Gaussian Rationals form Number Field

Theorem
The set of Gaussian rationals $\Q \sqbrk i$, under the operations of complex addition and complex multiplication, forms a number field.

Proof
By definition, a number field is a subfield of the field of complex numbers $\C$.

Recall the definition of the Gaussian rationals:
 * $\Q \sqbrk i = \set {z \in \C: z = a + b i: a, b \in \Q}$

From Complex Numbers form Field, $\C$ forms a field.

Thus it is possible to use the Subfield Test.

$\Q \sqbrk i$ is not empty, as (for example) $0 + 0 i \in \Q \sqbrk i$.

Let $a + b i, c + d i \in \Q \sqbrk i$.

Then we have $-\paren {c + d i} = -c - d i$, and so:

As $a, b, c, d \in \Q$ and $\Q$ is a field, it follows that $a - c \in \Q$ and $b - d \in \Q$, and hence $\paren {a - c} + \paren {b - d} i \in \Q \sqbrk i$.

Now consider $\paren {a + b i} \paren {c + d i}$.

By the definition of complex multiplication, we have:
 * $\paren {a + b i} \paren {c + d i} = \paren {a c - b d} + \paren {a d + b c} i$

As $a, b, c, d \in \Q$ and $\Q$ is a field, it follows that $a c - b d \in \Q$ and $ad + bc \in \Q$ and so $\paren {a + b i} \paren {c + d i} \in \Q \sqbrk i$.

Finally, let $z = x + y i \in \Q \sqbrk i_{\ne 0} = \Q \sqbrk i \setminus \set {0 + 0 i}$, that is such that $x + y i \ne 0 + 0 i$.

Then, by Inverse for Complex Multiplication:
 * $\dfrac 1 {x + y i} = \dfrac {x - y i} {x^2 + y^2}$

As $x$ and $y$ are not both zero, it follows that:
 * $x^2 + y^2 \ne 0$ and so $x^2 + y^2 \in \Q_{\ne 0}$

Thus it follows that either $\dfrac x {x^2 + y^2} \in \Q_{\ne 0}$ or $\dfrac y {x^2 + y^2} \in \Q_{\ne 0}$ (or both, or course).

Thus $\dfrac 1 {x + y i} \in \Q \sqbrk i_{\ne 0}$.

So by the Subfield Test, $\Q \sqbrk i$ is a subfield of $\C$.