 Solution

Write the Gaussian integer $\alpha =a+bi$, with $a,b\in \mathbb{Z}$, and set $R=\mathbb{Z}[\alpha ]$. Complex conjugation preserves $R$, because $\bar{\alpha }=2a-\alpha$. Choose a nonzero $\beta \in P$. Then $N=\beta \bar{\beta }$ is a positive integer in $P$. It cannot be one because $P$ is a proper prime ideal. Factoring $N$ into rational primes and applying primality repeatedly gives a prime $p\in P$.

The element $\alpha$ satisfies the monic polynomial $F(t)=t^2-2at+a^2+b^2$. Evaluation induces a ring homomorphism that is a surjectionThe source has $p^2$ elements by part (b), soIf $b=0$, use $F(t)=t-a$ instead and get at most $p$ elements. Moreover $R/P$ is an integral domain, so its finiteness makes it a field. This proves the finite residue-field theorem for Gaussian integer orders without incorrectly assuming that every Gaussian integer order is a principal ideal domain.