Gaussian Rationals form Number Field

Theorem
The set of Gaussian rationals $$\Q \left[{\imath}\right]$$, 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 $$\Q \left[{\imath}\right]$$ is not empty, as (for example) $$0 + 0 \imath \in \Q \left[{\imath}\right]$$.

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

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

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

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

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

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

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

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 $$\frac x {x^2 + y^2} \in \Q^*$$ or $$\frac y {x^2 + y^2} \in \Q^*$$ (or both, or course).

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

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