Show that $R = I_1 ⊕ I_2$ is an internal direct sum if and only if there exists $e ^2 = e ∈ C(R)$ such that $I_1 = eR$ and $I_2 = (1 − e)R$.

Question

Let R be a ring with $1 ∈ R$. Show that $R = I_1 ⊕ I_2$ is an internal direct sum if and only if there exists $e ^2 = e ∈ C(R)$ where $C(R)$ = {$e \in R | er = re, r \in R$} (i.e. $C(R)$ is the Center of $R$) such that $I_1 = eR$ and $I_2 = (1 − e)R$.

Here is what I know:

I know that I should show that $I_1 \cap I_2 = {0}$ which will then imply that $R$ is an internal sum. I have tried to use that properties "there exists $e ^2 = e ∈ C(R)$ such that $I_1 = eR$ and $I_2 = (1 − e)R$" to prove $I_1 \cap I_2 = {0}$. But it has not been successful. Please I would appreciate clear explanation on this. Thank you.

Answer 1

Hint: if $e^2 = e$ and $x \in eR \cap (1-e)R$, then you have $x = ey = (1-e)z$ for some $y$ and $z$. But then $x = ey = e^2y = e(1-e)z = (e - e^2)z = 0$.