The constant term of the minimal polynomial of $cos(2\pi /n)$ over ℚ

Abstract

Let $H({\lambda }_q)$ be the Hecke group associated to ${\lambda }_q=2cos\frac{\pi }{q}$ for $q\ge 3$ integer. In this paper, we determine the constant term of the minimal polynomial of ${\lambda }_q$ denoted by $P_q^{\ast }(x)$.

MSC:12E05, 20H05.

1 Introduction

The Hecke groups $H(\lambda )$ are defined to be the maximal discrete subgroups of $\mathit{PSL}(2,\mathbb{R})$ generated by two linear fractional transformations

$$T(z)=-\frac{1}{z}\text{and}S(z)=-\frac{1}{z+\lambda },$$

where λ is a fixed positive real number.

Hecke [1] showed that $H(\lambda )$ is Fuchsian if and only if $\lambda ={\lambda }_q=2cos\frac{\pi }{q}$ for $q\ge 3$ is an integer, or $\lambda \ge 2$. In this paper, we only consider the former case and denote the corresponding Hecke groups by $H({\lambda }_q)$. It is well known that $H({\lambda }_q)$ has a presentation as follows (see [2]):

$$H({\lambda }_q)=〈T,S\mid T^2=S^q=I〉.$$

(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 $H({\lambda }_3)=\mathrm{\Gamma }=\mathit{PSL}(2,\mathbb{Z})$ (the modular group), $H({\lambda }_4)=H(\sqrt{2})$, $H({\lambda }_5)=H(\frac{1+\sqrt{5}}{2})$, and $H({\lambda }_6)=H(\sqrt{3})$. It is clear from the above that $H({\lambda }_q)\subset \mathit{PSL}(2,\mathbb{Z}[{\lambda }_q])$, but unlike in the modular group case (the case $q=3$), the inclusion is strict and the index $[\mathit{PSL}(2,\mathbb{Z}[{\lambda }_q]):H({\lambda }_q)]$ is infinite as $H({\lambda }_q)$ is discrete, whereas $\mathit{PSL}(2,\mathbb{Z}[{\lambda }_q])$ is not for $q\ge 4$.

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 $cos(2\pi /n)$ 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 $n\in \mathbb{N}\cup \{0\}$, the n th Chebycheff polynomial, denoted by $T_n(x)$, is defined by

$$T_n(x)=cos(n\cdot arccosx),x\in \mathbb{R},|x|\le 1,$$

(3)

or

$$T_n(cos\theta )=cosn\theta ,\theta \in \mathbb{R}(\theta =arccosx+2k\pi ,k\in \mathbb{Z}).$$

(4)

Here we use Chebycheff polynomials.

For $n\in \mathbb{N}$, Cangul denoted the minimal polynomial of $cos(2\pi /n)$ over Q by ${\mathrm{\Psi }}_n(x)$. Then he obtained the following formula for the minimal polynomial ${\mathrm{\Psi }}_n(x)$.

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

Let $m\in \mathbb{N}$ and $n=[|m/2|]$. Then

• (a) If $m=1$, then ${\mathrm{\Psi }}_1(x)=x-1$, and if $m=2$, then ${\mathrm{\Psi }}_2(x)=x+1$.

• (b) If m is an odd prime, then

  $${\mathrm{\Psi }}_m(x)=\frac{T_{n+1}(x)-T_n(x)}{2^n(x-1)}.$$

  (5)

• (c) If $4\mid m$, then

  $${\mathrm{\Psi }}_m(x)=\frac{T_{n+1}(x)-T_{n-1}(x)}{2^{n/2}(T_{\frac{n}{2}+1}(x)-T_{\frac{n}{2}-1}(x)){\prod }_{d\mid m,d\ne m,d\mid \frac{m}{2}}^{q-1}{\mathrm{\Psi }}_d(x)}.$$

  (6)

• (d) If m is even and $m/2$ is odd, then

  $${\mathrm{\Psi }}_m(x)=\frac{T_{n+1}(x)-T_{n-1}(x)}{2^{n-n^{\mathrm{\prime }}}(T_{n^{\mathrm{\prime }}+1}(x)-T_{n^{\mathrm{\prime }}}(x)){\prod }_{d\mid m,d\ne m,d\mathit{\text{even}}}^{q-1}{\mathrm{\Psi }}_d(x)},$$

  (7)

where $n^{\mathrm{\prime }}=\frac{\frac{m}{2}-1}{2}$.

