Central Simple Algebras and the Brauer groupxviiicla/pdf_files/cft.pdf4 Central Simple Algebras and the Brauer group Theorem 3.2 (Wedderburn) Let Kbe a ﬁeld and let Abe a ﬁnite

Central Simple Algebras

and the Brauer group

XVIII Latin American Algebra Colloquium

Eduardo Tengan

(ICMC-USP)

Copyright c© 2009 E. Tengan

Permission is granted to make and distribute verbatim copies of this document provided the copyrightnotice is preserved on all copies.

The author was supported by FAPESP grant 2008/57214-4.

Chapter 1

CentralSimpleAlgebrasandthe

Brauergroup

1 Some conventions

Let A be a ring. We denote by Mn(A) the ring of n× n matrices with entries in A, and by GLn(A) itsgroup of units. We also write Z(A) for the centre of A, and Aop for the opposite ring, which is the ring

with the same underlying set and addition as A, but with the opposite multiplication: a×Aop bdf= b×A a

for a, b ∈ Aop. For instance, we have an isomorphism Mn(R)≈→ Mn(R)op for any commutative ring R,

given by M 7→MT where MT denotes the transpose of M .

A ring A is simple if it has no non-trivial two-sided ideal (that is, an ideal different from (0) or A).A left A-module M is simple or irreducible if has no non-trivial left submodules. For instance, for anyfield K, Mn(K) is a simple ring and Kn (with the usual action) is an irreducible left Mn(K)-module.If K is a field, a K-algebra D is called a division algebra over K if D is a skew field (that is, a“field” with a possibly non-commutative multiplication) which is finite dimensional over K and suchthat Z(D) = K.

Let K ⊂ L be an extension of fields and let A be an arbitrary K-algebra. We shall write AL for theL-algebra A⊗K L obtained by base change −⊗K L. We shall often refer to AL as the restriction of Ato L (in geometric language, base change with respect to SpecL→ SpecK corresponds to restriction inthe flat topology).

2 Cyclic Algebras

Let L ⊃ K be a cyclic extension, that is, a Galois extension of fields with cyclic Galois group G =Gal(L/K). Let n = |G| = [L : K]. An element χ ∈ Hom(G,Z/n) will be called a character. Notice

that to give a surjective character is the same as to give an isomorphism χ:G≈→ Z/n, i.e., to choose a

specific generator σ of G, characterised by χ(σ) = 1.

Now let a ∈ K× and let χ:G≈→ Z/n be a surjective character. Denote by σ the generator of G

chosen by χ. We now manufacture a K-algebra (χ, a) as follows. As an additive group, (χ, a) is ann-dimensional vector space over L with basis 1, e, e2, . . . , en−1:

(χ, a)df=

⊕

1≤i<n

Lei

Multiplication is given by the relations

e · λ = σ(λ) · e for λ ∈ Len = a

Some boring computations show that (χ, a) is, in fact, an associative K-algebra, which is called thecyclic algebra associated to the character χ and the element a ∈ K. Observe that since dimL(χ, a) =[L : K] = n we have that dimK(χ, a) = n2 is always a square.

Example 2.1 (Real cyclic algebras) Consider the cyclic extension C ⊃ R of degree n = 2 withG = Gal(C/R) = id, σ (here σ(z) = z is the conjugation automorphism) and let χ:G → Z/2 be thecharacter defined by χ(σ) = 1. Then, for a ∈ R×,

(χ, a) = z + w · e | z, w ∈ C

2 Central Simple Algebras and the Brauer group

where e2 = a and e · z = z · e for z ∈ C. For instance, we have that

(z + we) · (z − we) = zz − zwe+ wez − wewe= zz − zwe+ wze− wwe2

= |z|2 − |w|2a ∈ R

(∗)

If a > 0 is positive then (χ, a) ∼= M2(R). In fact, we have that the standard embedding

φ: C →M2(R)

α+ βi 7→(

α β−β α

)

α, β ∈ R

can be extended to a map of R-algebras φ: (χ, a)→M2(R) by setting φ(e) =

(√a 0

0 −√a

)

, that is,

φ(α+ βi+ γe+ δie) =

(

α+ γ√a β − δ√a

−β − δ√a α− γ√a

)

α, β, γ, δ ∈ R

and since this map is injective and both (χ, a) and M2(R) have dimension 4 over R, it must be anisomorphism.

