HOMOGENEOUS POLYNOMIALS AND THE MINIMAL POLYNOMIAL OF COS(2π/n)

David Surowski and Paul McCombs

Abstract. It is well known that if Φ_{n}(x) is the nth cyclotomic polynomial, then there is a factorization x^{n} − 1 = ∏ Φ_{d}(x), where the product is taken over the divisors d of n. Thus, one can obtain, by Möbius inversion, a product formula for each Φ_{n}(x) in terms of the various factors x^{d} − 1. The purpose of this note is two– fold. First, we show that the above factorization implies a similar factorization for the minimal polynomials of the algebraic numbers cos(2π/n), where n is a positive integer. Secondly, we give an explicit formula for the minimal polynomials of cos(2π/p), where p is prime.

Introduction. If ζ ∈ ℂ is the primitive nth root of unity ζ = e^{2πi/n}, and if Φ_{n}(x) is the minimal polynomial of ζ, then it is well known that Φ_{n}(x) is a monic polynomial with integer coefficients, has degree ϕ(n) (Euler ϕ–function), and satisfies the identity

x^{n} − 1 = ∏_{d|n} Φ_{d}(x).  (1)

From this, the cyclotomic polynomials can be computed via Möbius inversion:

Φ_{n}(x) = ∏_{d|n}(x^{d} − 1)^{μ( \frac{n}{d})},  (2)

where for any integer k,

μ(k) = \{ \begin{matrix}(−1)^{l} & if k factors into l distinct primes, \\ 0 & if not. \end{matrix}

In [3], W. Watkins and J. Zeitlin show that if Ψ_{n}(x) is the minimal polynomial of cos(2π/n), then in analogy with equation (1) one has identities of the form

T_{s+1}(x) − T_{s}(x) = 2^{s} ∏_{d|n} Ψ_{d}(x) if n = 2s + 1 is odd,  (3)

T_{s+1}(x) − T_{s−1}(x) = 2^{s} ∏_{d|n} Ψ_{d}(x) if n = 2s is even.  (4)

In the above expressions, T_{s}(x) is the sth Chebyshev polynomial, defined by setting T_{s}(cos θ) = cos sθ. Thus, one can, in principle, compute the polynomials Ψ_{n}(x) using the Chebyshev polynomials and Möbius inversion.

In this note, we shall show first that equations (3) and (4) are not merely analogs of equation (1), they are consequences of it. In the last section, we shall give an explicit formula for the minimal polynomial of cos(2π/p), where p is prime. This seems not to have been given in the literature.

1. Homogeneous Polynomials and Specialization. If D is an integral domain and f(x) ∈ D[x] is a polynomial of degree k, we shall denote by f(x, y) ∈ D[x, y] the corresponding homogeneous polynomial, also of degree k. Thus, if f(x) = ∑ a_{i}x^{i}, then f(x, y) = ∑ a_{i}x^{i}y^{k−i}. Clearly, if f(x), g(x) ∈ D[x], and if h(x) = f(x)g(x), then h(x, y) = f(x, y)g(x, y). This allows us to consider the homogeneous version of equation (1):

x^{n} − y^{n} = ∏_{d|n} Φ_{d}(x, y).  (5)

Next, as the roots of Φ_{n}(x) consist of all of the primitive nth roots of unity, and since these are closed under taking inverses, we see that if n > 1, the polynomial is “palindromic” in the sense that if k = ϕ(n) and Φ_{n}(x) = ∑_{i=0}^{k} a_{i}x^{i}, then a_{i} = a_{k−i}, i = 0, 1, …, k. If n > 2, then ϕ(n) is even: ϕ(n) = 2s, for some integer s. In this case we set

L_{n}(x, y) = ∑_{i=0}^{s} a_{i}(x^{i} + y^{i})(\sqrt{xy})^{s−i} ∈ ℤ[x, y, \sqrt{xy}].

Note that since the polynomials x^{i} + y^{i} are symmetric in x and y, then by the Fundamental Theorem on Symmetric Polynomials (FTSP) [1], x^{i} + y^{i} can be expressed as a polynomial in the “elementary symmetric polynomials” σ_{1} = x+y and σ_{2} = xy. Therefore, each L_{n}(x, y) ∈ ℤ[σ_{1}, \sqrt{σ_{2}}], L_{n}(x, y) = Λ_{n}(σ_{1}, \sqrt{σ_{2}}), where Λ_{n}(x, y) ∈ ℤ[x, y].

