Class Field Theory. Anna Haensch. Spring 2012

Transcription

1 Class Field Theory Anna Haensch Spring 202 These are my own notes put together from a reading of Class Field Theory by N. Childress [], along with other references, [2], [4], and [6]. Goals and Origins of Class Field Theory First we introduce some notation: K/Q a finite extension. O K the ring of algebraic integers over K. a a fractional ideal, i.e. a finitely generated O K -module with generators in K. A = A K is the multiplicative group of fractional ideals, O K = () is the identity element. If a and b are fractional ideals, with with α i, β j K, then a = (α, α 2,..., α t ) and b = (β, β 2,..., β s ) ab = (..., α i β j,...) is the O K -module generated by the products of the generators of a and b. By the arithmetic of K, we mean the study of K, its subgroups, its factor groups, groups isomorphic to subgroups of K, and certain ideals of K. Motivated by this notion have the following goals of class field theory:. Describe all abelian extensions of K in terms of the arithmetic of K. 2. Find a canonical way to describe Gal(L/K) in terms of the arithmetic of K, whenever L/K is abelian. 3. Describe the decomposition of a prime ideal from K to L in terms of the arithmetic of K, whenever L/K is abelian.. Some Terminology We will briefly review some concepts from algebraic number theory: L/K is a degree n Galois extension. G := Gal(L/K).

2 O K (resp. O L ) is the ring of algebraic integers of K (resp. L). p is a prime ideal in O K P i is a prime ideal in O L. L O L P e P g e g K O K In this case we use the following equivalent statements: p P i divides p P i p P occurs in the factorization of po L Definition. e i is called the ramification index of P i over p, and the degree of the field extension f i = [O L /P i : O K /p] is called the inertia degree (or residue degree) of P i over p. The following can be found in [5, Chapter, 9]. Proposition.. G acts transitively on the set of P i lying a bove p. Proof. Let P := P. Suppose for some i, that P i σp for any σ G. By the Chinese Remainder theorem, there exist x O L such that x 0 mod P i and x mod σp for all σ G. Then, N L/K (x) = σ G σ(x) P i O K = p. On the other hand, x σp for any σ G, hence σ(x) P for any σ G, consequently P Ok = p, σ G a contradiction. So, P i = σ i P for some suitable σ i G. Induced by σ i, we have the following isomorphism σ i : O L /P O L /σ i P a mod P σ i (a) mod σ i (P), so that f i = [O L /σ i (P) : O K /p] = [O L /P : O K /p] = f 2

3 for i =,..., g. Furthermore, σ i (po L ) = po L, P ν po L σ i (P ν ) σ i (po L ) (σ i P) ν po L, so e i = e for all i =,..., g. So in the Galois case, po L = (P P g ) e, and the fundamental identity states efg = n = [L : K]. Definition 2. Suppose L/K is Galois, and p splits from K to L into (P P g ) e, we say p is ramified in L e and p is unramified in L e = Definition 3. p splits completely from K to L if g = n. This means, e = f =, so in particular p is unramified in L. particular property? Why do we care about this Theorem.2. (Inclusion Theorem) L, L 2 Galois extenions on K. S, S 2 the set of primes which split completely from K to L, L 2. Then and S s 2 L L 2 S = s 2 L = L 2 ( means containment with finitely many exceptions). So primes that split completely capture (or describe) the Galois extension..2 The Historical Setting for Class Field Theory Gauss ( ): When does x 2 a 0 mod p have a solution x Z? here p is a prime with p a. Theorem.3. (Gauss Theorem of Quadratic Reciprocity) If p.q are odd prime p, q a, and p q mod 4a, then x 2 a 0 mod p has a solution if and only if x 2 a 0 mod q has a solution. This is more concisely expressed using the Legendre symbol which is defined: { ( a if x = p) 2 a 0 mod p is solvable in Z otherwise so Theorem.3 says: ( a ) ( a ) = =. p q 3

4 Theorem.4. Let L = Q( d) where d is a square free integer. p is an odd prime, p d. ( d p) = p splits completely from Q to L. Gauss relates the decomposition of primes, namely primes that split completely, to congruence conditions in Z. Example. Find all primes which split completely from Q to Q( 2). Q( 2) O Q( 2) First, which primes ramify? From a theorem of Minkowski, Q Z p ramifies in O L p d L (See [3, Chapter 2]). Recall, if O L has the basis {ω,..., ω n }, then d L =do L =d({ω,..., ω n }) = σ i (ω j ) 2. So, O Q( 2) = {a + b 2 : a, b Z} = Z + Z, and d Q( 2) = = ( 2 2) 2 = 8. So only (2) ramifies in Q( 2). Now, of the unramified primes, which split completely? For all odd primes, p, p, 3, 5, or 7 mod 8 (and notice that 7 mod 8). ) = since = 36 2 = 34 0 mod 7. ( 2 7 ( 2 7) = since = 9 2 = 7 7 mod 8. ( ( 2 3) = and 2 5) =. ( So, p, 7 mod 8, then by Theorem.3 2 p) =, and by Theorem.4 p splits completely from Q to Q( 2). So the following split completely:, + 8, + 2 8,... 4

