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Home , Cyclotomic field, Root of unity

LECTURE NOTES, WEEK 8 MATH 222A, ALGEBRAIC THEORY

MARTIN H. WEISSMAN

Abstract. We discuss cyclotomic fields, and the material in Chapter 6 of Milne’s notes.

1. Cyclotomic Fields Cyclotomic fields arise naturally in for many reasons. Histor- ically, they were some of the first number fields to be extensively studied (along with quadratic fields), because of their relevance to Fermat’s last theorem. Galois theoretically, they arise because every finite of Q is contained in a cyclotomic field. A in a field K is an element x ∈ K which satisfies xn = 1, for some 1 ≤ n ∈ Z. A primitive of unity is such an element x ∈ K, which satisfies xn = 1, and which does not satisfy xd = 1 for any 1 ≤ d < n (or equivalently, for any d properly dividing n). A cyclotomic field is an extension K/Q such that K = Q(ζ), for some root of unity ζ. Proposition 1.1. Cyclotomic fields are abelian extensions of Q. Proof. Suppose that K = Q(ζ), where ζ is a primitive nth root of unity. Then K is a splitting field for the P (X) = Xn − 1, since over K, we have: nY−1 P (X) = Xn − 1 = (X − ζi). i=0 Indeed, we see that the ζi are all distinct (since ζ is a primitive nth root of unity), and every ζi is a root of Xn − 1 (since ζin − 1 = (ζn)i − 1 = 0). Thus the ζi, for 0 ≤ i ≤ n − 1 form a complete set of roots for P (X). It follows that K is a of Q, hence Galois. Now, consider an element γ ∈ G = Gal(K/Q). We have γ(ζ) = ζi, for some 0 ≤ i ≤ n − 1; equivalently, for some i ∈ Z/nZ. Moreover, we have γ−1(ζ) = ζj, for some j ∈ Z/nZ. Thus we have: ζ = γγ−1(ζ) = γ(ζj) = ζij. Thus ij = 1 in Z/nZ. It follows that there is a unique homomorphism: α: G → (Z/nZ)×, satisfying: γ(ζ) = ζα(γ), for all γ ∈ G. The homomorphism α is injective, since a Galois element γ is completely de- termined by where it sends ζ (since K = Q(ζ)). Thus G can be identified with a subgroup of (Z/nZ)×. ¤ 1 2 MARTIN H. WEISSMAN

Among the nth roots of unity, there are primitive dth roots of unity, for every d dividing n. Define: Y 0 Φd(X) = (X − ζ ), ζ0 where the product is over all primitive dth roots of unity. These are the cyclotomic . They satisfy the following properties: (1) Y n X − 1 = Φd(X). d|n

(2) Φd ∈ Z[X], for every d, and has degree φ(d) (Euler’s φ function) (3) Every polynomial Φd is irreducible. Hence it is the minimal polynomial of th ζd (any primitive d root of unity). The first property is clear, since every nth root of unity is a primitive dth root of unity for some d dividing n, and every primitive dth root of unity for some d dividing n is an nth root of unity. th For the second property, if ζd is a primitive d root of unity, then the other th i × primitive d roots of unity have the form ζd for any i ∈ (Z/dZ) . Thus there are φ(d) such roots, and so Φd has degree φ(d). Assume, by induction, that Φe ∈ Z[X], for all e < d (note that Φ1 ∈ Z[X] trivially for a base step). Then we have: d X − 1 = Φd(X)f(X), where Y f(x) = Φe(X) ∈ Z[X]. e|d,e6=d

Thus, division of polynomials over Z[X] implies that Φd ∈ Z[X]. For irreducibility, we follow the argument from Dummit and Foote: suppose that we had a Φd = f · g, with f, g ∈ Z[X] monic polynomials. Without loss of generality, assume that f is irreducible (since Z[X] is a UFD). Thus f is the minimal polynomial for some primitive nth root of unity ζ. Suppose that p is a prime, and p does not divide n. Then p ∈ (Z/nZ)×, so ζp is a primitive nth root of unity; it follows that ζp is a root of f, or a root of g. If ζp is a root of g, then ζ is a root of g(Xp), so f(X) divides g(Xp) in Z[X], since f is the minimal polynomial of ζ. Thus g(Xp) = f(X)h(X), for some h ∈ Z[X]. Reducing mod p, and noting that “raising to the pth power” is a homomorphism in p, we see that: g¯(Xp) = (¯g(X))p = f¯(X)h¯(X). ¯ It follows that f andg ¯ have a common factor in the UFD Fp[X]. ¯ It now follows that Φ¯ d = fg¯ has a multiple root. Since p does not divide d, d the polynomial P¯d = X − 1 ∈ Fp[X] has no multiple roots. But Φ¯ d divides P¯d, a contradiction. Hence ζp must be a root of f(X). Applying this repeatedly, replacing ζ by ζp when necessary, we see that ζa is a root of f(X) for every a relatively prime th to d. Thus every primitive d root of unity is a root of f(X). Hence f(X) = Φd(X), and we are done. It follows that α is an : ∼ × Gal(Q(ζd)/Q) = (Z/dZ) . LECTURE NOTES, WEEK 8 MATH 222A, THEORY 3

