Math 223A - Algebraic Number Theory

Home , Group cohomology, Local class field theory

Math 223a - Algebraic Number Theory

Taught by Alison Beth Miller Notes by Dongryul Kim Fall 2017

This course was taught by Alison Beth Miller. The lectures were on Tuesdays and Thursdays at 11:30–1, and the main textbook was Algebraic number theory by Cassels and Fr¨olich. There were weekly problem sets and a ﬁnal paper, and there were 15 students enrolled. The course assistant was Lin Chen. Lecture notes can be found online on the course website.

Contents

1 August 31, 2017 4 1.1 Historic perspective: class fields ...... 4 1.2 Frobenius elements ...... 5 1.3 Infinite field extensions and their Galois groups ...... 5

2 September 5, 2017 7 2.1 Overview of global class field theory ...... 7 2.2 Overview of local class field theory ...... 8 2.3 Local field: evaluations ...... 9 2.4 Completion ...... 10

3 September 7, 2017 11 3.1 Hensel’s lemma ...... 11 3.2 Extending absolute values ...... 13

4 September 12, 2017 14 4.1 Ramiﬁcation and inertia ...... 14 4.2 Extending absolute values II ...... 14 4.3 Local ﬁelds ...... 16

5 September 14, 2017 17 5.1 Multiplicative group of local ﬁelds ...... 17 5.2 Exponential and logarithm ...... 18 5.3 Unramiﬁed extensions ...... 19

1 Last Update: November 4, 2018 6 September 19, 2017 21 6.1 Artin map for unramiﬁed extensions ...... 22 6.2 Decomposition group ...... 23 6.3 Inertia group ...... 24

7 September 21, 2017 25 7.1 Totally ramified extensions of local fields ...... 25 7.2 Ramification groups ...... 26 7.3 Tamely ramified extensions ...... 27

8 September 26, 2017 29 8.1 Category of G-modules ...... 29

9 September 28, 2017 33 9.1 Group cohomology ...... 33 9.2 Standard bar resolution ...... 34

10 October 3, 2017 36 10.1 Explicit descriptions of cohomology ...... 36 10.2 The ﬁrst cohomology and torsors ...... 37 10.3 Group homology ...... 39

11 October 5, 2017 40 11.1 Changing the group ...... 41 11.2 Inﬂation-restriction exact sequence ...... 42

12 October 10, 2017 44 12.1 Tate cohomology ...... 44 12.2 Restriction and corestriction ...... 46

13 October 12, 2017 48 13.1 Cup products ...... 48 13.2 Galois cohomology ...... 49 13.3 Cohomology of proﬁnite groups ...... 50

14 October 17, 2017 52 14.1 Kummer theory ...... 52 14.2 Second Galois cohomology of unramiﬁed extensions ...... 53 14.3 Some diagrams ...... 54

15 October 19, 2017 56 15.1 Second Galois cohomology of ﬁnite extensions ...... 56 15.2 Estimating the size ...... 57 15.3 Brauer group of a local ﬁeld ...... 58 15.4 Tate’s theorem ...... 59

2 16 October 24, 2017 61 16.1 Reciprocity map ...... 62 16.2 Group of universal norms ...... 64

17 October 26, 2017 65 17.1 Normic subgroups ...... 65 17.2 Quadratic reciprocity ...... 66 17.3 Reciprocity map on units ...... 68

18 October 31, 2017 69 18.1 Aﬃne group schemes ...... 69 18.2 Formal groups ...... 71

19 November 2, 2017 73 19.1 Morphisms between formal groups ...... 73 19.2 Lubin–Tate theory ...... 74

20 November 7, 2017 76 20.1 Torsion points in a formal group ...... 76 20.2 Field obtained by adjoining torsion points ...... 78

21 November 9, 2017 80 21.1 Characterization of the Artin map ...... 80 21.2 Independence on the uniformizers ...... 81

22 November 14, 2017 83 22.1 Artin map and the ramiﬁcation ﬁltration ...... 84 22.2 Brauer group and central simple algebras ...... 86

23 November 16, 2017 87 23.1 Tensor product of central simple algebras ...... 88 23.2 Structure theorem for central simple algebras ...... 89

24 November 21, 2017 91 24.1 Modules over simple algebras ...... 91 24.2 Extension of base ﬁelds ...... 92

25 November 28, 2017 95 25.1 Noether–Skolem theorem ...... 95 25.2 Classiﬁcation of central simple algebras ...... 96

26 November 30, 2017 98 26.1 Fields with trivial Brauer group ...... 98 26.2 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 field theory. For local class field theory, we are going to be proving stuff about Qp and their finite extensions. There is another local field, Fp((t)), which may be more comfortable depending on your background. The main tool we will use to analyze these is Galois cohomology. An example of a global field is Q, and we will of course look at their finite extensions, which are also called number fields. The polynomials Fp(t) is also a global 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–Fröhlich, which is actually a set of conference notes. This also includes Tate’s thesis. The book has a few errors and there is an errata online. Other recommended books are Neukirch’s Algebraic 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 Z and in L there is also a natural ring OL. One thing people were interested in was to take (p), lift it to pOL, and see how this ideal factorizes. In the case of a Galois extension, this becomes

e e pOl = p1 ··· pr 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 “splitting type 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 you are in? √For simplicity assume√ that n 6≡ 1 (mod 4). We are trying√ to check whether pZ[ 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 modulo 4n.

Definition 1.2. A Galois number field L/Q is a class field if there exists on N such that the splitting type of p depends only on p modulo N. For example, quadratic extensions are class fields, but general cubic extensions are not. If OL = Z[α], where the minimal polynomial of α is f(x), then the 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 related to the factorization of Φn(x) over Z/pZ. You can use this to show that this is a 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–Weber theorem. The reason this is not the end of the story is that we want to generalize this 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 decomposition 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 map ϕ is always surjective.

p Note that Gal(`/Fp) is always cyclic and so is generated by F : a 7→ a . So F has a unique lift to a Frobenius element Frobp ∈ Dp. If I had chosen some 0 −1 other p = gp for g ∈ Gal(L/Q), then Frobp0 = g Frobp g . This means that Frob is well-deﬁned a conjugacy class. In particular, if L/Q is abelian, we can just write Frobp ∈ Gal(L/Q).

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

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

p Now there is going to be a Frobenius element Frobp(ζ) ≡ ζ (mod p). Then it × has 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

ab [ Q = Q[ζ∞] = Q[ζn]. n≥1

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 inﬁnite group. But these are related to ﬁnite Galois groups by restriction maps

ab ∼ × Gal(Q /Q) Gal(Q[ζn]/Q) = (Z/nZ) . So we can understand this group by considering the map

ab Y Gal(Q /Q) ,→ Gal(Q(ζn)/Q). n

ab Injectivity comes from the fact that Q(ζn) ﬁlls Q . But this is not surjective 0 because gn ∈ Gal(Q(ζn)/Q) has to restrict to gn0 ∈ Gal(Q(ζn0 )/Q) if n | n. So this Galois group is the inverse limit

Gal( ab/ ) ∼ lim Gal( (ζ )/ ) ∼ lim( /n )× ∼ ˆ×, Q Q = ←− Q n Q = ←− Z Z = Z n n where ˆ = lim ( /n ). Actually these are topological groups and topological Z ←−n Z Z rings, but we’ll get there.

Deﬁnition 1.7. The p-adic integers Zp is deﬁned as = lim( /pi ). Zp ←− Z Z i

2 These are sort of like infinite series c0 + c1p + c2p + ··· . The following proposition follows from the Chinese remainder theorem. ˆ ∼ Q Proposition 1.8. Z = p Zp. Thus Y × ∼ ab Zp = Gal(Q /Q). p This is not true for other general fields. So we are going to define a crazy ring called the adele ring. This is

Y AQ,fin = (ap) ∈ Qp : all but finitely many ap are in Zp , AQ = AQ,fin×R. p

We can extend the isomorphism to

× >0 ∼ Y × ∼ ab AQ → AQ/(Q × R ) = Zp = Gal(Q /Q). p

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 abelian extension of Q: ab ∼ Y ∼ × >0 Gal(Q /Q) = Zp = AQ/Q × R . p Here

AQ = {(xp)p, (x∞): xp ∈ Zp for all but ﬁnitely p}. The group × is actually a topological group, with basis AQ

Y Y × Upi × Zp × UR, p1,...,pr all other p

× × × where Up ⊆ and U ⊆ are open sets. The fact that θ : → i Qpi R R AQ Gal(Qab/Q) contains Q× in the kernel is called Artin reciprocity, and implies quadratic reciprocity. For K a number ﬁeld and a prime p ⊆ OK , there is the completion Kp. In general, we deﬁne

× Y × × AK,ﬁn = {(xp) ∈ Kp : xp ∈ Op for all but ﬁnitely p}, p

× Y × Y × AK,inf = R × C , K,→R K,→C × × × AK = AK,ﬁn × AK,inf . Again, you can prove that

ab ∼ × × Gal(K /K) = AK /K · (AK )0, where (AK )0 is the connected component of the identity, and the connected Q >0 Q × component of 1 of R × C . This isomorphism is the main theorem of × × global class field theory. If we define the adelic class group as Ck = AK /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 corresponds to open finite index subgroups of CK , with L mapping to NCL. So in global class field theory, we are going to construct the Artin map θ : AK → Gal(L/K). Then we will show Artin reciprocity, and show the map is surjective. At the end of the semester, we are going to know how to construct the 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 × ab θK : K → Gal(K /K). This is not an isomorphism, but it is very close to; it is an injection with dense image. 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 finite abelian extensions L/K and finite index subgroups of K×, which is given by LN ×. Suppose K is a number field, and take a prime p ⊆ OK . Then Kp is a local field and so we have the local Artin map × ab θKp : Kp → Gal(Kp /Kp). On the other hand, we have the global Artin map × ab θK : AK → Gal(K /K). 0 If L/K is abelian, then we pick a prime p ⊆ L lying over p, and then Lp0 /Kp is an extension and we get a map

Gal(Lp0 /Kp) ,→ Gal(L/K), which is actually an isomorphism onto the decomposition group Dp0 . Putting these together, we get an injection ab ab Gal(Kp /Kp) ,→ Gal(K /K), where the image is a decomposition group. As we hope, the diagram

θ × Kp ab Kp Gal(Kp /Kp)

× θK ab AK Gal(K /K)

× × commutes. Because Kp generates AK topologically, we will be able to use this to 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 generalize Qp, Fp((t)). So we want to give a definition of local fields so that all local fields are finite extensions of Qp or Fp((t)).

Deﬁnition 2.1. A valuation on a ﬁeld K is a map v : K → Z ∪ {∞} such that (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 factorization of x. This is valuation. Example 2.3. Let O be a Dedekind domain, i.e., an integral domain with unique factorization of ideals. For p a prime ideal, deﬁne vp(x) to be the power of 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|.

v(x) If v is a valuation, then |x|v = a for a < 1 is an absolute value. This in fact satisﬁes (c0) |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 classify all absolute values.

Theorem 2.5 (Ostrowski). Any absolute value of Q is equivalent to |−|p for some p or to |−|R coming from the embedding Q ,→ R. a >0 Here, we say that |−|1 ∼ |−|2 if |−|1 = |−|2 for some a ∈ R .

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 valuation is |−|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 integers in this ﬁeld, but I’m arbitrarily going to pick O = Fp[t]. Now every prime ideal O gives a valuation, and also f(t) 7→ − deg(f(t)) is another valuation. There is an 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 PID that 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) for any π with v(π) = 1. Conversely, if O is a discrete valuation ring with maximal ideal 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 Kˆ as the topological completion of K as a metric space. (If |−| comes from a valuation v, then we write this as Kv.) Then Kˆ 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 embeddings into R or C. We use the notation v for a place, and then write Kv for the completion at the place v. Using this, we can formally define Y × AK = (xv) ∈ Kv :(xv) ∈ Ov for all but finite v . v places Math 223a Notes 11

3 September 7, 2017

We finished by talking about absolute values and valuations, and making the field complete. Now we are going to talk about fields that are complete. First let us talk about non-archemedian absolute values, i.e., absolute values with |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 that K is a ﬁeld complete with respect to a discrete absolute value, that is, im|−| is a discrete subgroup of R>0. Because of the previous theorem, |−| is non-archemedian, and this must come from exponentiating a valuation. That is, |x| = c−v(x) for some c > 1. Now the subring O = {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 cosets ` 2 of p, and write p as the disjoint union p + a = b(p + b). The upshot of this is that O = lim O/pnO = lim O/πnO, ←− ←− n n by completeness of K. Generally, let O be any Dedekind domain, p ⊆ O be a prime. Let K = Frac O and Kˆ be the completion with respect to vp. If Oˆ and pˆ are the completed ring and prime ideal, then O/pi =∼ Oˆ/pˆi. We can thus write

Oˆ = lim Oˆ/pˆi = lim Oˆ/pi. ←− ←− 3.1 Hensel’s lemma Let K be complete with respect to a discrete absolute value, and L/K be a ﬁnite extension. Is it always possible to extend |−|K to L? Is it unique, and does 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) and k = O/p. Suppose f ∈ O[x] and let f¯ ∈ k[x] be the reduction. If f¯ factors as f¯ =g ¯h¯, with g¯ and h¯ relatively prime, then there exists a factorization f = gh such that g reduces to g¯ and h reduces to h¯.

2 Without the relatively prime condition, a counterexample is x − p in Qp[x]. Math 223a Notes 12

r r Proof. The idea is to use O = lim O/π . It suﬃces to ﬁnd gr, hr ∈ (O/π )[x] r ←− 0 such that grhr = f (mod π ), and gr0 reduces to gr for r > r. Actually we also need degree conditions deg gr = dg = degg ¯ and deg hr ≤ dh = deg f − degg ¯, because the degree might blow up. ¯ We induct on r. If r = 1, we can just takeg ¯ = g1 and h = h1. Assume we r 0 0 have gr, hr ∈ (O/π )[x] which works. Choose arbitrary lifts gr+1 and hr+1. We want to correct this by writing

0 r 0 r gr+1 = gr+1 + π a, hr+1 = hr+1 + π b.

Then what we need is g0 h0 − f ah0 + bg0 = r+1 r+1 (mod π). r+1 r+1 πr 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 of polynomials 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 that the map is surjective. As an exercise, try to show that the polynomials g and h are unique up to multiplication by elements of O×. Corollary 3.3. If f ∈ O[x], and there exists some a¯ ∈ k such that f¯(¯a) = 0 but f¯0(¯a) 6= 0, then a¯ lifts to a unique root a ∈ O of f. Proof. Use Hensel’s lemma on f¯ = (x − a)¯g.

p−1 Example 3.4. Let K = Qp and O = Zp. Take f(x) = x − 1 so that ¯ p−1 × f(x) = (x − 1) ··· (x − p + 1). Then x − 1 has p − 1 roots in Zp , and these are distinct mod p.

