Special Cases of the Class Number Formula

Proof. Using that (Z/p)× is a cyclic group of order p − 1 (i.e. the existence of primitive roots), we see that there is a square root of −1 (that is, a non-trivial fourth root of 1) in (Z/p)× if and only if p ≡ 1 mod 4. Suppose now that p ≡ −1 mod 4, and suppose that α and β are two elements of Z[i] such that p|αβ. Then p = N(p)|N(α)N(β), and so (after relabelling if necessary) we may assume that p|N(α). Writing α = x+iy, we have x+y ≡ 0 mod p. If either x or y is not 0 mod p, we obtain a square root of −1, a contradication. Thus p|x, y, and so p|α. This proves that p is prime. Suppose instead that p ≡ 1 mod 4. Then we can solve x + 1 ≡ 0 mod p, and hence p|(x + i)(x − i). Since obviously p x ± i, we see that p is not prime, thus not irreducible, and so p has a proper factor in Z[i], call it π. Since N(p) = p, we see that N(π) = p. One easily sees that an element of prime norm is irreducible (hence prime), and so p = N(π) = ππ is the prime factorization of p.