Now suppose that a < 0. Then (χ, a) is a division algebra since for every z + we 6= 0 the real number|z|2 − |w|2a > 0 is nonzero and hence

(z + we)−1 =z

|z|2 − |w|2a −w

|z|2 − |w|2a · e

in (χ, a) by the computation (∗). We have that (χ, a) is actually isomorphic to the usual real quaternionalgebra H, which is the 4-dimensional real algebra

Hdf= R + Ri + Rj + Rk

with basis 1, i, j,k satisfying the relations

i2 = j2 = k2 = −1

ij = k, jk = i, ki = j

ij = −ji, jk = −kj, ik = −ki

In fact, an isomorphism φ: (χ, a)≈→ H is given by extending the inclusion φ: C → H given by φ(α+βi) =

α+ βi, α, β ∈ R, by setting φ(e) =√

|a| · j, as can be easily checked.

The above example shows that the cyclic algebra construction encompasses both regular matrixalgebras as well as quaternion algebras. Cyclic algebras can be thought of “twisted forms” of matrixalgebras: by a suitable base extension, they can be easily “trivialised.”

Theorem 2.2 (Splitting cyclic algebras) Let L ⊃ K be a cyclic extension of order n. Let a ∈ K×

and let χ:G≈→ Z/n be a surjective character. Then

(χ, a)⊗K L ∼= Mn(L)

Proof Let σ be the generator of G given by χ(σ) = 1. Define a morphism of L-algebras φ: (χ, a)⊗KL→Mn(L) by setting

φ(λ ⊗ 1) =

λ 0 0 · · · 0 00 σ(λ) 0 · · · 0 00 0 σ2(λ) · · · 0 0

...0 0 0 · · · 0 σn−1(λ)

for λ ∈ L

Central Simple Algebras 3

and φ(e⊗ 1) to be the transpose of the “companion matrix” of xn − a:

φ(e⊗ 1) =

0 1 0 · · · 0 00 0 1 · · · 0 0

...0 0 0 · · · 1 00 0 0 · · · 0 1a 0 0 · · · 0 0

It is easy to check that φ is a well-define morphism of L-algebras. Since dimL(χ, a)⊗KL = dimK(χ, a) =n2 = dimL Mn(L), in order to show that φ is an isomorphism it is enough to show that it is surjective.But by the lemma below one has

φ(L⊗K L) =

L 0 · · · 00 L · · · 0...0 0 · · · L

and therefore

φ(Le⊗K L) =

0 L 0 · · · 00 0 L · · · 0

...0 0 0 · · · LL 0 0 · · · 0

, φ(Le2 ⊗K L) =

0 0 L 0 · · · 00 0 0 L · · · 0

...0 0 0 0 · · · LL 0 0 0 · · · 00 L 0 0 · · · 0

, and so on

Hence imφ = Mn(L), as required.

Lemma 2.3 (Lemma ⊗K Lemma) Let L ⊃ K be a Galois field extension of degree n with G =Gal(L/K). Then we have an isomorphism of L-algebras

L⊗K L≈→ Maps(G,L) ∼= Ln

a⊗ b 7→ f(σ) = σ(a)b for σ ∈ G

Proof Let G = σ1, . . . , σn. Using the primitive element theorem, write L = K(θ) for some elementθ ∈ L. Let p(t) =

∏

1≤i≤n(t− σi(θ)) ∈ K[t] be the minimal polynomial of θ. We have isomorphisms of

L-algebras (where L⊗K L is viewed as an L-algebra via its second entry)

