Theorem. Let $(R,+,·)$ be a finite ring with no zero divisors such that $\left|R\right|\ge 2$. Then $R$ is a field.

Proof. Since R has no zero divisors, the set of non-zero elements, denoted $R^{\star }=R\setminus 0$, is closed under $(·)$. Thus, $(R^{\star },·)$ is a semigroup.

Let $R^{\star }=a_1,\dots ,a_n$. For an arbitrary $a\in R^{\star }$, none of the products $a{a}_1,\dots ,a{a}_n$ are zero, as $a$ is not a left zero divisor. Since $R^{\star }$ is finite and cancellation holds, the map $a\mapsto ax$ is a bijection on $R^{\star }$.

Therefore, the equation $ax=b$ has a unique solution in $R^{\star }$ for every $b\in R^{\star }$. Similarly, it can be verified that the equation $ya=b$ has a solution in $R^{\star }$ for every $b\in R^{\star }$.

It follows that $(R^{\star },·)$ is a group. Consequently, $R$ is a skew field. Since R is finite, by Wedderburn's Theorem, $R$ is commutative. Therefore, $R$ is a field. $\square$