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.