 # if e 1 </msub> , &#x2026;<!-- … --> , e n </msub> is a Augustus Acevedo 2022-07-01 Answered
if ${e}_{1},\dots ,{e}_{n}$ is a complete set of orthogonal idempotents in a commutative ring, then $R=R{e}_{1}×\cdots ×R{e}_{n}$ is a direct product decomposition.
How can be proved this?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it Shawn Castaneda
Define a map $f:R\to R{e}_{1}×\cdots ×R{e}_{n}$ by $f\left(x\right)=\left(x{e}_{1},\dots ,x{e}_{n}\right)$ and prove that $f$ is a surjective ring homomorphism with $\mathrm{ker}f=\left(0\right)$. (Note that $R{e}_{i}$ are unitary commutative rings and $f$ is a homomorphism of unitary rings.)
###### Not exactly what you’re looking for? Willow Pratt
You have ${1}_{R}=\sum _{i=1}^{n}{e}_{i}$ and ${e}_{i}{e}_{j}={\delta }_{ij}{e}_{i}$ for each $i,j.$. From here on in, it is just a matter of writing down consequences of that.