organically:

Every ascending chain of ideals $(\mathfrak{b}_i)_I$ converging upwards (or leading upwards) to the ideal $\mathfrak{b}$ stabilises. $(*)$

where the italicised terms are my own coinage, meaning that $(\mathfrak{b}_i)_I$ is an ascending chain of ideals* and $\bigcup_I \mathfrak{b}_i = \mathfrak{b}$, which it is also convenient to denote by $\mathfrak{b}_i \uparrow \mathfrak{b}$.

It is clear that $\mathfrak{b}$ finitely generated implies $(*)$. However, I am not sure about the converse.

Question: if $\mathfrak{b}$ satisfies $(*)$, is $\mathfrak{b}$ finitely generated? (AC allowed.)

There is an "obvious" proof to attempt. Suppose not, and let $\mathfrak{b}$ have generating set $\{r_j\}_J$ with $J$ infinite. Then, let $(J_i)$ be a chain of subsets (with respect to inclusion) converging to $J$, and set:

$$\mathfrak{b}_i = \langle J_i \rangle$$

However, this runs into two possible issues. Firstly, not every ideal has a minimal generating set, so it is not clear from this idea alone that the ascending chain does not stabilise. Secondly, if $J$ is uncountable, even with the well-ordering principle, it is not clear (to me, at least) how to construct the $(J_i)$ adequately. Nevertheless, it is perhaps helpful to note that any counterexample $\mathfrak{b}$ cannot be countably generated, and it must not have a minimal generating set.

Overall, I am quite unsure about whether or not $(*)$ implies finite generation. The only other remark that I can make is that I know that the jump between finite and uncountable is not unprecedented (cardinalities of power sets, dimensions of Banach spaces, etc.), so I'm not sure that even what I've found is an effective heuristic for intuition.

Thank you for reading.

Edit: for this question, $(\mathfrak{b}_i)_I$ is an ascending chain of ideals whenever $I$ is a total order and $i \le j \implies \mathfrak{b}_i \subseteq \mathfrak{b}_j$. We say that it stabilises if there exists some $i$ such that $\mathfrak{b} = \mathfrak{b}_i$.

Answer 1

More generally, we can ask: if $M$ is some $R$-module which is not the union of a chain of proper submodules, is $M$ finitely generated? This is in fact true. This was already asked at SE/246182, and even though the proof posted there is not correct, a comment refers to

Bodo Pareigis, Categories and functors, Academic Press, 1970

The result is Corollary 1 in section 4.10 (page 205). The proof is not really easy. The main technical ingredient is Lemma 1 in section 4.7, the proof uses ordinal numbers.

Answer 2

Let $M$ be a module (over anything). Let $\alpha$ be the smallest cardinal of a generating subset. Let $f:\alpha\to M$ be a map whose image generates $M$. For $\beta\in\alpha$ let $M_\beta$ be the submodule $\mathrm{Span}(f(\gamma):\gamma\le\beta)$. Let $I$ be the set of modules $M_\beta$ when $\beta\in\alpha$. Then $I$ is a chain.

The union of $I$ is $M$ (easy since $f(\alpha)$ generates).

If $M$ is not finitely generated, $\alpha$ is a limit ordinal. Hence for $\beta\in\alpha$, $\beta+1$ has cardinal $<\alpha$, so $M_\beta\neq M$. Thus $M$ is the union of a chain of proper submodules.