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$\DeclareMathOperator\Ext{Ext}\DeclareMathOperator\pd{pd}$Let A be a finite dimensional algebra over some field k and S a nonprojective simple left A-module. Suppose the projective dimension $\pd_A(S)$ of $S$ is finite. Let $n$ be a nonnegative integer such that $n+1\leq \pd_A(S)$. Then we can find 2 simple left $A$-modules $S_1, S_2$ such that $\Ext_A^{n}(S, S_1)\neq 0$, $\Ext_A^{n+1}(S, S_2)\neq 0$. My question is the following:

Can we always choose nonisomorphic simple $A$-modules $S_1, S_2$ in the above situation?

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Let $A$ be the path algebra of the quiver with three vertices $1$, $2$, $3$ with arrows $a:1\to 2$, $b:2\to 3$, $c:3\to 1$, modulo relations $bca=0$, $cabc=0$, so the indecomposable projectives are uniserial modules $$P_1=\begin{matrix}S_1\\S_2\\S_3\\S_1\end{matrix},\quad\quad\quad P_2=\begin{matrix}S_2\\S_3\\S_1\end{matrix},\quad\quad\quad P_3=\begin{matrix}S_3\\S_1\\S_2\\S_3\end{matrix}$$

Then $S_3$ has a finite minimal projective resolution $$0\rightarrow P_2\rightarrow P_1\rightarrow P_1\rightarrow P_3\rightarrow S_3\rightarrow0,$$

and so $\operatorname{pd}_A(S_3)=3$, and both $\operatorname{Ext}^1_A(S_3,S_i)$ and $\operatorname{Ext}^2_A(S_3,S_i)$ are nonzero only for $i=1$.

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    $\begingroup$ Should that be bca=0 $\endgroup$ – Benjamin Steinberg Jun 21 at 10:01
  • $\begingroup$ @BenjaminSteinberg Yes, thanks. I've fixed it now. I think I was using the alphabet mod $3$, which doesn't really work, since $3$ doesn't divide $26$. $\endgroup$ – Jeremy Rickard Jun 21 at 10:05
  • $\begingroup$ @JeremyRickard Very nice example! Thanks for your answer. $\endgroup$ – Master Gang Jun 21 at 13:00

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