# Show the $R$-module $R$ is isomorphic to $Rb \times R(1-b)$ where $b$ is an idempotent of a commutative ring with unity

Let $R$ be a commutative ring with unity and let $B(R)$ be the set of all idempotent elements in $R$.

Show for $b\in B(R)$, the $R$-modules $R$ and $Rb \times R(1-b)$ are isomorphic to one another.

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It is preferred for questions to be given in the form of a question. Have you tried to find an isomorphism? Any partial progress or context would improve your question. – Jonas Meyer Dec 13 '10 at 0:18
I have tried looking at what I know about R-Modules and idempotent elements and have hypothetically got my isomorphism function to be F: r-> (rb, r(1-b)) however this seems far too simple. I know that r(x+y)=rx+ry and (r+s)x=rx+sx for r,s in R and x,y in B(R). Also that the elements b in B(R) satisfy b*b=b. Sorry first time I've used the site thanks for the advice! – Lucy Marshall Dec 13 '10 at 0:18
I have also considered the elements of this set of the idempotent elements. Can I say that the set has only 0 and 1 in it? Thanks – Lucy Marshall Dec 13 '10 at 0:20
@lucy: Thanks. As for your $F$, that is a very good guess. You can try to show that it is an isomorphism by verifying each part of the definition. It should be a module homomorphism, it should be one-to-one, and it should be onto. Onto will be the least straight-forward. – Jonas Meyer Dec 13 '10 at 0:25
@lucy: And if you figure it out, consider writing it up as an answer; then people can comment on improvements in that answer, and you can eventually accept it yourself. – Arturo Magidin Dec 13 '10 at 0:33

Let $b\in R$ be idempotent. Let $c=1-b$ for convenience. Then $b+c=1$ and $bc=0=cb$. Consider now as you have suggested the natural map $f:R \to Rb \times Rc$ given by $f(x) = (xb,xc)$. It is easy to see that $f$ is an $R$-module homomorphism.

If $f(x)=0$ then $xb=0=xc$ and so $0=xb+xc=x(b+c)=x\cdot1=x$, which means that $f$ is injective.

Given $(ub,vc) \in b \times Rc$, we want $x\in R$ such that $xb=ub$ and $xc=vc$. If this is the case, then $x=x\cdot1=x(b+c)=xb+xc=ub+vc$. So, take $x=ub+vc$. Then $xb=ub^2+vcb=ub$, because $b^2=b$ and $cb=0$. Similarly, $xc=vc$. Thus, $(ub,vc)=f(ub+vc)$ and $f$ is surjective.

We have proved that $f$ is an isomorphism. (BTW, I don't think commutativity of $R$ is used here.)

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I've added a full answer because this question has been bumped automatically and has remained unanswered for a long time now. On the other hand, the OP has probably moved on... – lhf Jun 10 '11 at 0:44

I hope this hint isn't completely giving it away but to show surjectivity, you have to use the fact that b(1-b)=0 to construct a preimage for your arbitrary element. Please yell at me if I said too much.

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