Question

Let R be any semisimple ring. If a ∈ R is such that I = aR is an ideal in R, then I = Ra.

Answer 1

By the Wedderburn-Artin Theorem, we are reduced to the case when R = Mn(D) where D is a division ring.
Since R = M_n(D) is now a simple ring, we have either I = 0 or I = R. If I = 0, then a = 0 and Ra = I holds. If I = R, then ab = 1 for some b ∈ R. By Dedekind-finiteness of R, ba = 1, so Ra = I also holds.