Let $A$ be a noetherian ring, $\mathfrak a_{0}$ an ideal in $A$ and $$S=\{ n\in \mathbb{N}\mid \text{there exist ideals }(\mathfrak a_{i})_{i=1,\dots,n}\text{ such that }\mathfrak a_{0}\subset \mathfrak a_{1}\subset \dots \subset \mathfrak a_{n}\}.$$ Then, is $S$ finite?
Thanks in advance!
No, $S$ does not have to be finite. In fact, $S$ need not be finite even if we restrict ourselves to prime ideals $a_i$. Take Nagata's noetherian ring of infinite Krull dimension, which is constructed as follows:
Let $A=k[x_1,x_2,\ldots]$ be the polynomial ring over a field $k$ in countably infinitely many variables. Let $$\mathfrak{p}_1=(x_1),\quad \mathfrak{p}_2=(x_2,x_3),\quad\mathfrak{p}_3=(x_4,x_5,x_6),\quad\ldots$$ and let $S=A\setminus\bigcup_{i=1}^\infty\mathfrak{p}_i$, which is a multiplicatively closed subset of $A$. Then $S^{-1}A$ is noetherian, but we have strict inclusions $$(0)\subset S^{-1}(x_2)\subset S^{-1}\mathfrak{p}_2,\qquad (0)\subset S^{-1}(x_4)\subset S^{-1}(x_4,x_5)\subset S^{-1}\mathfrak{p}_3,\qquad\ldots$$ (here we're using $a_0=(0)$).
Certainly not:
$$(p_n\cdots 2)\subsetneq (p_{n-1}\cdots 2)\subsetneq\cdots\subseteq (2)$$
where $p_i$ is the $i^{\text{th}}$ prime.
More generally, in any Dedekind domain that is not a PID you can find infinitely many primes $\{\mathfrak{p}_i\}$ (because a Dedekind domain with finitely many primes is a PID) and so the chains of the form
$$\mathfrak{p}_n\cdots\mathfrak{p}_1\subsetneq \cdots\subsetneq \mathfrak{p}_1$$
show that $S$ is not finite.
Note that the reason of picking Dedekind domains is somewhat obvious. To get a decreasing chain of non-zero ideals (which we can always interpret as an increasing chain of ideals from $0$!) the obvious thing to do is to successively multiply ideals. But, we want a cheap reason to deduce that we have proper containment at each step. Thus, we'd like to be able to "cancel" ideals in an equality, and thus Dedekind domains are the obvious (only?) choice.
EDIT: I missed the first mention of $a_0$. But, taking $a_0=(0)$ shows the above still can make $S$ infinite. Of course, if we take $a_0$ to be maximal, then of course $S$ is finite. So, the answer is definitively "sometimes".
Take any commutative noetherian ring of positive dimension. Then it is not artinian, hence there is an infinite decreasing chain of ideals $\dotsc \subset \mathfrak{a}_2 \subset \mathfrak{a}_1 \subset \mathfrak{a}_0$. Then $S=\mathbb{N}$ by considering $\mathfrak{a}_n \subset \dotsc \subset \mathfrak{a}_0$.
For example, if we have an integral domain which is not a field, take $a$ to be a non-zero non-unit and $\mathfrak{a}=(a^n)$.
