Number of indecomposable direct summands of $A/AeA$ for a finite dimensional basic algebra $A$ and idempotent $e$
Solution 1:
Every indecomposable projective $A$-module is isomorphic to $Ae_i$ for some $i$ and every simple $A$-module is isomorphic to $Ae_i/\text{rad}\,Ae_i$ for some $i$. Because $A$ is basic, we have $Ae_i \cong Ae_j$ iff $i=j$ and $Ae_i/\text{rad}\,Ae_i \cong Ae_j/\text{rad}\,Ae_j$ iff $i=j$. Now $A/AeA$ is an $A$-$A$-bimodule, so we obtain a decomposition of $A/AeA$ as a left $A$-module via
$A/AeA = \bigoplus_{i=1}^n (A/AeA)e_i = \bigoplus_{i=1}^n (Ae_i+AeA)/AeA \cong \bigoplus_{i=1}^n Ae_i/(Ae_i\cap AeA) $.
For $1\leq i\leq k$ clearly $Ae_i \cap AeA = Ae_i$, so $Ae_i/(Ae_i\cap AeA) \cong 0$. For $i > k$ we will show that $Ae_i \cap AeA \subsetneq Ae_i$, so $Ae_i/(Ae_i\cap AeA)$ is a proper quotient of an indecomposable projective $A$-module and hence indecomposable. It follows that $m = n-k.$
Now the only thing that is left to show is $Ae_i\cap AeA \subsetneq Ae_i$ for $i>k$. For that we will prove the following claims:
- For any $i$ we obtain $e_i \notin \text{rad}\,A$.
- For $i\neq j$ we obtain $e_iAe_j \subseteq \text{rad}\,A$.
1: We have $\text{rad}\,A = \bigoplus_{i=1}^n \text{rad}\,(Ae_i).$ If $e_i \in \text{rad}\,A$ it would follow that $e_i \in \text{rad}\,Ae_i$. But then $Ae_i = \text{rad}\,Ae_i$, which is a contradiction.
2: For $i\neq j$ the simple modules $Ae_i/\text{rad}\,Ae_i$ and $Ae_j/\text{rad}\,Ae_j$ are not isomorphic. It follows that
$0 = \text{Hom}_A(Ae_i, Ae_j/\text{rad}\,Ae_j) = e_i (Ae_j/\text{rad}\,Ae_j) = (e_i Ae_j+\text{rad}\,Ae_j)/\text{rad}\,Ae_j$.
Hence $e_iAe_j \subseteq \text{rad}\,Ae_j \subseteq \text{rad}\,A$.
With 1 and 2 we have everything we need to prove $Ae_i\cap AeA \subsetneq Ae_i$ for $i>k$. Assume equality holds. Then $e_i \in AeA$, so $e_i = e_i^2 \in AeAe_i$. With 2 we obtain $eAe_i = \sum_{j=1}^k e_jAe_i \subseteq \text{rad}\,A$, so $e_i \in AeAe_i \subseteq \text{rad}\,A$. This is a contradiction to 1.