Gaussian Integers form Subring of Complex Numbers

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

This ring is an integral domain.

Proof
We will use the Subring Test.

This is valid, as the set of complex numbers $\C$ forms a field, which is by definition itself a ring.

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

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

Then we have $- \left({c + d i}\right) = -c - d i$, and so:

We have that $a, b, c, d \in \Z$ and $\Z$ is an integral domain, therefore by definition a ring.

So it follows that $a - c \in \Z$ and $b - d \in \Z$, and hence $\left({a - c}\right) + \left({b - d}\right)i \in \Z \left[{i}\right]$.

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 $a, b, c, d \in \Z$ and $\Z$ is a ring, it follows that $a c - b d \in \Z$ and $ad + bc \in \Z$ and so $\left({a + b i}\right) \left({c + d i}\right) \in \Z \left[{i}\right]$.

So by the Subring Test, $\Z \left[{i}\right]$ is a subring of $\C$.

The last statement is proved in Gaussian Integers form Integral Domain.

Note
$\Z \left[{i}\right]$ is not a subfield of $\C$, as $\Z \left[{i}\right]$ is not in itself a field.

This is demonstrated by counterexample, as follows.

Thus, although $2 + 0 i \in \Z \left[{i}\right]$, there is no $z \in \Z \left[{i}\right]$ such that $x \left({2 + 0 i}\right) = 1 + 0 i$.

The only such $z$ is $\dfrac 1 2 + 0 i \notin \Z \left[{i}\right]$.