2 Example 3.5. Let K = Qp and O = Zp with p odd. Take f(x) = x − a where × a ∈ Zp . If a is not a square mod p, then f(x) is irreducible, and if a is a square × mod p, then f(x) has two distinct solutions. Therefore, a ∈ Zp is a square if and only ifa ¯ is square in Fp. × i × More generally, if a ∈ Qp and a = p u where u ∈ Zp , a is a square if and only if i is even andu ¯ is a square in Fp. Lemma 3.6. Let K be a ﬁeld complete with respect to a valuation v. If f = n 0 anx +···+a0x ∈ 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 a degree m and a degree n − m polynomial. Then m = 0 or m = n. This generalizes to the theory of Newton polygons. Here you apply the theorem 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 q n |a|L = |NL/K a|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 are not 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

m m−1 Proposition 3.9. If a has minimal polynomial x +cm−1x +···+c0 ∈ K[x], n/m n/m then χma = f(x) and NL/K (a) = c0 .

Proof of Theorem 3.7. We first check that |a+b|L ≤ max{|a|L +|b|L}. Without loss of generality, let max = |a|L = 1. It suffices to show that |b|L ≤ 1 implies |1 + b|L ≤ 1. m m−1 Let f(x) = x + cm−1x + ··· + c0 ∈ O[x] be the minimal polynomial of m b. Then |c0| = |b| ≤ 1, and the previous lemma implies |ci| ≤ 1 for all i. Then f(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 |−|L 0 and |−|L. We can view them as a norms on L as a finite-dimensional vector space K. It is a theorem that any two such norms induce the same topology. n n 0 Then |a|L < 1 if and only if {a } → 0 in |−|L if and only if {a } → 0 in |−|L if 0 and only if |a|L < 1. For completeness, we can again use the fact that any norm induce the same topology. 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 a ﬁnite extension. Assume that L/K is a ﬁnite extension K is complete with respect to |−|. We have shown there exists a unique extension of |−| to L, and that 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 πK OL ⊆ OL is an ideal, and it has to be e πK OL = (πL) for some e. This e is called the ramiﬁcation index. This is the index of

× >0 × >0 im(|−| : K → R ) ⊆ im(|−| : L → R ), because the images are generated by |πK | and |πL|. We can also deﬁne the inertia 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 ∈ Zp 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 = . Consider the extension L = ( p), √ √ Qq Qp 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/K fL/K = [L : K] = n.

Proof. The quotient OL/πK OL is a vector space over k = OK /πK OK , and we compute its dimension in two diﬀerent ways. Firstly, OL is a free OK -module and is of rank [L : K] = n. So OL/πK OL is a free k-module of rank [L : K], and hence dimk OL/πK = n. e Secondly, we have πK OL = πLOL and so

e 2 dimk OL/π OL = dimk OL/πLOL + dimk πLOL/πLOL + ··· = e dimk ` = 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 . But let us not assume that K is complete. Now suppose that L/K is ﬁnite. How many ways can we |−|K extended to |−|L? Consider the case K = Frac OK where OK is a Dedekind domain. Assume that |−|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.) 0 Let p ⊆ OL be a factor of pOL with exponent e. Then for each a ∈ K, we have

vp0 (a) = evp(a).

After appropriate normalization, |−|p0 extends |−|p.

0 Theorem 4.4. Any absolute value |−| of L extending |−|p is equivalent to |−|p0 for some p0 | p. Proof. First we will show that |−| is discrete. Look at the completions Kˆ and Lˆ, where the both valuations will be discrete. This implies that |−|0 is discrete and non-archemedian. Consider 0 OL,v0 = {a ∈ L : |a| ≤ 1} which is a DVR. This ring contains OK and so contains OL. Let

0 p = pL,v0 ∩ OL, which is a prime ideal. Inside L we have (OL)p0 ⊆ OL,v0 , both DVRs. But if A ⊆ B ( L are DVRs, then A = B. So OL,v0 = (OL)p0 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 Kˆ is going to be R or C. Then the embedding K → Kˆ gives the valuation. If it is non-archemedian, then its restriction to Q gives |−|p for some p, and so it is |−|p for some p | pOK .

Now let K be a ﬁeld with an absolute value |−|v, and L be a ﬁnite separable extensions of K with absolute value |−|v0 extending |−|v. Now we have completions

K Kv

L Lv0 and furthermore Lv0 = Kv · L is the compositum. So Lv0 /Kv is ﬁnite. Con- 0 0 versely, if there exists some L /Kv ﬁnite with L = L · Kv, then we can extend 0 |−|v on Kv to L and restricting to L gives an absolute value |−|L0 on L. This gives a bijection

isom.classes of absolute values 0 0 ←→ . ﬁelds L with L = Kv · L of L extending |−|v Math 223a Notes 16

∼ Q Proposition 4.6. Kv ⊗K L = v0 extending v Lv0 .

Proof. To get the map, note that we have bilinear maps Kv × L → Kv · L = Lv0 for any v0. It is a general fact that if L/K is finite separable and K0/K is any extension, 0 ∼ Q 0 then K ⊗K L = Li where Li runs over isomorphism classes of fields L such that L0 = K0 · L. Using this fact and the previous isomorphism, we get the result. √ Example 4.7. Take K = Q, Kv = Q3 and L = Q[ 7]. By Hensel, 7 is a square 2 in Q3, and so x − 7 = (x − a)(x + a) in Q3[x]. Now 2 ∼ L ⊗ Q3 = Q3[x]/(x − 7) = Q3[x]/(x − a) × Q3[x]/(x + a). Q √ So we have two embeddings L = Q[ 7] ,→ Q3, and each give different absolute values. √ Example 4.8. Let K = Q, Kv = Q3 and L = Q[ 3]. In this case, 2 L ⊗Q Q3 = Q3[x]/(x − 3) is a field, and this is the unique extension of the absolute value. √ 3 ∼ Another example is L = Q[ 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 because n for a ∈ OK with |a| < 1, one of the sets a OK 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 has an open cover {a : a < c} for c < 1 and so has a finite subcover. Then there no absolute values in (c, 1) for some c < 1. Choose a π with |π| < 1 maximal. The quotient OK /(π) is compact and discrete. This implies that this is a finite field. Therefore non-archemedian local fields are field complete with respect to a discrete valuation with with finite residue field. If there is such a Ks, then O = lim O /(πn) K ←− K 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 and residue field k = O/π, the standard normalization is given by −v(a) |a|K = |k| . 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

5 September 14, 2017

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

k = OK /πK OK being finite. For example, Qp and Fp((t)) are both local fields, and finite extensions of local fields are local.

Theorem 5.1. In fact, all local fields are finite extensions of Qp or Fp((t)), or just 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), which complete 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×,

Y |a|v = 1, v where v runs over normalized absolute values.

Proof. For K = Q and K = Fp(t), we can manually check this. If the product formula 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

Y 0 |a |v = |NL/K a|v. v0 extending v ∼ Q (This claim follows from what we did last time, that L ⊗K Kv = Lv0 .)

5.1 Multiplicative group of local ﬁelds × Let K be a (non-archemedian) local ﬁeld, with discrete valuation K Z. We have an exact sequence

× × v 1 → O → K −→ Z → 0. This sequence splits, but not canonically, with the choice of a uniformizer π with 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 unity in O×. Then we can say that × ∼ ∼ × O = U1 × µq−1 = U1 × k . We can ﬁlter down even more. For any n we have

× n Un = {a ∈ O : a ≡ 1 (mod π )}, 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

+ n k → Un/Un+1; a 7→ [1 + π a]. This is an isomorphism and it is not hard to see that it is an isomorphism.

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

m Un → Un; x 7→ x is a bijection.

Proof. To show that this map is injective, take any a ∈ Un with a 6= 1. Let N be maximal with a ∈ UN − UN+1. Then [a] ∈ UN /UN+1 is not equal to 1, m m and so [a ] ∈ UN /UN+1 is not equal to 1. This shows a 6= 1 and so we have injectivity. For surjectivity, use Hensel’s lemma, or directly prove it using surjectivity m of x 7→ x on Un/Un+1. So if we look at the roots of unity with order prime to p, it must have trivial × × intersection with U1. It also have to map bijectively to O /U1 = k . So these are exactly µq−1, where k = Fq. On the other hand, if we have a root of unity a ∈ O× of order pi, it must be in a ∈ U1.

5.2 Exponential and logarithm

Let K be a local field of characteristic 0. (So it is a finite extension of Qp.) Let e pOK = (π) for some e. Definition 5.6. We define the exponential as X xn exp (x) = ∈ K[[x]]. p n! n≥0

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 the sum of base-p digits of n. Math 223a Notes 19

Deﬁnition 5.7. Deﬁne the logarithm as

X n+1 n logp(1 + z) = (−1) z . n≥1

This converges with v(z) > 0, equivalently 1 + z ∈ U1. Theorem 5.8. For any n > e/(p − 1), the maps

n n expp :(π) → Un, logp : Un → (π) are inverse homomorphisms.

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

× ∼ + + Corollary 5.9. For p > 2, we have Zp = µp × U1 = (Z/(p − 1)Z) × Zp , and × ∼ + + for p = 2, we have Z2 = µ2 × U2 = (Z/2Z) × Z2 . 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 Zp , but in general we won’t have something as nice as this. Here, we have some

∼ + ∼ [K:Qp] Un = O = Zp . Again we have 1 → 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 and k = OK /πK . Then we have the ramification index

eL/K πK OL = (πLOL) and the inertia degree fL/K = [` : k]. We have proved that n = [L : K] = eL/K fL/K .

Definition 5.10. L/K is unramified if eL/K = 1 and `/k is separable. (This second condition is automatic for local fields.) Lemma 5.11. Suppose L = K(a)/K be a finite extension, and K be complete with a discrete valuation. If there exists a monic polynomial f(x) ∈ OK [x] such that f(a) = 0 such that f(x) ∈ k[x] is separable, then L/K is unramified and OL = OK [a]. Math 223a Notes 20

Proof. I might as well take f to be the minimal polynomial of a so that f is irreducible. If follows that f¯ ∈ k[x] is also irreducible by Hensel’s lemma, since f¯ is separable. Nowa ¯ ∈ ` 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 that OK [a] = OL. Let L/K be unramiﬁed and let `/k be their residue ﬁelds. Choose any prime elementa ¯ ∈ ` with ` = k(¯a). Let f¯ be the monic minimal polynomial of f. Then ¯ we can lift f ∈ k[x] to a f ∈ OK [x]. Then we can lifta ¯ 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 lemma applies and we have L/K unramiﬁed and OL = OK [ζn]. m In particular, let n = p − 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

6 September 19, 2017

Let K be a local field and L/K be unramified. This means that eL/K = 1, or equivalently fL/K = [L : K]. We showed that any such L is of the form K(a) × ¯ with a ∈ OL , 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, for every n ≥ 1.

Proof. We constructed the extension L = K(ζqn−1) where q = |k|. For uniqueness, suppose we have L, L0 as above. We can write L0 = K(a) where f(x) is the minimal polynomial of a. The polynomial f has degree [L0 : K] = n, and so 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 finite field, f¯(x) ∈ k[x] splits in `. By Hensel’s lemma, f(x) has a root in L. That is, we have an inclusion L0 ,→ L; a 7→ root which must be an isomorphism because [L : K] = [L0 : K]. More generally, suppose L and L0 are unramified extensions of K with residue fields ` and `0. If there is a map ϕ : ` ,→ `0 that fixes k, then the same argument shows that we can lift the map uniquely to

ϕ˜ : L,→ L0 so that OL/πL → OL0 /πL0 is ϕ. In fact, we have an equivalence of categories

finite extensions finite unramified ←→ . of k extensions of K

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

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

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

unr [ [ K = (ﬁnite unramiﬁed extensions) = K(ζm). (m,p)=1

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 there exists 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 X v(Na) = v(ga) = nv(a). g∈Gal(L/K) × This means that v : NL nZ. So we have a short exact sequence × × × × + 1 → OK /NOL → K /NL → (Z/nZ) → 0. × × We need to show that OK = NOL . We have filtrations

× OK = UK,0 ⊇ UK,1 ⊇ UK,2 ⊇ · · · × OL = 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 : OL /UL,1 = ` → k = OK /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 + ∼ i ` = UL,i/UL,i+1; a 7→ [1 + πLa] and a similar identification for K. But note that we can choose πL = πK because L/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 ∈ OK there exists a b ∈ OL with Nb = a. × First ﬁnd a b1 ∈ OL such that

Nb1 ≡ a (mod UK,1). Math 223a Notes 23

× −1 Then there exists a c2 ∈ UL,1 such that Nc2 ≡ a(Nb1) 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 is an 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 v0 ∈ L be places with v0 extending v. We deﬁne decomposition group

Dv0 = Dv0 (L/K) = {g ∈ Gal(L/K): |ga|v0 = |a|v0 for all a ∈ L}.

We have shown that if K is complete, then Dv0 (L/K) = Gal(L/K). If K = Frac OK with OK a Dedekind domain and v = vp, then we have 0 0 shown that v = vp0 for some p dividing p. Then

0 0 Dv0 = Dp0 = {g ∈ Gal(L/K): gp = p }. If v is archemedian, then complex conjugation is always going to be in the decomposition group. If Kv and Lv0 are completions, then

Dv0 (L/K) = Dv0 (Lv0 /Kv0 ) = Gal(Lv0 /Kv0 ).

0 This is because if g ﬁxes v then it lifts to the completion, and if it acts on Lv0 then its restriction to L ﬁxes v0. In this case, let

0 Z = Z(v ) = ﬁxed ﬁeld of Dv0 (L/K).

0 0 We can restrict v to Z and get a place vZ on Z(v ).

L v0

0 Z(v ) vZ

K v

0 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-archemedian case, and can be worked out in the archemedian case.) Because Dv0 (L/K) ﬁxes 0 v , there is only one valuation extending vZ . Math 223a Notes 24

We now have ∼ ∼ Gal(L/Z) = Dv0 = Gal(Lv0 /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 on K. Suppose v0 on L extends v. Deﬁne the inertia group as

0 Iv0 = Iv0 (L/K) = {g ∈ Gal(L/K): v (ga − a) > 0 for all a ∈ OL}.

In other words, Iv0 is the kernel of the map

Dv0 → Gal(`/k). We then have a short exact sequence

1 → Iv0 → Dv0 → Gal(`/k) → 1 because Dv0 → Gal(`/k) is surjective. The proof of surjectivity goes like this.

