Intersection of Ideals of Ring contains Product

Theorem
Let $R$ be a ring.

Let $I$ be a right ideal of $R$.

Let $J$ be a left ideal of $R$.

Let $I J$ be their product.

Then $I J \subseteq I \cap J$.

Proof
Let $a_1, \ldots, a_n \in I$ and $b_1, \ldots, b_n \in J$ be arbitrary.

By definition of right ideal:

By definition of left ideal:

By definition of set intersection:
 * $\ds \sum_{k \mathop = 1}^n a_k b_k \in I \cap J$

As $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ were arbitrary, this completes the proof.