Open Access

The constant term of the minimal polynomial of cos ( 2 π / n ) over

Fixed Point Theory and Applications20132013:77

https://doi.org/10.1186/1687-1812-2013-77

Received: 21 January 2013

Accepted: 11 March 2013

Published: 29 March 2013

Abstract

Let H ( λ q ) be the Hecke group associated to λ q = 2 cos π q for q 3 integer. In this paper, we determine the constant term of the minimal polynomial of λ q denoted by P q ( x ) .

MSC:12E05, 20H05.

Keywords

Hecke groups minimal polynomial constant term

1 Introduction

The Hecke groups H ( λ ) are defined to be the maximal discrete subgroups of PSL ( 2 , R ) generated by two linear fractional transformations
T ( z ) = 1 z and S ( z ) = 1 z + λ ,

where λ is a fixed positive real number.

Hecke [1] showed that H ( λ ) is Fuchsian if and only if λ = λ q = 2 cos π q for q 3 is an integer, or λ 2 . In this paper, we only consider the former case and denote the corresponding Hecke groups by H ( λ q ) . It is well known that H ( λ q ) has a presentation as follows (see [2]):
H ( λ q ) = T , S T 2 = S q = I .
(1)

These groups are isomorphic to the free product of two finite cyclic groups of orders 2 and q.

The first few Hecke groups are H ( λ 3 ) = Γ = PSL ( 2 , Z ) (the modular group), H ( λ 4 ) = H ( 2 ) , H ( λ 5 ) = H ( 1 + 5 2 ) , and H ( λ 6 ) = H ( 3 ) . It is clear from the above that H ( λ q ) PSL ( 2 , Z [ λ q ] ) , but unlike in the modular group case (the case q = 3 ), the inclusion is strict and the index [ PSL ( 2 , Z [ λ q ] ) : H ( λ q ) ] is infinite as H ( λ q ) is discrete, whereas PSL ( 2 , Z [ λ q ] ) is not for q 4 .

On the other hand, it is well known that ζ, a primitive n th root of unity, satisfies the equation
x n 1 = 0 .
(2)
In [3], Cangul studied the minimal polynomials of the real part of ζ, i.e., of cos ( 2 π / n ) over the rationals. He used a paper of Watkins and Zeitlin [4] to produce further results. Also, he made use of two classes of polynomials called Chebycheff and Dickson polynomials. It is known that for n N { 0 } , the n th Chebycheff polynomial, denoted by T n ( x ) , is defined by
T n ( x ) = cos ( n arccos x ) , x R , | x | 1 ,
(3)
or
T n ( cos θ ) = cos n θ , θ R ( θ = arccos x + 2 k π , k Z ) .
(4)

Here we use Chebycheff polynomials.

For n N , Cangul denoted the minimal polynomial of cos ( 2 π / n ) over Q by Ψ n ( x ) . Then he obtained the following formula for the minimal polynomial Ψ n ( x ) .

Theorem 1 ([[3], Theorem 1])

Let m N and n = [ | m / 2 | ] . Then
  1. (a)

    If m = 1 , then Ψ 1 ( x ) = x 1 , and if m = 2 , then Ψ 2 ( x ) = x + 1 .

     
  2. (b)
    If m is an odd prime, then
    Ψ m ( x ) = T n + 1 ( x ) T n ( x ) 2 n ( x 1 ) .
    (5)
     
  3. (c)
    If 4 m , then
    Ψ m ( x ) = T n + 1 ( x ) T n 1 ( x ) 2 n / 2 ( T n 2 + 1 ( x ) T n 2 1 ( x ) ) d m , d m , d m 2 q 1 Ψ d ( x ) .
    (6)
     
  4. (d)
    If m is even and m / 2 is odd, then
    Ψ m ( x ) = T n + 1 ( x ) T n 1 ( x ) 2 n n ( T n + 1 ( x ) T n ( x ) ) d m , d m , d even q 1 Ψ d ( x ) ,
    (7)
     
where n = m 2 1 2 .
  1. (e)
    Let m be odd and let p be a prime dividing m. If p 2 m , then
    Ψ m ( x ) = T n + 1 ( x ) T n ( x ) 2 n n ( T n + 1 ( x ) T n ( x ) ) ,
    (8)
     
where n = m p 1 2 . If p 2 m , then
Ψ m ( x ) = T n + 1 ( x ) T n ( x ) 2 n n ( T n + 1 ( x ) T n ( x ) ) Ψ p ( x ) ,
(9)

where n = m p 1 2 .

