TITLE: Prove that the Gaussian Integers are an integral domain
QUESTION [5 upvotes]: We have the following Theorem: A non-zero commutative ring is an integral domain if and only if for all $a$,$b$ $\neq 0$ $\implies ab \neq 0$.
Now, we need to prove that the Gaussian integers form an integral domain.
Proof: Let $\Bbb Z[i]$ denote the Gaussian Integers, which is a commutative ring. Take $z,w \in \Bbb Z[i]$ s.t: $z,w \neq 0$ and $z = a + ib$, $w = c + id$.
Then, $zw = (ac - bd) + (ad + bc)i \in \Bbb Z[i]$. Since the elements of $\Bbb Z[i]$ are non-zero $\implies zw \neq 0$. QED.
I am wondering if this is correct? Thanks.

REPLY [2 votes]: Alternative; I'd use the fact that norm (or absolute value) of Gaussian numbers, $\mathbb Z[i]$, is multiplicative.
$$0 \not =a,b \in \mathbb Z[i] \implies |a|^2 \not = 0, |b|^2 \not=0$$
$$ |a|^2|b|^2 = |ab|^2 \not= 0 \implies ab \not=0$$