Let $R$ be a ring and let $R = I_1 ⊕ I_2$ be an internal direct sum. If $S$ is a subring of $R$ containing $I_1$ then $S = I_1 ⊕ (S ∩ I_2)$.
I am supposed to prove the that $S = I_1 ⊕ (S ∩ I_2)$ with the condition given above, however, to start with, I do not know what (what kind of elements) are in $S ∩ I_2$.
Here is what I know:
$I_1 ∩ I_2 = 0$ (by the definition of internal direct sum). But now the fact that $S = I_1 ⊕ (S ∩ I_2)$ means that $S ∩ I_2$ is not empty? So what is in $S ∩ I_2$?