Let $R$ be a semisimple ring
Show the following
(i) If $xy=1 \in R$, then $yx=1$.
(ii) If $x \in R$ is such that $xR$ is a left ideal of $R$, then $xR=Rx$.
I am pretty lost with the two items. I mean, the only think I can think of is that $R$ is a direct summand of finite simple ideals, but I don't see how that relates to these two properties. I would appreciate any suggestions. Thanks in advance.
The first question has been answered a few times before in other posts, so let's reduce duplication by pointing you to one. More generally, any Noetherian ring has the property that $xy=1$ implies $yx=1$ (a ring like this is called a Dedekind finite ring.)
But I don't actually recall seeing the second question before.
The second part is an easy conclusion from the fact that the ideals of $R$ are just products of the various simple components of $R$. For each ideal $I$, we have that $I=eR$ for some central idempotent $e$, and $I$ is a semisimple ring with identity $e$.
So in particular for $xR=eR$, we have that $xy=e$ for some $y\in eR$. By the first part, $yx=e$ also, and so $xR=eR=Re=Ryx\subseteq Rx$. Finally, the hypothesis says that $Rx\subseteq xR$. Thus $xR=Rx$.