[Math] How to show that a finite commutative ring without zero divisors is a field

Question

$R$ is finite commutative ring without zero divisors which has at least two elements. How to show that $R$ is field?

I'm just starting with abstract algebra and I'd really appreciate if someone could explain it to me.

Answer 1

It follows from the pigeon-hole principle.

First, we show that $R$ has an identity. (Sometimes the existence of an identity is included in the definition of a ring, but you do not need it here.)

For each $0 \ne a \in R$, consider the map $$ \phi_{a} : R \to R, \qquad x \mapsto a x. $$ Because $a$ is not a zero-divisor, the map $\phi_{a}$ is injective, thus surjective. So there is $e \in R$ such that $$a = \phi_{a}(e) = e a.\tag{1}$$ Again because the map $\phi_{a}$ is surjective, every $b \in R$ is of the form $$b = \phi_{a}(x_{b}) = a x_{b}$$ for some $x_{b}$, so multiplying (1) by $x_{b}$ you find $$ b = e b $$ for all $b \in R$, so $e$ is the identity.

Now use once more the fact that $\phi_{a}$ is surjective, to show that there is $c \in R$ such that $$a c = \phi_{a}(c) = e,$$ so $c$ is an inverse of $a$, $a$ being an arbitrary non-zero element.