Is the ring $R$ a commutative ring with identity in the Freyd–Mitchell embedding theorem?

Question

The Freyd–Mitchell embedding theorem says that every abelian category is a full subcategory of a category of modules over some ring $R$ and that the embedding is an exact functor. Does it mean $R$ is a commutative ring with identity?

Answer 1

No, the ring $R$ in question looks something like $End(P)$ for some projective generator $P$ and has no reason to be commutative in general (although it does have a unit !)