Sec. 2] generalization to number fields 19

For this, we write σ1(x), . . . , σd(x) ∈ for the elementary symmetric functions in

the conjugates of x. According to Vieta, x is a zero of the polynomial

Fx(T ) := T

d

− σ1(x)T

d−1

+ σ2(x)T

d−2

+ . . . +

(−1)dσd(x)

∈ [T ] .

Lemma 2.7 shows that

Hnaive

(

1 : σr(x)

)

≤

d

r

· Hnaive(1 :

x)rd

≤

d

r

·

Brd.

Thus, by Fact 1.3, we know that for each σr(x) there are only finitely many possi-

bilities. Hence, there are only finitely many possibilities for the polynomial Fx and,

therefore, only finitely many possibilities for x.

This completes the proof.

2.7. Lemma. Let K be number field of degree d that is Galois over .

For x ∈ K, denote by σ1(x), . . . , σd(x) ∈ the elementary symmetric functions

in the conjugates of x. Then

Hnaive

(

1 : σr(x)

)

≤

d

r

· Hnaive(1 :

x)rd.

Proof. We denote the conjugates of x by x1, . . . , xd. Let ν be any valuation of K.

Then

σr(x)

ν

=

1≤i1...ir≤d

xi1 · . . . · xir

ν

≤ C(d, r, ν) · max

1≤i1...ir≤d

xi1 · . . . · xir

ν

≤ C(d, r, ν) · max

i

xi

r

ν

by the triangle inequality. We have C(d, r, ν) =

d

r

for ν Archimedean and

C(d, r, ν) = 1 for ν non-Archimedean.

By consequence,

max{1, σr(x) ν} ≤ C(d, r, ν) · max

i

max{1, xi

ν}r

≤ C(d, r, ν) ·

i

max{1, xi

ν

}r

.

Multiplying over all valuations of K yields

Hnaive

(

1 : σr(x)

)d

≤

d

r

d

·

i

Hnaive(1 :

xi)rd

=

d

r

d

· Hnaive(1 :

x)rd2

since conjugate points have the same height. This inequality is equivalent to the as-

sertion.