Recently, I am reading a book "Elements of the Representation Theory of Associative Algebras".
Let $A$ be a basic and connected finite dimensional $~\mathbb{K}$-algebra and $\{e_1, e_2, \cdots, e_n\}$ a complete set of primitive orthogonal idempotents of $A$. The $($ordinary$)$ quiver of $A$, denote by $Q_A$, is defined as follows:
(a) The points of $Q_A$ are the numbers $1, 2, \cdots, n$, which are in bijective correspondence with the idempotents $e_1, e_2, \cdots, e_n$.
(b) Given two points $a, b \in (Q_A)_0$, the arrows $\alpha:a \to b$ are in bijective correspondence with the vectors in a basis of the $\mathbb{K}$-vector space $e_a(rad A/rad^2 A)e_b$.
If $A$ is a basic and non-connected associative algebra, then how to define $Q_A$? Thank you very much.
In the case that $A$ is basic and non-connected you can define the ordinary quiver (sometimes also called Gabriel quiver) using the same definition.
It will just turn out that the quiver is not connected. More precisely, every finite dimensional algebra $A$ can be written as $A\cong A_1\times\dots\times A_r$ with $A_i$ connected for all $i$. (The $A_i$ are called the blocks of $A$.) Then it turns out that $Q_A$ is the disjoint union of the $Q_{A_i}$. This also explains the name connected as it is equivalent to the quiver of $A$ being (path)connected.