$Ae$ indecomposable iff $e$ is a primitive idempotent

Question

I'm trying to prove the following statement.

"Let $A$ be a (finite dimensional?) algebra over some field $K$. Then $Ae$ is indecomposable if and only if the idempotent $e$ is primitive."

It is clear to me that if $e=e_1+e_2$ for some orthogonal idempotents, then this will yield $Ae=Ae_1\oplus Ae_2$. I am however, unsure about the other direction.

I assume that $Ae=P_1\oplus P_2$, so that $e=e_1+e_2$ for some uniquely determined $e_1\in P_1, e_2\in P_2$. I can't seem to derive the orthogonality or idempotency of $e_1, e_2$.

Answer 1

Let $e_i\in P_i$ with $e=e_1+e_2$. Then $e_1e_2\in P_1\cap P_2$. But $P_1\cap P_2=\{0\}$ and so $e_1e_2=0$. This completes the proof if $A$ is commutative.

Answer 2

$e_1 \in Ae$, so $e_1=re$ for some $r\in A$. Multiplying on the right by $e$ gives $e_1e = re^2= re = e_1$. Since $e=e_1+e_2$, this tells us $e_1^2+e_1e_2 = e_1$. It follows $e_1-e_1^2 = e_1e_2$.

Now $e_1-e_1^2\in P_1$ (because $e_1\in P_1$ and $P_1$ is a submodule) and $e_1e_2 \in P_2$ (because $e_2 \in P_2$ and $P_2$ is a submodule). Since they are equal, they lie in $P_1\cap P_2$ which is $\{0\}$, so they are both zero: $e_1e_2=0$ and $e_1^2=e_1$. Similarly $e_2^2=e_2$ and $e_2e_1=0$. We've established that the $e_i$ are orthogonal idempotents.