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08 Cyclotomic Polynomials

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08 Cyclotomic Polynomials 8. Cyclotomic polynomials 8.1 Multiple factors in polynomials 8.2 Cyclotomic polynomials 8.3 Examples 8.4 Finite subgroups of fields 8.5 Infinitude of primes p = 1 mod n 8.6 Worked examples 1. Multiple factors in polynomials There is a simple device to detect repeated occurrence of a factor in a polynomial with coefficients in a field. Let k be a field. For a polynomial n f(x) = cnx + ::: + c1x + c0 [1] with coefficients ci in k, define the (algebraic) derivative Df(x) of f(x) by n−1 n−2 2 Df(x) = ncnx + (n − 1)cn−1x + ::: + 3c3x + 2c2x + c1 Better said, D is by definition a k-linear map D : k[x] −! k[x] defined on the k-basis fxng by D(xn) = nxn−1 [1.0.1] Lemma: For f; g in k[x], D(fg) = Df · g + f · Dg [1] Just as in the calculus of polynomials and rational functions one is able to evaluate all limits algebraically, one can readily prove (without reference to any limit-taking processes) that the notion of derivative given by this formula has the usual properties. 115 116 Cyclotomic polynomials [1.0.2] Remark: Any k-linear map T of a k-algebra R to itself, with the property that T (rs) = T (r) · s + r · T (s) is a k-linear derivation on R. Proof: Granting the k-linearity of T , to prove the derivation property of D is suffices to consider basis elements xm, xn of k[x]. On one hand, D(xm · xn) = Dxm+n = (m + n)xm+n−1 On the other hand, Df · g + f · Dg = mxm−1 · xn + xm · nxn−1 = (m + n)xm+n−1 yielding the product rule for monomials. === A field k is perfect if either the characteristic of k is 0 [2] or if, in characteristic p > 0, there is a pth root a1=p in k for every a 2 k. [3] [1.0.3] Proposition: Let f(x) 2 k[x] with a field k, and P an irreducible polynomial in k[x]. If P e divides f then P divides gcd(f; Df). If k is perfect and e − 1 6= 0 in k, there is a converse: [4] if P e−1 divides both f and Df then P e divides f. Proof: On one hand, suppose f = P e · g with ≥ 2. By the product rule, Df = eP e−1DP · g + P e · Dg is a multiple of P e−1. [5] This was the easy half. On the other hand, for the harder half of the assertion, suppose P e−1 divides both f and Df. Write f=P e−1 = Q · P + R with deg R < deg P . Then f = QP e + RP e−1. Differentiating, Df = DQ P e + eQP e−1DP + DR P e−1 + R(e − 1)P e−2 DP By hypothesis P e−1 divides Df. All terms on the right-hand side except possibly R(e − 1)P e−2 DP are divisible by P e−1, so P divides R(e − 1)P e−2 DP . Since P is irreducible, either e − 1 = 0 in k, or P divides R, or P divides DP . If P divides R, P e divides f, and we're done. If P does not divide R then P divides DP . Since deg DP < deg P , if P divides DP then DP = 0. This would require that all the exponents of x occurring with non-zero coefficient are divisible by the characteristic p, which must be positive. So P is of the form pm p(m−1) p(m−2) 2p p P (x) = apmx + ap(m−1)x + ap(m−2)x + ::: + a2px + apx + a0 [2] as for Q, R, and C [3] As is the case for finite fields such as Z=p, by Fermat's Little Theorem. [4] In particular, this converse holds if the characteristic of k is 0. [5] This half does not need the irreducibility of P . Garrett: Abstract Algebra 117 th Using the perfect-ness of the field k, each ai has a p root bi in k. Because the characteristic is p > 0, (A + B)p = Ap + Bp Thus, P (x) is the pth power of n (m−1) (m−2) 2 bpmx + bp(m−1)x + bp(m−2)x + ::: + b2px + bpx + b0 If P is a pth power it is not irreducible. Therefore, for P irreducible DP is not the zero polynomial. Therefore, R = 0, which is to say that P e divides f, as claimed. === 2. Cyclotomic polynomials For b 6= 0 in a field k, the exponent of b is the smallest positive integer n (if it exists) such that bn = 1. n d That is, b is a root of x − 1 but not