Proof. By replacing with completions if necessary, we can assume that Dv0 = Gal(L/K). Pick somea ¯ ∈ ` and let its minimal polynomial be f¯(x). Lift it to a ∈ Ov0 . For Y h(x) = (x − ga) ∈ OK [x], g∈Gal(L/k) we will have thata ¯ is a root of Y h¯(x) = (x − ga) ∈ k[x]. g∈Gal(L/K) For anya ¯0 Galois conjugate toa ¯ in `, there exists a g ∈ Gal(L/K) such that ga =a ¯0.

Corollary 6.5. If L/K is complete, then Iv0 has size e.

Proof. This is because Dv0 = Gal(L/K) has size ef and Gal(`/k) has size f.

0 0 Proposition 6.6. Let T (v ) be the fixed field of Iv0 . Then T (v ) is the maximal unramified 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) 1 Gal(`/k).

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

7 September 21, 2017

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

eL/T = [L : T ], fL/T = 1, and eT/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 that eL/K = [L : K] = n.

Proposition 7.1. L = K(πL) and OL = OK [πL]. n−1 Proof. We will show that 1, πL, . . . , πL are linearly independent over K. In other words, that X i aiπL 6= 0 0≤i

The minimal polynomial f of πL has degree n. Let n n−1 f(x) = x + cn−1x + ··· + 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 Eisenstein polynomial. Conversely, if f(x) ∈ K[x] is any Eisenstein polynomial, i.e., vK (ci) > 0 and vK (c0) = 1, then L = K(a) = K[x]/f(x) is totally ramiﬁed. This is because 1/n |a|L = |c0|K . √ Example 7.2. Let K be an arbitrary local ﬁeld and let L = K( m π ). Then √ K πL = m πK has minimal polynomial m x − 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 v0 be a place of L lying over v of K. We have defined the decomposition group Dv0 (L/K) and the inertia group Iv0 (L/K). Then L/T is totally ramified and T/Z is entirely inert.

L {1}

total ram.

T Iv0 (L/K)

inert

Z Dv0 (L/K)

K Gal(L/K)

Now we are going to deﬁne a ﬁltration

Iv0 (L/K) = G0,v0 ⊇ G1,v0 ⊇ G2,v0 ⊃ · · · .

Definition 7.4. We define the ramification groups as

0 Gi,v0 (L/K) = {g ∈ Dv0 (L/K): v (ga − a) > i for all a ∈ Ov0 }.

Note that GN,v0 = {1} for all N 0. Also, gi,v0 is a subgroup of Dv0 acting i+1 trivially on Ov0 /(πL ). If we complete the ﬁeld, then

Gi,v0 (L/K) = Gi,v0 (Lv0 /Kv0 ).

Now let’s assume that L, K are non-archemedian local ﬁelds, and write vL = 0 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 is because we can replace K by T . Now in this case we have OL = OK [πL], and so 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 that this map is completely canonical, i.e., independent of the choice of πL. If we replace πL with uπL, then g(uπ ) gu gπ L = L uπL u πL 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 goes down to {1}, we deduce that G1 is a p-group. So G1 is a p-Sylow subgroup of I(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 for eL/K = |G0| is relatively prime to p.

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

× Gal(L/K) = (Z/8Z) = {g1, g3, g5, g5} i so that gi(ζ8) = ζ8. Let’s compute vL(giπL − πL) for each i. We have

vL(g1πL − πL) = vL(0) = ∞, 3 vL(g3πL − πL) = vL(ζ8 − ζ8) = vL(ζ8(ζ8 − 1)(ζ8 + 1)) = 2, 5 vL(g5πL − πL) = vL(ζ8 − ζ8) = vL(−2ζ8) = 4, 7 2 3 vL(g7πL − πL) = vL(ζ8 − ζ8) = vL(ζ8 (ζ8 − 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 and only 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 ramified. √ 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, we may assume that L/K is totally ramified. Then ∼ Gal(L/K) = G0(L/K) = G0/G1, 0 0 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(L0/K0) ,→ Gal(L/K) and so Gal(L0/K0) is cyclic of order d | n. Since K0 contains the n-root of unity, we can use Kummer theory to write √ L0 = K0( d a) 0 × r × for some a ∈ (K ) . If we write a = πK u for u ∈ OK0 , then √ √ √ 0 0 d d 00 d L ⊆ L ⊆ K ( u, πK ) ⊆ K ( πK ) for some unramified extension K00 of K. Before wrapping up the discussion of ramification, let me give a big picture of local class field theory over Qp. Over abelian extensions of Qp, the maximal unramified extension is Qp[ζprime to p] and the maximal tamely ramified 1/(p−1) extension is Qp[ζprime to p, p ].

ab Qp

+ Zp wild

1/(p−1) Qp[ζprime to p, p ]

× Fp

Qp[ζprime to p]

ˆ+ Z

Qp The Artin map is then

× ab × ∼= ab unr Qp → Gal(Qp /Qp), Zp −→ Gal(Qp /Qp ). Another remark is that the numbering of the ramification group is not the 0 0 best. For L /L/K, the groups Gi(L /K) and Gi(L/K) are not going to be related because they are defined in terms of different valuations. To fix this, there is an upper numbering. Define Z u dt φ(u) = , 0 [G0 : Gt] i i 0 and write G = Gφ−1(i). Then Herbrard showed that G (L /K) restricts to Gi(L/K), and Hasse–Arf showed that if L/K is abelian then jumps in Gi occur at integer points. 3 2 Example 7.9. For the case Q2(ζ8)/Q2, we have G = G4 and G = G2. Math 223a Notes 29

8 September 26, 2017

I am now going to move on to Galois cohomology. Roughly I will be following Cassels–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 vector spaces. + × × Example 8.2. For G = Gal(L/K), the abelian groups L , L , OL , µn(L), × × × Cl(OL), AL /L are all G-modules. We are going to be interested in L for L a × × local field this semester. Next semester we are going to be interested in AL /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 trivial action ga = a that makes A into a G-module. In many cases we are going to look at A = Z. A G-module is the same thing as Z[G]-modules, where Z[G] is the group ring, a non-commutative ring of formal linear combinations X agg, g∈G with multiplication done formally. If G = Cn = hti is a cyclic group, then

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

This is not the same thing as Z[ζn]. The ring Z[G] has an important ideal, called the augmentation ideal IG = ker , where X X : Z[G] Z; agg 7→ ag.

As a Z-module, IG is free and is spanned by g − 1. In the case G = Cn, the ideal IG is just (t − 1)Z[G]. P We can also consider the operator N = g∈G g. Then for any G-module A, we have nX o NA = ga : a ∈ A . g∈G Math 223a Notes 30

If A = L×, then N becomes the norm and NA = NL×. If A = L+, then N becomes 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−). Likewise we 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/ha − gai = A/IGA.

G The A is called the invariants and AG is called the coinvariants. It can be checked that these are both functions. These are actually special cases. If A, B ∈ G−Mod, then we have HomG(B,A) is an abelian group, but not generally a G-module. If B = Z, then

G HomG(Z,A) = A .

We can also take the tensor product. Generally, A ⊗R G makes sense if A is a right 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 the g−1 action instead of g. So

−1 A ⊗G B = A ⊗Z B/hg a ⊗ b − a ⊗ gbi. G Then AG = Z ⊗G A. On the other hand, we have HomG(B,A) = Hom(A, B) and A ⊗G B = (A ⊗ B)G. Proposition 8.4. The functor A → AG is left exact, i.e., 0 → A → B → C exact implies 0 → AG → BG → CG exact. Proof. Injectivity of AG → BG is clear. If b ∈ BG with ψ(b) = 0, then by exactness there exists a unique A such that ϕ(a) = b. This means that a = ga for any g ∈ G. So a ∈ AG. This is also a special case of the fact that Hom(B, −) is left exact for any abelian category.

Proposition 8.5. The functor A → AG is right exact, i.e., A → B → C → 0 exact implies AG → BG → CG → 0 exact. This is also a special case of the fact that B ⊗ − is right exact. You can use that 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 as g(a) = χ(g)a. We have the exact sequence

·2 0 → Zχ −→ Zχ → Z/2 → 0. If we take the invariants, we get

0 → 0 → 0 → Z/2. If we take the coinvarians, we get

·2 Z/2 −→ Z/2 → Z/2 → 0. Example 8.7. Let L/K be a totally ramiﬁed extension of local ﬁelds. To make √ this simple, let us use a quadratic extension, e.g., Qp( p)/Qp. There is an exact sequence i × v 1 → OL −→ L −→ Z → 0. If we take G-invariants,

× × v|K 1 → OK → 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 inﬁnitely many) 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, −) is exact means that F is projective: there always is a lift

F ϕ ϕ˜ B π C.

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

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

When G is ﬁnite, we have Hom(Z[G],X) =∼ Z[G] ⊗ X and so coinduction is 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. The coinduced modules are going to be “cohomologically trivial” objects, and the induced 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, and every A has a surjection from an induced G-module.

Proof. We take A0 = 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 a surjection 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, if 0 → 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 is i i i Hˆ (G, A) = H (G, A) if i ≥ 1 and Hˆ (G, A) = H−1−i(G, A) for i ≤ −2. Here, we will get Hˆ 0(G, A) = AG/NA. So if A = L×, then Hˆ 0(G, A) = K×/NL×. Math 223a Notes 33

9 September 28, 2017

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 as gϕ = g ◦ ϕ ◦ g−1, because we needed to make A have a right G-action. But note that Z[G] has a natural right action. So we can simply define the group action on 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 ≥ 0 with the following properties: (i) H0(G, A) = AG, j (ii) if 0 → A −→i B −→ C → 0 is a short exact sequence, then it induces a long exact sequence

j 0 → H0(G, A) −→i∗ H0(G, B) −→∗ 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 d2 d1 ··· −→ P2 −→ P1 −→ P0 −→ Z → 0, so that all Pi are free and the sequence is exact. Such a resolution exists, because for any G-module A, there exists a free G-module that surjects onto A. i Deﬁne a chain complex K = HomG(Pi,A). Then

1 2 0 d 1 d 2 K = HomG(P0,A) −→ K = HomG(P1,A) −→ K = 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

0 1 G H (G, A) = ker d = HomG(P0/ im d1,A) = HomG(Z,A) = A . Math 223a Notes 34

If 0 → A → B → C → 0 is a short exact sequence, for each q we have a short exact 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 apply cohomology and get the long exact sequence. This long exact sequence is further natural in the sense that a morphism of short exact sequences induce a morphism of long exact sequences. Finally, if A = CoindG(X), then q G K = HomG(Pq, Coind ) = 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 dimension shifting. Let A be any G-module, and take a short exact sequence 0 → A → A0 → A∗ → 0 where A0 is coinduced by A. This gives a long exact sequence

j∗ · · · → Hq−1(A0) −→→ Hq−1(A∗) → Hq(A) → Hq(A0) = 0 → · · · . So Hq(A) is the cokernel of j∗ : Hq−1(A0) → Hq−1(A∗) is determined by Hq−1.

Example 9.2. Take G = Cn = hti. We can take the projective resolution

×(t−1) ×N ×(t−1) t7→1 · · · → Z[G] −−−−→ Z[G] −−→ Z[G] −−−−→ Z[G] −−−→ Z → 0. Then our chain complex is

×(t−1) ×(t−1) 0 → A −−−−→ A −−→×N A −−−−→ A → · · · and we can now compute cohomology. We have H0(G, A) = ker(t − 1) = AG, 2q−1 H (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 the even things will be the same.

9.2 Standard bar resolution i+1 Let G be any group and let Pi = Z[G ] = Z[G] be the formal span of i+1 tuples (g0, . . . , gi) ∈ G . Let us make this a G-module by a diagonal action g(g0, . . . , gi) = (gg0, . . . , ggi). Then this is a free Z[G]-module with basis (1, g1, . . . , gi). Deﬁne

X j di : Pi → Pi−1; di(g0, . . . , gi) = (−1) (g0,..., gˆj, . . . , gi). j Math 223a Notes 35

We can easily check that di−1di = 0. To show exactness, suppose that some linear combination of (g0, . . . , gi) is in ker di. If we put an arbitrary s ∈ G in front 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

q+1 q ker(d : HomG(Pq,A) → HomG(Pq+1,A)) H (G, A) = q . im(d : HomG(Pq−1,A) → HomG(Pq,A))

We can interpret Hom(Pq,A) as homogeneous cohains C˜q(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 homogeneous 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 C˜0(A) is determined by 2 f(1) = a ∈ A. The function f : G → A that satisfy f(gg0, gg1) = gf(g0, g1) is determined by f(1, g) = ϕ(g). So the boundary map is

1 d : C˜0 → C˜1; 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 map is 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. Last time we defined homogeneous cochains i i+1 C˜ (G, A) = {(f : G → A): f(gg0, . . . , ggn) = gf(g0, . . . , gn)}. The differentials are defined as d : C˜i(G, A) → C˜i+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 C˜i(G, A) via i i C˜ (G, A) → C (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 i d ϕ(g1, . . . , gi+1) = g1ϕ(g2, . . . , gi+1) − ϕ(g1g2, g3, . . . , gi+1) + ϕ(g1, g2g3, . . . , gi+1) i i+1 − · · · + (−1) ϕ(g1, . . . , gi−1, gigi+1) + (−1) ϕ(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 1 1 2 d : C (G, A) → C (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 {(ϕ : G → A): ϕ(g g ) = ϕ(g ) + g ϕ(g )} H1(G, A) =∼ 1 2 1 1 2 . {(ϕa : G → A): ϕa(g) = ga − a}

These ϕ satisfying ϕ(g1g2) = ϕ(g1) + g1ϕ(g2) are called crossed homomorphisms. If G acted on A trivially, this would just be ϕ(g1g2) = ϕ(g1) + ϕ(g2). Then we are quotienting out by nothing, so 1 H (G, A) = HomGrp(G, A). Math 223a Notes 37

Deﬁnition 10.1. If G is a group and A is a G-module, we deﬁne the semidirect product G n A = G × A with multiplication

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

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

0 → A → G n A → G → 0. Now we can interpret Z1(G, A) as a splitting of this exact sequence, i.e., sections of G n A → G. This is because the map

G → G n A; g 7→ (g, ϕ(g)) is a group homomorphism if and only if ϕ is a crossed homomorphism. 1 On the other hand, ϕ1 and ϕ2 are equivalent modulo B (G, A) if and only if the splittings G → G n A are conjugate by some element of A. These is a similar interpretation for H2(G, A). This classiﬁes short exact sequences of groups 0 → 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 → G n A → G → 0. The idea is that given such 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. j Recall that if we have a short exact sequence of G-modules 0 → A −→i B −→ 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 with j(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, which is 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 the previous 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 to A as sets non-canonically.