5 and 7, 7 + 8, ,... and these progressions will contain infinitely many primes. So infinitely many ideals split completely, and there is a clear relationship between the decomposition of ideals, and congruence conditions in Z. Kronecker (82-89): Observed the following correspondence Elliptic Curves Automorphic Forms Abelian extensions of imaginary quadratic number fields given by adjoining certain values of the automorphic form. Does this method give us all abelian extensions of a given number field (Kronecker s Jugendtraum)? The followingt was conjectured by Kronecker and proved by Weber: Theorem.5. (Kronecker-Weber) Every abelian extension of Q is contained in a cyclotomic extension of Q. Hilbert (900): At the Paris, ICM, Hilbert posed the following two problems, which are the two main questions of class field theory: Hilbert s 9 th : To develop the most general reciprocity law in an arbitrary number field, generalizing Gauss law of quadratic reciprocity. Hilbert s 2 th : To Generalize Kronecker s Jugendtraum. 2 Dirichlet s Theorem on Primes in Arithmetic Progression 2. Characters of Finite Abelian Groups We will now begin a discussion of characters of finite abelian groups, which will lead us to Dirichlet characters, and eventually a proof of Dirichlet s Theorem on Primes in Arithmetic Progression. Definition 4. Suppose G is a finite abelian group. A character,, of G is a multiplicative group homomorphism : G C. The character group, denoted Ĝ, is the set of all characters on G. For any, ψ Ĝ, ψ : G C g (g)ψ(g) defines multiplication in the group Ĝ, and 0 : G C is the trivial character. g 5

6 Proposition 2.. If G is a finite abelian group, then G = Ĝ. Proof. Since G is a finite abelian group, it is a direct sum of Z/mZ, where m Z. So, Ĝ is a product of Ẑ/mZ, and for any Ẑ/mZ, : Z/mZ C (). Since Z/mZ is an additive group generated by, is completely determined by (). Further, only 0 sends to the multiplicative identity in C, so we have the following homomorphism Ẑ/mZ C () which has trivial kernel, and hence is injective. Since Z/mZ is finite and additive with idenitity 0, and for any Ẑ/mZ, (0) =. Consequently, } + + {{... + } = 0. m times = (0) = ( ) = () () so the image of this homomorphism is precisely the m th roots of unity in C, which is isomorphic to a cyclic group of order m. Therefore, and hence Ĝ = G. Z/m/Z = Z/mZ, Immediate from this, we have Ĝ = G, with the following canonical homomorphism sending any g G to g, defined by: g : Ĝ C Exercise 2.2. Show g g is a homomorphism. (g) Proposition 2.3. The map g g is an isomorphism G Ĝ. Proof. Suppose g G, and (g) = for all Ĝ (i.e. g is the trivial character). Let H = g < G. Then, every Ĝ can be expressed as a character of G/H, so Ĝ/H Ĝ, and by Proposition 2. G/H G, so H =. Therefore g =, and G Ĝ is injective. But from Proposition 2. we also know that so the map surjective, and G = Ĝ. G = Ĝ = Ĝ 6

7 Proposition 2.4. Let G be a finite abelian group. For H < G, let Then,. if H < G, then H = Ĝ/H. 2. if H < G, then Ĥ = Ĝ/H. 3. (H ) = H (if we identify Ĝ = G). H = { Ĝ : (h) = for all h H}. Proof. (Proof is repeated applications of Propositions 2. and 2.3. For a detailed proof, see [, Proposition.3].) The following proposition describes what are called the Orthogonality Relations of characters of finite abelian groups. Proposition Fix a character of the finite abelian group G. Then { 0 if 0 (g) =. G if = 0 g G 2. Fix an element g of the finite abelian group G. Then, { 0 if g (g) = G if g =. Ĝ Proof. Let h G. Then, so (g) = (gh) = (h) (g), g G g G g G ( (h)) (g) = 0 g G for every h G. If 0, then there exists h G such that (h), and then If = 0, then the expression becomes (g) = 0. g G g G 0 (g) = g G = G. For the second part, note that Ĝ (g) = g(). Now use part (). Ĝ 7

8 2.2 Dirichlet Characters Definition 5. Let n be a positive integer. A Dirichlet Character modulo n is a character of the abelian group (Z/nZ). That is, a multiplicative homomorphism of the form We call n the modulus of. : (Z/nZ) C. Example 2.. Let p be an odd prime. : (Z/mZ) C ( a a = p) { if a x 2 0 mod p for some x Z otherwise This particular character is called the Legendre Symbol. 2. Let i =. Define : (Z/5Z) C 2 i 3 i 4 Suppse is a Dirichlet character of modulus n and n m. Then, by the natural homomorphism, ϕ : (Z/mZ) (Z/nZ) define = ϕ. Here is a Dirichlet character of modulus m. We say is induced by. Definition 6. Let f be the minimal modulus for the character. That is, is not induced by any Dirichlet character of modulus less that f. Then we call f the conductor of. A Dirichlet character defined modulo its conductor is called primitive. Note, for any Dirichlet character, the following is always true, : (Z/nZ) C, and since for any a (Z/nZ), we know a ϕ(n) =, it follows that = (a ϕ(n) ) = (a) ϕ(n) so the image of is always a finite set of ϕ(n) th roots of unit in C. Example 3.. Define : (Z/2Z) C

9 Notice, (a + 3k) = (a), so in fact is induced by ψ is irreducible, and f = Define ψ : (Z/3Z) C 2. : (Z/2Z) C 5 7. Here is primitive so f = 2. To verify, just check possible maps on (Z/mZ) where m 2. Definition 7. Let 0 denote the trivial character, and ψ primitive Dirichlet characters of conductor f and f ψ. Let n = lcm(f, f ψ ). Then ψ is the primitive Dirichlet character that induces η defined η : (Z/nZ) C a (a)ψ(a). Note, f ψ f f ψ, and the set of Dirichlet characters is closed under this multiplication. Recall now, that for any root of unit in C, its complex conjugate is its inverse. Define : (Z/nZ) C a (a) = (a) = (a). Exercise 2.6. Show that the set of all Dirichlet characters form a commutative group under this notion of multiplication, with = 0 as the identity element. Let s see an example of how this η works. Example 4. Define and ψ as the following, and : (Z/2Z) C 5 7 ψ : (Z/4Z) C 3. 9

