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 $$\frac 1 2 + 0 i \notin \Z \left[{i}\right]$$.