Math 223a - Algebraic Number Theory€¦ · Math 223a - Algebraic Number Theory Taught by Alison...

Math 223a - Algebraic Number Theory

Taught by Alison Beth MillerNotes by Dongryul Kim

Fall 2017

This course was taught by Alison Beth Miller. The lectures were on Tuesdaysand Thursdays at 11:30–1, and the main textbook was Algebraic number theoryby Cassels and Frolich. There were weekly problem sets and a final paper, andthere were 15 students enrolled. The course assistant was Lin Chen. Lecturenotes can be found online on the course website.

Contents

1 August 31, 2017 41.1 Historic perspective: class fields . . . . . . . . . . . . . . . . . . . 41.2 Frobenius elements . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Infinite field extensions and their Galois groups . . . . . . . . . . 5

2 September 5, 2017 72.1 Overview of global class field theory . . . . . . . . . . . . . . . . 72.2 Overview of local class field theory . . . . . . . . . . . . . . . . . 82.3 Local field: evaluations . . . . . . . . . . . . . . . . . . . . . . . . 92.4 Completion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3 September 7, 2017 113.1 Hensel’s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.2 Extending absolute values . . . . . . . . . . . . . . . . . . . . . . 13

4 September 12, 2017 144.1 Ramification and inertia . . . . . . . . . . . . . . . . . . . . . . . 144.2 Extending absolute values II . . . . . . . . . . . . . . . . . . . . . 144.3 Local fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

5 September 14, 2017 175.1 Multiplicative group of local fields . . . . . . . . . . . . . . . . . 175.2 Exponential and logarithm . . . . . . . . . . . . . . . . . . . . . 185.3 Unramified extensions . . . . . . . . . . . . . . . . . . . . . . . . 19

1 Last Update: November 4, 2018

http://www.math.harvard.edu/~abmiller/

https://canvas.harvard.edu/courses/30968

6 September 19, 2017 216.1 Artin map for unramified extensions . . . . . . . . . . . . . . . . 226.2 Decomposition group . . . . . . . . . . . . . . . . . . . . . . . . . 236.3 Inertia group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

7 September 21, 2017 257.1 Totally ramified extensions of local fields . . . . . . . . . . . . . . 257.2 Ramification groups . . . . . . . . . . . . . . . . . . . . . . . . . 267.3 Tamely ramified extensions . . . . . . . . . . . . . . . . . . . . . 27

8 September 26, 2017 298.1 Category of G-modules . . . . . . . . . . . . . . . . . . . . . . . . 29

9 September 28, 2017 339.1 Group cohomology . . . . . . . . . . . . . . . . . . . . . . . . . . 339.2 Standard bar resolution . . . . . . . . . . . . . . . . . . . . . . . 34

10 October 3, 2017 3610.1 Explicit descriptions of cohomology . . . . . . . . . . . . . . . . . 3610.2 The first cohomology and torsors . . . . . . . . . . . . . . . . . . 3710.3 Group homology . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

11 October 5, 2017 4011.1 Changing the group . . . . . . . . . . . . . . . . . . . . . . . . . 4111.2 Inflation-restriction exact sequence . . . . . . . . . . . . . . . . . 42

12 October 10, 2017 4412.1 Tate cohomology . . . . . . . . . . . . . . . . . . . . . . . . . . . 4412.2 Restriction and corestriction . . . . . . . . . . . . . . . . . . . . . 46

13 October 12, 2017 4813.1 Cup products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4813.2 Galois cohomology . . . . . . . . . . . . . . . . . . . . . . . . . . 4913.3 Cohomology of profinite groups . . . . . . . . . . . . . . . . . . . 50

14 October 17, 2017 5214.1 Kummer theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5214.2 Second Galois cohomology of unramified extensions . . . . . . . . 5314.3 Some diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

15 October 19, 2017 5615.1 Second Galois cohomology of finite extensions . . . . . . . . . . . 5615.2 Estimating the size . . . . . . . . . . . . . . . . . . . . . . . . . . 5715.3 Brauer group of a local field . . . . . . . . . . . . . . . . . . . . . 5815.4 Tate’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

2

16 October 24, 2017 6116.1 Reciprocity map . . . . . . . . . . . . . . . . . . . . . . . . . . . 6216.2 Group of universal norms . . . . . . . . . . . . . . . . . . . . . . 64

17 October 26, 2017 6517.1 Normic subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . 6517.2 Quadratic reciprocity . . . . . . . . . . . . . . . . . . . . . . . . . 6617.3 Reciprocity map on units . . . . . . . . . . . . . . . . . . . . . . 68

18 October 31, 2017 6918.1 Affine group schemes . . . . . . . . . . . . . . . . . . . . . . . . . 6918.2 Formal groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

19 November 2, 2017 7319.1 Morphisms between formal groups . . . . . . . . . . . . . . . . . 7319.2 Lubin–Tate theory . . . . . . . . . . . . . . . . . . . . . . . . . . 74

20 November 7, 2017 7620.1 Torsion points in a formal group . . . . . . . . . . . . . . . . . . 7620.2 Field obtained by adjoining torsion points . . . . . . . . . . . . . 78

21 November 9, 2017 8021.1 Characterization of the Artin map . . . . . . . . . . . . . . . . . 8021.2 Independence on the uniformizers . . . . . . . . . . . . . . . . . . 81

22 November 14, 2017 8322.1 Artin map and the ramification filtration . . . . . . . . . . . . . 8422.2 Brauer group and central simple algebras . . . . . . . . . . . . . 86

23 November 16, 2017 8723.1 Tensor product of central simple algebras . . . . . . . . . . . . . 8823.2 Structure theorem for central simple algebras . . . . . . . . . . . 89

24 November 21, 2017 9124.1 Modules over simple algebras . . . . . . . . . . . . . . . . . . . . 9124.2 Extension of base fields . . . . . . . . . . . . . . . . . . . . . . . 92

25 November 28, 2017 9525.1 Noether–Skolem theorem . . . . . . . . . . . . . . . . . . . . . . 9525.2 Classification of central simple algebras . . . . . . . . . . . . . . 96

26 November 30, 2017 9826.1 Fields with trivial Brauer group . . . . . . . . . . . . . . . . . . . 9826.2 Brauer groups of local fields . . . . . . . . . . . . . . . . . . . . . 100

3

Math 223a Notes 4

1 August 31, 2017

In 223a we will be studying local class field theory and in 223b global class fieldtheory. For local class field theory, we are going to be proving stuff about Qpand their finite extensions. There is another local field, Fp((t)), which may bemore comfortable depending on your background. The main tool we will use toanalyze these is Galois cohomology.

An example of a global field is Q, and we will of course look at their finiteextensions, which are also called number fields. The polynomials Fp(t) is also aglobal field. The rings Z or Fp[t] have many prime ideals, while Zp and Fp[[t]]have few primes.

The textbook we are going to use is Cassels–Frohlich, which is actually aset of conference notes. This also includes Tate’s thesis. The book has a fewerrors and there is an errata online. Other recommended books are Neukirch’sAlgebraic Number Theory and Class Field Theory.

1.1 Historic perspective: class fields

Suppose we have a finite Galois extension L/Q. In Q there is a natural ring Zand in L there is also a natural ring OL. One thing people were interested inwas to take (p), lift it to pOL, and see how this ideal factorizes. In the case ofa Galois extension, this becomes

pOl = pe1 · · · per

and Gal(L/Q) acts transitively on the set {p1, . . . , pr}.One thing you should know is that e = 1 except for finitely many p. (These

are called ramified primes.) For p unramified, the data of r is called the “splittingtype of p”.

Example 1.1. Suppose L = Q[√n] is quadratic. There can be two splitting

types, and either pOL = p1 or pOL = p1p2. How can you tell which case youare in?

For simplicity assume that n 6≡ 1 (mod 4). We are trying to check whetherpZ[√n] is prime in Z[

√n], or equivalently, whether (Z/p)[

√n] is an integral

domain. Quadratic reciprocity then tells us that this only depends on p modulo4n.

Definition 1.2. A Galois number field L/Q is a class field if there exists onN such that the splitting type of p depends only on p modulo N .

For example, quadratic extensions are class fields, but general cubic exten-sions are not. If OL = Z[α], where the minimal polynomial of α is f(x), thenthe splitting type of p in L is given by the number of irreducible factors of f in(Z/pZ)[x].

Example 1.3. If L = Q(ζn), then OL = Z[ζn] so the splitting type is relatedto the factorization of Φn(x) over Z/pZ. You can use this to show that this isa class field.

Math 223a Notes 5

Here is the big main theorem people proved in class field theory over Q.

Theorem 1.4. The following are equivalent for a Galois extension L/Q:

• L is a class field.

• L/Q is abelian.

• L is contained in some Q(ζn).

The equivalence of the second and third is also called the Kronecker–Webertheorem. The reason this is not the end of the story is that we want to generalizethis to general number fields or function fields.

1.2 Frobenius elements

We want to build something called an Artin map over Q. Return to pOL =p1 · · · pr for an unramified prime p, and let p = p1. We define the decomposi-tion group

Dp = stabilzer of p ⊆ Gal(L/Q).

By transitivity of the action, we will know |Dp| = [L : Q]/r.Let ` = OL/p be the finite extension of Z/p. We have a homomorphism

ϕ : Dp → Gal(`/Fp)

Theorem 1.5. The map ϕ is injective if and only if p is unframified. The mapϕ is always surjective.

Note that Gal(`/Fp) is always cyclic and so is generated by F : a 7→ ap. SoF has a unique lift to a Frobenius element Frobp ∈ Dp. If I had chosen someother p′ = gp for g ∈ Gal(L/Q), then Frobp′ = g Frobp g

−1. This means thatFrob is well-defined a conjugacy class. In particular, if L/Q is abelian, we canjust write Frobp ∈ Gal(L/Q).

Example 1.6. Let L = Q(ζn) and OL = Z[ζn]. Take any p ∈ Z that isunramified (in this case, relatively prime to 2n). There is an isomorphism

Gal(L/Q)∼=−→ (Z/nZ)×; g 7→ a such that g(ζn) = ζan.

Now there is going to be a Frobenius element Frobp(ζ) ≡ ζp (mod p). Then ithas to be Frobp = p ∈ (Z/nZ)×.

1.3 Infinite field extensions and their Galois groups

We can state the Kronecker–Weber theorem in an alternative form as

Qab = Q[ζ∞] =⋃n≥1

Q[ζn].

Here, Qab is the maximal abelian extension of Q.

Math 223a Notes 6

We can try to look at the Galois group Gal(Qab/Q), which is an infinitegroup. But these are related to finite Galois groups by restriction maps

Gal(Qab/Q)� Gal(Q[ζn]/Q) ∼= (Z/nZ)×.

So we can understand this group by considering the map

Gal(Qab/Q) ↪→∏n

Gal(Q(ζn)/Q).

Injectivity comes from the fact that Q(ζn) fills Qab. But this is not surjectivebecause gn ∈ Gal(Q(ζn)/Q) has to restrict to gn′ ∈ Gal(Q(ζn′)/Q) if n′ | n. Sothis Galois group is the inverse limit

Gal(Qab/Q) ∼= lim←−n

Gal(Q(ζn)/Q) ∼= lim←−n

(Z/nZ)× ∼= Z×,

where Z = lim←−n(Z/nZ). Actually these are topological groups and topologicalrings, but we’ll get there.

Definition 1.7. The p-adic integers Zp is defined as

Zp = lim←−i

(Z/piZ).

These are sort of like infinite series c0 + c1p + c2p2 + · · · . The following

proposition follows from the Chinese remainder theorem.

Proposition 1.8. Z ∼=∏p Zp.

Thus ∏p

Z×p ∼= Gal(Qab/Q).

This is not true for other general fields. So we are going to define a crazy ringcalled the adele ring. This is

AQ,fin =

{(ap) ∈

∏p

Qp : all but finitely many ap are in Zp}, AQ = AQ,fin×R.

We can extend the isomorphism to

AQ → AQ/(Q× × R>0) ∼=∏p

Z×p ∼= Gal(Qab/Q).

This is called the Artin map.

Math 223a Notes 7

2 September 5, 2017

2.1 Overview of global class field theory

At the end of last class, we computed the Galois group of the maximal abelianextension of Q:

Gal(Qab/Q) ∼=∏p

Zp ∼= AQ/Q× × R>0.

HereAQ = {(xp)p, (x∞) : xp ∈ Zp for all but finitely p}.

The group A×Q is actually a topological group, with basis

∏p1,...,pr

Upi ×∏

all other p

Z×p × UR,

where Upi ⊆ Q×pi and UR ⊆ R× are open sets. The fact that θ : A×Q →Gal(Qab/Q) contains Q× in the kernel is called Artin reciprocity, and impliesquadratic reciprocity.

For K a number field and a prime p ⊆ OK , there is the completion Kp. Ingeneral, we define

A×K,fin = {(xp) ∈∏p

K×p : xp ∈ O×p for all but finitely p},

A×K,inf =∏K↪→R

R× ×∏K↪→C

C×,

A×K = A×K,fin × A×K,inf .

Again, you can prove that

Gal(Kab/K) ∼= A×K/K× · (AK)0,

where (AK)0 is the connected component of the identity, and the connectedcomponent of 1 of

∏R>0 ×

∏C×. This isomorphism is the main theorem of

global class field theory. If we define the adelic class group as Ck = A×K/K×,then this is also isomorphic to CK/(CK)0.

For finite extensions L/K, there is a reciprocity map

θL/K : CK/NCL∼=−→ Gal(L/K),

where NCL is the norm of CL. Then finite abelian extensions L/K correspondsto open finite index subgroups of CK , with L mapping to NCL.

So in global class field theory, we are going to construct the Artin mapθ : AK → Gal(L/K). Then we will show Artin reciprocity, and show the map issurjective. At the end of the semester, we are going to know how to constructthe Artin map.

Math 223a Notes 8

2.2 Overview of local class field theory

Let K be a local field, e.g., Qp or Fp((t)). Then there exists a local Artin map

θK : K× → Gal(Kab/K).

This is not an isomorphism, but it is very close to; it is an injection with denseimage. As with the global case, if L/K is a finite abelian extension,

θL/K : K×/NL×∼=−→ Gal(L/K)

is now an isomorphism. So again we have the correspondence between finiteabelian extensions L/K and finite index subgroups of K×, which is given byLN×.

Suppose K is a number field, and take a prime p ⊆ OK . Then Kp is a localfield and so we have the local Artin map

θKp: K×p → Gal(Kab

p /Kp).

On the other hand, we have the global Artin map

θK : A×K → Gal(Kab/K).

If L/K is abelian, then we pick a prime p′ ⊆ L lying over p, and then Lp′/Kp

is an extension and we get a map

Gal(Lp′/Kp) ↪→ Gal(L/K),

which is actually an isomorphism onto the decomposition group Dp′ . Puttingthese together, we get an injection

Gal(Kabp /Kp) ↪→ Gal(Kab/K),

where the image is a decomposition group.As we hope, the diagram

K×p Gal(Kabp /Kp)

A×K Gal(Kab/K)

θKp

θK

commutes. Because K×p generates A×K topologically, we will be able to use thisto construct θK .

In this course, we will cover

• basics of local fields

• ramification

• Galois cohomology

• proofs of local class field theory

• Lubi–Tate theory (explicit local class field theory)

• the Brauer group

• applications to global class field theory

Math 223a Notes 9

2.3 Local field: evaluations

Now I want to do this from scratch. The motivation is that I want to generalizeQp, Fp((t)). So we want to give a definition of local fields so that all local fieldsare finite extensions of Qp or Fp((t)).

Definition 2.1. A valuation on a field K is a map v : K → Z ∪ {∞} suchthat

(a) v(0) =∞,

(b) v : K× → Z is a group homomorphism,

(c) v(x+ y) ≥ min(v(x), v(y)).

Example 2.2. On Q, define vp(x) as the power of p in the prime factorizationof x. This is valuation.

Example 2.3. Let O be a Dedekind domain, i.e., an integral domain withunique factorization of ideals. For p a prime ideal, define vp(x) to be the powerof p in the prime factorization of (x) as a fractional ideal.

Definition 2.4. An absolute value on K is a map |−| : K → R≥0 such that

(a) |0| = 0,

(b) |−| : K× → R0 is a group homomorphism,

(c) |x+ y| ≤ |x|+ |y|.

If v is a valuation, then |x|v = av(x) for a < 1 is an absolute value. This infact satisfies

(c′) |x+ y| ≤ max(|x|, |y|).

These are called non-archemedian absolute values.Note that any embedding i : K ↪→ R or i : K ↪→ C will give an absolute

value |x| = |i(x)|. The nice thing that happens is that it is possible to classifyall absolute values.

Theorem 2.5 (Ostrowski). Any absolute value of Q is equivalent to |−|p forsome p or to |−|R coming from the embedding Q ↪→ R.

Here, we say that |−|1 ∼ |−|2 if |−|1 = |−|a2 for some a ∈ R>0.

Sketch of proof. If |−|1 is an absolute value, consider the set

{x ∈ Z : |x| < 1},

which turns out to be a prime ideal. Either it is (0), in which case the valuationis |−|R, or (p), in which case the valuation is equivalent to |−|p.

Math 223a Notes 10

Example 2.6. Consider Fp(t). Here there are choices for the ring of integersin this field, but I’m arbitrarily going to pick O = Fp[t]. Now every prime idealO gives a valuation, and also f(t) 7→ −deg(f(t)) is another valuation. There isan analogue Ostrowski’s theorem, which says that these are all the valuations,and you can exponentiate them to get all absolute values.

Definition 2.7. A discrete valuation ring, or DVR for short, is a local PIDthat is not a field.

If v is a (surjective) valuation of K, the set

Ov = {x ∈ K : v(x) ≥ 0}

is a discrete valuation ring with maximal ideal pv = {x ∈ k : v(x) ≥ 1} = (π) forany π with v(π) = 1. Conversely, if O is a discrete valuation ring with maximalideal p, then vp on K = Frac(O) is going to give O.

2.4 Completion

If K is a field with absolute value |−|, then define the completion K as thetopological completion of K as a metric space. (If |−| comes from a valuationv, then we write this as Kv.) Then K is a complete topological field.

Definition 2.8. A place of K is an equivalence class of absolute values of K:“finite places” come from valuations and “infinite places” come from embeddingsinto R or C. We use the notation v for a place, and then write Kv for thecompletion at the place v.

Using this, we can formally define

AK =

{(xv) ∈

∏v places

Kv : (xv) ∈ O×v for all but finite v

}.

Math 223a Notes 11

3 September 7, 2017

We finished by talking about absolute values and valuations, and making thefield complete. Now we are going to talk about fields that are complete.

First let us talk about non-archemedian absolute values, i.e., absolute valueswith

|x+ y| ≤ max{|x|, |y|}.

Theorem 3.1. If K is complete with respect to an archemedian absolute value,then K ∼= R or K ∼= C.

Now let’s assume thatK is a field complete with respect to a discrete absolutevalue, that is, im|−| is a discrete subgroup of R>0. Because of the previoustheorem, |−| is non-archemedian, and this must come from exponentiating avaluation. That is, |x| = c−v(x) for some c > 1.

Now the subringO = {a ∈ K : |a| ≤ 1}

is a DVR, and it is both open and closed in K. The maximal ideal is

p = πO = {a ∈ K : |a| < 1},

and we can cover O by cosets of p. We can do the same thing inside the cosetsof p, and write p as the disjoint union p + a =

∐b(p

2 + b). The upshot of thisis that

O = lim←−n

O/pnO = lim←−n

O/πnO,

by completeness of K.Generally, let O be any Dedekind domain, p ⊆ O be a prime. Let K =

FracO and K be the completion with respect to vp. If O and p are the completed

ring and prime ideal, then O/pi ∼= O/pi. We can thus write

O = lim←−O/pi = lim←−O/p

i.

3.1 Hensel’s lemma

Let K be complete with respect to a discrete absolute value, and L/K be afinite extension. Is it always possible to extend |−|K to L? Is it unique, anddoes this make L complete? The answer to all three questions is yes.

Lemma 3.2 (Hensel). Let O be a complete DVR with prime p = (π) andk = O/p. Suppose f ∈ O[x] and let f ∈ k[x] be the reduction. If f factors asf = gh, with g and h relatively prime, then there exists a factorization f = ghsuch that g reduces to g and h reduces to h.

Without the relatively prime condition, a counterexample is x2−p in Qp[x].

Math 223a Notes 12

Proof. The idea is to use O = lim←−O/πr. It suffices to find gr, hr ∈ (O/πr)[x]

such that grhr = f (mod πr), and gr′ reduces to gr for r′ > r. Actually we alsoneed degree conditions deg gr = dg = deg g and deg hr ≤ dh = deg f − deg g,because the degree might blow up.

We induct on r. If r = 1, we can just take g = g1 and h = h1. Assume wehave gr, hr ∈ (O/πr)[x] which works. Choose arbitrary lifts g′r+1 and h′r+1. Wewant to correct this by writing

gr+1 = g′r+1 + πra, hr+1 = h′r+1 + πrb.

Then what we need is

ah′r+1 + bg′r+1 =g′r+1h

′r+1 − fπr

(mod π).

We are now working in k[x], and we are trying to solve

ah+ bg = c

for some c ∈ k[x]. This we can do by linear algebra. Let Pm be the space ofpolynomials of degree at most m in k[x]. Then the map

Pdg × Pdh → Pdg+dh = Pdeg f ; (a, b) 7→ ah+ bg.

The kernel is generated by (g,−h), and so has dimension 1. This shows thatthe map is surjective.

As an exercise, try to show that the polynomials g and h are unique up tomultiplication by elements of O×.

Corollary 3.3. If f ∈ O[x], and there exists some a ∈ k such that f(a) = 0but f ′(a) 6= 0, then a lifts to a unique root a ∈ O of f .

Proof. Use Hensel’s lemma on f = (x− a)g.

Example 3.4. Let K = Qp and O = Zp. Take f(x) = xp−1 − 1 so thatf(x) = (x− 1) · · · (x− p+ 1). Then xp−1 − 1 has p− 1 roots in Z×p , and theseare distinct mod p.

Example 3.5. Let K = Qp and O = Zp with p odd. Take f(x) = x2−a wherea ∈ Z×p . If a is not a square mod p, then f(x) is irreducible, and if a is a squaremod p, then f(x) has two distinct solutions. Therefore, a ∈ Z×p is a square ifand only if a is square in Fp.

More generally, if a ∈ Q×p and a = piu where u ∈ Z×p , a is a square if andonly if i is even and u is a square in Fp.

Lemma 3.6. Let K be a field complete with respect to a valuation v. If f =anx

n+· · ·+a0x0 ∈ K[x] is irreducible, then min0≤i≤n v(ai) = min(v(an), v(a0)).

Math 223a Notes 13

Proof. By rescaling, we may assume that min0≤i≤n v(ai) = 0 so that f ∈ O[x].Let m be the maximal number with v(am) = 0, so that deg f = m. Apply

Hensel to the factorization f = f · 1, so that we have a factorization of f into adegree m and a degree n−m polynomial. Then m = 0 or m = n.

This generalizes to the theory of Newton polygons. Here you apply thetheorem to stuff like f(πkx).

3.2 Extending absolute values

Theorem 3.7. Let K be a field complete respect to a discrete absolute valuation|−|K . If L/K is finite with [L : K] = n, then there exists a unique extension of|−|K to an absolute value |−|L, given by

|a|L = n

√|NL/Ka|K .

Furthermore, L is complete with respect to |a|L.

Before I prove this I want to say a couple of things about norms. Here we arenot assuming L/K is Galois, so we can’t just multiply the Galois conjugates.