• (e) Let m be odd and let p be a prime dividing m. If $p^2\mid m$, then

  $${\mathrm{\Psi }}_m(x)=\frac{T_{n+1}(x)-T_n(x)}{2^{n-n^{\mathrm{\prime }}}(T_{n^{\mathrm{\prime }}+1}(x)-T_{n^{\mathrm{\prime }}}(x))},$$

  (8)

where $n^{\mathrm{\prime }}=\frac{\frac{m}{p}-1}{2}$. If $p^2\mid m$, then

$${\mathrm{\Psi }}_m(x)=\frac{T_{n+1}(x)-T_n(x)}{2^{n-n^{\mathrm{\prime }}}(T_{n^{\mathrm{\prime }}+1}(x)-T_{n^{\mathrm{\prime }}}(x)){\mathrm{\Psi }}_p(x)},$$

(9)

where $n^{\mathrm{\prime }}=\frac{\frac{m}{p}-1}{2}$.

For the first four Hecke groups Γ, $H(\sqrt{2})$ , $H({\lambda }_5)$, and $H(\sqrt{3})$, we can find the minimal polynomial, denoted by $P_q^{\ast }(x)$, of ${\lambda }_q$ over Q as ${\lambda }_3-1$, ${\lambda }_4^2-2$, ${\lambda }_5^2-{\lambda }_5-1$, and ${\lambda }_6^2-3$, respectively. However, for $q\ge 7$, the algebraic number ${\lambda }_q=2cos\frac{\pi }{q}$ is a root of a minimal polynomial of degree ≥3. Therefore, it is not possible to determine ${\lambda }_q$ for $q\ge 7$ as nicely as in the first four cases. Because of this, it is easy to find and study with the minimal polynomial of ${\lambda }_q$ instead of ${\lambda }_q$ itself. The minimal polynomial of ${\lambda }_q$ has been used for many aspects in the literature (see [5–8] and [9]).

Notice that there is a relation

$$P_q^{\ast }(x)=2^{\phi (2q)/2}\cdot {\mathrm{\Psi }}_{2q}\left(\frac{x}{2}\right)$$

between $P_q^{\ast }(x)$ and ${\mathrm{\Psi }}_m(x)$.

In [10], when the principal congruence subgroups of $H({\lambda }_q)$ for $q\ge 7$ prime were studied, we needed to know whether the minimal polynomial of ${\lambda }_q$ 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 $P_q^{\ast }(x)$ of ${\lambda }_q$. 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 $P_q^{\ast }(x)$

In this section, we calculate the constant term for all values of q. Let c denote the constant term of the minimal polynomial $P_q^{\ast }(x)$ of ${\lambda }_q$, i.e.,

$$c=P_q^{\ast }(0).$$

(10)

We know from [[4], Lemma, p.473] that the roots of $P_q^{\ast }(x)$ are $2cos\frac{h\pi }{q}$ with $(h,q)=1$, h odd and $1\le h\le q-1$. Being the constant term, c is equal to the product of all roots of $P_q^{\ast }(x)$:

$$c=\underset{\underset{h\mathrm{odd}}{\underset{(h,q)=1}{h=1}}}{\prod ^{q-1}}2cos\frac{h\pi }{q}.$$

(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 ${\prod }_{h=0}^{q-1}2sin(\frac{h\pi }{q}+\theta )=2sinq\theta$.

