Let $R$ be an Artinian Ring and suppose there exists $a,b\in R$ s.t. $I=aR=Rb$, then prove $I=bR=Ra$. (You may assume that a right Artinian Ring is Right Noetherian).
I've managed to get $Ra$, $bR$ contained in $I$ without using the fact that it's Artinian. The hint confuses me because I'm not sure how to make a useful ascending chain to use the fact that it's right Noetherian.