Properties of an ideal

Question

Let $S$ be a ring with $1$ such that $x \in S$ there is a $y \in S$ such that $xys = x$.

Let $I$ non-zero left ideal of $S$. Please help me to show that $I^2 $ is not equal to ${0\right\}$.

Answer 1

Let $a$ be a non-zero element of $I$. Take $b\in R$ such that $aba=a$. Since $I$ is a left-ideal, $ba\in I$, so $a=a(ba)\in I^2$ and hence $I^2$ contains a non-zero element.