Definition 3.8. If L/K is a finite extension, a ∈ L, then NL/K = detK(ma :L→ L).

In practice, what you use is

Proposition 3.9. If a has minimal polynomial xm+cm−1xm−1+· · ·+c0 ∈ K[x],

then χma = f(x)n/m and NL/K(a) = cn/m0 .

Proof of Theorem 3.7. We first check that |a+b|L ≤ max{|a|L+ |b|L}. Withoutloss of generality, let max = |a|L = 1. It suffices to show that |b|L ≤ 1 implies|1 + b|L ≤ 1.

Let f(x) = xm + cm−1xm−1 + · · ·+ c0 ∈ O[x] be the minimal polynomial of

b. Then |c0| = |b|m ≤ 1, and the previous lemma implies |ci| ≤ 1 for all i. Thenf(x) ∈ O[x], and so the minimal polynomial of b + 1 is also in O[x]. That is,NL/K(b+ 1) ∈ O and |b+ 1|L ≤ 1.

We now have to check uniqueness and completeness. Suppose I have |−|Land |−|′L. We can view them as a norms on L as a finite-dimensional vectorspace K. It is a theorem that any two such norms induce the same topology.Then |a|L < 1 if and only if {an} → 0 in |−|L if and only if {an} → 0 in |−|′L ifand only if |a|′L < 1.

For completeness, we can again use the fact that any norm induce the sametopology. Take the max norm on L ∼= Kn. This topological space is complete,and so L is complete with respect to |−|L.

Math 223a Notes 14

4 September 12, 2017

We’ve just shown that every absolute value extends to an absolute value on afinite extension. Assume that L/K is a finite extension K is complete withrespect to |−|. We have shown there exists a unique extension of |−| to L, andthat L is complete with respect to |−|.

4.1 Ramification and inertia

Now let OK and OL be the discrete valuation rings, with prime ideals pK =(πK) ⊆ OK and pL = (πL) ⊆ OL. Now πKOL ⊆ OL is an ideal, and it has tobe

πKOL = (πL)e

for some e. This e is called the ramification index. This is the index of

im(|−| : K× → R>0) ⊆ im(|−| : L× → R>0),

because the images are generated by |πK | and |πL|. We can also define theinertia degree as

f = fL/K = [OL/(πL) : OK/(πK)].

Example 4.1. Let p be odd and K = Qq. Consider the extension Qp[√u]

where u ∈ Z×p is not a square mod p. Here it is clear that e = 1 and f = 2,since p = πK = πL and so ` = Fp[

√u].

Example 4.2. Let p be odd and K = Qq. Consider the extension L = Qp(√p),

OL = Zp[√p]. Here πL =

√p and so e = 2. But ` = Fp and so f = 1.

Theorem 4.3. If L/K is as above, then eL/KfL/K = [L : K] = n.

Proof. The quotient OL/πKOL is a vector space over k = OK/πKOK , and wecompute its dimension in two different ways.

Firstly, OL is a free OK-module and is of rank [L : K] = n. So OL/πKOLis a free k-module of rank [L : K], and hence dimkOL/πK = n.

Secondly, we have πKOL = πeLOL and so

dimkOL/πeOL = dimkOL/πLOL + dimk πLOL/π2LOL + · · · = edimk ` = ef.

This shows that n = ef .

4.2 Extending absolute values II

Let K be a field with usual discrete non-archemedian absolute value |−|K . Butlet us not assume that K is complete. Now suppose that L/K is finite. Howmany ways can we |−|K extended to |−|L?

Consider the case K = FracOK where OK is a Dedekind domain. Assumethat |−|K = |−|p for some p ⊆ OK . One way of extending |−|p to L is to factor

Math 223a Notes 15

p ⊆ OL. (Here OL is the integral closure of OK in L, and is thus Dedekind.)Let p′ ⊆ OL be a factor of pOL with exponent e. Then for each a ∈ K, we have

vp′(a) = evp(a).

After appropriate normalization, |−|p′ extends |−|p.

Theorem 4.4. Any absolute value |−|′ of L extending |−|p is equivalent to |−|p′for some p′ | p.

Proof. First we will show that |−| is discrete. Look at the completions K andL, where the both valuations will be discrete. This implies that |−|′ is discreteand non-archemedian.

ConsiderOL,v′ = {a ∈ L : |a|′ ≤ 1}

which is a DVR. This ring contains OK and so contains OL. Let

p′ = pL,v′ ∩ OL,

which is a prime ideal.Inside L we have (OL)p′ ⊆ OL,v′ , both DVRs. But if A ⊆ B ( L are DVRs,

then A = B. So OL,v′ = (OL)p′ and the valuations on them must agree.

Corollary 4.5. Let K be a number field and |−| an absolute value. Then |−|is equivalent to

• an absolute value from K ↪→ R or C,

• or vp for some p ⊆ OK .

Proof. If |−| is archemedian, then its completion K is going to be R or C. Thenthe embedding K → K gives the valuation.

If it is non-archemedian, then its restriction to Q gives |−|p for some p, andso it is |−|p for some p | pOK .

Now let K be a field with an absolute value |−|v, and L be a finite sepa-rable extensions of K with absolute value |−|v′ extending |−|v. Now we havecompletions

K Kv

L Lv′

and furthermore Lv′ = Kv · L is the compositum. So Lv′/Kv is finite. Con-versely, if there exists some L′/Kv finite with L′ = L ·Kv, then we can extend|−|v on Kv to L′ and restricting to L gives an absolute value |−|L′ on L. Thisgives a bijection{

isom.classes offields L′ with L′ = Kv · L

}←→

{absolute values

of L extending |−|v

}.

Math 223a Notes 16

Proposition 4.6. Kv ⊗K L ∼=∏v′ extending v Lv′ .

Proof. To get the map, note that we have bilinear maps Kv×L→ Kv ·L = Lv′

for any v′.It is a general fact that if L/K is finite separable and K ′/K is any extension,

then K ′ ⊗K L ∼=∏Li where Li runs over isomorphism classes of fields L′ such

that L′ = K ′ · L. Using this fact and the previous isomorphism, we get theresult.

Example 4.7. Take K = Q, Kv = Q3 and L = Q[√

7]. By Hensel, 7 is a squarein Q3, and so x2 − 7 = (x− a)(x+ a) in Q3[x]. Now

L⊗Q Q3 = Q3[x]/(x2 − 7) ∼= Q3[x]/(x− a)×Q3[x]/(x+ a).

So we have two embeddings L = Q[√

7] ↪→ Q3, and each give different absolutevalues.

Example 4.8. Let K = Q, Kv = Q3 and L = Q[√

3]. In this case,

L⊗Q Q3 = Q3[x]/(x2 − 3)

is a field, and this is the unique extension of the absolute value.

Another example is L = Q[ 3√

2]. In this case L⊗Q Q3∼= Q3 ×K where K is

a quadratic extension. This is because L/Q is not Galois.

4.3 Local fields

Definition 4.9. A local field K is a locally compact complete valued field.

If K is a non-archemedian local field, then OK is compact. This is becausefor a ∈ OK with |a| < 1, one of the sets anOK is compact by local compactness.These sets are all homeomorphic to OK .

This also means that the absolute value on OK is discrete. Indeed, OK hasan open cover {a : a < c} for c < 1 and so has a finite subcover. Then there noabsolute values in (c, 1) for some c < 1.

Choose a π with |π| < 1 maximal. The quotient OK/(π) is compact anddiscrete. This implies that this is a finite field. Therefore non-archemedianlocal fields are field complete with respect to a discrete valuation with withfinite residue field. If there is such a Ks, then

OK = lim←−OK/(πn)

is compact.Archemedian local fields are either R or C, and these are actually locally

compact.

Definition 4.10. If K is a non-archemedian local field with valuation v andresidue field k = O/π, the standard normalization is given by

|a|K = |k|−v(a).

This works well, if you think K has a Haar measure on it. Multiplication by|a| rescales the measure by |a|K .

Math 223a Notes 17


We have been talking about local fields. A field K is a local field, if it is acomplete field (with respect to a absolute value) that is locally compact. If theabsolute value is discrete, is equivalent to

k = OK/πKOK

being finite. For example, Qp and Fp((t)) are both local fields, and finite exten-sions of local fields are local.

Theorem 5.1. In fact, all local fields are finite extensions of Qp or Fp((t)), orjust R or C

Definition 5.2. K is a global field if every completion of K is a local field.

Examples of global fields are Q, which complete to Qp or R, and Fp(t), whichcomplete to Fq((t)) for some q.

Theorem 5.3. All global fields are finite extensions of either Q or Fp(t).

If K is a global field, we have the product formula.

Proposition 5.4 (Product formula). For a ∈ K×,∏v

|a|v = 1,

where v runs over normalized absolute values.

Proof. For K = Q and K = Fp(t), we can manually check this. If the productformula holds for K, we want a product formula for L/K a finite extension.This follows from the fact that if v is a place of K, then∏

v′ extending v

|a′|v = |NL/Ka|v.

(This claim follows from what we did last time, that L⊗K Kv∼=∏Lv′ .)

5.1 Multiplicative group of local fields

Let K be a (non-archemedian) local field, with discrete valuation K× � Z. Wehave an exact sequence

1→ O× → K×v−→ Z→ 0.

This sequence splits, but not canonically, with the choice of a uniformizer πwith v(π) = 1.

Now there is another exact sequence

1→ U1 → O× → k× → 1,

Math 223a Notes 18

and this splits canonically due to Hensel lifting. There are q − 1 roots of unityin O×. Then we can say that

O× ∼= U1 × µq−1∼= U1 × k×.

We can filter down even more. For any n we have

Un = {a ∈ O× : a ≡ 1 (mod πn)},

and get a decreasing filter U1 ⊇ U2 ⊇ · · · .

Proposition 5.5. For n ≥ 1, Un/Un+1∼= k+ non-canonically.

Proof. Pick a uniformizer π and look at the map

k+ → Un/Un+1; a 7→ [1 + πna].

This is an isomorphism and it is not hard to see that it is an isomorphism.

This means that Upn ⊆ Un+1. On the other hand, if (m, p) = 1 then

Un → Un; x 7→ xm

is a bijection.

Proof. To show that this map is injective, take any a ∈ Un with a 6= 1. LetN be maximal with a ∈ UN − UN+1. Then [a] ∈ UN/UN+1 is not equal to 1,and so [am] ∈ UN/UN+1 is not equal to 1. This shows am 6= 1 and so we haveinjectivity.

For surjectivity, use Hensel’s lemma, or directly prove it using surjectivityof x 7→ xm on Un/Un+1.

So if we look at the roots of unity with order prime to p, it must have trivialintersection with U1. It also have to map bijectively to O×/U1 = k×. So theseare exactly µq−1, where k = Fq.

On the other hand, if we have a root of unity a ∈ O× of order pi, it must bein a ∈ U1.

5.2 Exponential and logarithm

Let K be a local field of characteristic 0. (So it is a finite extension of Qp.) LetpOK = (π)e for some e.

Definition 5.6. We define the exponential as

expp(x) =∑n≥0

xn

n!∈ K[[x]].

It is an exercise in analysis that this power series converges for v(x) > e/(p−1).The important fact here is that vp(n!) = (n − s(n))/(p − 1) where s(n) is thesum of base-p digits of n.

Math 223a Notes 19

Definition 5.7. Define the logarithm as

logp(1 + z) =∑n≥1

(−1)n+1zn.

This converges with v(z) > 0, equivalently 1 + z ∈ U1.

Theorem 5.8. For any n > e/(p− 1), the maps

expp : (π)n → Un, logp : Un → (π)n

are inverse homomorphisms.

Proof. You mainly just need to check that exp sends (πn) to Un and vice versa.Then as power series, they are inverses and homomorphisms, as we know fromour previous experiences.

Corollary 5.9. For p > 2, we have Z×p = µp × U1∼= (Z/(p− 1)Z)+ × Z+

p , and

for p = 2, we have Z×2 = µ2 × U2∼= (Z/2Z)+ × Z+

2 .

Proof. With p odd, we have e = 1 and so e/(p−1) < 1. Now apply the theorem.For p = 2, we have e = 1 and so apply to 2 = n > e/(p− 1) = 1.

This works out very nicely for Z×p , but in general we won’t have somethingas nice as this. Here, we have some

Un ∼= O+ ∼= Z[K:Qp]p .

Again we have1→ Un → U1 → U1/Un → 1,

where U1/Un is a finite p-group, but it may not split.

5.3 Unramified extensions

Let L/K be a finite extension of complete fields with discrete absolute values.Recall that we have OL over OK and also have residue fields ` = OL/πL andk = OK/πK . Then we have the ramification index

πKOL = (πLOL)eL/K

and the inertia degree fL/K = [` : k]. We have proved that n = [L : K] =eL/KfL/K .

Definition 5.10. L/K is unramified if eL/K = 1 and `/k is separable. (Thissecond condition is automatic for local fields.)

Lemma 5.11. Suppose L = K(a)/K be a finite extension, and K be completewith a discrete valuation. If there exists a monic polynomial f(x) ∈ OK [x] suchthat f(a) = 0 such that f(x) ∈ k[x] is separable, then L/K is unramified andOL = OK [a].

Math 223a Notes 20

Proof. I might as well take f to be the minimal polynomial of a so that f isirreducible. If follows that f ∈ k[x] is also irreducible by Hensel’s lemma, sincef is separable. Now a ∈ ` have minimal polynomial f . This means that

[` : k] ≥ [k(a) : k] = deg f = deg f = [L : K].

This means that [` : k] = [L : K].Note that we also have from this, ` = k(a). Apply Nakayama to OK [a] ⊆ OL

as OK-modules. We have that OK [a] + πLOL = OL and so we conclude thatOK [a] = OL.

Let L/K be unramified and let `/k be their residue fields. Choose any primeelement a ∈ ` with ` = k(a). Let f be the monic minimal polynomial of f . Thenwe can lift f ∈ k[x] to a f ∈ OK [x]. Then we can lift a to a ∈ OK with f(a) = 0,by Hensel. Then

[K(a) : K] = [k(a) : k] = [` : k] = [L : K],

and so L = K(a) and OL = OK [a] by our lemma.

Example 5.12. If L = K(ζn) for some n relatively prime to p, then the lemmaapplies and we have L/K unramified and OL = OK [ζn].

In particular, let n = pm − 1 and K = Qp. This extension has degree

[Qp(ζpm−1) : Qp] = [Fp(ζpm−1) : Fp] = [Fpm : Fp] = m.

So Qp(ζpm−1) is an unramified extension of degree m.

Math 223a Notes 21


Let K be a local field and L/K be unramified. This means that eL/K = 1, orequivalently fL/K = [L : K]. We showed that any such L is of the form K(a)

with a ∈ O×L , such that the minimal polynomial f(x) of a satisfies f(x) ∈ k[x]is irreducible.

Proposition 6.1. There is a unique unramified extension L/K of degree n, forevery n ≥ 1.

Proof. We constructed the extension L = K(ζqn−1) where q = |k|. For unique-ness, suppose we have L,L′ as above. We can write L′ = K(a) where f(x) isthe minimal polynomial of a. The polynomial f has degree [L′ : K] = n, andso f(x) is irreducible of degree n over k.

The residue field ` of L is a extension of k of degree n. Because k is a finitefield, f(x) ∈ k[x] splits in `. By Hensel’s lemma, f(x) has a root in L. That is,we have an inclusion

L′ ↪→ L; a 7→ root

which must be an isomorphism because [L : K] = [L′ : K].

More generally, suppose L and L′ are unramified extensions ofK with residuefields ` and `′. If there is a map ϕ : ` ↪→ `′ that fixes k, then the same argumentshows that we can lift the map uniquely to

ϕ : L ↪→ L′

so that OL/πL → OL′/πL′ is ϕ. In fact, we have an equivalence of categories{finite extensions

of k

}←→

{finite unramifiedextensions of K

}.

There is even a canonical construction called Witt vectors that goes from finiteextensions of k to finite unramified extensions of K.

Corollary 6.2. Every unramified extension of K is equal to K(ζm) for somem.

The compositum of two unramified extensions of K is unramified. So it ismeaningful to talk about the maximal unramified extension Kunr of K. This is

Kunr =⋃

(finite unramified extensions) =⋃

(m,p)=1

K(ζm).

This is even an abelian extension.

Math 223a Notes 22

6.1 Artin map for unramified extensions

Let L/K be a finite extension of local field. Recall that we’ve conjectured thereexists an Artin map

K×/NL×∼=−→ Gal(L/K).

We will verify this for unramified extensions by just computing both sides.If [L : K] = n, we have the isomorphism

Gal(L/K) ∼= Gal(`/k) ∼= (Z/nZ)+

by what we know about finite fields.The other side is harder to deal with. We have the valuation map v : K× →

Z, and this extends to v : L× → Z. Let’s look at what this does on the norm.We have

v(Na) =∑

g∈Gal(L/K)

v(ga) = nv(a).

This means that v : NL× � nZ. So we have a short exact sequence

1→ O×K/NO×L → K×/NL× → (Z/nZ)+ → 0.

We need to show that O×K = NO×L .We have filtrations

O×K = UK,0 ⊇ UK,1 ⊇ UK,2 ⊇ · · ·O×L = UL,0 ⊇ UL,1 ⊇ UL,2 ⊇ · · ·

with N : UL,i → UK,i. My claim is that

N : UL,i/UL,i+1 → UK,i/UK,i+1

is surjective. If i = 0, we have

N : O×L /UL,1 ∼= `× → k× ∼= O×K/UK,1.

This is something about finite fields, and you can check that this is surjective.Something similar happens for i > 0. There is an identification

`+ ∼= UL,i/UL,i+1; a 7→ [1 + πiLa]

and a similar identification for K. But note that we can choose πL = πK becauseL/K is unramified. Then what we are trying to show reduces to showing that

tr : `+ → k+

is surjective. This also can be checked.Now we want to show that if a ∈ O×K there exists a b ∈ O×L with Nb = a.

First find a b1 ∈ O×L such that

Nb1 ≡ a (mod UK,1).

Math 223a Notes 23

Then there exists a c2 ∈ U×L,1 such that Nc2 ≡ a(Nb1)−1 mod UK,2 so that

N(b1c2) ≡ a (mod UK,2).

We can keep on going to get b.

Theorem 6.3. If L/K is a unramified extension of local fields, then there isan isomorphism

K×/NL×∼=−→ Gal(L/K)

with both sides being isomorphic to (Z/nZ)+.

6.2 Decomposition group

Let L/K be a Galois extension, and v ∈ K and v′ ∈ L be places with v′

extending v. We define decomposition group

Dv′ = Dv′(L/K) = {g ∈ Gal(L/K) : |ga|v′ = |a|v′ for all a ∈ L}.

We have shown that if K is complete, then Dv′(L/K) = Gal(L/K).If K = FracOK with OK a Dedekind domain and v = vp, then we have

shown that v′ = vp′ for some p′ dividing p. Then

Dv′ = Dp′ = {g ∈ Gal(L/K) : gp′ = p′}.

If v is archemedian, then complex conjugation is always going to be in thedecomposition group.

If Kv and Lv′ are completions, then

Dv′(L/K) = Dv′(Lv′/Kv′) = Gal(Lv′/Kv′).

This is because if g fixes v′ then it lifts to the completion, and if it acts on Lv′

then its restriction to L fixes v′. In this case, let

Z = Z(v′) = fixed field of Dv′(L/K).

We can restrict v′ to Z and get a place vZ on Z(v′).

L v′

Z(v′) vZ

K v

Proposition 6.4. v′ is the only valuation of L extending vZ .

Proof. Note that Gal(L/Z) acts transitively on {valuations of L extending vZ}.(This is equivalent to the same statement about primes in the non-archemediancase, and can be worked out in the archemedian case.) Because Dv′(L/K) fixesv′, there is only one valuation extending vZ .

Math 223a Notes 24

We now haveGal(L/Z) ∼= Dv′

∼= Gal(Lv′/Kv),

and this really means that Z embeds in Kv and Z = Kv ∩ L.

6.3 Inertia group

Suppose I have a Galois extension L/K and v a discrete valuation onK. Supposev′ on L extends v. Define the inertia group as

Iv′ = Iv′(L/K) = {g ∈ Gal(L/K) : v′(ga− a) > 0 for all a ∈ OL}.

In other words, Iv′ is the kernel of the map

Dv′ → Gal(`/k).

We then have a short exact sequence

1→ Iv′ → Dv′ → Gal(`/k)→ 1

because Dv′ → Gal(`/k) is surjective. The proof of surjectivity goes like this.

Proof. By replacing with completions if necessary, we can assume that Dv′ =Gal(L/K). Pick some a ∈ ` and let its minimal polynomial be f(x). Lift it toa ∈ Ov′ . For

h(x) =∏

g∈Gal(L/k)

(x− ga) ∈ OK [x],

we will have that a is a root of

h(x) =∏

g∈Gal(L/K)

(x− ga) ∈ k[x].

For any a′ Galois conjugate to a in `, there exists a g ∈ Gal(L/K) such thatga = a′.

Corollary 6.5. If L/K is complete, then Iv′ has size e.

Proof. This is because Dv′ = Gal(L/K) has size ef and Gal(`/k) has size f .

Proposition 6.6. Let T (v′) be the fixed field of Iv′ . Then T (v′) is the maximalunramified subextension of L/K.

Proof. Let t be the residue field of T . Then Gal(L/T )� Gal(`/t) but

Gal(L/T ) Gal(`/t)

I(L/K) Gal(`/k).

∼=

1

This implies that ` = t and fL/T = 1. On the other hand, eL/T = [L : T ] =eL/K . This shows that eT/K = 1 and fT/K = fL/K .

Math 223a Notes 25


Last time we had L/K a Galois extension of local fields and defined T whichcorresponds to I(L/K) ⊆ Gal(L/K). Then

eL/T = [L : T ], fL/T = 1,

andeT/K = 1, fL/K = [L : K].

7.1 Totally ramified extensions of local fields

Let L/K be a totally ramified finite extension between local fields, so thateL/K = [L : K] = n.

Proposition 7.1. L = K(πL) and OL = OK [πL].

Proof. We will show that 1, πL, . . . , πn−1L are linearly independent over K. In

other words, that ∑0≤i<n

aiπiL 6= 0

for ai ∈ K not all zero. Note that

vL(aiπL)i = i+ nvK(ai),

and so

v

( ∑0≤i<n

aiπiL

)= min

0≤i<n(i+ nvK(ai)) <∞.

It can also be checked from this formula that∑

0≤i<n aiπiL is in OL if and only

if ai ∈ OK for all i.

The minimal polynomial f of πL has degree n. Let

f(x) = xn + cn−1xn−1 + · · ·+ c1x+ c0.

Then we get c0 = NL/K(πK) so vL(c0) = n. It follows that vK(c0) = 1.Likewise, we have ci as a symmetric polynomial of the Galois conjugates of πK ,so vK(ci) > 0 for 0 ≤ i ≤ n − 1. Such a polynomial is called an Eisensteinpolynomial.

Conversely, if f(x) ∈ K[x] is any Eisenstein polynomial, i.e., vK(ci) > 0 andvK(c0) = 1, then L = K(a) = K[x]/f(x) is totally ramified. This is because

|a|L = |c0|1/nK .

Example 7.2. Let K be an arbitrary local field and let L = K( m√πK). Then

πL = m√πK has minimal polynomial

xm − πK = 0.

Example 7.3. Let K = Qp and let L = Qp[ζpr ]. This time, πL = ζpr − 1,which you can check as an exercise.

Math 223a Notes 26

7.2 Ramification groups

Let L/K be an arbitrary finite extension, and let v′ be a place of L lying overv of K. We have defined the decomposition group Dv′(L/K) and the inertiagroup Iv′(L/K). Then L/T is totally ramified and T/Z is entirely inert.