We now want to obtain a similar formula for cosine. Replacing θ by $\frac{\pi }{2}-\theta$, we get

$$\prod _{h=0}^{q-1}2cos(\frac{h\pi }{q}-\theta )=2sinq(\frac{\pi }{2}-\theta ).$$

(12)

Let now μ denote the Möbius function defined by

$$\mu (n)=\{\begin{array}{cc}0\hfill & \text{if }n\text{ is not square-free},\hfill \\ 1\hfill & \text{if }n=1,\hfill \\ {(-1)}^k\hfill & \text{if }n\text{ has }k\text{ distinct prime factors},\hfill \end{array}$$

(13)

for $n\in \mathbb{N}$. It is known that

$$\sum _{d\mid n}\mu (d)=\{\begin{array}{cc}0\hfill & \text{if }n>1,\hfill \\ 1\hfill & \text{if }n=1.\hfill \end{array}$$

(14)

Using this last fact, we obtain

(15)

Therefore

$$\underset{\underset{(h,q)=1}{h=0}}{\prod ^{q-1}}2cos(\frac{h\pi }{q}-\theta )=\prod _{d\mid q}{(sind(\frac{\pi }{2}-\theta ))}^{\mu (q/d)}.$$

(16)

Finally, as $(0,q)\ne 1$, we can write (16) as

$$\underset{\underset{(h,q)=1}{h=1}}{\prod ^{q-1}}2cos(\frac{h\pi }{q}-\theta )=\prod _{d\mid q}{(sind(\frac{\pi }{2}-\theta ))}^{\mu (q/d)}.$$

(17)

Note that if q is even, then

$$\underset{\underset{(h,q)=1}{h=1}}{\prod ^{q-1}}2cos\left(\frac{h\pi }{q}\right)=\underset{\underset{h\mathrm{odd}}{\underset{(h,q)=1}{h=1}}}{\prod ^{q-1}}2cos\frac{h\pi }{q}=c,$$

(18)

while if q is odd, then

$$|\underset{\underset{(h,q)=1}{h=1}}{\prod ^{q-1}}2cos\left(\frac{h\pi }{q}\right)|=c^2,$$

(19)

as $cos(h-i)\frac{\pi }{q}=-cos\frac{i\pi }{q}$. Also note that

$$sind(\frac{\pi }{2}-\theta )=\{\begin{array}{cc}cosd\theta \hfill & \text{if }d\equiv 1mod4,\hfill \\ sind\theta \hfill & \text{if }d\equiv 2mod4,\hfill \\ -cosd\theta \hfill & \text{if }d\equiv 3mod4,\hfill \\ -sind\theta \hfill & \text{if }d\equiv 0mod4.\hfill \end{array}$$

(20)

To compute c, we let $\theta \to 0$ in (17). If d is odd, then $sind(\frac{\pi }{2}-\theta )\to \pm 1$ as $\theta \to 0$ by (20). So, we are only concerned with even d. Indeed, if q is odd, then the left-hand side at $\theta =0$ 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

$$c=\underset{\theta \to 0}{lim}\prod _{d\mid q}{(sind(\frac{\pi }{2}-\theta ))}^{\mu (q/d)}.$$

(22)

Proof Note that by (20), the right-hand side of (22) becomes a product of $\pm {(cosd\theta )}^{\pm 1}$’s and $\pm {(sind\theta )}^{\pm 1}$’s. Above we saw that we can omit the former ones as they tend to ±1 as θ tends to 0. Now, as ${\sum }_{d\mid n}\mu (d)=0$, there are equal numbers of the latter kind factors in the numerator and denominator, i.e., if there is a factor $sind\theta$ in the numerator, then there is a factor $sin{d}^{\mathrm{\prime }}\theta$ in the denominator. Then using the fact that

$$\underset{\theta \to 0}{lim}\frac{sink\theta }{sinl\theta }=\frac{k}{l},$$

(23)

we can calculate c.