Example 10.4. Suppose L/K is a ﬁeld extension, and suppose that µn ⊆ L. × n Let A = µn(L). For any c ∈ (L ) , 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 imagine a situation in which µn(L) ⊆ K but c has no nth roots in K. Then G acts trivially 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 can construct 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.

1 It can be checked that ϕ ∈ Z (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 homomorphism H0(G, C) = CG → H1(G, A). For c ∈ CG, the set j−1(c) ⊆ B is an A-torsor. The class of this torsor gives the connecting morphism. Example 10.6. In the short exact sequence

× j × n 0 → µn(L) → L −→ (L ) → 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 exact sequence

· · · → 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 with A. So we can use the standard resolution to get explicit descriptions. Class ﬁeld theory only uses a limited range of (co)homology groups, H0, H1, H2, Hˆ 0, −1 Hˆ , H0, H1. ∼ 2 ab Theorem 11.1. We have H1(G, Z) = IG/IG = G . Proof. We use the short exact sequence

0 → IG → Z[G] −→ Z → 0 of G-modules. This gives a long exact sequence

2 0 H1(G, Z[G]) = 0 → H1(G, Z) → H0(G, IG) = IG/IG −→ H0(G, Z[G]) = Z[G]/IG. ∼ 2 So we get H1(G, Z) = H0(G, IG) = IG/IG. For the next isomorphism, write 2 ag = g − 1. Note that IG is spanned by

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

2 This shows that IG/IG 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]/IH Z[G]. This shows that

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

2 It is a good exercise to check that this is just IH /IH . Math 223a Notes 41

11.1 Changing the group i Both the functors Hi(G, A) and H (G, A) depend on G and A. These are i covariant functors of A. Hi is also going to be covariant in G, and H will be contravariant in G. Deﬁnition 11.2. Two pairs (G, A) and (G0,A0) are compatible for cohomology if there exist morphisms ρ : G0 → G and λ : A → A0 such that

λ(ρ(g0), a) = g0λ(a).

Example 11.3. If G0 = H ⊆ G is a subgroup, then we can set ρ = i : H,→ G and λ = idA. Example 11.4. If H ⊆ G is a normal subgroup, then ρ : G → G/H and λ : AH → A gives a map from (G, A) to (G/H, AH ). Then ρ maps

0 0 q+1 q+1 ρ∗ : Pq(G ) = Z[(G ) ] → Pq(G) = Z[G ]. This gives a map

0 0 HomG(Pq(G),A) → HomG0 (Pq(G ),A ) of cochain complexes. So we get maps on homology

ρ∗ : Hq(G, A) → Hq(G0,A0).

This can be described explicitly in homogeneous cochains. It sends [ϕ] to [ϕ0] where 0 0 0 0 0 ϕ (g1, . . . , gq) = λ(φ(ρ(g1), . . . , ρ(gq))).

Example 11.5. If (G, A) and (H,A) are compatible by (i, idA), then this induces 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 that Res ◦ Inf = 0 for q ≥ 1, because this is induced by (G/H, AH ) and (H,A) with the map of groups being trivial. Deﬁnition 11.7. For pairs (G, A) and (G0,A0), a compatible map for homology is ρ : G → G0 and λ : A → A0 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 ).

0 0 As before, compatible maps give induced homomorphisms Hq(G, A) → Hq(G ,A ). So there is the corestriction

coRes : Hq(H,A) → Hq(G, A) and coinﬂation Coinf : 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 long exact 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 dimension shifting. ∼ 0 Hq(G, . . .) = Hq+1(G, . . .) 0

Res ∼ 0 Hq(H,...) = Hq+1(H,...) 0 That is, this is a “universal δ-functor”. Later, we will deﬁne

Res : Hˆ q(G, A) → Hˆ q(H,A), coRes : Hˆ q(H,A) → Hˆ q(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 going to 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 we have ϕ(H) = ha − a, which is constant for all h ∈ H. So a ∈ AH and ϕ ∈ B1(G/H, AH ). Math 223a Notes 43

Now we need to prove ker Res ⊆ im Inf. Take φ ∈ Z1(G, A) such that Res ϕ ∈ B1(H,A). Then φ(h) = ha − a for some a ∈ A. But ϕ is living in H1(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 homomorphism G/H → AH . Proposition 11.10. Suppose Hi(H,A) = 0 for 1 ≤ i < q. Then there exists an 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. Now suppose we know this for q − 1. Let A be a G-module satisfying the above, and let A∗ = CoindG(A) so that A,→ A∗. Then we have a short exact sequence

0 → A → A∗ → A0 → 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, A0) = Hq(G, A)

∼ Hq−1(H,A0) = Hq(H,A).

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

0 → AH → (A∗)H → (A0)H → H1(A, H) = 0.

So we get an compatible

∼ Hq−1(G/H, (A0)H ) = Hq(G/H, AH )

Inf Inf Hq−1(G, (A0)H ) Hq(G, AH ).

But A0 satisﬁes Hi(G, A0) = Hi+1(G, A) = 0 for i < q−1. So you can dimension shift. Math 223a Notes 44

12 October 10, 2017

Today we are going to deﬁne Tate cohomology.

12.1 Tate cohomology q We have defined H (G, A) and Hq(G, A) for q ≥ 0. We will define ( Hq(G, A) q ≥ 1 Hˆ q(G, A) = H−1−q(G, A) q ≤ −2 and we will fill in the module. These are called Tate cohomology. We are going to focus only on finite groups G. P For G finite, we have this operator N = g∈G g ∈ Z[G], and sends

X N : A → A; a 7→ Na = ga. g∈G

G ∗ G Then im N ⊆ A and ker N ⊇ IGA. So we actually have a map N : AG → A . This allows to deﬁne

−1 ∗ G 0 ∗ G Hˆ (G, A) = ker(N : AG → A ), Hˆ (G, A) = coker(N : AG → A ).

Because we have

··· H1(G, C) AG BG CG 0

∗ ∗ ∗ NA NB NC 0 AG BG CG H1(G, A) ··· .

Then a standard application of the snake lemma gives

· · · → Hˆ −2(G, C) → Hˆ −1(G, A) → Hˆ −1(G, B) → Hˆ −1(G, C) → Hˆ 0(G, A) → Hˆ 0(G, B) → Hˆ 0(G, C) → Hˆ 1(G, A) → · · · .

Proposition 12.1. For A a G-module that is induced (or equivalently, coinduced), Hˆ q(G, A) = 0 for all q. Proof. We know this except for q = 0 and q. So it suﬃces to show that N ∗ : G L AG → A is an isomorphism. Suppose A = g∈G gX where X ⊆ A is a Z- G submodule. We claim that both AG and A are isomorphic to X. An element P G g gxg is in A if and only if all xg are equal in X. So we get an isomorphism

X X → AG; x 7→ gx. g∈G Math 223a Notes 45

Conversely, we have a map

X X AG → X; gxg 7→ xg. g g

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

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

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

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

Now we can deﬁne

q q • Hˆ (G, A) = H (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 we take the G-invariants of both side, the left hand side is what we want, and the G ∼ right hand side is (Pi ⊗Z A) = (Pi ⊗Z A)G = Pi ⊗G A. n Example 12.2. Let G = Cn = ht : t = 1i. Recall that we have the projective resolution

×(t−1) N ×(t−1) · · · → Z[G] −−−−→ Z[G] −→ Z[G] −−−−→ Z[G] −→ Z → 0. Now dualizing gives

17→N ×(t−1−1) N 0 → Z −−−→ Z[G] −−−−−−→ Z[G] −→ Z[G] → · · · . So we can put them together to get

×(t−1) N ×(t−1−1) N · · · → Z[G] −−−−→ Z[G] −→ Z[G] −−−−−−→ Z[G] −→ Z[G] → · · · . Thus 2q G 2q−1 Hˆ = A /NA, Hˆ = (ker N)/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 : Hˆ q(G, A) → Hˆ q(H,A) for q ≥ 1, coRes : Hˆ q(H,A) → Hˆ q(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 Hˆ q+1(G, A) =∼ Hˆ q(G, A+). So if Resq+1 are already deﬁned, we can deﬁne Resq by

Res Hˆ q(G, A) q Hˆ q(G, A)

∼= ∼= Res Hˆ q+1(G, A+) q+1 Hˆ q+1(H,A+).

We can likewise deﬁne coResq for q ∈ Z by upwards induction. Both of these are functors, and are compatible with long exact sequences. Note that the only G H thing we needed is that Res0 is A → A and coRes0 = AH → AG. After that, we have only used dimension shifting. Let us just look at how the map

−1 −1 Res−1 : Hˆ (G, A) → Hˆ (H,A) is deﬁned. This is induced by the map

0 X −1 NG/H : AG → AH ;[a] 7→ [g a]. g∈G/H Proof. It is enough to show that if 0 → A → B → C → 0 is a short exact sequence, Hˆ −1(G, C) δ Hˆ 0(G, A)

0 NG/H Res0 Hˆ −1(H,C) δ Hˆ 0(H,A) commutes. Let b 7→ c. An element [c] ∈ Hˆ −1(G, C) maps to [Nb] ∈ AG/ im N ∗ 0 P in Hˆ (G, A), and this maps to NGb = gb. On the other hand [c] maps P g∈G downwards to g∈H\G[gc] and this maps to X X X X NH (gb) = hgb = gb. g∈H\G g∈H\G h∈H g∈G So the diagram commutes. Math 223a Notes 47

Likewise, 0 0 coRes0 : Hˆ (H,A) → Hˆ (G, A) comes from the map

H G X NG/H : A → A ; a 7→ ga. g∈G/H

Proposition 12.3. For all q, we have

Res coRes Hˆ q(G, A) −−−→q Hˆ q(H,A) −−−−→q Hˆ q(G, A).

The composite is multiplication by [G : H].

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

G Res H coRes G A /NGA −−→ A /NH A −−−→ A /NGA. P This comes from a 7→ a 7→ g∈G/H ga = [G : H]a.

Corollary 12.4. If |G| = n, then Hˆ q(G, A) are n-torsion. Proof. We that Hˆ q(G, A) → Hˆ q({1},A) → Hˆ q(G, A) is multiplication by n. But Hˆ q({1},A) = 0.

Corollary 12.5. For G finite and A a finitely generated Z[G]-module, Hˆ q(G, A) is also finite. Proof. This is because Hˆ q(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, then Hˆ q(G, A) = 0. Proof. Then n : Hˆ q(G, A) → Hˆ q(G, A) is an isomorphism.

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

q q Res : (Hˆ (G, A))p → Hˆ (Gp,A) is injective.

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

q q Corollary 12.8. If Hˆ (Gp,A) = 0 for all p, then Hˆ (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 p q p+q ` : H (G, A) × H (G, B) → H (G, A ⊗ B) for p, q ≥ 0, and if G is ﬁnite, there is a map ˆ p ˆ q ˆ p+q ` : H (G, A) × H (G, B) → H (G, A ⊗ B) for all p, q ∈ Z. This satisfy the axioms: • The product is natural in both A and B. G G G • If p, q = 0, the map is ` : A × B → (A × B) ; a ` b = a ⊗ b. • This is compatible with exact sequences. That is, if 0 → A → A0 → A00 → 0 is an exact sequence such that 0 → A ⊗ B → A0 ⊗ B → A00B → 0 is also exact, then for a00 ∈ Hp(G, A00) and b ∈ Hq(G, B) 00 00 δ(a ` b) = (δa ) ` b. This axioms uniquely determine the cup product on Hˆ , and the proof is by dimension-shifting. (This is in the Bonn lectures of Neukirch.) You can also explicitly construct this using resolutions. There are explicit formulas for the cup product. Let f ∈ Cp(G, A) and g ∈ 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 Frölich wirte down explicit formulas in all dimensions. The Bonn lectures 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). pq ∼ • a ` b = (−1) b ` 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, the last formula is, for a ∈ AH and b ∈ BG, X X NG/H (a ⊗ b) = g(a ⊗ b) = ga ⊗ b = (NG/H a) ⊗ b. g∈G/H g∈G/H For p = 0 and q arbitrary, the map q q a ` − : H (G, B) → H (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 = Z normally. The group H2(L/K, L×) is called the Brauer group of L/K, and we are going to get an explicit isomorphism

2 × ∼= 1 H (L/K, L ) −→ n Z/Z

1 where n = [L : K]. Now there is the preferred generator n on the right hand side, and we will call the preimage uL/K . This will give a cup product

ˆ −2 ˆ 0 × − ` uL/K : H (L/K, Z) → H (L/K, L ).

Note that the left hand side is just the abelianization Gal(L/K)ab. But the right hand side is K×/NL×. This map is what we call the Artin map. To get this, we will need to compute some Galois cohomology. Let L/K be a ﬁnite extension. For motivation, I can ask, what is Hˆ q(L/K, L+)? Does this have any interesting Galois cohomology? The answer is no, because L+ 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 to talk 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 that ϕ(gh) = ϕ(g) · gϕ(h). We need to show that there exists some a ∈ L× such that ϕ(g) = a/ga. Let x ∈ L, and we are going to let

X a = ϕ(g) · gx 6= 0. g∈G

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

X X a = ϕ(g0) · g0x = ϕ(gg0) · (gg0x) g0∈G gg0∈G X X = ϕ(g) · gϕ(g0) · gg0x = ϕ(g) g(ϕ(g0) · g0x) = ϕ(g) · ga. g0∈G g0∈G Math 223a Notes 50

Here is the original Hilbert 90. Let L/K be a cyclic extension and let Gal(L/K) = hgi. Then we have

ker(N : L× → L×) H1(L/K, 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 let x ∈ Q(i) be such that Nx = 1. That is, x is on the unit circle. Then there exists a y ∈ Q(i) such that x =y/y ¯ . If we write y = a + bi, then a + bi (a2 − b2) + 2abi x = = . a − bi 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 be written as lim G with each G finite. For example, ←−i∈I i i Gal(K/K) = lim Gal(L/K) ←− L/K finite is profinite. We can give a natural topology on G by setting ker(G → Gi) as the neighborhood basis at 1. Theorem 13.4. A topological group G is profinite if and only if G is compact and totally disconnected. G is profinite if and only if G = lim G/U where U ←− runs 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 discrete topology) with a group action G × A → A such that this map is continuous, or equivalently, all stabilzers are open in G, or equivalently, [ A = AU . U⊆G open

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

q U q V InfU,V : H (G/U, A ) → H (G/V, A ).

