Gaussian Rationals form Number Field

Theorem
The set of Gaussian rationals $\Q \left[{i}\right]$, under the operations of complex addition and complex multiplication, forms a subfield of the set of complex numbers $\C$.

Proof
We will use the Subfield Test.

This is valid, as the set of complex numbers $\C$ forms a field.

We note that $\Q \left[{i}\right]$ is not empty, as (for example) $0 + 0 i \in \Q \left[{i}\right]$.

Let $a + b i, c + d i \in \Q \left[{i}\right]$.

Then we have $- \left({c + d i}\right) = -c - d i$, 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 $\left({a - c}\right) + \left({b - d}\right)i \in \Q \left[{i}\right]$.

Now consider $\left({a + b i}\right) \left({c + d i}\right)$.

By the definition of complex multiplication, we have:
 * $\left({a + b i}\right) \left({c + d i}\right) = \left({a c - b d}\right) + \left({ad + bc}\right) i$

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 $\left({a + b i}\right) \left({c + d i}\right) \in \Q \left[{i}\right]$.

Finally, let $z = x + y i \in \Q \left[{i}\right]^* = \Q \left[{i}\right] - \left\{{0 + 0 i}\right\}$, i.e. such that $x + y i \ne 0 + 0 i$.

Then, from Multiplicative Group of Complex Numbers, $\dfrac 1 {x + y i} = \dfrac {x - y i} {x^2 + y^2}$.

As $x$ and $y$ are not both zero, it follows that $x^2 + y^2 \ne 0$ and so $x^2 + y^2 \in \Q^*$.

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

Thus $\dfrac 1 {x + y i} \in \Q \left[{i}\right]^*$.

So by the Subfield Test, $\Q \left[{i}\right]$ is a subfield of $\C$.