QUASI-FROBENIUS RINGS

I’ll sketch a solution to the following interesting problem, where R is commutative and Noetherian. (A full treatment would work with non-commutative rings.) I once assigned this problem. I misremembered at the time and imagined that it was easier than it is. I think that it should be easier, and I invite somebody to come up with a simpler proof, probably making use of homological algebra.

(1) R is said to be quasi-Frobenius if R is injective as an R-module. Prove that the following are equivalent.

(a) R is quasi-Frobenius.

(b) Every projective R-module is injective.

(c) Every injective R-module is projective.

Trivially, (b) implies (a), and (a) implies (b) since direct sums of injectives in a Noetherian ring are injective, as are direct summands of injectives. The equivalence of these conditions with (c) is a non-trivial result of Carl Faith and Elbert Walker, Direct–sum representations of injective modules, J. Algebra 5(1967), 203–221.

(a) ⟹ (c): R is Artinian (you can prove this) and any module M is a direct sum of indecomposable modules, injective if M is injective. Thus to show that injectives are projective, it suffices to show that an indecomposable injective module M is projective. Any module N has an injective hull (or envelope) N̂. This is an injective module containing N such that if the intersection of N and a submodule N′ is zero, then N′ is zero; N̂ embeds in any other injective module containing N. If C is the submodule of M generated by a nonzero element, then Ĉ = M. A less hard characterization of R being quasi–Frobenius is that every ideal I is the annihilator of another ideal J. (The annilator of an annihilator is the ideal you started with). A cyclic module is isomorphic to R/I for some I, and if I = ann(J), where J is generated by a subset {x_{1}, ⋯, x_{n}} of R, then r ↦ (rx_{1}, …, rx_{n}) induces an embedding R/I ⟶ R^{n} since I is the annhilator of {x_{i}}. Thus our C embeds in some R^{n}. Since R is injective, so is the free R-module R^{n}, and the embedding extends to an embedding of Ĉ = M. Since R^{n} is injective, the inclusion splits and M is a direct summand of R^{n}. Therefore M is projective.

(c) ⟹ (a): Since any module embeds in an injective module and injective modules are projective, any module embeds in a free module. It is a fact — not too difficult to prove — that if every module embeds in a direct sum of cyclic modules, then R is Artinian. A module C is said to be a “cogenerator” if every module embeds in a Cartesian product of copies of C. The main theorem of Faith and Walker is that if a ring admits a finitely generated cogenerator C, and if R/rad(R) is semi-simple, then C and R are injective. In our case, R itself is a cogenerator (direct sums embed in direct products) and R/rad(R) is semi-simple (since R is Artinian). Therefore R is injective, which means that R is quasi-Frobenius.