Extension-algebras of $A_4$ Consider the quiver $Q\colon 1\xrightarrow{\alpha} 2\xrightarrow{\beta} 3\xrightarrow{\gamma} 4$ and the algebra $A=k[Q]/(\gamma\beta\alpha)$. Denote the simple $A$-modules by $L(-)$ and let $M$ be the direct sum of the simple submodules.
What is the algebra $\operatorname{Ext}^*(M,M)$, and, more importantly, how can it be computed? I'd guess that the simples have extensions
$$\begin{aligned}\operatorname{Ext}^1(L(2),L(1))\colon\quad & 0 \to L(1) \xrightarrow{\alpha} e_2A =\langle e_2, \alpha\rangle \to L(2)\to 0\\
\operatorname{Ext}^1(L(3),L(2))\colon\quad & 0\to L(2) \xrightarrow{\beta} e_3A =\langle e_3, \beta\rangle \to L(3)\to 0\end{aligned}$$
and so on, but I have the impression that with this approach, I don't end up with the correct algebra. For example, how do I know that $\operatorname{Ext}^1(L(3), L(1))=0$ and does not contain e.g.
$$0\to L(1)\to \langle e_3, \alpha\beta\rangle \to L(3)\to 0?$$
I guess it's because the middle term is no $A$-module… but still: even if I find some extensions, how do I know I find all, and then, their relations?
 A: Ext is an additive functor, so it suffices to know the$\operatorname{Ext}^*(L(i), L(j))$ and their compositions. We can compute Ext-algebras using projective resolutions. We choose
$$\begin{aligned}
0\to P(1) \to L(1)\to 0\\
0 \to P(1) \to P(2) \to L(2) \to 0\\
0 \to P(2) \to P(3) \to L(3) \to 0\\
0 \to P(1) \to P(3) \to P(4) \to L(4) \to 0
\end{aligned}$$
as projective resolutions of the respecitve simple modules. Then, for instance 
$$\operatorname{Hom}^2(P(4), P(1)) = \left\langle
\begin{matrix} P(1) & \to & P(3) & \to & P(4)\\
\downarrow\\
P(1)\end{matrix}\right\rangle$$
and all other $\operatorname{Hom}^*(P(4), P(1))$ are zero. This yields that $\operatorname{Ext}^*(P(4), P(1))$ is concentrated in degree 2.
As another example, one sees that 
$$\operatorname{Hom}^2(P(1), P(4)) = \left\langle\begin{matrix}
& & & & P(1)\\
& & & \swarrow & \downarrow\\
P(1) & \to & P(3) & \to & P(4)
\end{matrix}\right\rangle$$
is spanned by a single degree-0-map that factors through the indicated degree-1-map and thus is null-homotopic. Taking homology gives that $\operatorname{Ext}^*(P(1), P(4))=0$.
All other Ext-spaces are obtained similarly. Composition is given by composition of these diagrams.
