If DR is odd and in the above notation N is even, then the primes lying over 2 and . 2 + . and . 2 + . (1 - ). When N is odd, then the only prime lying over 2 is 2R. Now, if p is an odd prime that does not divide the discriminant then either p[T]/(P(T)) is isomorphic to p2 or it splits into two p factors. The former case occurs when DR is not a square modulo p and in this case the prime lying over p is just the ideal pR; which is principal. In the second case DR is a square modulo p and we obtain two primes Pp and Qp, lying over p; both these have norm p and their product (and intersection) is pR. Let cp be a number between 1 and p - 1 so that cp2 = DR(mod p); then ap = (1 + cp)/2 satisfies the equation modulo p in the DR odd case and ap = (cp)/2 satisfies the equation modulo p in the DR even case. Thus we can pick a solution ap of the equation modulo p in each case and declare that Pp = . p + . ( - ap). The primes Pp and Qp are interchanged by the involution.