For the first four Hecke groups Γ, H ( 2 ) , H ( λ 5 ) , and H ( 3 ) , we can find the minimal polynomial, denoted by P q ( x ) , of λ q over Q as λ 3 1 , λ 4 2 2 , λ 5 2 λ 5 1 , and λ 6 2 3 , respectively. However, for q 7 , the algebraic number λ q = 2 cos π q is a root of a minimal polynomial of degree ≥3. Therefore, it is not possible to determine λ q for q 7 as nicely as in the first four cases. Because of this, it is easy to find and study with the minimal polynomial of λ q instead of λ q itself. The minimal polynomial of λ q has been used for many aspects in the literature (see [58] and [9]).

Notice that there is a relation
P q ( x ) = 2 φ ( 2 q ) / 2 Ψ 2 q ( x 2 )

between P q ( x ) and Ψ m ( x ) .

In [10], when the principal congruence subgroups of H ( λ q ) for q 7 prime were studied, we needed to know whether the minimal polynomial of λ q is congruent to 0 modulo p for prime p and also the constant term of it modulo p.

In this paper, we determine the constant term of the minimal polynomial P q ( x ) of λ q . We deal with odd and even q cases separately. Of course, this problem is easier to solve when q is odd.

2 The constant term of P q ( x )

In this section, we calculate the constant term for all values of q. Let c denote the constant term of the minimal polynomial P q ( x ) of λ q , i.e.,
c = P q ( 0 ) .
(10)
We know from [[4], Lemma, p.473] that the roots of P q ( x ) are 2 cos h π q with ( h , q ) = 1 , h odd and 1 h q 1 . Being the constant term, c is equal to the product of all roots of P q ( x ) :
c = q 1 h = 1 ( h , q ) = 1 h odd 2 cos h π q .
(11)

Therefore we need to calculate the product on the right-hand side of (11). To do this, we need the following result given in [11].

Lemma 1 h = 0 q 1 2 sin ( h π q + θ ) = 2 sin q θ .

We now want to obtain a similar formula for cosine. Replacing θ by π 2 θ , we get
h = 0 q 1 2 cos ( h π q θ ) = 2 sin q ( π 2 θ ) .
(12)
Let now μ denote the Möbius function defined by
μ ( n ) = { 0 if  n  is not square-free , 1 if  n = 1 , ( 1 ) k if  n  has  k  distinct prime factors ,
(13)
for n N . It is known that
d n μ ( d ) = { 0 if  n > 1 , 1 if  n = 1 .
(14)
Using this last fact, we obtain
(15)
Therefore
q 1 h = 0 ( h , q ) = 1 2 cos ( h π q θ ) = d q ( sin d ( π 2 θ ) ) μ ( q / d ) .
(16)
Finally, as ( 0 , q ) 1 , we can write (16) as
q 1 h = 1 ( h , q ) = 1 2 cos ( h π q θ ) = d q ( sin d ( π 2 θ ) ) μ ( q / d ) .
(17)
Note that if q is even, then
q 1 h = 1 ( h , q ) = 1 2 cos ( h π q ) = q 1 h = 1 ( h , q ) = 1 h odd 2 cos h π q = c ,
(18)
while if q is odd, then
| q 1 h = 1 ( h , q ) = 1 2 cos ( h π q ) | = c 2 ,
(19)
as cos ( h i ) π q = cos i π q . Also note that
sin d ( π 2 θ ) = { cos d θ if  d 1 mod 4 , sin d θ if  d 2 mod 4 , cos d θ if  d 3 mod 4 , sin d θ if  d 0 mod 4 .
(20)

To compute c, we let θ 0 in (17). If d is odd, then sin d ( π 2 θ ) ± 1 as θ 0 by (20). So, we are only concerned with even d. Indeed, if q is odd, then the left-hand side at θ = 0 is equal to ±1. Therefore we have the following result.

Theorem 2 Let q be odd. Then
| c | = 1 .
(21)

Proof It follows from (19) and (20). □

Let us now investigate the case of even q. As ( h , q ) = 1 , h must be odd. So, by a similar discussion, we get the following.

Theorem 3 Let q be even. Then
c = lim θ 0 d q ( sin d ( π 2 θ ) ) μ ( q / d ) .
(22)
Proof Note that by (20), the right-hand side of (22) becomes a product of ± ( cos d θ ) ± 1 s and ± ( sin d θ ) ± 1 s. Above we saw that we can omit the former ones as they tend to ±1 as θ tends to 0. Now, as d n μ ( d ) = 0 , there are equal numbers of the latter kind factors in the numerator and denominator, i.e., if there is a factor sin d θ in the numerator, then there is a factor sin d θ in the denominator. Then using the fact that
lim θ 0 sin k θ sin l θ = k l ,
(23)

we can calculate c.

