Semi-simple commutative algebra

Question

Let $A$ be a semi-simple commutative algebra over a field $F$, and $F$ is algebraically closed. The proposition is that we can express $A=Fe_1 \oplus ... \oplus Fe_n$, where $e_i$ are orthogonal idempotents. How to prove this?

I know that by Artin-Wedderburn theorem $A$ is isomorphic as a ring to direct product of fields.

Answer 1

Central idempotents in rings with identity are directly tied to ring decompositions into a product of ideals. (In commutative rings, these are all idempotents, of course.)

You can easily prove this lemma:

If $I$ and $J$ are ideals of a ring $R$ with identity such that $I\oplus J=R$, then there exist two central idempotents $e,f$ such that $eR=I$, $fR=J$. (Hint: the obvious candidates arise from $e+f=1$.)

You can easily state and prove a version for $\bigoplus_{i=1}^n I_i=R$ if you can solve this special case.

That pretty much answers how "the idempotents appear."

For the question about orthogonality, you should see that if $I\oplus J$ is a direct sum of ideals, then $ij=0$ for any $i\in I$, $j\in J$, because $ij\in I\cap J=\{0\}$.