Then we have Hq(G, A) = lim Hq(G/U, AU ). −→ UCG Math 223a Notes 51

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

Hi(K,A) = lim Hi(Gal(L/K),A). −→ L/K ﬁnite Galois

In particular, we have

× H1(K, K ) = lim H1(L/K, L×) = 0. −→ L/K ﬁnite Galois

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. A map 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 ⊆ G closed. We don’t have Tate cohomology because the groups are not ﬁnite, but we 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 if char K = 0. This what we should use if we want to do Galois theory. sep Deﬁnition 14.1. If A is a GK = Gal(K /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 char K. There is a short exact sequence of Gal(Ksep/K)-modules

sep sep × n sep × 0 → µn(K ) → (K ) −→ (K ) → 0, where the last map is surjective because Ksep has all n-th roots because xn − a is always irreducible for a 6= 0. The associated long exact sequence is × n × 1 1 sep × 0 → µn(K) → K −→→ K → H (K, µn) → H (K, (K ) ) = 0 → · · · . 1 × × Thus H (K, µn) is just the cokernel of n : K → K . That is,

× × n δ 1 K /(K ) −→ H (K, µn) is an isomorphism. sep Suppose that µn ⊆ K. Then µn has trivial Gal(K /K)-action, and so we have 1 sep H (K, µn) = Homcts(Gal(K /K) → µn). sep If ϕ : Gal(K /K) → µn is a continuous homomorphism, then ker ϕ is an open subgroup of Gal(Ksep/K), which is Gal(Ksep/L) for some L. So this map ϕ factors as sep ϕ : Gal(K /K) Gal(L/K) ,→ µn. The connecting homomorphism δ can be described explicitly as √ g( n a) δ(a) = ϕa : g 7→ √ ∈ µn. n a Suppose I have some cyclic Galois L/K of degree n, where I still assume µn ⊆ K. Take some

∼= 1 1 sep (ϕ : Gal(L/K) −→ µn) ∈ H (L/K, µn) ,→ H (K /K, µn). × × n 1 Because δ : K /(K ) → H (K, µn) is an isomorphism, this is inf ϕ = ϕa for some a ∈ K×/(K×)n. Then √ sep sep n Gal(K /L) = ker ϕa = Gal(K /K( 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 second cohomology H2(L/K) = H2(L/K, L×). Note that we have already classified all unramified extensions, and it turned out that Gal(L/K) is cyclic. Then

∼ ˆ 2 × ∼ ˆ 0 × × × vK ,= 1 H (L/K, L ) = H (L/K, L ) = K /NL −−−→ Z/nZ = n Z/Z. But this is not very satisfactory. If L0/L/K is unramified, we can try to think about what Inf : Hˆ 2(L/K, L×) → Hˆ 2(L0/K, L0×). But inflation is not ˆ 0 1 1 defined on H . It turns out that this is just n Z/Z ,→ N Z/Z, but I’m getting ahead of myself. ˆ q × Lemma 14.2. H (L/K, OL ) = 0 for all q provided that L/K is unramified. Proof. Because L/K is cyclic, we can check this for Hˆ 0 and Hˆ 1. We have shown that ˆ 0 × × × H (L/K, OL ) = OK /NOL = 0. ˆ 1 × For H (L/K, OL ) is almost Hilbert’s theorem 90. We have a short exact se- × × vL quence 0 → OL → L −−→ Z → 0, so × × 1 × 1 × 0 → OK → K → Z → H (L/K, OL ) → H (L/K, L ) = 0 → · · · . × Because L/K is unramified, vK : K → Z is surjective and then we conclude ˆ 1 × H (L/K, OL ) = 0. × Now we know that all stuff coming from OL is trivial, and so we see that 2 × 2 (vL)∗ : H (L/K, L ) → H (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] is an isomorphism Q → Q. So we get a connecting homomorphism

1 ∼= 2 δ : H (L/K, Q/Z) −→ H (L/K, Z) which is an isomorphism. Now the ﬁrst cohomology can be computed as 1 ∼ 1 H (L/K, Q/Z) = Hom(Gal(L/K), Q/Z) = n Z/Z, because there is a preferred generator FrobK ∈ Gal(L/K). Now we have a canonical identiﬁcation

2 × v∗ 2 δ 1 1 invL/K : H (L/K, L ) −→ H (L/K, Z) ←− H (L/K, Q/Z) → n Z/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 H2(L/K, L×) = lim H2(K /K, K×), −→ −→ n n L/K unram. n ﬁn. Gal. Math 223a Notes 54

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

−1 2 × (vKn )∗ 2 δ 1 1 H (Kn/K, Kn ) H (Kn/K, Z) H (Kn/K, Q/Z) n Z/Z

Inf Inf Inf

(v ) −1 2 × KN ∗ 2 δ 1 1 H (KN /K, KN ) H (KN /K, Z) H (KN /K, Q/Z) N Z/Z, because the restriction of the Frobenius is still the Frobenius. So we get

inv ,∼= H2(Kunr/K, (Kunr)×) = lim H2(K /K, K×) −−−−−→K lim 1 / =∼ / . −→ n n −→ n Z Z Q Z n n Ultimately, we want to compute the Brauer group

2 sep sep × Br(K) = H (K /K, (K ) ) =∼ Q/Z. This is going to take more time. What we need to show is that if L/K is any Galois extension with [L : K] = n,

2 × 1 inv : H (L/K, L ) → n Z/Z is an isomorphism. To show this, we compare the two invariant maps

2 unr unr × invL 2 unr unr × invK H (L /L, (L ) ) −−−→ Q/Z,H (K /K, (K ) ) −−−→ Q/Z. Note that Lunr = L · Kunr because the unramiﬁed closure is constructed by adjoining 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

invK H2(Kunr/K, (Kunr)×) Q/Z

Res ×n

invL H2(Lunr/L, (Lunr)×) Q/Z commutes. We are actually in the section titled “Some diagrams” in Serre’s local ﬁelds part in Cassels–Fr¨ohlich. Math 223a Notes 55

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

evK (Kunr)× Z

vL (Lunr)× Z.

So we get

−1 e(vK )∗ δ FrobK H2(Kunr/K, (Kunr)×) H2(Kunr/K, Z) H2(Kunr/K, Q/Z) Q/Z

Res Res Res f

−1 (vL)∗ δ FrobL H2(Lunr/L, (Lunr)×) H2(Lunr/L, Z) H2(Lunr/L, Q/Z) Q/Z.

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

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, then H2(L/K, L×) =∼ 1 2 × n Z/Z. First we will find an element uL/K ∈ H (L/K, L ). If L/K, we can just −1 1 set uL/K = inv ( n ). If L/K is ramified, we can set Kn/K to be the extension of the same degree, and then compare the inflation maps

2 × Inf 2 × 0 H (Kn/K, Kn ) H (Ln/K, Ln )

2 × Inf 2 × 0 H (L/K, L ) H (Ln/K, Ln ).

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 of degree n = [Kn : K], then there is an isomorphism

2 × ∼= 1 inv : H (Kn/K, Kn ) −→ n Z/Z. There is also a proﬁnite version

2 unr unr × ∼= inv : H (K /K, (K ) ) −→ Q/Z. The goal of today is to get, for any ﬁnite Galois extension L/K of local ﬁelds with n = [L : K], 2 × ∼= 1 inv : H (L/K, L ) −→ n Z/Z. This would imply

2 sep sep × ∼= inv : H (K /K, (K ) ) −→ Q/Z.

15.1 Second Galois cohomology of ﬁnite extensions 2 × 1 Because inv : H (Kn/K, Kn ) → n Z/Z is an isomorphism, we can look at the 1 2 × inverse image un of n . We want to ﬁnd the element uL/K ∈ H (L/K, L ) of order 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 restriction map 2 × 2 unr unr × inv H (Ln/L, Ln ) H (L /L, (L ) ) Q/Z

Res Res ×n

2 × 2 unr unr × inv H (Kn/K, Kn ) H (K /K, (K ) ) Q/Z.

1 Because un maps to n in Q/Z, this shows that un maps to 0 in the upper right Q/Z. That is, Res un = 0. Now note that we have a inﬂation-restriction exact sequence

2 × H (Ln/L, Ln ) ResL

2 × Res H (Ln/K, Ln ) InfL

InfK 2 × n 2 × H (Kn/K, Kn ) H (L/K, L ).

2 × 2 × So we can push un to H (Ln/K, Ln ) and then because it vanishes at H (Ln/L, Ln ), 2 × 2 × it comes from H (L/K, L ). Let us call this uL/K ∈ H (L/K, L ). This is going to be the order of element n we want. 2 × To show that H (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×) = Hˆ 0(L/K, L×) =∼ K×/NL×.

This can be shown on the nose. One way of doing it is to use the Herbrand quotient. You can use the following lemma. Lemma 15.1. If L/K is a ﬁnite Galois extension of local ﬁelds, there exists × an Gal(L/K)-invariant open subgroup V ⊆ OL such that

Hˆ q(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+. We know that L is coinduced as a Gal(L/K)-module. That is, there L L exists an a ∈ L such that L = g∈Gal(L/K) K · (ga). Without loss of generality, r we may assume that a ∈ (πL) by rescaling. (We are going to choose r later.) Now define M r A = OK · (ga) ⊆ (πL) . g∈Gal(L/K) r Now take r large enough such that expp : πLOL → UL,r is an isomorphism × of topological groups. This V = expp(A) is now an open subgroup of OL . Also V is coinduced because A is coinduced. Recall from the homework that if G is cyclic and A is a G-module, then the Herbrand quotient is |Hˆ 0(G, A)| h(A) = |Hˆ 1(G, A)| if both are defined. This satisfies • 0 → A → B → C → 0 exact implies h(B) = h(A)h(C) if any two are defined, • h(A) = 1 if A is finite, • h(Z) = |G|. × Proposition 15.2. h(OL ) = 1. Proof. There is a short exact sequence

× × 0 → V → OL → OL /V → 0

× or Gal(L/K)-modules. Then h(V ) = 1 and h(OL/V ) = 1 because OL /V is × ﬁnite. This shows that h(OL ) = 1. Proposition 15.3. h(L×) = n. Math 223a Notes 58

Proof. Again we have a short exact sequence

× × 0 → OL → L → Z → 0 and look at the Herbrand quotients. So we have |Hˆ 0(L/K, L×)| = n |Hˆ 1(L/K, L×)| 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 case is covered by the cyclic case. Otherwise Gal(L/K) is solvable so there exists a H normal subgroup H C Gal(L/K). Let M = L . 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]. 2 × Proof. There is an element uL/K ∈ H (L/K, L ) of order n, and we have the estimate |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 we 2 × just need surjectivity. The Brauer group is limL/K H (L/K, L ), and so any 2 × element of Br(K) is represented by some element auL/K ∈ H (L/K, L ). But recall that we have

2 × H (Ln/K, Ln ) InfKn InfL H2(K/K, K×) H2(L/K, L×).

2 × So auL/K is also represented by aun ∈ H (Kn/K, Kn ) 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:

invK Br(K) Q/Z

Res ×n

invL Br(L) Q/Z.

Proof. Just use the deﬁnition Br(K) =∼ H2(Kunr/K, (Kunr)×) and the result last time. Corollary 15.8. Let E/L/K be a tower of local ﬁelds and assume E/K is Galois. The restriction map

2 × 2 × ResL/K : H (E/K, E ) → H (E/L, E ) sends uE/K to uE/L. Proof. Look at the diagram

H2(E/K, E×) Res H2(E/L, E×)

invK invL

1 [L:K] 1 [E:K] Z/Z [E:L] Z/Z 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. Assume that for all H ⊆ G, H1(H,A) = 0 and H2(H,A) is cyclic of order |H|. If a generates H2(G, A), then ˆ q ˆ q+2 − ` a : H (G, Z) → H (G, A) is an isomorphism for all q ∈ Z. Let G = Gal(L/K) and A = L×. This satisﬁes the assumptions by Hilbert’s theorem 90 and the theorem we have just proven. Apply the theorem to q = −2. Then we get

ab ∼= 0 × × × Gal(L/K) = Hˆ 2 (L/K, Z) −→ Hˆ (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 exact sequence

q − a q+2 q+2 q+1 · · · → Hˆ (G, Z) −−−→` Hˆ (G, A) → Hˆ (G, M) → Hˆ (G, Z) → · · · . Then we will show that Hq(H,M) = 0 for all H ⊆ G and some two consecutive q. This implies (by the homework) that Hq(H,M) = 0 for all H ⊆ G and all q. 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 that for all H ⊆ G we have H1(H,A) = 0 and H2(H,A) is cyclic of order |H|. If a is a generator of H2(G, A), then

ˆ q ˆ q+2 − ` a : H (G, Z) → H (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 Hˆ q(G, B) =∼ Hˆ q+2(G, A). Let b ∈ Hˆ 0(G, B) be a preimage of a ∈ Hˆ 2(G, A) under this isomorphism. Then

− ` a Hˆ q(G, Z) Hˆ q+2(G, A)

∼= − ` b Hˆ q(G, B) commutes. The condition can be reformulated as

−1 0 Hˆ (H,B) = 0, Hˆ (H,B) =∼ Z/|H|Z for all H ⊆ G. If b is a generator of Hˆ 0(G, B), we want to show that

ˆ q ˆ q − ` b : H (G, Z) → H (G, B) is an isomorphism. This maps are induced by Z → B with n 7→ nb. Letting B0 = B ⊕ Z[G], we get an isomorphism Hˆ q(G, B) =∼ Hˆ q(G, B0) induced by B,→ B0. Deﬁne

0 i : Z → B = B ⊕ Z[G]; n 7→ (nb, nN), and we get a short exact sequence

i 0 j 0 → Z −→ B −→ M = coker i → 0 of G-modules. This is analogous to the mapping cylinder construction in topology. This gives a long exact sequence

−1 0 −1 δ 0 · · · → Hˆ (H,B ) = 0 → Hˆ (H,M) −→ Hˆ (H, Z) − b 0 0 j∗ 1 −−−→` Hˆ (H,B) → Hˆ (H,M) −→ Hˆ (H, Z) = 0 → · · · . Math 223a Notes 62

Here, the map Hˆ 0(H, Z) → Hˆ 0(H,B) is given by [n] 7→ nb. Here, b is a generator 0 of B/NGB so b is a generator of B/NH B = Hˆ (H,B). Because both are cyclic ˆ 0 ˆ 0 groups of order |H|, the map − ` b : H (H, Z) → H (H,B) is an isomorphism. This shows that Hˆ −1(H,M) = Hˆ 0(H,M) = 0, for all H ⊆ G. The homework problem shows that all cohomology of M are Hˆ q(G, M) = 0. Then the long exact sequence implies that ˆ q ˆ q − ` b : H (G, Z) → H (G, B) are isomorphisms.

