Finite Ring with No Proper Zero Divisors is Field

Theorem
Let $\left({R,+,\circ}\right)$ be a finite ring with unity and no proper zero divisors.

Then $R$ is a field.

Proof
Let $1_R$ denote the unity of $R$.

Consider the two maps from $R$ to itself, for each nonzero $a \in R$:


 * $x \mapsto a \circ x$
 * $x \mapsto x \circ a$

By Ring Element is Zero Divisor iff not Cancellable, all nonzero elements in $R$ are cancellable. Thus:


 * $a \circ x = a \circ y \implies x=y$
 * $x \circ a = y \circ a \implies x=y$

Therefore, both maps are by definition injective. By Same Cardinality Bijective Injective Surjective, the maps are then also surjective.

Since the maps are surjective, it follows that $a \circ x = 1_R$ for some $x$, and $y \circ a = 1_R$ for some $y$. Recalling that the maps were defined for each nonzero $a \in R$, this means that every nonzero element of $R$ has both a left inverse and a right inverse.

By Left Inverse and Right Inverse is Inverse, each nonzero element of $R$ has an inverse, so $R$ is by definition a division ring.

It now follows by Wedderburn's Theorem that $R$ is a field.