CLASS FIELD THEORY NOTES HAO (BILLY) LEE Abstract. These are notes I took in class, taught by Professor Frank Calegari. I claim no credit to the originality of the contents of these notes. Nor do I claim that they are without errors, nor readable. 1. Galois Theory Let L=K be a nite Galois extension with Galois group G. Then L is a K[G]-module. Problem 1.1. What is L as a K[G]-module? Claim 1.2. It's just IndhGi or just . 1 1 K [G] Proof. By primitive element theorem, L = K[x] = K(α) for some α 2 L. 2. Kummer Theory Understanding Z=pZ extensions for a prime p. Case : contains a -th root of unity . Let be a -extension with Galois group . Since ∼ , 1 K p ζp L=K Z=pZ G = hσi L = K[G] we know the map σ : L ! L has characteristic polynomial xp − 1 and has all primitive roots of unity as eigenvalues (also from σ −1 6= 0). Since ζp 2 K, we know that σ −ζp has a non-trivial kernel. That is, there exists θ 2 L such that σθ = ζpθ for some θ 6= 0. p p p p σθ = (σθ) = (ζpθ) = θ : p 1=p p Then we get θ 2 K and θ2 = K and so L = K(θ) = K γ where γ = θ 2 K. This means that every Z=pZ-extension is adjoining a p-th root. Constructive pseudo-proof: Take a random element z 2 L. If z 6= 0 and σz = ζpz then we are done. Else, let −1 −2 2 1−p p−1 y = z + ζp σz + ζp σ z + ::: + ζp σ z then σy = ζy: Just have to make sure y 6= 0. × Case 2: K general (charK 6= p). Since the Galois group of K(ζ)=K is a subgroup of (Z=pZ) which is prime to p, we see that is still a extension. Therefore, we can reduce the problem to the previous case. L(ζp)=K(ζp) Z=pZ Let be the absolute Galois group of . We want to understand . GK K hom (GK ; Z=pZ) Claim 2.1. There exists an isomorphism (assuming ζp 2 K) × ×p ∼ ∼ K =K = hom (GK ; µp) = hom (GK ; Z=pZ) : However, the last isomorphism depends on the choice of root of unity (that maps to 1). The map is σ(γ1=p) γ 7! σ 7! 2 µ : γ1=p p Proof. We have exact sequence × p × 1 ! µp ! K¯ ! K¯ ! 1: 1 CLASS FIELD THEORY NOTES 2 Take invariance under GK . This gives × p × 1 1 × K ! K ! H (GK ; µp) ! H GK ; K¯ [p] ! 0: Since 1 ¯ × by Hilbert's theorem . Therefore, we have our desired isomorphism. H GK ; K = 0 90 In general (ζp 2= K), K(ζ )× hom (G ; =p ) ,! hom G ; =p ∼ hom G ; µ ∼ p : K Z Z K(ζp) Z Z = K(ζp) p = ×p K(ζp) Which Galois extensions are possible? We know that 1 ! GK(ζp) ! GK ! G ! 1 and the action of on is just lift then conjugate. Since elements of are -invariant, everything G GK(ζp) hom (GK ; Z=pZ) G lands in the -invariant (here −1 ). The next map however is not -equivariant. We G hom GK(ζp); Z=pZ g (x) = (gxg ) G know that × and acts by a. Then the image in are just the elements where G ⊆ (Z=pZ) a 2 G [a] ζ = ζ hom GK(ζp); µp [a] c = ac. a Claim 2.2. × ×p [a]x=x . hom (GK ; Z=pZ) = (K(ζp) =K(ζp) ) Example 2.3. Let K = Q and p = 3. p p × × ×3 −1 −1 We know that G = (Z=3Z) = {±1g = hσi : Then we want x 2 Q −3 =Q −3 where σx = x (ie. σx = x modulo cubes). For example, take −1 ( is of degree , then take invariance under conjugation). Then L = Q ζ9 + ζ9 Q (ζ9) 6 1 1 −1 3 3 Q(ζ9) = Q ζ9 + ζ9 ; ζ3 = Q ζ3; γ = Q ζ3; ζ3 : Now try −1 1=3 Q ζ7 + ζ7 ; ζ3 = Q ζ3; γ : We need 1=3 −1 −1 2 2 (so the form −1 as before). γ = ζ7 + ζ7 + ζ3 ζ7 + ζ7 z + ζ σz Now take the base eld to be some Q(t). One degree 3 extension is K(t) where K=Q is degree 3 (it's usually called the isotrivial extension. The way to think about it, is that if degree 3 extensions are viewed as a family, you can have a curve that curls over 3 times, or 3 lines that look the same). p p 3 3 2.1. S3-extension of F . If n 2 F , and −3 2= F , then x − n (if n is not a cube) has splitting eld F ( n; ζ3) has Galois group S3. These all have the property that the quadratic subeld is F (ζ3). Does there exists an S3 extension of F with arbitrary quadratic extension E=F . We have extensions p 3 L L(ζ3) E(ζ3; α) Z=3Z E E(ζ3) Z=2Z F p p 3 the extension 3 is given by σp α ∼ in Galois cohomology. E(ζ3; α)=E(ζ3) σ 7! 