LECTURE NOTES, WEEK 7 MATH 222A, ALGEBRAIC NUMBER THEORY

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1 LECTURE NOTES, WEEK 7 MATH 222A, ALGEBRAIC NUMBER THEORY MARTIN H. WEISSMAN Abstract. We discuss the connection between quadratic reciprocity, the Hilbert symbol, and quadratic class field theory. We also introduce the adelic perspective. 1. Quadratic Class Field Theory The main theorems of (local and global) quadratic class field theory are essentially proven in Serre s course in arithmetic. We begin by making this connection explicit and clear Local quadratic class field theory. Serre proves the following result about the (quadratic) Hilbert symbol over Q p or R: Theorem 1.1. The Hilbert symbol (, ) v is a non-degenerate bilinear form on the F 2 -vector space Q v /Q 2 Let S v = Q v /Q 2 If x Q v, then we write x for its image in the vector space S This is a finite-dimensional F 2 -vector space. Its dimension is 1, 2, or 3, depending on whether v =, 2 < v <, or v = 2, respectively. It is not a deep theorem that S v is isomorphic to its dual vector space this is true for every finite-dimensional vector space, over any field. The deep theorem (non-degeneracy of the Hilbert symbol) is that the Hilbert symbol induces such a duality. Let H v : S v S v (where S v is the dual vector space) denote the linear map given by: H v (a)(b) = (a, b) Non-degeneracy means that H v is an isomorphic of F 2 -vector spaces; this is what is proven in Serre. The following is the statement of local class field theory (over Q p or R): Theorem 1.2. Let G v = Gal( Q v /Q v ). Then, if A is any finite abelian group, then there is a group isomorphism: rec (A) v : Hom(G v, A) Hom(Q v, A). Of course, to prove this theorem, it suffices to assume that A is cyclic, by the structure theorem for finite abelian groups. We can use the Hilbert symbol to prove this theorem in the quadratic case: Theorem 1.3. Let G v = Gal( Q v /Q v ). Then there is a group isomorphism: rec (2) v : Hom(G v, Z/2Z) Hom(Q v, Z/2Z). Proof. The construction of the reciprocity map rec v goes as follows: 1

2 2 MARTIN H. WEISSMAN Begin with a nontrivial homomorphism γ : G v Z/2Z. Equivalently, we may begin with a subgroup G v of index two in G Associated to the subgroup G v of index two, we have a field extension K v /Q v of degree two. Associated to the field extension K v /Q v of degree two, we have the subgroup NK v Q Note that Q 2 v NK v If K v = Q v ( a), with a Q v, then b NK v if and only if (b, a) v = 1 (Chapter III, Section 1, Proposition 1 of Serre). In other words, NK v /Q 2 v = ker(h v (ā)). By non-degeneracy of the Hilbert symbol, the kernel of the nonzero linear functional H v (ā) has codimension 1. Thus NK v has index 2 in Q Associated to the subgroup NK v of index two in Q v, there is a unique homomorphism rec v (γ) from Q v to Z/2Z, with kernel NK v. We construct an inverse to the reciprocity map as follows: Suppose that η Hom(Q v, Z/2Z) is a nontrivial homomorphism. Equivalently, we may begin with a subgroup E of Q v of index 2. Since E has index two, E Q 2 Thus, E is uniquely determined by an index two subgroup of Q v /Q 2 Equivalently, E is uniquely determined by a codimension 1 subspace O v of S By the non-degeneracy of the Hilbert symbol, every codimension 1 subspace of S v arises as the kernel of H v (ā) for some 0 ā S Such an element ā lifts to an element a of Q v, unique up to multiplication by Q 2 Thus we get a quadratic field extension K v = Q v ( a). Every quadratic field extension corresponds to a subgroup of G v of index 2, and hence to a homomorphism from G v to Z/2Z. To prove the theorem, it must be shown that this is an inverse to the function rec v, and that rec v is a group homomorphism. We leave both of these facts as exercises for the reader Adeles and Ideles. The statement of global class field theory is significantly more difficult than local class field theory. One side is expected: if G = Gal( Q/Q), then we are interested in the group Hom(G, A), for any finite abelian group A. But we do not find an isomorphism from Hom(G, A) to Hom(Q, A). In this case, we must replace Q with the idele class group, which we explain here. Recall that for every place v (a prime number or ), we have a field Q v (either Q p or R). Within the multiplicative group Q p, we have the compact open subgroup Z p, and Q p /Z p = Z as abelian groups. We write Z = Q = R, for convenience. The ring of adeles (with an a, and denoted A), is defined to be the subring of v Q v (the infinite product), whose elements (x v ) satisfy: x v Z v, for all but finitely many indices v. One may check easily that A is a ring. It squeezes between the following two rings: Ẑ R = Z v A Q

