Quadratic Integers over 3 form Integral Domain

Theorem
Let $\R$ denote the set of real numbers.

Let $\Z \sqbrk {\sqrt 3} \subseteq \R$ denote the set of quadratic integers over $3$:
 * $\Z \sqbrk {\sqrt 3} = \set {a + b \sqrt 3: a, b \in \Z}$

Then $\struct {\Z \sqbrk {\sqrt 3}, +, \times}$ is an integral domain.

Proof
From Real Numbers form Integral Domain we have that $\struct {\R, +, \times}$ is an integral domain.

Hence to demonstrate that $\struct {\Z \sqbrk {\sqrt 3}, +, \times}$ is an integral domain, we can use the Subdomain Test.

We have that the unity of $\struct {\R, +, \times}$ is $1$.

Then we note:
 * $1 = 1 + 0 \times \sqrt 3$

and so $1 \in S$.

Thus property $(2)$ of the Subdomain Test is fulfilled.

It remains to demonstrate that $\struct {\Z \sqbrk {\sqrt 3}, +, \times}$ is a subring of $\struct {\R, +, \times}$, so fulfilling property $(2)$ of the Subdomain Test.

Hence we use the Subring Test.

We note that $\Z \sqbrk {\sqrt 3} \ne \O$ as $1 \in \Z \sqbrk {\sqrt 3}$.

This fulfils property $(1)$ of the Subring Test.

Let $x, y \in \Z \sqbrk {\sqrt 3}$ such that:


 * $x = a + b \sqrt 3$


 * $y = c + d \sqrt 3$

Then:

This fulfils property $(2)$ of the Subring Test.

Then:

This fulfils property $(3)$ of the Subring Test.

Hence the result.