I have proven for $e$ an idempotent in a commutative ring $R$ that there exists an isomorphism $\phi:R\rightarrow R/eR\times R/(1-e)R$ given by $r\mapsto(r,r-re)$.
Now I would like to prove that the idempotents of $R$ correspond bijectively to the 'decompositions' of $R$ as a product of $R=R_1\times R_2$. However I don't fully understand what is meant by this or how to do this. I understood that somehow I have to do something with the sets $\{(e,1-e):\text{e idempotent}\}$ and $\{(R_1,R_2):R=R_1\times R_2\}$ and the maps $R_1\times R_2\mapsto((0,1),(1,0))$ and $(e,1-e)\mapsto R/eR\times R/(1-e)R$.
Extra: I also want to prove that the set of idempotents in $R$ are a ring with the 'usual' multiplication and addition defined by $x+y=(x-y)^2$. Do I have to use the above for this proof (if yes, how) or is this 'easily' done without it?
Best Answer
Expressing it as $\phi:R\rightarrow R/eR\times R/(1-e)R$ is a little weird considering that $R=eR\oplus (1-e)R$ as a direct sum of ideals. No isomorphism is needed. The two things match since $eR\cong R/(1-e)R$ and $(1-e)R\cong R/eR$ as rings. I guess if you are armed with the Chinese remainder theorem, the isomorphism is give for free. Hi it does not really help you understand decompositions.
Really, you would like to prove that for ideals $R_1$ and $R_2$ of $R$, then $R=R_1\oplus R_2$ iff there is an idempotent $e$ in $R$ such that $eR=R_1$ and $(1-e)R=R_2$.
It's easy to show that $R=eR\oplus (1-e)R$. Conversely, suppose $R=R_1\oplus R_2$ and think about what the identity $1$ must look like in terms of $R_1\oplus R_2$. This establishes the link between the idempotent and a particular splitting of $R$ into two pieces.
This is the same thing as considering $R=R_1\times R_2$ as a direct product of rings since you can just view $R_1\times \{0\}$ and $\{0\}\times R_2$ as two ideals which add up to $R$.
The proof above seems unrelated. Just verify that the set of idempotents is a group under the new addition operation, and that multiplication appropriately distributes.
Generally