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Chapter IV. Cyclotomic Fields and Applications

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Chapter IV. Cyclotomic Fields and Applications Matematisk Institut Mat 3AL 4.1 Chapter IV. Cyclotomic fields and applications ROOTS OF UNITY AND CYCLOTOMIC POLYNOMIALS. In this chapter we shall consider an important class of normal extensions of the rational number field Q. Historically these were the first algebraic extensions of Q which were the subject of thorough investigations. First some remarks about roots of unity. n 2πi k The complex solutions of the equation x = 1, i.e. the numbers e n ,0 k<n, are called the n-th roots of unity. They form a multiplicative cyclic group of order n.Ann-th root of unity ε is called a primitive n-th root of unity, if it generates the group of n-th roots of unity. Therefore an n-th root of unity ε is a primitive n-th root of unity if and only if εk =1 , 0 <k<n. The following lemma is quite elementary Lemma 4.1. Let ε be a primitive n-th root of unity. The following conditions are equivalent 1) εk is a primitive n-th root of unity. 2) (k, n)=1 ( i.e. k and n are relatively prime). Proof. Just an easy exercise in group theory.(Cf. Remark 1.82) 2πi k In particular e n , 1 ≤ k<n,(k, n) = 1, are exactly all the primitive n-th roots of unity. 2πi k Definition 4.2. The polynomial Fn(x)= 1≤k≤n x − e n is called the n-th (k,n)=1 cyclotomic polynomial. The roots of Fn(x) are exactly the primitive n-th roots of unity. The degree of Fn(x)isϕ(n), where ϕ(n) is the so-called ”Eulers ϕ-function”, defined as the number of residue classes modulo n prime to n. A priori it is only clear that Fn(x) has coefficients in C. However, as we shall see in the following theorem, Fn(x) has integer coefficients. Theorem 4.3. Fn(x) is a polynomial in Z[x]. To prove this we need the following n Lemma 4.4. x − 1= d|n Fd(x),whered runs through the positive divisors of n. n Proof. The monic polynomials x − 1and d|n Fd(x) have no multiple roots (note that Fd1 (x)andFd2 (x) have no common roots if d1 = d2). Therefore it suffices to n show that the two polynomials x − 1and d|n Fd(x) have exactly the same roots. Matematisk Institut Mat 3AL 4.2 Indeed, an n-th root of unity is a primitive d-th root of unity for exactly one divisor d of n.Conversely,ifε is a primitive d-th root of unity for a divisor d of n,thenε is certainly an n-th root of unity. Proof of Theorem 4.3. By induction on n.SinceF1(x)=x − 1 the assertion is clear for n = 1. Assume it has been proved that Fm(x) ∈ Z[x] for all m<n.Wemust then prove that Fn(x) lies in Z[x]. n By the above lemma we know x − 1=Fn(x) ·{ d|n Fd(x)},where d|n Fd(x) d<n d<n by the induction assumption is a monic polynomial with integer coefficients. The division algorithm now shows that Fn(x) has integer coefficients. We now give an explicit formula for Fn(x). n μ(d) Theorem 4.5. Fn(x)= d|n(x d − 1) ,whereμ denotes the M¨obius function introduced in Chap. II. Proof. By lemma 4.4 and Theorem 2.83 in Chap. II we get: n μ(d) μ(d) (x d − 1) = ( Fδ(x)) = n d|n d|n δ| d P μ(d) d| n μ(d) Fδ(x) = Fδ(x) δ = Fn(x) . d,δ,d·δ|n δ|n Since Fn(x) has degree ϕ(n), where ϕ(n)isEulersϕ-function, by comparison of degrees we obtain n 1 Corollary 4.6. ϕ(n)= d|n d μ(d)=n · p|n(1 − p ),wherep runs through the distinct prime divisors of n. n μ(d) d − Proof. By identifying the degrees of Fn(x)andof d|n(x 1) we see that n n 1 ϕ(n)= d|n d μ(d). It remains then to show that d|n d μ(d)=n · p|n(1 − p ). Since μ(d)=0ifd is divisible by the square of a prime, it suffices to consider a1 ar the square-free divisors of n.Letn = p1 ···pr be the prime decomposition of n where p1,...,pr are the distinct prime divisors of n. We consider first the divisor 