Finite Integral Domain is Galois Field/Proof 2

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

Let $R^*$ denote $R \setminus \set 0$, the set $R$ without the zero.

As $R$ is finite, we may enumerate the elements of $R$ as:
 * $x_0 = 0, x_1 = 1, x_2, x_3, \ldots, x_n$

Let $R^*$.

Consider the elements:


 * $x_k x_1, x_k x_2, \ldots, x_k x_n$

which are members of $R$ by closure of the ring product.

Since $R$ is an integral domain, by definition it has no proper zero divisors.

Therefore all of $x_k x_1, x_k x_2, \ldots, x_k x_n$ are elements of $R^*$.

Recall:


 * $x_k x_p = x_k x_q \implies x_p = x_q$

by the Cancellation Law for Ring Product of Integral Domain.

So the list of products consists of $n$ distinct elements and is therefore equal to the whole $R^*$.

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

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

As the underlying set of $R$ is finite, $R$ is by definition a Galois field.