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basis of ideal in algebraic number field∗
pahio†
2013-03-22 0:28:33
Theorem. Let OK be the maximal order of the algebraic number field K
of degree n. Every ideal a of OK has a basis, i.e. there are in a the linearly
independent numbers α1 , α2 , . . . , αn such that the numbers
m1 α1 + m2 α2 + . . . + mn αn ,
where the mi ’s run all rational integers, form precisely all numbers of a. One
has also
a = (α1 , α2 , . . . , αn ),
i.e. the basis of the ideal can be taken for the system of generators of the ideal.
Since {α1 , α2 , . . . , αn } is a basis of the field extension K/Q, any element
of a is uniquely expressible in the form m1 α1 + m2 α2 + . . . + mn αn .
It may be proven that all bases of an ideal a have the same discriminant
∆(α1 , α2 , . . . , αn ), which is an integer; it is called the discriminant of the ideal.
The discriminant of the ideal has the minimality property, that if β1 , β2 , . . . , βn
are some elements of a, then
∆(β1 , β2 , . . . , βn ) = ∆(α1 , α2 , . . . , αn )
or
∆(β1 , β2 , . . . , βn ) = 0
But if ∆(β1 , β2 , . . . , βn ) = ∆(α1 , α2 , . . . , αn ), then also the βi ’s form a basis
of the ideal a.
√
√
Example. The integers of the quadratic field Q( 2) are l + m 2 with
l, m √
∈ Z. Determine
a basis
√
√ {α1 , α2 } and the discriminant of the ideal a)
(6−6 2, 9+6 2), b) (1−2 2).
a) The ideal may be seen√
to be the principal ideal (3), since the both √
gener2)
·
3
and
on
the
other
side,
3
=
0
·
(6
−
6
2) +
ators are
of
the
form
(l
+
m
√
√
(3 − 2 2)(9 + 6 2). Accordingly, any element of the ideal are of the form
√
√
(m1 + m2 2) · 3 = m1 · 3 + m2 · 3 2
∗ hBasisOfIdealInAlgebraicNumberFieldi
created: h2013-03-2i by: hpahioi version:
h40329i Privacy setting: h1i hTheoremi h12F05i h11R04i h06B10i
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where m1 and m2 are rational integers. Thus we can infer that
√
α1 = 3, α2 = 3 2
is a basis of the ideal concerned. So its discriminant is
√ 3 3 2 2
√ = 648.
∆(α1 , α2 ) = 3 −3 2
√
b) All elements of the ideal (1 − 2 2) have the form
√
√
√
α = (a + b 2)(1 − 2 2) = (a − 4b) + (b − 2a) 2
with a, b ∈ Z.
(1)
Especially the rational integers of the ideal satisfy b − 2a = 0, when b = 2a
and thus α = a − 4 · 2a = −7a. This means that in the presentation α =
m1 α1 + m2 α2 we can assume α1 to be 7. Now the rational portion a − 4b in
the form (1) of α should be splitted into two parts so that the first would be
always divisible by 7 and the second by b − 2a, i.e. a − 4b = 7m1 + (b − 2a)x;
this equation may be written also as
(2x + 1)a − (x + 4)b = 7m1 .
By experimenting, one finds the simplest value x = 3, another would be x = 10.
The first of these yields
√
√
α = 7(a − b) + (b − 2a)(3 + 2) = m1 · 7 + m2 (3 + 2),
i.e. we have the basis
α1 = 7,
α2 = 3 +
√
2.
The second alternative x = 10 similarly would give
√
α1 = 7, α2 = 10 + 2.
For both alternatives, ∆(α1 , α2 ) = 392.
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