Let $A = kQ/\rho $, $Q$ is the quiver \begin{align} 1 \overset{a}{\underset{a^*}{\rightleftarrows}} 2 \end{align} $\rho$ is the relation $a a^* - a^* a = 0$. Question: compute $Ext_A^{1}(S_1, S_2)$.
My solution: Since $aa^*$ and $a^*a$ start from different vertices, $\rho$ is the same as $a a^*=0$, $a^* a=0$. We have the following indecomposable modules of $A$: \begin{align} S_1 = \mathbb{C} \overset{0}{\underset{0}{\rightleftarrows}} 0, \quad S_2 = \mathbb{0} \overset{0}{\underset{0}{\rightleftarrows}} \mathbb{C}, \quad P(1) = \mathbb{C} \overset{1}{\underset{0}{\rightleftarrows}} \mathbb{C}, \quad P(2) = \mathbb{C} \overset{0}{\underset{1}{\rightleftarrows}} \mathbb{C} \end{align} Let $\dim(M) = (d_i)_{i\in I}$ be the dimension vector of $M$. The following exact sequence is a projective resolution of $M$: \begin{align} 0 \to \bigoplus_{\alpha \in Q_1} d_{s(\alpha)} P(t(\alpha)) \to \bigoplus_{i \in Q_0} d_{i} P(i) \to M \to 0 \end{align} For $M=S_2$, we have $\dim(S_1) = (1, 0)$ and an exact sequence \begin{align} 0 \to P(2) \to P(1) \to S_1 \to 0 \end{align} After applying $\hom(S_2, -)$ to the above exact sequence, we obtain a left exact sequence \begin{align} 0 \to \hom(S_2, S_1) \to \hom(S_2, P(1)) \overset{f_1}{\to} \hom(S_2, P(2)) \overset{f_2}{\to} 0 \end{align} I think that $Ext_A^1(S_1, S_2) = \ker(f_2)/im(f_1) = \hom(S_2, P(2))/im(f_1)$. But I don't konw $im(f_1)$.
How to compute $Ext_A^{1}(S_2, S_1)$ and how to show that $Ext_A^{1}(S_1, S_2)=1$? Thank you very much.