# Gaussian Integers form Integral Domain

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## 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 $50$