16.1 Reciprocity map Last time, for L/K a Galois extension of local ﬁeld, we used Tate’s theorem to give an isomorphism

ab −2 uL/K 0 × × × Gal(L/K) = Hˆ (L/K, Z) −−−→ Hˆ (L/K, L ) = K /NL . This gives the reciprocity map

× × ab θL/K : K /NL → Gal(L/K) . People also write this as θL/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 am going to give another way of describing this map, which may be more convenient in some contexts.

× × ab Proposition 16.2. For [a] ∈ K /NL , its image θL/K ([a]) ∈ G is charac- δ,∼= terized by: for any χ ∈ Hom(G, Q/Z) = H1(L/K, Q/Z) −−→ 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 to do part of it and leave part of it on the homework.

Proof. By deﬁnition of θL/K , we have

[a] = [θL/K (a)] ` uL/K . So

inv([a] ` δχ) = inv(θL/K (a) ` uL/K ` δχ) = inv((θL/K (a) ` χ) ` uL/K ) = inv(δ(χ ` θL/K (a)) ` uL/K ). ˆ 1 ˆ −2 What is χ ` θL/K (a)? Here χ ∈ H (G, Q/Z) and θL/K (a) ∈ H (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 δ / =∼ Hˆ −1(G, / ) −→ Hˆ 0(G, ) =∼ /|G| |G|Z Z Q Z Z Z Z is just ×|G|. So

inv([a] ` δχ) = inv([L : K]χ(θL/K (a))uL/K ) 1 = [L : K]χ(θ (a)) = χ(θ (a)). L/K [L : K] L/K

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. There ab ab × is a natural quotient map πE/L : Gal(E/K) → Gal(L/K) . 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 toχ ˜ : 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

θ K×/NE× E/K Gal(E/K)ab

θ K×/NL× L/K Gal(L/K)ab.

So we get a map

θ : K× → lim Gal(L/K)ab = lim Gal(L0/K) = Gal(Kab/K) /K ←− ←− L/K ﬁn. Gal. L0/K ﬁn. abe. that has dense image. Math 223a Notes 64

16.2 Group of universal norms × ab What is the kernel of θ/K : K → Gal(K /K)? This is going to be simply the intersection of the kernels, and is

\ × ker θ/K = NL/K L . L/K

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 Galois × L/K such that A = NL/K L . We’ll show that any finite index open subgroup of K× is normic. For profinite groups, open subgroups have finite index. But note that K× is not profi- × nite/compact, and in particular contains OK . On the other hand, finite index subgroups of K× are open. This is because (K×)n contains an open ball around 1.

Proposition 16.6. If L/K is Galois, and L0/K is the maximal abelian subex- × 0 × tension, then NL/K L = NL0/K (L ) .

× 0 × Proof. The inclusion NL/K L ⊆ NL0/K (L ) is clear. On the other hand,

× ∼ ab 0 ∼ 0 × K/NL/K L = Gal(L/K) = Gal(L /K) = K/NL0/K (L ) .

× 0 × This means that the quotient map K/NL/K L → K/NL0/K (L ) is an isomorphism. Math 223a Notes 65

17 October 26, 2017

Recall that at the end of last time we deﬁned that a subgroup A ⊆ K× is × normic if there exists some Galois extension L/K such that A = NL/K L . 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 ⊆ L0 implies NL0 ⊆ NL. By local class ﬁeld theory, [L : K] = [K× : NL×]. × ab This map is clearly surjective. Because we have θ/K : K → Gal(K /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 have N(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 that × θLM/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 and × K /A, and this will correspond to some θL/K (B/A) ⊆ Gal(L/K). Take M = LθL/K (B/A). Then

θ K×/A L/K Gal(L/K) ∼=

θ K×/NM × M/K Gal(M/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 finite index (and thus open), then A is normic. Proof. Note that if A ⊆ K× is finite index, then A ⊇ (K×)n for some n. So it is enough to show that (K×)n are normic. Let us first do this in the case when µn ⊆ K. Let L be the maximal exponent n of an abelian extension of K. By Kummer theory, √ L = K({ n a : a ∈ K×/(K×)n}) √ = composium of all extensions {K( n a)}.

Then \ √ NL× = NK( n a)× ⊇ (K×)n. a∈K×/(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

ab 1 ab ∼ × × n Hom(Gal(L/K), µn) = Hom(Gal(K /K), µn) = H (K /K, µn) = K /(K ) .

This shows that Gal(L/K) and K×/(K×)n have the same order. 0 We now generalize to the case when K might not contain µn. Let K = K(µn). Then 0 × 0 × n NL0/K0 (L ) = ((K ) ) . We don’t know if L0/K is Galois, so take L/K such that L ⊇ L0 and L/K is Galois. Then

× 0 × 0 × 0 × n × n NL/K L ⊆ NL0/K (L ) = NK0/K NL0/K0 (L ) = NK0/K ((K ) ) ⊆ (K ) .

This implies that (K×)n is normic. This gives us

Gal(Kab/K) = lim (K×/NL×) = lim K×/A = K×. ←− ←− b L/K abelian [K×:A]<∞

This group Kb × is called the proﬁnite completion of K. × × × Z We can be more explicit in describing Kb . We have K = OK × π . So

× × ˆ× × Zb Kb = ObK × Z = OK × π .

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

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

−1 2 × v∗ 2 δ 1 evalFrob 1 H (L/K, L ) −→ H (L/K, Z) −−→ H (L/K, Q/Z) −−−−−→ n Z/Z.

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

× × 2 a ∈ Qp /(Qp ) = {1, u, p, up}

× u where u ∈ Zp is a unit with ( p ) = −1. So there are three cases to consider. × If a = u, then L/K is unramified and NL = {x : vp(x) even}, which is an × × 2 index 2 subgroup of Qp , generated by (Qp ) and u. If a = p, then L/K is totally ramified. Then we can say that NL× is × 2 generated by (Qp ) and −p. Likewise, if a = up then L/K is totally ramified and NL× is generated by × 2 (Qp ) and −up. Let us look a global example.

∗ (`−1)/2 Example 17.6. Let√` be an odd prime. Take ` = (−1) ` (≡ 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, K / ) = θ (a) ∈ Gal(K / ) ,→ Gal(K/ ) ∼ {±1} w Qv Kw/Qv w Qv Q = where w is a place of K above v. If v = `, we have total ramiﬁcation and

× × 2 ∗ NLw = (Q` ) × {1, (−` )}.

× a So for a ∈ Z` if and only ( ` ) = 1, i.e., a (a, K / ) = . w Q` ` Let v = p be an odd prime with p 6= `. Thre are two cases. If `∗ is a square in Qp, then Kw = Qp and so

(a, Kw/Qp) = 1 Math 223a Notes 68

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

vp(a) (a, Kw/Qp) = (−1) . In general, we see that

∗ ` vp(a) (a, K / ) = . w Qp a

∗ ∗ × 2 Let us now consider the case p√= 2. Because ` ≡ 1 mod 4, either ` ∈ (Q2 ) 2 ∗ ∗ if ` ≡ 1 mod 8 or Q2(` ) = Q2( 5) if ` ≡ 5 mod 8. Since a is a unit modulo 2, 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, but the global reciprocity law tells us that Y (a, Kw/Q) = 1. v Say a = p. Then p`∗ = 1. ` p 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 (OK ) = I(L/K). × × Proof. We ﬁrst prove θL/K (OK ) ⊆ I(L/K). Note that θT/K (OK ) = {1}. Then

θ K× L/K Gal(L/K)

θT /K Gal(T/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 × vK (a) some u ∈ OK . If θL/K ([a]) ∈ IL/K then θT/K ([a]) = FrobT/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 local Kronecker–Weber for Qp from the problem set. Let K be a local ﬁeld. We have an Artin map × × Z ab θ/K : K = OK × π → Gal(K /K) This map is not surjective, but we can extend it to

× × Zb ab θ/K : Kb = OK × π → Gal(K /K). So we can decompose

Kab × OK πZb

Kunr Kπ

× b πZ OK K

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

ab Qp = Qp(ζ∞)

unr Qp = Qp(ζ(n,p)=1) Qp(ζp∞ )

Qp as you’ve computed in your problem set. Now we want to see what θ (a) is. /Qp r unr r Let a = p u. On Qp , we know that θL/K (a) = FrobL/K if L/K is unramiﬁed. pr So it is going to map ζn 7→ ζn . On the other hand, it is harder to ﬁgure out u−1 what ζpk is mapped to. It turns out that ζpk 7→ ζpk . k There is a global proof of this fact. If we use global reciprocity on Q(ζp )/Q, u−1 the images should cancel out so it has to be ζpk . But we want to do everything locally, 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 group under addition. This is called the additive group and we say B = Ga(B). Math 223a Notes 70

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

−1 n ∼ Example 18.3. Set R = A[t, t ]/(t −1). Then Hom(R,B) = µn(B) and this is 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

−1 R = A[(ti,j)1≤i,j≤n, (det[tij]) ].

The answer is that we want R to be a Hopf algebra. This means that there exist maps • comultiplication ∆ : R → R ⊗ R • coinversion S : R → R • coidentity : R → A with axioms • coassociativity R ∆ R ⊗ R

∆ 1⊗∆ R ⊗ R ∆⊗1 R ⊗ R ⊗ R

• identity R R ⊗ R 1⊗ R

• coinversion R ⊗ R 1⊗s R ⊗ R

∆ R A R

∆ R ⊗ R s⊗1 R ⊗ R

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 : A[t] → A is t 7→ 0. This is the right map 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, : A[t, t−1] → A; t 7→ 1, t−1 7→ 1.

Why are group schemes useful? Suppose we have an affine abelian group scheme G defined over a field K. Assume that it comes from some Hopf algebra R, and also assume that R is finite over A (i.e., is a finite dimensional K-vector space). Then we can look at G(K) = Hom(R, K). This is going to be a finite group 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 of ﬁnite 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 can be proven that over K of characteristic 0, the only affine abelian group schemes are Ga, Gm, µn, and their products. There are two ways to go from here: • Drop the affine condition. Here we may consider elliptic curves, abelian varieties, etc. These all have a lot of subgroup schemes. If you have an elliptic 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 not necessarily abelian. So we need some extra structure on G(K) that rigidifies the structure.

18.2 Formal groups Let A be any ring. We are going to consider the ring A[[x]] and we want to make 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 πK OK because we are looking only at continuous maps. We need something like comultiplication, but we are not going to look at A[[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. The comultiplication 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. Then F (x, y) = x + y + xy. The power series 1 i (x) = − 1 = −x + x2 − x3 + x4 − · · · F 1 + x works as an inverse.

If A = OK and L/K is a ﬁnite extension, then πLOL using the group law given 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 actually redundant), • 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 such that 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 then sep F (πLOL) is πLOL made into a group by a +F b = F (a, b). If K /K is the separable closure, then we can look at its maximal ideal

sep K ⊇ 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 homomorphisms h : 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 Gb a(x, y) = x + y and Gb m(x, y) = x + y + xy. The power series h = expp −1 ∈ Qp[[x]] is then a formal group homomorphism because

Gb m(h(x), h(y)) = expp(x + y) − 1 = h(Gb a(x, y)).

−1 There is also an inverse h (y) = logp(1 + y). So Gb a and Gb m are isomorphic over Qp. These are not isomorphic over Zp. For instance, L = Qp(ζp) has ζp − 1 ∈ πLOL and so Qb m(πLL) has p-torsion. But Qb a(πLL) never has torsion.

Example 19.3. Let us look at Gb m over Zp. For n ∈ Z, deﬁne X n h (x) = (x + 1)n − 1 = xk. n k k≥1

Then hn ∈ End(Gb m). 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, with h1+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 + hm = hn+m, hn ◦ hm = hnm. Gbm So we have a ring homomorphism

Zp → End(Gb m). In this case, this is actually and isomorphism although I am not going to prove this. Suppose F and G be formal groups over OK . Last time I talked about getting cyclotomic ﬁeld from group schemes. Given h : F → G, how can we think of ker h? We have a group homomorphism

F (pKsep ) → G(pKsep ). and so we can look at the kernel of this homomorphism. Often we are going to be in the case when h is a monic polynomial. Then the kernel is the set of solutions 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 scheme over OK .

Example 19.4. For instance, if K = Qp and F = Gb m, with hp, we would get p the ring OK [[x]]/(x + 1) − 1. Then our formal group obtained by taking the kernel is µp(OK ).

19.2 Lubin–Tate theory sep We can look at Gal(K /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 uniformizer. The Lubin–Tate series is given as

q Fπ = {f(x) ∈ OK [[x]] : f(x) ≡ πx mod deg 2, f(x) ≡ x mod π}.

p Example 19.6. For K = Qp, the polynomial f(x) = (x + 1) − 1 is a Lubin– Tate series. Actually constructing them are easy, because you can take anything like f(x) = xp + πx.

Theorem 19.7. (a) For any f ∈ Fπ, there exist a unique formal group Ff such 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 unique polynomial 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 Φk already. Then any Φk+1 satisfying the condition must reduce to Φk modulo degree k + 1. Then Φk+1 = Φk + Q where Q is a homogeneous polynomial of degree 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)) k+1 ≡ Φk(g(x1), . . . , g(xr)) + π Q(x1, . . . , xr) (mod deg(k + 2)). This shows that there exists a unique Q. But we still need to show that Q has coeﬃcients in OK . Working modulo π, we have q q q f(Φk(x1, . . . , xn)) ≡ Φk(x1, . . . , xr) ≡ Φk(x1, . . . , xr) ≡ Φk(g(x1), . . . , g(xr)). k+1 So Q actually has coeﬃcients in OK because π − π is π times a unit.

Now we can start proving the theorem. We construct Ff as the unique power series 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 series in x that commute with f. Again uniqueness shows that Ff (x, 0) = x. This proves (a). Math 223a Notes 76

20 November 7, 2017

Recall our setup

q Fπ = {f ∈ OK [[x]] : f(x) ≡ πx mod deg 2, f(x) ≡ x mod π}.

We proved the lemma saying that if f, g ∈ Fπ and Φ1 ∈ OK [x1, . . . , xr] is a homogeneous 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 there exists 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 to ∼ show 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 ∈ OK , we have

−1 [a]f,g ◦ [a ]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 along with 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, then F (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 formal group, we deﬁne n Ef,n = {x ∈ pKsep :[π ]f (X) = 0}. These are the “πn-torsion of f”.

