Minimal polynomial of 2cos(2pi/n)

Template:Short description In number theory, the real parts of the roots of unity are related to one-another by means of the minimal polynomial of $2\cos (2\pi /n).$ The roots of the minimal polynomial are twice the real part of the roots of unity, where the real part of a root of unity is just $\cos (2k\pi /n)$ with $k$ coprime with $n.$

Formal definition

For an integer $n\ge 1$, the minimal polynomial ${\mathrm{\Psi }}_n(x)$ of $2\cos (2\pi /n)$ is the non-zero monic polynomial of smallest degree for which ${\mathrm{\Psi }}_n(2\cos (2\pi /n))=0$.

For every Template:Mvar, the polynomial ${\mathrm{\Psi }}_n(x)$ is monic, has integer coefficients, and is irreducible over the integers and the rational numbers. All its roots are real; they are the real numbers $2\cos (2k\pi /n)$ with $k$ coprime with $n$ and either $1\le k<n$ or $k=n=1.$ These roots are twice the real parts of the [[primitive root of unity|primitive Template:Mvarth roots of unity]]. The number of integers $k$ relatively prime to $n$ is given by Euler's totient function $\phi (n);$ it follows that the degree of ${\mathrm{\Psi }}_n(x)$ is $1$ for $n=1,2$ and $\phi (n)/2$ for $n\ge 3.$

The first two polynomials are ${\mathrm{\Psi }}_1(x)=x-2$ and ${\mathrm{\Psi }}_2(x)=x+2.$

The polynomials ${\mathrm{\Psi }}_n(x)$ are typical examples of irreducible polynomials whose roots are all real and which have a cyclic Galois group.

Examples

The first few polynomials ${\mathrm{\Psi }}_n(x)$ are

$\begin{array}{cc}{\mathrm{\Psi }}_1(x)& =x-2\\ {\mathrm{\Psi }}_2(x)& =x+2\\ {\mathrm{\Psi }}_3(x)& =x+1\\ {\mathrm{\Psi }}_4(x)& =x\\ {\mathrm{\Psi }}_5(x)& =x^2+x-1\\ {\mathrm{\Psi }}_6(x)& =x-1\\ {\mathrm{\Psi }}_7(x)& =x^3+x^2-2x-1\\ {\mathrm{\Psi }}_8(x)& =x^2-2\\ {\mathrm{\Psi }}_9(x)& =x^3-3x+1\\ {\mathrm{\Psi }}_{10}(x)& =x^2-x-1\\ {\mathrm{\Psi }}_{11}(x)& =x^5+x^4-4x^3-3x^2+3x+1\\ {\mathrm{\Psi }}_{12}(x)& =x^2-3\\ {\mathrm{\Psi }}_{13}(x)& =x^6+x^5-5x^4-4x^3+6x^2+3x-1\\ {\mathrm{\Psi }}_{14}(x)& =x^3-x^2-2x+1\\ {\mathrm{\Psi }}_{15}(x)& =x^4-x^3-4x^2+4x+1\\ {\mathrm{\Psi }}_{16}(x)& =x^4-4x^2+2\\ {\mathrm{\Psi }}_{17}(x)& =x^8+x^7-7x^6-6x^5+15x^4+10x^3-10x^2-4x+1\\ {\mathrm{\Psi }}_{18}(x)& =x^3-3x-1\\ {\mathrm{\Psi }}_{19}(x)& =x^9+x^8-8x^7-7x^6+21x^5+15x^4-20x^3-10x^2+5x+1\\ {\mathrm{\Psi }}_{20}(x)& =x^4-5x^2+5\end{array}$

Explicit form if n is odd

If $n$ is an odd prime, the polynomial ${\mathrm{\Psi }}_n(x)$ can be written in terms of binomial coefficients following a "zigzag path" through Pascal's triangle:

