International Journal of Pure and Applied Mathematics
Volume 74 No. 1 2012, 43-54
ISSN: 1311-8080 (printed version)
url: http://www.ijpam.eu
AP
ijpam.eu
THE DISCRIMINANT OF A TRINOMIAL
Fernando Barrera Mora1 § , Alexander Clemente-Torres2
Adalberto Garcı́a-Máynez3 , Rubén Mancio-Toledo4
1 Universidad
Autónoma del Estado de Hidalgo
Área Académica de Matemáticas y Fı́sica
Carretera Pachuca Tulancingo Km 4.5
Colonia Carboneras, 42182, Hidalgo, MÉXICO
2 Departamento de Control Automático
CINVESTAV del IPN
Apartado Postal 14-740
México D.F., 07000, MÉXICO
3 Instituto de Matemáticas
Universidad Nacional Autónoma de México
Área de la Investigación Cientı́fica, Circuito Exterior
Ciudad Universitaria, Distrito Federal, 04510, MÉXICO
4 Escuela Superior de Fı́sica y Matemáticas
Instituto Politécnico Nacional
Edificio No. 9, Unidad Profesional Adolfo López Mateos
Colonia Lindavista, México D.F., 07738, MÉXICO
4 e-mail: rmancio@esfm.ipn.mx
Abstract: In this paper we calculate the discriminant of a trinomial using
two different methos and as an application we compute the discriminant of a
cyclotomic number field.
AMS Subject Classification: 12E10, 11R18
Key Words: discriminant, trinomials, cyclotomic number field
Received:
August 24, 2011
§ Correspondence
author
c 2012 Academic Publications, Ltd.
url: www.acadpubl.eu
44
F.B. Mora, A. Clemente-Torres, A. Garcı́a-Máynez, R. Mancio-Toledo
1. Introduction
One of the main problems in algebraic number theory is to compute the discriminant δK of a given number field K. However, this is not an easy task. In
approaching this problem, it is useful to compute the discriminant of the polynomial defining such a field, since this is a square multiple of δK . In general,
it is difficult to compute the discriminant of a polynomial, however, for special
cases this can be done effectively in terms of the coefficients of the polynomial
under consideration, that is the case for a trinomial. In this paper we compute
the discriminant of a trinomial using two different methods.
2. First Method
For a given field K, let us denote by K a fixed algebraic closure of K and let
f (x) = an xn + · · · + a1 x + a0 = an (x − α1 ) · · · (x − αn ) and g(x) = bm xm +
· · · + b1 x + b0 = bm (x − β1 ) · · · (x − βm ) be polynomials in K[x], with αi and
βj elements of K for every i = 1, . . . , n and j = 1, . . . , m. We recall one of the
equivalent definition of the resultant of f and g which is a useful concept to
identify common roots of polynomials. Let us denote the resultant of f and g
by R(f, g). Then:
n
Y
m
g(αi ).
R(f, g) := an
i=1
A straightforward calculation shows that
R(g, f ) = (−1)mn R(f, g).
(1)
If the polynomial
f is as above, then the discriminant of f is defined by
Q
Df := an2n−2 (αi − αj )2 . From the very definition of Df , it is clear that f has
i<j
multiple roots ⇐⇒ Df = 0. Also, it is well known that the derivative of f
provides a criterion to decide if a polynomial has multiple roots, actually the
relationship between the discriminant of a polynomial and the resultant of f
and its derivative is obtained in the following lines.
One has
n Y
X
′
f (x) = an
(x − αi ),
j=1 j6=i
hence
f ′ (αj ) = an
Y
(αj − αi ).
i6=j
THE DISCRIMINANT OF A TRINOMIAL
45
Using this last equation and the definition of the resultant we have:
R(f, f ′ ) = ann−1
n
Y
f ′ (αj )
j=1
= ann−1
n
Y
j=1

