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 ∈ 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 ∈ K, we know that σ −ζp has a non-trivial kernel. That is, there exists θ ∈ 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 ∈ K and θ∈ / K and so L = K(θ) = K γ where γ = θ ∈ K. This means that every Z/pZ-extension is adjoining a p-th root.

Constructive pseudo-proof: Take a random element z ∈ 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 ∈ 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→ ∈ 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 ∈/ 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 gψ(x) = ψ(gxg ) G know that × and acts by a. Then the image in
are just the elements where G ⊆ (Z/pZ) a ∈ 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. √ √ ×
×
×3 −1 −1 We know that G = (Z/3Z) = {±1} = hσi . Then we want x ∈ 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). √ √ 3 3 2.1. S3-extension of F . If n ∈ F , and −3 ∈/ 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 √ 3 L L(ζ3) E(ζ3, α)

Z/3Z

E E(ζ3)

Z/2Z F √ √ 3 the extension 3 is given by σ√ α ∼ in Galois cohomology. E(ζ3, α)/E(ζ3) σ 7→ 3 ∈ µ3 = Z/3Z √ √ α √ √ √ √ √ √ 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 √ √ √ α σ 3 α 3 trivially on √ so it is in the base eld. Similarly, √ = ζ and σ√ β = ζ or ζ2. Then our condition is either β 3 α 3 β √ √ 3 α 3 α2 σ xes √ or √ . 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] ∈ G acts by [a] x ≡ x−1 mod cubes. √ √ √ 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 ∈

E(ζ3) with some β conjugate to β. Example 2.5. extensions of S3 Q √  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 ∈ 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˜ √  Then -extensions of containing (for square) are S3 Q Q D D 6= −3 · √ √ √  3  L(ζ3) = Q D, −3, x √ where y ,
is not a cube. We identify -invariant under , so x = σy y ∈ Q −3D L G √ 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 √ Example 2.6. Let and . Suppose . Then
. Then extensions correspond to F = Fp E = Fp2 ζ3 ∈/ 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 ∈ 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 ∈ Qp √ The quadratic extensions are are either √
or √
. 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. -extensions of . We will do this by using the fact that it's a solvable group, rather than it is the A4 Q alternating group (can take splitting eld of polynomial with square discriminant otherwise). Then we need to nd E

2 (Z/2Z) −1
L = Q ζ9 + ζ9

Z/3Z Q √ We can then nd a quadratic subextension of L, which is some L ( α). We should be able to just take the Galois closure √ of L ( α) and will be ne. Let hσi = Gal (L/Q), then the Galois closure is √ √ √  E = L α, σα, σ2α ,

3 which is at most (Z/2Z) extension of L. The one we want is the one complement to the (i, i, i) subgroup. This is a 2 (Z/2Z) where σ acts by permutation of the coordinates, which is what we want. √ √ What we deduce, is that if × ×2 and then Gal ∼ . α 6= Q L N(α) = 1 (L ( α, σα) /Q) = A4

3. Dirichlet's Theorem and Units

If has sign then × ∼ r1+r2−1. If is Galois, then × is a Galois module. What can you say F/Q (r1, r2) OF = µF ⊕ Z F/Q OF about it?

 |G|−1 if is totally real × Q F OF ⊗ Q = . Q|G|/2−1 if F is complex Proof idea: log|·| × × × Y × Y × Y Y r1+r2 OF ,→ F → F ⊗ R = R × C → R ⊕ R = R . r1 r2 But it really lands into Rr1+r2−1 modulo scalar. By Minkowski, should we suitablly compact. Let be totally real. Then × |G| and this is still Galois equivariant, and is really just a map to . Ignoring F OF ,→ R R[G] the diagonal part, we get × ∼ OF ⊗ R = R[G]/R. The question is: do we also have the same isomorphism when we do ⊗Q. log|·| If F is complex, O× ⊗ → Q × → Q and this image is [G/ hci] / . This is essentially just the permutation F Q r2 C r2 R R R reputation, but complex conjugation acts trivially.

Example 3.1. Let F/Q be Galois and totally imaginary, with Galois group G. Let V be an irreducible representation of C [G].  0 if V = trivial ×
 µ (V ) = dim hom V, OF ⊗ C = dim V c=1 else (this works for F totally real too).

[L:K] Question: is ∼ ∼ (not obvious). It is true that ∼ ∼ [L:Q]. OL = OK [G] = OK OL = Z[G] = Z ∼ First: rank 1 modules of OK are classied by the class group. Rank 2, you have things like I ⊕ J = IJ ⊕ O. Basically, the classication of modules over are just n . OK OK ⊕ I CLASS FIELD THEORY NOTES 5

4. Parametrizing G-extensions

n Take the space (stack) A with acting by permutation. Then a ¯ -point, is all the -conjugates. This space G G,→ Sn Q G parametrizes polynomial with G-conjugates.

2 Example 4.1. Consider A , then Z/2Z /2 Q [x, y]Z Z = Q [xy, x + y] =∼ Q [u, v] . This essentially parametrizes all quadratic extensions. Similarly for G = Z/3Z,

/3 Q [x, y, z]Z Z = Q [x + y + z, xy + xz + yz, xyz, ∆ = (x − y)(x − z)(y − z)] 2 = Q [p, q, r, ∆] /∆ − (...). This is Sym2 Q ⊕ Vtaut ⊕ Vtaut ⊕ ...

n Example 4.2. Then A is parametrized by [r1, ..., rn] where ri's are symmetric polynomials. Sn Q

n Theorem 4.3. A Spec of something nitely presented. G = Let 1 → A → Γ → G → 1 be an exact sequence where A is abelian with exponent p. Assume G = Gal (K/Q) and Γ = Gal (L/Q), ∆ = Gal (K(ζ)/K). Want to construct × ×p
∆ A ⊆ K(ζp) /K(ζp) want the action of G on both sides to coincide. Need to understand this quotient as a Gal (K(ζ)/Q)-module. In general, let G = Gal (E/Q) and want to understand E×/E×p as a G-module.

Claim 4.4. There exists n × ×p. Fp [G] ⊆ E /E

Proof. Suppose splits completely in × so . Essentially, it follows from the fact that × generates p E , p = π1...π[E:Q] π1 ∈ E . Fp [G] 

Consider Γ = Z/4, which we know 1 → Z/2Z → Γ → Z/2Z → 1 doesn't split. The trick is to nd two Z/4Z-extensions. Their compositum is then an abelian degree 8 extension, which can't be Z/8Z (has two Z/4Z). Then it must be Z/4Z ⊕ Z/2Z, then take the xed. This is an example of: if we don't get things split in the correct way (the quadratic extension is not the one we wanted) we can do tricks with compositums.