Basic finite dimensional K-algebra isomorphic to bound quiver algebra

I am currently working my way through the book 'elements of the represenation theory of associative algebras' by Assem, Simson and Skowronski.

In chapter 2, theorem 3.7 states:

Let $A$ be a basic and connexted finite dimensional $K$-algebra. there exists an admissable ideal I $KQ_A$ such that $A \cong KQ_A/I$.

(Definition basic algebra): Assume that $A$ is a $K$-algebra with a complete set of primitive orthogonal idempotents {e_1, \dots, e_n}. The algebra $A$ is called basic if $e_iA \not\cong e_j A$, $\forall i \neq j$.

In this book (or at least for this chapter) $K$ is an algebraically closed field.

To my knowledge, the requirement for $A$ to be connected is not completely necessary, this only would imply that the associated quiver would be connected.

My question is the following: Why do we only do this for basic algebra's? In other words; where would the string of proofs fail for this theorem if we don't require our algebra to be basic?

My guess would be that one would fail to prove that the associated quiver is well defined in the first place, but I am not entirely sure.

Any help would be greatly appreciated.

I don't think there should be a problem with defining the quiver $Q_A$ of an arbitrary finite-dimensional $K$-algebra $A$, take a complete set of primitive orthogoanl idempotents $\{e_1, \dots, e_n\}$ as points and $\dim_K(e_i rad(A)/rad^2(A) e_j)$ arrows from $e_i$ to $e_j$. This is independent of the choices of the $e_i$ for the same reasons as for basic algebras. However, I believe this construction wouldn't be of much use in the general case.
Say, you take $A = K^{n \times n}$, then you have a complete set $\{e_1,\dots, e_n\}$ of orthogonal primitive idempotents. The problem now is that in $Q_A$ we cannot see that the modules $A e_i \cong K^n$ are all isomorphic. Note that in this example $(Q_A)_1$ would be empty since $rad(A)$ is trivial. You would have $K Q_A \cong K^n$ and no useful morphism $K Q_A \to A$ to work with.
On the other hand, for arbitrary $A$, you can take a set of orthogonal primitive idempotents ${e_1, \dots, e_n}$ which is not necessarily complete, such that every isomorphism type of the $A e_i$ appears exactly once in this set. Then $e A e$ is a basic algebra with $e = \sum_{i = 1}^n e_i$ and at least the representation theories of $A$ and $e A e$ are the same (the algebras are Morita equivalent). Then you have a morphism $K Q_{eAe} \to eAe \to A$. In general it will not be unital, of course, but I do not think you can do much better.
• Could you perhaps elaborate on your second alinea? I don't see what you mean by saying 'we cannot see that the modules $Ae_i \cong K^n$ are all isomorphic'. – M.v.Roozendaal May 27 '17 at 16:19
• I mean that from the information encoded in $Q_A$, you are not able to tell that the $A e_i$ are all isomorphic. In general, $Q_A$ only tells you about the number of idempotents occurring in a complete set of orthogonal primitive idempotents and the dimension for the spaces of morphisms $\text{Hom}(A e_i, rad(A e_j)/rad^2(A e_j))$. For basic algebras, this tells you everything because you already know that $A e_i$ and $A e_j$ are not isomorphic, but in general, you would also need information about the dimensions of spaces like $\text{Hom}(A e_i, A e_j/rad(A e_j))$, for example. – Matthias Klupsch May 28 '17 at 5:59