Proposition (Prime in \( {\mathbb Z [i] } \)) Primes in \( {\mathbb Z [i] } \) are either of the form a +bi, where \( a^2 +b^2 \) is an integer prime, or, q and its associate, where q is an integer prime that is not the sum of two squares.
This proposition will be clear once we prove the following lemma.
Lemma Let p be a prime in \( {\mathbb Z [i] } \). Then there exists an integer prime q such that either \( N (p) = q \) or \( N (p) = q^2 \). In the latter case p is an associate of q. Moreover, \( N (p) = q \) if and only if q is the sum of two squares.
Proof. \( N(p) \in \mathbb{Z}^* \implies p \overline {p} = N(p)=q_1q_2\cdots q_m \) where \( q_i \) are integer primes.
Note \( q_i \in {\mathbb Z [i] } \implies \exists q_s : p \mid q_s \implies \exists t \in {\mathbb Z [i] } : pt=q_s \implies N(p) N(t)= N(q_s) \)
\( \implies N(p) \mid q_s^2 \implies N(p)= q_s \, \lor \, q_s^2 \)
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Claim 1 If \( N (p) = q^2 \), then p is an associate of q.
\( N(p) N(t)= N(q)=q^2 \implies N(t)=1 \implies t \) is a unit
\( \implies pu=q \implies \) \( p \) is an associate of \( q \)
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Claim 2 \( N (p) = q \) if and only if q is the sum of two squares.
- \( N (p) = N(a+bi)=a^2+b^2=p \)
- Assume \( p \) is a prime in \( {\mathbb
Z [i] } \) and \( N (p) = q^2 \). Based on Claim 1,
\( q \) is an associate of \( p
\). Thus, \( q
\) is a prime in \( {\mathbb Z [i] } \).
Note \( \exists a,b \in \mathbb Z : a^2+b^2=p \implies \exists r=a+bi \in {\mathbb Z [i] } : r\overline {r}=p \) \( \implies q \) is not a prime in \( {\mathbb Z [i] } \). Contradiction, \( N(p) \) must be \( q \)
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