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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.2010-12-13
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    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!2010-12-13
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    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? Thanks2010-12-13
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    @lucy: Your last comment is not true. For example, in boolean rings, every element is an idempotent.2010-12-13
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    @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.2010-12-13
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    @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.2010-12-13

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