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 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.