In fact the calculations show that there are three possibilities:

(i) Let $q=2^{{\alpha }_0}$, ${\alpha }_0\ge 2$. Then the only divisors of q such that $\mu (q/d)\ne 0$ are $d=2^{{\alpha }_0}$ and $2^{{\alpha }_0-1}$. Therefore

$$\begin{array}{rcl}c& =& \underset{\theta \to 0}{lim}\frac{sin{2}^{{\alpha }_0}(\frac{\pi }{2}-\theta )}{sin{2}^{{\alpha }_0-1}(\frac{\pi }{2}-\theta )}\\ =& \{\begin{array}{cc}2\hfill & \text{if }{\alpha }_0>2,\hfill \\ -2\hfill & \text{if }{\alpha }_0=2.\hfill \end{array}\end{array}$$

(24)

(ii) Secondly, let $q=2p^{\alpha }$, $\alpha \ge 1$, p odd prime. Then the only divisors of q such that $\mu (q/d)\ne 0$ are $d=2p^{\alpha }$, $2p^{\alpha -1}$, $p^{\alpha }$ and $p^{\alpha -1}$. Therefore

$$\begin{array}{rcl}c& =& \underset{\theta \to 0}{lim}\frac{sin2p^{\alpha }(\frac{\pi }{2}-\theta )\cdot sin{p}^{\alpha -1}(\frac{\pi }{2}-\theta )}{sin{p}^{\alpha }(\frac{\pi }{2}-\theta )\cdot sin2p^{\alpha -1}(\frac{\pi }{2}-\theta )}\\ =& \underset{\theta \to 0}{lim}ϵ\cdot \frac{sin2p^{\alpha }\theta \cdot cos{p}^{\alpha -1}\theta }{cos{p}^{\alpha }\theta \cdot sin2p^{\alpha -1}\theta }\\ =& ϵ\cdot p,\end{array}$$

(25)

where

$$ϵ=\{\begin{array}{cc}1\hfill & \text{if }p\equiv 1mod4,\hfill \\ -1\hfill & \text{if }p\equiv -1mod4.\hfill \end{array}$$

(26)

(iii) Let q be different from above. Then q can be written as

$$q=2^{{\alpha }_0}{p}_1^{{\alpha }_1}\cdots p_k^{{\alpha }_k},$$

(27)

where $p_i$ are distinct odd primes and ${\alpha }_i\ge 1$, $0\le i\le k$.

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

Let $k=1$, i.e., let $q=2^{{\alpha }_0}{p}_1^{{\alpha }_1}$. We have already discussed the case ${\alpha }_0=1$. Let ${\alpha }_0>1$. Then the only divisors d of q with $\mu (q/d)\ne 0$ are $d=2^{{\alpha }_0}{p}_1^{{\alpha }_1}$, $2^{{\alpha }_0-1}{p}_1^{{\alpha }_1}$, $2^{{\alpha }_0}{p}_1^{{\alpha }_1-1}$ and $2^{{\alpha }_0-1}{p}_1^{{\alpha }_1-1}$. Therefore

$$\begin{array}{rcl}c& =& \underset{\theta \to 0}{lim}\frac{sin{2}^{{\alpha }_0}{p}_1^{{\alpha }_1}(\frac{\pi }{2}-\theta )\cdot sin{2}^{{\alpha }_0-1}{p}_1^{{\alpha }_1-1}(\frac{\pi }{2}-\theta )}{sin{2}^{{\alpha }_0-1}{p}_1^{{\alpha }_1}(\frac{\pi }{2}-\theta )\cdot sin{2}^{{\alpha }_0}{p}_1^{{\alpha }_1-1}(\frac{\pi }{2}-\theta )}\\ =& 1.\end{array}$$

(28)