Y
an (αj − αi )

i6=j
= (−1)n(n−1)/2 an Df .
Therefore, from this and Equation 1 we have
n(n−1)
2
(−1)
Df =
an
R(f ′ , f )
(2)
Our first method to compute the discriminant of a trinomial is based on
Equation 2.
Theorem 1. Let K be a field, f (x) = xn + Axk + B ∈ K[x] a trinomial.
Assume that the characteristic of K does not divide n(n − k) and let d be the
gcd of n and n − k. Then the discriminant of f is given by:
n
Df = n (−1)
n(n−1)
2
B
k−1
"
B
n−k
d
n
d
− (−1) A
n
d
k
1−
n
n−k k #d
d
k d
.
n
(3)
Proof. (First Proof) In order to apply Equation 2, we need to compute
the roots of f ′ (x) = nxn−1 + kAxk−1 = 0, and this can be attained
q by a
j
n−k
− kA
straightforward computation, that is, the roots of f ′ are: βj = ζn−k
n ,
j = 0, . . . , n − k − 1, and βj = 0 for j = n − k, . . . , n − 2, where ζn−k denotes a
primitive (n − k)−th root of unity. From this we have for j = 0, . . . , n − k − 1:
f (βj ) =
=
=
=
1 !n
1 !k
kA n−k
kA n−k
j+ 21
+ A ζn−k
+B
n
n
k k
k
k
kA n−k kA
kA n−k
kj
kj
2
2
A+B
−ζn−k ζn−k
+ ζn−k ζn−k
n
n
n
k k
kA n−k
k
kj
2
ζn−k ζn−k
− +1 A+B
n
n
k k
kA n−k n − k
kj
2
ζn−k ζn−k
A + B.
n
n
j+ 1
ζn−k2
46
F.B. Mora, A. Clemente-Torres, A. Garcı́a-Máynez, R. Mancio-Toledo
k kA n−k n − k
B
A and B ′ = ′ , then we have: f (βj ) =
Set
= ζn−k
n
n
A
kj
A′ (ζn−k
+B ′ ), for j = 0, . . . , n − k − 1 and f (βj ) = B, for j = n − k, . . . , n − 2.
From above and the definition of the resultant we have:
A′
k
2
′
R(f , f ) =
n−2
Y
f (βj )
j=0
=
n−k−1
Y
′
(A
kj
(ζn−k
′
+ B ))
j=0
B
j=n−k
′ n−k
= (A )
n−2
Y
B
k−1
n−k−1
Y
kj
(ζn−k
+ B ′ ).
(4)
j=0
The product
n−k−1
Q
j=0
kj
(ζn−k
+B ′ ) can be obtained as follows: let d = gcd(k, n−
n−k
we have j = q( n−k
d )+r,
d
n−k
with r ∈ {0, 1, . . . , d − 1}. Notice that if j ∈ {0, 1, . . . , n − k − 1} then q varies
from 0 to d − 1 in the previous representation. If r is fixed and q is an integer
k), then by the division algorithm applied to j and
k( n−k q+r)
kj
between 0 and d − 1, then ζn−k
= ζn−kd
kr , hence we can compute the
= ζn−k
desired product by considering the polynomial g(x) := (x − B ′ )
kj
kr + B ′ , r = 0, . . . , n−k − 1.
roots are ζn−k
+ B ′ = ζn−k
d
From this we obtain
n−k
−1
d
Y
n−k
d
n−k
−1
d
jk
(ζn−k
+ B ′) =
Y
rk
+ B ′)
(ζn−k
r=0
j=0
= (−1)
n−k
d
= (−1)
n−k
d
= (B ′ )
n−k
d
g(0)
n−k
n−k
(−1) d (B ′ ) d − 1
− (−1)
n−k
d
.
Consequently,
n−k−1
Y
kj
(ζn−k
+ B ′ ) = [(B ′ )
n−k
d
− (−1)
j=0
From Equation 4 and the previous, one has
R(f, f ′ ) = B k−1 (A′ )n−k [(B ′ )
n−k
d
− (−1)
n−k
d
]d
n−k
d
]d .
− 1, whose
THE DISCRIMINANT OF A TRINOMIAL
= B k−1 [(A′ B ′ )
"
= B k−1 B
n−k
d
= B
k−1
"
B
n−k
d
= B
k−1
"
B
n−k
d
n−k
d
n−k
d
47
n−k
(A′ ) d ]d
#d
n−k k
n−k
n−k
kA d k2 n−k
n−k d
ζn−kd
− (−1) d A d
n
n
#d
n−k k
n−k
k
n
k d
k d
(−1) d
− (−1) d A d 1 −
n
n
n−k k #d
n
n
k d
k d
− (−1) d A d 1 −
.
n
n
− (−1)
Finally, from Equation 2 we have
Df = nn (−1)
n(n−1)
2
"
B k−1 B
n−k
d
n
d
− (−1) A
n
d
k
1−
n
n−k k #d
d
k d
,
n
as claimed.
3. Second Method
Here we present a second method to compute the discriminant of an irreducible
trinomial based on a result for evaluating determinants of special matrices,
