**Theorem.** Let $$(R,+,\cdot)$$ be a finite ring with no zero divisors such that $$| R | \geq 2$$. 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 $$(\cdot)$$. Thus, $$(R^\star,\cdot)$$ is a semigroup.

Let $$R^\star = {a_1,\ldots,a_n}$$. For an arbitrary $$a\in R^\star$$, none of the products $$aa_1,\ldots,aa_n$$ 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 $$(R^\star,\cdot)$$ 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. $$\square$$