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136 6. Proof of Mordell’s Finite Generation Theorem

Proof. It suffices to show that (1/2)M · M ≤ M ≤ 2M · M , where M =

max {|ww |, |wx + xw |, |xx |}, M = max {|w|, |x|},and M = max {|w |, |x |}.

If |w|≤|x|, |w |≤|x |, then M · M =|x|·|x |≤ M .If |w|≤|x|= M and

|x |≤|w |= M , then for M/2 ≤|w| we have MM /2 ≤|w||w |≤ M and

for M /2 ≤|x | we have MM /2 ≤|x ||x|≤ M . Finally, if |x |≤ M /2and

|w|≤ M/2, then |wx |≤ (1/4)|w x|= MM /4, and (3/4)MM ≤|wx + w x|≤

M . The other cases result from these by switching variables. The other inequality

results from |wx + xw |≤ 2 max {|wx |, |xw |}. The proposition itself follows now

by applying the log to the above inequality.
Now we return to showing that the canonical height satisfies the basic property
with respect to Q-morphisms.
(6.6) Lemma. Let ϕ be a form of degree d in y 0 ,..., y m . Then there exists a positive
d
constant c(ϕ) such for Z-reduced y ∈ P m (Q) we have |ϕ(y)|≤ c(ϕ)H(y) .

Proof. Decompose ϕ(y) = a α m α (y), where the index α counts off the monomi-
als m α (y) of degree d. Then clearly we have

d d
|ϕ(y)|≤ |a α |·|m α (y)|≤ |a α | (max {|y 0 |,..., |y m |}) = c(ϕ)H(y) ,
α α

where c(ϕ) = |a α |. This proves the lemma.
α
This lemma will give an upper estimate for H( f (y)) or (h( f (y)), but for a lower
estimate we need the following assertion which is a consequence of the Hilbert Null-
stellensatz.
(6.7) Assertion. A sequence of forms ( f 0 ,..., f m ) of degree d in Z[y 0 ,..., y m ]
defines a Q-morphism, i.e., they have no common zero in P m (Q), if and only if there
¯
exists a positive integer s, and integer b, and polynomials g ij (y) ∈ Z[y 0 ,..., y m ]
such that
s+d
g ij f j = by for all i = 0,..., m.
i
0≤ j≤m
The Nullstellensatz says that f 0 ,..., f m have no common zeros in P m (Q) if and
¯
s
only if the ideal I generated by f 0 ,..., f m contains a power (y 0 ,..., y m ) of the
ideal generated by y 0 ,..., y m , that is, if and only if there exist forms g ij (y) over Q
such that
s+d
g ij f j = y for all i = 0,..., m.
i
0≤ j≤m
The integer b is a common denominator of the g ij (y).
For m = 1 the condition on ( f 0 (y 0 , y 1 ), f 1 (y 0 , y 1 )) to be a Q-morphism is that
f 0 and f 1 be relatively prime. In this case (6.7) can be verified directly without the
Nullstellensatz.