# Example of Minimal Projective Resolutions

I am reading "Elements of the representation theory of associative algebras"'s book of Skowronski, Simson and Assem. I want to compute the global dimension of the example 2.5 c), of chapter 3, page 80.

By Auslander's theorem, I only need to compute the projective dimension of the simple modules.

Lets start with $S(1)$, since the quotient map $P(1) \to S(1)$ is a projective cover, it is the start of a minimal projective resolution, and this map has kernel $rad\big(P(1) \big)$. Now:

and:

I want to know how to continue with the resolution. I already know it has infinite global dimension because of the Cartan matrix.

That is a good start. What you need to do now is compute a projective cover of $rad(P(1))$. On this small example, it is not so hard to convince yourself that there is a short exact sequence $$0\to S(1) \to P(1) \to rad(P(1)) \to 0.$$ To find the next term in the projective resolution, you would then need to compute (again) a projective cover of $S(1)$; thus the projective resolution you are looking for will be periodic, and have the form $$\ldots \stackrel{\cdot\lambda^2}{\to} P(1) \stackrel{\cdot\lambda}{\to} P(1) \stackrel{\cdot\lambda^2}{\to} P(1) \stackrel{\cdot\lambda}{\to} P(1) \to S(1) \to 0.$$ Hence $S(1)$ has infinite projective dimension, as you guessed.
• One way to compute the projective cover of a module $M$ is to first compute its top, that is, $M/rad(M)$. This is a semisimple module, and its projective cover lifts to the projective cover of $M$. – Pierre-Guy Plamondon Nov 11 '15 at 23:13