Eisenstein Integers form Integral Domain

Theorem
The ring of Eisenstein integers $\struct {\Z \sqbrk \omega, +, \times}$ is an integral domain.

Proof
By Eisenstein Integers form Subring of Complex Numbers we know that $\struct {\Z \sqbrk \omega, +, \times}$ is a subring of the complex numbers $\C$.

Let $1_\C$ be the unity of $\C$.

Let $1_\omega$ be the unity of $\Z \sqbrk \omega$.

By the Subdomain Test it suffices to show that $1_\C = 1_\omega$.

By Unity of Ring is Unique it suffices to show that $1_\C$ is a unity of $\Z \sqbrk \omega$.

First we note that:
 * $\Z \sqbrk \omega = \set {a + b\omega: a, b \in \Z}$

In particular:
 * $1_\C \in \Z \sqbrk \omega$

Moreover, by definition, $\Z \sqbrk \omega$ inherits its ring product from $\C$.

For any $\alpha \in \Z \sqbrk \omega$:
 * $1_\C \alpha = \alpha 1_\C = \alpha$

in $\C$.

Therefore this identity holds in $\Z \sqbrk \omega$ as well.