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.

Expert's answer

(2) *⇒** *(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.

is semisimple, and equal to

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