Gaussian Integers form Subring of Complex Numbers

Theorem
The ring of Gaussian integers:
 * $\struct {\Z \sqbrk i, +, \times}$

forms a subring of the set of complex numbers $\C$.

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

By definition, a field is a ring.

Thus it is possible to use the Subring Test.

We note that $\Z \sqbrk i$ is not empty, as (for example) $0 + 0 i \in \Z \sqbrk i$.

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

Then we have $-\paren {c + d i} = -c - d i$, 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:
 * $\paren {a - c} + \paren {b - d} i \in \Z \sqbrk i$

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 $a, b, c, d \in \Z$ and $\Z$ is a ring, it follows that:
 * $a c - b d \in \Z$ and $ad + bc \in \Z$

So:
 * $\paren {a + b i} \paren {c + d i} \in \Z \sqbrk i$

So by the Subring Test, $\Z \sqbrk i$ is a subring of $\C$.

Also see

 * Gaussian Integers form Integral Domain

Note
$\Z \sqbrk i$ is not a subfield of $\C$, as $\Z \sqbrk i$ is not in itself a field.

This is demonstrated by counterexample, as follows.

Thus, although $2 + 0 i \in \Z \sqbrk i$, there is no $z \in \Z \sqbrk i$ such that $x \paren {2 + 0 i} = 1 + 0 i$.

The only such $z$ is $\dfrac 1 2 + 0 i \notin \Z \sqbrk i$.