Is the connected assumption necessary in the following theorem? I'm reading about the following facts from the book 'Elements of the Representation Theory of
Associative Algebras' by Assem, Simpson, and Skrowronski. Note that the algebras involved are not assumed to be commutative. I would like to find a proof of the following fact, usually attributed to Gabriel.
Theorem.(?) Let $A$ be a finite dimensional $K$-algebra, where $K$ is an algebraically closed field. Then, there is an equivalence of categories $A$-mod $\to$ $KQ/I$-mod where $Q$ is a quiver and $I$ is an admissible ideal in the path algebra $KQ$.
My first question is whether or not the above theorem is indeed true.
The way to aforementioned book handles this fact is to first reduce to the case that $A$ is basic using theorem I.6.8. This theorem asserts the existence of a basic algebra $A^b$ and an equivalence of categories $A$-mod $\to$ $A^b$-mod.  In the next chapter, the book proves the following theorem.
Theorem. Let $A$ satisfy the assumptions of the previous theorem, in addition to it being basic and connected. Then there is a quiver $Q$ and an admissible ideal $I \subset KQ$ so that $A \cong KQ/I$.
Is the above theorem true without the connected assumption? If $A$ is connected, do we know $A^b$ is connected? Perhaps a connectedness assumption is missing in my first theorem?
Thank you!
This question and answer lead me to believe the connectedness assumption is unnecessary but I have not found a reference for this.
 A: The case where $A$ is not necessarily connected is deduced from the connected case.  Indeed, let $A$ be any finite-dimensional $K$-algebra.  Then $A = A_1 \times A_2 \times \cdots \times A_r$, with each $A_i$ a connected algebra.  In that case, $A-\operatorname{mod}$ is equivalent to $A_1-\operatorname{mod} \times \cdots \times A_r-\operatorname{mod}$.  Replacing each $A_i$ with a Morita-equivalent basic algebra $A_i^b$, we get that $A$ is Morita-equivalent to $A_1^b \times \cdots \times A_r^b$.  By the theorem you cite, for each $A^b_i$, there is a quiver $Q_i$ and and admissible ideal $I_i$ such that $A^b_i$ is isomorphic to $KQ_i/I_i$.  Taking $Q$ to be the disjoint union of the $Q_i$ and $I$ to be the ideal generated by all the $I_i$ in $KQ$, we get that $A^b_1\times \cdots \times A^b_r$ isomorphic to $KQ/I$.  Thus $A$ is Morita-equivalent to $KQ/I$.  This finishes the proof.
Note that if a quiver $Q$ is not connected and $I$ is an admissible ideal of $KQ$, then $KQ/I$ is not a connected algebra.  Indeed, if $Q$ is the disjoint union of the non-empty quivers $Q_1$ and $Q_2$, then $KQ/I \cong KQ_1/(I\cap KQ_1) \times KQ_2/(I\cap KQ_2)$.
