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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?

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However I don't fully understand what is meant by this or how to do this.

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 $R$.

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?

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

  • For any idempotent in $R$, $eRe$ is a subring of $R$ with identity $e$.
  • If $e$ is a central idempotent, $eRe=eR$ and you get this splitting of the ring. Splittings of the ring correspond to central idempotents.
  • For a noncentral idempotent, you get $R=eR\oplus (1-e)R$ as right $R$ modules, still. (And a similar statement as left $R$ modules.) Splittings as a module correspond to idempotents.
  • For any idempotent $e$, you have $R\cong \begin{bmatrix}eRe&eR(1-e)\\(1-e)Re&(1-e)R(1-e)\end{bmatrix}$ with formal matrix multiplication. As you can see, when $e$ is central the off-diagonal entries are zero, and you're looking at a "block sum" of two rings $eRe$ and $(1-e)R(1-e)$.
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  • $\begingroup$ The $\phi$-isomorphism part is done already, that's not a problem. But can you help me with the bijective correspondence part and those maps? $\endgroup$ – user235238 Sep 18 '15 at 19:05
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    $\begingroup$ @TonyStrong That is the purpose of the entire first half of this post. For every idempotent there is a decomposition, for every decomposition there is an idempotent. $\endgroup$ – rschwieb Sep 18 '15 at 19:52
  • $\begingroup$ But how do I use those maps i described? $\endgroup$ – user235238 Sep 18 '15 at 21:07
  • $\begingroup$ @TonyStrong I don't think it is a good use of time to use that map to the product of quotients. The point is that in $R_1\times R_2$, $e=(1,0)$ and $(1-e)=(0,1)$ are idempotents, and $R=eR\oplus (1-e)R$ and $eR\cong R_1$ and $(1-e)R\cong R_2$. You can go further and show $eR\cong R/(1-e)R$, but I don't see what advantage there is. $\endgroup$ – rschwieb Sep 19 '15 at 3:12

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