proposed by Greenfield and Drucker [1].
Assume that L/K is a finite separable extension of degree n. Then L =
K(α), for some α ∈ L. Let f (x) be the irreducible polynomial of α over K.
Claim. If NL/K denotes the norm map from L → K then
Df = (−1)
In fact, write f (x) =
n
Q
n(n−1)
2
NL/K (f ′ (α)).
(5)
(x−αj ), where α = α1 . The n different embeddings
j=1
of L into an algebraic closure of K Q
can be chosen so that σi (α1 ) = αi . It is
(αj − αi ) and σk (f ′ (α)) = f ′ (αk ). From
straightforward to see that f ′ (αj ) =
i6=j
this remarks we have:
σ1 (f ′ (α)) = (α1 − α2 )(α1 − α3 ) · · · (α1 − αn )
48
F.B. Mora, A. Clemente-Torres, A. Garcı́a-Máynez, R. Mancio-Toledo
σ2 (f ′ (α)) = (α2 − α1 )(α2 − α3 ) · · · (α2 − αn )
σ3 (f ′ (α)) = (α3 − α1 )(α3 − α2 ) · · · (α3 − αn )
..
.
σn (f ′ (α)) = (αn − α1 )(αn − α2 ) · · · (αn − αn−1 )
Notice that each factor of f ′ (α1 ) appears exactly once in f ′ (αi ) with opposite sign, i > 1. All but the first factor of f ′ (α2 ) appear in f ′ (αi ) exactly once
n
Q
with opposite sign, i > 2, so when multiplying out
σj (f ′ (α)), one obtains,
j=1
N (f ′ (α)) =
n
Q
Q
σj (f ′ (α)) = (−1)l (αi − αj )2 , where l is the number of factors
j=1
i6=j
that have changed sign and can be counted as follows: in σ1 (f ′ (α)) there were
n(n − 1)
n − 1; in σ2 (f ′ (α)), there were n − 2, and so, l = 1+ 2+ · · · + (n − 1) =
,
2
proving the claim.
3.1. Evaluating Norms of Elements
With the notation and assumptions as above, if β ∈ L, then β induces a Klineal transformation Tβ : L → L given by Tβ (θ) = βθ. We shall denote the
characteristic polynomial of Tβ by fβ (x) = det(Tβ − xI). Set M = K(β)
and let τ1 , . . . , τm be the different K-embeddings of M in K, then each τi has
exactly k = [K(α) : K(β)] extensions to K(α) into K. More precisely {τij },
i = 1, . . . , m, j = 1, . . . , k are the K-embeddings of K(α) in K and τij = τi
when restricted to M ; the equality holds for every j = 1, . . . , k.
if NL/K denotes the norm map from L → K, then from above we have
NL/K (β) = [NM/K (β)]k .
We also know, from basic linear algebra, that the characteristic and the minimum polynomial of a linear transformation have the same irreducible factors.
Hence fβ (x) = [irr(β, K)]k .
For a ∈ K we have
Y
NL/K (β + a) =
τij (β + a)
= [(τ1 (β) + a)]k · · · [τm (β) + a]k
= (−1)n g(−a)k
= (−1)n fβ (−a),
where g(x) = irr(β, K) is the irreducible polynomial of β over K.
(6)
THE DISCRIMINANT OF A TRINOMIAL
49
Let f (x) = xn + Axk + B be an irreducible trinomial in K[x] and let α be
a root of f . From Equation 5, one obtains:
n(n−1)
k
(7)
Df = (−1) 2 (−1)n(k−1) nn (B)k−1 NL/K αn−k + A
n
If fαn−k (x) denotes the characteristic polynomial of Tαn−k , then, from Equations 6 y 7, we have:
n(n−1)
k
n(k−1)
n
k−1
Df = (−1) 2 (−1)
(8)
n B
fαn−k − A
n
3.2. Calculating Df
Let B = 1, α, ..., αn−1 be the power basis of L over K generated by α and
consider the transformation Tαn−k : L → L, Tαn−k (x) = αn−k x. In order to
calculate the discriminant of f (x) = xn + Axk + B we shall find the matrix
associated to the transformation Tαn−k respect to the base B, [Tαn−k ]B . We
have:


