$\newcommand\rad{\operatorname{rad}}$When reading about the construction of Gabriels Ext-quiver, I often read somethink like the following. Let $A$ be a f.d. algebra over an algebraically closed field. Then its Ext quiver $Q$ is either
- the quiver with vertices corresponding to isoclasses of simple $A$-modules $S$ and $\dim \operatorname{Ext}^1(S_j,S_i)$ many arrows $S_i \to S_j$, or
- the quiver with vertices corresponding to isoclasses of indecomposable projective $A$-modules $P$ and $\dim \operatorname{Hom}(P_i, \rad P_j/\rad^2 P_j)$ many arrows $P_i \to P_j$.
I assume that $P_i$ is the projective cover of $L_i$ for all $i$. The statement that both dimensions coincide is stated e.g. here.
My question is why $\dim \operatorname{Ext}^1(S_j,S_i) = \dim \operatorname{Hom}(P_i, \rad P_j/\rad^2 P_j)$. Are they isomorphic as vector spaces?
Additionally, I'd like to know a reference to Gabriel's theorem, saying that every finite dimensional (basic?) algebra is isomorphic to its Ext-quiver modulo an ideal generated by paths of length at least two. The theorem is also stated here – alas, I cannot find a copy of his indecomposable representations ii.
Both statements together made me wonder if $\operatorname{Ext}^*(\bigoplus_i S_i) \cong \operatorname{Hom}(\bigoplus_i P_i)$ as algebras. Is that true for e.g. basic $A$?