CYCLOTOMIC POLYNOMIALS


 Ethelbert O’Neal’
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1 CYCLOTOMIC POLYNOMIALS 1. The Derivative and Repeated Factors The usual definition of derivative in calculus involves the nonalgebraic notion of limit that requires a field such as R or C (or others) where limits are sensible for analytic reasons. However, polynomial differentiation is an algebraic notion over any field. One can simply define the polynomial derivative to be as it is from calculus, ( n ) n 1 D a i X i = (i + 1)a i+1 X i. i=0 Alternatively (and working a bit casually here), one can introduce a second indeterminate T and observe that the polynomial i=0 f(x + T ) f(x) k[x][t ] is satisfied by T = 0, and so it takes the form f(x + T ) f(x) = T g(x)(t ), g k[x][t ]. The derivative of f is then defined as the T constant term of g(x)(t ), f (X) = g(x)(0) k[x]. The familiar facts that the derivative is linear, i.e., (f + cg) = f + cg, and that the derivative satisfies the product rule, i.e., (fg) = fg + f g, can be rederived from this purely algebraic definition of the derivative. In fact, yet a third approach is to define X = 1 and then stipulate that differentiation is the unique extension of this rule that is linear and satisfies the product rule. (For example, 1 = X = (X 1) = X 1 + X 1 = X 1 + 1, so 1 = 0.) Proposition 1.1. Let k be a field, and let f k[x] be a polynomial. Then: If gcd(f, f ) = 1 then f has no repeated factors. Proof. To establish the contrapositive, suppose that f = g 2 h in k[x]. The product rule gives f = 2gg h + g 2 h = g(2g h + gh ). Thus g f, and so g gcd(f, f ), and so gcd(f, f ) Definition of the Cyclotomic Polynomials Working in Z[X] we want to define cyclotomic polynomials Provisionally, define Φ n, n Z 1. Φ n (X) = Xn 1 Φ d (X), n 1. d<n 1
2 2 CYCLOTOMIC POLYNOMIALS Here it is understood that for n = 1 the denominator is the empty product, i.e., Φ 1 (X) = X 1. Certainly each Φ n lies in the quotient field Z(X) of Z[X] but we will show considerably more. In fact, always Φ n lies in Z[X] and is irreducible. For example, repeatedly using the identity X n 1 = Φ d(x) in various ways, Φ 2 (X) = X2 1 X 1 = X + 1 = Φ 1( X), Φ p (X) = Xp 1 X 1 = Xp X + 1, p prime, Φ p e(x) = Xpe 1 X pe 1 1 = Φ p(x p e 1 ), p prime, X 2dp 1 Φ 2d p(x) = (X 2d 1 p 1)Φ 2 d(x) = X2 + 1 X 2d = Φ p( X 2 d 1 ), p > 2 prime, Φ 2 d p e(x) = X 2d p e 1 (X 2d 1 p e 1)Φ 2 d(x)φ 2d p(x) Φ 2d p e 1(X) = X 2d 1 p e + 1 Φ 1 ( X 2d 1 )Φ p ( X 2d 1 ) Φ p e 1( X 2d 1 ) X2d 1 p e + 1 = X 2d 1 p e = Φ p( X 2 d 1 p e 1 ), (X 15 1) Φ 15 (X) = (X 5 1)Φ 3 (X) = X10 + X X 2 + X + 1 = X 8 X 7 + X 5 X 4 + X 3 X + 1, Φ 2m (X) = Φ m ( X), Φ 21 (X) = X 21 1 (X 7 1)Φ 3 (X) = X14 + X X 2 + X + 1 = X 12 X 11 + X 9 X 8 + X 6 X 4 + X 3 X + 1. (induction on e) p > 2 prime, m odd, These identities give Φ n for all n 32. We begin with two observations about the polynomial X n 1 where n Z 1. X n 1 has no repeated factors in Z[X] for any n Z 1. Indeed, working in the larger ring Q[X] we have (X n 1) (1/n)X (X n 1) = (X n 1) (1/n)X nx n 1 = 1 Q, so that gcd(x n 1, (X n 1) ) = 1. As explained above, X n 1 therefore has no repeated factors in Q[X], much less in Z[X]. (For future reference, we note that this argument also works in (Z/pZ)[X] where p is prime, so long as p n.) For any n, m Z 1, gcd(x n 1, X m 1) = X gcd(n,m) 1. Indeed, taking n > m, the calculation X n 1 X n m (X m 1) = X n m 1
