13 Heights

For simplicity, take K=. Write Pn(Q) as P=(a0:a1::an) where a0,,an and gcd(a0,an)=1.

Definition (Height of a point). H(P)=max0in|ai|.

Lemma 13.1. Assuming that:

  • f1,f2[X1,X2] coprime homogeneous polynomials of degree d

  • let

    F:11(x1:x2)(f1(x1,x2):f2(x1,x2))

Then there exist c1,c2>0 such that for all P1(),
c1H(P)dH(F(P))c2H(P)d.

Proof. Without loss of generality f1,f2[X1,X2].

Upper bound: Write P=(a1:a2), a1,a2 coprime.

H(F(P))max(|f1(a1,a2)|,|f2(a1,a2)|)c2max(|a1|d,|a2|d)

where

c2=maxi=1,2(sum of absolute values of coefficients of fi).

Therefore H(F(P))c2H(P)d.

Lower bound: We claim that there exists gij[X1,X2] homogeneous of degree d1 and κ>0 such that

j=12gijfj=κXi2d1,i=1,2.(†)

Indeed, running Euclid’s algorithm on f1(X,1) and f2(X,1) gives r,s[X] of degree <d such that

r(X)f1(X,1)+s(X)f2(X,1)=1.

Homogenising and clearing denominators gives () for i=2. Likewise for i=1.

Write P=(a1:a2) with a1,a2 coprime. () gives

j=12gij(a1,a2)fj(a1,a2)=κai2d1,i=1,2.

Therefore gcd(f1(a1,a2),f2(a1,a2)) divides gcd(κa12d1,κa22d1)=κ.

|κai2d1|maxj=1,2|fj(a1,a2)|κH(F(P))j=12|gij(a1,a2)|γiH(D)d1

where

γi=j=12(sum of absolute values of coefficients of gij).

Therefore

κ|ai|2d1κH(F(P))γiH(P)d1

so

H(P)2d1max(γ1,γ2)H(F(P))H(P)d1

so

1max(γ1,γ2)=c1H(P)dH(F(P)).

Notation. For x, H(X)=H((x:1))=max(|u|,|v|), where x=uv, u,v coprime.

Definition (Height of a point). Let E be an elliptic curve, y2=x3+ax+b.

Define the height

H:E()1P{H(x)if P=(x,y)1if P=0E

Alsdefine logarithmic height

h:E()0PlogH(P)

Lemma 13.2. Assuming that:

Then there exists c>0 such that
|h(ϕ(P))(degϕ)h(P)|cPE().

Note. c depends on E and E, but not P.

Proof. Recall (Lemma 5.4)

  E       E ′

   1       1
ϕxxξ ℙ       ℙ
degϕ= deg ξ (=d say). Lemma 13.1 tells us that there exists c1,c2>0 such that

c1H(P)dH(ϕ(P))c2H(P)dPE().

Taking logs gives

|h(ϕ(P))dh(P)|max( log c2 log c1)=c

Example. ϕ=[2]:EE. Then there exists c>0 such that

|h(2P)4h(P)|cPE().

Definition (Canonical height of a point). The canonical height is

ĥ(P)=limn14nh(2nP).

We check convergence:

Let mn. Then

|14mh(2mP)14nh(2nP)|r=nm1|14r+1h(2r+1P)14rh(2rP)|=r=nm114r+1|h(2(2rP))4h(2rP)|<cr=n14r+1=c4n+11114=c3×4n0

as n. So the sequence is Cauchy, ĥ(P) exists.

Lemma 13.3. |h(P)ĥ(P)| is bounded for PE().

Proof. Put n=0 in above calculation to get

|14mh(2mP)h(P)|<c3.

Take limit m.

Lemma 13.4. Assuming that:

  • B>0

Then
#{PE()|ĥ(P)B}<.

Proof. ĥ(P) bounded means we have a bound on h(P) (by Lemma 13.3). So only finitely many possibilities for x. Each x gives 2 choices for y.

Lemma 13.5. Assuming that:

  • ϕ:EE an isogeny defined over

Then
ĥ(ϕP)=(degϕ)ĥ(P)PE().

Proof. By ?? 66, there exists c>0 such that

|h(ϕP)(degϕ)h(P)|<cPE().

Replace P by 2nP, divide by 4n and take limit n.

Remark.

Lemma 13.6. Assuming that:

Then there exists c>0 such that for all P,QE() with P,Q,P+Q,PQ0, we have
H(P+Q)H(PQ)cH(P)2H(Q)2.

Proof. Let E have Weierstrass equation y2=x3+ax+b, a,b. Let P,Q,P+Q,PQ have x coordinates x1,,x4. By Lemma 5.8, there exists W0,W1,W2[x1,x2] of degree 2 in x2 such that

(1:x3+x4:x3x4)=(W0=(x1x2)2:W1:W2).

Write xi=risi with ri,si coprime.

(s3s4:r3s4+r4s3:r3r4)gcd=1=((r1s2r2s1)2:)H(P+Q)H(PQ)=max(|r3|,|s3|)max(|r4|,|s4|)2max(|s3s4|,|r3s4+r4s3|,|r3r4|)2max(|r1s2r2s1|2,)cmax(|r1|2,|s1|)2max(|r2|,|s2|)2=cH(P)2H(Q)2

where c depends on E, but not on P and Q.

Theorem 13.7. ĥ:E()0 is a quadratic form.

Proof. Lemma 13.6 and |h(2P)4h(P)| bounded gives that there exists c such that

h(P+Q)+h(PQ)2h(P)+2h(Q)+cP,QE().

Replacing P, Q by 2nP, 2nQ, dividing by 4n and taking the limit n gives

ĥ(P+Q)+ĥ(PQ)2ĥ(P)+2ĥ(Q).

Replacing P,Q by P+Q, PQ and ĥ(2P)=4ĥ(P) gives the reverse inequality. Therefore ĥ satisfies the parallelogram law, and hence ĥ is a quadratic form.

Remark. For K a number field and P=(a0:a1::an)n(K), define

H(P)=vmax0in|ai|v

where the product is over all places v, and the absolute values are normalised such that

v|λ|v=1λK.

Using this definition, all results in this section generalise when is replaced by a number field K.