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properties of discriminant in algebraic
number field∗
pahio†
2013-03-22 3:22:08
Theorem 1. Let α1 , α2 , . . . , αn and β1 , β2 , . . . , βn be elements of the
algebraic number field Q(ϑ) of degree n. If they satisfy the equations
αi =
n
X
cij βj
for i = 1, 2, . . . , n,
j=1
where all coefficients cij are rational numbers, then the discriminants are connected via the equation
∆(α1 , α2 , . . . , αn ) = det(cij )2 · ∆(β1 , β2 , . . . , βn ).
As a special case one obtains the
Theorem 2. If
αi = ci1 + ci2 ϑ + . . . + cin ϑn−1
for i = 1, 2, . . . , n
(1)
are the canonical forms of the elements αi in Q(ϑ), then
∆(α1 , α2 , . . . , αn ) = det(cij )2 · ∆(1, ϑ, . . . , ϑn−1 ),
where ∆(1, ϑ, . . . , ϑn−1 ) is a Vandermonde determinant thus having the product form

2
Y
Y
∆(1, ϑ, . . . , ϑn−1 ) = 
(ϑj − ϑi ) =
(ϑi − ϑj )2
(2)
1≤i≤j
1≤i≤j
where ϑ1 = ϑ, ϑ2 , . . . , ϑn are the algebraic conjugates of ϑ.
∗ hPropertiesOfDiscriminantInAlgebraicNumberFieldi
created: h2013-03-2i by: hpahioi
version: h42062i Privacy setting: h1i hResulti h11R29i
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Since the discriminant (2) is also the polynomial discriminant of the irreducible minimal polynomial of ϑ, the numbers ϑi are inequal. It follows the
Theorem 3. When (1) are the canonical forms of the numbers αi , one has
∆(α1 , α2 , . . . , αn ) = 0
⇐⇒
det(cij ) = 0.
The powers 1, ϑ, . . . , ϑn−1 of the primitive element form a basis of the field
extension Q(ϑ)/Q (see the canonical form of element of number field). By the
theorem 3 we may write the
Theorem 4. The numbers α1 , α2 , . . . , αn of Q(ϑ) are linearly independent
over Q if and only if ∆(α1 , α2 , . . . , αn ) 6= 0.
Theorem 5. Q(α) = Q(ϑ) ⇐⇒ ∆(1, α, α2 , . . . , αn−1 ) 6= 0. Here, the
the discriminant is the discriminant of the algebraic number α.
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