3 LECTURE NOTES, WEEK 7 MATH 222A, ALGEBRAIC NUMBER THEORY 3 Note that Q is a subring of A. Indeed, if q Q, then we may consider the sequence (q v ), with q v = i v (q), where i v : Q Q v is the unique inclusion of fields, for all v. Note that (q v ) is in A, since for every prime p not dividing the denominator of q, i p (q p ) Z p. Let i: Q A denote the resulting embedding. Note that: i(q) (Ẑ R) = Z. The group of invertible elements of A is called the group of ideles (with an i), and denoted A (though some people use I). The group A can be seen as the subgroup of sequences (x v ) v Q v, satisfying: x v Z v for all but finitely many indices v. The ideles squeeze between the following two abelian groups: Ẑ R = Z v A Q The ring embedding i: Q A restricts to an embedding of abelian groups i: Q A. The quotient group A /Q is called the idele class group. In order to understand it, we first prove the following: Proposition 1.4. The group of ideles satisfies: A = (Ẑ R )i(q ). Proof. Suppose that x = (x v ) A. Let S be the set of places at which x v Z Then S is finite, and S, since x is an idele. Let n p = val(x p ), for all p S. Let q = p S p n p. Then we see that: Thus xi(q) Ẑ R, and we are done. val(x p i p (q)) = 0, = p V. Corollary 1.5. There is an isomorphism: A /i(q ) = (Ẑ R )/i(z ) Global Class Field Theory. The main theorem of global class field theory may now be stated, using the idele class group. We abuse notation, and just view Q as a subgroup of A, rather using the embedding i. Theorem 1.6. Let G = Gal( Q/Q). Then, if A is any finite abelian group, then there is a group isomorphism: rec (A) : Hom(G, A) Hom(A /Q, A). In the quadratic case, this follows from Theorems 3 and 4, of Chapter III, of Serre: Theorem 1.7. Let G = Gal( Q/Q). Then there is a group isomorphism: rec (2) : Hom(G, Z/2Z) Hom cont (A /Q, Z/2Z). Proof. We begin by defining the homomorphism rec (2). Suppose that γ Hom(G, Z/2Z) is nontrivial. Equivalently, we may choose a subgroup G G, of index 2. Equivalently, we may choose a quadratic extension K of Q.

4 4 MARTIN H. WEISSMAN Choose a generator of K as a quadratic extension. Namely, we can write K = Q( a), for an element nonsquare element a Q, uniquely determined up to Q 2. Suppose that x = (x v ) A. Then we may define: η(x) = v (a, x v ) This infinite product is well-defined, since for all but finitely many places, a Z v and x v Z v, and hence (a, x v ) v = 1 (if v 2, ). This follows from Theorem 1, of Chapter III of Serre. If x Q, then we have η(x) = 1, by the Product Formula (Theorem 3, of Chapter III of Serre). Hence η, originally a homomorphism from A to Z/2Z, descends to a homomorphism η from A /Q to Z/2Z. η depends only on a modulo squares in Q 2, and hence only on the original choice of γ. Hence, associated to every γ Hom(G, Z/2Z), we have a well-defined element: rec (2) (γ) = η : A /Q Z/2Z. Now, we show that rec (2) has an inverse: Suppose that we are given a nontrivial continuous homomorphism: η : A /Q Z/2Z. Then, η is uniquely determined by its restriction to Ẑ R, and it is trivial on Z = {±1}. It follows that η is uniquely determined by its restriction ˆη to Ẑ = p Z p. There exists a family of homomorphisms η p, one for each prime number, such that: ˆη(ˆx) = p η p (x p ), for every ˆx = (x p ) Ẑ. Moreover, there exists a finite set S, such that η p is nontrivial, if and only if p S. Let {a i } Z be a finite set of rational numbers, such that every square class in Z p, for every p S, is represented by at least one element a i. Let ɛ i,p = η p (a i ), for every prime number p. Let ɛ i, be the unique number, ±1, such that the product formula holds. Namely, we have: η(a i ) = ɛ i,v = 1. The three conditions of Theorem 4 of Chapter III of Serre are satisfied. Almost all of the ɛ i,p are equal to 1, since η p is trivial for all but finitely many p. The product formula holds by construction. Finally, for every v V, there exists an x v Q v with (a i, x v ) = ɛ i,v, by the non-degeneracy of the local Hilbert symbol. By Theorem 4 of Chapter III of Serre, there exists x Q, such that: ɛ i,v = (a i, x v ) v, for all i, v.

5 LECTURE NOTES, WEEK 7 MATH 222A, ALGEBRAIC NUMBER THEORY 5 Let η(y) = v (y v, x v ) v, for all y A. Then we see that η : A /Q Z/2Z is a well-defined homomorphism. As such, it is uniquely determined by its restriction to Ẑ R, and hence by the family of homomorphisms η p = (y p, x p ) p. Since η p is determined by its values on representatives for square classes, we have: η p = η p, for all p S. It follows that, for all i I, we have: η(a i ) = v (a i, x) By choosing the x Q more carefully (a subtlety that I haven t quite figured out), one should be able to further have: η(y) = v (y v, x v ) v, for all y A. The element x Q, which is not in Q 2 since η is nontrivial, yields a quadratic extension K/Q, and hence a homomorphism from G to Z/2Z. Exercise 1.8. Suppose that k is a field of characteristic zero. Let G = Gal( k/k), for some algebraic closure of k. Suppose that γ 1, γ 2 : G Z/2Z are two group homomorphisms with nontrivial image. Let H 1, H 2 be the kernels of γ 1, γ 2, respectively. Let K 1, K 2 be the quadratic extensions of k, corresponding to H 1, H 2, respectively. Choose generators x 1, x 2 k such that K i = k( x i ) for i = 1, 2. Let γ = γ 1 + γ 2. Let K be the extension of k corresponding to ker(γ). Prove that K = k( x 1 x 2 ). 361B Baskin, Department of Mathematics, University of California, Santa Cruz, CA address: weissman@ucsc.edu