L {1}

T Iv′(L/K)

Z Dv′(L/K)

K Gal(L/K)

total ram.

inert

Now we are going to define a filtration

Iv′(L/K) = G0,v′ ⊇ G1,v′ ⊇ G2,v′ ⊃ · · · .

Definition 7.4. We define the ramification groups as

Gi,v′(L/K) = {g ∈ Dv′(L/K) : v′(ga− a) > i for all a ∈ Ov′}.

Note that GN,v′ = {1} for all N � 0. Also, gi,v′ is a subgroup of Dv′ actingtrivially on Ov′/(πi+1

L ). If we complete the field, then

Gi,v′(L/K) = Gi,v′(Lv′/Kv′).

Now let’s assume that L,K are non-archemedian local fields, and write vL =v′ and vK = v. We can rewrite

Gi(L/K) = {g ∈ Gal(L/K) : vL(ga− a) > i for all a ∈ OL}.

If a0 generates OL/OK , i.e., OL = OK [a0], then we can just check for a0:

Gi(L/K) = {g ∈ Gal(L/K) : vL(ga0 − a0) > i}.

Proposition 7.5. Gi(L/K) = {g ∈ I(L/K) : vL(gπL − πL) > i}.

Proof. It suffices to show this in the case when L/K is totally ramified. This isbecause we can replace K by T . Now in this case we have OL = OK [πL], andso take a0 = πL.

For each i ≥ 0, we have a map

ϕi : Gi/Gi+1 ↪→ Ui,L/Ui+1,L; g 7→ [gπL/πL].

Math 223a Notes 27

It can be shown that these maps are group homomorphisms. It turns out thatthis map is completely canonical, i.e., independent of the choice of πL. If wereplace πL with uπL, then

g(uπL)

uπL=gu

u

gπLπL

where gu/u ∈ Ui+1,L.Recall that U0,L/U1,L

∼= `× and Ui,L/Ui+1,L∼= `+ for i ≥ 1. We conclude

that G0/G1 is cyclic of order relatively prime to char(`) = p and Gi/Gi+1

is an abelian group of exponent of p. From the fact that the filtration goesdown to {1}, we deduce that G1 is a p-group. So G1 is a p-Sylow subgroup ofI(L/K) = G0.

We call G0/G1 the tame inertia group and G1 the wild inertia group.We say that L/K is tamely ramified if G1 = {1}. This is equivalent foreL/K = |G0| is relatively prime to p.

Example 7.6. Let K = Q2 and L = K[ζ8]. Then [L : K] = 4 = eL/K and wecan take the uniformizer as πL = ζ8 − 1. We have

Gal(L/K) = (Z/8Z)× = {g1, g3, g5, g5}

so that gi(ζ8) = ζi8. Let’s compute vL(giπL − πL) for each i. We have

vL(g1πL − πL) = vL(0) =∞,vL(g3πL − πL) = vL(ζ3

8 − ζ8) = vL(ζ8(ζ8 − 1)(ζ8 + 1)) = 2,

vL(g5πL − πL) = vL(ζ58 − ζ8) = vL(−2ζ8) = 4,

vL(g7πL − πL) = vL(ζ78 − ζ8) = vL(ζ2

8 (ζ38 − ζ3)) = 2.

So

{g1, g3, g5, g7} = G0 = G1 ) G2 = {g1, g5} = G3 ) G4 = {g1} = G5 = · · · .

7.3 Tamely ramified extensions

Let L/K be a Galois extension of local fields. This is tamely ramified if andonly if (eL/K , p) = 1. Clearly, unramified extensions are tamely ramified!

Note that we can take the definition of tamely ramified as (eL/K , p) = 1,and then we don’t need to assume the Galois extension. Here are two facts:

• Compositum of tamely ramified extensions are tamely ramified.

• L/K tamely ramified implies that the Galois closure of L is tamely rami-fied.

Example 7.7. The field K[ζm, d√πK ] for (m, p) = 1 and d | m is a Galois

extension of K and is tamely ramified.

Proposition 7.8. Let L/K be Galois and tamely ramified. Then L ⊆ K[ζm, d√πK ]

for some (m, p) = 1 and d | m.

Math 223a Notes 28

Proof. By replacing K with its maximal unramified extension of K in L, wemay assume that L/K is totally ramified. Then

Gal(L/K) = G0(L/K) ∼= G0/G1,

is a cyclic group of order n with (n, p) = 1. Let K ′ = K(ζn) and L′ = L(ζn).Then there is an inclusion

Gal(L′/K ′) ↪→ Gal(L/K)

and so Gal(L′/K ′) is cyclic of order d | n.Since K ′ contains the n-root of unity, we can use Kummer theory to write

L′ = K ′( d√a)

for some a ∈ (K ′)×. If we write a = πrKu for u ∈ O×K′ , then

L ⊆ L′ ⊆ K ′( d√u, d√πK) ⊆ K ′′( d

√πK)

for some unramified extension K ′′ of K.

Before wrapping up the discussion of ramification, let me give a big pictureof local class field theory over Qp. Over abelian extensions of Qp, the max-imal unramified extension is Qp[ζprime to p] and the maximal tamely ramifiedextension is Qp[ζprime to p, p

1/(p−1)].

Qabp

Qp[ζprime to p, p1/(p−1)]

Qp[ζprime to p]

Qp

Z+p wild

F×p

Z+

The Artin map is then

Q×p → Gal(Qabp /Qp), Z×p

∼=−→ Gal(Qabp /Qunr

p ).

Another remark is that the numbering of the ramification group is not thebest. For L′/L/K, the groups Gi(L

′/K) and Gi(L/K) are not going to berelated because they are defined in terms of different valuations.

To fix this, there is an upper numbering. Define

φ(u) =

∫ u

0

dt

[G0 : Gt],

and write Gi = Gφ−1(i). Then Herbrard showed that Gi(L′/K) restricts toGi(L/K), and Hasse–Arf showed that if L/K is abelian then jumps in Gi occurat integer points.

Example 7.9. For the case Q2(ζ8)/Q2, we have G3 = G4 and G2 = G2.

Math 223a Notes 29


I am now going to move on to Galois cohomology. Roughly I will be followingCassels–Frohlich, but it can be a bit terse. Other references include Neukirch,Class Field Theory—Bonn lectures or Dummit–Foote.

8.1 Category of G-modules

Let G be a finite group, in most cases, G = Gal(L/K).

Definition 8.1. A G-module A is an abelian group A with an action of G,satisfying g(a+ b) = ga+ gb.

This is like representations, but with abelian groups rather than vectorspaces.

Example 8.2. For G = Gal(L/K), the abelian groups L+, L×, O×L , µn(L),Cl(OL), A×L/L× are all G-modules. We are going to be interested in L× for L alocal field this semester. Next semester we are going to be interested in A×L/L×for L a global field.

Also, for any commutative algebraic group G over K, you can look at G(L)as a Gal(L/K)-module. For instance, if E is an elliptic curve over K, then E(L)is a G-module. This is important if you study Selmer groups.

Example 8.3. For G any group and A any abelian group, there is a trivialaction ga = a that makes A into a G-module. In many cases we are going tolook at A = Z.

A G-module is the same thing as Z[G]-modules, where Z[G] is the groupring, a non-commutative ring of formal linear combinations∑

g∈Gagg,

with multiplication done formally. If G = Cn = 〈t〉 is a cyclic group, then

Z[G] = Z⊕ Zt⊕ · · · ⊕ Ztn−1 = Z[t]/(tn − 1).

This is not the same thing as Z[ζn].The ring Z[G] has an important ideal, called the augmentation ideal IG =

ker ε, where

ε : Z[G]� Z;∑

agg 7→∑

ag.

As a Z-module, IG is free and is spanned by g − 1. In the case G = Cn, theideal IG is just (t− 1)Z[G].

We can also consider the operator N =∑g∈G g. Then for any G-module A,

we have

NA ={∑g∈G

ga : a ∈ A}.

Math 223a Notes 30

If A = L×, then N becomes the norm and NA = NL×. If A = L+, then Nbecomes the trace map.

G-modules form an abelian category, i.e., we can take kernels, cokernels,images, direct sums, etc. If A and B are G-modules, we can take

Hom(A,B) = HomZ(A,B)

has a natural structure of a G-module. This works by g·φ = gφ(g−1−). Likewisewe can take tensor products

A⊗B = A⊗Z B

and the action is g(a⊗ b) = ga⊗ gb.There are also functors G−Mod→ Z−Mod. If A is a G-module, we have

AG = {a ∈ A : ga = a for all g ∈ G},AG = A/〈a− ga〉 = A/IGA.

The AG is called the invariants and AG is called the coinvariants. It can bechecked that these are both functions.

These are actually special cases. IfA,B ∈ G−Mod, then we have HomG(B,A)is an abelian group, but not generally a G-module. If B = Z, then

HomG(Z, A) = AG.

We can also take the tensor product. Generally, A ⊗R G makes sense if A is aright R-module and B is a left R-module. So in order for us to take A ⊗G B,we need to look at a A as a right Z[G]-module. This is done by looking at theg−1 action instead of g. So

A⊗G B = A⊗Z B/〈g−1a⊗ b− a⊗ gb〉.

Then AG = Z⊗G A. On the other hand, we have HomG(B,A) = Hom(A,B)G

and A⊗G B = (A⊗B)G.

Proposition 8.4. The functor A → AG is left exact, i.e., 0 → A → B → Cexact implies 0→ AG → BG → CG exact.

Proof. Injectivity of AG → BG is clear. If b ∈ BG with ψ(b) = 0, then byexactness there exists a unique A such that ϕ(a) = b. This means that a = gafor any g ∈ G. So a ∈ AG.

This is also a special case of the fact that Hom(B,−) is left exact for anyabelian category.

Proposition 8.5. The functor A → AG is right exact, i.e., A → B → C → 0exact implies AG → BG → CG → 0 exact.

This is also a special case of the fact that B⊗− is right exact. You can usethat this is the left adjoint functor to Hom(B,−) to show this.

Math 223a Notes 31

Example 8.6. Let G = C1 = {1, t}. Consider the Zχ = Z as an abelian group,but with the G-action by the character χ : G → {1,−1}, 1 7→ 1, t 7→ −1 asg(a) = χ(g)a. We have the exact sequence

0→ Zχ·2−→ Zχ → Z/2→ 0.

If we take the invariants, we get

0→ 0→ 0→ Z/2.

If we take the coinvarians, we get

Z/2 ·2−→ Z/2→ Z/2→ 0.

Example 8.7. Let L/K be a totally ramified extension of local fields. To makethis simple, let us use a quadratic extension, e.g., Qp(

√p)/Qp. There is an exact

sequence

1→ OLi−→ L×

v−→ Z→ 0.

If we take G-invariants,

1→ O×K → K×v|K−−→ Z,

but the image of K× → Z is now 2Z.

We say that a G-module is free if F is a direct sum of (possibly infinitelymany) copies of Z[G]. If F is free, then HomG(F,−) is exact because HomG(Z[G],−)is just the identity. Likewise, F ⊗G − is exact. The fact that HomG(F,−) isexact means that F is projective: there always is a lift

F

B C.

ϕϕ

π

That F ⊗G − is exact is that F is flat.

Definition 8.8. AG-module is coinduced if it is of the formA = Hom(Z[G], X)for X an abelian group. Likewise A is called induced if A = Z[G]⊗X for someabelian group X.

When G is finite, we have Hom(Z[G], X) ∼= Z[G] ⊗ X and so coinductionis the same thing as induction. This is sort of a generalization of freeness,because if X is a free module, then we would get a free Z[G]-module. Thecoinduced modules are going to be “cohomologically trivial” objects, and theinduced modules are going to be “homologically trival”.

Proposition 8.9 (Baby Frobeinus reciprocity). HomG(Z[G]⊗X,B) ∼= Hom(X,B)and HomG(B,Hom(Z[G], X)) ∼= Hom(B,X).

Math 223a Notes 32

Proposition 8.10. Every G-module A injects into a coinduced G-module, andevery A has a surjection from an induced G-module.

Proof. We take A′ = Hom(Z[G], A) where A is taken to have a trivial G-action.Then the map a 7→ φa given by φa(g) = ga is injective. Likewise, we have asurjection Z[G]⊗A→ A.

Next time, we are going to construct group cohomology functors Hi(G,A)that are Z-modules. These will be derived functors of A → AG. That is, if0→ A→ B → C → 0 then

0→ AG → BG → CG → H1(G,A)→ H1(G,B)→ H1(G,C)→ H2(G,A)→ · · · .

We will also construct Hi(G,A) that gives

· · · → H2(G,C)→ H1(G,A)→ H1(G,B)→ H1(G,C)→ AG → BG → CG → 0.

It turns out we can splice these together to get Tate cohomology, which isHi(G,A) = Hi(G,A) if i ≥ 1 and Hi(G,A) = H−1−i(G,A) for i ≤ −2. Here,we will get

H0(G,A) = AG/NA.

So if A = L×, then H0(G,A) = K×/NL×.

Math 223a Notes 33


The goal for today is to define group cohomology. But there is a correc-tion I want to make. When we defined coinduction, we define CoindG(X) =Hom(Z[G], X). In general, we defined Hom(A,B) for G-modules A and B asgϕ = g ◦ϕ ◦ g−1, because we needed to make A have a right G-action. But notethat Z[G] has a natural right action. So we can simply define the group actionon CoindG(X) as

(gϕ)(b) = ϕ(bg).

Actually, the results are isomorphic, i.e., there exists an isomorphism Hom(Z[G], X)→CoindG(X).

9.1 Group cohomology

Theorem 9.1. There exist functors Hq(G,−) : G−Mod → Z−Mod for q ≥ 0with the following properties:

(i) H0(G,A) = AG,

(ii) if 0→ Ai−→ B

j−→ C → 0 is a short exact sequence, then it induces a longexact sequence

0→ H0(G,A)i∗−→ H0(G,B)

j∗−→ H0(G,C)δ−→ H1(G,A)

i∗−→→ · · · .

(iii) Hq(G,A) = 0 for q ≥ 1 if A is coinduced.

Moreover, such Hq(G,−) are unique up to natural isomorphism.

This is a special case of derived functors.

Proof. For existence, let us take a free resolution

· · · d3−→ P2d2−→ P1

d1−→ P0ε−→ Z→ 0,

so that all Pi are free and the sequence is exact. Such a resolution exists, becausefor any G-module A, there exists a free G-module that surjects onto A.

Define a chain complex Ki = HomG(Pi, A). Then

K0 = HomG(P0, A)d1−→ K1 = HomG(P1, A)

d2−→ K2 = HomG(P2, A)→ · · · ,

and let Hi(G,A) be the cohomology of this complex,

Hi(G,A) = ker di+1/ im di.

Let us verify that this satisfies the properties. We have

H0(G,A) = ker d1 = HomG(P0/ im d1, A) = HomG(Z, A) = AG.

Math 223a Notes 34

If 0 → A → B → C → 0 is a short exact sequence, for each q we have a shortexact sequence

0→ HomG(Pq, A)→ HomG(Pq, B)→ HomG(Pq, C)→ 0,

and so we have a short exact sequence of chain complexes. Then we can applycohomology and get the long exact sequence. This long exact sequence is furthernatural in the sense that a morphism of short exact sequences induce a morphismof long exact sequences. Finally, if A = CoindG(X), then

Kq = HomG(Pq,CoindG) = HomZ(Pq, X).

Because Pq is a free Z-module, this is exact.Now let us show uniqueness of Hq(G,A) by induction on q. First we see that

the case q = 0 is given by the axiom. For q > 0, we are going to use dimensionshifting. Let A be any G-module, and take a short exact sequence

0→ A→ A′ → A∗ → 0

where A′ is coinduced by A. This gives a long exact sequence

· · · → Hq−1(A′)j∗−→→ Hq−1(A∗)→ Hq(A)→ Hq(A′) = 0→ · · · .

So Hq(A) is the cokernel of j∗ : Hq−1(A′)→ Hq−1(A∗) is determined by Hq−1.

Example 9.2. Take G = Cn = 〈t〉. We can take the projective resolution

· · · → Z[G]×(t−1)−−−−→ Z[G]

×N−−→ Z[G]×(t−1)−−−−→ Z[G]

t 7→1−−−→ Z→ 0.

Then our chain complex is

0→ A×(t−1)−−−−→ A

×N−−→ A×(t−1)−−−−→ A→ · · ·

and we can now compute cohomology. We have

H0(G,A) = ker(t− 1) = AG,

H2q−1(G,A) = ker(N)/(t− 1)A = ker(N)/IGA,

H2q(G,A) = ker(t− 1)/ im(N) = AG/NA.

If we extend to Tate cohomology, all the odd things will be the same and all theeven things will be the same.

9.2 Standard bar resolution

Let G be any group and let Pi = Z[Gi+1] = Z[G] be the formal span oftuples (g0, . . . , gi) ∈ Gi+1. Let us make this a G-module by a diagonal ac-tion g(g0, . . . , gi) = (gg0, . . . , ggi). Then this is a free Z[G]-module with basis(1, g1, . . . , gi). Define

di : Pi → Pi−1; di(g0, . . . , gi) =∑j

(−1)j(g0, . . . , gj , . . . , gi).

Math 223a Notes 35

We can easily check that di−1di = 0. To show exactness, suppose that somelinear combination of (g0, . . . , gi) is in ker di. If we put an arbitrary s ∈ G infront of every such tuple, then we check

di+1(s, g0, . . . , gi) = (g0, . . . , gi)− (s, di(g0, . . . , gi)).

So this is going to be in the image.Now we have

Hq(G,A) =ker(dq+1 : HomG(Pq, A)→ HomG(Pq+1, A))

im(dq : HomG(Pq−1, A)→ HomG(Pq, A)).

We can interpret Hom(Pq, A) as homogeneous cohains Cq(A), as maps f :Gi+1 → A such that

f(gg0, . . . , ggi) = gf(g0, . . . , gi).

The kernel of dq+1 are homogeneous cocyles and the image of dq are homoge-neous boundaries.

But this is a cumbersome condition, and we can make a change of variables.For example, take q = 1. The function f : G → A in C0(A) is determined byf(1) = a ∈ A. The function f : G2 → A that satisfy f(gg0, gg1) = gf(g0, g1) isdetermined by f(1, g) = ϕ(g). So the boundary map is

d1 : C0 → C1; a 7→ ((1, g) 7→ a− ga).

That is, the image is ϕ(g) = a−ga. These are the inhomogeneous 1-coboundaries.For q = 2, we are going to let (g0, g1, g2) = (1, g, gh). Then the coboundary mapis

d2 : ϕ 7→ ((g, h) 7→ gϕ(h)− ϕ(gh) + ϕ(g)).

Math 223a Notes 36

10 October 3, 2017

The plan for today is to continue on ways to compute group cohomology. Lasttime we defined homogeneous cochains

Ci(G,A) = {(f : Gi+1 → A) : f(gg0, . . . , ggn) = gf(g0, . . . , gn)}.

The differentials are defined as

d : Ci(G,A)→ Ci+1(G,A);

f 7→ (df : (g0, . . . , gi+1) 7→ f(g1, . . . , gi+1)− f(g0, g2, . . . , gi+1) + · · · ).

This is now very explicitly defined.We are now going to change variables. Inhomogeneous cochains

Ci(G,A) = {ϕ : Gi → A}

and identify with Ci(G,A) via

Ci(G,A)→ Ci(G,A); f 7→ (ϕ : (g1, . . . , gi) 7→ f(1, g1, g1g2, . . . , g1 · · · gi)).

Through this bijection, we get the differentials

di : Ci(G,A)→ Ci+1(G,A),

which maps ϕ ∈ Ci(G,A) to

diϕ(g1, . . . , gi+1) = g1ϕ(g2, . . . , gi+1)− ϕ(g1g2, g3, . . . , gi+1) + ϕ(g1, g2g3, . . . , gi+1)

− · · ·+ (−1)iϕ(g1, . . . , gi−1, gigi+1) + (−1)i+1ϕ(g1, . . . , gi).

10.1 Explicit descriptions of cohomology

We can use these efficient cochains to compute the cohomologies. We have

d0 : C0(G,A)→ C1(G,A); a 7→ (ϕ : g 7→ ga− a).

Then the next differential is

d1 : C1(G,A)→ C2(G,A); ϕ 7→ ((g1, g2) 7→ g1ϕ(g2)− ϕ(g1g2) + ϕ(g1)),

and the next one is

d2 : C2(G,A)→ C3(G,A);

ϕ 7→ ((g1, g2, g3) 7→ g1ϕ(g2, g3)− ϕ(g1g2, g3) + ϕ(g1, g2g3)− ϕ(g1, g2)).

From this, we can immediately read

H1(G,A) ∼={(ϕ : G→ A) : ϕ(g1g2) = ϕ(g1) + g1ϕ(g2)}

{(ϕa : G→ A) : ϕa(g) = ga− a}.

These ϕ satisfying ϕ(g1g2) = ϕ(g1) + g1ϕ(g2) are called crossed homomor-phisms. If G acted on A trivially, this would just be ϕ(g1g2) = ϕ(g1) + ϕ(g2).Then we are quotienting out by nothing, so

H1(G,A) = HomGrp(G,A).

Math 223a Notes 37

Definition 10.1. IfG is a group and A is aG-module, we define the semidirectproduct GnA = G×A with multiplication

(g1, a1)(g2, a2) = (g1g2, a1 + g1a2).

This is a group, and fits in the short exact sequence

0→ A→ GnA→ G→ 0.

Now we can interpret Z1(G,A) as a splitting of this exact sequence, i.e., sectionsof GnA→ G. This is because the map

G→ GnA; g 7→ (g, ϕ(g))

is a group homomorphism if and only if ϕ is a crossed homomorphism.On the other hand, ϕ1 and ϕ2 are equivalent modulo B1(G,A) if and only

if the splittings G→ GnA are conjugate by some element of A.These is a similar interpretation for H2(G,A). This classifies short exact

sequences of groups0→ A→ X → G→ 0,

such that the G-action on A by conjugation agrees with the G-module structure.Here, 0 ∈ H2 corresponds to 0→ A→ GnA→ G→ 0. The idea is that givensuch an X, take a section (as a set) s : G→ X and compare s(gh) and s(g)s(h).This lies in the kernel, so there exists a unique ϕ(g, h) such that

s(gh) = ϕ(g, h)s(g)s(h).

This is a cocycle, after using associativity.

Recall that if we have a short exact sequence of G-modules 0 → Ai−→ B

j−→C → 0, we get a connecting homomorphism

H0(G,C)→ H1(G,A).

This can be described explicitly with cocycles. If c ∈ CG, choose b ∈ B withj(b) = c and then

g 7→ gb− b ∈ A

is an element of Z1(G,A). This is well-defined up to elements like ga−a, whichis B1(G,A). So we get this boundary map.

10.2 The first cohomology and torsors

Let A be an abelian group.

Definition 10.2. An A-torsor is a set X with a simply transitive action of A.(Simple means that stabilizers are trivial.)

Now let A be a G-module.

Math 223a Notes 38

Definition 10.3. An A-torsor is a G-set X such that X is an A-torsor in theprevious sense and

g(a+ x) = ga+ gx.

(If you want to use multiplication as the action, use g(ax) = (ga)(gx).)

A is an A-torsor. If A is just a group, then every A-torsor is isomorphic toA as sets non-canonically.

Example 10.4. Suppose L/K is a field extension, and suppose that µn ⊆ L.Let A = µn(L). For any c ∈ (L×)n, let

X = {a ∈ L : an = c}.

