$R$ is isomorphic to product of ideals $(a)\times (1-a)$

Question

Following a question of mine on here, the answer says that if $a\in R$ (a commutative ring) is idempotent ($a^2=a$), then: $$R\simeq (a)\times (1-a)$$ I am trying to make sense of this by proving it.

Ideas: I have shown that $a$ and $1-a$ "split" $R$ into distinct parts, that is: $$xa=x(1-a)\Rightarrow x=0$$ This is proved by multiplying the equality by $a$ and $1-a$ to get $ax=0, x(1-a)=0$ from which we get $x=0$. So I then wanted to show that each $r\in R$ can be uniquely represented as: $$r=xa-y(1-a)$$ so that I could pair up each $r\in R$ with $\big(ax,y(1-a)\big)$ to get an isomorphism. But this doesn't work since $x=-y=r$ satisfies the equation, but does not account for all of $(a)\times (1-a)$...

Am I missing something obvious?

Answer 1

To answer your comment $x\mapsto (xa,(1-a)x)$ is surjective. Indeed let $(ya,(1-a)z)$ be the most general element in $(a)\times(1-a)$. Define :

$$x:=ya+(1-a)z $$

Then

$$xa=ya^2+(1-a)za=ya+(a-a^2)z=ya+(a-a)z=ya$$

$$x(1-a)=ya(1-a)+(1-a)(1-a)z=y(a-a^2)+(1-2a+a^2)z=y(a-a)+(1-a)z=(1-a)z$$

So $(xa,(1-a)x)=(ya,(1-a)z)$.

Answer 2

Let $a$ be a (central) idempotent. There is a map $f:R\to Ra\times R(1-a)$ that sends $x\to (xa,x(1-a))$, and a map $g:Ra\times R(1-a)\to R$ that sends $(sa,r(1-a))\to sa+r(1-a)$. You should check both maps are ring morphisms and are in fact inverse isomorphisms.

Answer 3

Alternatively, by the second isomorphism theorem

$\frac{R}{aR}=\frac{aR\oplus (1-a)R}{aR}\cong (1-a)R$

and

$\frac{R}{(1-a)R}=\frac{aR\oplus (1-a)R}{(1-a)R}\cong aR$

Then noting that $aR$ and $(1-a)R$ are comaximal, by the Chinese remainder theorem, you have that $(1-a)R\times aR\cong\frac{R}{aR}\times \frac{R}{(1-a)R}\cong \frac{R}{aR\cap (1-a)R}=\frac{R}{\{0\}}\cong R$