How does Euclidean Domain imply Unique Factorization domain for Gaussian Integers?

Question

I know, generally, it is hard to directly prove that ED $\rightarrow$ UFD. However, is it easier to prove it specifically for Gaussian Integers?

Note: to be a Euclidean domain means that there is a defined application (often called norm) that verifies these two conditions:

• $\forall a, b \in \mathbb{Z}[i] \backslash {0} \hspace{2 mm} N(a) \leq N (ab)$
• $\forall a, b \in \mathbb{Z}[i] \hspace{2 mm} b \neq 0 \rightarrow \exists c,r \in \mathbb{Z}[i] \hspace {2 mm}$ so that $\hspace{2 mm} a = bc + r \hspace{2 mm} \text{and} \hspace{2 mm} (r = 0 \hspace{2 mm} \text{or} \hspace{2 mm} r \neq 0 \hspace{2 mm} N(r) \lt N (b) )$

Where $N(a) = |a|^2$ for Gaussian integers.

Answer 1

Existence of factorization follows directly by induction from

• $N(ab)=N(a)N(b)$.

• $a$ is a unit iff $N(a)=1$.

exactly as for rational integers and the norm $|a|$.

As in the rational integers, uniqueness is another matter. I think it requires the knowledge of how rational primes factor in the Gaussian integers: $2$ is a square, $p$ remains prime if $p\equiv 3 \bmod 4$ or splits as $p=(x+yi)(x-yi)$ for $p\equiv 1 \bmod 4$. But I don't recall the details right now.