Show that $eR\,\cap\,(1-e)R=\{0\}$ if $e$ is an idempotent in R 
For a ring $R$, and an idempotent $e \in R$, show that $eR\,\cap\,(1-e)R=\{0\}$, and that $eR$ and $(1-e)R$ are subrings of R with multiplicative identity.

Clearly I am supposed to use the definition of idempotent ($e^2=e)$ somehow, but I do not see the trick.
I have tried rewriting $(1-e)R$ in different ways with no luck. All I really know so far is that $R$ cannot be a field. 
 A: Let $a \in eR \cap (1-e)R$. Then $a=er$ and $a=(1-e)s$ for some $s,r \in R$. This means,
$$er=(1-e)s \implies e^2r=e(1-e)s \overbrace{\implies}^{\because \,\, e^2=e} er=(e-e)s=0s=0.$$
Thus $a=0$.
A: One of the parts is done. Although, I have not been able to prove that $eR$ and $\left( 1 - e \right)R$ are subrings.
You can easily observe that $\left\lbrace 0 \right\rbrace \subseteq eR \cap \left( 1 - e \right)R$, because $0 = e0 = \left( 1 - e \right)0$. Now, let $x \in eR \cap \left( 1 - e \right)R$. Then, $\exists r_1, r_2 \in R$ such that $x = er_1 = \left( 1 - e \right) r_2$.
From this, you can get $r_2 = e \left( r_1 + r_2 \right)$. When you substitute for r_2 above, you shall get
$$er_1 = e \left( r_1 + r_2 \right) - e^2 \left( r_1 + r_2 \right) = 0$$
since $e^2 = e$. Hence, $x = er_1 = 0$.
A: Suppose
$r \in eR \cap (1 - e)R; \tag 1$
then
$\exists s, t \in R, \; r = es = (1 - e)t; \tag 2$
we note that
$r = es = e^2s = e(es) = er; \tag 3$
thus,
$r = er = e(1 - e)t = (e - e^2)t = 0 \cdot t = 0, \tag 4$
which shows that
$eR \cap (1 - e)R = \{0\}. \tag 5$ 
It is clear that $eR$ and $(1 - e)R$ form additive subgroups of $R$; for example, with
$es, \; et \in eR, \tag 6$
$es + et = e(s + t) \in eR; \tag 7$
that is, $eR$ is closed under "$+$"; furthermore, 
$(es)(et) = e(set) \in eR, \tag 8$
so $eR$ is multiplicatively closed as well; it is likewise clear that $(1 - e)R$ is closed under both ring operations $+$ and $\cdot$; the remaining ring axioms are inherited directly from $R$ and verification they bind in $eR$ and $(1 - e)R$ is a simple task left to the reader.  
We see that $e$ is a left identity in $eR$:
$e(es) = e^2s = es, \tag 9$
as is $1 - e$ in $(1 - e)R$, since
$(1 - e)^2 = (1 - e)(1 - e) = 1 - 2e + e^2 = 1 - 2e + e = 1 - e; \tag{10}$
that is $1 - e$ is also idempotent, whence
$(1 - e)((1 - e)t) = (1 - e)^2t = (1 - e)t, \tag{11}$
and $1 - e$ is a left identity in $(1 - e)R$.
If $R$ is a commutative ring, then obviously $e$ and $1 - e$ are two-sided identities for $eR = Re$ and $(1 - e)R = R(1 - e)$, respectively.
But I don't know how to deal with right identities in $eR$ and $(1 - e)R$ for non-commutative $R$ quite yet . . . more to follow!!!