Next, FTSP also says [1] that the polynomials are algebraically independent over ℤ, from which it follows easily that σ_{1} and \sqrt{σ_{2}} are also algebraically independent. Therefore, if R ⊇ ℤ is any integral domain containing ℤ, and if r_{1}, r_{2} ∈ R are arbitrary elements, then the evaluation σ_{1} ↦ r_{1}, \sqrt{σ_{2}} ↦ r_{2} determines a unique homomorphism ℤ[σ_{1}, \sqrt{σ_{2}}] → R, f(σ_{1}, \sqrt{σ_{2}}) ↦ f(r_{1}, r_{2}). Notice that if n > 2, and ϕ(n) = 2s, then x^{−s}Φ_{n}(x) = L_{n}(x, x^{−1}) = Λ_{n}(x + x^{−1}, 1). We now define the polynomials Θ_{n}(x) = Λ_{n}(x, 1). Clearly, Θ_{n}(x) is a monic polynomial of degree s having 2 cos(2π/n) as a root. Since it is a simple matter to show that the minimal polynomial of 2 cos(2π/n) must have degree s [2,3], one concludes that Θ_{n}(x) is the minimal polynomial of 2 cos(2π/n). From this, we see easily that if Ψ_{n}(x) is the minimal polynomial of cos(2π/n), then we must have Ψ_{n}(x) = 2^{−s}Θ_{n}(2x).

\underline{Lemma 1.1}. For each n > 2, L_{n}(x, y) = Φ_{n}(\sqrt{x}, \sqrt{y}).

\underline{Proof}. As Φ_{n}(\sqrt{x}, \sqrt{y}) = ∏(\sqrt{x} − ω\sqrt{y}), where the product is taken over the primitive nth roots of unity, and since L_{n}(x, y), Φ_{n}(\sqrt{x}, \sqrt{y}) have the same degree as polynomials in \sqrt{x}, \sqrt{y}, it suffices to prove that L_{n}(ω^{2}y, y) = 0 for any primitive nth root of unity. We have

\begin{aligned}L_{n}(ω^{2}y, y) & = ∑_{i=0}^{s} a_{i}(ω^{2i}y^{i} + y^{i})(ωy)^{s−i} \\ & = y^{s}ω^{−s} ∑_{i=0}^{s} a_{i}(ω^{i} + ω^{−i}) = 0. \end{aligned}

Next, we set L_{1}(x, y) = x + y − 2\sqrt{xy}, L_{2}(x, y) = x + y + 2\sqrt{xy}. Thus, L_{1}(x, y) = Λ_{1}(σ_{1}, \sqrt{σ_{2}}) := σ_{1} − 2\sqrt{σ_{2}}, L_{2}(x, y) = Λ_{2}(σ_{1}, \sqrt{σ_{2}}) := σ_{1} + 2\sqrt{σ_{2}}, and Λ_{i}(x, 1) = Θ_{i}(x), i = 1, 2. Note also that L_{1}(x, y)L_{2}(x, y) = (x − y)^{2}.

\underline{Proposition 1.2}. For any integer n ≥ 1,

(x − y)(x^{n} − y^{n}) = ∏_{d|2n} L_{d}(x, y).

\underline{Proof}. We have

x^{n} − y^{n} = ∏_{d|2n} Φ_{d}(\sqrt{x}, \sqrt{y});

as (x − y)Φ_{1}(\sqrt{x}, \sqrt{y})Φ_{2}(\sqrt{x}, \sqrt{y}) = (x − y)^{2} = L_{1}(x, y)L_{2}(x, y), we may multiply both sides of the above equation by (x − y) and apply Lemma 1.1 to obtain the result.

\underline{Theorem 1.3}. With the notation as above, we have

x^{s+1} + y^{s+1} − \sqrt{xy}(x^{s} + y^{s}) = ∏_{d|n} L_{d}(x, y), if n = 2s + 1 is odd,

x^{s+1} + y^{s+1} − \sqrt{xy}(x^{s−1} + y^{s−1}) = ∏_{d|n} L_{d}(x, y), if n = 2s is even.

