Let R be a Noetherian ring, and $\mathcal{I}$ a set of ideals of R. Let $J = \sum_{I\in \mathcal{I}} I$ be the set of all finite sums $i_1 + \cdots + i_n$ where $i_k \in I_k \in \mathcal{I},$ for all $k \leq n.$
Suppose $\mathcal{I}$ is countable, and enumerate the ideals in $\mathcal{I}$. Let $S_n = \sum_{i=1}^n I_k,$ where $I_k \in \mathcal{I}.$
Since R is Noetherian, the chain of ideals $$S_1 \subset S_2\subset S_3 \subset \cdots $$
satisfies the ascending chain condition, i.e. there exits $n\in \mathbb{N}$ such that $S_n = S_{n+1} = \cdots$ .
Question 1: Does this imply that $J = S_n$?
Question 2: If $\mathcal{I}$ is an uncountable set of ideals, can we still say that J is the sum of finitely many ideals in $\mathcal{I}$? Can you prove this using the ascending chain condition?
Best Answer
Question 1: Observe that $\forall j\in J$, $j=i_{k_1}+i_{k_2}+\cdots+i_{k_n}\in S_{k_n}$ (for some $k_1<k_2<\dots<k_n$ that may depend on $j$). So $J\subseteq \bigcup_{k=1}^\infty S_k=S_n$. But also $J\supseteq S_n$ for obvious reasons. It follows that $J=S_n$ $\Box$
Question 2: Consider the following (non-deterministic) algorithm $\mathcal A$. At step $0$ we have the ideal $J_0=(0)\subseteq J$. At step $t+1$ we choose an arbitrary $I\in \mathcal I$ such that $I\not\subseteq J_t$, and add it to $J_t\subseteq J$ to obtain $J_{t+1}=I+J_t\subseteq J$. If such an $I$ does not exist, the process terminates and outputs $J_t$, which by the above considerations would necessarily equal $J$. Suppose that $\mathcal A$ does not necessarily terminate. Then there would necessarily exist an infinite strictly ascending chain of ideals $$(0)=J_0\subsetneq J_1\subsetneq J_2\subsetneq \dots \subsetneq J$$ A contradiction to the assumption that $R$ is Noetherian $\Box$