0 0 . . . 0 −B
0
0 ...
0
 0 0 ... 0
0 −B
0 ...
0 


 .. ..
..
..
..
..
.. ..
.. 
 . .
.
.
.
.
. .
. 


 0 0 . . . 0 −A
0
0 ...
0 

.
[Tαn−k ]B = 

1
0
.
.
.
0
0
−A
0
.
.
.
−B


 0 1 ... 0
0
0 −A 0
0 


 . .
..
..
..
..
.. ..
.. 
 .. ..
.
.
.
.
. .
. 
0 0 ...
1
0
0
0 ...
−A
This is an n × n tridiagonal matrix with coefficients in K whose supradiagonal and subdiagonal, in general, are not symmetric respect to the main
diagonal.
The value fαn−k − nk A is given by the following determinant
50
F.B. Mora, A. Clemente-Torres, A. Garcı́a-Máynez, R. Mancio-Toledo
k
nA
0
..
.
0
∆ = 0
1
0
.
..
0
0 ...
k
nA . . .
..
..
.
.
0 ...
0 ...
0 ...
1 ...
..
..
.
.
0
...
0
0
..
.
k
nA
0
0
0
..
.
−B
0
..
.
k
nA
1
0
−A
0
0
..
.
0
0
−B
..
.
0
0
..
.
...
...
..
.
0
0
k
nA − A
0
..
.
0
0
0
k
A
n −A
..
.
···
...
...
...
..
.
0
0
...
0
0
..
.
0 −B .
0 0 ..
.
k
A
−
A
n
k
Set [Tαn−k ]B + AIn = (bij ), then applying Greenfield and Drucker method
n
to evaluate determinants we have:
∆=


n−k
n
n
−1
−1
−1
d
d
d
d
Y
Y
YY
n

bdt+s,dt+s + (−1) d −1
btd+s,td+s+k
btd+s,td+s−(n−k)  , (9)

