# Are RHom and Ext the same thing?

The right-derived functor of the Hom functor is called the Ext functor or the RHom functor. The choice between calling it Ext or RHom doesn't seem to depend on anything which I can see.

Are Ext and RHom the same thing? Is there a reason to pick one over the other depending on context, or existence/non-existence of extra structure?

• I often see $\operatorname{Ext}$ when we want the $i$-th derived functor, that is $\operatorname{Ext}^i$ while $R\operatorname{Hom}$ is more often used as the total derived functor, in other words an object of the derived cateogry (a complex up to quasi-isomorphism). The relationship is thus $\operatorname{Ext}^i=H^i(R\operatorname{Hom})$ (or something similar if instead of the derived category, we have the homotopy category of some model category). I guess some author do not use this choice of notation (I won't even say convention because it isn't), but this is what I see the most often. – Roland Oct 26 '20 at 12:33
• I guess, we don't use $\operatorname{Ext}$ for the total derived functor because this notation is more often use for the group of extension in an abelian category (this is the case for instance in the linked nLab page for the Ext functor), so that we are able to claim that there is a natural isomorphism $\operatorname{Ext}=\operatorname{Ext}^1$. – Roland Oct 26 '20 at 12:41
• Thank you! I will happily accept these comments as an answer if you'd like to copy them into an answer :) – Matt Oct 26 '20 at 14:24

I often see $$\operatorname{Ext}$$ when we want the i-th derived functor, that is $$\operatorname{Ext}^i$$ while $$R\operatorname{Hom}$$ is more often used as the total derived functor, in other words an object of the derived cateogry (a complex up to quasi-isomorphism). The relationship is thus $$\operatorname{Ext}^i=H^i(R\operatorname{Hom})$$ (or something similar if instead of the derived category, we have the homotopy category of some model category). I guess some author do not use this choice of notation (I won't even say convention because it isn't), but this is what I see the most often.
I guess, we don't use $$\operatorname{Ext}$$ for the total derived functor because this notation is more often use for the group of extensions in an abelian category (this is the case for instance in the linked nLab page for the Ext functor), so that we are able to claim that there is a natural isomorphism $$\operatorname{Ext}=\operatorname{Ext}^1$$.