π,n Let K be the ﬁeld generated by Ef,n, and write

[ π [ π,n Ef = Ef,n,K = K . n 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.

n Proof. Note that x ∈ Ef,n if and only if [π ]f (x) = 0 if and only if

n n 0 = [1]g,f [π ]f (x) = [π ]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)

π,n,f because K(x) is still a complete ﬁeld. Then K = K(Ef,n) ⊇ Eg,n and so Kπ,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. ∼ n Theorem 20.4. As OK -modules, Ef,n = OK /π OK , non-canonically. Proof. Without loss of generality, let f be a monic polynomial of degree q, e.g., q k (k) x + πx. Note here that [π]f = f and [π ]f = f ◦ · · · ◦ f = f . Then

(n) Ef,n = {x ∈ pKsep : f (x) = 0}. Note that f (n) | f (n+1) because x | f and so we can factor

f f (2) f (3) f (n) f (n) = x · · · ····· . x f f (2) f (n−1)

We claim that all f (k)/f (k−1) are Eisenstein polynomials. We just check

k k−1 k k−1 g ≡ xq /xq ≡ xq −q (mod π) Math 223a Notes 78 and the constant term is π. So they are all irreducible, and all of its roots have absolute value less than 1. Then all of these roots lie in pKsep . Therefore n |Ef,n| = q whatever it is. Note that Ef,n is a module over OK , which is a DVR. So Ef,n is going to be a direct sum ∼ d1 dm Ef,n = OK /π ⊕ · · · ⊕ OK /π .

But the π-torsion submodule of Ef,n is Ef,1 which has order q. So m = 1 and ∼ n Ef,n = OK /π .

n Ef,n is not canonically isomorphic to OK /π OK , but it is true that

n × ∼ (OK /π OK ) = AutOK (Ef,n); a 7→ [a]f . S Corollary 20.5. Ef = n Ef,n is isomorphic to K/OK as OK -modules.

Proof. Choose a generator α1 ∈ Ef,1 and then for n ≥ 2 inductively choose αn such that [π]f αn = αn−1. Then glue them together using Ef → K/OK by −n αn 7→ π . ∼ Again, Ef = K/OK is not canonical but

Aut (E ) =∼ lim Aut (E ) = lim(O /πnO )× =∼ O× OK f ←− OK f,n ←− K K K n n is canonical.

20.2 Field obtained by adjoining torsion points π,n Proposition 20.6. The field K = K(Ef,n) = K(αn) is a Galois extension of 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 show that it is generated by αn, note that for any x ∈ Ef,n, there exists an a ∈ OK such that x = [a]f (αn). But [a]f is an infinite series with coefficients in OK , so x ∈ K(αn) because K(αn) is complete.

(n) (n−1) Here αn is a root of f but is not a root of f , so that αn is a root of (n) (n−1) f /f . So this is the minimal polynomial for αn. π,n Now we’d like to understand Gal(K /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

π,n ∼ n × Gal(K /K) → AutOK (Ef,n) = (OK /π OK ) .

Proposition 20.7. This map is an isomorphism. Math 223a Notes 79

π,n Proof. The map is clearly injective, because Ef,n generates K . For surjectivity, 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

n × n n−1 |(OK /π OK ) | = q − q because we have to remove that non-units, which are multiplies of π. So this is an isomorphism.

Corollary 20.8. Gal(Kπ/K) = lim Gal(Kπ,n/K) = Aut (E ) =∼ O× . ←− OK f K So we’ve constructed this field Kπ by our bare hands, without using any class field theory. Now I would like to show that this is the same as the Kπ we have defined before, using the reciprocity map.

π,n × Proposition 20.9. π ∈ NKπ,n/K (K ) . Proof. We have qn−qn−1 Nαn = (−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

× π rπ : K → Gal(L /K)

× given by, for u ∈ OK ,

−1 rπ(u)|Kunr = id |Kunr , rπ(u)(x) = [u ]f (x) where x ∈ Ef , and rπ(π)|Kπ = id |Kπ , rπ(π)|Kunr = Frob . π Next time, we are going to prove that L and rπ does not depend on π. Then π ab we will show that rπ = θLπ /K . This will give us that L is actually K . 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

× π rπ : K → Gal(L /K) 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 all π. (Here we’re assuming that Lπ is independent of the choice of π.) Proposition 21.1. Let r : K× → Gal(L/K) be a homomorphism such that the composition

× r unr K −→ Gal(L/K) Gal(K /K)

vk(x) is the reciprocity map x 7→ Frob and for any uniformizer π, r(π)|Kπ is the π identity on K . 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 that r(π)|Kunr = Frob = θL/K (π) and r(π)|Kπ = idKπ . π,n × Here, note that we have shown π ∈ N(K ) 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 independent of π, and also that rπ = r$ for uniformizers π and $. But assuming all 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

π,n × × π,n N(K ) = ker(θKπ,n/K : K → Gal(K /K)). Here, for an arbitrary πru ∈ K×,

r −1 −1 θKπ,n/K (π u)(x) = rπ(u)(x) = [u ]f (x) = (u ) · x. This is going to be the identity if and only if u−1 ≡ 1 (mod πn), because this is just multiplication. So this is true if and only if u ≡ 1 (mod πn). Hence

π,n × × π,n Z N(K ) = ker(θKπ,n/K : K → Gal(K /K)) = π · Un. Math 223a Notes 81

Theorem 21.2. Any finite abelian extension K0/K is contained in L = Kπ · Kunr. Proof. Note that N(K0)× is a finite index subgroup of K× by class field theory. f So it must contain π Z · Un for some f and n. Then

N(K0)× ⊇ N(Kπ,n)× ∩ N(Kunr,f ) = N(Kπ,n · Kunr,f ).

This implies that K0 ⊆ Kπ,n · Kunr,f . Therefore K0 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, $. Let Ff and Gg be the Lubin–Tate formal groups for f and g. The problem is that Ff and Gg are not isomorphic over OK . So we need to work over a larger field. The natural thing to do is to work over Kunr, because Kπ · Kunr = K$ · Kunr. But a problem here is that Kunr is not complete. So we are going to look at its unr unr completion Kb and its ring of integers ObK . 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 it is a countable-dimensional vector space and all Banach spaces are uncountable- dimensional vector spaces. Then we can form the completion Lb.

Example 21.3. Let K = Qp and L = Qp. Then we can complete it, and it is generally referred to as Lb = Qb p = Cp.

Any g ∈ Gal(L/K) extends continuously to g ∈ Aut(L/Kb ). For example, we can deﬁne Frob ∈ Aut(Kb unr/K).

Proposition 21.4. Let K be local and L/K algebraic. Take any x ∈ Lb such that x is separable over L. Then x ∈ L.

Proof. Let L0 be the separable closure of L in Lb. We want to show that L0 is a separable extension of L. Then it is normal so L0/L is Galois. Any element g ∈ Gal(L0/L) preserves the norm because L0/L is algebraic. Because L0 ⊆ Lb, it also preserves the absolute value on L0. Then g is continuous. Because L is 0 0 dense in L , we have g = idL0 . Therefore L = L. Let $ = uπ.

Lemma 21.5. There exists a non-unique α(x) ∈ Obunr[[x]] such that α(x) = x mod deg 2 with ∈ (Obunr)× and

ϕ (a) Frob(α) = α = α ◦ [u]f , (b) αϕ ◦ f = g ◦ α,

(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 . Because we want (a), we need

ϕ ϕ x = (x) = (x) ◦ [u]f = ux (mod deg 2).

This is guaranteed by homework, and we can choose an ∈ Obunr such that ϕ × / = u. You will even show that this is unique up to multiplication by OK . We now build α satisfying (a) inductively. We want polynomials αn of degree ϕ n such that αn = αn ◦ [u]f (mod deg(n + 1)). We already have our basecase. n+1 Induction step goes like, if I have αn we write αn+1 = αn + cx . Then we have

ϕ ϕ ϕ n+1 αn+1 = αn + c x , n+1 n+1 n+1 αn+1 ◦ [u]f = (αn + cx ) ◦ [u]f ≡ αn ◦ [u]f + cu x . We are good if we can always solve d = cun+1 − cϕ. To solve this, change to b = cn+1. What we want is d = b(u)n+1 − bϕ(ϕ)n+1 or b − bϕ = d/(u)n+1. This exist by homework. unr ϕ Now we have have a power series α ∈ Ob such that α = α ◦ [u]f . We want αϕ ◦ f = g ◦ α to hold, but it probably doesn’t. So we will have to modify 0 0 α 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

0 ϕ −1 −1 g = α ◦ f ◦ α = α[u]f ◦ f ◦ α . 0 0 We want to show that g ∈ OK [[x]]. I also want to show that g F$. It suﬃces to show that (g0)ϕ = g. But

0 ϕ ϕ −1 ϕ −1 −1 0 (g ) = α ◦ [u]f ◦ f ◦ (α ) = α ◦ [u]f ◦ [u]f ◦ f ◦ [u]f ◦ α = g because f and [u]f are endomorphisms of Ff . On the other hand, we have g0 ≡ αϕ ◦ xq ◦ α−1 ≡ αϕ ◦ (α−1)q (mod m = (π) = ($)).

But note that h((xq))ϕ ≡ hq for h ∈ Obunr[[x]]. So we get g0 ≡ αϕ ◦ (α−1)ϕ ◦ xq = xq (mod $).

0 This shows that g ∈ F$. Note that we already have F$. So there exists a power series [1]g,g0 ∈ OK [[x]] 0 −1 such that [1]g,g0 ◦ g ◦ [1]g,g0 − g. Now take 0 α = [1]g,g0 ◦ α. Then g = (α0)ϕ ◦ f ◦ (α0)−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–Tate unr series for π and $, with corresponding formal groups Ff and Gg. Let Ob be the ring of integers in Kb unr. Let u = $/π. Last time we constructed an α ∈ Obunr[[x]] such that

φ α = 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 Lbπ = Lb$. Then from the lemma last time, Lπ is the separable closure of L in Lbπ and likewise L$ is the separable closure of L in Lb$ so we deduce that Lπ = L$. Consider the bijection

∼= Ef −→ Eg; x 7→ α(x).

We get that

$,n π,n unr K = K(Eg,n) = K({α(x): x ∈ Ef,n}) ⊆ K · Kb .

By symmetry, we get K$,n · Kb unr = Kπ,n · Kb unr. Taking the union over n gives K$ · Kb unr = Kπ · Kb unr and then taking completion gives Lb$ = Lbπ. So the ﬁeld Lπ is independent of π. Now we show that the candidate Artin map is also independent of π. Note that we deﬁned the map

× π rπ : K → Gal(L /K)

Theorem 22.2. The two maps rπ and r$ are equal. Proof. Because K× is generated by uniformizers, it suﬃces to to show that rπ(y) = r$(y) for any uniformizers y ∈ OK . Then it suﬃces to show that rπ(y) = ry(y) for any uniformizer y ∈ OK . Rename y = $ and we are going to show 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 torsion 0 0 element. Then there exists an x ∈ Ef such that x = α(x ). Because rπ($) acts continuously it extends to Kb unr and so we can write

0 0 rπ($)(x) = rπ($)(α(x )) = (rπ($)(α))(rπ($)(x )) φ φ 0 φ −1 0 = α (rπ(u) ◦ rπ(π)(x)) = α (rπ(u)x ) = α ([u]f x ) = x.

$ So we get rπ($) = r$($) on K . At this point we are done, because last time we have shown that if Lπ and π ab rπ are independent of choice of π, then L = K 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), OK I(L/K).

× Recall that there is a ﬁltration on both OK and I(L/K), given by

× m UK,m = {a ∈ OK : a ≡ 1 (mod πK )} i+1 Gi(L/K) = {g ∈ I(L/K): g(πL) ≡ πL (mod πL )}. 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 : OK → Gal(L/K) = End(Ef,n) = OK /UK,n,

−1 −1 with the map being u 7→ [u ]f 7→ u . So the kernel of θL/K is actually just to be UK,n. This shows that  Gal(L/K) i = 0  m−1 m Gi = θL/K (UK,m) q ≤ i < q 1 qn ≤ i. Math 223a Notes 85

Now let us switch to the upper numbering:

Z u dt φ(u) = 0 [G0 : Gt] and we deﬁned i G (L/K) = Gφ−1(i)(L/K). In the case L = Kπ,n, we have ( (q − 1)qm−1 m ≤ n [G0 : Gi] = [UK,0 : UK,m] = (q − 1)qn−1 m > n.

So we will get φ−1(i) = qi−1 for i = 1, 2, . . . , n + 1. Then i G (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 L infinite, by Gi(L/K) = lim Gi(L0/K). ←− L0 π S π,n Example 22.4. Taking L = K = n K gives

i π G (K /K) = θL/K (UK,i).

× π Because θKπ /K : OK → Gal(K /K) is an isomorphism by the construction of Lubin–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)

∼=

× × × OK K OK .

Then Gi(Kab/K) in I(Kab/K) corresponds to Gi(L/K) in Gal(Kπ/K). Then

i ab G (K /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 going to give a different definition of the Brauer group. Let K be a field. Some non-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 interested in algebras A/K such that A ⊗K K is isomorphic to some Mn(K). We are then looking at 1 sep H (K, PGLn(K )), sep sep where PGLn(K ) is a non-abelian Gal(K /K)-module. We can deﬁne this, and there is a connecting homomorphism

1 sep 2 sep × H (K, PGLn(K )) → H (K, (K ) ) = Br(K).

Deﬁnition 22.6. An (non-commutative) algebra A/K is simple if A has no nontrivial 2-sided ideals.

× Example 22.7. Mn(K) is simple. For a, b ∈ K , the algebra

Hha, bi = Khi, ji/(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 has center K.

We will show that central simple algebras over K are twists of Mn(K), and are classiﬁed by H2(K, (Ksep)×). Math 223a Notes 87

23 November 16, 2017

Recall that a (non-commutative) K-algebra A is • simple if A has no 2-sided ideals other than 0 and A, which is equivalent to 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 not a 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 such that aij 6= 0. We can rescale to get aij = 1. If we multiply by the elementary matrices, 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 simple over 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 is Khi, ji 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-algebra generated by L and an additional element γ with

γn = a, γb = (gb)γ

2 for b ∈ L. You can check that Aa is central simple with dimension n over K. If a0 = Nb for some b ∈ L, then

id −1 Aa → Aaa0 ; L −→ L, γa 7→ b γa0 is an isomorphism. So the isomorphism class of Aa depends only on the class of 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 isomorphism 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 K to 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 either I ⊗K L = 0 or I ⊗K L = A ⊗K L.

The converse is not true. Consider A = C as a simple R-algebra. Then ∼ A ⊗R C = C × C is not simple.

