Quadratic Integers over 3 form Integral Domain

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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:

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

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


Then:

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

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


Hence the result.

$\blacksquare$


Sources

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