Math 223b : Algebraic Number Theory notes Alison Miller 1 January 22 1.1 Where we are and what’s next Last semester we covered local class field theory. The central theorem we proved was Theorem 1.1. If L/K is a finite Galois extension of local fields there exists a canonical map, the local Artin map × × ab θL/K : K /NL (Gal(L/K)) We proved this by methods of Galois cohomology, intepreting both sides as Tate ! cohomology groups. To prove this, we needed the following two lemmas: • H1(L/K, L×) =∼ 0 • H2(L/K, L×) is cyclic of order [L : K]. This semester: we’ll do the analogous thing for L and K global fields. We will need × × × to replace K with CK = AK /K . Then the analogues of the crucial lemmas are true, but harder to prove. Our agenda this semester: • start with discussion of global fields and adeles. • then: class formations, axiomatize the parts of the argument that are common to the global and local cases. • algebraic proofs of global class field theory • then we’ll take the analytic approach e.g. L-functions, class number formula, Ceb- otarev density, may mention the analytic proofs • finally talk about complex multiplication, elliptic curves. 1 1.2 Global fields, valuations, and adeles Recall: have a notion of a global field K. equivalent definitions: • every completion of K is a local field. • K is a finite extension of Q (number field) or of Fp((t)) (function field). The set of global fields is closed under taking finite extensions. We’ll primarily focus on the number field case in this class. Definition. A place v of a global field K is an equivalence class of absolute values on K. We say that v is finite / nonarchimedean if it comes from a discrete absolute value, ∼ otherwise v is infinite /archimedean (and Kv = R or C). We can pick out a distinguished element of this equivalence class, the normalized absolute value j · jv by requiring -1 jπvjv = jkvj 2 if Kv is nonarchimedean, and jajR = a, jajC = jaj . Recall from last semester: Theorem 1.2 (Product Formula). If a 2 K, then v jajv = 1. Sketch of how this was done last semester. Check forQ K = Q, Fp[t], then show that if the product formula holds for K, it holds for any finite extension L/K.( v0 extends v jajv0 = jajv.) Q + Remark. Kv has a Haar measure, unique up to scaling (if v is finite, normalize by µ(Ov) = 1, if v is infinite use the standard normalization on R and C.) + × For any measurable set E ⊂ Kv and any K 2 Kv we have µ(aE) = jajvµ(E). Definition. Suppose that fXigi2I are topological groups, and for each i we have a sub- group Yi ⊂ Xi. Then the restricted topological product of the Xi with respect to the Yi is given by Xi = f(xi)i2I j xi 2 Yi for all but finitely many ig. i aY The group structure here is clear. We take as a basis of open sets those sets of the form Ui × Yi i2S Y Yi2/S where S ⊂ I is an arbitrary finite set, and Ui ⊂ Xi are arbitrary open sets. (Note this is not the same as the subspace topology coming from viewing i Xi as a subspace of X .) i i `Q Q 2 Then we define + + AK = AK = Kv v aY + with respect to the family of subgroups Ov and × × AK = Kv v aY × with respect to the family of subgroups Ov . + × Exercise: AK = AK is a topological ring with group of units AK . 2 January 24 2.1 Adeles Definition. If K is a number field, define the topological ring AK as the restricted topo- logical product v Kv of the Kv with respect to the Ov. This ring has open cover given by the sets `Q AK,S = Kv × Ov v2S Y Yv2/S for any finite set S containing the infinite places, and the subspace topology on each AK,S ⊂ AK agrees with the product topology. × × Likewise define AK as v Kv , where now the restricted product is with respect to O×. Define A× likewise. v K,S `Q × Here AK is equal to the group of units of AK, so the notation is consistent. × Caution! The topology on AK is not the subspace topology inherited from AK: indeed, it shouldn’t be, because the map x x-1 is not continuous in the subspace topology. In general, if you have a topological ring R, the correct topology to put on R× × -1 comes from the embedding of R , R × R given! by x 7 (x, x ). × × × (Note though that if R = Kv or Ov , this topology on R agrees with the subspace topology coming from R. It’s only! because of the infinite! restricted product that we have issues.) Proposition 2.1. AK is a locally compact topological ring. Proof. all AK,S are locally compact by Tychonoff. + Hence AK has a Haar measure also, which can be described as the product of the local Haar measures. + + × × Note that K embeds into AK and K embeds into AK , diagonally. 