Show that $eR\,\cap\,(1-e)R=\{0\}$ if $e$ is an idempotent in R

Question

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.

Answer 1

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 $a=0$.

Answer 2

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, $x = er_1 = 0$.

Answer 3

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!!!