L⊗K L ∼= K[t](

p(t)) ⊗K L ∼= L[t]

(

p(t))

CRT∼=∏

1≤i≤n

L[t](

t− σi(θ))

∼= Maps(G,L) ∼= Ln

where CRT stands for Chinese remainder theorem. Following the above isomorphisms, if λ = h(θ) ∈ Lwith h(t) ∈ K[t], we have that

λ⊗ 1 = h(θ) ⊗ 1 7→(

h(σ1(θ)), h(σ2(θ)), . . . , h(σn(θ)))

= (σ1(λ), σ2(λ), . . . , σn(λ))

Cyclic algebras are the simplest examples of central simple algebras, which are nothing else otherthan “twisted forms” of matrices. That is our next topic.

3 Central Simple Algebras

We now introduce a very important class of algebras which, in some sense, are not too distant relativesof matrix algebras:

Definition 3.1 Let K be a field and let Ω be an algebraically closed field extension of K. We say thata finite dimensional K-algebra A is a central simple algebra over K (CSA for short) if it satisfiesone (hence all) of the following equivalent conditions:

1. A is a simple ring with centre K;

2. there is an isomorphism A ∼= Mn(D) where D is a division algebra with centre K;

3. AΩ is isomorphic to the matrix ring Md(Ω) for some d;

4. the canonical map φ:A⊗KAop → EndK-mod(A) is an isomorphism (φ sends a⊗b to theK-vector

space endomorphism x 7→ axb of A).

Before showing that the above are indeed equivalent, let us recall the well-known


Theorem 3.2 (Wedderburn) Let K be a field and let A be a finite dimensional simple K-algebra (i.e.A has no non-trivial two-sided ideals). Then A ∼= Mn(D) for some division algebra D and some integern. Moreover D is uniquely determined up to isomorphism as D = EndA(I) for any (non-zero) minimalideal I of A (they are all isomorphic to Dn).

Proof First, some pep talk. Secretly, we know that A ∼= Mn(D) for some division algebra D. How canwe recover D intrinsically in terms of A? The idea is to notice that the minimal non-zero left ideals ofMn(D) are given by “column matrices” (as can be easily seen by a straightforward computation):

I1 =

D 0 · · · 0D 0 · · · 0...D 0 · · · 0

I2 =

0 D · · · 00 D · · · 0...0 D · · · 0

· · · In =

0 · · · 0 D0 · · · 0 D...0 · · · 0 D

Being minimal, they are irreducible as left Mn(D)-modules. Besides they are all isomorphic to the n-dimensional D-vector space Dn and therefore we have Mn(D) ∼= EndD(I1). Finally, we can identify Dwith the set of all A-endomorphisms of I1 via d 7→ Rd, where Rd: I1 → I1 is the right multiplication byd ∈ D. In fact, Rd is clearly an A-endomorphism of the left A-module I1; conversely, given φ ∈ EndA(I1),consider

vdf=

1 0 · · · 01 0 · · · 0...1 0 · · · 0

∈ I1

Since I1 is irreducible, Mn(D)v = I1 and therefore since

φ(av) = aφ(v) for all a ∈Mn(D)

we conclude that φ is completely determined by the value of φ(v), which is of the form φ(v) = vd = Rd(v)for some d ∈ D, as one can see by taking a ∈Mn(D) to be permutation matrices in the above formula.

Now we turn the above reasoning upside-down. Starting with a simple finite dimensional K-algebraA, let

Idf= any minimal non-zero ideal of A

Ddf= EndA(I)

Notice that I exists since A is artinian (it is finite dimensional over a field) and that D is a division ringby the so-called Schur’s lemma: given any non-zero d ∈ D, we must have ker d = 0 and im d = I since Iis irreducible as a left A-module, hence d is an automorphism of I.