3 2 µ3 = Z=3Z p p α p p p p p p 3 3 When does E(ζ3)( α) = E (ζ3) β ? Well, if E( α) = E( β) then σ α = − α and σ β = − β so σ acts p p p α σ 3 α 3 trivially on p so it is in the base eld. Similarly, p = ζ and σp β = ζ or ζ2. Then our condition is either β 3 α 3 β p p 3 α 3 α2 σ xes p or p : 3 β 3 β This means that α = β · cube or α = β2 · cube. CLASS FIELD THEORY NOTES 3 We know that extensions of are parametrized by × ×3. Those that come from extensions Z=3Z E(ζ3) E(ζ3) =E(ζ3) Z=3Z of are those such that Gal × acts by a. E G = (E(ζ3)=E) ,! (Z=3Z) [a] x = x We know that Gal (E(ζ3)=F ) = Gal (F (ζ3)=F ) × Gal (E=F ) when E 6= F (ζ3). × ×3 Claim 2.4. An S3-extension of F contains a quadratic eld E 6= F (ζ3) i there is a line in E(ζ3) =E(ζ3) such that [a] 2 G acts by [a] x ≡ x−1 mod cubes. p p p Alternatively, if 3 is Galois, with Galois groups . Then Gal 2. 3 3 E (ζ3; α) =F S3 × Z=2Z (E(ζ3)=F ) = (Z=2Z) α ! β 2 E(ζ3) with some β conjugate to β. Example 2.5. extensions of S3 Q p Let . Special for (and doesn't happen for . 2 acts on × ×3, and want E = Q D p = 3 p > 3) ∆ = (Z=2Z) E(ζ3) =E(ζ3) × ×3 to nd x 2 E(ζ3) =E(ζ3) such that it is an eigenvector for ∆. Write ∆ = hσ1i × hσ2i. Let x~ be any lift, then x~ σ2x~ xσ~ 1σ2x~ (1 − [σ1]) (1 − [σ2]) x = ÷ = : σ1x~ σ2σ1x~ σ1xσ~ 2x~ p Then -extensions of containing (for square) are S3 Q Q D D 6= −3 · p p p 3 L(ζ3) = Q D; −3; x p where y , is not a cube. We identify -invariant under , so x = σy y 2 Q −3D L G p r y rσy L = D; 3 + 3 : Q σy y For example, taking the splitting eld of 3 3 1 and 3 y σy . x − 3x = θ + θ3 x − 3x = σy + y p Example 2.6. Let and . Suppose . Then . Then extensions correspond to F = Fp E = Fp2 ζ3 2= F E = Fp −3 S3 G=triv [−1]x=x−1 × ×3 and extensions correspond to × ×3 By looking at the order (without descending Fp2 =Fp2 Z=6Z Fp2 =Fp2 : to eigenspace) there are possibilities, and so just unique extension. Z=3Z 1 S3 σx=x−1 σx=x−1 Now suppose . Suppose . Then extensions correspond to × ×3 ∼ × ×3 ζ3 2 F p ≡ 1 mod 3 S3 Fp2 =Fp2 = Fp2 =Fp2 = trivial. Example 2.7. extension of and (so ). S3 Qp p ≡ 1 mod 3 ζ3 2 Qp p The quadratic extensions are are either p or p . We want to pick so that Qp Qp p ; Qp ( u) Qp up E σx=x−1 E×=E×3 is non-trivial: Recall that × ∼ × Z ∼ Z E = OE × πE = (1 + πEOE) × µE × πE: Then × E 2 (1 + mE) ×3 = µE/µE × 3 × Z=3Z: E (1 + mE) Since is cyclic of order divisible by , the quotient is isomorphic to . The expression of cubes can be written as µE 3 Z=3Z a power series, which will converge on 1 + mE. That is, 1=3 X 1=3 1 + uπk = unπkn : E E n Therefore, E× =∼ ( =3 )2 : E×3 Z Z Now, we need to see how Galois acts. In the Z , the Galois will change uniformizers, but the valuation does not change. πE That is, it does nothing to the quotient Z=3Z. CLASS FIELD THEORY NOTES 4 The is either × or × . In the rst case, Galois acts trivially because there's no new roots of unity. In the section µE Fp Fp2 case, we did in the previous example and found that the Galois group acts trivially. Hence, Galois acts trivially on × ×3. Therefore, there are no extensions of . E =E S3 Qp Example 2.8.