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.

Alternative Proof
Let $R$ be a finite integral domain with unity $1$ and zero $0$.

Then we may enumerate the elements as:
 * $x_0 = 0,\ x_1 = 1,\ x_2,\ x_3,\ ... ,\ x_n$

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

Consider the elements:
 * $x_k x_1,\ x_k x_2,\ ... ,\ x_k x_n$ which are certainly members of $R$ by closure.

We have:
 * $x_k x_p = x_k x_q \implies x_p = x_q$ by the cancellation property for integral domains.

So the list of products consists of $n + 1$ distinct elements and is therefore equal to the whole group $R$.

Since $1 \in R$, we have that $x_k x_{k^*} = 1$ for some $k^*$

Also, $x_{k^*} x_k = 1$ by integral domain commutativity.

Therefore, there exists an inverse for every non-zero element, so $R$ is a field.