Then this is a torsor for A = µn(L). We can also bring in a group G =Gal(L/K). If c ∈ (K×)n then the G-action is trivial. But we can also imaginea situation in which µn(L) ⊆ K but c has no nth roots in K. Then G actstrivially on A but not on X.

Theorem 10.5. H1(G,A) classifies A-torsors with the G-action.

Proof. I will construct maps in both directions. Given [ϕ] ∈ H1(G,A), we canconstruct a torsor X with X = A as a set but with G-action

g ∗ϕ x = gx+ ϕ(g).

This really is a group action because

g ∗ϕ (h ∗ϕ x) = g(hx+ ϕ(h)) + ϕ(g) = ghx+ gϕ(h) + ϕ(g) = (gh)x+ ϕ(gh).

Now we want a map from A-torsors to H1(G,A). Given X an A-torsor,choose an element x0 ∈ X and define

g(x0) = ϕ(g) + x0.

It can be checked that ϕ ∈ Z1(G,A), and if we replace x0 by a+ x0, then

g(a+ x0) = g(a) + g(x0) = (ga− a) + ϕ(g) + (a+ x0).

This gives a map to H1(G,A).

Using torsors, there is another way of describing the connecting homomor-phism H0(G,C) = CG → H1(G,A). For c ∈ CG, the set j−1(c) ⊆ B is anA-torsor. The class of this torsor gives the connecting morphism.

Example 10.6. In the short exact sequence

0→ µn(L)→ L×j−→ (L×)n → 0,

j−1(c) is the nth roots of c.

Math 223a Notes 39

10.3 Group homology

Recall we have the functor G−Mod→ Z−Mod; A 7→ AG. This is defined as

AG = A/IGA = A/(ga− a) = A⊗G Z.

This is right exact but not left exact.

Definition 10.7. There exist homology functors Hq(G,A) for q ≥ 0, satisfying

(a) H0(G,A) = AG,

(b) Hq(G,A) = 0 for q ≥ 1 if A is induced,

(c) if 0 → A → B → C → 0 is a short exact sequence, there is a long exactsequence

· · · → H1(G,B)→ H1(G,C)→ H0(G,A)→ H0(G,B)→ H0(G,C)→ 0.

The proof of this is virtually identical. Take a flat (free) resolution

· · · → F2 → F1 → F0 → Z→ 0

and take the homology of

· · · → F2 ⊗G A→ F1 ⊗G A→ F0 ⊗G A→ 0.

Uniqueness is again by dimension shifting.

Math 223a Notes 40

11 October 5, 2017

Last time we defined group homology by

(a) H0(G,A) = AG = A/IGA,

(b) Hq(G,A) = 0 for q ≥ 1 if G is induced,

(c) a short exact sequence gives a long exact sequence.

We can construct it as the homology of the free resolution of Z tensored withA. So we can use the standard resolution to get explicit descriptions. Classfield theory only uses a limited range of (co)homology groups, H0, H1, H2, H0,H−1, H0, H1.

Theorem 11.1. We have H1(G,Z) ∼= IG/I2G = Gab.

Proof. We use the short exact sequence

0→ IG → Z[G]ε−→ Z→ 0

of G-modules. This gives a long exact sequence

H1(G,Z[G]) = 0→ H1(G,Z)→ H0(G, IG) = IG/I2G

0−→ H0(G,Z[G]) = Z[G]/IG.

So we get H1(G,Z) ∼= H0(G, IG) = IG/I2G. For the next isomorphism, write

ag = g − 1. Note that I2G is spanned by

(g − 1)(h− 1) = gh− g − h+ 1 = (gh− 1)− (g − 1)− (h− 1) = agh − ag − ah.

This shows that IG/I2G is generated by ag, with relations agh = ag + ah. This

is precisely Gab.

Let H ⊆ G. There is a short exact sequence

0→ IG → Z[G]→ Z→ 0

of H-modules. Also Z[G] is a free Z[H]-module. So

H1(H,Z[G]) = 0→ H1(H,Z)

→ H0(H,HG) = IG/IGIH → H0(H,Z[G]) = Z[G]/IHZ[G].

This shows that

H1(H,Z) = ker(IG/IGIH → Z[G]/IHZ[G]) = IHZ[G]/IGIH .

It is a good exercise to check that this is just IH/I2H .

Math 223a Notes 41

11.1 Changing the group

Both the functors Hi(G,A) and Hi(G,A) depend on G and A. These arecovariant functors of A. Hi is also going to be covariant in G, and Hi will becontravariant in G.

Definition 11.2. Two pairs (G,A) and (G′, A′) are compatible for coho-mology if there exist morphisms ρ : G′ → G and λ : A→ A′ such that

λ(ρ(g′), a) = g′λ(a).

Example 11.3. If G′ = H ⊆ G is a subgroup, then we can set ρ = i : H ↪→ Gand λ = idA.

Example 11.4. If H ⊆ G is a normal subgroup, then ρ : G → G/H andλ : AH → A gives a map from (G,A) to (G/H,AH).

Then ρ maps

ρ∗ : Pq(G′) = Z[(G′)q+1]→ Pq(G) = Z[Gq+1].

This gives a map

HomG(Pq(G), A)→ HomG′(Pq(G′), A′)

of cochain complexes. So we get maps on homology

ρ∗ : Hq(G,A)→ Hq(G′, A′).

This can be described explicitly in homogeneous cochains. It sends [ϕ] to [ϕ′]where

ϕ′(g′1, . . . , g′q) = λ(φ(ρ(g′1), . . . , ρ(g′q))).

Example 11.5. If (G,A) and (H,A) are compatible by (i, idA), then this in-duces restriction

Res : Hq(G,A)→ Hq(H,A)

is just done by restricting the function.

Example 11.6. Consider (G,A) and (G/H,AH). This gives the inflation map

Inf : Hq(G/H,AH)→ Hq(G,A),

which is done by replacing φ by Inf φ(g1, . . . , gq) = φ(g1H, . . . , gqH).

One thing we see is thatRes ◦ Inf = 0

for q ≥ 1, because this is induced by (G/H,AH) and (H,A) with the map ofgroups being trivial.

Definition 11.7. For pairs (G,A) and (G′, A′), a compatible map for ho-mology is ρ : G→ G′ and λ : A→ A′ such that λ(ga) = ρ(g)λ(a).

Math 223a Notes 42

Example 11.8. For H ⊆ G a subgroup, there is a map from (H,A) to (G,A).For H ⊆ G a normal subgroup, there is (G,A)→ (G/H,AH).

As before, compatible maps give induced homomorphismsHq(G,A)→ Hq(G′, A′).

So there is the corestriction

coRes : Hq(H,A)→ Hq(G,A)

and coinflationCoinf : Hq(G,A)→ Hq(G/H,AH).

The functors Res and coRes are compatible with long exact sequence maps.If 0 → A → B → C → 0 is a short exact sequence of G-modules, the longexact sequence for Hq(H,−) and Hq(G,−) commute with the restriction maps.Likewise, the long exact sequence for homology commute with corestriction.

Also note that Res is completely determined by the property that Res :AG → AH and compatibility with the long exact sequence, using dimensionshifting.

0 Hq(G, . . .) Hq+1(G, . . .) 0

0 Hq(H, . . .) Hq+1(H, . . .) 0

Res

∼=

∼=

That is, this is a “universal δ-functor”. Later, we will define

Res : Hq(G,A)→ Hq(H,A), coRes : Hq(H,A)→ Hq(G,A)

for all q ∈ Z by dimension shifting.

11.2 Inflation-restriction exact sequence

Let H C G be a normal subgroup. If q ≥ 1, we can look at the sequence

Hq(G/H,AH)Inf−−→ Hq(G,A)

Res−−→ Hq(H,A).

Is this exact?

Theorem 11.9. The sequence

0→ H1(G/H,AH)Inf−−→ H1(G,A)

Res−−→ H1(H,A)

is exact.

In general, there is a spectral sequence for higher degree, but we are goingto just do this explicitly.

Proof. First let us prove that Inf : H1(G/H,AH) → H1(G,A) is injective.Suppose [ϕ] ∈ ker Inf, so that ϕ(gH) = ga − a for all g ∈ G. Now wehave ϕ(H) = ha − a, which is constant for all h ∈ H. So a ∈ AH andϕ ∈ B1(G/H,AH).

Math 223a Notes 43

Now we need to prove ker Res ⊆ im Inf. Take φ ∈ Z1(G,A) such thatResϕ ∈ B1(H,A). Then φ(h) = ha − a for some a ∈ A. But ϕ is living inH1(G,A), so we can subtract g 7→ (ga − a) from φ and take this as φ instead.Then φ(h) = 0 for all h ∈ H. Now

φ(gh) = gφ(h) + φ(g) = φ(g)

for all g ∈ G and h ∈ H. So φ is defined on G/H and

φ(g) = φ(hg) = hφ(g) + φ(h) = hφ(g).

That is, φ maps to AH . This shows that φ comes from a crossed homomorphismG/H → AH .

Proposition 11.10. Suppose Hi(H,A) = 0 for 1 ≤ i < q. Then there existsan exact sequence

0→ Hq(G/H,AH)Inf−−→ Hq(G,A)

Res−−→ Hq(H,A).

Proof. This goes by dimension shifting. If q = 1, we already know this. Nowsuppose we know this for q− 1. Let A be a G-module satisfying the above, andlet A∗ = CoindG(A) so that A ↪→ A∗. Then we have a short exact sequence

0→ A→ A∗ → A′ → 0.

Because A∗ is coinduced as a G-module, it is also coinduced as a H-module.This is because Z[G] is a free Z[H]-module. So

Hq−1(G,A′) Hq(G,A)

Hq−1(H,A′) Hq(H,A).

∼=

∼=

If we further take H-invariants of our short exact sequence. Here, we get

0→ AH → (A∗)H → (A′)H → H1(A,H) = 0.

So we get an compatible

Hq−1(G/H, (A′)H) Hq(G/H,AH)

Hq−1(G, (A′)H) Hq(G,AH).

∼=

Inf Inf

But A′ satisfies Hi(G,A′) = Hi+1(G,A) = 0 for i < q−1. So you can dimensionshift.

Math 223a Notes 44

12 October 10, 2017

Today we are going to define Tate cohomology.

12.1 Tate cohomology

We have defined Hq(G,A) and Hq(G,A) for q ≥ 0. We will define

Hq(G,A) =

{Hq(G,A) q ≥ 1

H−1−q(G,A) q ≤ −2

and we will fill in the module. These are called Tate cohomology. We aregoing to focus only on finite groups G.

For G finite, we have this operator N =∑g∈G g ∈ Z[G], and sends

N : A→ A; a 7→ Na =∑g∈G

ga.

Then imN ⊆ AG and kerN ⊇ IGA. So we actually have a map N∗ : AG → AG.This allows to define

H−1(G,A) = ker(N∗ : AG → AG), H0(G,A) = coker(N∗ : AG → AG).

Because we have

· · · H1(G,C) AG BG CG 0

0 AG BG CG H1(G,A) · · · .

N∗A N∗B N∗C

Then a standard application of the snake lemma gives

· · · → H−2(G,C)→ H−1(G,A)→ H−1(G,B)→ H−1(G,C)

→ H0(G,A)→ H0(G,B)→ H0(G,C)→ H1(G,A)→ · · · .

Proposition 12.1. For A a G-module that is induced (or equivalently, coin-duced), Hq(G,A) = 0 for all q.

Proof. We know this except for q = 0 and q. So it suffices to show that N∗ :AG → AG is an isomorphism. Suppose A =

⊕g∈G gX where X ⊆ A is a Z-

submodule. We claim that both AG and AG are isomorphic to X. An element∑g gxg is in AG if and only if all xg are equal in X. So we get an isomorphism

X → AG; x 7→∑g∈G

gx.

Math 223a Notes 45

Conversely, we have a map

AG → X;∑g

gxg 7→∑g

xg.

These two are isomorphisms and N∗ is just the composite.

There is a slightly different perspective on how to splice these sequences. Let

· · · → P2 → P1 → P0 → Z→ 0

be a (finite rank) free resolution. Then we can dualize and get

0→ Z→ Hom(P0,Z) = P−1 → Hom(P1,Z) = P−2 → · · · .

This is still exact because all objects are free Z-modules. Now we can composethese can get a complete resolution

· · · → P2 → P1 → P0 → P−1 → P−2 → · · · .

Now we can define

Hq(G,A) = Hq(K• = HomG(P•, A)).

To show this for q ≥ 0, we only need to check that

HomG(Hom(Pq,Z), A) = HomG(P−1−q, A) ∼= Pq ⊗G A.

This is done by noticing that HomZ(HomZ(Pq,Z), A) ∼= Pi ⊗Z A. Then if wetake the G-invariants of both side, the left hand side is what we want, and theright hand side is (Pi ⊗Z A)G ∼= (Pi ⊗Z A)G = Pi ⊗G A.

Example 12.2. Let G = Cn = 〈t : tn = 1〉. Recall that we have the projectiveresolution

· · · → Z[G]×(t−1)−−−−→ Z[G]

N−→ Z[G]×(t−1)−−−−→ Z[G]

ε−→ Z→ 0.

Now dualizing gives

0→ Z 17→N−−−→ Z[G]×(t−1−1)−−−−−−→ Z[G]

N−→ Z[G]→ · · · .

So we can put them together to get

· · · → Z[G]×(t−1)−−−−→ Z[G]

N−→ Z[G]×(t−1−1)−−−−−−→ Z[G]

N−→ Z[G]→ · · · .

ThusH2q = AG/NA, H2q−1 = (kerN)/IGA.

Math 223a Notes 46

12.2 Restriction and corestriction

We have defined restriction for cohomology and corestriction for homology.These are defined for

Res : Hq(G,A)→ Hq(H,A) for q ≥ 1, coRes : Hq(H,A)→ Hq(G,A) for q ≤ −2.

We want these for all q ∈ Z.We can construct these by dimension shifting. If we define A+ as the cokernel

of A→ coIndG(A), then we get a short exact

0→ A→ coIndG(A)→ A+ → 0.

Then we have Hq+1(G,A) ∼= Hq(G,A+). So if Resq+1 are already defined, wecan define Resq by

Hq(G,A) Hq(G,A)

Hq+1(G,A+) Hq+1(H,A+).

Resq

∼= ∼=

Resq+1

We can likewise define coResq for q ∈ Z by upwards induction. Both of theseare functors, and are compatible with long exact sequences. Note that the onlything we needed is that Res0 is AG → AH and coRes0 = AH → AG. After that,we have only used dimension shifting.

Let us just look at how the map

Res−1 : H−1(G,A)→ H−1(H,A)

is defined. This is induced by the map

N ′G/H : AG → AH ; [a] 7→∑

g∈G/H

[g−1a].

Proof. It is enough to show that if 0 → A → B → C → 0 is a short exactsequence,

H−1(G,C) H0(G,A)

H−1(H,C) H0(H,A)

δ

N ′G/H Res0

δ

commutes. Let b 7→ c. An element [c] ∈ H−1(G,C) maps to [Nb] ∈ AG/ imN∗

in H0(G,A), and this maps to NGb =∑g∈G gb. On the other hand [c] maps

downwards to∑g∈H\G[gc] and this maps to∑g∈H\G

NH(gb) =∑

g∈H\G

∑h∈H

hgb =∑g∈G

gb.

So the diagram commutes.

Math 223a Notes 47

Likewise,coRes0 : H0(H,A)→ H0(G,A)

comes from the map

NG/H : AH → AG; a 7→∑

g∈G/H

ga.

Proposition 12.3. For all q, we have

Hq(G,A)Resq−−−→ Hq(H,A)

coResq−−−−→ Hq(G,A).

The composite is multiplication by [G : H].

Proof. By dimension shifting, it suffices to show this for q = 0. The groups are

AG/NGARes−−→ AH/NHA

coRes−−−→ AG/NGA.

This comes from a 7→ a 7→∑g∈G/H ga = [G : H]a.

Corollary 12.4. If |G| = n, then Hq(G,A) are n-torsion.

Proof. We that Hq(G,A) → Hq({1}, A) → Hq(G,A) is multiplication by n.But Hq({1}, A) = 0.

Corollary 12.5. For G finite and A a finitely generated Z[G]-module, Hq(G,A)is also finite.

Proof. This is because Hq(G,A) is finitely generated by the explicit description.Then it is torsion, so it is finite.

Corollary 12.6. If |G| = n and if n : A → A is an isomorphism, thenHq(G,A) = 0.

Proof. Then n : Hq(G,A)→ Hq(G,A) is an isomorphism.

Corollary 12.7. If Gp is a Sylow p-subgroup of G, then

Res : (Hq(G,A))p → Hq(Gp, A)

is injective.

Proof. The map coRes ◦Res : (Hq(G,A))p → (Hq(G,A))p is injective.

Corollary 12.8. If Hq(Gp, A) = 0 for all p, then Hq(G,A) = 0.

Math 223a Notes 48

13 October 12, 2017

13.1 Cup products

If A and B are G-modules, there exists a map

` : Hp(G,A)×Hq(G,B)→ Hp+q(G,A⊗B)

for p, q ≥ 0, and if G is finite, there is a map

` : Hp(G,A)× Hq(G,B)→ Hp+q(G,A⊗B)

for all p, q ∈ Z. This satisfy the axioms:

• The product is natural in both A and B.

• If p, q = 0, the map is ` : AG ×BG → (A×B)G; a` b = a⊗ b.• This is compatible with exact sequences. That is, if 0→ A→ A′ → A′′ →

0 is an exact sequence such that 0→ A⊗B → A′⊗B → A′′B → 0 is alsoexact, then for a′′ ∈ Hp(G,A′′) and b ∈ Hq(G,B)

δ(a′′` b) = (δa′′)` b.

This axioms uniquely determine the cup product on H, and the proof is bydimension-shifting. (This is in the Bonn lectures of Neukirch.) You can alsoexplicitly construct this using resolutions.

There are explicit formulas for the cup product. Let f ∈ Cp(G,A) andg ∈ Cq(G,B) be homogeneous cochains. Then

(f ` g)(g0, . . . , gp+q) = f(g0, . . . , gp)⊗ g(gp, . . . , gp+q).

If ϕ ∈ Cp(G,A) and g ∈ Cq(G,B) are inhomogeneous cochains, then

(ϕ`ψ)(g1, . . . , gp+q) = ϕ(g1, . . . , gp)⊗ g1 · · · gpψ(gp+1, . . . , gp+q).

Cassels and Frolich wirte down explicit formulas in all dimensions. The Bonnlectures do some explicit computations in small dimensions.

Cup product satisfies

• (a` b)` c = a`(b` c) under the identification (A⊗B)⊗C ∼= A⊗(B⊗C).

• a` b = (−1)pqb` a under A⊗B ∼= B ⊗A.

• Res(a` b) = (Res a)`(Res b).

• coRes(a`Res b) = coRes(a)` b.

To prove this, check for p = q = 0 and then dimension shift. For instance, thelast formula is, for a ∈ AH and b ∈ BG,

NG/H(a⊗ b) =∑

g∈G/H

g(a⊗ b) =

( ∑g∈G/H

ga

)⊗ b = (NG/Ha)⊗ b.

For p = 0 and q arbitrary, the map

a`− : Hq(G,B)→ Hq(G,A⊗B)

is just tensoring with a.

Math 223a Notes 49

13.2 Galois cohomology

Let L/K be any Galois extension of local fields. We are going to use the notation

Hq(L/K,A) = Hq(Gal(L/K), A).

This is called Galois cohomology, and we are going to use A = L× and A = Znormally.

The group H2(L/K,L×) is called the Brauer group of L/K, and we aregoing to get an explicit isomorphism

H2(L/K,L×)∼=−→ 1

nZ/Z

where n = [L : K]. Now there is the preferred generator 1n on the right hand

side, and we will call the preimage uL/K . This will give a cup product

−`uL/K : H−2(L/K,Z)→ H0(L/K,L×).

Note that the left hand side is just the abelianization Gal(L/K)ab. But theright hand side is K×/NL×. This map is what we call the Artin map. To getthis, we will need to compute some Galois cohomology.

Let L/K be a finite extension. For motivation, I can ask, what is Hq(L/K,L+)?Does this have any interesting Galois cohomology? The answer is no, becauseL+ is an induced Gal(L/K)-module. This follows from the following theorem.

Theorem 13.1 (Normal basis theorem). There exists a x ∈ L such that {gx}g∈Gal(L/K)

is a basis for L as a K-vector space.

However multiplicative group is more interesting. We are mainly going totalk about the second cohomology, but what about H1?

Theorem 13.2 (Hilbert 90). H1(L/K,L×) = 0.

Proof. We use inhomogeneous cochains. Suppose that ϕ ∈ Z1(L/K,L×) so thatϕ(gh) = ϕ(g) · gϕ(h). We need to show that there exists some a ∈ L× such thatϕ(g) = a/ga. Let x ∈ L, and we are going to let

a =∑g∈G

ϕ(g) · gx 6= 0.

We can choose x so that a 6= 0, by linearly independence of characters. Now

a =∑g′∈G

ϕ(g′) · g′x =∑gg′∈G

ϕ(gg′) · (gg′x)

=∑g′∈G

ϕ(g) · gϕ(g′) · gg′x = ϕ(g)∑g′∈G

g(ϕ(g′) · g′x) = ϕ(g) · ga.

Math 223a Notes 50

Here is the original Hilbert 90. Let L/K be a cyclic extension and letGal(L/K) = 〈g〉. Then we have

H1(L/K,L×) =ker(N : L× → L×)

im(g − 1 : L× → L×; x 7→ gx/x).

So the theorem is saying that if x ∈ L× with Nx = 1, then there exists a y ∈ L×such that x = gy/y.

Example 13.3. Here is a nice application. Suppose L = Q(i)/K = Q and letx ∈ Q(i) be such that Nx = 1. That is, x is on the unit circle. Then thereexists a y ∈ Q(i) such that x = y/y. If we write y = a+ bi, then

x =a+ bi

a− bi=

(a2 − b2) + 2abi

a2 + b2.

This parametrizes Pythagorean triples.

13.3 Cohomology of profinite groups

Let’s say that G is a group. Recall that G is profinite group if it can bewritten as lim←−i∈I Gi with each Gi finite. For example,

Gal(K/K) = lim←−L/K finite

Gal(L/K)

is profinite. We can give a natural topology on G by setting ker(G → Gi) asthe neighborhood basis at 1.

Theorem 13.4. A topological group G is profinite if and only if G is compactand totally disconnected. G is profinite if and only if G = lim←−G/U where Uruns over open finite index subgroups of G.

So an open subgroup U ⊆ G necessarily has finite index.

Definition 13.5. A discrete G-module A is a Z-module A (with the discretetopology) with a group action G×A→ A such that this map is continuous, orequivalently, all stabilzers are open in G, or equivalently,

A =⋃

U⊆G open

AU .

Let U ⊆ G be open and normal. If V ⊆ U , we can inflate

InfU,V : Hq(G/U,AU )→ Hq(G/V,AV ).

Then we haveHq(G,A) = lim−→

UCG

Hq(G/U,AU ).

Math 223a Notes 51

For any K, we write Hi(K,A) = Hi(Gal(K/K), A). This is

Hi(K,A) = lim−→L/K finite Galois

Hi(Gal(L/K), A).

In particular, we have

H1(K,K×

) = lim−→L/K finite Galois

H1(L/K,L×) = 0.

This is now a profinite form of Hilbert 90.Because direct limit commute with exactness, we still have our long exact

sequences. We can also define cohomology in terms of continuous cochains. Amap ϕ : Gn → A is continuous if and only if it factors as

ϕ : Gn → (G/U)n → A

for some open subgroup U . Inflation and restriction also works for H ⊆ Gclosed. We don’t have Tate cohomology because the groups are not finite, butwe do have cup products.

