Numbers of Type Integer a plus b root 3 form Integral Domain

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Theorem

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

Let $S \subseteq \R$ denote the set of set of real numbers of the form $a + b \sqrt 3$ where $a$ and $b$ are integers:

$S = \set {a + b \sqrt 3: a, b \in \Z}$


Then $\struct {S, +, \times}$ 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 $\struct {S, +, \times}$ 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 $\struct {S, +, \times}$ is a subring of $\struct {\R, +, \times}$, so fulfilling property $(2)$ of the Subdomain Test.


Hence we use the Subring Test.

We note that $S \ne \O$ as $1 \in S$.

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


Let $x, y \in S$ such that:

$x = a + b \sqrt 3$
$y = c + d \sqrt 3$


Then:

\(\displaystyle x + \paren {-y}\) \(=\) \(\displaystyle \paren {a + b \sqrt 3} - \paren {c + d \sqrt 3}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {a - c} + \paren {b \sqrt 3 - d \sqrt 3}\) Definition of Real Addition
\(\displaystyle \) \(=\) \(\displaystyle \paren {a - c} + \paren {b - d} \sqrt 3\)
\(\displaystyle \) \(\in\) \(\displaystyle S\)

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


Then:

\(\displaystyle x \times y\) \(=\) \(\displaystyle \paren {a + b \sqrt 3} \paren {c + d \sqrt 3}\)
\(\displaystyle \) \(=\) \(\displaystyle a c + a d \sqrt 3 + b c \sqrt 3 + 3 b d\) Definition of Real Multiplication
\(\displaystyle \) \(=\) \(\displaystyle \paren {a c + 3 b d} + \paren {a d + b c} \sqrt 3\)
\(\displaystyle \) \(\in\) \(\displaystyle S\)

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


Hence the result.

$\blacksquare$


Sources