Question #16893

Let R be a (left) semisimple ring. Show that, for any right ideal I and any left ideal J in R, IJ = I ∩ J. If I, J, K are ideals in R, prove the following distributive law: I ∩ (J + K) = (I ∩ J) + (I ∩ K)

Expert's answer

First, it is clear that *IJ **⊆** I *and *IJ **⊆** J*, so *IJ **⊆** I ∩ J*. To prove the reverse inclusion, write *J *= *Rf *where *f *= *f*^{2} *∈** J*. For any *a **∈** I ∩ J*, write *a *= *rf **∈** I*, where *r **∈** R*. Then *a *= *rf · f **∈** IJ*, as desired. Assuming that *I, J, K *are ideals, we have

*I ∩ *(*J *+ *K*) = *I*(*J *+ *K*) = *IJ *+ *IK *= (*I ∩ J*) + (*I ∩ K*)

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