In Hatcher, Lemma 3.3.1 the author shows that two free resolutions $F,F'$ of an abelian group $H$ are homotopy equivalent. He later notes that this is still true for $R$-modules $H$ over any principal ideal ring $R$ and this allows him to define the functor $\operatorname{Ext}_R(\,\_\,,H)$.
Diving a bit further into the theory, I noticed that in general people use projective resolutions in order to define $\operatorname{Ext}^n_R(\,\_\,,H)$ since there is an analogous statement to Lemma 3.3.1 for projective resolutions of modules over arbitrary rings.
However, it seems to me that for the definition of $\operatorname{Ext}^n_R(\,\_\,,H)$ we actually don't have to use the greater generality of the statement since we know that there is always a free resolution and that every free resolution is projective. My question is: would it make any difference to define $\operatorname{Ext}^n_R(\,\_\,,H)$ via free instead of projective resolutions in the general case? If not, why do people define it via projectives when they compute it via free resolutions anyway (most of the time, at least that's my impression).
EDIT: Taking into account Pedro's comment I would like to rephrase the question as follows:
Can we define a model structure on $\operatorname{Ch}_{\bullet \geq 0}(\operatorname{Mod}_R)$ by taking free resolutions to be the cofibrant objects so that the corresponding homotopy category is the same as for the projective model structure?
Best Answer
Indeed, it doesn't make a difference whether you use free resolutions or projective resolutions in the definition, since free resolutions are a special case of projective resolutions and they always exist. The notion of projective resolution generalizes better to broader situations, though. For instance, it makes sense to talk about projective resolutions in an arbitrary abelian category and use them to define Ext (and other derived functors), but an abstract abelian category does not have any notion of a "free object".