Intersection of Ideals in a ring

Question

It is clear that $I = \langle a \rangle \cap \langle b \rangle $ is an ideal. I am trying to prove that $a_1b_1 - a_2b_2 \in I \ \forall \ a_1b_1,a_2b_2 \in I $

It seems so easy but I am stuck. Thanks

Answer 1

Note that by assumption $a_1b_1, a_2b_2 \in \langle a \rangle$, since $\langle a \rangle \supseteq I$. Multiplying by $-1$, we find that $(-1)a_2b_2 \in \langle a \rangle$, using the property that ideals absorb multiplication. Since $\langle a \rangle$ is an ideal, then it is closed under addition, so $a_1b_1 + (-1)a_2b_2 = a_1b_1- a_2b_2 \in \langle a \rangle$. Similarly, $a_1b_1 - a_2b_2 \in \langle b \rangle$. So $a_1b_1 - a_2b_2 \in \langle a \rangle \cap \langle b \rangle$.