We can now view I as a D-vector space via d · v df= d(v) for d ∈ D and v ∈ I. Everything is finite

dimensional over K, a fortiori I is finite dimensional over D. Hence I ∼= Dn (non-canonically since itdepends on the choice of a basis) for some n and EndD(I) ∼= Mn(D). We can now define a ring morphism

ρ:A→ EndD(I) ∼= Mn(D)

a 7→ La

where La: I → I denotes the left multiplication by a, which is a D-endomorphism since La(d · v) =a · d(v) = d(av) = d · La(v) for all a ∈ A, d ∈ D and v ∈ I. Now it is easy to check that ρ is a ringmorphism, which must be injective since its kernel is a proper two-sided ideal of the simple ring A.

We are left to show that ρ is also surjective. For that it is enough to show that ρ(A) is a left idealof EndD(I) since ρ(A) contains the unity element id = L1 of EndD(I). First we show that ρ(I) is a leftideal. For w ∈ I, right multiplication Rw by w belongs to D = EndA(I), hence

f(vw) = f(Rwv) = Rwf(v) = f(v)w for all f ∈ EndD(I) and v, w ∈ I,

that is,f Lv = Lf(v) ⇐⇒ f · ρ(v) = ρ(f(v)) for all f ∈ EndD(I) and v ∈ I,

showing that ρ(I) is a left ideal of EndD(I). Since A is simple, the two-sided ideal IA coincides with A,hence ρ(A) = ρ(I)ρ(A) is also a left ideal. This completes the proof that A ∼= Mn(D). This also provesthat all minimal ideals I of A are isomorphic, and therefore D is uniquely determined by the equationD = EndA(I) (up to isomorphism).

Central Simple Algebras 5

We also need a

Lemma 3.3 Let K be a field and A and B be two finite dimensional K-algebras.

1. Z(A⊗K B) = Z(A)⊗K Z(B);

2. if A and B are simple rings and Z(A) = K, then A⊗K B is also simple with centre Z(B).

Proof 1. Clearly Z(A ⊗K B) ⊃ Z(A) ⊗K Z(B). Conversely, let ω1, . . . , ωn be a basis of B over K.Since

A⊗K B = A⊗K

(

⊕

1≤i≤n

Kωi

)

=⊕

1≤i≤n

Aωi

as K-vector spaces, any element z ∈ A ⊗K B can be uniquely written as z = a1 ⊗ ω1 + · · · + an ⊗ ωn

with ai ∈ A. In particular, if z ∈ Z(A⊗K B), then

(a⊗ 1) · z = z · (a⊗ 1) ⇐⇒ aa1 ⊗ ω1 + · · ·+ aan ⊗ ωn = a1a⊗ ω1 + · · ·+ ana⊗ ωn

for any a ∈ A, hence aai = aia for all i by the uniqueness of the above representation. Hence allai ∈ Z(A), that is, z ∈ Z(A) ⊗K B ⊂ A ⊗K B. Now switching the roles of the first and second entries,we conclude that z ∈ Z(A)⊗K Z(B) ⊂ Z(A)⊗K B, as was to be shown.

2. Let I be a non-zero two-sided ideal of A⊗KB. Suppose first that there is a “simple non-zero tensor”a⊗ b ∈ I. Since A is simple, the two-sided ideal generated by a 6= 0 equals A, hence there exist ai anda′i in A such that

∑

1≤i≤n

aiaa′i = 1

Hence 1⊗ b =∑

1≤i≤n(ai ⊗ 1) · (a⊗ b) · (a′i ⊗ 1) ∈ I. Reversing the roles of A and B, we conclude that1⊗ 1 ∈ I as well, i.e., I = A⊗K B.

Now letx = a1 ⊗ b1 + · · ·+ an ⊗ bn ∈ I

be an element with smallest n. We may assume that the bi are linearly independent over K, otherwisewriting (say) b1 = λ2b2 + · · ·+ λnbn as a linear combination of the other bi (with λi ∈ K) and pluggingin in the above expression, we can rewrite it to make it shorter. Similarly, we can assume that the ai arealso linearly independent over K. Moreover, applying the reasoning of the special case above, we mayassume that a1 = 1 as well.