Putting $n=2m+1$ and

$\begin{array}{cc}{\chi }_n(x):& =(\genfrac{}{}{0ex}{}{m}{0})x^m+(\genfrac{}{}{0ex}{}{m-1}{0})x^{m-1}-(\genfrac{}{}{0ex}{}{m-1}{1})x^{m-2}-(\genfrac{}{}{0ex}{}{m-2}{1})x^{m-3}+(\genfrac{}{}{0ex}{}{m-2}{2})x^{m-4}+(\genfrac{}{}{0ex}{}{m-3}{2})x^{m-5}--++\cdots \\ & =\sum _{k=0}^m(-1{)}^{\lfloor k/2\rfloor }(\genfrac{}{}{0ex}{}{m-\lfloor (k+1)/2\rfloor }{\lfloor k/2\rfloor })x^{m-k}\\ & =(\genfrac{}{}{0ex}{}{m}{m})x^m+(\genfrac{}{}{0ex}{}{m-1}{m-1})x^{m-1}-(\genfrac{}{}{0ex}{}{m-1}{m-2})x^{m-2}-(\genfrac{}{}{0ex}{}{m-2}{m-3})x^{m-3}+(\genfrac{}{}{0ex}{}{m-2}{m-4})x^{m-4}+(\genfrac{}{}{0ex}{}{m-3}{m-5})x^{m-5}--++\cdots \\ & =\sum _{k=0}^m(-1{)}^{\lfloor (m-k)/2\rfloor }(\genfrac{}{}{0ex}{}{\lfloor (m+k)/2\rfloor }{k})x^k,\end{array}$

then we have ${\mathrm{\Psi }}_p(x)={\chi }_p(x)$ for primes $p$.

If $n$ is odd but not a prime, the same polynomial ${\chi }_n(x)$, as can be expected, is reducible and, corresponding to the structure of the cyclotomic polynomials ${\mathrm{\Phi }}_d(x)$ reflected by the formula $\prod _{d\mid n}{\mathrm{\Phi }}_d(x)=x^n-1$, turns out to be just the product of all ${\mathrm{\Psi }}_d(x)$ for the divisors $d>1$ of $n$, including $n$ itself:

$\prod _{\genfrac{}{}{0ex}{}{d\mid n}{d>1}}{\mathrm{\Psi }}_d(x)={\chi }_n(x).$

This means that the ${\mathrm{\Psi }}_d(x)$ are exactly the irreducible factors of ${\chi }_n(x)$, which allows to easily obtain ${\mathrm{\Psi }}_d(x)$ for any odd $d$, knowing its degree $\frac{1}{2}\phi (d)$. For example,

$\begin{array}{cc}{\chi }_{15}(x)& =x^7+x^6-6x^5-5x^4+10x^3+6x^2-4x-1\\ & =(x+1)(x^2+x-1)(x^4-x^3-4x^2+4x+1)\\ & ={\mathrm{\Psi }}_3(x)\cdot {\mathrm{\Psi }}_5(x)\cdot {\mathrm{\Psi }}_{15}(x).\end{array}$

Explicit form if n is even

From the below formula in terms of Chebyshev polynomials and the product formula for odd $n$ above, we can derive for even $n$

$\prod _{\genfrac{}{}{0ex}{}{d\mid n}{d>1}}{\mathrm{\Psi }}_d(x)=({\chi }_{n+1}(x)+{\chi }_{n-1}(x)).$

Independently of this, if $n=2^k$ is an even prime power, we have for $k\ge 2$ the recursion (see Template:OEIS2C)

${\mathrm{\Psi }}_{2^{k+1}}(x)=({\mathrm{\Psi }}_{2^k}(x){)}^2-2$,

starting with ${\mathrm{\Psi }}_4(x)=x$.

Roots

The roots of ${\mathrm{\Psi }}_n(x)$ are given by $2\cos \left(\frac{2\pi k}{n}\right)$, where $1\le k<\frac{n}{2}$ and $\gcd (k,n)=1$. Since ${\mathrm{\Psi }}_n(x)$ is monic, we have

${\mathrm{\Psi }}_n(x)=\prod _{\begin{array}{c}1\le k<\frac{n}{2}\\ \gcd (k,n)=1\end{array}}(x-2\cos \left(\frac{2\pi k}{n}\right)).$

Combining this result with the fact that the function $\cos (x)$ is even, we find that $2\cos \left(\frac{2\pi k}{n}\right)$ is an algebraic integer for any positive integer $n$ and any integer $k$.