Now let $k=2$, i.e., let $q=2^{{\alpha }_0}{p}_1^{{\alpha }_1}{p}_2^{{\alpha }_2}$, ($p_1<p_2$). Similarly, all divisors d of q such that $\mu (q/d)\ne 0$ are $d=2^{{\alpha }_0}{p}_1^{{\alpha }_1}{p}_2^{{\alpha }_2}$, $2^{{\alpha }_0-1}{p}_1^{{\alpha }_1}{p}_2^{{\alpha }_2}$, $2^{{\alpha }_0}{p}_1^{{\alpha }_1-1}{p}_2^{{\alpha }_2}$, $2^{{\alpha }_0}{p}_1^{{\alpha }_1}{p}_2^{{\alpha }_2-1}$, $2^{{\alpha }_0}{p}_1^{{\alpha }_1-1}{p}_2^{{\alpha }_2-1}$, $2^{{\alpha }_0-1}{p}_1^{{\alpha }_1-1}{p}_2^{{\alpha }_2-1}$, $2^{{\alpha }_0-1}{p}_1^{{\alpha }_1}{p}_2^{{\alpha }_2-1}$ and $2^{{\alpha }_0-1}{p}_1^{{\alpha }_1-1}{p}_2^{{\alpha }_2}$. Therefore

$$\begin{array}{rcl}c& =& \underset{\theta \to 0}{lim}\frac{sin{2}^{{\alpha }_0-1}{p}_1^{{\alpha }_1-1}{p}_2^{{\alpha }_2-1}(\frac{\pi }{2}-\theta )\cdot sin{2}^{{\alpha }_0}{p}_1^{{\alpha }_1}{p}_2^{{\alpha }_2-1}(\frac{\pi }{2}-\theta )}{sin{2}^{{\alpha }_0}{p}_1^{{\alpha }_1-1}{p}_2^{{\alpha }_2-1}(\frac{\pi }{2}-\theta )\cdot sin{2}^{{\alpha }_0-1}{p}_1^{{\alpha }_1}{p}_2^{{\alpha }_2-1}(\frac{\pi }{2}-\theta )}\\ \times \underset{\theta \to 0}{lim}\frac{sin{2}^{{\alpha }_0}{p}_1^{{\alpha }_1-1}{p}_2^{{\alpha }_2}(\frac{\pi }{2}-\theta )\cdot sin{2}^{{\alpha }_0-1}{p}_1^{{\alpha }_1}{p}_2^{{\alpha }_2}(\frac{\pi }{2}-\theta )}{sin{2}^{{\alpha }_0-1}{p}_1^{{\alpha }_1-1}{p}_2^{{\alpha }_2}(\frac{\pi }{2}-\theta )\cdot sin{2}^{{\alpha }_0}{p}_1^{{\alpha }_1}{p}_2^{{\alpha }_2}(\frac{\pi }{2}-\theta )}\\ =& 1.\end{array}$$

(29)

Finally, $k\ge 3$, i.e., let

$$q=2^{{\alpha }_0}{p}_1^{{\alpha }_1}\cdots p_k^{{\alpha }_k}\text{with }{p}_1<p_2<\cdots <p_k.$$

In this case the proof is similar, but rather more complicated. In fact, the number of all divisors d of q such that $\mu (q/d)\ne 0$ is $2^{k+1}$. There is $\left(\begin{array}{c}k+1\\ 0\end{array}\right)=1$ divisor of the form

$$d=2^{{\alpha }_0}{p}_1^{{\alpha }_1}\cdots p_k^{{\alpha }_k}.$$

There are $\left(\begin{array}{c}k+1\\ 1\end{array}\right)=k+1$ divisors of the form

$$d=2^{{\alpha }_0-1}{p}_1^{{\alpha }_1}\cdots p_k^{{\alpha }_k},2^{{\alpha }_0}{p}_1^{{\alpha }_1-1}\cdots p_k^{{\alpha }_k},\dots ,2^{{\alpha }_0}{p}_1^{{\alpha }_1}\cdots p_k^{{\alpha }_k-1}.$$