3 CYCLOTOMIC POLYNOMIALS 3 shows that X n 1, X m 1 = X n m 1, X m 1, and, letting n = qm + r where 0 r < n and repeating the calculation q times, X n 1, X m 1 = X r 1, X m 1. That is, carrying out the Euclidean algorithm on X n 1 and X m 1 in Q(X) slowly carries out the Euclidean algorithm on n and m in Z. (If the Euclidean algorithm on n and m takes k divisionsteps, with quotients q 1 through q k, then the Euclidean algorithm on X n 1 and X m 1 takes q q k steps.) The result follows. Recall the definition Note that Φ n (X) = Xn 1 Φ d (X), n 1. d<n Φ 1 Z[X], and (vacuously) no Φ k and Φ m for distinct k, m < 1 share a factor. The two bullets are the base case of an induction argument. Fix some n > 1 and assume that all Φ m for m n lie in Z[X], and no Φ k and Φ m for distinct k, m < n share a factor. Let m < n, and let k = gcd(m, n) m < n. Any common factor of Φ m and Φ n is a common factor of X m 1 and X n 1, hence a factor of X k 1. However, the calculation Φ n (X) = Xn 1 X n 1 d<n Φ d (X) d k Φ d (X) = Xn 1 X k 1 shows that no factor of Φ n is a factor of X k 1. In sum, the strict inequality in the second bullet weakens: no Φ k and Φ m for distinct k, m n share a factor. Equivalently, no Φ k and Φ m for distinct k, m < n + 1 share a factor. Now, for each proper divisor d of n + 1, each factor of Φ d is a factor of X d 1, hence a factor of X n+1 1. And the product +1 d<n+1 Φ d (X) contains no repeat factors of X n+1 1. Thus Φ n+1 Z[X]. In sum, all Φ m for m n + 1 lie in Z[X], and no Φ k and Φ m for distinct k, m < n + 1 share a factor. By induction, Φ n Z[X] for all n Z 1, and no two Φ n share a factor. The totient function identity ϕ(d) = n quickly combines with the formula Φ d(x) = X n 1 and a little induction to show that the degree of the cyclotomic polynomial is the totient function of its index, deg(φ n ) = ϕ(n), n 1.
4 4 CYCLOTOMIC POLYNOMIALS Substitute X = 0 in the identity X n 1 = Φ d (X) to see that the constant d n coefficient of Φ d for all divisors d of n is ±1. In particular, the constant coefficient of Φ n is ±1. If we view the integers Z as a subring of the complex number field C then the nth cyclotomic polynomial factors as Φ n (X) = (X ζn), e where ζ n = e 2πi/n. 0 e<n gcd(e,n)=1 Indeed, the formula produces the monic polynomial in C[X] whose roots are the complex numbers z such that z n = 1 but z m 1 for all 1 m < n. This polynomial s constant term is ( 1) ϕ(n) 0 e<n ζn, e and for n = 1 this is ( 1) 1 = gcd(e,n)=1 1, for n = 2 it is ( 1) ( 1) = 1, and for n 3 it is 1 1 = 1. That is the constant coefficient of Φ n is 1 for n = 1 but is 1 for all n 2. The previous paragraph notwithstanding, we do not want to think of the cyclotomic polynomials innately in complex terms (even though the word cyclotomic literally refers to dividing the circle), because the integers are also compatible with other algebraic structures that are incompatible with the complex numbers in turn. Rather, we want to think of the roots of Φ n as the elements having order exactly n in whatever environment is suitable. In the next example, the environment is Z/pZ where p is a prime that does not divide n. Earlier we observed that our arguments apply here as well. 3. Application: An Infinite Congruence Class of Primes Theorem 3.1. Let n > 1 be a positive integer. There exist infinitely many primes p such that p = 1 mod n. Proof. Suppose that the only primes equal to 1 mod n are p 1,, p t. The idea is to find some other prime p and some integer a such that Φ n (a) = 0 mod p and p n. Granting the display, it says that the multiplicative order of a modulo p is exactly n. Since (Z/pZ) is cyclic of order p 1, we thus have n p 1, i.e., p = 1 mod n. So the original list of such primes was not exhaustive after all, and hence there is no such finite list. To carry out the idea, define for any positive integer l, Then a = a l = l n p 1 p t. Φ n (a l ) > 1 for all large enough l. Take such l, and with l chosen, omit it as a subscript of a. Consider any prime divisor p of Φ n (a). Since Φ n (a) = ±1 mod a, we have p a. That is, since n a, Φ n (a) = 0 mod p and p n. Thus the proof is complete, as explained in the previous paragraph,
