A left-sided zero divisor in a ring is a non-zero element $a$ for which there is a a non-zero $b$ such that $a\cdot b = 0$. In a finite ring is it true that we can find a non-zero $c$ such that $c \cdot a = 0$ and vice-versa?
(What I'm thinking to do is constructing the element $c$ somehow, because the ring is finite, but I'm not sure how.)
If there is no nonzero $c$ such that $ca=0$, then the additive group homomorphism $x \mapsto xa$ is injective (check the kernel), so it is also surjective (by finiteness), so there is $d$ such that $da = 1$.
If $ab=0$, then $dab=0$, so $b=0$.