# Finite Ring with No Proper Zero Divisors is Field

## Theorem

Let $\left({R, +, \circ}\right)$ be a finite non-null ring with no proper zero divisors.

Then $R$ is a field.

## Proof

As $R$ is non-null, there is at least one nonzero element in $R$.

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

- $\varphi_R: x \mapsto a \circ x$
- $\varphi_L: x \mapsto x \circ a$

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 Equivalence of Mappings between Sets of Same Cardinality, the maps are then also surjective.

First, we show that $R$ has a unity.

Since $\varphi_R$ is surjective, any element $b \in R$ can be written in the form:

- $b = a \circ x_b$

for some $x_b \in R$.

As $\varphi_L$ is surjective, there is some $x_L \in R$ such that:

- $a = x_L \circ a$

We now have:

\(\displaystyle b\) | \(=\) | \(\displaystyle a \circ x_b\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x_L \circ a \circ x_b\) | as $a = x_L \circ a$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x_L \circ b\) | as $b = a \circ x_b$ |

Thus, $x_L$ is a multiplicative left identity.

By similar arguments, $R$ also has a multiplicative right identity, which can be denoted $x_R$.

By Left and Right Identity are the Same, it follows that $x_L = x_R$, so they are the unity of $R$.

Now we show that $R$ is a division ring, i.e. that each nonzero element of $R$ has an inverse.

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

Since $\varphi_R$ is surjective, it follows that:

- $a \circ y_R = 1_R$

for some $y_R \in R$.

Since $\varphi_L$ is surjective, it follows that:

- $y_L \circ a = 1_R$

for some $y_L \in R$.

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.

$\blacksquare$