Class Field Theory Peter Stevenhagen Class field theory is the study of extensions Q K L K ab K = Q, where L/K is a finite abelian extension with Galois grou G. 1. Class Field Theory for Q First we discuss the situation where K = Q. In this case, we have the cyclotomic extension K = Q L = Q(ζ m ), and (Z/mZ) = Gal(L/Q) a mod n (σ a : ζ m ζ a m) It is the theorem of Kronecker-Weber that Gal(Q ab /Q) = lim Gal(Q(ζ n )/Q) n = lim(z/nz) = Ẑ = n In addition to an exlicit descrition of the Galois grou, we also can describe the slitting behavior of a rime in Q Q(ζ m ) it is determined by mod m. For examle, a rime = x 2 + y 2 = (x + iy)(x iy) Z[i] = Z[ζ 4 ] if and only if = 2 or 1 (mod 4). In articular: (a) If ramifies in Q(ζ m ), then m; the ramification index e of the rimes over is φ( k ) = ( 1) k 1 if k m. (b) The rime is wildly ramified in Q(ζ m ) if and only if 2 m. (c) If m, then is unramified. In the ring of integers Z[ζ m ] of Q(ζ m ), we obtain O L = g i=1 i. The rimes i have residue class degree f = [F (ζ m ) : F ], which (by looking at the Frobenius x x acting on F (ζ m ) = k i ) is equal to the order of mod m in (Z/mZ). The number of such rimes is g = #{ i } = [(Z/mZ) : mod m ]. In articular, a rime slits cometely if and only if 1 (mod m). If Q L H L = Q(ζ m ) is a subfield given by a subgrou H (Z/mZ), then one has the Artin ma A : (Z/mZ) /H Gal(L/Q). If m is the minimal such number, it is called the conductor of L. Z.
2 Peter Stevenhagen 2. Weak Recirocity We would like to generalize this situation to a general number field K. Let K L be an abelian extension of number fields. We first would like to make sense of the statement that the slitting behavior of rimes of K in L is determined by mod m. Let G = Gal(L/K) be the Galois grou of L over K, and let K and L have ring of integers O K and O L, resectively. Let O K be an unramified rime. We see that O L / = g i=1 O L/q i, where O L /q i over O K / is a finite extension of finite fields. This extension has a cyclic Galois grou with a distinguished generator Frob, which mas x x N. We see that f = # Frob G, and g = #{q i } = [G : Frob ]. Lemma 2.1. If L/K is an abelian extension of number fields and is unramified in L/K, then the Frobenius Frob is locally induced by a unique element in G = Gal(L/K). We let I K be the grou of (fractional) O K -ideals, which is exactly I K = Z. O K rime Let D be the discriminant of L/K. We denote by I K (D) the subgrou of I K generated by ideals which are corime to D. We then define the Artin ma I K (D) G = Gal(L/K) Frob. As for the case of Q, we would like that the kernel of this ma contains all rimes which are 1 mod m. We must be careful: we have ideals, so we must restrict to the case of rincial ideals, and we must take care of units of the field. From this, we have the following first weak form of recirocity: Theorem 2.1. There exists an ideal m O K (divisible by all ramifying rimes) such that ker A {(π) : π 1 (O K /m), π totally ositive}. 3. Cycles We invent cycles to allow us to talk about these congruences in a uniform way. A cycle in K is m = m 0 m, where m 0 is an ideal in O K and m is a set of real rimes of K. We also write m = m(). so that m() Z 0 for all, m() is zero for almost all, and m() is 0 or 1 for a real rime, 0 for a comlex rime.
