## In a commutative ring, the product of two finitely generated ideals is finitely generated

Let $R$ be a commutative ring. Suppose $I = (A)$ and $J = (B)$, where $A = \{a_1, \ldots, a_n \}$ and $B = \{b_1, \ldots, b_m \}$. Prove that $IJ = (C)$, where $C = \{ab \ |\ a \in A, b \in B \}$.

$(\subseteq)$ Let $x = \sum_i r_is_i \in IJ$, where $r_i = \sum_j t_{i,j}a_j$ and $s_i = \sum_k u_{i,k} b_k$. Then we have $x = \sum_i (\sum_j t_{i,j}a_j)(\sum_k u_{i,k}b_k)$ $= \sum_i \sum_j \sum_k t_{i,j}u_{i,k}a_jb_k \in (C)$ since $R$ is commutative.

$(\supseteq)$ Let $x = \sum r_{i,j}a_ib_j \in (C)$. Since $I$ is an ideal, $x \in IJ$.