# Answer to Question #16893 in Abstract Algebra for Irvin

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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