3 ∼ Proposition 2.2. If L/K is a finite extension, then AL = AK ⊗K L as topological rings. (Here ∼ n ∼ n we topologize AK ⊗K L as follows: by choosing a basis, identify L = K , so AK ⊗K L = AK, and n use the product topology on AK. One can check that this doesn’t depend on choice of basis.) ∼ L Proof. (Sketch): Use fact that Kv ⊗K L = v0 extends v Lv0 . See Chapter 2, Section 14 of Cassels-Fröhlich for details. (In fact, this is also an isomorphism of Gal(L/K)-modules set up correctly.) + + + + Proposition 2.3. For K a global field, K is discrete (hence closed) in AK and AK /K is compact. Proof. By the previous proposition, it’s enough to check this for K = Q and K = Fp(t). We’ll do Q; the proof for Fp(t) is similar. Discrete: the set U = v6=R Zv × (-1, 1) is open and Q \ U = Z \ (-1, 1) = f0g. Cocompact: Let D = Zv × [-1/2, 1/2]. We wish to show that D + Q = A . Let Qv6=R K a 2 A+ be arbitrary: we want to show that a is congruent mod Q+ to some element of K Q D. Then there are finitely many primes p such that (a)p 2/ Zp. For each such p, there exists rp 2 Q such that rp ≡ ap (mod Z)p. Then let r = rp 2 Q. Then a - r 2 Zv × R. By subtracting off an appropriate s 2 Z, get a - r - s 2 Zv × v6=R P v6=R [-1/2, 1/2], as desired. Q Q 3 January 26 + + Last time, we showed that AK /K is compact. + Note that AK has a Haar measure. It can be described by µ( Ev) = µv(Ev) v v Y Y where Ev ⊂ Kv are measurable sets with Ev = Ov for almost all v. + + + Since K is discrete inside AK , we can push forward the measure on AK to get a + + measure on AK /K . To be more precise, let D be a measurable fundamental domain + + for the action of K on AK . (For example with K = Q we can take D = v6=R Zv × [-1/2, 1/2).) Then for E ⊂ A+ Borel define K Q -1 µAK/K(E) = µAK (π (E) \ D) where π : AK AK/K is the projection map. × Definition. If a = (av) 2 A , then we define the content c(a) = javj (this is defined ! K v because javj = 1 for all but finitely many v. Q 4 × >0 Exercise: The map c : AK R is a continuous homomorphism. Proposition 3.1. For any measurable set E ⊂ A+, and any a 2 A×, then µ(aE) = aµ(E). ! K K Proof. This follows from the description of µ as a product of local measures, and the local fact that µv(aEv) = jajvµv(Ev). As a corollary, we get another proof of the product formula: if a 2 K×, then mul- + tiplication by a gives an automorphism of AK which scales the Haar measure by c(a), and also sends K+ to itself. Hence multiplication by a scales the Haar measure on the + + quotient AK/K by c(a). On the other hand, µ(AK/K ) has finite measure, so c(a) must be 1. × × Now we move to the multiplicative situation: To show K is discrete in AK , use the × injection AK AK × AK. × × × × >0 Next question: is AK /K compact? No: c is a continuous map from AK /K R , and the latter! is not compact. However, that’s the only obstruction: ! Definition. Let 1 × × >0 AK = ker(c : AK /K R ) be the subgroup of multiplicative adeles of content 1. ! 1 × × 1 Then AK is a closed subgroup of AK , and K is discrete in AK. 1 × We plan to show that AK/K is compact, and use this to deduce the finiteness of class group and Dirichlet’s units theorem. But first we’re going to develop the adelic version of Minkowski’s theorem. × + Definition. For a 2 AK , let Sa ⊂ AK be the subset defined by + Sa = fx 2 AK j jxvjv ≤ javjv for all vg Theorem 3.2 (Adelic Minkowski). There exists a constant C > 1, depending on K, such that × × for every a = (av) 2 AK with c(a) > C, there exists some x 2 K \ Sa (in particular, x 6= 0).