\underline{Proof}. Assume first that n = 2s + 1 is odd. Then

(x^{s+1} + y^{s+1} − \sqrt{xy}(x^{s} + y^{s}))(x^{s+1} + y^{s+1} + \sqrt{xy}(x^{s} + y^{s})) = (x − y)(x^{n} + y^{n})

= ∏_{d|2n} L_{d}(x, y)

= ∏_{d|n} L_{d}(x, y) ∏_{d|n} L_{2d}(x, y).

Since the polynomials L_{d}(x, y) are pairwise relatively prime in ℤ[\sqrt{x}, \sqrt{y}], it suffices to show that L_{d}(x, y)|(x^{s+1} + y^{s+1} − \sqrt{xy}(x^{s} + y^{s})) in ℤ[\sqrt{x}, \sqrt{y}], whenever d|n. If d|n, d ≠ 1, 2, then L_{d}(x, y) = Φ_{d}(\sqrt{x}, \sqrt{y}) which has factors of the form (\sqrt{x}−ω\sqrt{y}) where ω is a primitive dth root of unity. On the other hand, if we set \sqrt{x} = ω\sqrt{y} in x^{s+1} + y^{s+1} − \sqrt{xy}(x^{s} + y^{s}), we obtain

ω^{2s+2}y^{s+1} + y^{s+1} − ωy(ω^{2s}y^{s} + y^{s}) = y^{s+1}(ω^{2s+2} + 1 − ω^{2s+1} − ω)

= y^{s+1}(ω^{n+1} + 1 − ω^{n} − ω) = 0;

since d|n implies that ω^{n} = 1. Finally, note that

x^{s+1} + y^{s+1} − \sqrt{xy}(x^{s} + y^{s}) = (\sqrt{x} − \sqrt{y})((\sqrt{x})^{2s+1} − (\sqrt{y})^{2s+1})

and so L_{1}(x, y) = x + y − 2\sqrt{xy} = (\sqrt{x} − \sqrt{y})^{2} divides x^{s+1} + y^{s+1} − \sqrt{xy}(x^{s} + y^{s}), as well.

Next, assume that n = 2s is even. As above, we assume that d|n and that ω is a primitive dth root of unity. Setting \sqrt{x} = ω\sqrt{y} yields

x^{s+1} + y^{s+1} − \sqrt{xy}(x^{s−1} + y^{s−1}) = y^{s+1}(ω^{2s+2} + 1 − ω^{2s} − ω^{2}) = 0.

This proves that if d|n, L_{d}(x, y)|(x^{s+1} + y^{s+1} − \sqrt{xy}(x^{s−1} + y^{s−1})). Also, L_{1}(x, y)L_{2}(x, y) = (x + y − 2\sqrt{xy})(x + y + 2\sqrt{xy}) = (x + y)^{2} − 4xy = (x − y)^{2} and x^{s+1} + y^{s+1} − \sqrt{xy}(x^{s−1} + y^{s−1}) = (x − y)(x^{s} − y^{s}) and hence, is divisible by (x − y)^{2}. Finally, the argument is concluded in both cases by observing that as a polynomial in ℤ[\sqrt{x}, \sqrt{y}], ∏_{d|n} L_{d}(x, y) has degree n + 2.

2. The Minimal Polynomial of cos(2π/n). To relate the work of Section 1 with that of Watkins and Zeitlin [3], we recall the definition of the Chebyshev polynomial T_{n}(cos θ) = cos nθ. Equivalently, if ζ = e^{2πiθ}, then cos θ = \frac{1}{2}(ζ + ζ^{−1}), cos nθ = \frac{1}{2}(ζ^{n} + ζ^{−n}) and so the coefficients of T_{s}(x) are obtained by writing \frac{1}{2}(ζ^{n} + ζ^{−n}) as a polynomial in \frac{1}{2}(ζ + ζ^{−1}). That this can be done is an easy inductive argument. On the other hand, using FTSP one writes x^{n} + y^{n} as a polynomial in σ_{1} = x + y and σ_{2} = xy. For example, we have

x^{3} + y^{3} = (x + y)^{3} − 3xy(x + y) = σ_{1}^{3} − 3σ_{2}σ_{1}.

Thus, if x^{n} + y^{n} = S_{n}(σ_{1}, σ_{2}) then one sees easily that T_{n}(x) = \frac{1}{2}S_{n}(2x, 1). Next, the polynomials occurring in Theorem 1.3 are all in ℤ[σ_{1}, \sqrt{σ_{2}}]; we may then specialize σ_{1} ↦ x, \sqrt{σ_{2}} ↦ 1. We have already observed that the polynomials L_{d}(x, y) specialize to Θ_{d}(x), the minimal polynomial of 2 cos(2π/d). The following is immediate from which equations (3) and (4) follow easily.