In fact the calculations show that there are three possibilities:

(i) Let q = 2 α 0 , α 0 2 . Then the only divisors of q such that μ ( q / d ) 0 are d = 2 α 0 and 2 α 0 1 . Therefore
c = lim θ 0 sin 2 α 0 ( π 2 θ ) sin 2 α 0 1 ( π 2 θ ) = { 2 if  α 0 > 2 , 2 if  α 0 = 2 .
(24)
(ii) Secondly, let q = 2 p α , α 1 , p odd prime. Then the only divisors of q such that μ ( q / d ) 0 are d = 2 p α , 2 p α 1 , p α and p α 1 . Therefore
c = lim θ 0 sin 2 p α ( π 2 θ ) sin p α 1 ( π 2 θ ) sin p α ( π 2 θ ) sin 2 p α 1 ( π 2 θ ) = lim θ 0 ϵ sin 2 p α θ cos p α 1 θ cos p α θ sin 2 p α 1 θ = ϵ p ,
(25)
where
ϵ = { 1 if  p 1 mod 4 , 1 if  p 1 mod 4 .
(26)
(iii) Let q be different from above. Then q can be written as
q = 2 α 0 p 1 α 1 p k α k ,
(27)

where p i are distinct odd primes and α i 1 , 0 i k .

Here we consider the first two cases k = 1 and k = 2 .

Let k = 1 , i.e., let q = 2 α 0 p 1 α 1 . We have already discussed the case α 0 = 1 . Let α 0 > 1 . Then the only divisors d of q with μ ( q / d ) 0 are d = 2 α 0 p 1 α 1 , 2 α 0 1 p 1 α 1 , 2 α 0 p 1 α 1 1 and 2 α 0 1 p 1 α 1 1 . Therefore
c = lim θ 0 sin 2 α 0 p 1 α 1 ( π 2 θ ) sin 2 α 0 1 p 1 α 1 1 ( π 2 θ ) sin 2 α 0 1 p 1 α 1 ( π 2 θ ) sin 2 α 0 p 1 α 1 1 ( π 2 θ ) = 1 .
(28)
Now let k = 2 , i.e., let q = 2 α 0 p 1 α 1 p 2 α 2 , ( p 1 < p 2 ). Similarly, all divisors d of q such that μ ( q / d ) 0 are d = 2 α 0 p 1 α 1 p 2 α 2 , 2 α 0 1 p 1 α 1 p 2 α 2 , 2 α 0 p 1 α 1 1 p 2 α 2 , 2 α 0 p 1 α 1 p 2 α 2 1 , 2 α 0 p 1 α 1 1 p 2 α 2 1 , 2 α 0 1 p 1 α 1 1 p 2 α 2 1 , 2 α 0 1 p 1 α 1 p 2 α 2 1 and 2 α 0 1 p 1 α 1 1 p 2 α 2 . Therefore
c = lim θ 0 sin 2 α 0 1 p 1 α 1 1 p 2 α 2 1 ( π 2 θ ) sin 2 α 0 p 1 α 1 p 2 α 2 1 ( π 2 θ ) sin 2 α 0 p 1 α 1 1 p 2 α 2 1 ( π 2 θ ) sin 2 α 0 1 p 1 α 1 p 2 α 2 1 ( π 2 θ ) × lim θ 0 sin 2 α 0 p 1 α 1 1 p 2 α 2 ( π 2 θ ) sin 2 α 0 1 p 1 α 1 p 2 α 2 ( π 2 θ ) sin 2 α 0 1 p 1 α 1 1 p 2 α 2 ( π 2 θ ) sin 2 α 0 p 1 α 1 p 2 α 2 ( π 2 θ ) = 1 .
(29)
Finally, k 3 , i.e., let
q = 2 α 0 p 1 α 1 p k α k with  p 1 < p 2 < < p k .
In this case the proof is similar, but rather more complicated. In fact, the number of all divisors d of q such that μ ( q / d ) 0 is 2 k + 1 . There is ( k + 1 0 ) = 1 divisor of the form
d = 2 α 0 p 1 α 1 p k α k .
There are ( k + 1 1 ) = k + 1 divisors of the form
d = 2 α 0 1 p 1 α 1 p k α k , 2 α 0 p 1 α 1 1 p k α k , , 2 α 0 p 1 α 1 p k α k 1 .
There are ( k + 1 2 ) = k ( k + 1 ) 2 divisors of the form
d = 2 α 0 1 p 1 α 1 1 p 2 α 2 p k α k , 2 α 0 1 p 1 α 1 p 2 α 2 1 p k α k , , 2 α 0 1 p 1 α 1 p 2 α 2 p k α k 1 , 2 α 0 p 1 α 1 1 p 2 α 2 1 p k α k , , 2 α 0 p 1 α 1 1 p 2 α 2 p k α k 1 , , 2 α 0 p 1 α 1 p k 1 α k 1 1 p k α k 1 .