Math 223a Notes 52

14 October 17, 2017

For K any field, we can look at the separable closure Ksep. This is just K ifcharK = 0. This what we should use if we want to do Galois theory.

Definition 14.1. If A is a GK = Gal(Ksep/K)-module, then

Hq(K,A) = Hq(Gal(Ksep/K), A).

Now the correct Hilbert theorem 90 is

H1(K, (Ksep)×) = 0.

14.1 Kummer theory

Let n be an integer relatively prime to charK. There is a short exact sequenceof Gal(Ksep/K)-modules

0→ µn(Ksep)→ (Ksep)×n−→ (Ksep)× → 0,

where the last map is surjective because Ksep has all n-th roots because xn − ais always irreducible for a 6= 0. The associated long exact sequence is

0→ µn(K)→ K×n−→→ K× → H1(K,µn)→ H1(K, (Ksep)×) = 0→ · · · .

Thus H1(K,µn) is just the cokernel of n : K× → K×. That is,

K×/(K×)nδ−→ H1(K,µn)

is an isomorphism.Suppose that µn ⊆ K. Then µn has trivial Gal(Ksep/K)-action, and so we

haveH1(K,µn) = Homcts(Gal(Ksep/K)→ µn).

If ϕ : Gal(Ksep/K)→ µn is a continuous homomorphism, then kerϕ is an opensubgroup of Gal(Ksep/K), which is Gal(Ksep/L) for some L. So this map ϕfactors as

ϕ : Gal(Ksep/K)� Gal(L/K) ↪→ µn.

The connecting homomorphism δ can be described explicitly as

δ(a) = ϕa : g 7→ g( n√a)

n√a∈ µn.

Suppose I have some cyclic Galois L/K of degree n, where I still assumeµn ⊆ K. Take some

(ϕ : Gal(L/K)∼=−→ µn) ∈ H1(L/K, µn) ↪→ H1(Ksep/K, µn).

Because δ : K×/(K×)n → H1(K,µn) is an isomorphism, this is inf ϕ = ϕa forsome a ∈ K×/(K×)n. Then

Gal(Ksep/L) = kerϕa = Gal(Ksep/K( n√a)).

This shows that L = K( n√a) by the Galois correspondence.

Math 223a Notes 53

14.2 Second Galois cohomology of unramified extensions

When L/K is an unramified extension of local fields, we want to study the secondcohomology H2(L/K) = H2(L/K,L×). Note that we have already classified allunramified extensions, and it turned out that Gal(L/K) is cyclic. Then

H2(L/K,L×) ∼= H0(L/K,L×) = K×/NL×vK ,∼=−−−→ Z/nZ = 1

nZ/Z.

But this is not very satisfactory. If L′/L/K is unramified, we can try tothink about what Inf : H2(L/K,L×) → H2(L′/K,L′×). But inflation is notdefined on H0. It turns out that this is just 1

nZ/Z ↪→ 1NZ/Z, but I’m getting

ahead of myself.

Lemma 14.2. Hq(L/K,O×L ) = 0 for all q provided that L/K is unramified.

Proof. Because L/K is cyclic, we can check this for H0 and H1. We have shownthat

H0(L/K,O×L ) = O×K/NO×L = 0.

For H1(L/K,O×L ) is almost Hilbert’s theorem 90. We have a short exact se-

quence 0→ O×L → L×vL−−→ Z→ 0, so

0→ O×K → K× → Z→ H1(L/K,O×L )→ H1(L/K,L×) = 0→ · · · .

Because L/K is unramified, vK : K× → Z is surjective and then we concludeH1(L/K,O×L ) = 0.

Now we know that all stuff coming from O×L is trivial, and so we see that

(vL)∗ : H2(L/K,L×)→ H2(L/K,Z)

is an isomorphism. We can further dimension shift this using 0 → Z → Q →Q/Z → 0. Here, Q has trivial cohomology because multiplication by [L : K] isan isomorphism Q→ Q. So we get a connecting homomorphism

δ : H1(L/K,Q/Z)∼=−→ H2(L/K,Z)

which is an isomorphism. Now the first cohomology can be computed as

H1(L/K,Q/Z) = Hom(Gal(L/K),Q/Z) ∼= 1nZ/Z,

because there is a preferred generator FrobK ∈ Gal(L/K).Now we have a canonical identification

invL/K : H2(L/K,L×)v∗−→ H2(L/K,Z)

δ←− H1(L/K,Q/Z)→ 1nZ/Z.

This is called the invariant map.Now I want to compute H2(Kunr/K, (Kunr)×). This is defined as the direct

limit

H2(Kunr/K, (Kunr)×) = lim−→L/K unram.

fin. Gal.

H2(L/K,L×) = lim−→n

H2(Kn/K,K×n ),

Math 223a Notes 54

where Kn/K is the unique unramified extension of degree n. Then we need toknow what the connected homomorphisms are, for n | N . This can be checkedby the diagram

H2(Kn/K,K×n ) H2(Kn/K,Z) H1(Kn/K,Q/Z) 1

nZ/Z

H2(KN/K,K×N ) H2(KN/K,Z) H1(KN/K,Q/Z) 1

NZ/Z,

Inf

(vKn )∗

Inf

δ−1

Inf

(vKN )∗ δ−1

because the restriction of the Frobenius is still the Frobenius. So we get

H2(Kunr/K, (Kunr)×) = lim−→n

H2(Kn/K,K×n )

invK ,∼=−−−−−→ lim−→n

1nZ/Z ∼= Q/Z.

Ultimately, we want to compute the Brauer group

Br(K) = H2(Ksep/K, (Ksep)×) ∼= Q/Z.

This is going to take more time. What we need to show is that if L/K is anyGalois extension with [L : K] = n,

inv : H2(L/K,L×)→ 1nZ/Z

is an isomorphism.To show this, we compare the two invariant maps

H2(Lunr/L, (Lunr)×)invL−−−→ Q/Z, H2(Kunr/K, (Kunr)×)

invK−−−→ Q/Z.

Note that Lunr = L · Kunr because the unramified closure is constructed byadjoining all roots of unity of order prime to p. So we get Gal(Lunr/L) ↪→Gal(Kunr/K). This map gives a restriction map

Res : H2(Kunt/K, (Kunr)×)→ H2(Lunr/L, (Lunr)×).

14.3 Some diagrams

We claim that the diagram

H2(Kunr/K, (Kunr)×) Q/Z

H2(Lunr/L, (Lunr)×) Q/Z

invK

Res ×n

invL

commutes. We are actually in the section titled “Some diagrams” in Serre’slocal fields part in Cassels–Frohlich.

Math 223a Notes 55

First note that e 6= 1 now, and so we have

(Kunr)× Z

(Lunr)× Z.

evK

vL

So we get

H2(Kunr/K, (Kunr)×) H2(Kunr/K,Z) H2(Kunr/K,Q/Z) Q/Z

H2(Lunr/L, (Lunr)×) H2(Lunr/L,Z) H2(Lunr/L,Q/Z) Q/Z.

e(vK)∗

Res

δ−1

Res

FrobK

Res f

(vL)∗ δ−1 FrobL

Here, the last square commutes because FrobK corresponds to x 7→ x|k| whileFrobL corresponds to x 7→ x|`| = (x|k|)f .

Proposition 14.3. If L/K is a finite extension, then n · invK = invL ◦Res.

We want to show that if L/K is finite Galois with degree n, thenH2(L/K,L×) ∼=1nZ/Z. First we will find an element uL/K ∈ H2(L/K,L×). If L/K, we can just

set uL/K = inv−1( 1n ). If L/K is ramified, we can set Kn/K to be the extension

of the same degree, and then compare the inflation maps

0 H2(Kn/K,K×n ) H2(Ln/K,L

×n )

0 H2(L/K,L×) H2(Ln/K,L×n ).

Inf

Inf

Then we will show |H2(L/K,L×)| ≤ n by induction.

Math 223a Notes 56

15 October 19, 2017

Last time, if Kn/K is an unramified finite Galois extension of local field ofdegree n = [Kn : K], then there is an isomorphism

inv : H2(Kn/K,K×n )

∼=−→ 1nZ/Z.

There is also a profinite version

inv : H2(Kunr/K, (Kunr)×)∼=−→ Q/Z.

The goal of today is to get, for any finite Galois extension L/K of local fieldswith n = [L : K],

inv : H2(L/K,L×)∼=−→ 1

nZ/Z.

This would imply

inv : H2(Ksep/K, (Ksep)×)∼=−→ Q/Z.

15.1 Second Galois cohomology of finite extensions

Because inv : H2(Kn/K,K×n ) → 1

nZ/Z is an isomorphism, we can look at theinverse image un of 1

n . We want to find the element uL/K ∈ H2(L/K,L×) oforder n. Let Ln = Kn · L so that Ln/L is a Galois extension with [Ln : L] | n.We have an inclusion Gal(Ln/L) ↪→ Gal(Kn/K) and so we have a restrictionmap

H2(Ln/L,L×n ) H2(Lunr/L, (Lunr)×) Q/Z

H2(Kn/K,K×n ) H2(Kunr/K, (Kunr)×) Q/Z.

inv

Res

inv

Res ×n

Because un maps to 1n in Q/Z, this shows that un maps to 0 in the upper right

Q/Z. That is, Resun = 0.Now note that we have a inflation-restriction exact sequence

H2(Ln/L,L×n )

H2(Ln/K,L×n )

H2(Kn/K,K×n ) H2(L/K,L×).

ResL

InfKn

ResInfL

So we can push un toH2(Ln/K,L×n ) and then because it vanishes atH2(Ln/L,L

×n ),

it comes from H2(L/K,L×). Let us call this uL/K ∈ H2(L/K,L×). This isgoing to be the order of element n we want.

To show that H2(L/K,L×) is actually generated by un, we will show that|H2(L/K,L×)| ≤ n. We induct on [L : K].

Math 223a Notes 57

15.2 Estimating the size

In the base case, assume that L/K is cyclic. Then

H2(L/K,L×) = H0(L/K,L×) ∼= K×/NL×.

This can be shown on the nose. One way of doing it is to use the Herbrandquotient. You can use the following lemma.

Lemma 15.1. If L/K is a finite Galois extension of local fields, there existsan Gal(L/K)-invariant open subgroup V ⊆ O×L such that

Hq(L/K, V ) = 0

for all q ∈ Z.

Proof. We’ll choose V to be coinduced. Let V = expp(A) for some coinduced

A ⊆ O+L . We know that L is coinduced as a Gal(L/K)-module. That is, there

exists an a ∈ L such that L =⊕

g∈Gal(L/K)K · (ga). Without loss of generality,

we may assume that a ∈ (πL)r by rescaling. (We are going to choose r later.)Now define

A =⊕

g∈Gal(L/K)

OK · (ga) ⊆ (πL)r.

Now take r large enough such that expp : πrLOL → UL,r is an isomorphism

of topological groups. This V = expp(A) is now an open subgroup of O×L . AlsoV is coinduced because A is coinduced.

Recall from the homework that if G is cyclic and A is a G-module, then theHerbrand quotient is

h(A) =|H0(G,A)||H1(G,A)|

if both are defined. This satisfies

• 0 → A → B → C → 0 exact implies h(B) = h(A)h(C) if any two aredefined,

• h(A) = 1 if A is finite,

• h(Z) = |G|.

Proposition 15.2. h(O×L ) = 1.

Proof. There is a short exact sequence

0→ V → O×L → O×L /V → 0

or Gal(L/K)-modules. Then h(V ) = 1 and h(OL/V ) = 1 because O×L /V isfinite. This shows that h(O×L ) = 1.

Proposition 15.3. h(L×) = n.

Math 223a Notes 58

Proof. Again we have a short exact sequence

0→ O×L → L× → Z→ 0

and look at the Herbrand quotients.

So we have|H0(L/K,L×)||H1(L/K,L×)|

= n

and by Hilbert’s theorem 90, H1(L/K,L×) = 0. This shows that |H2(L/K,L×)| =n.

Theorem 15.4. Let L/K be any Galois extension. Then |H2(L/K,L×)| ≤n = [L : K].

Proof. We induct on [L : K]. The base case is when [L : K] is prime. This caseis covered by the cyclic case. Otherwise Gal(L/K) is solvable so there exists anormal subgroup H C Gal(L/K). Let M = LH .

We have a inflation-restriction sequence

0→ H2(M/K,M×)→ H2(L/K,L×)→ H2(L/M,L×).

Then we have

|H2(L/K,L×)| ≤ |H2(M/K,M×)||H2(L/M,L×)| ≤ [M : K][L : M ] = [L;K]

by the inductive hypothesis.

Theorem 15.5. If L/K is any Galois extension of local fields, H2(L/K,L×)is cyclic of order n = [L : K].

Proof. There is an element uL/K ∈ H2(L/K,L×) of order n, and we have theestimate |H2(L/K,L×)| ≤ n.

15.3 Brauer group of a local field

Theorem 15.6. If K is a local field, the inflation map

H2(Kunr/K, (Kunr)×)→ H2(Ksep/K, (Ksep)×) = Br(K)

is an isomorphism.

Proof. If is injective because inflation is injective. (All H1 vanish.) Now wejust need surjectivity. The Brauer group is limL/K H

2(L/K,L×), and so anyelement of Br(K) is represented by some element auL/K ∈ H2(L/K,L×). Butrecall that we have

H2(Ln/K,L×n )

H2(K/K,K×) H2(L/K,L×).

InfKn InfL

So auL/K is also represented by aun ∈ H2(Kn/K,K×n ) in the direct limit.

Math 223a Notes 59

So we have an isomorphism

invK : Br(K)∼=−→ Q/Z.

Proposition 15.7. For L/K a finite extension of local fields of degree n = [L :K], the following commutes:

Br(K) Q/Z

Br(L) Q/Z.

invK

Res ×n

invL

Proof. Just use the definition Br(K) ∼= H2(Kunr/K, (Kunr)×) and the resultlast time.

Corollary 15.8. Let E/L/K be a tower of local fields and assume E/K isGalois. The restriction map

ResL/K : H2(E/K,E×)→ H2(E/L,E×)

sends uE/K to uE/L.

Proof. Look at the diagram

H2(E/K,E×) H2(E/L,E×)

1[E:K]Z/Z

1[E:L]Z/Z

invK

Res

invL

[L:K]

and track where uE/K and uE/L maps to.

Similarly, if L/K is Galois, then Inf(uE/K) = [L : K]uL/K .

15.4 Tate’s theorem

Theorem 15.9 (Tate). Let G be a finite group and A be a G-module. Assumethat for all H ⊆ G, H1(H,A) = 0 and H2(H,A) is cyclic of order |H|. If agenerates H2(G,A), then

−` a : Hq(G,Z)→ Hq+2(G,A)

is an isomorphism for all q ∈ Z.

Let G = Gal(L/K) and A = L×. This satisfies the assumptions by Hilbert’stheorem 90 and the theorem we have just proven. Apply the theorem to q = −2.Then we get

Gal(L/K)ab = H2(L/K,Z)∼=−→ H0(L/K,L×) = K×/NL×,

This which comes from taking the cup product with uL/K .

Math 223a Notes 60

Idea of proof. We construct some module M such that there exists a long exactsequence

· · · → Hq(G,Z)−` a−−−→ Hq+2(G,A)→ Hq+2(G,M)→ Hq+1(G,Z)→ · · · .

Then we will show that Hq(H,M) = 0 for all H ⊆ G and some two consecutiveq. This implies (by the homework) that Hq(H,M) = 0 for all H ⊆ G and allq.

Math 223a Notes 61

16 October 24, 2017

We need to prove Tate’s theorem.

Theorem 16.1 (Tate). Let G be a finite group and A a G-module such thatfor all H ⊆ G we have H1(H,A) = 0 and H2(H,A) is cyclic of order |H|. If ais a generator of H2(G,A), then

−` a : Hq(G,Z)→ Hq+2(G,A)

is an isomorphism for all q ∈ Z.

Proof. Last time I gave the idea. Let us set this up. By dimension-shifting twice,there exists a B = (A+)+ such that Hq(G,B) ∼= Hq+2(G,A). Let b ∈ H0(G,B)be a preimage of a ∈ H2(G,A) under this isomorphism. Then

Hq(G,Z) Hq+2(G,A)

Hq(G,B)

−` a

−` b∼=

commutes.The condition can be reformulated as

H−1(H,B) = 0, H0(H,B) ∼= Z/|H|Z

for all H ⊆ G. If b is a generator of H0(G,B), we want to show that

−` b : Hq(G,Z)→ Hq(G,B)

is an isomorphism. This maps are induced by Z→ B with n 7→ nb.Letting B′ = B ⊕ Z[G], we get an isomorphism Hq(G,B) ∼= Hq(G,B′)

induced by B ↪→ B′. Define

i : Z→ B′ = B ⊕ Z[G]; n 7→ (nb, nN),

and we get a short exact sequence

0→ Z i−→ B′j−→M = coker i→ 0

of G-modules. This is analogous to the mapping cylinder construction in topol-ogy.

This gives a long exact sequence

· · · → H−1(H,B′) = 0→ H−1(H,M)δ−→ H0(H,Z)

−` b−−−→ H0(H,B)→ H0(H,M)j∗−→ H1(H,Z) = 0→ · · · .

Math 223a Notes 62

Here, the map H0(H,Z)→ H0(H,B) is given by [n] 7→ nb. Here, b is a generatorof B/NGB so b is a generator of B/NHB = H0(H,B). Because both are cyclicgroups of order |H|, the map −` b : H0(H,Z)→ H0(H,B) is an isomorphism.

This shows that H−1(H,M) = H0(H,M) = 0, for all H ⊆ G. The home-work problem shows that all cohomology of M are Hq(G,M) = 0. Then thelong exact sequence implies that

−` b : Hq(G,Z)→ Hq(G,B)

are isomorphisms.

16.1 Reciprocity map

Last time, for L/K a Galois extension of local field, we used Tate’s theorem togive an isomorphism

Gal(L/K)ab = H−2(L/K,Z)uL/K−−−→ H0(L/K,L×) = K×/NL×.

This gives the reciprocity map

θL/K : K×/NL× → Gal(L/K)ab.

People also write this asθL/K(a) = (a, L/K).

This is called the norm residue symbol and can be thought of as the obstruc-tion of a being a norm. It is reminiscent of quadratic reciprocity. Now I amgoing to give another way of describing this map, which may be more convenientin some contexts.

Proposition 16.2. For [a] ∈ K×/NL×, its image θL/K([a]) ∈ Gab is charac-

terized by: for any χ ∈ Hom(G,Q/Z) = H1(L/K,Q/Z)δ,∼=−−→ H2(L/K,Z),

χ(θL/K([a])) = invL/K([a]` δχ) ∈ Q/Z.

The proof involves a lot of homological algebra chasing, and I am going todo part of it and leave part of it on the homework.

Proof. By definition of θL/K , we have

[a] = [θL/K(a)]ùL/K .

So

inv([a]` δχ) = inv(θL/K(a)ùL/K ` δχ) = inv((θL/K(a)`χ)ùL/K)

= inv(δ(χ` θL/K(a))ùL/K).

What is χ` θL/K(a)? Here χ ∈ H1(G,Q/Z) and θL/K(a) ∈ H−2(G,Z) ∈Gab. We claim that

χ`[g] = χ(g) ∈ Q/Z.

Math 223a Notes 63

The proof of this fact is homework. Given this fact, we have

inv([a]` δχ) = inv(δ(χ(θL/K(a)))`uL/K).

We also claim that the map

1

|G|Z/Z ∼= H−1(G,Q/Z)

δ−→ H0(G,Z) ∼= Z/|G|Z

is just ×|G|. So

inv([a]` δχ) = inv([L : K]χ(θL/K(a))uL/K)

= [L : K]χ(θL/K(a))1

[L : K]= χ(θL/K(a)).

This is the desired result.

The consequence of this is the following compatibility.

Proposition 16.3. Let E/L/K be a tower with E/K and L/K Galois. Thereis a natural quotient map πE/L : Gal(E/K)ab → Gal(L/K)ab. For any a ∈ K×,

πE/K(θE/K([a])) = θL/K([a]).

Proof. It is enough to show that

χ(πE/L(θE/K([a]))) = invL/K([a]` δχ).

But we can always lift χ to χ : Gal(E/K)→ Q/Z and then the left hand side is

χ(θE/K([a])) = invE/K([a]` δχ) = invL/K([a]` δχ)

by the compatibility of Brauer groups.

The upshot is that we have a commutative diagram

K×/NE× Gal(E/K)ab

K×/NL× Gal(L/K)ab.

θE/K

θL/K

So we get a map

θ/K : K× → lim←−L/K fin. Gal.

Gal(L/K)ab = lim←−L′/K fin. abe.

Gal(L′/K) = Gal(Kab/K)

that has dense image.

Math 223a Notes 64

16.2 Group of universal norms

What is the kernel of θ/K : K× → Gal(Kab/K)? This is going to be simply theintersection of the kernels, and is

ker θ/K =⋂L/K

NL/KL×.

This is called the group of universal norms.

Proposition 16.4. The only universal norm in K× is 1.

Definition 16.5. A subgroup A ⊆ K× is called normic if there exists a GaloisL/K such that A = NL/KL

×.

We’ll show that any finite index open subgroup of K× is normic. For profi-nite groups, open subgroups have finite index. But note that K× is not profi-nite/compact, and in particular contains O×K . On the other hand, finite indexsubgroups of K× are open. This is because (K×)n contains an open ball around1.

Proposition 16.6. If L/K is Galois, and L′/K is the maximal abelian subex-tension, then NL/KL

× = NL′/K(L′)×.

Proof. The inclusion NL/KL× ⊆ NL′/K(L′)× is clear. On the other hand,

K/NL/KL× ∼= Gal(L/K)ab = Gal(L′/K) ∼= K/NL′/K(L′)×.

This means that the quotient map K/NL/KL× → K/NL′/K(L′)× is an isomor-

phism.

Math 223a Notes 65

17 October 26, 2017

Recall that at the end of last time we defined that a subgroup A ⊆ K× isnormic if there exists some Galois extension L/K such that A = NL/KL

×.Our goal is to prove that finite index subgroups of K× are normic.

17.1 Normic subgroups

We showed that we may only take L/K abelian. So we have a map

{abelian extensions L/K} −→ {normic subgroups of K×}.

This is order-reversing, i.e., L ⊆ L′ implies NL′ ⊆ NL. By local class fieldtheory,

[L : K] = [K× : NL×].

This map is clearly surjective. Because we have θ/K : K× → Gal(Kab/K),there is also an inverse map

A 7→ (Kab)θ/K(A).

To checked that this is indeed an inverse, we need to check that (Kab)θ/K(NL×) =

L. The inclusion (Kab)θ/K(NL×) ⊇ L is easily verified, and equality follows from[L : K] = [K× : NL×].

Proposition 17.1. If A = NL× and B = NM×, then A ∩B = N(LM)×.

Proof. We clearly haveN(LM)× ⊆ NL×∩NM×. If a ∈ K× such that a ∈ NL×and a ∈ NM× then θL/K([a]) = 1L and θM/K([a]) = 1M . This implies thatθLM/K([a]) = 1LM . That is, a ∈ N(LM)×.

Proposition 17.2. If A = NL× is normic and B ⊇ A, then B is normic.

Proof. We have Gal(L/K) ∼= K×/A. We may consider B/A as a subgroup andK×/A, and this will correspond to some θL/K(B/A) ⊆ Gal(L/K).

Take M = LθL/K(B/A). Then

K×/A Gal(L/K)

K×/NM× Gal(M/K).

