Prove that Z[i]/nZ[i] is a field iff n is Gaussian prime.
=> If Z[i]/nZ[i] is a field then n is a Gaussian prime.
Suppose by the contrary that n is not a Guassian prime, so n = a*b where a,b belong to Z[i]. Witness that n = a*b = 0 mod n, so either a = 0 or b = 0. Suppose that a=0 then a = n*(x + yi). However notice that Norm(a) = Norm(n)*Norm(x + yi) >= Norm(n), impossible. Contradiction.
=> If n is a Gaussian prime then Z[i]/nZ[i] is a field.
Suppose that n is a Gaussian prime and a belongs to Z[i]. Let gcd(a,n) = 1 so there exists r,j in Z[i] s.t. ar + nj = 1 and ar = 1 - nj = 1 mod n thus a has an inverse. This shows that Z[i]/nZ[i] is a field.
=> If Z[i]/nZ[i] is a field then n is a Gaussian prime.
Suppose by the contrary that n is not a Guassian prime, so n = a*b where a,b belong to Z[i]. Witness that n = a*b = 0 mod n, so either a = 0 or b = 0. Suppose that a=0 then a = n*(x + yi). However notice that Norm(a) = Norm(n)*Norm(x + yi) >= Norm(n), impossible. Contradiction.
=> If n is a Gaussian prime then Z[i]/nZ[i] is a field.
Suppose that n is a Gaussian prime and a belongs to Z[i]. Let gcd(a,n) = 1 so there exists r,j in Z[i] s.t. ar + nj = 1 and ar = 1 - nj = 1 mod n thus a has an inverse. This shows that Z[i]/nZ[i] is a field.
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The second part looks good, but I would simplify the first part a little:
Suppose n isn't prime. Factor n into two proper divisors n = a*b, ie, neither a nor b = 0 mod n. But since Z[i]/nZ[i] is a field, a*b = o mod n implies one of the numbers a or b must = 0 mod n. Contradiction.
Suppose n isn't prime. Factor n into two proper divisors n = a*b, ie, neither a nor b = 0 mod n. But since Z[i]/nZ[i] is a field, a*b = o mod n implies one of the numbers a or b must = 0 mod n. Contradiction.