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Answer on Abstract Algebra Question 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 &cap; J. If I, J, K are ideals in R, prove the following distributive law: I &cap; (J + K) = (I &cap; J) + (I &cap; K)
First, it is clear that IJ &sube; I and IJ &sube; J, so IJ &sube; I &cap; J. To prove the reverse inclusion, write J = Rf where f = f2 &isin; J. For any a &isin; I &cap; J, write a = rf &isin; I, where r &isin; R. Then a = rf &middot; f &isin; IJ, as desired. Assuming that I, J, K are ideals, we have
I &cap; (J + K) = I(J + K) = IJ + IK = (I &cap; J) + (I &cap; K)

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