10 Then n = 2 = lcm(2, 4), and we get the following map: η : (Z/2Z) C ()ψ() = 5 (3)ψ(3) = 7 (5)ψ(5) = ()ψ() = We can see that η is imprimitive, since it is induced by the map (Z/3Z) C 2. Since this map is primitive (clearly) and induces η, it must be the product ψ. Note that this definition does not mean that (x)ψ(x) = ψ(x), since ψ(2) =, but and ψ are not defined for 2. Definition 8. The order of a Dirichlet character is its order as an element in the group of all Dirichlet characters. This order will always be finite, and if has conductor n, then the order of must divide ϕ(n). A Dirichlet character of order 2 is called a quadratic Dirichlet character. For any Dirichlet character, we must have ( ) = ±. We say is even ( ) = is odd ( ) =. Exercise 2.7. The set of even Dirichlet characters is a subgroup of the group of all Dirichlet characters under multiplication. 2.3 Dirichlet Characters and Q(ζ n ) Fix an integer n. The set of all Dirichlet characters such that the conductor divides n form a group. In fact, G := Gal(Q(ζ n )/Q) = (Z/nZ), so the group of Dirichlet characters Definition 9. Let mod n can be seen as the characters of that Galois group. : G C, and let K be the fixed field of the kernel of. Then K is the field associated to. Example 5. Let G := Gal(Q(ζ 2 )/Q), and define as follows: : G C σ σ 5 σ 7 σ 0

11 where σ j : ζ 2 ζ j 2. Then ker() = {σ = id, σ 7 }, and we have the following Galois correspondence: Q(ζ 2 ) {} Q(ζ 3 ) σ 7 Note that σ 7 (ζ 3 ) = ζ 7 3 = ζ3 3 ζ3 3 ζ 3 = ζ 3, so Q(ζ 3 ) is the fixed field of σ 7. Since we know Q Gal(Q(ζ 2 )/Q(ζ 3 ) = [Q(ζ 2 ) : Q(ζ 3 )] = 4 2 = 2 G ker() = Gal(Q(ζ 2 )/Q(ζ 3 )), so Q(ζ 3 ) is the field associate to. So we may view as the following: since Note that f = 3. More generally, let : G/ ker() C G/ ker() = Gal(Q(ζ 3 )/Q) = (Z/3Z). X = {finite group of Dirichlet characters}, let n = lcm(f xi ). Now X Ĝ, G = Gal(Q(ζ n)/q). Let H = ker, so H G. Let K be the fixed field of H, then we call K the fixed field associated to X. Notice, X = H ker(). So, X {field associated to } {field associated to X}. If X =, then {field associated to } = {field associated to X}. Q(ζ n ) {field associated to X} {field associated to } {} H ker() Q G

12 Example 6. : (Z/5Z) C Here ker() = 4, so if K is the fixed field of ker(), then [K : Q] = 2. Let G = Gal(Q(ζ 5 )/Q), and consider : G C. By above, we know σ 4 ker(), so σ 4 (k) = k for any k K. But σ 4 corresponds to complex conjugation, so its fixed field must be real. Therefore, K = Q( 5), since this is the only real subfield in Q(ζ 5 ). Note: d K/Q = 5, and = ϕ, where since is primitive, f = 5. Example 7. Let : (Z/5Z) C ψ : (Z/5Z) C

13 and θ : (Z/5Z) C In each of these, ker(ψ) = ker(θ) = 4, so the field associated to each of these characters must be a degree 2 extension of Q. Immediately, we notice that ψ factors through ψ, where ψ : (Z/3Z) C 2. So, f ψ = 3. It is not too difficult to see now that θ is primitive, so f θ = 5. So what are the quadratic fields associated to these character groups? Since σ a+3k is in the kernel of ψ, a reasonable conclusion is that K ψ = Q( 3). Now there is only one remaining quadratic extension, so K θ = Q( 5). Notice, there seems to be some connection between the conductor of a character, and the discriminant of its associated field. Let s explore that further. But first, one more example. Example 8. Let G := Gal(Q(ζ n )/Q) = (Z/nZ). Let X be the set of all even characters in Ĝ. Then X < Ĝ, by Exercise 2.7, and in particular it is an index 2 subgroup, since the product of any two odd characters is an even character. Further, ( ) = for every X, and so σ is in the kernel of (here we have identified (Z/nZ) = G). Since σ is complex conjugation, the field associated to it must be real. Notice, that for any character, the field associated to is real if and only if is even. Here, X is the largest subgroup made entirely of even characters, so a good guess would be that is associated field is the maximal real subfield Q(ζ n + ζn ) of Q(ζ n ). We know, by definition, that the field associated to X is the fixed field of H = ker(). X We know {σ, σ } H. Now, suppose σ r H, and σ r is non-trivial. Then σ r given by G Ĝ is nontrivial, by Proposition 2.3. Then, there exists some ψ Ĝ such that ψ(σ r). But then ψ X, or else σ r ker(ψ). So ψ is odd, and ψ 2 is even, and so ψ 2 (σ r ) =, and hence ψ(σ r ) = (since ψ(σ r ) ). Now, since [Ĝ : X] = 2, any odd character has the form ψ for some X. So, σ r : Ĝ C ψ ψ(σ r )(σ r ) = (σ ) =, 3

14 So σ r = σ. Therefore, {σ, σ } = H. Now the field associated to X is precisely the fixed field of H, so it must be real, and it must be of index 2 in Q(ζ n ). So there is only one choice, namely Q(ζ n + ζ n ). Theorem 2.8. (Conductor Discriminant Formula) Let X be a finite group of Dirichlet characters of K its associated field. Then, d K/Q = ( ) r 2 f, X where r 2 is the number of pairs of imaginary embeddings of K. Exercise Let p > 2 be prime. Use Theorem 2.8 to find the dicriminant of the maximal real subfield of Q(ζ n ). (Hint: remember that only the prime p can ramify here, if at all.) 2. Let p > 2 be prime. Use Theorem 2.8 to find the discriminant of Q(ζ p n) over Q, where n > 0. Our next goal is to establish a full Galois Type correspondence between character groups and associated fields. From the fundamental theorem of Galois Theory, we already have the following: K {} L H = Gal(K/L) Q G = Gal(K/Q) Let X K be the group of characters of Gal(K/Q), so X K = Ĝ. Then, we may define the following subset of X K { X K : (g) =, g Gal(K/L)} = Gal(K/L) = Gal(K/Q)/Gal(K/L) = Gal(L/Q)(by the F.T. of G.T.) = X L. So X L X K. But from this definition, we see that given any X L, (g) = for every g Gal(K/L), so in particular, X L ker() = Gal(K/L) and by the fundamental theorem of Galois Theory, L is the fixed field associated to Gal(K/L). So, for every subgroup of X K, we have an associated fixed field in K. Suppose Y X K, and L is the fixed field of Y. Then Y = Gal(K/L) by the fundamental theorem of Galois Theory. So, (Y ) = (Gal(K/L)) = X L 4

