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.

Then:

\(\ds \forall k \in \set {1, \ldots, n}: \, \)

\(\ds a_k b_k\)

\(\in\)

\(\ds I\)

Definition of Right Ideal of Ring

\(\ds \leadsto \ \ \)

\(\ds \sum_{k \mathop = 1}^n a_k b_k\)

\(\in\)

\(\ds I\)

and:

\(\ds \forall k \in \set {1, \ldots, n}: \, \)

\(\ds a_k b_k\)

\(\in\)

\(\ds J\)

Definition of Left Ideal of Ring

\(\ds \leadsto \ \ \)

\(\ds \sum_{k \mathop = 1}^n a_k b_k\)

\(\in\)

\(\ds J\)

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.

$\blacksquare$