Class Field Theory 3 If α K, then we say α 1 (mod m) if and only if α 1 (O/m 0 O) and α > 0 for all real rimes. This says that: For m finite, ord (α 1) m(); For, given by σ : K R, then σ(α) > 0. For examle, 1/5 1 (mod 6). In this language, the theorem becomes: ker A {αo K : α 1 (mod m)} for a suitable cycle m. If m is a cycle in K, we define I(m) = {a I K : ord (a) = 0 if m 0 }. The denote by P (m) I(m) the subgrou of rincial ideals with this roerty. We also have the ray modulo m, R(m) = {αo K : α 1 (mod m)}, with the containment R(m) P (m). Since any element of the ideal class grou of K is generated by an ideal rime to the discriminant, we see that I(m) Cl K. We define the ray class grou of m, I(m)/R(m) = Cl m. It admits a surjection Cl m Cl K with kernel P (m)/r(m) = (O K /m) = (O K /m 0 ) m 1. All together, we have the exact sequence O K (O K /m) Cl m Cl K 0. For an examle, consider again Q. Then any cycle is m = (m) or m = (m). In the first case, we have the ray class grou (Z/mZ) /{±1}, and in the other case simly (Z/mZ), which is what we would exect from the descrition above. 4. Class Field Theory We now can state the main theorem of class field theory. Theorem 4.1. If K L is a finite abelian extension of number fields, then: (a) There exists a cycle m divisible by all ramifying rimes (a real rime ramifies if it has a comlex extension) such that the Artin ma A : I(D) Gal(L/K) factors through Cl m and the ma is surjective. Frob A : Cl m Gal(L/K) [] Frob
4 Peter Stevenhagen (b) There exists a minimal such cycle f, called the conductor of L/K, with the roerty that f if and only if is ramified in L/K, and 2 f if and only if is wildly ramified in L/K. (c) (Existence) Given a cycle m, there exists a ray class field H m K ab which is maximal in the sense that R(m) ker A, and Cl m Gal(Hm /K). Therefore, K ab = f H f Q. This construction is highly nontrivial. For examle, for m = (1), we see that from the exact sequence that Cl m = Cl K gives us H = H 1, the ray class field mod 1, which is usually called the Hilbert class field. It has the roerty that Cl K Gal(H/K), and H/K is unramified at all rimes of K (including the infinite rimes), which is often called totally unramified. For examle, the field K = Q( 5) has class grou Cl K = Z/2Z. In this case, the Hilbert class field is H = K( 5). Also, for K = Q( 23), Cl K = Z/3Z, with H = K(α) where α 3 α 1 = 0. In each case, these extensions are in fact unramified everywhere. 5. Idèles However, we would like to deal with the maximal abelian extension K ab /K in one stroke. We have seen that Gal(K ab /K) = lim Cl f, a rojective limit of ray class grous. For examle, for K = Q we get Gal(Q ab /Q) = lim(z/nz) = Ẑ. To realize this in general, we need to define the idèles. Let K be the comletion of K at, so that if is real or comlex, K = R or C, resectively. We denote by A K the valuation ring A = {x K : x 1} with unit grou A = {x K : x = 1}. Then we define the idèle grou as the restricted direct roduct J = K = {(x ) K : x A for almost all }. Each of these factors has a toology, and we make J into a locally comact grou by defining the oen subgrous U to be, for S a finite set of rimes, U = S where O K is an oen subgrou. We have a filtration K A = U (0) A S O, 1 + 1 + 2 1 + n = U (n)....
