Finite Integral Domain is Galois Field

Theorem
A finite integral domain is a field.

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 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 we are done.