s=1
t=0
t=0
t= n−k
d
where d = gcd(n − k, k) = gcd(n, k). The following remarks are useful to
evaluate the above product.
1. The determinant has exactly n − k elements on the supradiagonal, all
equal to −B.
2. The subdiagonal has exactly k elements, all equal to 1.
3. By (2) above, we have
n
−1
d
Q
btd+s,td+s−(n−k) = 1.
t= n−k
d
4. On the main diagonal, the first k elements are equal to
maining n − k elements are equal to:
(n − k)A
k
A − A = (−1)
.
n
n
k
nA
and the re-
From the remarks above, the second term inside parenthesis in Equation 9 is
n−k
n−k
equal to (−1) d B d ; the first is obtained from (4.) and by noting that it is a
product of nd elements of the main diagonal. In the second term just obtained,
(n−k)
the product must have exactly n−k
d factors of type (−1) d A, consequently
THE DISCRIMINANT OF A TRINOMIAL
it has
is:
n
d
− n−k
d =
(−1)
n−k
d
k
d
51
factors equal to nk A. Then, the first term inside parenthesis
(n − k)
n−k
d
A
n−k
d
n−k
k
n−k k
d
d
n−k
k d n
k
k
= (−1) d
Ad .
A
1−
n
n
n
n d
Using all these remarks we have:
!d
n−k k
d
n−k
n−k
n
k d n
−1
∆ = (−1)
A d + (−1) d (−1) d B d
n
!d
n−k k
d
n−k
k+d
n−k
k
k d n
= (−1) d
1−
A d + (−1) d B d
n
n
!d
n−k k
d
n−k
n
k
k d n
k+d
B d + (−1) d 1 −
= (−1)
.
Ad
n
n
n−k
d
k
1−
n
Hence, from Equation 8 and using the fact that nk + k + d − n is even, we
have:
"
n−k #d
n(n−1)
n−k
n
n
k
k d
n
k−1
,
Df = n (−1) 2 B
B d − (−1) d A d 1 −
n
n
as claimed.
4. An Application
The importance of computing the discriminant of the cyclotomic field Q(ζm )
is well known; it is also known that its ring of integers is Z[ζm ], hence the
discriminant of Q(ζm ) is the same as the discriminant of the m-th cyclotomic
polynomial Φm (x). We also know that Q(ζm ) is the composita of the pn -th
cyclotomic fields, where pn is the exact power of p dividing m, p a rational
prime. From Theorem 7Q [4], to compute the discriminant of Q(ζm ), it is
enough to compute the discriminant of Q(ζpn ), for p a rational prime.
4.1. The Discriminant of Q(ζpn )
For a given field K, let f (x), f1 (x), f2 (x), . . . , fk (x) ∈ K[x] be monic polynomials such that
k
Y
fi (x),
f (x) =
i=1
52
F.B. Mora, A. Clemente-Torres, A. Garcı́a-Máynez, R. Mancio-Toledo
then the roots of each fi (x) can be arranged in order to compute the discriminant of f (x) and obtain the next relationship between Df and the discriminants
Dfi for each fi (x)
k
Y
Y
Dfi
Df =
R(fi , fj )2 .
(10)
i=1
1≤i,j≤k
A binomial is a particular case of a trinomial, hence applying Equation 3
to compute the discriminant of fm (x) = xm − 1 one obtains
Dfm = (−1)
m(m−1)
2
(−1)m−1 mm .
Set m = pn , and considering that pn − 1 is even we have
Dfpn = (−1)
pn (pn −1)
2
Using the fact that fm (x) = xm − 1 =
Q
n
pnp .
Φd (x) and applying it to fpn (x)
d|m
one obtains
Y
fpn (x) =
Φd (x)
d|pn
= Φpn (x)
Y
Φd (x)
d|pn−1
= Φpn (x)(xp
n−1
− 1),
hence, fpn (x) = Φpn (x)fpn−1 (x), and from this
n−1
Φpn (x) =
n
(xp )p − 1
= Φp (xp −1 ) .
n−1
p
x
−1
In order to find the resultant of Φpn (x) and fpn−1 (x), let α1 , α2 , . . . , αpn−1
be the roots of fpn−1 (x), then αpi
follows
n−1
R(fpn−1 (x), Φpn (x)) =
= 1. The resultant can be obtained as
n−1
pY
Φpn (αi ) =
=
i=1
Φp (αpi
i=1
i=1
n−1
pY
n−1
pY
Φp (1) = pp
n−1
.
n−1
)
THE DISCRIMINANT OF A TRINOMIAL
53
Using Equation (10) for fpn (x) = Φpn (x)fpn−1 (x) one has:
Dfpn = DΦpn Dfpn−1 (R(fpn−1 (x), Φpn (x)))2
then
(−1)
pn (pn −1)
2
n
pnp = DΦpn (−1)
pn−1 (pn−1 −1)
2
p(n−1)p
n−1
p2p
n−1
and finally
DΦpn
= (−1)
pn−1 (p−1)
2
= (−1)
ϕ(pn )
2
pp
pnϕ(p
p
n−1 (np−n−1)
n)
ϕ(pn )
p−1
.
From the result above and Theorem 7Q [4], we have for an integer m:
Theorem 2. The discriminant of the cyclotomic field Q(ζm ) is given by:
δm = (−1)
ϕ(m)
2
mϕ(m)
Q ϕ(m) .
p p−1
p|m
Acknowledgments
The first author acknowledges the support received from CONACyT México
through research project number 61996, SNI and from the PIFI 3.5 funds,
“Cuerpos Académicos”.
References
[1] G.R. Greenfield, D. Drucker, On the discriminant of a Trinomial, Linear
Algebra and its Applications, 62 (1984), 105-112.
[2] G.J. Janusz, Algebraic Number Fields, American Mathematical Society,
USA (1996).
[3] R.S. Mancio-Toledo, The Discriminant of Number Fields Defined by Trinomials, M.S. Thesis, Instituto Politécnico Nacional, México (2001), In
Spanish.
54
F.B. Mora, A. Clemente-Torres, A. Garcı́a-Máynez, R. Mancio-Toledo
[4] R. Ribenboim, Algebraic Numbers, Wiley-Interscience, Volume XXVII,
Wiley and Sons (1972).
[5] R. Swan, Factorization of polynomials over finite fields, Pacific J. Math.,
12 (1962), 1099-1106.