Answer to Question #24882 in Abstract Algebra for john.george.milnor

Question

Let I ⊆ R be a right ideal containing no nonzero nilpotent right ideals of R. (For instance, I may be any right ideal in a semiprime ring.) Show that the following are equivalent:
(1) IR is an artinian module;
(2) IR is a finitely generated semisimple module. In this case, show that
(3) I = eR for an idempotent e ∈ I.

Answer 1

(2) ⇒ (1) is clear. Conversely, assume (1). If I is nonzero , itcontains a minimal right ideal A1. Since A1 cannot be nilpotent,Brauer’s Lemma implies that A1 = e1R for some idempotent e1.Let B1 = I ∩ (1 − e1)R, so I = A1 ⊕ B1. If B1 <> 0, the sameargument shows that B1 contains a minimal right ideal A2 = e2Rwhere e₂²= e₂. Since e2∈ (1 − e1)R, we have e1e2 = 0. By astraightforward calculation, e'2 : = e1 + e2(1 − e1)is an idempotent, and A1 ⊕ A2 = e'2R.Now let B2 = I ∩ (1 − e'2)R. Since 1 − e'2 =(1 − e1) + (1 − e1)e2(1 − e1) ∈ (1 − e1)R, we have B2 ⊆ B1. If B2 <> 0, we can takea minimal right ideal A3 = e3R ⊆ B2, and continue this construction. Since B1> B2 > · · · , this construction process must stop in afinite number of steps. If, say, Bn = 0, then
IR = A1 ⊕· · ·⊕An
is semisimple, and equal to e'nR forthe idempotent e'n, as desired.