15 where the last equality comes from above. So to enhance our existing picture, K {} X K = Ĝ L H = Gal(K/L) X L = H = Ĝ/H Q G = Gal(K/Q) { 0 } = G So this is a containment preserving correspondence of given by or by {Subgroups of X K } {Subfields of K} Y {Fixed field of Y } X L = Gal(K/L) L So let s revisit Example 8. Here X was the set of all even Dirichlet characters whose conductors divide n. We know that the field associated to X is real (since it is fixed by complex conjugation). Suppose L is any real subfield of Q(ζ n ), then L is fixed by σ. By the correspondence given above, L is precisely the fixed field of X L, so σ X L. But or equivalently, So, X L = { g Ĝ : g() = X L } X L = {g G : (g) = X L }. σ X L = ( ) = X L, in other words, every element of X L is even. But then, X L X, so the field associated to X must be the maximal real subfield, since any other real subfield corresponds to a subgroup of X. 2.4 Ramification Just as we describe the Galois correspondence for finite extensions of Q in terms of Dirichlet characters, we can describe the ramification behavior of finite extensions. Definition 0. Let n be a positive integer, and suppose n = where p j are distinct primes, and a j > 0. Then m j= p a j j, (Z/nZ) m = (Z/p a j j Z), j= 5

16 so any Dirichlet character modulo n, can be written as a m = pj, j= where pj is a Dirichlet character modulo p a j j. For any group of characters X mod n, we will let X pj = { pj : X}. Notice, that for any pj X pj, the conductor is some power of p j, so in particular, p always divides the conductor. Example 9. Consider the character This may be rewritten as θ = θ 3 θ 5, where and θ : (Z/5Z) C θ 3 : (Z/3Z) C 2 θ 5 : (Z/5Z) C In our example above, if X = {θ}, then X 5 = {θ 5 } and X 3 = {θ 3 }. Theorem 2.0. Suppose X is a group of Dirichlet characters with associated field K. If p Z is a prime, then the ramification index of p in K/Q is e = X p. Proof. (See [, Theorem 2.2]) Sometimes it will help us to view a Dirichlet character as a function : Z C given by { (a mod f ) if (a, f ) = (a) = 0 if (a, f ). But note that this is equivalent to our previous construction, except now (f ) = 0. 6

17 Corollary 2.. Let X be a group of Dirichlet characters and let K be its associated field. A prime p is unramified in K/Q if and only if (p) 0 for all X. Proof. Suppose p ramifies in K/Q. We then have e = X p. So X p contains some non-trivial element. Since X p, we know p f, so in particular, (p) = 0. Conversely, if (p) = 0 for some element of X, then the conductor of must be divisible by p. Thus, X p must be non-trivial. But this implies that e = X p >, so p is ramified. Example 0. Consider K = Q(ζ 2 ) and L = Q(i), so L K. By our previous discussion, L has the associated field X L = Gal(K/L) = { (Z/2Z) : (σ) = σ Gal(K/L)}. But, Gal(K/L) must have size 2, since [K : Q] = 4. So Gal(K/L) = {, σ}, and σ fixes L. Consider the element i = ζ2 3 L. Suppose σ is an element in Gal(K/Q)) fixing this element, then σ : (Z/2Z) (Z/2Z) ζ 2 ζ 5 2 ζ 3 2 (ζ 3 2) 5 = ζ 5 2 = ζ 3 2 = i. So, (σ) = if and only if (5) = (recall σ = σ 5 and 5 correspond under the canonical isomorphism Gal(K/Q) = (Z/2Z).) So, X L = {, θ}, where θ : (Z/2Z) C However, this factors through θ in the following way 5 7. θ : (Z/2Z) (Z/4Z) C so f θ = 4. We know θ(4) = 0, so by Corollary 2., we know that 2 is the only prime that ramifies in Q(i)/Q (thus, confirming something we already know...but in a much cooler way.) Before proving the following theorem, let s recall what we know about inertia and decomposition groups from algebraic number theory. Suppose we have an abelian extension, K/L and G := Gal(K/L). Suppose p is a prime in L, which splits in K as po K = (P P n ) e. Recall from the first lecture, we know that the ramification index is identical for each prime sitting above p. We define the decomposition group of P over L as G P = {σ G : σ(p) = P}. 7

18 Recall, that since the primes over p are Galois conjugates of one another, and it is easy to show that τ G σp τ σg P σ. Also, since G P is the stabilizer of P under the action of the Galois group, the index of G P in G is precisely the size of the orbit of containing P. So we know that so from the fundamental identity, we conclude [G : G P ] = g, [K : L] = efg G P = n g = ef. The fixed field of G P, denoted Z P is the decomposition field. Define, κ(p) = O K /P and κ(p) = O L /p, then [κ(p) : κ(p)] = f, the residue degree. But κ(p)/κ(p) is a finite extension of a finite field, and as such, we can easily identify the Galois group of this extension. The following two lemmas will be stated without proof, but the reader may refer to [5, 9]. Lemma 2.2. Fixing a prime P above p, let P Z = P Z P. Then,. P is the only prime above P Z. 2. The ramification index and the inertia degree of P Z over p are both equal to. Lemma 2.3. The map G P Gal(κ(P)/κ(p)) is a surjective homomorphism. The kernel of this homomorphism is called the inertia group, denoted I. It is known that the fixed field of I, which we call the inertia field, and denote T P, is the largest intermediate field such that e =. Now we can prove the theorem. Theorem 2.4. Let X be a group of Dirichlet characters with associated field K. Let p be prime, and define the subgroups Y = { X : (p) 0} Then, Y = { X : (p) = }.. X/Y is isomorphic to the inertia subgroup for p. 2. X/Y is isomorphic to the decomposition group for p. 3. Y/Y is cyclic of order f. Proof. [, Theorem 2.4] 8