θL/K∼=

θM/K∼=

Here,

θL/K(B/A) ∼= Gal(L/M) = ker(Gal(L/K)→ Gal(M/K))

= θL/K(ker(K×/A→ K×/NM×)) = θL/K(NM×/A).

So B = NM×.

Math 223a Notes 66

Theorem 17.3. Suppose K is local of characteristic 0. If A ⊆ K× is finiteindex (and thus open), then A is normic.

Proof. Note that if A ⊆ K× is finite index, then A ⊇ (K×)n for some n. So itis enough to show that (K×)n are normic.

Let us first do this in the case when µn ⊆ K. Let L be the maximal exponentn of an abelian extension of K. By Kummer theory,

L = K({ n√a : a ∈ K×/(K×)n})

= composium of all extensions {K( n√a)}.

ThenNL× =

⋂a∈K×/(K×)n

NK( n√a)× ⊇ (K×)n.

To show that NL× = (K×)n, it is enough to show that [K× : (K×)n] = [K× :NL×] = |Gal(L/K)|.

We are going look at the Pontryagin dual

Hom(Gal(L/K), µn) = Hom(Gal(Kab/K), µn) = H1(Kab/K, µn) ∼= K×/(K×)n.

This shows that Gal(L/K) and K×/(K×)n have the same order.We now generalize to the case when K might not contain µn. Let K ′ =

K(µn). ThenNL′/K′(L

′)× = ((K ′)×)n.

We don’t know if L′/K is Galois, so take L/K such that L ⊇ L′ and L/K isGalois. Then

NL/KL× ⊆ NL′/K(L′)× = NK′/KNL′/K′(L

′)× = NK′/K((K ′)×)n ⊆ (K×)n.

This implies that (K×)n is normic.

This gives us

Gal(Kab/K) = lim←−L/K abelian

(K×/NL×) = lim←−[K×:A]<∞

K×/A = K×.

This group K× is called the profinite completion of K.We can be more explicit in describing K×. We have K× = O×K × πZ. So

K× = O×K × Z× = O×K × πZ.

17.2 Quadratic reciprocity

Let us get back to general local field of any characteristic. Let L/K be unram-ified.

Proposition 17.4. For a ∈ K×, θL/K([a]) = (FrobL/K)v(a).

Math 223a Notes 67

Proof. From last time, θL/K([a]) is characterized by

χ(θL/K([a])) = inv([a]` δχ).

Recall that the invariant map was defined as

H2(L/K,L×)v∗−→ H2(L/K,Z)

δ−1

−−→ H1(L/K,Q/Z)evalFrob−−−−−→ 1

nZ/Z.

By functoriality, v∗ maps [a]` δχ to v∗([a])` δχ = v(a) · δx. Then δ−1 sends it

to v(a)χ and then evaluating at the Frobenius gives v(a)·χ(Frob ) = χ(Frobv(a)).This give the right thing.

We are now going to look at quadratic extensions, because these are thesimplest examples. Let L/K be a quadratic extension so that Gal(L/K) ∼=Z/2Z. Then K×/NL× ∼= Z/2Z.

Example 17.5. Let K = Qp and L = Qp(√a). We have

a ∈ Q×p /(Q×p )2 = {1, u, p, up}

where u ∈ Z×p is a unit with (up ) = −1. So there are three cases to consider.

If a = u, then L/K is unramified and NL× = {x : vp(x) even}, which is anindex 2 subgroup of Q×p , generated by (Q×p )2 and u.

If a = p, then L/K is totally ramified. Then we can say that NL× isgenerated by (Q×p )2 and −p.

Likewise, if a = up then L/K is totally ramified and NL× is generated by(Q×p )2 and −up.

Let us look a global example.

Example 17.6. Let ` be an odd prime. Take `∗ = (−1)(`−1)/2` (≡ 1 mod 4)and take K/Q = Q(

√`∗)/Q. Let a ∈ Z with a > 0 and (a, 2`) = 1. For each

place v of Q, consider the function

a 7→ (a,Kw/Qv) = θKw/Qv (a) ∈ Gal(Kw/Qv) ↪→ Gal(K/Q) ∼= {±1}

where w is a place of K above v.If v = `, we have total ramification and

NL×w = (Q×` )2 × {1, (−`∗)}.

So for a ∈ Z×` if and only (a` ) = 1, i.e.,

(a,Kw/Q`) =(a`

).

Let v = p be an odd prime with p 6= `. Thre are two cases. If `∗ is a squarein Qp, then Kw = Qp and so

(a,Kw/Qp) = 1

Math 223a Notes 68

for all a ∈ Z×p . If not, then Kw = Qp(√`∗) is unramified. So we just have

(a,Kw/Qp) = (−1)vp(a).

In general, we see that

(a,Kw/Qp) =(`∗a

)vp(a)

.

Let us now consider the case p = 2. Because `∗ ≡ 1 mod 4, either `∗ ∈ (Q×2 )2

if `2 ≡ 1 mod 8 or Q2(`∗) = Q2(√

5) if `∗ ≡ 5 mod 8. Since a is a unit modulo2, we always get

(a,Kw/Q2) = 1.

Finally, consider the case v =∞. Either Kw = R in which case (a,Kw/R) =1, or Kw = C in which case the function takes signs. Because a > 0, we get(a,Kw/C) = 1.

Now we have computed all the symbols. We have not talked about this, butthe global reciprocity law tells us that∏

v

(a,Kw/Q) = 1.

Say a = p. Then (p`

)(`∗p

)= 1.

This is a form of the standard quadratic reciprocity.

17.3 Reciprocity map on units

Let L/K be a Galois extension of local fields. Let T/K be the maximal unram-ified subextension. We have the characterization T = LI(L/K).

Proposition 17.7. We have θL/K(O×K) = I(L/K).

Proof. We first prove θL/K(O×K) ⊆ I(L/K). Note that θT/K(O×K) = {1}. Then

K× Gal(L/K)

Gal(T/K)

θL/K

θT/K

shows the inclusion.Now let us prove the other direction. We know that θL/K : K× → Gal(L/K)

is surjective. It is enough to show that if θL/K(x) ∈ I(L/K) then [x] = [u] for

some u ∈ O×K . If θL/K([a]) ∈ IL/K then θT/K([a]) = FrobvK(a)T/K = 1 and so

[T : K] = f divides v(a). Take b = a ·N(πL)−vk(a)/f . This is a unit.

Math 223a Notes 69

18 October 31, 2017

Today we are moving to our next topic, Lubin–Tate theory. We have localKronecker–Weber for Qp from the problem set. Let K be a local field. We havean Artin map

θ/K : K× = O×K × πZ → Gal(Kab/K)

This map is not surjective, but we can extend it to

θ/K : K× = O×K × πZ → Gal(Kab/K).

So we can decompose

Kab

Kunr Kπ

K

O×K πZ

πZ O×K

When K = Qp, we can actually describe these fields. Note that Kπ dependson π, so we choose π = p ∈ Qp. Then

Qabp = Qp(ζ∞)

Qunrp = Qp(ζ(n,p)=1) Qp(ζp∞)

Qp

as you’ve computed in your problem set. Now we want to see what θ/Qp(a) is.Let a = pru. On Qunr

p , we know that θL/K(a) = FrobrL/K if L/K is unramified.

So it is going to map ζn 7→ ζpr

n . On the other hand, it is harder to figure out

what ζpk is mapped to. It turns out that ζpk 7→ ζu−1

pk .

There is a global proof of this fact. If we use global reciprocity on Q(ζkp )/Q,

the images should cancel out so it has to be ζu−1

pk . But we want to do everythinglocally, and there is a proof using Lubin–Tate theory.

18.1 Affine group schemes

Here is one motivating question. Let A be any ring. For which (commutative)A-algebras R is HomA−Alg(R,B) naturally a group for all A-algebras B?

Example 18.1. If R = A[t], then Hom(R,B) ∼= B as sets, so it is a groupunder addition. This is called the additive group and we say B = Ga(B).

Math 223a Notes 70

Example 18.2. If R = A[t, t−1], then Hom(R,B) ∼= B× = Gm(B). This iscalled the multiplicative group.

Example 18.3. Set R = A[t, t−1]/(tn−1). Then Hom(R,B) ∼= µn(B) and thisis a group scheme µn.

Example 18.4. Here is a non-abelian example. We want Hom(R,B) ∼= GLn(B).To achieve this, we can take

R = A[(ti,j)1≤i,j≤n, (det[tij ])−1].

The answer is that we want R to be a Hopf algebra. This means that thereexist maps

• comultiplication ∆ : R→ R⊗R• coinversion S : R→ R

• coidentity ε : R→ A

with axioms

• coassociativity

R R⊗R

R⊗R R⊗R⊗R

∆

∆ 1⊗∆

∆⊗1

• identity

R R⊗R R

id

1⊗ε

• coinversion

R⊗R R⊗R

R A R

R⊗R R⊗R

1⊗s

ε

∆

∆

s⊗1

This is exactly what we need to get associativity

Hom(R,B)×Hom(R,B)×Hom(R,B) Hom(R,B)×Hom(R,B)

Hom(R,B)×Hom(R,B) Hom(R,B).

Math 223a Notes 71

Example 18.5. Let R = A[t]. The comultiplication map is

∆ : A[t]→ A[t]⊗A[t]; t 7→ 1⊗ t+ t⊗ 1.

Then S = A[t] → A[t] is t 7→ −t and ε : A[t] → A is t 7→ 0. This is the rightmap because

∆∗ : Hom(A[t]⊗A[t], B)→ Hom(A[t], B)

is given by (ϕ,ψ) 7→ ϕ+ ψ because t 7→ 1⊗ t+ t⊗ 1 is sent to ψ + ϕ.

Example 18.6. Let R = A[t, t−1]. Here,

∆ : A[t, t−1]→ A[t, t−1]⊗A[t, t−1]; t 7→ t⊗ t, t−1 7→ t−1 ⊗ t−1,

S : A[t, t−1]→ A[t, t−1]; t 7→ t−1, t−1 7→ t,

ε : A[t, t−1]→ A; t 7→ 1, t−1 7→ 1.

Why are group schemes useful? Suppose we have an affine abelian groupscheme G defined over a field K. Assume that it comes from some Hopf algebraR, and also assume that R is finite over A (i.e., is a finite dimensional K-vectorspace). Then we can look at G(K) = Hom(R,K). This is going to be a finitegroup with an action of Gal(K/K).

If we take G = µn, then we are going to get µn(K) = G(K). So we get

Gal(K/K)→ Aut(µn(K)),

and µn(K) ∼= Z/nZ. This sounds like a great plan, but we need a source offinite group schemes. So we would like a group scheme with lots of subschemes.Gm is a good example. Note that µn is actually

ker(×[n] : Gm → Gm).

The problem is that there aren’t many affine abelian group schemes. It canbe proven that over K of characteristic 0, the only affine abelian group schemesare Ga, Gm, µn, and their products. There are two ways to go from here:

• Drop the affine condition. Here we may consider elliptic curves, abelianvarieties, etc. These all have a lot of subgroup schemes. If you have anelliptic curve E, the n-torsion points E[n] are group schemes.

• Take the completion. This works only over local fields.

There is also another problem. In general, Aut of an abelian group is notnecessarily abelian. So we need some extra structure on G(K) that rigidifiesthe structure.

18.2 Formal groups

Let A be any ring. We are going to consider the ring A[[x]] and we want tomake this into something like a Hopf algebra. Say, e.g., A = OK . I can look at

Homcts(A[[x]],OK),

Math 223a Notes 72

where A[[x]] has the x-adic topology. This group is isomorphic to πKOK becausewe are looking only at continuous maps.

We need something like comultiplication, but we are not going to look atA[[x]]→ A[[x]]⊗A A[[x]]. Instead, we look at

A[[x]]→ A[[x1, x2]].

This is because

Homcts(A[[x1, x2]],OL)→ Homcts(A[[x1]],OL)×Homcts(A[[x2]],OL)

is bijective.

Definition 18.7. A formal group law F over A is a power series F (x, y) ∈A[[x, y]] such that

• F (x, y) ≡ x+ y (mod deg 2),

• F (x, y) = F (y, x),

• F (x, F (y, z)) = F (F (x, y), z),

• there exists an iF (x) ∈ A[[x]] with F (x, iF (x)) = 0,

• F (0, y) = y, F (x, 0) = x.

Example 18.8. The additive group Ga still works as a formal group. Thecomultiplication x 7→ x⊗ 1 + 1⊗ x now is interpreted as F (x, y) = x+ y.

,

Example 18.9. For the multiplicative group, we need to shift the origin. ThenF (x, y) = x+ y + xy. The power series

iF (x) =1

1 + x− 1 = −x+ x2 − x3 + x4 − · · ·

works as an inverse.

If A = OK and L/K is a finite extension, then πLOL using the group lawgiven by F .

Math 223a Notes 73

19 November 2, 2017

We were doing formal groups.

Definition 19.1. A formal group over A is a power series F (x, y) ∈ A[[x, y]]such that

• F (x, y) ≡ x+ y (mod deg 2),

• F (x, y) = F (y, x),

• F (x, F (y, z)) = F (F (x, y), z),

• there exists an iF (x) ∈ A[[x]] such that F (x, iF (x)) = 0 (this is actuallyredundant),

• F (0, y) = y and F (x, 0) = x.

19.1 Morphisms between formal groups

A homomorphism h : F → G is a power series h ∈ A[[x]] with h(0) = 0 suchthat

G(h(x), h(y)) = h(F (x, y)) ∈ A[[x, y]].

Recall that if F is a formal group over OK and L/K is a finite extension thenF (πLOL) is πLOL made into a group by a +F b = F (a, b). If Ksep/K is theseparable closure, then we can look at its maximal ideal

Ksep ⊇ OKsep ⊇ pKsep ,

which is not a principal ideal now. So we can define F (pKsep).If h : F → G is a formal group homomorphism, we get group homomor-

phismsh : F (πLOL)→ G(πLOL); a 7→ h(a),

or F (pKsep)→ G(pKsep).

Example 19.2. Take A = Qp. The additive and multiplicative groups are

Ga(x, y) = x + y and Gm(x, y) = x + y + xy. The power series h = expp−1 ∈Qp[[x]] is then a formal group homomorphism because

Gm(h(x), h(y)) = expp(x+ y)− 1 = h(Ga(x, y)).

There is also an inverse h−1(y) = logp(1 + y). So Ga and Gm are isomorphicover Qp. These are not isomorphic over Zp. For instance, L = Qp(ζp) has

ζp − 1 ∈ πLOL and so Qm(πLL) has p-torsion. But Qa(πLL) never has torsion.

Example 19.3. Let us look at Gm over Zp. For n ∈ Z, define

hn(x) = (x+ 1)n − 1 =∑k≥1

(n

k

)xk.

Then hn ∈ End(Gm). Actually this can be defined for all n ∈ Zp. The coeffi-cients are still going to be integers in Zp.

Math 223a Notes 74

If F and G Are formal groups, then Hom(F,G) is an abelian group, withh1+Gh2 = G(h1, h2). Then End(F ) = Hom(F, F ) is a (possibly non-commutative)ring with addition +F and multiplication composition. In the above example,you can show that

hn +Gm hm = hn+m, hn ◦ hm = hnm.

So we have a ring homomorphism

Zp → End(Gm).

In this case, this is actually and isomorphism although I am not going to provethis.

Suppose F and G be formal groups over OK . Last time I talked aboutgetting cyclotomic field from group schemes. Given h : F → G, how can wethink of kerh? We have a group homomorphism

F (pKsep)→ G(pKsep).

and so we can look at the kernel of this homomorphism. Often we are goingto be in the case when h is a monic polynomial. Then the kernel is the set ofsolutions to h(X) = 0. The ring

OK [[x]]/h(x)

is then going to be a finite OK-algebra, and this gives a finite group schemeover OK .

Example 19.4. For instance, if K = Qp and F = Gm, with hp, we would getthe ring OK [[x]]/(x + 1)p − 1. Then our formal group obtained by taking thekernel is µp(OK).

19.2 Lubin–Tate theory

We can look at Gal(Ksep/K) acting on {x ∈ pKsep : h(X) = 0} = ker(F (pKsep)→G(pKsep)).

Definition 19.5. Let K be a local field with |k| = q and π ∈ OK be a uni-formizer. The Lubin–Tate series is given as

Fπ = {f(x) ∈ OK [[x]] : f(x) ≡ πx mod deg 2, f(x) ≡ xq mod π}.

Example 19.6. For K = Qp, the polynomial f(x) = (x + 1)p − 1 is a Lubin–Tate series. Actually constructing them are easy, because you can take anythinglike f(x) = xp + πx.

Theorem 19.7. (a) For any f ∈ Fπ, there exist a unique formal group Ffsuch that f ∈ End(Ff ).

Math 223a Notes 75

(b) For any a ∈ OK , there exists a unique endomorphism [a]f of Ff such that[a]f commutes with f and [a]f ≡ aX (mod deg 2). Then the map

OK → End(Ff ); a 7→ [a]f

is an isomorphism of rings.

(c) If f, g ∈ Fπ then Ff ∼= Fg as formal groups over OK .

We’re going to have a “workhorse lemma”.

Lemma 19.8. For f, g ∈ Fπ and a linear polynomial Φ1(x1, . . . , xr) over OK ,the equation

f(Φ(x1, x2, . . . , xr)) = Φ(g(x1), g(x2), . . . , g(xr))

will have a unique solution Φ ∈ OK [[x1, . . . , xr]] such that Φ ≡ Φ1 (mod deg 2).

Proof. We do this by induction. It is enough to show that there is a uniquepolynomial Φk ∈ OK [x1, . . . , xr] of degree k such that Φk ≡ Φ1 (mod deg 2)and

f(Φk(x1, . . . , xr)) ≡ Φk(g(x1), . . . , g(xr)) (mod deg(k + 1)).

We already have our base case k = 1. Assume we have our unique Φkalready. Then any Φk+1 satisfying the condition must reduce to Φk modulodegree k + 1. Then Φk+1 = Φk + Q where Q is a homogeneous polynomial ofdegree k + 1. Here

f(Φk+1(x1, . . . , xr)) ≡ f(Φk(x1, . . . , xr)) + πQ(x1, . . . , xr) (mod deg(k + 2))

and

Φk+1(g(x1), . . . , g(xr)) = Φk(g(x1), . . . , g(xr)) +Q(g(x1), . . . , g(xr))

≡ Φk(g(x1), . . . , g(xr)) + πk+1Q(x1, . . . , xr) (mod deg(k + 2)).

This shows that there exists a unique Q. But we still need to show that Q hascoefficients in OK .

Working modulo π, we have

f(Φk(x1, . . . , xn)) ≡ Φk(x1, . . . , xr)q ≡ Φk(xq1, . . . , x

qr) ≡ Φk(g(x1), . . . , g(xr)).

So Q actually has coefficients in OK because π − πk+1 is π times a unit.

Now we can start proving the theorem. We construct Ff as the unique powerseries in OK [[x, y]] such that

Ff (f(x), f(y)) = f(Ff (x, y))

and Ff (x, y) ≡ x+ y (mod deg 2).We need to show that Ff is a commutative formal group. Commutativity

follows from the uniqueness. We also get associativity because Ff (Ff (x, y), z)and Ff (x, Ff (y, z)) both have the same linear term and satisfy the same func-tional equation. To show that Ff (x, 0) = x, we note that both are power seriesin x that commute with f . Again uniqueness shows that Ff (x, 0) = x. Thisproves (a).

Math 223a Notes 76

20 November 7, 2017

Recall our setup

Fπ = {f ∈ OK [[x]] : f(x) ≡ πx mod deg 2, f(x) ≡ xq mod π}.

We proved the lemma saying that if f, g ∈ Fπ and Φ1 ∈ OK [x1, . . . , xr] is ahomogeneous linear polynomial, then

f(Φ(x1, . . . , xr)) = Φ(g(x1), . . . , g(xr))

has a unique solution Φ ∈ OK [x1, . . . , xr] such that Φ ≡ Φ1 mod deg 2.

Proof of Theorem 19.7. We have shown (a) last time. If f ∈ Fπ then thereexists a unique formal group Ff such that f is an endomorphism of Ff .

Next we want to show that F has lots of endomorphisms and we want toshow that Ff ∼= Fg for all f, g ∈ Fπ. We accomplish both of this by defining[a]g,f for a ∈ OK such that

[a]g,f ◦ f = g ◦ [a]g,f

and [a]g,h ≡ aX mod deg 2. Uniqueness immediately shows that

[a]g,f +Fg [b]g,f = [a+ b]g,f , [a]h,g ◦ [b]g,f = [ab]h,f .

Note that [a]g,h is a formal group homomorphism Ff → Fg because

Ff ◦ ([a]g,f , [a]g,f ), [a]g,f ◦ Fg

both have the property that Φ ◦ (f × f) = g ◦Φ and have linear terms ax+ ay.So they are equal by uniqueness.

Specializing to f = g and writing [a]f = [a]f,f shows that the map

OK → End(Ff ); a 7→ [a]f

is a ring homomorphism. For any f, g ∈ Fπ and a ∈ O×K , we have

[a]f,g ◦ [a−1]g,h = x

and so [a]f,g is an isomorphism between Ff and Fg.

20.1 Torsion points in a formal group

Definition 20.1. A formal OK-module F is a formal group F over OK alongwith a ring homomorphism

OK → End(F ); a 7→ [a]F

such that [a]F ≡ ax mod deg 2.

Math 223a Notes 77

For example Ff is a formal OK-module. If F is a formal OK-module, thenF (pKsep) is an actual OK-module with

a · x = [a]f (x) ∈ pKsep .

Definition 20.2. If f is a Lubin–Tate series and Ff the corresponding formalgroup, we define

Ef,n = {x ∈ pKsep : [πn]f (X) = 0}.

These are the “πn-torsion of f”.

Let Kπ,n be the field generated by Ef,n, and write

Ef =⋃n

Ef,n, Kπ =⋃n

Kπ,n.

Proposition 20.3. Let f, g ∈ Fπ. There exists an isomorphism

Ef,n∼=−→ Eg,n; x 7→ [1]g,f (x).

Then field Kπ,n does not depend on the choice of f .

Proof. Note that x ∈ Ef,n if and only if [πn]f (x) = 0 if and only if

0 = [1]g,f [πn]f (x) = [πn]g[1]g,f .

This happens if and only if [1]g,f (x) ∈ Eg,n.Now we show that the field is independent of f . Suppose x ∈ Ef,n. Then

[1]g,f (x) ∈ K(x)

because K(x) is still a complete field. Then Kπ,n,f = K(Ef,n) ⊇ Eg,n and soKπ,n,f ⊇ Kπ,n,g. Symmetry shows that Kπ,n,f = Kπ,n,g.

Note that Ef,n is an OK-submodule of pKsep (with the formal group law)annihilated by πn.

Theorem 20.4. As OK-modules, Ef,n ∼= OK/πnOK , non-canonically.

Proof. Without loss of generality, let f be a monic polynomial of degree q, e.g.,xq + πx. Note here that [π]f = f and [πk]f = f ◦ · · · ◦ f = f (k). Then

Ef,n = {x ∈ pKsep : f (n)(x) = 0}.

Note that f (n) | f (n+1) because x | f and so we can factor

f (n) = x · fx· f

(2)

f· f

(3)

f (2)· · · · · f (n)

f (n−1).

We claim that all f (k)/f (k−1) are Eisenstein polynomials. We just check

g ≡ xqk

/xqk−1

≡ xqk−qk−1

(mod π)

Math 223a Notes 78

and the constant term is π. So they are all irreducible, and all of its rootshave absolute value less than 1. Then all of these roots lie in pKsep . Therefore|Ef,n| = qn whatever it is.

