Section 13.6. Let ζn be a primitive n-th root of unity, i.e. any generator of the group of roots of unity. The goal of this lecture is to prove that [Q(ζn): Q] = φ(n); where φ(n) is the Euler's function, that is equal to the number of positive integers k < n such that (k; n) = 1. Let us denote the group of n-th roots of unity by µn and recall that µn ' Z=nZ. Note that µd ⊂ µn () d j n: Indeed, if d j n and ζ is a d-th root of unity, that is ζd = 1 then ζn = 1. Conversely, if ζ is a primitive d-th root of unity and ζ 2 µn, the order of ζ is d, but its order has to divide the order of µn which is n. We know define cyclotomic polynomials which will be minimal polynomials for primitive roots of unity. Definition 1. The n-th cyclotomic polynomial Φn(x) is defined by Y Φn(x) = (x − ζ); where the product is taken over all primitive n-th roots of 1. Note that Y xn − 1 = (x − ζ): ζn=1 If we group together factors (x − ζ) where the order of ζ is d, that is ζ is a primitive d-th root of unity, we get n Y x − 1 = Φd(x): (∗) djn Comparing the degrees of polynomials in both sides of the equation we derive the following interesting equality: X n = φ(n): djn Equality (∗) allows us to compute polynomials Φn(x) by induction on n, for example, Φ1(x) = x − 1 and 2 x − 1 = Φ1(x)Φ2(x) = (x − 1)Φ2(x); hence Φ2(x) = x + 1. Similarly, 3 x − 1 = Φ1(x)Φ3(x) = (x − 1)Φ3(x); 2 hence Φ3(x) = x + x + 1, 4 x − 1 = Φ1(x)Φ2(x)Φ4(x) = (x − 1)(x + 1)Φ4(x); 2 hence Φ4(x) = x + 1, and so on. To show that the degree [Q(ζn): Q] of the cyclotomic extension Q(ζn) over Q is φ(n), it is enough to show that Φn(x) is an irreducible polynomial in Q[x] of degree φ(n). Theorem 2. The cyclotomic polynomial Φn(x) is an irreducible monic polynomial in Z[x] of degree φ(n). 1 2 Proof. First of all, it is obvious from the definition that Φn(x) is a monic polynomial of degree φ(n). We shall show that Φn(x) 2 Z[x] by induction. Clearly, Φ1(x) 2 Z[x], assume Φd(x) 2 Z[x] for all 1 6 d < n. Then n Y x − 1 = f(x)Φn(x) where f(x) = Φd(x): d j n 1<d<n By inductive assumption, f(x) is a monic polynomial with integer coefficients. We now that n n f(x) divides x −1 in Q(ζn)[x], but since both f(x) and x −1 have coefficients in the field Q and Euclidean algorithm does not depend on field extensions, we conclude that f(x) divides n n x − 1 in Q. Now, by Gauss' Lemma we have that f(x) divides x − 1 over Z, which proves that Φn(x) 2 Z[x]. It remains to show that Φn(x) is irreducible. Let us write Φn(x) = f(x)g(x); where f(x) is irreducible. Let ζ be a primitive root of 1, such that f(ζ) = 0, and p be a prime which does not divide n. Then, ζp is again a primitive root of 1, and hence is either a root of f(x) or a root of g(x). Assume that g(ζp) = 0, then ζ is a root of g(xp), and f(x) j g(xp) as f(x) is a minimal polynomial of ζ. Then, g(xp) = f(x)h(x) p p for some h(x) 2 Z[x]. Let us consider the above equality in Fp[x]. There we haveg ¯(x ) =g ¯(x) , and g¯(x)p = f¯(x)h¯(x) ¯ thereforeg ¯(x) and f(x) have a common factor in Fp[x]. This in turn implies that the poly- ¯ ¯ n nomial Φn(x) = f(x)¯g(x) has a multiple root over F, therefore, so does x − 1. But as we know, xn − 1 is separable over any field whose characteristic does not divide n. Thus, for any root ζ of f(x) and any integer a co-prime with n, ζa is also a root of f(x). But the latter implies that any primitive n-th root of 1 is a root of f(x), hence Φn(x) = f(x), which shows that Φn(x) is irreducible. Corollary 3. The degree of the cyclotomic field Q(ζn) over Q equals [Q(ζn): Q] = φ(n):.