19 2.5 Dirichlet Series Definition. A Dirichlet series is a series of the form f(s) = where a n C for all n and s is a complex variable. We will first introduce some elementary facts about Dirichlet series, beginning with Abel s Lemma. Exercise 2.5. Prove Abel s Lemma: Let (a n ) and (b n ) be sequences, and for r m put A m,r = r n=m = a n and S m,r = r n=m a nb n. Then, S m,r = r n=m n= a n n s A m,n (b n b n+ ) + A m,r b r. Exercise 2.6. Let A be am open subset of C and let (f n ) be a sequence of holomorphic functions on A that converge uniformly on every compact subset to a function f. Show that f is holomorphic on A and the sequence of derivatives (f n) converges uniformly on all compact subsets to f. Lemma 2.7. If f(s) = n= an n converges for s = s s 0, then it converges uniformly in every domain of the form {s : Re(s s 0 ) 0, Arg(s s 0 ) θ} with θ < π 2. Proof. Translating if necessary, we may assume s 0 = 0. Then we have that n= converges, and we must show that f(s) converges uniformly in every domain of the form {s : Re(s) 0, Arg(s) θ} for θ < π 2. Equivalently, we must show that f(s) converges uniformly in every domain of the form s {s : Re(s) 0, Re(s) M}. Let ɛ > 0, and let A m,r = as in Abel s Lemma. Since n= a n converges, there is some sufficiently large number N, so that if r > m N, then we have A m,r < ɛ. Now, let b n = n s and apply Abel s Lemma to get S m,r = r n=m a n r n=m a n A m,n (n s (n + ) s ) + A m,r r s. First, we notice the following equivalence obtained by taking the real integral: s d c e ts dt = e cs e ds. 9

20 So, taking the absolute value, we get e cs e ds = s d c e tre(s) dt = s Re(s) (e cre(s) e dre(s) ). So, taking c = ln(n), that is, n = e c, and similarly, d = ln(n + ).we obtain S m,r ɛ r s + r n=m ( r ɛ r s + ( ɛ + s Re(s) n=m r n=m But, this sum is telescoping, so simplifying, we get and m, r N and Re(s) > 0, so in fact ɛ n s (n + ) s ) n s (n + ) s ) (n Re(s) (n + ) Re(s) ). S m,r ɛ( + M(m Re(s) r Re(s) )), S m,r ɛ( + M). Of course, S m,r is just a difference of partial sums of our Dirichlet series, so the uniform convergence s of f(s) on {s : Re(s) 0, Re(s) M} follows. Theorem 2.8. If the Dirichlet series f(s) = n= an n converges for s = s s 0, then it converges (not necessarily absolutely) for Re(s) > Re(s 0 ) to a function that is holomorphic there. Proof. Clear from Lemma 2.7 and Exercise 2.6. Corollary 2.9. Let f(s) = n= an n s be a Dirichlet series. i. If the a n are bounded, then f(s) converges absolutely for Re(s) >. ii. If A n = a a n is a bounded sequence, then f(s) converges (though not necessarily absolutely) for Re(s) > 0. iii. If f(s) = n= an n converges at s = s s 0, then it converges absolutely for Re(s) > Re(s 0 ) +. Proof. i. Suppose the a n are bounded, so say that a n M. Also suppose that Re(s) <. Then, r a n r n s a n n s n= n= r M n s = M = M n= r n Re(s) (n i ) Im(s) n= r n Re(s). But since Re(s) >, this n= n Re(s) converges, and the result follows. n= 20

21 ii. Suppose that the A m are bounded. For r > m let A m,r = r n=m a n as in Abel s Lemma. We have that the A m,r are bounded, let s say A m,r M. Applying Abel s Lemma, with b n = n s, as in the proof of Lemma 2.7, we get r ( S m,r = A m,n n s ) ( ) (n + ) s + A m,r r s n=m r ( A m,n n s ) (n + ) s + ) m,r( A r s n=m ( r M n s (n + ) s + ) r s. n=m By Theorem 2.8, it suffices to show that f(σ) converges for a real value of σ, since then we may conclude that f(s) converges whenever Re(s) > σ. So, suppose s = σ is real, then we have S m,r M m σ, which is a Cauchy sequence whenever σ > 0. So f(s) converges whenever Re(s) > σ > 0, and we are done. iii. Suppose f(s) = n= an n s converges for s = s 0. Let g(s) = f(s + s 0 ) = ( a n n s 0 )( n s ). Since f(s 0 ) converges, b n = an n s 0 approaches 0 as n. Therefore, {b n} and hence {b n ( n s )} is bounded if Re(s) >. By part (i), g(s) converges absolutely for Re(s) >. Thus, f(s) = g(s s 0 ) converges absolutely for Re(s s 0 ) >, that is, Re(s) > + Re(s 0 ). Definition 2. Let : associated to is The Riemann zeta function is ( Z/mZ) C, be a Dirichlet character. The Dirichlet L-function L(s, ) = ζ(s) = n= n= (n) n s. Suppose 0. Let A n = () (n) and write n = mk + r where 0 r m. Then, A n = [()+...+(m)]+[(m+)+...+(2m)]+...+[(km+)+...+(km+r)] = [(km+)+...+(km+r)] by our orthogonality relations. Therefore, n s. A n (km + ) (km + r) = r < m. Now, use Corollary 2.9. If 0, then we ve shown A n is a bounded sequence, so L(s, ) converges for Re(s) > 0, by part (ii). Then by part (iii), L(s, ) converges absolutely for any Re(s) >. 2

