Answer to Question #16893 in Abstract Algebra for Irvin
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)
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)
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