Gaussian Integers form Integral Domain

Theorem
The ring of Gaussian integers $\left({\Z \left[{i}\right], +, \times}\right)$ is an integral domain.

Proof
The set of complex numbers $\C$ forms a field, which is by definition a division ring.

We have that $\Z \left[{i}\right] \subset \C$.

So from Cancellable Element is Cancellable in Subset, all non-zero elements of $\Z \left[{i}\right]$ are cancellable for complex multiplication.

The identity element for complex multiplication is $1 + 0 i$ which is in $\Z \left[{i}\right]$.

We also have that complex multiplication is commutative.

From Cancellable Monoid Identity of Submonoid, the identity element of $\left({\Z \left[{i}\right]^*, \times}\right)$ is the same as for $\left({\C^*, \times}\right)$.

So we have that:
 * $\left({\Z \left[{i}\right], +, \times}\right)$ is a commutative ring with unity
 * All non-zero elements of $\left({\Z \left[{i}\right], +, \times}\right)$ are cancellable.

Hence the result from definition of integral domain.