22 Lemma L(s, ) has a so-called Euler product: for Re(s) >. L(s, ) = p prime Proof. Fix s, with Re(s) >. We want to show: lim N p N ( (p)p s ) ( (p) ) p s = L(s, ). Say p,..., p N are the primes less than N. Then, we have the following infinite geometric series k i= ( (p ) i) p s = i = = k i= k i= ( + (p i) p i (p i) m ) (p s +... i )m ( + (p i) (pm i ) ) p i p sm +... i m,...,m k 0 = n J n (n) n s, (p m p m k k ) (p m p m k k )s where J n = {n Z : n > 0 and n is not divisible by any prime p > N}. Now we have L(s, ) ( (p) ) p s = (n) n s. p N n (Z + \J n) Taking absolute values and applying the triangle inequality, we get (n) n s n (Z + \J n) (n) n s n (Z + \J n) = n s n (Z + \J n) = n (Z + \J n) n N n σ, n σ where σ = Re(s) since clearly (Z + \ J n ) {n N}. However, the series n>n n σ converges for σ >, n N n σ 0, 22

23 as N. So, and so we are done. lim N L(s, ) p N ( (p) ) p s = 0, Note that when Re(s) >, then we have L(s, ) = p prime ( (p)p s ) = p prime p s p s (p) 0 since p s 0. Taking the log of L(s, ) for Re(s) >, we get log(l(s, )) = log ( (p) p prime p s ) = p log( (p)p s ). Let this be the branch of log defined on the upper half plane, such that log(l(s, )) 0 as s. Recall, for a complex valued T with Re(T ) <, we have the following Taylor series expansion Since (p) p s = p s Now, where σ = Re(s). So, log( T ) = n= T n n. (p) when Re(s) >, we may use this expansion. Letting T = p, we get s p log(l(s, )) = p n = p = p ( log (p) ) p s (p) n p ns n n n (p) n np ns. (p) n np ns p ns = p nσ, (p) n np ns p n p nσ m m σ, () which converges for σ >. Therefore, the above expression for log(l(s, )) is absolutely convergent for Re(s) > (since the sum of the absolute values is convergent). So we may rearrange the terms to get log(l(s, )) = p n (p) n np ns = p (p) p s + p n 2 (p) n np ns. 23

24 Let β(s, ) = p n 2 (p) n np ns. From equation () we know that β(s, ) is absolutely convergent Re(s) > 2, so in particular, it takes a finite value whenever Re(s) >. 2.6 Dirichlet s Theorem on Primes in Arithmetic Progressions Consider Euler s classical proof of the existence of infinitely many primes, which uses the Riemann zeta function. We have the Euler product ζ(s) = n= n s = p prime ( p s ). Suppose that there are only finitely many primes, say p,..., p n Z. Then, n n ζ(s) = ( p s j ) =. j= j= p s j Taking the limit as s, we get n ( lim ζ(s) = s j= p j ), which is rational. However, it is clear that lim s n= n s =, which is a contradiction. The proof of Dirichlet s theorem of primes in arithmetic progression will be a generalization of this idea, except here the Riemann zeta function is replaced by the Dirichlet L-function. Theorem 2.2. (Dirichlet s Theorem on Primes in Arithmetic Progressions) If m is a positive integer and s is an integer for which (a, m) =, then there are infinitely many primes p satisfying p a mod m. Proof. Let (a, m) =, and consider the set of all Dirichlet characters modulo m. Then, (a) log(l(s, )) = (Z/mZ) (a) ( = p = p p (p) ) p s + β(s, ) p s (a) (p) + (a) β(s, ) p s (pa ) + (a) β(s, ). 24

25 Recall from our orthogonality relations, (pa ) = { ϕ(m) if pa mod m 0 if pa mod m, in other words, So, (pa ) = { ϕ(m) if p a mod m 0 otherwise. (a) log(l(s, )) = ϕ(m) p a mod m p s + ( ). (2) Here ( ) = (a) β(s, ), and as shown previously, for Re(s) > 2, β(s, ) is finite, say M, so (a) β(s, ) β(s, ) (a) = M. So, ( ) is something that converges absolutely for Re(s) > 2. Now let s, then on the right hand side of (2), we have lim ϕ(m) s p a mod m p s + ( a finite constant ), which would be finite if there were only finitely many primes p with p a mod m. Now, the proof will be complete if we can show that the left side of (2) is infinite as s. If = 0 (with modulus m), then L(s, 0 ) = p ( 0(p) p s ) = ζ(s) ( p s ) as s. p m So, L(s, 0 ) as s. Now, suppose 0. We have and any partial sum, L(s, ) = n= (n) n s. () + (2) (n) () + (2) (n) n is bounded, so from (ii) of Corollary 2.9 we know that L(s, ) converges for any Re(s) > 0. Further, L(s, ) is analytic, from our complex analysis fact. We know that L(, ) is defined for 0, so log(l(s, )) will be finite if we can show that L(, ) 0 for 0. Given this, we ll have (a) log(l(s, )) as s +, and the proof will be complete. Now we need to show that L(s, ) 0, when 0. 25

