Let R be a finite dimensional $\mathbb{Q}$-algebra that has no elements $r \in R\setminus\{0\}$ such that $rxr=0$ for all $x \in R$. Can it be determined whether $R$ is left semisimple from this information?
My thoughts:
I believe $_{R} R$ is indeed semisimple, though I don't have much more than a hunch at this point.
Since R is a finite dimensional $\mathbb{Q}$-algebra, it is a left and right artinian ring, and a left and right noetherian ring: $_{R} R$ is a finite dimensional $\mathbb{Q}$-vector space, and $R$-submodules are in particular $\mathbb{Q}$-vector subspaces, but no infinite chains of vector subspaces can exist in a finite dimensional vector space.
I am familiar with the fact that a ring R is left semisimple if and only if R is left artinian and the Jacobson radical is zero. I am also familiar with the fact that there are many equivalent characterizations of the Jacobson radical, so my hunch is that the given condition is related to one of these numerous characterizations.
Suppose that the Jacobson radical $J$ is not zero. Since $J$ is nilpotent, there exists an $n\geq1$ such that $J^n\neq0$ and $J^{n+1}=0$. Pick any $r\in J^n\setminus0$.
If $x\in R$, then $rxr\in J^nRJ^n\subseteq J^{n+1}=0$.