Finite Integral Domain is Galois Field/Proof 1

Proof
Let $R$ be a finite integral domain whose unity is $1$ and whose zero is $0$.

Let $a \in R$ such that $a \ne 0$.

We wish to show that $a$ has a product inverse in $R$. So consider the map $f: R \to R$ defined by $f: x \mapsto a x$.

We first show that the kernel of $f$ is trivial.

Consider that:
 * $\ker \left({f}\right) = \left\{{x \in R: f \left({x}\right) = 0}\right\} = \left\{{x \in R: a x = 0}\right\}$

Since $R$ is an integral domain, it has no proper zero divisors and thus $a x = 0$ means that $a = 0$ or $x = 0$.

Since, by definition, $a \ne 0$, then it must be true that $x = 0$.

Therefore, $\ker \left({f}\right) = \left\{{0}\right\}$ and so $f$ is injective.

Next, the Pigeonhole Principle gives us that an injective mapping from a finite set onto itself is surjective.

Since $R$ is finite, the mapping $f$ is surjective.

Finally, since $f$ is surjective and $1 \in R$, we have:
 * $\exists \, x \in R: f \left({x}\right) = a x = 1$

So this $x$ is the product inverse of $a$ and the proof is complete.