Note that Ef,n is a module over OK , which is a DVR. So Ef,n is going tobe a direct sum

Ef,n ∼= OK/πd1 ⊕ · · · ⊕ OK/πdm .

But the π-torsion submodule of Ef,n is Ef,1 which has order q. So m = 1 andEf,n ∼= OK/πn.

Ef,n is not canonically isomorphic to OK/πnOK , but it is true that

(OK/πnOK)× ∼= AutOK (Ef,n); a 7→ [a]f .

Corollary 20.5. Ef =⋃nEf,n is isomorphic to K/OK as OK-modules.

Proof. Choose a generator α1 ∈ Ef,1 and then for n ≥ 2 inductively chooseαn such that [π]fαn = αn−1. Then glue them together using Ef → K/OK byαn 7→ π−n.

Again, Ef ∼= K/OK is not canonical but

AutOK (Ef ) ∼= lim←−n

AutOK (Ef,n) = lim←−n

(OK/πnOK)× ∼= O×K

is canonical.

20.2 Field obtained by adjoining torsion points

Proposition 20.6. The field Kπ,n = K(Ef,n) = K(αn) is a Galois extensionof K, where αn is any generator of αn.

Proof. The field Kπ,n is Galois because it is the splitting field of f (n). To showthat it is generated by αn, note that for any x ∈ Ef,n, there exists an a ∈ OKsuch that x = [a]f (αn). But [a]f is an infinite series with coefficients in OK , sox ∈ K(αn) because K(αn) is complete.

Here αn is a root of f (n) but is not a root of f (n−1), so that αn is a root off (n)/f (n−1). So this is the minimal polynomial for αn.

Now we’d like to understand Gal(Kπ,n/K). The Galois group acts on Ef,n,because it is just the set of roots of a polynomial. It also preserves the OK-module structure. So we have a homomorphism

Gal(Kπ,n/K)→ AutOK (Ef,n) ∼= (OK/πnOK)×.

Proposition 20.7. This map is an isomorphism.

Math 223a Notes 79

Proof. The map is clearly injective, because Ef,n generates Kπ,n. For surjec-tivity, we just count the orders of both side. The size of the Galois group is

|Gal(Kπ,n/K)| = deg(f (n)/f (n−1)) = qn − qn−1

and the size of the units on the right hand side is

|(OK/πnOK)×| = qn − qn−1

because we have to remove that non-units, which are multiplies of π. So this isan isomorphism.

Corollary 20.8. Gal(Kπ/K) = lim←−Gal(Kπ,n/K) = AutOK (Ef ) ∼= O×K .

So we’ve constructed this field Kπ by our bare hands, without using anyclass field theory. Now I would like to show that this is the same as the Kπ wehave defined before, using the reciprocity map.

Proposition 20.9. π ∈ NKπ,n/K(Kπ,n)×.

Proof. We have

Nαn = (−1)qn−qn−1

π = π

unless q is even and n = 1, in which case N(−αn) = π.

We can construct the field

Lπ = Kπ ·Kunr

and this is a candidate for Kab. There is a map

rπ : K× → Gal(Lπ/K)

given by, for u ∈ O×K ,

rπ(u)|Kunr = id |Kunr , rπ(u)(x) = [u−1]f (x) where x ∈ Ef ,

andrπ(π)|Kπ = id |Kπ , rπ(π)|Kunr = Frob .

Next time, we are going to prove that Lπ and rπ does not depend on π. Thenwe will show that rπ = θLπ/K . This will give us that Lπ is actually Kab.

Math 223a Notes 80

21 November 9, 2017

Last time we were able to use Lubin–Tate theory to construct Kπ and

Lπ = Kunr ·Kπ.

This is the candidate for Kab. We also define the map


which is the candidate for θKab/K .

21.1 Characterization of the Artin map

How do we show that something is the Artin map? Let L = Kunr ·Kπ for allπ. (Here we’re assuming that Lπ is independent of the choice of π.)

Proposition 21.1. Letr : K× → Gal(L/K)

be a homomorphism such that the composition

K×r−→ Gal(L/K)� Gal(Kunr/K)

is the reciprocity map x 7→ Frobvk(x) and for any uniformizer π, r(π)|Kπ is theidentity on Kπ. Then r = θL/K .

Proof. Note that K× is generated by the uniformizers, because u = (uπ)π−1.So it is enough to check that r(π) = θL/K(π) for all uniformizers π.

But note thatr(π)|Kunr = Frob = θL/K(π)

andr(π)|Kπ = idKπ .

Here, note that we have shown π ∈ N(Kπ,n)× for all n and so θKπ,n/K(π) =idKπ,n . Since this is true for all n, we get θKπ/K = idKπ .

To use this characterization, we need to show that Lπ = Kunr ·Kπ is inde-pendent of π, and also that rπ = r$ for uniformizers π and $. But assumingall this, we have

r : K× → Gal(L/K) = θL/K .

We want next want to show that L = Kab. How do we do this? First

N(Kπ,n)× = ker(θKπ,n/K : K× → Gal(Kπ,n/K)).

Here, for an arbitrary πru ∈ K×,

θKπ,n/K(πru)(x) = rπ(u)(x) = [u−1]f (x) = (u−1) · x.

This is going to be the identity if and only if u−1 ≡ 1 (mod πn), because thisis just multiplication. So this is true if and only if u ≡ 1 (mod πn). Hence

N(Kπ,n)× = ker(θKπ,n/K : K× → Gal(Kπ,n/K)) = πZ · Un.

Math 223a Notes 81

Theorem 21.2. Any finite abelian extension K ′/K is contained in L = Kπ ·Kunr.

Proof. Note that N(K ′)× is a finite index subgroup of K× by class field theory.So it must contain πfZ · Un for some f and n. Then

N(K ′)× ⊇ N(Kπ,n)× ∩N(Kunr,f ) = N(Kπ,n ·Kunr,f ).

This implies that K ′ ⊆ Kπ,n ·Kunr,f . Therefore K ′ is contained in L.

21.2 Independence on the uniformizers

Let π and $ be uniformizers, and let f, g be Lubin–Tate polynomials for π,$.Let Ff andGg be the Lubin–Tate formal groups for f and g. The problem is thatFf and Gg are not isomorphic over OK . So we need to work over a larger field.The natural thing to do is to work over Kunr, because Kπ ·Kunr = K$ ·Kunr.But a problem here is that Kunr is not complete. So we are going to look at itscompletion Kunr and its ring of integers Ounr

K .Let me say a bit about infinite algebraic extensions. Let K be a local field

and L/K be an infinite algebraic extension. Then it is not complete, because itis a countable-dimensional vector space and all Banach spaces are uncountable-dimensional vector spaces. Then we can form the completion L.

Example 21.3. Let K = Qp and L = Qp. Then we can complete it, and it is

generally referred to as L = Qp = Cp.

Any g ∈ Gal(L/K) extends continuously to g ∈ Aut(L/K). For example,we can define

Frob ∈ Aut(Kunr/K).

Proposition 21.4. Let K be local and L/K algebraic. Take any x ∈ L suchthat x is separable over L. Then x ∈ L.

Proof. Let L′ be the separable closure of L in L. We want to show that L′ isa separable extension of L. Then it is normal so L′/L is Galois. Any element

g ∈ Gal(L′/L) preserves the norm because L′/L is algebraic. Because L′ ⊆ L,it also preserves the absolute value on L′. Then g is continuous. Because L isdense in L′, we have g = idL′ . Therefore L′ = L.

Let $ = uπ.

Lemma 21.5. There exists a non-unique α(x) ∈ Ounr[[x]] such that α(x) =

εx mod deg 2 with ε ∈ (Ounr)× and

(a) Frob(α) = αϕ = α ◦ [u]f ,

(b) αϕ ◦ f = g ◦ α,

(c) α : Ff → Gg is a formal group isomorphism,

Math 223a Notes 82

(d) α ◦ [a]f = [a]g ◦ α for all a ∈ OK .

Proof. I think this is the Lubin–Tate strategy. I am following Milne’s notes.First we figure out what we want to take as ε. Because we want (a), we need

εϕx = (εx)ϕ = (εx) ◦ [u]f = uεx (mod deg 2).

This is guaranteed by homework, and we can choose an ε ∈ Ounr such thatεϕ/ε = u. You will even show that this is unique up to multiplication by O×K .

We now build α satisfying (a) inductively. We want polynomials αn of degreen such that αϕn = αn ◦ [u]f (mod deg(n + 1)). We already have our basecase.Induction step goes like, if I have αn we write αn+1 = αn + cxn+1. Then wehave

αϕn+1 = αϕn + cϕxn+1,

αn+1 ◦ [u]f = (αn + cxn+1) ◦ [u]f ≡ αn ◦ [u]f + cun+1xn+1.

We are good if we can always solve

d = cun+1 − cϕ.

To solve this, change to b = cεn+1. What we want is

d = b(uε)n+1 − bϕ(εϕ)n+1

or b− bϕ = d/(uε)n+1. This exist by homework.

Now we have have a power series α ∈ Ounr such that αϕ = α ◦ [u]f . Wewant αϕ ◦ f = g ◦α to hold, but it probably doesn’t. So we will have to modifyα 7→ α′. This is done by α′ = h ◦ α for some OK [[x]].

I want to figure out how badly this fails. Define an auxiliary power series

g′ = αϕ ◦ f ◦ α−1 = α[u]f ◦ f ◦ α−1.

We want to show that g′ ∈ OK [[x]]. I also want to show that g′F$. It sufficesto show that (g′)ϕ = g. But

(g′)ϕ = αϕ ◦ [u]f ◦ f ◦ (α−1)ϕ = α ◦ [u]f ◦ [u]f ◦ f ◦ [u]−1f ◦ α

−1 = g′

because f and [u]f are endomorphisms of Ff . On the other hand, we have

g′ ≡ αϕ ◦ xq ◦ α−1 ≡ αϕ ◦ (α−1)q (mod m = (π) = ($)).

But note that h((xq))ϕ ≡ hq for h ∈ Ounr[[x]]. So we get

g′ ≡ αϕ ◦ (α−1)ϕ ◦ xq = xq (mod $).

This shows that g′ ∈ F$.Note that we already have F$. So there exists a power series [1]g,g′ ∈ OK [[x]]

such that [1]g,g′ ◦ g′ ◦ [1]−1g,g′ − g. Now take

α′ = [1]g,g′ ◦ α.

Then g = (α′)ϕ ◦ f ◦ (α′)−1 and α also satisfies (a).

Math 223a Notes 83

22 November 14, 2017

Let K be local fields and π,$ be uniformizers. Let f and g be Lubin–Tateseries for π and $, with corresponding formal groups Ff and Gg. Let Ounr be

the ring of integers in Kunr.Let u = $/π. Last time we constructed an α ∈ Ounr[[x]] such that

αφ = α ◦ [u]f

and α is an isomorphism of formal OK-modules Ff → Gg. Recall that

Lπ = Kπ ·Kunr

and similarly L$ = K$ ·Kunr.

Theorem 22.1. The two fields Lπ and L$ are actually equal.

Proof. We are first going to show that the Lπ = L$. Then from the lemma lasttime, Lπ is the separable closure of L in Lπ and likewise L$ is the separableclosure of L in L$ so we deduce that Lπ = L$.

Consider the bijection

Ef∼=−→ Eg; x 7→ α(x).

We get that

K$,n = K(Eg,n) = K({α(x) : x ∈ Ef,n}) ⊆ Kπ,n · Kunr.

By symmetry, we get K$,n · Kunr = Kπ,n · Kunr. Taking the union over n givesK$ · Kunr = Kπ · Kunr and then taking completion gives L$ = Lπ.

So the field Lπ is independent of π. Now we show that the candidate Artinmap is also independent of π. Note that we defined the map


as rπ(v)|Kunr = id |Kunr and rπ(v)(x) = [v]f (x) for x ∈ Ef on v ∈ O×K , andrπ(π)|Kunr = Frob and rπ(π)|Kπ = id |Kπ .

Theorem 22.2. The two maps rπ and r$ are equal.

Proof. Because K× is generated by uniformizers, it suffices to to show thatrπ(y) = r$(y) for any uniformizers y ∈ OK . Then it suffices to show thatrπ(y) = ry(y) for any uniformizer y ∈ OK . Rename y = $ and we are going toshow that rπ($) = r$($).

We know that

rπ($)|Kunr = rπ(u)|Kunr · rπ(π)|Kunr = Frob = r$($)|Kunr .

Math 223a Notes 84

Now we need to check that

rπ($)|K$ = r$($)|K$ = idK$ .

I need to get my hands dirty with formal groups. Let x ∈ Eg be a torsionelement. Then there exists an x′ ∈ Ef such that x = α(x′). Because rπ($) acts

continuously it extends to Kunr and so we can write

rπ($)(x) = rπ($)(α(x′)) = (rπ($)(α))(rπ($)(x′))

= αφ(rπ(u) ◦ rπ(π)(x)) = αφ(rπ(u)x′) = αφ([u]−1f x′) = x.

So we get rπ($) = r$($) on K$.

At this point we are done, because last time we have shown that if Lπ andrπ are independent of choice of π, then Lπ = Kab and rπ = θ/K .

22.1 Artin map and the ramification filtration

Suppose L/K is abelian. We have the Artin map

θL/K : K× � Gal(L/K), O×K � I(L/K).

Recall that there is a filtration on both O×K and I(L/K), given by

UK,m = {a ∈ O×K : a ≡ 1 (mod πmK )}Gi(L/K) = {g ∈ I(L/K) : g(πL) ≡ πL (mod πi+1

L )}.

We know the successive quotient

UK,0/UK,1 ∼= k×, UK,m/UK,m+1∼= k+

for m ≥ 1, and also

G0/G1 ↪→ k×, Gi/Gi+1 ↪→ k+

for i ≥ 1. So we might think that the Artin map relates these two.Let L = Kπ,n so that L/K is totally ramified. We have

θL/K : O×K → Gal(L/K) ∼= End(Ef,n) ∼= O×K/UK,n,

with the map being u 7→ [u−1]f 7→ u−1. So the kernel of θL/K is actually justto be UK,n. This shows that

Gi =

Gal(L/K) i = 0

θL/K(UK,m) qm−1 ≤ i < qm

1 qn ≤ i.

Math 223a Notes 85

Now let us switch to the upper numbering:

φ(u) =

∫ u

0

dt

[G0 : Gt]

and we definedGi(L/K) = Gφ−1(i)(L/K).

In the case L = Kπ,n, we have

[G0 : Gi] = [UK,0 : UK,m] =

{(q − 1)qm−1 m ≤ n(q − 1)qn−1 m > n.

So we will getφ−1(i) = qi−1 for i = 1, 2, . . . , n+ 1.

ThenGi(L/K) = Gqi−1(L/K) = θL/K(UK,i).

Proposition 22.3. If E/L/K is a tower, then

Gi(E/K) I(E/K)

Gi(L/K) I(L/K).

You can find this is Neukirch. We can use this to define Gi(L/K) for Linfinite, by

Gi(L/K) = lim←−L′Gi(L′/K).

Example 22.4. Taking L = Kπ =⋃nK

π,n gives

Gi(Kπ/K) = θL/K(UK,i).

Because θKπ/K : O×K → Gal(Kπ/K) is an isomorphism by the construction ofLubin–Tate, we actually have that the two filtrations agree.

Example 22.5. Let us take L = Kab = Kπ ·Kunr now. We have

I(Kab/K) Gal(Kab/K) Gal(Kπ/K)

O×K K× O×K .

∼=

Then Gi(Kab/K) in I(Kab/K) corresponds to Gi(L/K) in Gal(Kπ/K). Then

Gi(Kab/K) = θKab/K(UK,i).

Math 223a Notes 86

22.2 Brauer group and central simple algebras

We have been looking at the Brauer group

Br(K) = H2(Gal(Ksep/K), (Ksep)×).

This came up in the context of non-commutative algebra. We are now goingto give a different definition of the Brauer group. Let K be a field. Somenon-commutative algebras over K we might care are

• n× n matrices Mn(K) over K,

• H over R,

• HC = H⊗R C, which actually is isomorphic to M2(C).

In this sense, H is a “twist” of M2(R) because H ∼= M2(R). We are interestedin algebras A/K such that A⊗KK is isomorphic to some Mn(K). We are thenlooking at

H1(K,PGLn(Ksep)),

where PGLn(Ksep) is a non-abelian Gal(Ksep/K)-module. We can define this,and there is a connecting homomorphism

H1(K,PGLn(Ksep))→ H2(K, (Ksep)×) = Br(K).

Definition 22.6. An (non-commutative) algebra A/K is simple if A has nonontrivial 2-sided ideals.

Example 22.7. Mn(K) is simple. For a, b ∈ K×, the algebra

H〈a, b〉 = K〈i, j〉/(ij = −ji, i2 = a, j2 = b)

is a 4-dimensional simple algebra. If K = R, then H(−1,−1) = H.

Definition 22.8. A central simple algebra is a simple algebra that hascenter K.

We will show that central simple algebras over K are twists of Mn(K), andare classified by H2(K, (Ksep)×).

Math 223a Notes 87


Recall that a (non-commutative) K-algebra A is

• simple if A has no 2-sided ideals other than 0 and A, which is equivalentto any homomorphism A→ B (with B 6= 0) being injective,

• central if and only if the center is Z(A) = K.

For example, if L/K is an extension, then L is a simple K-algebra but L is nota central K-algebra.

Proposition 23.1. The matrices Mn(K) is central simple.

Proof. Consider a nonzero ideal I ⊆Mn(K). Pick a nonzero a ∈ I, and i, j suchthat aij 6= 0. We can rescale to get aij = 1. If we multiply by the elementarymatrices, we get

eiiaejj = eij ∈ I.

Then we can get ekl = ekieijejl ∈ I and they generate everything.If a ∈ Mn(K) commutes with everything, then aeii = eiia implies that a is

diagonal, and then aeij = eija implies that all diagonal entries are equal.

If D is a division algebra (i.e., x 6= 0 implies x−1 ∈ D), then D is simpleover K. This D may or may not be central over K.

Example 23.2. On the current homework you will consider H(a, b), which isK〈i, j〉 quotiented out by i2 = a, j2 = b, and ij = −ji.

Example 23.3. Let L/K be a cyclic extension and [L : K] = n and g ∈Gal(L/K) a generator. We can define the cyclic algebra Aa, a K-algebragenerated by L and an additional element γ with

γn = a, γb = (gb)γ

for b ∈ L. You can check that Aa is central simple with dimension n2 over K.

If a′ = Nb for some b ∈ L, then

Aa → Aaa′ ; Lid−→ L, γa 7→ b−1γa′

is an isomorphism. So the isomorphism class of Aa depends only on the classof a ∈ K×/NL× = H2(L/K,L×). Also, for a = 1 we have

A1∼= EndK(L) ∼= Mn(K); ` 7→ m`, γ 7→ g.

This is injective by linear independence of automorphisms and then is an iso-morphism by a dimension argument.

Math 223a Notes 88

23.1 Tensor product of central simple algebras

One useful fact is that if L/K is an extension, then we can base change from Kto L.

Proposition 23.4. If A ⊗K L is a simple L-algebra, then A is a simple K-algebra.

Proof. If I ⊆ A is an ideal, then I ⊗K L ⊆ A ⊗K L is an ideal and so eitherI ⊗K L = 0 or I ⊗K L = A⊗K L.

The converse is not true. Consider A = C as a simple R-algebra. ThenA⊗R C ∼= C× C is not simple.

Proposition 23.5. If A is central simple, then A ⊗K L is simple. (In fact, itis central simple over L.)

Theorem 23.6. If A,B are K-algberas, with A central simple and B simple,then A⊗K B is simple.

Proof. Let us take an ideal I ⊆ A ⊗K B and assume I 6= 0. We need to showthat I = A⊗K B. Take an element a1 ⊗ b1 + · · ·+ an ⊗ bn ∈ I that is nonzero,so that the number of terms is fewest possible.

First of all we can take a1 = 1. This is because Aa1A = A by simplicity ofA and so we can scale. The resulting thing is nonzero because we have startedout with a minimal presentation. Let a ∈ A be arbitrary. Then

(a⊗ 1)(1⊗ b1 + · · ·+ an ⊗ bn)− (1⊗ b1 + · · ·+ an ⊗ bn)(a⊗ 1)

= (aa2 − a2a)⊗ b2 + · · ·+ (aan − ana)⊗ bn ∈ I.

Because b2, . . . , bn are linearly independent, we have that a2, . . . , an all commutewith a. Then a2, . . . , an ∈ Z(A) = K and so we get n = 1.

Now 1⊗b ∈ I and 1 ⊇ 1⊗BbB = 1⊗B becauseB is simple. So I = A⊗B.

You can show as an exercise that if A,B are K-algebra, then Z(A⊗K B) =Z(A)⊗K Z(B).

Corollary 23.7. If A and B are central simple, then A⊗K B is central simple.That is, (finite-dimensional) central simple algebras form an abelian monoidunder ⊗.

The matrices algebras {Mn(K)} form a sub-monoid, because Mn(K) ⊗KMm(K) ∼= Mnm(K).

Definition 23.8. We define the Brauer group as

Br(K) = (f.d. CSAs /K)/(matrix algebras /K).

To show that this is indeed a group, we consider the opposite algebra Aop

such that a ·Aop b = b ·A a. Clearly A is central/simple if and only if Aop iscentral/simple.

Math 223a Notes 89

Proposition 23.9. For finite-dimensional central simple A over K, A⊗K Aop

is isomorphic to EndK(A) where A is regarded as K-vector spaces.

Proof. We define

ϕ : A⊗K Aop → EndK(A); a⊗ 1 7→ la, 1⊗ a 7→ ra

where laa′ = aa′ and raa

′ = a′a. Because A ⊗ Aop is central simple, ϕ isinjective. To show that ϕ is surjective, we just count dimension.

Corollary 23.10. The Brauer group Br(K) is actually a group, with [A]−1 =[Aop].

23.2 Structure theorem for central simple algebras

We define the Brauer group, we don’t have any machinery to compute it. Ournext goal is to show that if A is a central simple algebra overK, then A = Mn(D)where D is a division algebra over K. If we have this, then in the Brauer group,

[A] = [Mn(K)⊗K D] = [D].

Definition 23.11. Let A be a K-algebra and M be a left A-module. We saythat M is

• simple if M has no submodules other than 0 and M .

• semisimple if M =⊕Mi where Mi are simple.

• decomposable if M = M1 ⊕M2 for some proper M1,M2 (M .

If A is a K-algebra, then A can be considered as a left A-module over itself,as AA. The algebra A being simple does not imply that AA is simple as a leftA-module. This is because submodules are just left ideals.

Lemma 23.12 (Schur’s lemma). If S is a simple A-module, then EndA(S) isa division algebra.

Proof. Suppose ϕ ∈ EndA(S) is nonzero. Then ker(ϕ) = 0 and im(ϕ) = S. Soϕ is invertible.

Proposition 23.13. Let D be a division algebra. Then all finitely generatedD-modules are isomorphic to Dn.

Proof. You can basically do the same proof as for vector spaces over a field.

Let us try to figure out EndA(AA). We have

EndA(AA) ∼= {a 7→ ab : b ∈ A} ∼= Aop.

As a consequence, if M is a free A-module of rank n, then

EndA(M) = EndA(AA⊕n) ∼= Mn(A)op ∼= Mn(Aop).

Math 223a Notes 90

Theorem 23.14 (Double centralizer). Let V be a finite-dimensional vectorspace over K. Let A ⊆ EndK(V ) be a simple subalgebra. We define the cen-tralizer as

C(A) = {b ∈ EndK(V ) : ba = ab for all a ∈ A}.

Then C(C(A)) = A.

Proof. See Milne’s notes on class field theory.

