Finite Ring with No Proper Zero Divisors is Field
Theorem
Let $\struct {R, +, \circ}$ 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 Finite 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:
| \(\ds b\) | \(=\) | \(\ds a \circ x_b\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds x_L \circ a \circ x_b\) | as $a = x_L \circ a$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds x_L \circ b\) | 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.
$\blacksquare$