5 CYCLOTOMIC POLYNOMIALS 5 4. Application: Wedderburn s Theorem The cyclotomic polynomials combine with the counting formulas of group actions to provide a lovely proof of the following result. Theorem 4.1. Let D be a finite division ring, i.e., a field except that multiplication might not commute. Then D is a field. Proof. The center of D is a finite field k; let q denote its order. Since D is a vector space over k, the order of D is therefore q n for some n. We want to show that n = 1. For any x D, the centralizer D x = {y D : yx = xy} is again a division ring containing k, and its multiplicative group D x is a subgroup of D. Hence D x = q nx where q nx 1 divides q n 1, so that n x divides n. And if x / k then the divisibilities are proper. Let D act on D by conjugation. The class formula gives D = k + x D / D x where the sum adds the sizes of the nontrivial orbits, or q n = q + x q n 1 q nx 1 summing over representatives of nontrivial orbits. Recall that each n x arising from the sum is a proper divisor of n. The formulas Φ n (q) Φ d (q) = q n 1, Φ n (q) d n x Φ d (q) = qn 1 q nx 1, (in which all Φ (q)values are integers) show that Φ n (q) divides the left side and each term of the sum in the formula q n 1 = q 1 + x q n 1 q nx 1. Consequently Φ n (q) divides q 1. But viewing Z as a subring of the complex number system C lets us show that Φ n (q) can not divide q 1 unless n = 1, Φ n (q) 2 ( = (q cos(2πk/n)) 2 + (sin(2πk/n)) 2) 0 k<n gcd(k,n)=1 If n > 1 then each term of the product is greater than (q 1) 2, and so Φ n (q) > q 1. This completes the proof.
6 6 CYCLOTOMIC POLYNOMIALS 5. Irreducibility of PrimePower Cyclotomic Polynomials The irreducibility of primepower cyclotomic polynomials Φ p e(x) in Z[X] (and hence in Q[X]) can be established by an argument set entirely in Z[X] and its quotientstructures. To avoid interrupting the pending argument, we establish a small preliminary result. Lemma 5.1. Let p be prime. For all ε Z 0, (X 1) pε = X pε 1 Proof. The result is immediate if ε = 0. in (Z/pZ)[X], (X 1) pε+1 = ((X 1) pε ) p = (X pε 1) p = And for the induction step, working p i=0 ( ) p X ipε ( 1) p i = X pε+1 1, i because the binomial coefficients for 1 i p 1 vanish modulo p and because ( 1) p = 1 mod p for all p, including p = 2. Again let p be prime, and let e 1. We show that the cyclotomic polynomial Φ p e is irreducible in Z[X], thereby making it irreducible in Q[X] by Gauss s Lemma. Recall that Φ p e(x) = Φ p (X pe 1 ) = Xpe 1 X pe 1 1 That is, (X pe 1 1)Φ p e(x) = X pe 1 in Z[X], and so reducing the coefficients modulo p gives and then by the lemma, and therefore Also, (X pe 1 1)Φ p e(x) = X pe 1 in (Z/pZ)[X], (X 1) pe 1 Φ p e(x) = (X 1) pe in (Z/pZ)[X], Φ p e(x) = (X 1) pe 1 (p 1) Φ p e(1) = Φ p (1 e ) = Φ p (1) = p 0 in Z/p 2 Z. The previous two displays show precisely that Φ p e(x) is irreducible in Z[X] by Schönemann s Criterion and hence is irreducible in Q[X] by Gauss s Lemma. 6. Irreducibility of General Cyclotomic Polynomials In contrast to the previous section, quickly showing the irreducibility of the general cyclotomic polynomial Φ n (X) in Z[X] (and hence in Q[X]) requires an argument that extends beyond Z and Q. Again we establish a preliminary result. Lemma 6.1. Let n > 1 be an integer and let p n be prime. Then the cyclotomic polynomial Φ n (X) has no repeated factors modulo p. Proof. Indeed, Φ n (X) divides X n 1, which is coprime to its derivative nx n 1 modulo p because p n. Hence X n 1 has no repeated factors modulo p, and hence neither does Φ n (X).
