Answer to Question #23870 in Abstract Algebra for Hym@n B@ss
Show that any quasi-Frobenius ring is left (and right) Kasch.
For such a ring, we have always thefollowing Double Annihilator Property: if I is any left ideal, then annl (annr(I))= I. Assuming this, consider now anysimple left R-module V , say V = R/m, where m is amaximal left ideal. By the property quoted above, we see that annr(m) <>0. Fix a nonzero element a ∈ R such thatma = 0, and define ϕ : R → R by ϕ(r) = ra. Then ϕ is a nonzero endomorphism of RR,with m ⊆ker(ϕ). Since m is a maximal left ideal,we have m = ker(ϕ). Therefore ϕ(R) ∼R/m = V. So ϕ(R) is the minimal left ideal wesought.