The structure of cyclotomic fields can be reduced to the study of cyclotomic fields Q(ζ), where ζ is a primitive (pn)th root of unity, for some prime p. Indeed, we have:

Proposition 1.2. Suppose that 1 < m, n ∈ Z, and GCD(m, n) = 1. Let ζk denote a primitive kth root of unity in C. Then we have:

Q(ζmn) = Q(ζm)Q(ζn). Moreover, the Galois canonically splits: ∼ Gal(Q(ζmn)/Q) = Gal(Q(ζm)/Q) × Gal(Q(ζn)/Q). Proof. If we have a primitive (mn)th root of unity, then raising it to the m or th th n power yields a primitive n or m root of unity, respectively. Thus Q(ζmn) contains Q(ζm)Q(ζn). Conversely, since GCD(m, n) = 1, one may see that ζmζn th is a primitive mn root of unity. Thus Q(ζm)Q(ζn) contains Q(ζmn), proving the first result. th th Any element γ ∈ Gal(Q(ζmn)/Q) permutes the m roots of unity, and the n roots of unity. This yields a natural homomorphism:

σ : Gal(Q(ζmn)/Q) → Gal(Q(ζm)/Q) × Gal(Q(ζn)/Q). i j Moreover, if [σ(g)](ζm) = ζm, and [σ(g)](ζn) = ζn, then we have: i j k [σ(g)](ζmζn) = ζmζn = (ζmζn) , where k is any integer, mod mn satisfying: k ≡ i mod m, and k ≡ j mod n. Here we use the Chinese remainder theorem and the fact that GCD(m, n) = 1 to find such a k. Furthermore, the above process shows that σ makes the following diagram commute: σ Gal(Q(ζmn)/Q) / Gal(Q(ζm)/Q) × Gal(Q(ζn)/Q)

αmn αm×αn   (Z/mnZ)× / (Z/mZ)× × (Z/nZ)× CRT Since the bottom arrow is an isomorphism, the top arrow is an isomorphism. ¤ From the above theorem, we arrive at a theorem in infinite : Proposition 1.3. Define the field Q(µ) to be the smallest subfield of C which contains every root of unity. Then there is an isomorphism: Y ∼ ˆ× ∼ × Gal(Q(µ)/Q) = Z = Zp . p 2. Number Theory in Q(ζ) The previous section was primarily field theory. Now, we study cyclotomic fields from the standpoint of number theory. Fixing a positive integer d, we have a number field Q(ζd). The number-theoretic questions are:

• What is the ring of (the integral closure of Z) O in Q(ζd)? • Given a prime ` of Z, how does the prime ideal (`) factor in O? • What is the structure of Q(ζd) ⊗Q R? In other words, how many real and complex embeddings does the cyclotomic field have? 4 MARTIN H. WEISSMAN

• What is the different and discriminant of Q(ζd)? • What is the volume of Q(ζd) ⊗Q R/O? • What are the units, O×? • What is the regulator R = V ol(H/L(O×))? • What is the structure of the H(O)? In these notes, we focus especially on the case when d = pr is a prime power. Suppose that d = pr hereafter, and let ζ be a primitive dth root of unity. A central role is played by the following two elements:

1 − ζi u = , and π = 1 − ζ. ij 1 − ζj

Here we assume that i, j ∈ (Z/dZ)×, so that ζi and ζj are primitive roots of unity. Our first result is the following:

× Proposition 2.1. uij ∈ O .

Proof. Note that O ⊃ Z[ζ]; we will prove equality later. If we can prove that uij ∈ × −1 O, for all i, j ∈ (Z/dZ) , then these elements are clearly units, since uij = uji. There exists an element k ∈ (Z/dZ)×, such that kj = i. Hence we have:

1 − ζi u = ij 1 − ζj 1 − ζjk = 1 − ζj = 1 + ζj + ζ2j + ··· + ζ(k−1)j.