There are $\left(\begin{array}{c}k+1\\ 2\end{array}\right)=\frac{k(k+1)}{2}$ divisors of the form

$$\begin{array}{rcl}d& =& 2^{{\alpha }_0-1}{p}_1^{{\alpha }_1-1}{p}_2^{{\alpha }_2}\cdots p_k^{{\alpha }_k},2^{{\alpha }_0-1}{p}_1^{{\alpha }_1}{p}_2^{{\alpha }_2-1}\cdots p_k^{{\alpha }_k},\dots ,2^{{\alpha }_0-1}{p}_1^{{\alpha }_1}{p}_2^{{\alpha }_2}\cdots p_k^{{\alpha }_k-1},\\ 2^{{\alpha }_0}{p}_1^{{\alpha }_1-1}{p}_2^{{\alpha }_2-1}\cdots p_k^{{\alpha }_k},\dots ,2^{{\alpha }_0}{p}_1^{{\alpha }_1-1}{p}_2^{{\alpha }_2}\cdots p_k^{{\alpha }_k-1},\dots ,2^{{\alpha }_0}{p}_1^{{\alpha }_1}\cdots p_{k-1}^{{\alpha }_{k-1}-1}{p}_k^{{\alpha }_k-1}.\end{array}$$

If we continue, we can find other divisors d of q, similarly. Finally, there is $\left(\begin{array}{c}k+1\\ k+1\end{array}\right)=1$ divisor of the form $2^{{\alpha }_0-1}{p}_1^{{\alpha }_1-1}{p}_2^{{\alpha }_2-1}\cdots p_k^{{\alpha }_k-1}$. Thus, the product of all coefficients d in the factors $sind(\frac{\pi }{2}-\theta )$ in the numerator is equal to the product of all coefficients e in the factors $sine(\frac{\pi }{2}-\theta )$ 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 $\mu (8/d)\ne 0$ are $d=8$ and 4. Therefore

$$\begin{array}{rcl}c& =& \underset{\theta \to 0}{lim}\frac{sin8(\frac{\pi }{2}-\theta )}{sin4(\frac{\pi }{2}-\theta )}\\ =& 2.\end{array}$$

(ii) Let $q=14=2\cdot 7$. The only divisors of 14 such that $\mu (14/d)\ne 0$ are $d=14,2,7$ and 1. Therefore

$$\begin{array}{rcl}c& =& ϵ\cdot \underset{\theta \to 0}{lim}\frac{sin14(\frac{\pi }{2}-\theta )\cdot sin(\frac{\pi }{2}-\theta )}{sin7(\frac{\pi }{2}-\theta )\cdot sin2(\frac{\pi }{2}-\theta )}\\ =& -7,\end{array}$$

since $p\equiv -1mod4$.

(iii) Let $q=24=2^3\cdot 3$. The only divisors of 24 such that $\mu (24/d)\ne 0$ are $d=24,12,8$ and 4. Therefore

$$\begin{array}{rcl}c& =& \underset{\theta \to 0}{lim}\frac{sin24(\frac{\pi }{2}-\theta )\cdot sin4(\frac{\pi }{2}-\theta )}{sin12(\frac{\pi }{2}-\theta )\cdot sin8(\frac{\pi }{2}-\theta )}\\ =& 1.\end{array}$$

(iv) Let $q=30=2\cdot 3\cdot 5$. The only divisors of 30 such that $\mu (30/d)\ne 0$ are $d=30,15,10,6,5,3,2$ and 1. Therefore

$$\begin{array}{rcl}c& =& \underset{\theta \to 0}{lim}\frac{sin(\frac{\pi }{2}-\theta )\cdot sin6(\frac{\pi }{2}-\theta )\cdot sin10(\frac{\pi }{2}-\theta )\cdot sin15(\frac{\pi }{2}-\theta )}{sin2(\frac{\pi }{2}-\theta )\cdot sin3(\frac{\pi }{2}-\theta )\cdot sin5(\frac{\pi }{2}-\theta )\cdot sin30(\frac{\pi }{2}-\theta )}\\ =& 1.\end{array}$$