Gaussian Integers form Integral Domain

Theorem

The ring of Gaussian integers $\struct {\Z \sqbrk i, +, \times}$ is an integral domain.

Proof

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

We have that $\Z \sqbrk i \subset \C$.

So from Cancellable Element is Cancellable in Subset, all non-zero elements of $\Z \sqbrk i$ are cancellable for complex multiplication.

The identity element for complex multiplication is $1 + 0 i$ which is in $\Z \sqbrk i$.

We also have that Complex Multiplication is Commutative.

From Identity of Cancellable Monoid is Identity of Submonoid, the identity element of $\struct {\Z \sqbrk i^*, \times}$ is the same as for $\struct {\C^*, \times}$.

So we have that:

$\struct {\Z \sqbrk i, +, \times}$ is a commutative ring with unity

All non-zero elements of $\struct {\Z \sqbrk i, +, \times}$ are cancellable.

Hence the result from definition of integral domain.

$\blacksquare$

Sources

• 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $6$: Polynomials and Euclidean Rings: $\S 26$. Divisibility: Example $50$