7 CYCLOTOMIC POLYNOMIALS 7 Let n > 1. Let ζ n = e 2πi/n. The complex roots of Φ n (X) are {ζ e n : gcd(e, n) = 1}. Any root of Φ n (X) can be obtained starting from ζ n and repeatedly raising to various primes p n. Suppose that Φ n (X) = g(x)h(x) in Z[X]. We may assume that g(ζ n ) = 0. To show that all roots of Φ n are roots of g it suffices to show that: For any root ρ of g and for any prime p n, also ρ p is a root of g. To establish the displayed statement, let ρ be a root of g and let p n be prime. We need to show that h(ρ p ) 0. If instead h(ρ p ) = 0 then work modulo p and compute (h(ρ)) p = h(ρ p ) = 0 mod p and so h(ρ) = 0 mod p. Also, certainly g(ρ) = 0 mod p. It follows that g(x) and h(x) share a factor k(x) in (Z/pZ)[X]; indeed, a relation G(X)g(X) + H(X)h(X) = 1 would yield the impossibility 0 = 1 upon substituting ρ for X. Thus Φ n (X) = g(x)h(x) has a repeated factor k(x) 2 This contradicts the lemma. And so, h(ρ p ) 0, forcing g(ρ p ) = 0, and the argument is complete. 7. Reducibility of Cyclotomic Polynomials Modulo p Proposition 7.1. Let p be prime, let e 1, and let m 1 with gcd(p, m) = 1. Then Φ p e m(x) = Φ m (X) ϕ(pe ) Here ϕ is Euler s totient function, so that ϕ(p e ) = p e (1 1/p). Proof. Working modulo p, precompute X pem 1 X pe 1 m 1 = (Xm 1) p e = (X m 1) ϕ(p e ) (X m 1) pe 1 Now, breaking the proper divisors of p e m into two groups gives X pem 1 Φ p e m(x) = (X pe 1 m 1) d m = (Xm 1) ϕ(pe ) d m Φ p e d(x) Φ pe d(x) If m = 1 then the product in the denominator is 1, and we have shown Φ p e(x) = (X 1) ϕ(pe) = Φ 1 (X) ϕ(pe ) If m > 1 then by induction on m, the product in the denominator is Φ p e d(x) = Φ d (X) ϕ(pe ) in (Z/pZ)[X], d m d m and so overall, Φ p e m(x) = (Xm 1) ϕ(pe ) Φ d (X) = Φ ϕ(pe ) m(x) ϕ(p d m e )
8 8 CYCLOTOMIC POLYNOMIALS Now the question is how Φ m (X) factors in (Z/pZ)[X] where p m. Let e be the order of p modulo m, i.e., p e = 1 mod m and e is the smallest such positive integer. Then Φ m (X) factors into ϕ(m) linear terms over the field of p e elements, and these gather as ϕ(m)/e terms of degree e over Z/pZ. Here we are using basic facts about finite fields, and their consequences, without complete explanation: Up to isomorphism, for each primepower q = p f, a unique field F q of that order exists, and its multiplicative group F q is cyclic of order q 1. Again with q = p f, the group of automorphisms of F q that fix F p pointwise is cyclic of order f, generated by the automorphism σ : x x p. For each divisor d of f, the subfield of F q fixed by σ d is the field F p d over F p. For example, taking m = 20, Φ 20 (X) = 1 X 2 + X 4 X 6 + X 8. of degree d For p = 5, we first have Φ 20 (X) = φ 4 (X) 4 mod 5, and then φ 4 (X) = X factors into linear terms modulo 5 because 5 1 = 1 mod 4. Similarly Φ 20 (X) factors into quartic terms modulo 7 because 7 4 = 2401 = 1 mod 20, and 4 is the lowest such exponent of 7. Computer algebra confirms these factorizations and others, Φ 20 (X) = (1 + X + X 2 + X 3 + X 4 ) 2 mod 2, Φ 20 (X) = (1 + 2X + X 3 + X 4 )(1 + X + 2X 3 + X 4 ) mod 3, Φ 20 (X) = (2 + X) 4 (3 + X) 4 mod 5, Φ 20 (X) = (1 + 4X + 4X 2 + 3X 3 + X 4 )(1 + 3X + 4X 2 + 4X 3 + X 4 ) mod 7, Φ 20 (X) = (3 + X 2 )(4 + X 2 )(5 + X 2 )(9 + X 2 ) mod 11, Φ 20 (X) = (1 + 8X + 12X 2 + 5X 3 + X 4 )(1 + 5X + 12X 2 + 8X 3 + X 4 ) mod 13, Φ 20 (X) = (1 + 13X + 16X 2 + 4X 3 + X 4 )(1 + 4X + 16X X 3 + X 4 ) mod 17, Φ 20 (X) = (1 + 6X + X 2 )(1 + 8X + X 2 )(1 + 11X + X 2 )(1 + 13X + X 2 ) mod 19, Φ 20 (X) = (1 + 15X + 20X 2 + 8X 3 + X 4 )(1 + 8X + 20X X 3 + X 4 ) mod 23.
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