Proposition 23.5. If A is central simple, then A ⊗K L is simple. (In fact, it is 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 show that 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 of A and so we can scale. The resulting thing is nonzero because we have started out 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 commute with a. Then a2, . . . , an ∈ Z(A) = K and so we get n = 1. Now 1⊗b ∈ I and 1 ⊇ 1⊗BbB = 1⊗B because B 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 monoid under ⊗.

The matrices algebras {Mn(K)} form a sub-monoid, because Mn(K) ⊗K ∼ Mm(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 op such that a ·Aop b = b ·A a. Clearly A is central/simple if and only if A is central/simple. Math 223a Notes 89

op Proposition 23.9. For ﬁnite-dimensional central simple A over K, A ⊗K A is isomorphic to EndK (A) where A is regarded as K-vector spaces. Proof. We deﬁne

op ϕ : A ⊗K A → EndK (A); a ⊗ 1 7→ la, 1 ⊗ a 7→ ra

0 0 0 0 op where laa = aa and raa = a a. Because A ⊗ A is central simple, ϕ is injective. 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. Our next goal is to show that if A is a central simple algebra over K, 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 say that M is • simple if M has no submodules other than 0 and M. L • 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 left A-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) is a division algebra.

Proof. Suppose ϕ ∈ EndA(S) is nonzero. Then ker(ϕ) = 0 and im(ϕ) = S. So ϕ is invertible. Proposition 23.13. Let D be a division algebra. Then all ﬁnitely generated D-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 ∼ ∼ op EndA(AA) = {a 7→ ab : b ∈ A} = A .

As a consequence, if M is a free A-module of rank n, then

⊕n ∼ op ∼ op EndA(M) = EndA(AA ) = Mn(A) = Mn(A ). Math 223a Notes 90

Theorem 23.14 (Double centralizer). Let V be a ﬁnite-dimensional vector space over K. Let A ⊆ EndK (V ) be a simple subalgebra. We deﬁne the centralizer 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 nonzero submodule 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

∼ op A = C(D) = EndD(S) = Mn(D ).

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 isomorphic to Mn(D) for some division algebra D. Math 223a Notes 91

24 November 21, 2017

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 =∼ op EndA(S) .

24.1 Modules over simple algebras Simple algebras are the nicest things you could have after fields. There is going to be a unique finitely generated simple module S over A, and every module is going to be Sn for some n. First step in proving this is to decompose AA as a direct sum of simple submodules. Note that Z(A) is always a field, and so the classification gives ∼ A = Mn(D) for some division ring D. Then ∼ ∼ AA = Mn(D) = S1 ⊕ · · · ⊕ Sn where 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 S0 is any other simple A-module. Take a nonzero s0 ∈ S0. There is a map 0 0 ϕ : AA → S ; a 7→ as ∼ L which is a left A-module homomorphism. But note that AA = i Si. This shows that 0 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 to an isomorphism Sm → M for some m. Proposition 24.1. Let A be a simple algebra. There is a unique finitely generated simple module S over A, and any finitely generated module over A is isomorphic 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 over K. Note that K(x) is a commutative subalgebra. By assumption that D is a division 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 of division algebras over R. The only ones are R, C, and H. But C doesn’t count because 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 map Br(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). This condition 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 we get 2 dimK A = dimK A ⊗K K = dimK Mn(K) = n . Proposition 24.6. There always exists a finite extension L/K such that A is split by L.

Proof. We have an isomorphism Mn(K) → A ⊗K K, and the standard bases P eij 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. So [ [ Br(K) = Br(L/K) = Br(L/K). L/K fin. L/K fin. Gal. 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 over K = 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 being n 2 2 2 of the form 4 (8a + 1). In this case, you can show that x + y + z = 0√ has no 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 let B ⊆ A be a simple algebra. Then C = C(B) is simple and C(C) = B. Also we have [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. Then B ⊗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 centralizer theorem. For (b) to (c), consider L0 a maximal commutative K-subalgebra containing L. Then

[L0 : K]2 ≤ [L0 : K][C(L0): K] = [A : K] = [L : K]2.

So L0 = 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 maximal commutative subfield of A. In other words, all maximal subfields of D have degree p[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 op ∼ A ⊗K L = EndL(V )

op for some vector space L of dimension dimL V = n. Then dimK A = dimK A = 2 op op n . Let B be the centralizer of A inside EndK (V ). Because A and EndK (V ) are both central simple over K, we have that B is central simple over K. Then op ∼ A ⊗K B = EndK (V ) and so [B] = −[Aop] = [A] in Br(K). We certainly have that L ⊆ B because L op op commutes with everything in A , and [B : K] = [EndK (V ): K]/[A : K] = [L : K]2. Math 223a Notes 94

To prove the other direction, it is enough to show that L splits B. Let 2 ∼ [L : K] = n and [B : K] = n . We want B ⊗K L = EndL(V ) for some V , with dimL 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 L is central simple over L, so this should be injective. Dimension counting shows that this is an isomorphism. Math 223a Notes 95

25 November 28, 2017

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] = p[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 be a central simple algebra over K. Then any two homomorphism f, g : A → B are 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/K is Galois, we can precompose L,→ B by any element of Gal(L/K). They are going to be related by a conjugation by some element in B.

∼ n Proof of Noether–Skolem. First assume that B = Mn(K) = EndK (K ). Let n f, g : A → B be maps. Deﬁne two A-modules M1 and M2, both K 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 completely classiﬁed 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, this ϕ 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 op ∼ enlarge B to make it a matrix algebra. We know that B ⊗K B = EndK (B). We have two maps

op op ∼ f ⊗ 1, g ⊗ 1 : A ⊗K B → B ⊗K B = EndK (B). Math 223a Notes 96

The domain is still simple and the codomain is a matrix algebra. So these must op be conjugate by some x ∈ B ⊗K B so that

x(f(a) ⊗ b0)x−1 = g(a) ⊗ b0 for all a ∈ A and b0 ∈ Bop. Setting a = 1 shows that x commutes with 1 ⊗ b0 for all b ∈ Bop, so x ∈ B ⊗ 1 because Bop is central. If we write x = b ⊗ 1 then b 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

2 × Z (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 the Noether–Skolem theorem, for any g ∈ G there exists an ag ∈ A such that

−1 g(x) = agxag

2 for all x ∈ L. How unique is ag? Recall that if [A : K] = [L : K] then × C(L) = L. So ag is unique up to left multiplication by elements of L . Let us compare agh and agah. Here, we have

−1 −1 aghxagh = g(h(x)) = agahx(agah) .

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 ag1 ag2 ag3 , then

ag1 (ϕ(g2, g3)ag2g3 ) = g1(ϕ(g2, g3))ag1 ag2g3 = g1(ϕ(g2, g3))γ(g1, g2g3)ag1g2g3 equals

ϕ(g1, g2)ag1g2 ag3 = ϕ(g1, g2)ϕ(g1g2, g3)ag1g2g3 . Math 223a Notes 97

This is precisely that ϕ is a cocycle. × 0 On the other hand, ag are unique up to multiplication by L . So if ag is 0 another choice of such elements, we should be able to write ag = ψ(g)ag. It turns out that ϕa0 /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 I know exactly what to do. Suppose I have received an arbitrary cocycle ϕ ∈ 2 × H (L/K, L ) and I want to deﬁned a K-algebra Aφ by M Aϕ = L · eg g∈G 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 that this only depends on the cohomology class [ϕ] ∈ H2(L/K, L×). So we have a candidate for a map H2(L/K, L×) → A(L/K), but we need to show that it is central 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 ideal I ( 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 we must have k ≥ 2. Without loss of generality assume that g1 = 1. Take l ∈ L such 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 homomorphism Aϕ → A given by sending eϕ 7→ aϕ. Because Aϕ is central simple, it is an isomorphism by dimension counting. Math 223a Notes 98

26 November 30, 2017

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 add them and get ϕ1 + ϕ2, and we know how to construct Aϕ1 ⊗ Aϕ2 and 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×)

inf ∼ Br(E/K) = H2(E/K, E×) commutes.

In particular, we have [ Br(K) = Br(L/K) = lim Br(L/K) −→ L/K Gal L/K ∼ lim H2(L/K, L×) = H2(K, (Ksep)×). = −→ L/K

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 separably closed, 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 over K is K. The tool for studying division algebras is reduced norm. This might be familiar in the context of quaternions:

× × 2 2 2 2 H → R ; a + bi + cj + dk → a + b + c + d . Math 223a Notes 99

Suppose D is a central division algebra over K. Let’s say that L is a Galois extension such that L splits D. Then there is an isomorphism ϕ D ⊗ L −→ Mn(L). Then we can deﬁne × Nrd(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, any two L-algebra maps A → Mn(L) are conjugate by some b ∈ Mn(L). Take A = D ⊗ L, and consider them map

−1 ϕg : D ⊗ L → Mn(L); d ⊗ ` 7→ gϕ(d ⊗ g l) So if we evaluate at d ⊗ 1 and take determinant, we get

Nrd(d) = det ϕ(d ⊗ 1) = det gϕ(d ⊗ 1) = gNrd(d). Note that so far everything works for D not only a division algebra but also 2 any central simple algebra. Let [D : K] = n and consider d1, . . . , dn2 a K-basis for D. We can take the reduced norm

n2 X f(c1, . . . , cn2 ) = Nrd cidi ∈ K[c1, . . . , cn2 ] i=1 which is a homogeneous polynomial in the cis of degree n. For a nonzero v = n2 (v1, . . . , vn2 ) ∈ K , we have f(v) 6= 0 because D is a division algebra. So we have a low degree homogeneous polynomial with many variables, such that it has no nontrivial zeros. Deﬁnition 26.6. A ﬁeld K is called quasi-algebraically closed if any homogeneous polynomial f ∈ K[x1, . . . , xN ] of degree d < N has a nonzero solution in KN . If K is quasi-algebraically closed, then any central division algebra D/K should 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 in Fq[x1, . . . , xN ] of degree d < N, then the number of solutions is a multiple of p = char Fq. So there should be at least one solution other than 0. The idea is that X (1 − f(v)q−1) n v∈Fq is congruent to the number of solutions modulo p. Then you expand this out and 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-algebraically closed. More generally, the fraction field of any complete DVR with algebraically closed 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/K is also quasi-algebraically closed. If K is quasi-algebraically closed, then H2(K, (Ksep)×) = 0. For L/K, we also have H2(L, (Ksep)×) = 0. It is also true that the corresponding H1 vanish by Hilbert 90, and this is also true for all subextensions. So by Tate’s theorem, we have that all Hˆ q(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 2 × 0 × × × H (R, C ) =∼ Hˆ (R, C ) = R /NC =∼ {±1}. Previously we have computed this invoking the classiﬁcation of central simple algebras over R, but we can now do this using cohomology. Suppose now that K is a non-archemedian local ﬁeld. We should have

2 sep × inv : Br(K) =∼ H (K, (K ) ) =∼ Q/Z. This should give, in some sense, the classiﬁcation of central simple algebras. For any 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 multiplication by 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×) =∼ Hˆ 0(K,L×) =∼ K×/NL×. But we can understand also from the division algebra side. Let D be a central division 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

1/n |a|D = |Nrn(a)|K . Then this is a non-archemedian absolute value. We can likewise extend 1 v : D× → nZ and deﬁne ramiﬁcation degree e = [v(D×): v(K×)]. Math 223a Notes 101

We can deﬁne

1 OD = {a ∈ D : v(a) ≥ 0}, pD = {a ∈ D : v(a) ≥ n }.

Then this is a principal idea pD = πDOD = ODπD. Then kD = OD/pD is a ∼ division ring that is finite over k = OK /pK = Fq. In particular, kD/k is a finite extension 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.

× 1 But note that e | n, because v : D → n Z. Also, we have f ≤ n, because if a is a primitive element of kD then a ∈ OD has minimal polynomial degree at least f, but a maximal subﬁeld of D has degree n. So f ≤ n and e | n. We therefore get e = f = n.

Now when a generates kD/k, and a lifts a, the ﬁeld K(a) is unramiﬁed and splits D. By Noether–Skolem there exists some b ∈ D such that

−1 FrobK(a)/K (x) = bxb .

Then another way to deﬁne the invariant map is

1 inv([D]) = [v(b)] ∈ n Z/Z ⊆ Q/Z. What happens when K is a global ﬁeld? Here the Brauer group is computed too, and there exists a short exact sequence M 0 → Br(K) → Br(Kv) → Q/Z → 0. v

We are going to do this next semester. In particular, if [H] ∈ Br(K) is some quaternion algebra, this says that the number of v such that H⊗K Kv is not split is finite and even. This really quadratic reciprocity, when applied to H(p, q). Index absolute value, 9 Herbrand quotient, 57 non-archemedian, 9 Hilbert 90, 49 additive group, 69 homogeneous cochain, 35 adele ring, 6 adelic class group, 7 indecomposable module, 89 Artin map, 6, 49 induction, 31 augmentation ideal, 29 inertia degree, 14, 101 inertia group, 24 bar resolution, 34 tame, 27 Brauer group, 49, 88, 92 wild, 27 inflation, 41 central simple algebra, 86 inhomogeneous cochain, 36 centralizer, 90 invariant map, 53 Chevalley–Warning theorem, 99 Kronecker–Weber theorem, 5 class field, 4 coinduction, 31 local field, 16 coinflation, 42 logarithm, 19 compatible pair, 41 Lubin–Tate series, 74 completion, 10 corestriction, 42 multiplicative group, 70 crossed homomorphism, 36 Noether–Skolem theorem, 95 norm, 13 decomposition group, 23 norm residue symbol, 62 discrete valuation ring, 10 Normal basis theorem, 49 normic, 64 Eisenstein polynomial, 25 exponential, 18 Ostrowski’s theorem, 9 formal group p-adic integer, 6 homomorphism, 73 place, 10 formal group law, 72 product formula, 17 formal module, 76 profinite completion, 66 free module, 31 profinite group, 50 Frobenius element, 5 quasi-algebraically closed, 99 G-module, 29 ramification group, 26 discrete, 50 ramification index, 14, 100 Galois cohomology, 49 restriction, 41 global field, 17 group cohomology, 33 semidirect product, 37 group homology, 39 semisimple module, 89 group ring, 29 simple algebra, 86 simple module, 89 Hensel’s lemma, 11 standard normalization, 16

102 Math 223a Notes 103 standard resolution, 34 universal norm, 64 unramiﬁed extension, 19 tamely ramiﬁed, 27 Tate cohomology, 44 upper numbering, 28 Tate’s theorem, 59 torsor, 37 Tsen’s theorem, 100 valuation, 9