Eisenstein Integers form Integral Domain
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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.
$\blacksquare$