Answer to Question #16893 in Abstract Algebra for Irvin

Question

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)

Answer 1

First, it is clear that IJ ⊆ I and IJ ⊆ J, so IJ ⊆ I ∩ J. To prove the reverse inclusion, write J = Rf where f = f² ∈ 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)