Let $A$ be a finite dimensional basic algebra over an algebraically closed field $k$. We denote by $D$ the standart duality $D(A) = \text{Hom}_k(A,k)$. The repetitive algebra $\hat{A}$ of $A$ is a Frobenius algebra with underlying vector space $$\hat{A} = \left(\bigoplus_{i\in\mathbb{Z}}A\right)\oplus\left(\bigoplus_{i\in\mathbb{Z}}D(A)\right)$$
and multiplication
$$(a_i,\phi_i)_{i\in\mathbb{Z}}(b_i, \psi_i)_{i\in\mathbb{Z}} = (a_ib_i, a_{i+1}\psi_i + b_i\phi_i)_{i\in\mathbb{Z}}.$$
A finite dimensional $\hat{A}$-module can be viewed as a sequence of $A$-modules $(M_i,f_i)_{i\in\mathbb{Z}}$ with almost all being zero, together with maps
$$f_i : D(A)\otimes M_i \longrightarrow M_{i+1}$$
such that $f_{i+1}(1\otimes f_i) = 0$.
Let $\delta_{pi}$ be Kroneker delta. Corresponding to the decomposition of the local unit $1_p = (\delta_{ip},0)_{i\in\mathbb{Z}}$ in primitive, pairwise orthogonal idempotens, $e_{pj} = (\delta_{ip}e_j,0)_{i\in\mathbb{Z}}$ we get left ideals
$$\hat{A}e_{pj} = \begin{pmatrix}\dots\\0\\Ae_j\oplus D(A)e_j\\0\\ \dots\end{pmatrix},$$ where the non-zero entry is on the $p$-th position. And here is where my problem lies: This module $\hat{A}e_{pj}$ should be indecomposable, but I don't understand why $$\begin{pmatrix}\dots\\0\\Ae_j\\0\\ \dots\end{pmatrix}$$ is not a direct summand. And by that I mean that we find a split mono
$$\iota: \begin{pmatrix}\dots\\0\\Ae_j\\0\\ \dots\end{pmatrix} \longrightarrow \begin{pmatrix}\dots\\0\\Ae_j\oplus D(A)e_j\\0\\ \dots\end{pmatrix}$$
given by locally split monos that are compatible with the morphisms $f_i$ from the definition.
I've been using the book Triangulated Categories in the Representation Theory of Finite Dimensional Algebras by Dieter Happel, so any further literature suggestions or explenation as to why $\hat{A}e_{pj}$ is indecomposable would be extremely appreciated!
I'm trying to understand the simple $\hat{A}$-modules so any literature suggestions regarding this will also be highly appreciated.
Locally we have the algebra $$ \begin{pmatrix}A&0\\D(A)&A\end{pmatrix} $$ and your module is $$ \begin{pmatrix}Ae_j\\D(A)e_j\end{pmatrix}. $$ In other words we have $M_p=Ae_j$, $M_{p+1}=D(A)e_j$, and $f_p$ is just the natural multiplication map $D(A)\otimes Ae_j\xrightarrow{\sim}D(A)e_j$.
Now there is a natural epimorphism to the module $$ \begin{pmatrix}Ae_j\\0\end{pmatrix} $$ which has $M_p=Ae_j$ and $M_{p+1}=0$, but there is no non-zero module homomorphism in the other direction, since this require morphisms yielding the vertical maps in the commutative square $$ \require{AMScd}\begin{CD} D(A)\otimes Ae_j @>>> 0\\ @VVV @VVV\\ D(A)\otimes Ae_j @>{\sim}>> D(A)e_j \end{CD} $$
As for the finite dimensional simple modules for the repetitive algebra, there will be a maximal index $p$ such that $M_p\neq0$. Then every $A$-submodule of $M_p$ will give a submodule of the module for the repetitive algebra, so will be everything. Thus the simple modules for $\hat A$ are the simple $A$-modules supported in a single degree.