Basic properties of ideal products

Let $R$ be a commutative ring with 1 and let $A,B,C \subseteq R$ be ideals. Prove the following. (1) $A(BC) = (AB)C$, (2) $AB = BA$, (3) If $a \in A$ and $b \in B$ then $ab \in AB$, and (4) $AB \subseteq A$.

Let $a \in A$ and $b \in B$. Certainly then by our equivalent characterization of ideal products, $ab \in AB$.

By this previous exercise, if $z \in A(BC)$ then $z = \sum_i a_i (\sum_j b_{i,j}c_{i,j}) = \sum_i \sum_j a_i b_{i,j} c_{i,j}$. Each $a_ib_{i,j}$ is in $AB$, so that $(a_ib_{i,j})c_{i,j} \in (AB)C$. So $A(BC) \subseteq (AB)C$. The reverse inclusion is similar.

Suppose $\sum a_ib_i \in AB$. Since $A$ is an ideal, each $a_ib_i$ is in $A$. Then $\sum a_ib_i \in A$.

Since $R$ is commutative, $AB = \{\sum_{i=1}^n a_ib_i \ |\ n \in \mathbb{N}, a_i \in A, b_i \in B\}$ $= \{\sum_{i=1}^n b_ia_i \ |\ n \in \mathbb{N}, a_i \in A, b_i \in B\} = BA$.