# Can we always choose 2 nonisomorphic simple modules to satisfy the following nonvanishing extension conditions?

$$\DeclareMathOperator\Ext{Ext}\DeclareMathOperator\pd{pd}$$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?

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 $$i=1$$.
• @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 $26$. – Jeremy Rickard Jun 21 at 10:05