Let $R$ be a ring. An element $e \in R$ is called idempotent iff $e^2=e$.
If $R$ is an integral domain, then $0,1$ are the only idempotent elements.
If $1=e_1+\cdots e_s$ with elements $e_i \neq 0$ and $e_ie_j =0$ for $i \neq j$, the elements $e_i$ are idempotent and $$ R \cong \prod_{i=1}^{s} R_i,$$ with the rings $R_i:=Re_i$.
How to prove the isomorphism ?
If I define the map $f: R \to \prod_{i=1}^{s} R_i=\prod_{i=1}^{s} Re_i$ by $$f(x)=f(x \cdot 1)=f(x(e_1+\cdots+e_s))=(xe_1,xe_2, \cdots, xe_s).$$ Clearly this map is a homomorphism.
Also $f(x)=0$ implies $(xe_1,xe_2, \cdots, xe_s)=0 \Rightarrow x=0$ because $e_i \neq 0$.
So $f$ is injective.
How to show $f$ is surjective ?
We have $1e_i=(e_1+..+e_n)e_i=e_i^2=e_i$ since $e_ie_j=0, i\neq j$.
Let $(x_1,...,x_n)$, write $x=x_1+...+x_n$, $xe_i=x_1e_i+...+x_ne_i$, we have $x_j=y_je_j$ implies that $x_je_i=y_je_ie_j=0, i\neq j$ and $x_ie_i=y_ie_ie_i=y_ie_i=x_i$. We deduce that $xe_i=x_i$.