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

Abstract

Let H( λ q ) be the Hecke group associated to λ q =2cos π q for q3 integer. In this paper, we determine the constant term of the minimal polynomial of λ q denoted by P q (x).

MSC:12E05, 20H05.

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 andS(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 =2cos π q for q3 is an integer, or λ2. 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 q4.

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(2π/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 nN{0}, the n th Chebycheff polynomial, denoted by T n (x), is defined by

T n (x)=cos(narccosx),xR,|x|1,
(3)

or

T n (cosθ)=cosnθ,θR(θ=arccosx+2kπ,kZ).
(4)

Here we use Chebycheff polynomials.

For nN, Cangul denoted the minimal polynomial of cos(2π/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 mN and n=[|m/2|]. Then

  1. (a)

    If m=1, then Ψ 1 (x)=x1, 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 4m, 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 q7, the algebraic number λ q =2cos π q is a root of a minimal polynomial of degree ≥3. Therefore, it is not possible to determine λ q for q7 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 q7 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 2cos h π q with (h,q)=1, h odd and 1hq1. 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 2cos 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 2sin( h π q +θ)=2sinqθ.

We now want to obtain a similar formula for cosine. Replacing θ by π 2 θ, we get

h = 0 q 1 2cos ( h π q θ ) =2sinq ( 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 nN. 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 2cos ( 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 2cos ( 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 2cos ( h π q ) = q 1 h = 1 ( h , q ) = 1 h odd 2cos h π q =c,
(18)

while if q is odd, then

| q 1 h = 1 ( h , q ) = 1 2cos ( h π q ) |= c 2 ,
(19)

as cos(hi) π q =cos i π q . Also note that

sind ( 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 θ0 in (17). If d is odd, then sind( π 2 θ)±1 as θ0 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 sindθ 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 , α1, 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, 0ik.

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, k3, 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 sind( π 2 θ) in the numerator is equal to the product of all coefficients e in the factors sine( 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=27. 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 p1mod4.

(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=235. 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 .

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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.

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Correspondence to Ismail Naci Cangül.

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Demirci, M., Cangül, I.N. The constant term of the minimal polynomial of cos(2π/n) over . Fixed Point Theory Appl 2013, 77 (2013). https://doi.org/10.1186/1687-1812-2013-77

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