of x − 1 for any smaller d. We construct polynomials Φn(x) 2 Z[x] such that Φn(b) = 0 if and only if b is of exponent n These polynomials Φn are cyclotomic polynomials. [2.0.1] Corollary: The polynomial xn − 1 has no repeated factors in k[x] if the field k has characteristic not dividing n. Proof: It suffices to check that xn − 1 and its derivative nxn−1 have no common factor. Since the characteristic of the field does not to divide n, n · 1k 6= 0 in k, so has a multiplicative inverse t in k, and (xn − 1) − (tx) · (nxn−1) = −1 and gcd(xn − 1; nxn−1) = 1. === th Define the n cyclotomic polynomial Φn(x) by Φ1(x) = x − 1 and for n > 1, inductively, xn − 1 Φ (x) = n lcm of all xd − 1 with 0 < d < n, d dividing n with the least common multiple monic. [2.0.2] Theorem: [6] • Φn is a monic polynomial with integer coefficients. n t • For α in the field k,Φn(α) = 0 if and only if α = 1 and α 6= 1 for all 0 < t < n. • gcd(Φm(x); Φn(x)) = 1 for m < n with neither m nor n divisible by the characteristic of the field k. • The degree of Φn(x) is '(n) (Euler's phi-function) • Another description of Φn(x): xn − 1 Φn(x) = Q 1≤d<n;djn Φd(x) [6] More properly, if the ambient field k is of characteristic 0, then the coefficients lie in the copy of Z inside the prime field Q inside k. If the ambient field is of positive characteristic, then the coefficients lie inside the prime field (which is the natural image of Z in k). It would have been more elegant to consider the cyclotomic polynomials as polynomials in Z[x], but this would have required that we wait longer. 118 Cyclotomic polynomials • xn − 1 factors as n Y x − 1 = Φd(x) 1≤d≤n;djn Proof: We know that djn (and d > 0) implies that xd −1 divides xn −1. Therefore, by unique factorization, the least common multiple of a collection of things each dividing xn − 1 also divides xn − 1. Thus, the indicated lcm does divide xn − 1. t t For α in k, x − α divides Φn(x) if and only if Φn(α) = 0. And α = 1 if and only if x − α divides x − 1. The definition xn − 1 Φ (x) = n lcm of all xd − 1 with 0 < d < n, d dividing n n t shows first that Φn(α) = 0 implies α = 1. Second, if α = 1 for any proper divisor t of n then x − α divides xt − 1, and thus x − α divides the denominator. But xn − 1 has no repeated factors, so x − α dividing the denominator would prevent x − α dividing Φn(x), contradiction. That is, Φn(α) = 0 if and only if α is of order n. To determine the gcd of Φm and Φn for neither m nor n divisible by the characteristic of k, note that Φm m n divides x − 1 and Φn divides x − 1, so m n gcd(Φm; Φn) divides gcd(x − 1; x − 1) We claim that for m, n two integers (divisible by the characteristic or not) gcd(xm − 1; xn − 1) = xgcd(m;n) − 1 Prove this claim by induction on the maximum of m and n. Reduce to the case m > n, wherein xm − 1 − xm−n · (xn − 1) = xm−n − 1 For g a polynomial dividing both xm − 1 and xn − 1, g divides xm−n − 1. By induction, gcd(xm−n − 1; xn − 1) = xgcd(m−n;n) − 1 But gcd(m; n) = gcd(m − n; n) and xm − 1 = xm−n · (xn − 1) + xm−n − 1 so gcd(xm − 1; xn − 1) = gcd(xm−n − 1; xn − 1) and induction works. Thus, gcd(xm − 1; xn − 1) = xgcd(m;n) − 1 Since d ≤ m < n d is a proper divisor of n. Thus, from xn − 1 Φ (x) = n lcm of all xd − 1 with 0 < d < n, d dividing n n d n we see that Φn(x) divides (x − 1)=(x − 1). Since x − 1 has no repeated factors, Φn(x) has no factors in d common with x − 1. Thus, gcd(Φm; Φn) = 1. Garrett: Abstract Algebra 119 Next, use induction to prove that n Y x − 1 = Φd(x) 1≤d≤n; djn For n = 1 the assertion is true. From the definition of Φn, n d x − 1 = Φn(x) · lcmfx − 1 : djn; 0 < d < ng By induction, for d < n d Y x − 1 = Φe(x) 0<e≤d;ejd Since for m < n the gcd of Φm and Φn is 1, d Y lcmfx − 1 : djn; 0 < d < ng = Φd(x) djn;d<n Thus, n Y x − 1 = Φn(x) · Φd(x) djn;d<n as claimed.
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