Class Field Theory 5 If is archimedean, we let U (0) = K, and if is real, we let U (1) = R >0. From this we see that H J is an oen subgrou if and only if H U n() with n() = 0 for almost all, n() {0, 1} for real, n() = 0 for comlex. In articular, if f = n() is a cycle, it corresonds to an oen subgrou W f = U n(), and H is oen in J if and only if H W f for some cycle f. We ma K J by x (x). We then obtain the idèle class grou C K = J/K. Lemma 5.1. There exists an isomorhism A : J/(K W ) Cl f π [], ( f). Corollary 5.1 (Idèlic Main Theorem). If L/K is abelian, the Artin ma gives a surjection C K Gal(L/K) 0 π Frob, ( f). This yields a surjection C K Gal(K ab /K), and we obtain a bijection between oen subgrous of C K and finite abelian extensions L/K inside an algebraic closure K, where L corresonds to N L/K C L. 6. Local Class Field Theory If is a finite rime, given an abelian extension L/K of number fields, we obtain for a rime q which lies over a corresonding local extension L q /K, with G = {σ Gal(L/K) : σq = q} = Gal(L q /K ) Gal(L/K). Then in the Artin surjection J Gal(L/K), if we restrict to the factor K, this mas surjectively onto exactly the factor Gal(L q /K ). We have 1 I Gal(L q /K ) Gal(k q /k ) = Frob 1, where I is the inertia grou, a factor which is nontrivial only if ramifies at q. We then obtain mas: J Gal(L/K) K Gal(L q /K ) A I
6 Peter Stevenhagen Theorem 6.1. Let F = K be a local field. Given E F a finite abelian extension, there is an Artin ma In fact, ker A = N E/F (E ) F, so π A F = F Gal(E/F ) π Frob modi A F I F /NE Gal(E/F ). One can show in fact that F Gal(F ab /F ), with dense image. (The discreancy occurs by relacing π = Z maing into the grou Ẑ = Gal(F unr /F ) consisting of owers of the Frobenius.) Local class field theory allows us to comute conductors f of L/K. We see that f = n(), is unramified if and only if n() = 0, is tamely ramified if and only if n() = 1, and is wildly ramified if and only if n() 2. From the exlicit descrition of the kernel of the Artin ma in the local case, we see that the exonent n() is the smallest integer k such that U (k) in U (k) N Lq/K (L q). A K is contained Examle 6.2. Let us find all quadratic extensions of Q. If we use class field theory, we see that every such extension corresonds to subgrous H = N E/Q E Q of index 2, with Q /H = Gal(E/Q ). We have that Q = Z, and (Q ) 2 = 2 (Z ) 2, and for 2 in between there are exactly three subfields of index 2, giving the three quadratic extensions of Q. It is in fact easy to find these fields directly: the extensions corresond to Q 2 Q ( x) for x Q /(Q ) 2, with x =, x = a, x = a, for a Z such that a F \ (F ) 2. More recisely: N(Q ( ) = (Z ) 2 (ramified) N(Q ( a) = a (Z ) 2 (ramified) N(Q ( a) = 2 (Z ) 2 (unramified) Exercises In the first three exercises, we let K be an algebraic number field, Cl K its ideal class grou, and h K = # Cl K its class number. Let K be an algebraic closure of K, and H(K) the Hilbert class field of K, which is the maximal abelian totally unramified extension of K inside K. It is a consequence of the main theorem of class field theory that H(K) is a finite abelian extension of K of degree h K, and that the Artin symbol induces a grou isomorhism Cl K Gal(H(K)/K). The three exercises below illustrate how roerties of the Hilbert class field can be used to obtain information about the class number.
Class Field Theory 7 Exercise 2.1. (a) Let E be a finite extension of K. Prove that H(K) H(E), and that h K divides h E [E : K]. (b) Let E, F be two finite extensions of Q inside Q. Prove: if h E = h F = 1, then h E F = 1. Exercise 2.2. (a) Let E be a finite extension of K, and denote by L the maximal abelian totally unramified extension of K inside E. Show that the cokernel of the norm ma N E/K : Cl E Cl K is isomorhic to Gal(L/K). (b) Let n be a ositive integer, and denote by ζ n a rimitive n-th root of unity. Prove that the class number of Q(ζ n + ζn 1 ) divides the class number of Q(ζ n ). Exercise 2.3. The Hilbert class field tower of K is the sequence of fields H (0) (K) = K H (1) (K) = H(K) H (2) (K) = H(H(K))... H (i) (K)..., where each H (i) (K) is the Hilbert class field of H (i 1) (K). It is said to be finite if H (i+1) (K) = H (i) (K) for some i. Prove that the Hilbert class field tower of K is finite if and only if there is a finite extension E of K with h E = 1. [Golod and Shafarevich roved in 1964 that there exist number fields K for which the Hilbert class field tower is infinite.] Exercise 2.4. Show that the Hilbert class field of Q( 5) equals Q( 5, 5). Exercise 2.5. Let α be a zero of X 3 X 1 Z[X] in an algebraic closure of Q, and write K = Q( 23), L = K(α). (a) Prove that L is the Galois closure of Q(α) over Q, and that K L is an abelian extension of degree 3. (b) Prove that exactly two rimes of Q(α) are ramified over Q, and that they lie over 23 and. Prove that in both cases the ramification index equals 2. (c) Prove that K L is totally unramified. (d) Prove that L is the Hilbert class field of K. Exercise 2.6. For a rime number, let m be the number of distinct zeros of X 3 X 1 in F. Prove the following: (a) m = 0 if and only if ( ) 23 = 1 and cannot be written as = a 2 + 23b 2 with a, b Z. (b) m = 1 if and only if ( 23) = 1. (c) m = 2 if and only if = 23. (d) m = 3 if and only if can be written as = a 2 + 23b 2 with a, b Z, a 0. Exercise 2.7.
