Proof that a finite integral domain is a field

Proof

1. Let $R$ be an integral domain with finitely many elements. Then $R$ is commutative and $1 \neq 0$ in $R$.
2. By the definition of a field, it remains to show that each $a \in R$ with $a \neq 0$ has some $b \in R$ with $a b = 1$.
3. Fix $a \in R$ with $a \neq 0$. Define $m \colon R \to R$ by $m(x) = a x$.
4. The map $m$ is injective. Indeed, if $a x = a y$, then $x = y$ by cancellation in an integral domain, since $a \neq 0$.
5. We claim that $m$ is surjective. Suppose not, and let $B = m(R)$ be its image. Then $B$ is a proper subset of the finite set $R$, so $\abs{B} < \abs{R}$.
6. Consider $m$ as a function from $R$ to $B$. Since $\abs{R} > \abs{B}$, the pigeonhole principle gives $x \neq y$ in $R$ with $m(x) = m(y)$.
7. This contradicts step 4. So $m$ is surjective.
8. In particular, $1 = m(b) = a b$ for some $b \in R$.
9. So every nonzero element of $R$ has an inverse, and $R$ is a field.