× Thus uij ∈ O for all such i, j, and hence uij ∈ O . These are called the cyclotomic units (when i 6= j). ¤ Next, we prove:

Proposition 2.2. (π) is a prime ideal of O.

Proof. The Φd is the monic polynomial whose roots are all primitive (pr)th roots of unity. The condition of primitivity can be rephrased: a primitive (pr)th root of unity is a (pr)th root of unity which is not a (pr−1)th root of unity. Hence the cyclotomic polynomial can be written:

Ppr (X) Φd(X) = Ppr−1 (X) r Xp − 1 = Xpr−1 − 1 r−1 = 1 + t + t2 + · + tp−1, where t = Xp .

In particular, Φd(1) = p. LECTURE NOTES, WEEK 8 MATH 222A, 5

On the other hand, we compute: Y i Φd(1) = (1 − ζ ) i∈(Z/dZ)×) Y 1 − ζi = (1 − ζ) 1 − ζ Y = ui1(1 − ζ) = u · πφ(d) for a u. Thus (p) = (π)φ(d) as ideals in O. Since Q(ζd) is a of Q, the ideal (p) factors in O as: e e (p) = q1 ····· qg, for some integers g, e, such that efg = [Q(ζd): Q] = φ(d), for some integer f. Since (p) = (π)φ(d), we see that (π) cannot factor any further, so that (p) totally ramifies in O, and (π) is the unique prime ideal of O lying above (p). ¤ Note that the previous proof also implies the following: Proposition 2.3. Let p = (π) be the prime ideal generated by π in O. Then O/p =∼ Z/pZ. Now, we compute the “inverse different” d−1 of O. Recall its definition: d−1 = {x ∈ K such that ∀z ∈ O, T r(xz) ∈ Z}. We compute it directly, when d = p is a prime (not just a prime power). The prime power case is not much more difficult. When d = p, and K = Q(ζp), we have [K : Q] = p − 1. Let A = Z[ζ], and ζ = ζp, for now. Define the dual to A, with respect to the trace pairing by: A0 = {x ∈ K such that ∀z ∈ A, T r(xz) ∈ Z}. Then we prove: Proposition 2.4. A0 contains A, and [A0 : A] is a prime power. Proof. We may alternately characterize A0 as follows: 0 p−1 i A = {a = a0 + a1ζ + ··· + ap−1ζ ∈ K : ∀i, T r(aζ ) ∈ Z}. Note that: X i T r(aζ ) = (p − 1)ak − aj, j6=k where 0 ≤ i ≤ p − 1 and k = p − i. Let I denote the identity matrix, square of size p − 1. Let M denote the square matrix of size p−1, all of whose entries are 1. Then we see that A0 may be identified with “vectors” of rational ~a = (ai), such that: ( − M) · ~a ∈ Zp−1. Since Zp−1 is identified here with A, we see that A0 contains A, with index: [A0 : A] = |det(pI − M)|. If Q(X) = det(XI − M) is the characteristic polynomial of M, then det(pI − M) = Q(p). The characteristic polynomial may be computed by finding the eigenvalues of M. 6 MARTIN H. WEISSMAN

The vector (1,..., 1) is an eigenvector of M, of eigenvalue p − 1. Clearly, M projects every vector onto the one-dimensional eigenspace spanned by (1,..., 1). Hence, all eigenvalues are zero, except for p − 1. Thus we compute: Q(X) = Xp−1 − (p − 1)Xp−2. Therefore: Q(p) = pp−1 − (p − 1)pp−2 = pp−2. Hence [A0 : A] = pp−2. ¤ It follows that we have inclusions: A0 ⊃ d−1 ⊃ O ⊃ A. Since [A0 : A] = pp−2, we have [idiff : O] = |Disc| is a prime power as well as [OO : A]. In other words, our guess A is a sublattice of O of prime power index. Now, it is possible to prove the following: Proposition 2.5. A = O. In other words, O = Z[ζ]. Proof. Since O/(π) =∼ Z/pZ, we have: O = πO + Z. Since Z[ζ] contains Z, we have: O = πO + A. Multiplying by π yields: πO = π2O + πA. Continuing, and substituting yields: O = πnO + A. ¤

361B Baskin, Department of Mathematics, University of California, Santa Cruz, CA 95064 E-mail address: [email protected]