Let us now prove the classification theorem. Let A be a central simple K-algebra. Choose a nonzero simple A-module S. (Consider the minimal nonzerosubmodule of AA.) This gives an embedding, because (A is simple),

A ↪→ EndK(S); a 7→ la.

The centralizer is going to be

D = C(A) = EndA(S)

a division algebra by Schur. By the double centralizer theorem, we get

A = C(D) = EndD(S) ∼= Mn(Dop).

This is because S over D should be isomorphic to some Dn.

Theorem 23.15 (Structure theorem). If A is central simple algebra, it is iso-morphic to Mn(D) for some division algebra D.

Math 223a Notes 91


Last time we showed that if A is a central simple algebra over K then A ∼=Mn(D). In fact, n and D are uniquely determined up to isomorphism. Re-call that we have constructed D as, when S is a simple module over A, D ∼=EndA(S)op.

24.1 Modules over simple algebras

Simple algebras are the nicest things you could have after fields. There is goingto be a unique finitely generated simple module S over A, and every module isgoing to be Sn for some n.

First step in proving this is to decompose AA as a direct sum of simplesubmodules. Note that Z(A) is always a field, and so the classification givesA ∼= Mn(D) for some division ring D. Then

AA ∼= Mn(D) ∼= S1 ⊕ · · · ⊕ Snwhere Si is the matrices with 0 off the ith columns. These are all simple modules,and so it is a direct sum of n copies of the simple module S.

Assume that S′ is any other simple A-module. Take a nonzero s′ ∈ S′.There is a map

ϕ : AA→ S′; a 7→ as′

which is a left A-module homomorphism. But note that AA ∼=⊕

i Si. Thisshows that

ϕi : Si → S′

is nonzero for some i. Then Schur’s lemma tells us that ϕi is an isomorphism.For a general finitely generated A-module, we use a similar argument. Take

a surjective morphism An →M , and we can show that this can be restricted toan isomorphism Sm →M for some m.

Proposition 24.1. Let A be a simple algebra. There is a unique finitely gen-erated simple module S over A, and any finitely generated module over A isisomorphic to Sn for some n.

So the S that is simple over A is unique, and so D is uniquely determined.Then n2 = [A : K]/[D : K] and so n is determined.

Example 24.2. We have Br(K) = 0 if K is algebraically closed. To prove this,it suffices to show that if D/K is a finite dimensional division algebra over K,then D = K. Take x ∈ D and let f(x) be the minimal polynomial of x overK. Note that K(x) is a commutative subalgebra. By assumption that D is adivision algebra, f is linear and so x ∈ K.

Example 24.3. We have Br(Fq) = 0. This follows from Wedderburn’s theorem,which states that any finite division algebra is a field.

Example 24.4. We have Br(R) ∼= Z/2Z. The proof again is by classification ofdivision algebras over R. The only ones are R, C, and H. But C doesn’t countbecause it is not central.

Math 223a Notes 92

24.2 Extension of base fields

This is something we observed last time. We have a map

{CSAs over K} → {CSAs over L}; A 7→ A⊗K L.

This descends to a mapBr(K)→ Br(L).

Definition 24.5. The Brauer group Br(L/K) is defined as ker(Br(K) →Br(L)). This is the subgroup of classes such that A ⊗K L ∼= Mn(L). Thiscondition is called “A is split by L”, and is only dependent on the class of A.

Any central simple algebra A/K is split by K because Br(K) = 0. So weget

dimK A = dimK A⊗K K = dimKMn(K) = n2.

Proposition 24.6. There always exists a finite extension L/K such that A issplit by L.

Proof. We have an isomorphism Mn(K) → A ⊗K K, and the standard baseseij will be sent to some linear combination

∑k aijk ⊗ xijk. Then we can take

K(xijk) and then the isomorphism restricts to Mn(L)→ A⊗K L.

SoBr(K) =

⋃L/K fin.

Br(L/K) =⋃

L/K fin. Gal.

Br(L/K).

If L/K is finite Galois, our goal is to prove Br(L/K) ∼= H2(L/K,L×).If A is a central simple algebra over K, how can we tell which L/K split A?

There is going to be a nice answer to this. Let’s look at the quaternions overK = Q. This is A = H(−1,−1) over Q. If L/Q is finite, then

A⊗Q L = HL(−1,−1).

You have worked out in the homework how to tell if HL(−1,−1) is split or not.This is split if and only if the quadratic form

x2 + y2 + z2 = 0

has a nontrivial solution, or equivalently,

x2 + y2 + z2 + w2 = 0

has a nontrivial solution. Let us consider only L = Q[√D]. If D > 0, then

L ↪→ R and so these have no nontrivial solutions in L. If D < 0 and −D =x2 + y2 + z2 for x, y, z ∈ Q, then x2 + y2 + z2 + (

√D)2 = 0 is a nontrivial

solution and so L splits A. Actually this condition is equivalent to D not beingof the form 4n(8a + 1). In this case, you can show that x2 + y2 + z2 = 0 hasno nontrivial solution by working 2-adically. So the conclusion is that Q[

√D]

splits H(−1,−1) if and only if −D is a sum of three squares.

Math 223a Notes 93

Theorem 24.7 (Double centralizer). Let A be central simple algebra, and letB ⊆ A be a simple algebra. Then C = C(B) is simple and C(C) = B. Also wehave [A : K] = [B : K][C : K].

If B ⊆ A is not simple, there is a counterexample. Let A = Mn(K) and B ⊆A be the upper triangular matrices. Then C = K and C(C) = Mn(K) 6= B.

Proof. Again, this is in Milne.

Corollary 24.8. If Z(B) = K then Z(C) = K and A ∼= B ⊗ C.

Proof. In general, Z(B) = B ∩ C. But then Z(C) = B ∩ C = Z(B) = K.For the next part, note that B ⊗K C is a central simple algebra over K. ThenB ⊗K C ↪→ A and then this is an isomorphism by looking at dimension.

Corollary 24.9. If A is a central simple algebra over K and L ⊆ A is a field.The following are equivalent:

(a) L = C(L)

(b) [L : K]2 = [A : K]

(c) L is a maximal commutative sub K-algebra of A.

Proof. (a) to (b) is implied by the dimension part of the double centralizertheorem. For (b) to (c), consider L′ a maximal commutative K-subalgebracontaining L. Then

[L′ : K]2 ≤ [L′ : K][C(L′) : K] = [A : K] = [L : K]2.

So L′ = L and L is maximal. For (c) to (a), we have that for all x ∈ C(L), L[x]is commutative and so L[x] = L. This implies that C(L) = L.

If D is a division algebra, then (c) is equivalent to that L is a maximalcommutative subfield of A. In other words, all maximal subfields of D havedegree

√[D : K].

Theorem 24.10. The field L splits A if and only if there exists a B such that[B] = [A] in Br(K), B ⊇ L, and [B : K] = [L : K]2.

Proof. If L splits A, it also splits Aop. Then

Aop ⊗K L ∼= EndL(V )

for some vector space L of dimension dimL V = n. Then dimK Aop = dimK A =

n2. Let B be the centralizer of Aop inside EndK(V ). Because Aop and EndK(V )are both central simple over K, we have that B is central simple over K. Then

Aop ⊗K B ∼= EndK(V )

and so [B] = −[Aop] = [A] in Br(K). We certainly have that L ⊆ B because Lcommutes with everything in Aop, and [B : K] = [EndK(V ) : K]/[Aop : K] =[L : K]2.

Math 223a Notes 94

To prove the other direction, it is enough to show that L splits B. Let[L : K] = n and [B : K] = n2. We want B ⊗K L ∼= EndL(V ) for some V , withdimL V = n. We use V = B, where the action of L is by right multiplication(l · b = bl). Consider

B ⊗K L→ EndL(V ); b⊗ 1 7→ lb, 1⊗ l 7→ rl

where l• and r• are left and right multiplication maps. We know that B ⊗K Lis central simple over L, so this should be injective. Dimension counting showsthat this is an isomorphism.

Math 223a Notes 95


Last time we showed the following theorem.

Theorem 25.1. The field L splits A if and only if there exists a B such that[B] = [A] in Br(K), B ⊇ L, and [B : K] = [L : K]2.

Corollary 25.2. Let D be a division algebra with center K. If [L : K] =√[D : K], then L splits D if and only if L embeds in D.

Here the only thing that has correct dimension is D.

25.1 Noether–Skolem theorem

I have one more algebraic input before we can classify central simple algebras.

Theorem 25.3 (Noether–Skolem). Let A be a simple algebra over K and B bea central simple algebra over K. Then any two homomorphism f, g : A → Bare conjugate, i.e., there exists a b ∈ B such that f(a) = bg(a)b−1 for all a ∈ A.

Example 25.4. For example, we can take A = B central simple and g = id :B → B. Then any automorphism f : B → B is conjugation by an element of b,or inner.

Example 25.5. Let L/K be a finite extension assume L splits B and [B : K] =[L : K]2. If f, g are embeddings L ↪→ B, then f and g are conjugate. If L/Kis Galois, we can precompose L ↪→ B by any element of Gal(L/K). They aregoing to be related by a conjugation by some element in B.

Proof of Noether–Skolem. First assume that B = Mn(K) ∼= EndK(Kn). Letf, g : A → B be maps. Define two A-modules M1 and M2, both Kn as K-modules but

M1 : a ∗1 v = f(a)v, a ∗2 v = g(a)v

for a ∈ A. But because A is simple, recall that A-modules are completelyclassified by their dimension. So there exists an isomorphism

ϕ : M1 →M2.

Because ϕ is K-linear, we can view ϕ as an element of Mn(K) = B. Here, thisϕ being an isomorphism means that

f(a)(ϕv) = ϕ(g(a)v)

So f(a) = ϕg(a)ϕ−1 and take b = ϕ.Now let B be an arbitrary central simple algebra over K. We are going to

enlarge B to make it a matrix algebra. We know that B ⊗K Bop ∼= EndK(B).We have two maps

f ⊗ 1, g ⊗ 1 : A⊗K Bop → B ⊗K Bop ∼= EndK(B).

Math 223a Notes 96

The domain is still simple and the codomain is a matrix algebra. So these mustbe conjugate by some x ∈ B ⊗K Bop so that

x(f(a)⊗ b′)x−1 = g(a)⊗ b′

for all a ∈ A and b′ ∈ Bop. Setting a = 1 shows that x commutes with 1 ⊗ b′for all b ∈ Bop, so x ∈ B ⊗ 1 because Bop is central. If we write x = b⊗ 1 thenb has the desired property.

25.2 Classification of central simple algebras

Let L/K be a finite Galois extension. Let

A(L/K) = {central simple A/K : A is split by L, [A : K] = [L : K]2}/isomorphism

Then we have a bijection

A(L/K)→ Br(L/K) ⊆ Br(K); A 7→ [A].

Now we are going to construct a bijection

A(L/K)→ H2(L/K,L×).

Write G = Gal(L/K). Recall the inhomogeneous cocycles and coboundaries

Z2(L/K,L×) = {g1ϕ(g2, g3) · ϕ(g1, g2g3) = ϕ(g1g2, g3) · ϕ(g1, g2)}B2(L/K,L×) = {γ(g, h) = (gψ(h) · ψ(g))/ψ(gh)}.

For an arbitrary A ∈ A(L/k), pick an embedding i : L ↪→ K. By theNoether–Skolem theorem, for any g ∈ G there exists an ag ∈ A such that

g(x) = agxa−1g

for all x ∈ L. How unique is ag? Recall that if [A : K] = [L : K]2 thenC(L) = L. So ag is unique up to left multiplication by elements of L×.

Let us compare agh and agah. Here, we have

aghxa−1gh = g(h(x)) = agahx(agah)−1.

So agh and agah are the same up to some element of L. Define ϕ(g, h) so that

agah = φ(g, h)agh.

The cocycle condition comes from associativity. If we expand ag1ag2ag3 ,then

ag1(ϕ(g2, g3)ag2g3) = g1(ϕ(g2, g3))ag1ag2g3 = g1(ϕ(g2, g3))γ(g1, g2g3)ag1g2g3

equalsϕ(g1, g2)ag1g2ag3 = ϕ(g1, g2)ϕ(g1g2, g3)ag1g2g3 .

Math 223a Notes 97

This is precisely that ϕ is a cocycle.On the other hand, ag are unique up to multiplication by L×. So if a′g is

another choice of such elements, we should be able to write a′g = ψ(g)ag. Itturns out that

ϕa′/ϕa = dψ

is a coboundary. So we have a well-defined cohomology class

A(L/K)→ H2(L/K,L×).

Now I want to construct an inverse. This is straightforward because Iknow exactly what to do. Suppose I have received an arbitrary cocycle ϕ ∈H2(L/K,L×) and I want to defined a K-algebra Aφ by

Aϕ =⊕g∈G

L · eg

with relations eg · x = g(x)eg for all x ∈ L and

eg · eh = ϕ(g, h)egh.

This defines a K-algebra, with identity element e1/ϕ(1, 1). It can be shown thatthis only depends on the cohomology class [ϕ] ∈ H2(L/K,L×). So we have acandidate for a map H2(L/K,L×) → A(L/K), but we need to show that it iscentral simple over K.

Proposition 25.6. Aϕ is central simple.

Proof. The central part is straightforward, because if something commutes L,it must be in L · e1, and then if it commutes with eg, it should be in K.

Now we need to show that it is simple. Suppose we have a nonzero idealI ( Aϕ. Take nonzero a ∈ I with

a = x1eg1 + · · ·+ xkegk

for nonzero xi such that k is minimal. If k = 1, we would get everything so wemust have k ≥ 2. Without loss of generality assume that g1 = 1. Take l ∈ Lsuch that g2l 6= l. Then

al − la = x1e2l + x2eg2 l + · · · − lx1e1 − lx2eg2 − · · · = x2(g2(l)− l)eg2 + · · · ∈ I

has fewer terms and is nonzero.

So we have maps A(L/K) → H2(L/K,L×) and H2(L/K,L×) → A(L/K).We check that they are inverses. One direction is straightforward. If A ∈A(L/K) and ϕA is the cocycle I get from A, then we have an algebra homo-morphism Aϕ → A given by sending eϕ 7→ aϕ. Because Aϕ is central simple, itis an isomorphism by dimension counting.

Math 223a Notes 98


At the end of last time, we have reached the dramatic conclusion of this topic.

Proposition 26.1. There is a natural bijection

Br(L/K) ←→ H2(L/K,L×)

via the set A(L/K) of algebras split by L of dimension [L : K]2.

Theorem 26.2. This bijection preserves the group structure.

Proof. The proof is a bit painful. For cocycles ϕ1 and ϕ2, we know how to addthem and get ϕ1 + ϕ2, and we know how to construct Aϕ1

⊗ Aϕ2and Aϕ1+ϕ2

.We can show that

Aϕ1 ⊗Aϕ2∼= Aϕ1+ϕ2 ⊗Mn(K).

This is messy.

Proposition 26.3. Assume E/L/K with E/K Galois. Then the diagram

Br(L/K) H2(L/K,L×)

Br(E/K) H2(E/K,E×)

∼=

inf

∼=

commutes.

In particular, we have

Br(K) =⋃

L/K Gal

Br(L/K) = lim−→L/K

Br(L/K)

∼= lim−→L/K

H2(L/K,L×) = H2(K, (Ksep)×).

Corollary 26.4. If [L : K] = n, then Br(L/K) ∼= H2(L/K,L×) is n-torsion.So Br(K) is also torsion.

26.1 Fields with trivial Brauer group

We have seen a couple of examples already. We’ve seen that if K is separablyclosed, then Br(K) = 0. The other example I have given was K = Fq.

Note that Br(K) = 0 is equivalent to that any central division algebra overK is K. The tool for studying division algebras is reduced norm. This mightbe familiar in the context of quaternions:

H× → R×; a+ bi+ cj + dk → a2 + b2 + c2 + d2.

Math 223a Notes 99

Suppose D is a central division algebra over K. Let’s say that L is a Galoisextension such that L splits D. Then there is an isomorphism

D ⊗ L ϕ−→Mn(L).

Then we can defineNrd(d) = det(ϕ(d⊗ 1)) ∈ L×.

Proposition 26.5. For g ∈ Gal(L/K), we have g(Nrd(d)) = Nrd(d). That is,Nrd(d) ∈ K×.

Proof. We use the Noether–Skolem theorem. For A a simple L-algebra, anytwo L-algebra maps A → Mn(L) are conjugate by some b ∈ Mn(L). TakeA = D ⊗ L, and consider them map

ϕg : D ⊗ L→Mn(L); d⊗ ` 7→ gϕ(d⊗ g−1l)

So if we evaluate at d⊗ 1 and take determinant, we get

Nrd(d) = detϕ(d⊗ 1) = det gϕ(d⊗ 1) = gNrd(d).

Note that so far everything works for D not only a division algebra but alsoany central simple algebra. Let [D : K] = n2 and consider d1, . . . , dn2 a K-basisfor D. We can take the reduced norm

f(c1, . . . , cn2) = Nrd

( n2∑i=1

cidi

)∈ K[c1, . . . , cn2 ]

which is a homogeneous polynomial in the cis of degree n. For a nonzero v =(v1, . . . , vn2) ∈ Kn2

, we have f(v) 6= 0 because D is a division algebra. So wehave a low degree homogeneous polynomial with many variables, such that ithas no nontrivial zeros.

Definition 26.6. A field K is called quasi-algebraically closed if any homo-geneous polynomial f ∈ K[x1, . . . , xN ] of degree d < N has a nonzero solutionin KN .

If K is quasi-algebraically closed, then any central division algebra D/Kshould have degree n = n2 and so D = K.

Proposition 26.7 (Chevalley–Warning theorem). Finite fields Fq are quasi-algebraically closed.

Proof. The original theorem states that if f is a homogeneous polynomial inFq[x1, . . . , xN ] of degree d < N , then the number of solutions is a multiple ofp = charFq. So there should be at least one solution other than 0.

The idea is that ∑v∈Fnq

(1− f(v)q−1)

is congruent to the number of solutions modulo p. Then you expand this outand write things as sums of characters.

Math 223a Notes 100

Clearly algebraically closed field are quasi-algebraically closed.

Theorem 26.8 (Tsen’s theorem). The power series ring C((t)) is quasi-algebraicallyclosed. More generally, the fraction field of any complete DVR with algebraicallyclosed residue field is quasi-algebraically closed.

Theorem 26.9. If K is a local field, then Kunr is quasi-algebraically closed.

Theorem 26.10 (Lang). If K is quasi-algebraically closed, any algebraic L/Kis also quasi-algebraically closed.

If K is quasi-algebraically closed, then H2(K, (Ksep)×) = 0. For L/K, wealso have H2(L, (Ksep)×) = 0. It is also true that the corresponding H1 vanishby Hilbert 90, and this is also true for all subextensions. So by Tate’s theorem,we have that all

Hq(K, (Ksep)×) = 0.

This shows that for any L/K finite, the map N : L× → K× is surjective.

26.2 Brauer groups of local fields

Let K = R. We can compute the Brauer group

H2(R,C×) ∼= H0(R,C×) = R×/NC× ∼= {±1}.

Previously we have computed this invoking the classification of central simplealgebras over R, but we can now do this using cohomology.

Suppose now that K is a non-archemedian local field. We should have

inv : Br(K) ∼= H2(K, (Ksep)×) ∼= Q/Z.

This should give, in some sense, the classification of central simple algebras. Forany x ∈ Br(K), there exists an unramified extension L such that

x ∈ Br(L/K) = ker(Br(K)Res−−→ Br(L)).

This is because, under the identification of Br(K) ∼= Q/Z, it is multiplicationby n = [L : K]. So x ∈ Br(K/L) for some sufficiently large n = [L : K]. Here,L/K is cyclic, so we can understand H2(K,L×) ∼= H0(K,L×) ∼= K×/NL×.

But we can understand also from the division algebra side. LetD be a centraldivision algebra over K, and let L be its maximal subfield. Then [D : K] = n2

and [L : K] = n. We can extend the absolute value |−|K to D, by

|a|D = |Nrn(a)|1/nK .

Then this is a non-archemedian absolute value. We can likewise extend

v : D× → 1

nZ

and define ramification degree e = [v(D×) : v(K×)].

Math 223a Notes 101

We can define

OD = {a ∈ D : v(a) ≥ 0}, pD = {a ∈ D : v(a) ≥ 1n}.

Then this is a principal idea pD = πDOD = ODπD. Then kD = OD/pD is adivision ring that is finite over k = OK/pK ∼= Fq. In particular, kD/k is a finiteextension of finite fields. So we can define inertia degree f = [kD : k].

Using the exactly same proof as in the case of fields, we can show that

ef = [D : K] = n2.

But note that e | n, because v : D× → 1nZ. Also, we have f ≤ n, because if a is

a primitive element of kD then a ∈ OD has minimal polynomial degree at leastf , but a maximal subfield of D has degree n. So f ≤ n and e | n. We thereforeget

e = f = n.

Now when a generates kD/k, and a lifts a, the field K(a) is unramified andsplits D. By Noether–Skolem there exists some b ∈ D such that

FrobK(a)/K(x) = bxb−1.

Then another way to define the invariant map is

inv([D]) = [v(b)] ∈ 1nZ/Z ⊆ Q/Z.

What happens when K is a global field? Here the Brauer group is computedtoo, and there exists a short exact sequence

0→ Br(K)→⊕v

Br(Kv)→ Q/Z→ 0.

We are going to do this next semester. In particular, if [H] ∈ Br(K) is somequaternion algebra, this says that the number of v such that H⊗KKv is not splitis finite and even. This really quadratic reciprocity, when applied to H(p, q).

Index

absolute value, 9non-archemedian, 9

additive group, 69adele ring, 6adelic class group, 7Artin map, 6, 49augmentation ideal, 29

bar resolution, 34Brauer group, 49, 88, 92

central simple algebra, 86centralizer, 90Chevalley–Warning theorem, 99class field, 4coinduction, 31coinflation, 42compatible pair, 41completion, 10corestriction, 42crossed homomorphism, 36

decomposition group, 23discrete valuation ring, 10

Eisenstein polynomial, 25exponential, 18

formal grouphomomorphism, 73

formal group law, 72formal module, 76free module, 31Frobenius element, 5

G-module, 29discrete, 50

Galois cohomology, 49global field, 17group cohomology, 33group homology, 39group ring, 29

Hensel’s lemma, 11

Herbrand quotient, 57Hilbert 90, 49homogeneous cochain, 35

indecomposable module, 89induction, 31inertia degree, 14, 101inertia group, 24

tame, 27wild, 27

inflation, 41inhomogeneous cochain, 36invariant map, 53

Kronecker–Weber theorem, 5

local field, 16logarithm, 19Lubin–Tate series, 74

multiplicative group, 70

Noether–Skolem theorem, 95norm, 13norm residue symbol, 62Normal basis theorem, 49normic, 64

Ostrowski’s theorem, 9

p-adic integer, 6place, 10product formula, 17profinite completion, 66profinite group, 50

quasi-algebraically closed, 99

ramification group, 26ramification index, 14, 100restriction, 41

semidirect product, 37semisimple module, 89simple algebra, 86simple module, 89standard normalization, 16

102

Math 223a Notes 103

standard resolution, 34

tamely ramified, 27Tate cohomology, 44Tate’s theorem, 59torsor, 37Tsen’s theorem, 100

universal norm, 64

unramified extension, 19

upper numbering, 28

valuation, 9

Math 223a - Algebraic Number Theory€¦ · Math 223a - Algebraic Number Theory Taught by Alison...

Documents

Transcript of Math 223a - Algebraic Number Theory€¦ · Math 223a - Algebraic Number Theory Taught by Alison...