1, then the prime divisors of n, then all products of two distinct prime divisors, then all products of three distinct prime divisors, etc. For the sum we then find n n n n 1 μ(d)=n − + − + ... = n · (1 − ). where the pi’s d|n d p i pipj pipj pk pi|n p i are the distinct prime divisors of n,thepipj ’s are the products of two distinct prime divisors, pipjpk’s are the products of three distinct prime divisors etc. Matematisk Institut Mat 3AL 4.3 Some remarks concerning the coefficients of the cyclotomic polynomials. Remark 4.7. For n>2 the degree of Fn(x) is an even number and the constant term is 1. Remark 4.8. The coefficient of the next highest term (i.e. the coefficient of xϕ(n)−1) is −μ(n), since the sum of the primitive n-th roots of unity is μ(n) (this is an exercise provable by induction). Remark 4.9. Looking at the first cyclotomic polynomials one might have the temp- tation to conjecture that all the coefficents of the cyclotomic polynomials were 0 or 1or−1. This, however, is not true. The first counterexample is F105 where the coefficient of x7 is -2.But it can be proved that if n is divisible by at most two dis- tinct odd prime numbers, then the coefficients of Fn(x)are0,1or−1. (The proof is elementary, but not trivial.) i Remark 4.10. Fn(x) is ”reciprocal” for n>1, i.e. if ai is the coefficient of x then ai = aϕ(n)−i for 0 ≤ i ≤ ϕ(n). p p−1 Remark 4.11. For a prime number p one has Fp(x)=(x − 1)/(x − 1) = x + ···+ x +1. 2 Numerical examples 4.12. F1(x)=x − 1,F2(x)=x +1,F3(x)=x + x + 2 4 3 2 2 1,F4(x)=x +1,F5(x)=x + x + x + x +1,F6(x)=x − x +1,F8(x)= 4 6 3 4 3 2 4 2 x +1,F9(x)=x +x +1,F10(x)=x −x +x −x+1,F12(x)=x −x +1,F15(x)= x8 − x7 + x5 − x4 + x3 − x +1. Explicit computations of some roots of unity. 2πi Let εn be the n-th root of unity e n . It is clear that ε1 =1andε2 = −1andε4√= i. Similarly it is straightforward to 2π 2π check that ε8 =cos(8 )+isin( 8 )=(1+i)/ 2. ε3 is√ the root of F3(x) lying in the upper half complex plane. Therefore ε3 = (−1+i 3)/2. ε6 is the√ root of F6(x) lying in the first quadrant of the complex plane; hence ε6 =(1+i 3)/2. As for ε5 we write −2 2 1 −1 −2 2 x F5(x)=x + x +1+x + x =(x +1/x) +(x +1/x) − 1 2π −1 which implies that 2 cos( 5 )=ε5 + ε5 is the (positive) root of z2 + z − 1=0 √ √ 2π − 2π − hence 2 cos( 5 )=( 5 1)/2andcos(5 )=( 5 1)/4. √ 2π − 2 2π 1 Now sin( 5 )= 1 cos ( 5 )= 4 10 + 2 5sothat 1 √ √ ε5 = 5 − 1+i 10 + 2 5 . 4 Matematisk Institut Mat 3AL 4.4 CYCLOTOMIC POLYNOMIALS ARE IRREDUCIBLE. For the proof of the irreducibility of cyclotomic polynomials in Q[x] we need the following lemmas. Lemma 4.13.. If f(x) and g(x) are monic polynomials in Q[x] for which f(x)·g(x) ∈ Z[x],thenf(x) ∈ Z[x] and g(x) ∈ Z[x]. Proof. There are natural numbers a and b such that af(x)andbg(x) are primi- tive polynomials in Z[x]. According to Gauss’s lemma the product af(x)bg(x)= abf(x)g(x) is also a primitive polynomial in Z[x]. But then ab must be +1, and therefore a and b must also be 1. This implies that f(x)andg(x) are polynomials in Z[x]. Lemma 4.14. If p is a prime number and g(x) is a polynomial in Zp[x]=Fp[x] then g(xp)=(g(x))p. Proof. This follows from Freshmans Dream and the fact that αp = α for every α in the finite field Zp with p elements. Theorem 4.15. Fn(x) is irreducible in Q[x]. Proof. Let f(x) be a monic irreducible polynomial in Q[x]. We prove: 1◦ If ε is a primitive n-th root of unity and p is a prime number that does not divide n, then: f(ε)=0⇒ f(εp)=0. Proof of 1◦:Sincef(x) = Irr(ε, Q) the polynomial f(x) must divide xn − 1inside Q[x]. By lemma 4.13 it follows that f(x) must be a polynomial in Z[x]. We now consider g(x) = Irr(εp, Q). As before we see that g(x) ∈ Z[x]. The polynomial g(xp)hasε as a root.
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