26 Definition 3. Let K be an algebraic number field, and let a very through the nonzero integral ideals O K, (so that we may view Na as a positive integer). Define the Dedekind zeta function of K as ζ K (s) = N a K/Q (a s ) By an argument similar to the one for the L-function, we have ζ k (s) = p ( N K/Q (p s ), where p runs over the prime ideals in O K. (The proof uses uniques factorization of prime ideals.) It is easy to see that γ n ζ k (s) = n s where γ n = #{a : N K/Q (a) = n}. Exercise Show that ζ K (s) is absolutely convergent for Re(s) >. (compare this to (i) of Corollary are the γ n bounded?) n= Now we will state without proof a theorem of Dedekind: Theorem ζ K (s) can be analytically continued to C {} with a simple pole at s =, i.e. Moreover, where ζ K (s) = ρ(k) + { something entire } s ρ(k) = 2r (2π) r 2 h K R K ω K d K/Q r = # real embeddings of K r 2 = # pairs of complex embeddings of K h K = #C K = Class number of K R K = regulator of K ω k = # roots of unity in K d K/Q = the discriminant Now we can use the Dedekind zeta funtion to show that L(, ) 0 when 0, completing the proof of Primes in Arithmetic Progressions. Take K = Q(ζ m ), where m is the modulus in Dirichlet s theorem, so also the modulus of the characters. We have ζ K (s) = ( N K/Q (p s ) p = ( N K/Q (p s ) p p p = ( N K/Q (p s )) ( N K/Q (p s )). p m p p p m p p 26

27 Now, recall that N K/Q (p) = p f. Since K/Q is Galois, ϕ(m) = fg, and f = #Z(p) is the smallest integer so that p f mod m, and g = ϕ(m) f. Lemma If p m then ( T f ) ϕ(m) f = mod m ( (p)t ). Proof. Let G = Gal(K/Q), where K = Q(ζ m ) as before. Let Z = Z(p) be the decomposition group for p in K/Q and define a map First observe that Then, where Ĝ Ẑ Z. { mod m} = { (Z/mZ) } = { Ĝ}. (p)t ) = Ĝ( (p)t ) = ψ Ẑ Z=ψ( Ĝ, ψ Ẑ l(ψ) =#{ Ĝ : Z ψ} =# ker(ĝ Ẑ) = #Ĝ #Ẑ = ϕ(m) f =g. ( ψ(p)t ) l(ψ) Thus, mod m( (p)t ) = Ĝ( (p)t ) = ψ Ẑ ( ψ(p)t ) g, and we know g = ϕ(m) f. It only remains to be shown that T f = ψ Ẑ( ψ(p)t ). As a subgroup of G, Z is generated by the unique Frobenius element, F (p p), where F (p p)(a) a p mod p. Alternatively, we know that in the unramified case, Z = Gal(F p n/f p ), a cyclic group generated by σ p : F p n F p n a a p. So this generator must pull back to a generator of Z, giving us our unique Frobenius element, σ p. Viewing Z as a subgroup of (Z/mZ), Z is generated by p mod m. Consider the following map: Ẑ µ f = {f th roots of unity } ψ ψ(p). 27

28 Since p is a generator for Ẑ, this map is an isomorphism. Thus, ( ψ(p)t ) = ( ηt ) = T f. η µ f So we are done. ψ Ẑ Now, in Lemma 2.24, put T = p s, giving ( p sf ) ϕ(m)/f = mod m ( (p)p s ). taking the product over all primes p not dividing m, we get p p m( sf ) ϕ(m)/f = ( (p)p s ). p m Now, if p m, then (p) = 0, so Putting this all together, L(s, ) = p L(s, ) = = = ( (p)p s ) =. p m ( (p)p s ) ( (p)p s ) p ( (p)p s ) ( (p)p s ) p m p m p m( (p)p s ) = p m( p sf ) ϕ(m)/f. On the other hand, ( N(p) s ) = ( p sf ) p p p p =( p sf ) ϕ(m)/f, since there are g = ϕ(m)/f primes p contained in p. This gives us ( N(p) s ) p m( p sf ) ϕ(m)/f = p m p p =ζ k (s) p m ( N(p s ), p p 28

29 thus ζ k (s) ( N(p s )) = L(s, ) p m p p =L(s, 0 ) L(s, ) 0 =ζ(s) p m ( p s ) 0 L(s, ). We get Now, and 0 L(s, ) = ζ k (s) p m p p ( N(p s )) ζ(s) p m (. p s ) ( N(p s )) p m p p ( p s ) p m are non-zero constants, while each of the ζ K (s) and ζ(s) has a simple pole at s =. Let s, the expression on the left side approaches a constant, hence 0 L(s, ) does too. This shows that L(, ) 0 for all 0, completing the proof of Dirichlet s Theorem. 2.7 Dirichlet Density Suppose f(s) and g(s) are defined for s R, s >. We will write f(s) g(s) if f(s) g(s) is bounded as s +. Using this notation, we will reformulate our proof of Dirichlet s Theorem. For any, log(l(s, )) = (p) p s + {something converging for Re(s) > 2 } p (see the equality on the bottom of page 23). So, log(l(s, )) p (p) p s = {something converging for Re(s) > 2 }, which is bounded as s +. So, log(l(s, )) p (p) p s. Assuming we feel comfortable with L(, ) 0 when 0, we have that log(l(, )) is finite when 0, so (a ) log(l(s, )) = log(l(s, 0 )) + (a ) log(l(s, )). 0 29

30 As s +, the term on the far right is finite, so we conclude that (a ) log(l(s, )) log(l(s, 0 )). Furthermore, (a ) log(l(s, )) = p = p = p a (a ) (p) p s p p s (a p) p s mod m { ϕ(m) if p a mod m 0 otherwise ϕ(m) p s. Combining these expressions, we get log(l(s, 0 )) p a mod m ϕ(m) p s. Using our well-known Euler product, we have L(s, 0 ) = ( ) 0(p) p s = ζ(s) p p p m( s ). Taking the log of both sides, we get log(l(s, 0 )) = log(ζ(s)) + log p m( p s ), and since the term on the far right is clearly finite, So, we get Clearly, p a log(l(s, 0 )) log(ζ(s)). mod m lim log(ζ(s)) = log s + p s ϕ(m) log(ζ(s)). ( n= ). n So, p a mod m p diverges, which proves Dirichlet s theorem. s It is a nontrivial fact that lim )ζ(s) =. s +(s 30

