Equivalence in definition of quasi-Frobenius ring

Question

A quasi-Frobenius ring is defined to be a ring which satisfies the following conditions (I'm phrasing it with left modules, but it's the same with right modules):

(i) $R$ is Noetherian and self-injective (as a left $R$-module)

(ii) All projective left $R$-modules are injective

(iii) All injective left $R$-modules are projective.

It is claimed on wikipedia and other sources that those 3 conditions are equivalent.

I know how to prove that (i) implies (ii), but I'm quite clueless about the rest.

I have some specific situations in mind (e.g. the group algebra of a finite group over a field) where I can prove that they're all satisfied, but the situations are especially nice : for instance, assuming (ii) in some situations I can get that any injective module is a submodule of some projective module, in which case (iii) follows easily, but beyond that I have no clue.

Maybe some duality theory could help me, by considering say $\hom(M,I)$ for some injective module $I$, but I'm not getting anywhere with this.

I'm also wondering how specific to rings this is :

Let $C$ be an abelian category where all projectives are injective. Is it necessarily the case that all injectives are projective ? What if we have enough injectives/projectives ? Maybe a projective generator ?

And independently :

How can we see that conditions (i) through (iii) are equivalent for a ring ?

Answer 1

To show that (ii)+(iii) implies (i), firstly since $R$ is a free module over itself it is projective, so therefore injective by (ii). It is a theorem that if arbitrary direct sums of injective left modules are injective then $R$ is left Noetherian. This is proven on page 471 here. Since the direct sum of projective modules is projective, and injectivity and projectivity are equivalent in our case, we are done.

If we look at the characterisations of projectivity and injectivity via the splitting of exact sequences then the symmetry and equivalence of (ii) and (iii) becomes clear.

Then since you can show that (i) implies (ii) we are done.

I'm sorry I can't be more helpful as far as the categorical questions are concerned.

Edit: My reasoning for the equivalence of (ii) and (iii) was incorrect, but a proof of the equivalence of these characterisations is given as Theorem 24.20 in Algebra II Ring Theory by Carl Faith.