\underline{Corollary 2.1 [Watkins–Zeitlin]}. The following are polynomial identities.

S_{s+1}(x) − S_{s}(x) = ∏_{d|n} Θ_{d}(x) if n = 2s + 1 is odd;

S_{s+1}(x) − S_{s−1}(x) = ∏_{d|n} Θ_{d}(x) if n = 2s is even.

3. The Minimal Polynomial of cos(2π/p), Where p is Prime. In this section, we assume that p is prime, and that ζ = e^{2πi/p}. As in the earlier sections, we shall continue to focus on 2 cos(2π/p) = ζ + ζ^{−1} and its minimal polynomial Θ_{p}(x), and derive information on Ψ_{p}(x) as a consequence.

Naturally, one approach to this problem is to write p = 2s + 1 (the case p = 2 being trivial: Θ_{2}(x) = x + 2) and use Corollary 2.1. This yields

(x − 2)(S_{s+1}(x) − S_{s}(x)) = Θ_{p}(x).

This will generate recurrence relations on the coefficients of Θ_{p}(x). However, we prefer a more direct approach.

\underline{Theorem 3.1}. Let p = 2s + 1 be an odd prime. If Θ_{p}(x) is the minimal polynomial of 2 cos(2π/p), then

Θ_{p}(x) = ∑_{i=0}^{s}(−1)^{i}σ_{i}x^{s−i},

where

\begin{aligned}σ_{2k} & = (−1)^{k}\binom{s − k}{k}, k = 0, 1, …, ⌊\frac{s}{2}⌋; \\ σ_{2k+1} & = (−1)^{k}\binom{s − k}{k − 1}, k = 1, …, ⌊\frac{s + 1}{2}⌋. \end{aligned}

\underline{Proof}. If

f(x) = ∑_{k=0}^{⌊ \frac{s}{2} ⌋}(−1)^{k}\binom{s − k}{k}x^{s−2k} − ∑_{k=1}^{⌊ \frac{s+1}{2} ⌋}(−1)^{k}\binom{s − k}{k − 1}x^{s−(2k−1)},

then since deg f(x) = deg Θ_{p}(x), it suffices to show that f(ζ + ζ^{−1}) = 0. We have

f(ζ + ζ^{−1}) = ∑_{k=0}^{⌊ \frac{s}{2} ⌋}(−1)^{k}\binom{s − k}{k}(ζ + ζ^{−1})^{s−2k}

− ∑_{k=1}^{⌊ \frac{s+1}{2} ⌋}(−1)^{k}\binom{s − k}{k − 1}(ζ + ζ^{−1})^{s−2k+1}

= ∑_{k=0}^{⌊ \frac{s}{2} ⌋} ∑_{l=0}^{s−2k}(−1)^{k}\binom{s − k}{k}\binom{s − 2k}{l}ζ^{−s+2k+2l}

− ∑_{k=1}^{⌊ \frac{s+1}{2} ⌋} ∑_{l=0}^{s−2k+1}(−1)^{k}\binom{s − k}{k − 1}\binom{s − 2k + 1}{l}ζ^{−s+2k+2l−1}.

In the sum,

∑_{k=0}^{⌊ \frac{s}{2} ⌋} ∑_{l=0}^{s−2k}(−1)^{k}\binom{s − k}{k}\binom{s − 2k}{l}ζ^{−s+2k+2l},

the coefficient of ζ^{−s+2r}, 0 ≤ r ≤ ⌊\frac{s}{2}⌋ is given by

\begin{matrix}\binom{s}{0}\binom{s}{r} − \binom{s − 1}{1} & \binom{s − 2}{r − 1} + ⋯ + (−1)^{r}\binom{s − r}{r}\binom{s − 2r}{0} \\ = ∑_{m=0}^{r}(−1)^{m}\binom{s − m}{m}\binom{s − 2m}{r − m}. \end{matrix}

Next note that

\binom{s − m}{m}\binom{s − 2m}{r − m} = \frac{(s − m)!}{m!(s − m − m)!} \frac{(s − 2m)!}{(r − m)!(s − 2m − r + m)!}

= \frac{(s − m)!(s − 2m)!}{m!(s − 2m)!(r − m)!(s − m − r)!} = \frac{r!(s − m)!}{m!(r − m)!r!(s − m − r)!}

= \frac{r!}{m!(r − m)!} \frac{(s − m)!}{r!(s − m − r)!} = \binom{r}{m}\binom{s − m}{r}.