Now suppose that n > 1. We have that a2 /∈ K (otherwise a1 and a2 would be linearly dependentover K and again we could shorten the above expression). Since Z(A) = K there exists a ∈ A such thataa2 6= a2a. Consider the element

(a⊗ 1) · x− x · (a⊗ 1) = (aa2 − a2a)⊗ b2 + · · ·+ (aan − ana)⊗ bn ∈ I

Since the bi are linearly independent over K and aa2 − a2a 6= 0, this element is not 0, which contradictsthe minimality of n. Therefore n = 1 and by the special case proven, we are done.

Now we are ready to show:

Theorem 3.4 The above definition makes sense.

Proof We have 1 ⇐⇒ 2: the implication⇒ is just Wedderburn’s theorem and it is easy to check thatMn(D) is simple with centre Z(D) = K.

Next we show that there is no non-trivial division algebra D over an algebraically closed field Ω.Since dimΩD < ∞, for any a ∈ D, the subfield Ω(a) of D generated by a is finite dimensional over Ω,and hence algebraic over Ω, that is, Ω(a) = Ω⇒ a ∈ Ω. This proves that D = Ω.

Now let A be a simple ring with centre K. By the lemma, AΩ is also simple with centre Ω. By whatwe have just proven together with Wedderburn’s theorem we conclude that AΩ

∼= Md(Ω), i.e., 1⇒ 3.

Next we prove that 3 ⇒ 1. First observe that Ω is free as a K-algebra, hence Ω is faithfully flatover K. Recall that this means that − ⊗K Ω is an exact functor and that M ⊗K Ω = 0 ⇐⇒ M = 0for any K-module M . In particular, −⊗K Ω preserves injective maps. Hence if there were a non-trivialtwo-sided ideal a of A, then a⊗K Ω would be a non-trivial two-sided ideal of A⊗K Ω ∼= Mn(Ω), but that


would contradict the fact that Mn(Ω) is simple. Therefore A must be simple. On the other hand, bythe lemma Z(A)⊗K Ω = Z(A⊗K Ω) = Z(Mn(Ω)) = Ω, hence Z(A) = K, as required.

We now show that 1⇒ 4. Observe that φ is a ring morphism since φ(1⊗ 1) = id and

φ(

(a⊗ b)(a′ ⊗ b′))

= φ(aa′ ⊗ b′b) = φ(a⊗ b) φ(a′ ⊗ b′)

Since kerφ is a two-sided ideal of A ⊗K Aop, a simple ring by the lemma, we conclude that kerφ = 0,that is, φ is injective. But since dimK A ⊗K Aop = dimK EndK-mod(A) = (dimK A)2 (observe thatEndK-mod(A) is isomorphic to a matrix algebra), the map of K-vector spaces φ must also be surjective.

To prove 4⇒ 1, observe that by the lemma Z(A)⊗K Z(Aop) = Z(A⊗K Aop) = Z(EndK-mod(A)) =K, hence Z(A) = K. Besides EndK-mod(A) is simple, hence so is A, for any non-trivial two-sided ideala of A would give rise to a non-trivial two-sided ideal a ⊗K Aop of A ⊗K Aop (observe that Aop is freehence faithfully flat over K).

Corollary 3.5 If A is a CSA over a field K, then dimK A is a square.

Proof With the notation of the theorem, we have dimK A = dimΩAΩ = dimΩMn(Ω) = n2.

Definition 3.6 The degree of a CSA A over a field K is defined to be the square root of dimK A.

Just for the record, we specialise the previous lemma restating it as

Lemma 3.7 Let K be a field and let A and B be a CSA over K.

1. (Stability under tensor products) A⊗K B is a CSA over K.

