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

Question #24882

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.

(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.

Expert's answer

(2)

is semisimple, and equal to

*⇒**(1) is clear. Conversely, assume (1). If**I*is nonzero , itcontains a minimal right ideal*A*1. Since*A*1 cannot be nilpotent,Brauer’s Lemma implies that*A*1 =*e*1*R*for some idempotent*e*1.Let*B*1 =*I ∩*(1*− e*1)*R,*so*I*=*A*1*⊕**B*1. If*B*1*<>*0, the sameargument shows that*B*1 contains a minimal right ideal*A*2 =*e*2*R*where*e*_{2}^{2}=*e*_{2}. Since*e*2*∈**(1**− e*1)*R*, we have*e*1*e*2 = 0. By astraightforward calculation,*e'*2 : =*e*1 +*e*2(1*− e*1)is an idempotent, and*A*1*⊕**A*2 =*e'*2*R*.Now let*B*2 =*I ∩*(1*− e'*2)*R*. Since 1*− e'*2 =(1*− e*1) + (1*− e*1)*e*2(1*− e*1)*∈**(1**− e*1)*R,*we have*B*2*⊆**B*1. If*B*2*<>*0, we can takea minimal right ideal*A*3 =*e*3*R**⊆**B*2, and continue this construction. Since*B*1>*B*2 >*· · ·*, this construction process must stop in afinite number of steps. If, say,*Bn*= 0, then*IR*=*A*1*⊕**· · ·**⊕**An*is semisimple, and equal to

*e'nR*forthe idempotent*e'n*, as desired.
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