31 This is a consequence of ζ(s) having a simple pole at s =, a fact which we discussed earlier. So, we can see that ( ) ( ) (s )ζ(s) log(ζ(s)) = log = log((s )ζ(s)) + log. (s ) s So, and Therefore, and ϕ(m) log(ζ(s)) = ( ( )) log((s )ζ(s)) + log, ϕ(m) s ( ( )) lim log(ζ(s)) log = 0. s + ϕ(m) s p a lim = s + mod m p s ( ) ϕ(m) log, s p a mod m p s Definition 4. Let S be any set of primes. If lim = s + ( ) = bounded. log s p S p s ( ) = δ log s exists, then we say that S has Dirichlet density δ = δ(s). Theorem Let K/Q be Galois, and let Then δ(s K ) = [K:Q]. S K = {p Z : p splits completely in K/Q}. Proof. Let ζ K (s) = p ( Np s ) for Re(s) >, be the Dedekind zeta function for K. Consider s R, s >. We have log(ζ K (s)) = log( Np s ) = n Np ns, p p n and so ( ) log(ζ K (s)) = log((s )ζ K (s)) + log, s ( ) log ζ K (s) log s n Np ns p n p p Np s + p Np s. n=2 n Np ns 3

32 since p n=2 n Np ns is bounded as s +. Then, ( ) log ζ K (s) log s p = Np s f(p p)==e(p p) p s + f(p p)> p fs + f(p p)=,e(p p)> but the second term is bounded as s +, and the third term is finite, since only finitely many primes ramify in any given extension. So, ( ) log ζ K (s) log = g(p)p s, s p S K where g(p) counts the number of primes p sitting over p. For p S K, we know that g(p) = [K : Q], so ( ) log ζ K (s) log [K : Q]p s. s p S K Therefore, ( ) log = [K : Q] p s + b(s), s p S K where b(s) is something bounded as s +. Now, computing S K, δ(s K ) = lim s + = lim s + = lim s + = [K : Q]. p S K p s ( ) log s ( ) log s p S K p s ( [K : Q] p S K p s + b(s) p S K p s ) p s, More generally, we could let S be the set of prime ideals of O F, where F is a number field. Then, if lim s + exists, then S has Dirichlet density δ = δ F (S). p S p s ( ) = δ log s Corollary Let K/F be Galois, and S K/F = {p O F : p splits completely in K/F }, then δ F (S K/F ) = [K : F ]. 32

33 This proof follows from the previous theorem. Let S and T be sets of prime ideals in O F, then we say S T δ F (S \ T ) = 0 and S T S T S. Theorem E and K are number fields, both Galois over Q. Then, S K S E E K. Proof. =) obvious. = ) For this direction, suppose we have two sets of primes S and T, so that S T =. Then But we have and and So, δ(s T ) = lim s + So, from Theorem 2.25, we get so E K. 3 Ray Class Groups p S T p s ( ) = lim log s + s p S p s ( ) + log s S KE = S K S E, (S K S E ) (S K \ S E ) =, (S K S E ) (S K \ S E ) = S K. p T p s ( ) = δ(s) + δ(t ). log s δ(s KE ) = δ (S K S E ) = δ (S K S E ) + δ (S K \ S E ) = δ(s K ). [KE : Q] = δ(s KE ) = δ(s K ) = [K : Q], 3. Approximation Theorems and Infinite Primes Theorem 3.. Let n be non-trivial pairwise inequivalent absolute values on a number field F, and let β,..., β n be non-zero elements of F. For any ɛ > 0, there is an element α F such that α β j < ɛ, for each j =,..., n. Recall, given a number field F/Q of degree n, there are d real embeddings of F into R, given by the real roots of the minimal polynomial for F/Q. For any such embedding, σ i : F R, we get the absolute value σi : F R x σ i (x), where is the usual absolute value on R. To each of these places, we will associate the following formal object, i, given by α β mod i σ i ( α β ) > 0, and we call this an infinite real prime. 33

34 Definition 5. A modulus (or divisor) m of F is a formal product m = m 0 m where m 0 is a product of finite primes, and m is a product of infinite primes. In fact, m 0 = p p ordp(m 0), can just be thought of as a finite non-zero integral ideal of O F. Similarly, we define the formal product m = i i t where t = 0,. Example. If F = Q, then there are only two types of moduli, m = (n), where n Z, or m = (n), where is the unique embedding of Q into R. 3.2 Ray Class Groups and the Universal Norm-Index Inequality Recall the following I F = { nonzero fractional ideals of F }, and P F = { principal fractional ideals in I F }. From this, we get the ideal class group C F = I F /P F. Now define I F (m) = { free abelian group generated by finite places not dividing m}. So, in particular, if m = (), then and if m m, then Next, we define the ray I F (m) = I F, I m I m. (i) for all finite p m, ord p (α ) ord p (m), (ii) for all i m, σ i (α) > 0. P F (m) = {(α) : such that (i) and (ii) hold} 34

35 and the narrow ray P + F (m) = {(α) : such that (i) holds, and σ i(α) > 0 i}, observing that P + F (m) P F (m) I F (m). Definition 6. We define the ray class group of F for m as R F (m) = I F (m)/p F (m), and the narrow ray class group as R + F (m) = I F (m)/p + F (m), Notice, if we let m = i i, for some F, and let m = m 0 m, then P + F (m 0) = P F (m). References [] N. Childress, Class Field Theory, Universitext, Springer-Verlag, New York, [2] D. Garbanati, Class Field Theory Summarized, Rocky Mountain Journal of Mathematics,, No. 2, (98) [3] D. Marcus, Number Fields, Universitext, Springer-Verlag, New York, 977. [4] J. S. Milne, Class Field Theory, v [5] J. Neukirch, Algebraic Number Theory, Springer-Verlag, 999. [6] T. Shemanski, An Overview of Class Field Theory, expository-papers/tex/cft.pdf 35