2. (Stability under base change) AL is a CSA over L for any field extension L of K.

Example 3.8 Matrix algebras and cyclic algebras are CSA’s, and so are tensor products thereof. Hencewe know how to construct a plethora of CSA’s. A very deep theorem by Merkurjev and Suslin (K-cohomology of Severi-Brauer varieties and the norm residue homomorphism (Russian), Izv. Akad. NaukSSSR Ser. Mat. 46 (1982), no. 5, 1011–1046, 1135–1136) says that, in the presence of enough roots ofunity, all central simple algebras are “stably isomorphic” to tensor products of cyclic algebras. More onthat when we talk about the Brauer group.

4 Splitting Fields

Of great importance will be the study of the so-called splitting fields of CSA.

Theorem 4.1 Let A be a CSA over a field K. A field L ⊃ K is called a splitting field for A if ittrivialises (or splits) A, i.e., AL

∼= Mn(L) for some n.

Example 4.2 The algebraic closureKalg of K is always a splitting field for any CSA A overK. Actually,we can be more economical: since AKalg

∼= Mn(Kalg) and both rings are finite dimensional overKalg, thisisomorphism (and its inverse) is defined by a finite amount of data, namely the images of the elementsof a basis. Hence we can define this isomorphism over a finite extension L over K. To sum up, for anyCSA A there exists a finite extension L ⊃ K that splits A.

Example 4.3 If L ⊃ K is a cyclic extension, then any cyclic algebra (χ, a) (for a ∈ K× and χ asurjective character of Gal(L/K)) is split by L.

The last example is a particular instance of

Theorem 4.4 Let K be a field and A be a CSA over K of degree n. If L is a field contained in A suchthat n = [L : K] then L splits A.

Proof The restriction of the isomorphism φ:A⊗K Aop ≈→ EndK-mod(A) to A⊗K L induces an injectivemap φ:A ⊗K L → EndL-mod(A): in fact, the image under φ of d ⊗ λ for d ∈ A and λ ∈ L is the

(right) L-linear map x 7→ dxλ. Since dimL A⊗K L = dimK A = n2 = dimL EndL-mod(A), φ:A ⊗K L≈→

EndL-mod(A) must be an isomorphism.

Splitting Fields 7

Remark 4.5 It can be shown, using the double centraliser theorem, that for a division algebra D overK, the subfields L ⊂ D with [L : K] = degD are maximal subfields of D (with respect to the inclusionorder). Therefore we shall refer to such splitting fields of degree as maximal ones, even though we arenot going to use the fact that they are actually maximal.

Do such maximal subfields exist? In fact, any division algebra is packed with them! We can evenfind maximal subfields which are separable over the base field! Recall that a finite extension of fieldsL ⊃ K is separable if it satisfies any of the following three equivalent properties:

1. every element λ ∈ L has a separable minimal polynomial overK, i.e., a polynomial with distinctroots;

2. there are exactly n = [L : K] K-immersions φ:L → Kalg;

3. L⊗K Kalg ∼= (Kalg)n where n = [L : K].

Any finite extension of a perfect field (for instance, a finite field or a field of characteristic 0)is automatically separable. Hence separability may only pose an issue for infinite fields of positivecharacteristic.

Theorem 4.6 (Separable maximal subfields) Let D be a division algebra over a field K. Thenthere exists an element θ ∈ D such that K[θ] ⊃ K is a separable field extension of degree degD. Inparticular, D can be split by a separable extension of degree degD.

Proof Let n = degD and choose a basis ω1, . . . , ωn2 of D over K. Let L be the algebraic closure of K

and so that there exists an isomorphism φ:D⊗K L≈→Mn(L). View the elements a1ω1 + · · ·an2ωn2 ∈ D,

ai ∈ K, as points (a1, . . . , an2) of the affine space Kn2

, and similarly regard the elements of Mn(L) as

points of the affine space Ln2

.

