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.
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