If we continue, we can find other divisors d of q, similarly. Finally, there is ( k + 1 k + 1 ) = 1 divisor of the form 2 α 0 1 p 1 α 1 1 p 2 α 2 1 p k α k 1 . Thus, the product of all coefficients d in the factors sin d ( π 2 θ ) in the numerator is equal to the product of all coefficients e in the factors sin e ( π 2 θ ) in the denominator implying c = 1 . Therefore the proof is completed. □

Now we give an example for all possible even q cases.

Example 1 (i) Let q = 8 = 2 3 . The only divisors of 8 such that μ ( 8 / d ) 0 are d = 8 and 4. Therefore
c = lim θ 0 sin 8 ( π 2 θ ) sin 4 ( π 2 θ ) = 2 .
(ii) Let q = 14 = 2 7 . The only divisors of 14 such that μ ( 14 / d ) 0 are d = 14 , 2 , 7 and 1. Therefore
c = ϵ lim θ 0 sin 14 ( π 2 θ ) sin ( π 2 θ ) sin 7 ( π 2 θ ) sin 2 ( π 2 θ ) = 7 ,

since p 1 mod 4 .

(iii) Let q = 24 = 2 3 3 . The only divisors of 24 such that μ ( 24 / d ) 0 are d = 24 , 12 , 8 and 4. Therefore
c = lim θ 0 sin 24 ( π 2 θ ) sin 4 ( π 2 θ ) sin 12 ( π 2 θ ) sin 8 ( π 2 θ ) = 1 .
(iv) Let q = 30 = 2 3 5 . The only divisors of 30 such that μ ( 30 / d ) 0 are d = 30 , 15 , 10 , 6 , 5 , 3 , 2 and 1. Therefore
c = lim θ 0 sin ( π 2 θ ) sin 6 ( π 2 θ ) sin 10 ( π 2 θ ) sin 15 ( π 2 θ ) sin 2 ( π 2 θ ) sin 3 ( π 2 θ ) sin 5 ( π 2 θ ) sin 30 ( π 2 θ ) = 1 .

Declarations

Acknowledgements

Dedicated to Professor Hari M Srivastava.

Both authors are supported by the Scientific Research Fund of Uludag University under the project number F2012/15 and the second author is supported under F2012/19.

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Arts and Science, Uludag University

References

  1. Hecke E: Über die bestimmung dirichletscher reihen durch ihre funktionalgleichungen. Math. Ann. 1936, 112: 664–699. 10.1007/BF01565437MathSciNetView ArticleGoogle Scholar
  2. Cangul IN, Singerman D: Normal subgroups of Hecke groups and regular maps. Math. Proc. Camb. Philos. Soc. 1998, 123: 59–74. 10.1017/S0305004197002004MathSciNetView ArticleGoogle Scholar
  3. Cangul IN:The minimal polynomials of cos ( 2 π / n ) over . Probl. Mat. - Wyż. Szk. Pedagog. Bydg. 1997, 15: 57–62.MathSciNetGoogle Scholar
  4. Watkins W, Zeitlin J:The minimal polynomial of cos ( 2 π / n ) . Am. Math. Mon. 1993, 100(5):471–474. 10.2307/2324301MathSciNetView ArticleGoogle Scholar
  5. Arnoux P, Schmidt TA: Veech surfaces with non-periodic directions in the trace field. J. Mod. Dyn. 2009, 3(4):611–629.MathSciNetView ArticleGoogle Scholar
  6. Beslin S, De Angelis V:The minimal polynomials of sin ( 2 π / p ) and cos ( 2 π / p ) . Math. Mag. 2004, 77(2):146–149.MathSciNetGoogle Scholar
  7. Rosen R, Towse C: Continued fraction representations of units associated with certain Hecke groups. Arch. Math. 2001, 77(4):294–302. 10.1007/PL00000494MathSciNetView ArticleGoogle Scholar
  8. Schmidt TA, Smith KM: Galois orbits of principal congruence Hecke curves. J. Lond. Math. Soc. 2003, 67(3):673–685. 10.1112/S0024610703004113MathSciNetView ArticleGoogle Scholar
  9. Surowski D, McCombs P:Homogeneous polynomials and the minimal polynomial of cos ( 2 π / n ) . Mo. J. Math. Sci. (Print) 2003, 15(1):4–14.MathSciNetGoogle Scholar
  10. Ikikardes S, Sahin R, Cangul IN: Principal congruence subgroups of the Hecke groups and related results. Bull. Braz. Math. Soc. 2009, 40(4):479–494. 10.1007/s00574-009-0023-yMathSciNetView ArticleGoogle Scholar
  11. Keng HL, Yuan W: Applications of Number Theory to Numerical Analysis. Springer, Berlin; 1981.View ArticleGoogle Scholar

Copyright

© Demirci and Cangül; licensee Springer. 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.