8 Peter Stevenhagen (a) Prove that the field Q( 5) has class number 1, and that the grou of units of its ring of integers is generated by 1 and (1 + 5)/2. (b) Let be a rime number. Prove that there exists a field K satisfying [K : Q] = 4, 5 K, K/Q = 25 if and only if 2, 3 mod 5. Prove also that if such a field exists, it is uniquely determined by, u to isomorhism. We denote this field by K (). (c) Prove that among all fields K (), the only one that is Galois over Q is the field K (5). Can you embed K (5) in a cyclotomic extension of Q? Exercise 2.8. A number field is called totally real if it has no comlex rimes, totally comlex if it has no real rimes, and mixed if it is neither totally real nor totally comlex. The Fibonacci sequence (F n ) n=0 is inductively defined by F 0 = 0, F 1 = 1, F n+2 = F n+1 + F n. Let be a rime number with 1 or 4 mod 5, and K () as in the revious exercise. (a) Prove that K () is mixed if and only if 3 mod 4. (b) Suose that 1 mod 8. Prove that K () is totally real if divides F ( 1)/4, and totally comlex otherwise. (c) Suose that 5 mod 8. Prove that K () is totally comlex if divides F ( 1)/4, and totally real otherwise. Exercise 2.9. Let be a rime number with 11 or 19 mod 20, and let K () be as above. Prove that K () has exactly one rime lying over 5 if 11 mod 20, and exactly two rimes lying over 5 if 19 mod 20. In the next two exercises, we let K L be a finite abelian extension of number fields, and F(L/K) the conductor of L over K; this is the greatest common divisor of all cycles m of K for which L is contained in the ray class field modulo m of K. Denote by m the exonent to which a rime of K aears in F. It is art of the main theorem of class field theory that m 1 if and only if is ramified in L, and that m 2 if and only if is finite and wildly ramified in L. We want to find an uer bound for m. We may and do assume that is finite. Denote by be the rime of Q over which is lying, and let e = e(/) be the ramification index of over. Exercise 2.10. For an integer i > 0, denote by U i the oen subgrou 1 + i of K. Prove the following assertions. (a) If i, j are ositive integers with j 0 mod, then the ma U i U i sending every x to x j is an isomorhism. (b) If i > e/( 1), then there is an isomorhism U i U i+e sending every x to x. (c) If j is a ositive integer, then (K) j is an oen subgrou of K, and it contains U e +ke, where e denotes the least integer > e/( 1) and k is the number of factors in j.
Class Field Theory 9 (d) If K E is a finite extension, then N E/K [E ] is an oen subgrou of K, and it contains U e +ke, with e as in (c) and k the number of factors in [E : K ]. Exercise 2.11. (a) Prove that m e + ke, where e denotes the least integer > e/( 1) and k is the number of factors in [L : K]. (b) More recisely, rove that m e + ke, with e as before, but with k now equal to the number of factors in the exonent of the inertia grou of in Gal(L/K). Mathematisch Instituut, Universiteit Leiden, Postbus 9512, 2300 RA Leiden, The Netherlands E-mail address: sh@math.leidenuniv.nl