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LECTURE NOTES, WEEK 8 MATH 222A, ALGEBRAIC NUMBER THEORY 1. Cyclotomic Fields Cyclotomic Fields Arise Naturally in Number Theory

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LECTURE NOTES, WEEK 8 MATH 222A, ALGEBRAIC NUMBER THEORY 1. Cyclotomic Fields Cyclotomic Fields Arise Naturally in Number Theory LECTURE NOTES, WEEK 8 MATH 222A, ALGEBRAIC NUMBER THEORY MARTIN H. WEISSMAN Abstract. We discuss cyclotomic ¯elds, and the material in Chapter 6 of Milne's notes. 1. Cyclotomic Fields Cyclotomic ¯elds arise naturally in number theory for many reasons. Histor- ically, they were some of the ¯rst number ¯elds to be extensively studied (along with quadratic ¯elds), because of their relevance to Fermat's last theorem. Galois theoretically, they arise because every ¯nite abelian extension of Q is contained in a cyclotomic ¯eld. A root of unity in a ¯eld K is an element x 2 K which satis¯es xn = 1, for some 1 · n 2 Z. A primitive nth root of unity is such an element x 2 K, which satis¯es xn = 1, and which does not satisfy xd = 1 for any 1 · d < n (or equivalently, for any d properly dividing n). A cyclotomic ¯eld is an extension K=Q such that K = Q(³), for some root of unity ³. Proposition 1.1. Cyclotomic ¯elds are abelian extensions of Q. Proof. Suppose that K = Q(³), where ³ is a primitive nth root of unity. Then K is a splitting ¯eld for the polynomial P (X) = Xn ¡ 1, since over K, we have: nY¡1 P (X) = Xn ¡ 1 = (X ¡ ³i): i=0 Indeed, we see that the ³i are all distinct (since ³ is a primitive nth root of unity), and every ³i is a root of Xn ¡ 1 (since ³in ¡ 1 = (³n)i ¡ 1 = 0). Thus the ³i, for 0 · i · n ¡ 1 form a complete set of roots for P (X). It follows that K is a normal extension of Q, hence Galois. Now, consider an element γ 2 G = Gal(K=Q). We have γ(³) = ³i, for some 0 · i · n ¡ 1; equivalently, for some i 2 Z=nZ. Moreover, we have γ¡1(³) = ³j, for some j 2 Z=nZ. Thus we have: ³ = γγ¡1(³) = γ(³j) = ³ij: Thus ij = 1 in Z=nZ. It follows that there is a unique homomorphism: ®: G ! (Z=nZ)£; satisfying: γ(³) = ³®(γ); for all γ 2 G. The homomorphism ® is injective, since a Galois element γ is completely de- termined by where it sends ³ (since K = Q(³)). Thus G can be identi¯ed 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. De¯ne: Y 0 ©d(X) = (X ¡ ³ ); ³0 where the product is over all primitive dth roots of unity. These are the cyclotomic polynomials. They satisfy the following properties: (1) Y n X ¡ 1 = ©d(X): djn (2) ©d 2 Z[X], for every d, and has degree Á(d) (Euler's Á function) (3) Every polynomial ©d is irreducible. Hence it is the minimal polynomial of th ³d (any primitive d root of unity). The ¯rst 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 ³d is a primitive d root of unity, then the other th i £ primitive d roots of unity have the form ³d for any i 2 (Z=dZ) . Thus there are Á(d) such roots, and so ©d has degree Á(d). Assume, by induction, that ©e 2 Z[X], for all e < d (note that ©1 2 Z[X] trivially for a base step). Then we have: d X ¡ 1 = ©d(X)f(X); where Y f(x) = ©e(X) 2 Z[X]: ejd;e6=d Thus, division of polynomials over Z[X] implies that ©d 2 Z[X]. For irreducibility, we follow the argument from Dummit and Foote: suppose that we had a factorization ©d = f ¢ g, with f; g 2 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 ³. Suppose that p is a prime, and p does not divide n. Then p 2 (Z=nZ)£, so ³p is a primitive nth root of unity; it follows that ³p is a root of f, or a root of g. If ³p is a root of g, then ³ is a root of g(Xp), so f(X) divides g(Xp) in Z[X], since f is the minimal polynomial of ³. Thus g(Xp) = f(X)h(X), for some h 2 Z[X]. Reducing mod p, and noting