Minimal ideal in a ring which is generated by an idempotent element.

Question

Let $R$ be a commutative ring with unity and $M$ be a minimal ideal of $R$ such that $M = Re$ where $e$ is an idempotent element in $R$. Then $R = Re \oplus R(1-e) $

I am not able to see, in order to write $R = Re \oplus R(1-e) $ , where is the fact that $e$ is an idempotent element is used.

Thanks in advance!

Answer 1

To have $R=I\oplus J$, we need to know that $I+J = R$ and $I\cap J = (0)$.

$eR + (e-1)R=R$ is true for any $e\in R$, since $e-(e-1)=1$.

$eR\cap (e-1)R=(0)$ is where we use idempotence. If $x=ae=b(e-1)$, then $ex = ae^2=ae=x$, but $ex=b(e-1)e=b(e^2-e)=b(e-e)=0$, and therefore $x=0$.

Answer 2

$Re\cap R(1-e)=\{0\}$:

Indeed if $x$ lies in the intersection, it can be written both as $x=ae=b(1-e)$ for some $a, b\in R$. But then $$x=ae=ae^2=b(1-e)e=b(e-e^2)=b\cdot0=0.$$

Answer 3

This is the Pierce decomposition, which says that for an idempotent $e\in R$ we have $$ R=eRe \oplus eR(1-e) \oplus (1-e)Re \oplus (1-e)R(1-e). $$ For commutative rings it simplifies to $R=Re\oplus R(1-e)$. One uses $e^2=e$ several times in the proof, see the above link, e.g., $e(1-e)=e-e^2=0$.