Let A be an algebra. We denote by by A-proj the full subcategory of A-mod consisting of projective modules. An A-module T is called a tilting module if $proj.dim(\_{A}T)=n < \infty$, $Ext\_{A} ^{j} (T,T) =0$ for all $j > 0$, and there is an exact sequence $ 0 \rightarrow \_A A \rightarrow X\_0 \rightarrow X\_1 \rightarrow \cdots \rightarrow X\_n \rightarrow 0$ in A-mod with all $X\_j \in add(T)$.


Let T be a tilting A-module of projective dimension $n \geq 1$, then fix a minimal projective resolution of T as follows: $ 0 \rightarrow P\_n \rightarrow P\_{n-1} \rightarrow \cdots \rightarrow P\_1 \rightarrow P\_0 \rightarrow T \rightarrow 0$.


Then can anyone tell me why A-proj =$ add(\oplus \_{i=0} ^n P\_i)$? Thank you.



The projective modules of an algebra and the tilting module?
A more general argument is that if you delete T from the minimal projetive resolution, you get a tilting complex. (see for example the book by Happel)
But by definition, for a tilting complex M, add(M) generates the bounded homotopy category of finitely generated projective modules, and this is here only possible if the A-proj =$add(\bigoplus P\_i)$ condition is fullfilled.


