Let $A$ be a finite dimensional commutative algebra over a field $K$ which can be written as a direct product of simple algebras $A\cong \prod_{i=1}^r A_i.$
My question:
Is it possible to write any idempotent of $A$ as a sum of the primitive idempotents associated with this decomposition?
Thank you very much for your help.
Sure. By the Artin-Wedderburn theorem, each of the $A_i$ are finite field extensions of $K$.
Any idempotent of $\prod_{i=1}^r A_i$ would have to be of the form $(a_1,\ldots,a_r)$ where $a_i$ is an idempotent in $A_i$. There are only two idempotents in each field, though, $\{0,1\}$.
I'll use $e_i$ to denote the element of the product that's $1$ on the $i$th coordinate and zero elsewhere. According to the above description, it's obvious that every idempotent is just a sum of these $e_i$'s (the idempotent $0$ being interpreted as an empty sum.)