I am working on the following exercise (Exercise VIII.3.9 in Hungerford's Algebra).
Let $R$ be Noetherian and let $B$ be an $R$-module. If $P$ is a prime ideal such that $P=\text{ann }x$ for some nonzero $x\in B$ (see Exercise VIII.3.7), then $P$ is called an associated prime of $B$.
(a) If $B\neq 0$, then there exists an associated prime of $B$. [Hint: use Exercise VIII.3.8.]
(b) If $B\neq 0$ and $B$ satisfies the ascending chain condition on submodules, then there exist prime ideals $P_1, ..., P_{r-1}$ and a sequence of submodules $B=B_1\supseteq B_2\supseteq \cdots \supseteq B_r=0$ such that $B_i/B_{i+1}\cong R/P_i$ for each $i<r$.
My Solution: By (a), there exists an associated prime $P_1=\text{ann }x$ of $B$. Set $B=B_1=Rx$ and $B_2=0$. Then $B_1/B_2=Rx/0\cong Rx\cong R/\text{ann }x=R/P_1$.
My Question: I am wondering why we need the ascending chain condition on submodules of $B$?
You're not allowed to just set $B=B_1=Rx$: what if $B\neq Rx$? What you can do is set $Rx$ to be the last submodule in the chain before $0$ (i.e., $B_{r-1}$). You'll have to then iterate the construction to climb your way up to the whole module $B$, though, and you need to use the ascending chain condition to guarantee that you reach it in finitely many steps.