In the sum

∑_{k=1}^{⌊ \frac{s+1}{2} ⌋} ∑_{l=0}^{s−2k+1}(−1)^{k+1}\binom{s − k}{k − 1}\binom{s − 2k + 1}{l}ζ^{−s+2k+2l−1}

the coefficient of ζ^{−s+2r−1}, 1 ≤ r ≤ ⌊\frac{s+1}{2}⌋ is given by

\begin{aligned}\binom{s − 1}{0}\binom{s − 1}{r − 1} & − \binom{s − 2}{1}\binom{s − 3}{r − 2} + ⋯ + (−1)^{r}\binom{s − r}{r − 1}\binom{s − 2r + 1}{0} \\ & = ∑_{m=0}^{r−1}(−1)^{m}\binom{s − m − 1}{m}\binom{s − 2m − 1}{r − m − 1}. \end{aligned}

Now

\binom{s − m − 1}{m}\binom{s − 2m − 1}{r − m − 1}

= \frac{(s − m − 1)!}{m!(s − m − 1 − m)!} \frac{(s − 2m − 1)!}{(r − m − 1)!(s − 2m − 1 − r + m + 1)!}

= \frac{(s − m − 1)!(s − 2m − 1)!}{m!(s − m − 1)!(r − m − 1)!(s − m − r)!}

= \frac{(r − 1)!(s − m − 1)!}{m!(r − 1 − m)!(r − 1)!(s − m − r)!}

= \frac{(r − 1)!}{m!(r − 1 − m)!} \frac{(s − m − 1)!}{(r − 1)!(s − m − 1 − r + 1)!}

= \binom{r − 1}{m}\binom{s − m − 1}{r − 1}.

Therefore, since ∑_{m=−s}^{s} ζ^{m} = 0, it suffices to prove that

∑_{m=0}^{r}(−1)^{m}\binom{r}{m}\binom{s − m}{r} = 1,

and that

∑_{m=0}^{r−1}(−1)^{m}\binom{r − 1}{m}\binom{s − m − 1}{r − 1} = 1.

Thus, it suffices to prove the following.

\underline{Lemma 3.2}. For any integer s, we have

∑_{m=0}^{r}(−1)^{m}\binom{r}{m}\binom{s − m}{r} = 1.

\underline{Proof}. We shall prove the stronger result that

∑_{m=0}^{r}(−1)^{m}\binom{r}{m}\binom{s − m}{r′} = \{ \begin{matrix}1, & if r = r′; \\ 0, & if r′ < r. \end{matrix}

We use induction on s.

\begin{aligned}∑_{m=0}^{r}(−1)^{m}\binom{r}{m}\binom{s − m}{r} & = ∑_{m=0}^{r}(−1)^{m}\binom{r}{m}[\binom{s − m − 1}{r} + \binom{s − m − 1}{r − 1}] \\ & = 1 + 0 = 1. \end{aligned}

Now, if r′ < r,

\begin{aligned}∑_{m=0}^{r}(−1)^{m}\binom{r}{m}\binom{s − m}{r′} & = ∑_{m=0}^{r}(−1)^{m}\binom{r}{m}[\binom{s − m − 1}{r′} + \binom{s − m − 1}{r′ − 1}] \\ & = 0 + 0 = 0. \end{aligned}

This concludes the proof of Theorem 3.1.

\underline{References}

1. N. Jacobson, Basic Algebra I, W. H. Freeman and Company, New York, 1985.

2. D. H. Lehmer, “A Note on Trigonometric Algebraic Numbers,” American Mathematical Monthly, 40 (1933), 165–166.

3. W. Watkins and J. Zeitlin, “The Minimal Polynomial of cos(2π/n),” American Mathematical Monthly, 100 (1993), 471–474.

David Surowski Department of Mathematics Kansas State University Manhattan, KS 66506–2602 email: dbski@math.ksu.edu

Paul McCombs Department of Mathematics Olney Central College Olney, IL 62450 email: mccombsp@iecc.cc.il.us