Relation to the cyclotomic polynomials

For a positive integer $n$, let ${\zeta }_n=\exp \left(\frac{2\pi i}{n}\right)=\cos \left(\frac{2\pi }{n}\right)+\sin \left(\frac{2\pi }{n}\right)i$, a primitive $n$-th root of unity. Then the minimal polynomial of ${\zeta }_n$ is given by the $n$-th cyclotomic polynomial ${\mathrm{\Phi }}_n(x)$. Since ${\zeta }_n^{-1}=\cos \left(\frac{2\pi }{n}\right)-\sin \left(\frac{2\pi }{n}\right)i$, the relation between $2\cos \left(\frac{2\pi }{n}\right)$ and ${\zeta }_n$ is given by $2\cos \left(\frac{2\pi }{n}\right)={\zeta }_n+{\zeta }_n^{-1}$. This relation can be exhibited in the following identity proved by Lehmer, which holds for any non-zero complex number $z$:

${\mathrm{\Psi }}_n(z+z^{-1})=z^{-\frac{\phi (n)}{2}}{\mathrm{\Phi }}_n(z)$

Relation to Chebyshev polynomials

In 1993, Watkins and Zeitlin established the following relation between ${\mathrm{\Psi }}_n(x)$ and Chebyshev polynomials of the first kind.

If $n=2s+1$ is odd, thenTemplate:Check source

$\prod _{d\mid n}{\mathrm{\Psi }}_d(2x)=2(T_{s+1}(x)-T_s(x)),$

and if $n=2s$ is even, then

$\prod _{d\mid n}{\mathrm{\Psi }}_d(2x)=2(T_{s+1}(x)-T_{s-1}(x)).$

If $n$ is a power of $2$, we have moreover directly

${\mathrm{\Psi }}_{2^{k+1}}(2x)=2T_{2^{k-1}}(x).$

Absolute value of the constant coefficient

The absolute value of the constant coefficient of ${\mathrm{\Psi }}_n(x)$ can be determined as follows:

$|{\mathrm{\Psi }}_n(0)|=\{\begin{array}{cc}0& \text{if}n=4,\\ 2& \text{if}n=2^k,k\ge 0,k\ne 2,\\ p& \text{if}n=4p^k,k\ge 1,p>2\text{prime,}\\ 1& \text{otherwise.}\end{array}$

Generated algebraic number field

The algebraic number field $K_n=\mathbb{Q}({\zeta }_n+{\zeta }_n^{-1})$ is the maximal real subfield of a cyclotomic field $\mathbb{Q}({\zeta }_n)$. If ${𝒪}_{K_n}$ denotes the ring of integers of $K_n$, then ${𝒪}_{K_n}=\mathbb{Z}[{\zeta }_n+{\zeta }_n^{-1}]$. In other words, the set $\{1,{\zeta }_n+{\zeta }_n^{-1},\dots ,{({\zeta }_n+{\zeta }_n^{-1})}^{\frac{\phi (n)}{2}-1}\}$ is an integral basis of ${𝒪}_{K_n}$. In view of this, the discriminant of the algebraic number field $K_n$ is equal to the discriminant of the polynomial ${\mathrm{\Psi }}_n(x)$, that is

$D_{K_n}=\{\begin{array}{cc}{2}^{(m-1)2^{m-2}-1}& \text{if}n=2^m,m>2,\\ p^{(m{p}^m-(m+1)p^{m-1}-1)/2}& \text{if}n=p^m\text{or}2p^m,p>2\text{prime},\\ {\left(\prod _{i=1}^{\omega (n)}{p}_i^{e_i-1/(p_i-1)}\right)}^{\frac{\phi (n)}{2}}& \text{if}\omega (n)>1,k\ne 2p^m.\end{array}$