I'm having trouble proving the following:
Suppose that $R$ is a commutative ring (with or without identity) with ideals $A$ and $B$ such that $AB = \{\sum_{i=1}^n a_ib_i \, :\,a_i \in A ; b_i \in B; n \in \mathbb{N}\}$ is a principal ideal. Then $A$ and $B$ are finitely generated.
I want to use the fact that $AB$ is a principal ideal to get finite generating sets for $A$ and $B$. My initial thought is that since $AB$ is a principal ideal there exists $x = \sum_{j = 1}^k a_jb_j$ such that $AB = \{rx + nx \,:\, r \in R; n \in \mathbb{N} \} = \langle x \rangle$. Then maybe $X = \{a_1, \dots, a_k\}$ and $Y = \{b_1, \dots, b_k\}$ might serve as generating sets. If I pick $a \in A$ and $b \in B$ then I would like to write $a = \sum_{j = 1}^k (r_ja_j + n_ja_j)$ and $b = \sum_{j = 1}^k (t_jb_j + m_jb_j)$ for some $r_j,t_j \in R$ and $n_j, m_j \in \mathbb{N}$. But this is where I'm stuck, and now I am starting to think that I've gotten my hands on the wrong sets, in which case I'm even more lost.
Consider $\oplus_{n\geq 0}\mathbb{Z}$ endowed with the product structure. Let $A=(n,0,...0)$, $B=(0,n_1,n_2,...)$, $AB=0$ and $B$ is not finitely generated.