4 Mordell’s Theorem 565and hence
hˆ(P 1 )≤2[hˆ(P)+C]/m^2 ≤[hˆ(P)+C]/ 2.ButP 1 ∈/E′,sinceP∈/E′, and hencehˆ(P 1 )≥hˆ(P). It follows thathˆ(P)≤C,which
is a contradiction. HenceE′=E.
Proposition 12 shows that to complete the proof of Mordell’s theorem it is enough
to show that the factor groupE/ 2 Eis finite.We will prove this only for the case when
E contains an element of order2. A similar proof may be given for the general case,
but it requires some knowledge of algebraic number theory.
The assumption thatEcontains an element of order 2 means that there is a rational
point(x 0 , 0 ),wherex 0 is a root of the polynomialX^3 +aX+b.Sinceaandbare taken
to be integers, and the polynomial has highest coefficient 1,x 0 must also be an integer.
By changing variable fromXtox 0 +X, we replace the cubicCa,bby a cubicCA,B
defined by a polynomial
Y^2 −(X^3 +AX^2 +BX),whereA,B∈Z. The non-singularity conditiond:= 4 a^3 + 27 b^2 =0 becomes
D:=B^2 ( 4 B−A^2 )= 0 ,but this is the only restriction onA,B. The chord joining two rational points ofCA,B
is given by the same formulas as forCa,bin§3, but the tangent toCA,Bat the finite
pointP 1 =(x 1 ,y 1 ) is now the affine line
Y−mX−c,where
m=( 3 x 12 + 2 Ax 1 +B)/ 2 y 1 , c=−x 1 (x 12 −B)/ 2 y 1.The geometrical interpretation of the group law remains the same as before. We will
now denote byEthe group of all rational points ofCA,B. Our change of variable has
made the pointN=( 0 , 0 )an element ofEof order 2.
LetP=(x,y)be a rational point ofCA,Bwithx=0. We are going to show that,
in a sense which will become clear, there are only finitely many rational square classes
to whichxcan belong.
Writex =m/n,y = p/q,wherem,n,p,qare integers withn,q >0and
(m,n)=(p,q)=1. Then
p^2 n^3 =(m^3 +Am^2 n+Bmn^2 )q^2 ,which implies bothq^2 |n^3 andn^3 |q^2. Thusn^3 =q^2 .Fromn^2 |q^2 we obtainn|q. Hence
q=enfor some integere, and it follows thatn=e^2 ,q=e^3. Thus
x=m/e^2 , y=p/e^3 , wheree>0and(m,e)=(p,e)= 1.Moreover,
p^2 =m(m^2 +Ame^2 +Be^4 ).