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.

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