I'm looking for an example of a finite dimensional algebra of infinite global dimension with a non projective module $X$ with no self-extensions, that is $\text{Ext}_A^i(X,X) = 0$ for $i>0$. The only algebra that came to my mind was $k[X]/x^n$ for a field $k$, but I didn't find any module with no self-extensions.
Any help will be very much appreciated!
Take the path algebra of the quiver with two vertices $1,2$, two arrows $a\colon 1\to 2$ and $b\colon 2\to 2$, and one relation $b^2=0$.
Then the indecomposable projective $P_1$ has vector space $k$ at vertex 1, $k^2$ at vertex 2, and where $a$ and $b$ act via the matrices $$ a=\begin{pmatrix}0\\1\end{pmatrix}, \quad b=\begin{pmatrix}0&1\\0&0\end{pmatrix}. $$ The indecomposable projective $P_2$ is the submodule of $P_1$ having vector space 0 at vertex 1 and $k^2$ at vertex 2.
The projective resolutions of the simples $S_1$ and $S_2$ are now $$ 0 \to P_2 \to P_1 \to S_1 \to 0 \qquad\textrm{and}\qquad \cdots \to P_2 \to P_2 \to S_2 \to 0. $$ We see that $\mathrm{p.dim}\,S_1=1$, $\mathrm{p.dim}\,S_2=\infty$, so that the algebra has infinite global dimension, but $\mathrm{Ext}^i(S_1,S_1)=0$ for all $i>0$.
(Note that $S_1$ is injective, and this example is a 'one-point extension' of $k[x]/x^2$.)