that \raising to the pth power" is a ring homomorphism in characteristic p, we see that: g¹(Xp) = (¹g(X))p = f¹(X)h¹(X): ¹ It follows that f andg ¹ have a common factor in the UFD Fp[X]. ¹ It now follows that ©¹ d = fg¹ has a multiple root. Since p does not divide d, d the polynomial P¹d = X ¡ 1 2 Fp[X] has no multiple roots. But ©¹ d divides P¹d, a contradiction. Hence ³p must be a root of f(X). Applying this repeatedly, replacing ³ by ³p when necessary, we see that ³a is a root of f(X) for every integer a relatively prime th to d. Thus every primitive d root of unity is a root of f(X). Hence f(X) = ©d(X), and we are done. It follows that ® is an isomorphism: » £ Gal(Q(³d)=Q) = (Z=dZ) : LECTURE NOTES, WEEK 8 MATH 222A, ALGEBRAIC NUMBER THEORY 3 The structure of cyclotomic ¯elds can be reduced to the study of cyclotomic ¯elds Q(³), where ³ is a primitive (pn)th root of unity, for some prime p. Indeed, we have: Proposition 1.2. Suppose that 1 < m; n 2 Z, and GCD(m; n) = 1. Let ³k denote a primitive kth root of unity in C. Then we have: Q(³mn) = Q(³m)Q(³n): Moreover, the Galois group canonically splits: » Gal(Q(³mn)=Q) = Gal(Q(³m)=Q) £ Gal(Q(³n)=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(³mn) contains Q(³m)Q(³n). Conversely, since GCD(m; n) = 1, one may see that ³m³n th is a primitive mn root of unity. Thus Q(³m)Q(³n) contains Q(³mn), proving the ¯rst result. th th Any element γ 2 Gal(Q(³mn)=Q) permutes the m roots of unity, and the n roots of unity. This yields a natural homomorphism: σ : Gal(Q(³mn)=Q) ! Gal(Q(³m)=Q) £ Gal(Q(³n)=Q): i j Moreover, if [σ(g)](³m) = ³m, and [σ(g)](³n) = ³n, then we have: i j k [σ(g)](³m³n) = ³m³n = (³m³n) ; 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 ¯nd such a k. Furthermore, the above process shows that σ makes the following diagram commute: σ Gal(Q(³mn)=Q) / Gal(Q(³m)=Q) £ Gal(Q(³n)=Q) ®mn ®m£®n ² ² (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 in¯nite Galois theory: Proposition 1.3. De¯ne the ¯eld Q(¹) to be the smallest sub¯eld of C which contains every root of unity. Then there is an isomorphism: Y » ^£ » £ Gal(Q(¹)=Q) = Z = Zp : p 2. Number Theory in Q(³) The previous section was primarily ¯eld theory. Now, we study cyclotomic ¯elds from the standpoint of number theory. Fixing a positive integer d, we have a number ¯eld Q(³d). The number-theoretic questions are: ² What is the ring of integers (the integral closure of Z) O in Q(³d)? ² Given a prime ` of Z, how does the prime ideal (`) factor in O? ² What is the structure of Q(³d) ­Q R? In other words, how many real and complex embeddings does the cyclotomic ¯eld have? 4 MARTIN H. WEISSMAN ² What is the di®erent and discriminant of Q(³d)? ² What is the volume of Q(³d) ­Q R=O? ² What are the units, O£? ² What is the regulator R = V ol(H=L(O£))? ² What is the structure of the ideal class group 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 ³ be a primitive dth root of unity. A central role is played by the following two elements: 1 ¡ ³i u = ; and ¼ = 1 ¡ ³: ij 1 ¡ ³j Here we assume that i; j 2 (Z=dZ)£, so that ³i and ³j are primitive roots of unity. Our ¯rst result is the following: £ Proposition 2.1. uij 2 O . Proof. Note that O ⊃ Z[³]; we will prove equality later. If we can prove that uij 2 £ ¡1 O, for all i; j 2 (Z=dZ) , then these elements are clearly units, since uij = uji. There exists an element k 2 (Z=dZ)£, such that kj = i. Hence we have: 1 ¡ ³i u = ij 1 ¡ ³j 1 ¡ ³jk = 1 ¡ ³j = 1 + ³j + ³2j + ¢ ¢ ¢ + ³(k¡1)j: £ Thus uij 2 O for all such i; j, and hence uij 2 O . These are called the cyclotomic units (when i 6= j). ¤ Next, we prove: Proposition 2.2. (¼) is a prime ideal of O.
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