Finite Ring with No Proper Zero Divisors is Field
Then $R$ is a field.
- $\varphi_R: x \mapsto a \circ x$
- $\varphi_L: x \mapsto x \circ a$
- $a \circ x = a \circ y \implies x=y$
- $x \circ a = y \circ a \implies x=y$
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.
- $b = a \circ x_b$
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 $x_R$.
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 $y_L \in R$.