Number Theory: An Introduction to Mathematics

2 Quadratic Fields 149

(i) pOd=PP′and P=P′;
(ii) pOd=P=P′;
(iii)pOd=P^2 and P=P′.

Proof IfPis a prime ideal inOd,thenPP′ =lOdfor some positive integerl.
Moreoverl >1, sincel ∈ P.Ifl =mn,wheremandnare positive integers
greater than 1, thenPdivides eithermOdornOd. By repeating the argument it follows
thatPdividespOdfor some prime divisorpofl. The prime numberpis uniquely
determined by the prime idealPsince, by the B ́ezout identity, ifPcontained distinct
primes it would also contain their greatest common divisor 1.

Now letpbe any prime number and let the factorisation ofpOdinto a product of
positive powers of distinct prime ideals be

pOd=P 1 e^1 ···Pses.

If we putQj=P′j( 1 ≤j≤s),thenalso

pOd=Qe 11 ···Qess.

ButPjQj=njOdfor some integernj>1 and hence

p^2 =ne 11 ···ness.

Evidently the only possibilities are

(i)′s=2,n 1 =n 2 =p,e 1 =e 2 =1;
(ii)′s=1,n 1 =p^2 ,e 1 =1;
(iii)′s=1,n 1 =p,e 1 =2.

Since the factorisation is unique apart from order, this yields the three possibilities in
the statement of the proposition.

Proposition 21 does not tell us which of thethree possibilitiesholds for a given
primep.Forp = 2, the next result gives an answer in terms of the Legendre
symbol.

Proposition 22Let p be an odd prime. Then, in the statement of Proposition 21 ,(i),
(ii),or(iii)holds according as

p
d and(d/p)= 1 , p
d and(d/p)=− 1 , or p|d.

Proof Suppose first thatp
dand that there existsa∈Zsuch thata^2 ≡dmodp.
Thenp
aanda^2 −d = pbfor someb ∈ Z.IfP = (p,a+

√

d),then
P′=(p,a−

√

d)and

PP′=p(p,a+

√

d,a−

√

d,b).

Since(p,a+

√

d,a−

√

d,b)contains 2a, which is relatively prime top,italso
contains 1. HencePP′ = pOd.FurthermoreP = P′,sinceP = P′ would
imply 2a∈Pand hence 1∈P. We do not need to prove thatPis a prime ideal,
since what we have already established is incompatible with cases (ii) and (iii) of
Proposition 21.