Proposition The ring \( \mathbb Z[i] \) is a Euclidean domain.
Proof. Assume \( \exists a,b,c,d \in \mathbb Z : abcd \neq 0\, \land \, (a+bi)\cdot (c+di)=0 \)
\( \implies (ac-bd)+(ad+bc)i=0 \implies ac-bd=0 \, \land \, ad+bc=0 \)
\( \implies \frac dc = \frac ab = -\frac cd \implies c^2=-d^2 \implies c=d=0 \)
Contradict with the assumption therefore \( \mathbb Z[i] \) is an integral domain.
\( \forall a,b \in \mathbb Z[i] \) (WLOG, assume \( N(a) \leq N(b) \)) \( \exists c=c_1+c_2i \in \mathbb C : ac=b \, \land \, c_1,c_2 \in \mathbb R \)
Set \( q=q_1+q_2i \in \mathbb Z[i]: |q_1-c_1|\leq 1/2 \, \land \, |q_2-c_2|\leq 1/2 \)
\( |q_1-c_1|^2+|q_2-c_2|^2< 1 \implies N(c-q)<1 \implies N({b}/{a}-q) \leq 1 \implies N(b-aq)\leq N(a) \)
Thus, \( \mathbb Z[i] \) is a Euclidean domain.
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