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condition for power basis∗
pahio†
2013-03-22 0:25:29
Lemma. If K is an algebraic number field of degree n and the elements
α1 , α2 , . . . , αn of K can be expressed as linear combinations

α1 = c11 β1 + c12 β2 + . . . + c1n βn



α = c β + c β + . . . + c β
2
21 1
22 2
2n n

·
·
·



αn = cn1 β1 + cn2 β2 + . . . + cnn βn
of the elements β1 , β2 , . . . , βn of K with rational coefficients cij , then the discriminants of αi and βj are related by the equation
∆(α1 , α2 , . . . , αn ) = det(cij )2 · ∆(β1 , β2 , . . . , βn ).
Theorem. Let ϑ be an algebraic integer of degree n. The set {1, ϑ, . . . , ϑn−1 }
is an integral basis of Q(ϑ) if the discriminant d(ϑ) := ∆(1, ϑ, . . . , ϑn−1 ) is
square-free.
Proof. The adjusted canonical basis
ω1 = 1,
a21 +ϑ
ω2 =
,
d2
a31 +a32 ϑ+ϑ2
ω3 =
,
d3
..
..
..
.
.
.
an1 +an2 ϑ+ . . . +an,n−1 ϑn−2 +ϑn−1
ωn =
dn
of Q(ϑ) is an integral basis, where d2 , d3 , . . . , dn are integers. Its discriminant
is the fundamental number d of the field. By the lemma, we obtain
∗ hConditionForPowerBasisi created: h2013-03-2i by: hpahioi version: h40302i Privacy
setting: h1i hTheoremi h11R04i
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d = ∆(ω1 , ω2 , . . . , ωn ) = 1
0
a21
d2
1
d2
an1
dn
an2
dn
..
.
..
.
...
..
.
..
.
...
2
0 0 d(ϑ)
n−1
)=
.
∆(1, ϑ, . . . , ϑ
(d2 d3 · · · dn )2
0 1 dn
Thus (d2 d3 · · · dn )2 d = d(ϑ), and since d(ϑ) is assumed to be square-free,
we have (d2 d3 · · · dn )2 = 1, and accordingly d(ϑ) equals the discriminant of
the field. This implies (see minimality of integral basis) that the numbers
1, ϑ, . . . , ϑn−1 form an integral basis of the field Q(ϑ).
2