As we shall see, there are no non-trivial division algebras over a finite field, hence we may assume Kis infinite. Now consider the discriminant d(x1, . . . , xn2) ∈ L[x1, . . . , xn2 ] of the characteristic polynomialof a matrix in Mn(L) with entries x1, . . . , xn2 . By the general position lemma below, we can find ai ∈ Ksuch that the element θ = a1ω1 + · · · an2ωn2 ∈ D satisfies d φ(θ ⊗ 1) 6= 0. In other words, thecharacteristic polynomial p(x) of matrix φ(θ ⊗ 1) ∈ Mn(L) will have distinct roots, hence φ(θ ⊗ 1) willhave n distinct eigenvalues and will therefore be diagonalisable. Now let E = K[θ] be the subfield of Dgenerated by θ. By construction, this field has the property that E⊗K L is isomorphic to the subalgebraof Mn(L) generated by φ(θ ⊗ 1). But φ(θ ⊗ 1) is diagonalisable, hence E ⊗K L ∼= Ln. This proves that[E : K] = n = degD and that E is separable over K.

Lemma 4.7 (General Position) Let K ⊂ L be an extension of fields and suppose that K is an infinitefield. Let p(x1, . . . , xn) ∈ L[x1, . . . , xn] be a non-zero polynomial. Then there exist infinitely many points(a1, . . . , an) in Kn such that p(a1, . . . , an) 6= 0.

Proof We proceed by induction on n. If n = 1, the result is obvious since p(x1) has only finitely manyroots. Assume n > 1 and think of p(x1, . . . , xn) as a polynomial in xn with coefficients in L[x1, . . . , xn−1].By induction hypothesis, there exists (a1, . . . , an−1) ∈ Kn−1 such that at least one of the coefficients ofp(x1, . . . , xn) does not vanish, i.e., such that p(a1, . . . , an−1, xn) ∈ L[xn] is not identically 0. But nowthere exist infinitely many an ∈ K such that p(a1, . . . , an−1, xn) 6= 0, and we are done.

Corollary 4.8 (Galois Splitting) Any CSA A over a field K admits a splitting field L which is afinite Galois extension of K.

Proof We may suppose that K is infinite, since any splitting field of a CSA over a finite field K isautomatically a Galois (even cyclic) extension of K. Then, since A ∼= Mn(D) for some division algebraD over K, and D has a separable splitting field by the theorem, we conclude that A can also be split bya finite separable extension of K. By taking the normal closure, we we may assume this extension to beGalois as well.

Remark 4.9 Even though a division algebra admits a Galois splitting field, it does not necessarilycontain a maximal Galois subfield. Division algebras without Galois maximal subfields are called non-crossed products, and are quite hard to come about. The first examples of such bizarre algebras werediscovered by Amitsur (On central division algebras, Israel J. Math. 12 (1972), 408–420). Non-crossedproducts do not exist for division algebras over Q or Qp, but already appear for Q(t) and Q((t)) (seeBrussel’s construction in Noncrossed products and nonabelian crossed products over Q(t) and Q((t)),Amer. J. Math. 117 (1995), no. 2, 377–393).


5 Faithfully Flat Descent and Galois Cohomology

As we just saw, any degree n CSA A over K can be split or “trivialised” by restricting it to some finiteGalois extension L of K, hence A should be thought of a “twisted form” of Mn(L). The natural questionis: how to get all such forms?

In topology, we have a very similar situation. Recall that a (real) vector bundle of rank n overa topological space X is a family of “continuously varying” n-dimensional real vector spaces over X .More precisely, we have a continuous map π:E ։ X from a topological space E to X such that the fibreφ−1x ⊂ E of each point x ∈ X has the structure of an n-dimensional R-vector space, and π is locallyisomorphic to the first projection map p1:U ×Rn

։ U : each x ∈ X has an open neighbourhood U ⊂ Xtogether with a homeomorphism ξU :π−1U

≈→ U × Rn making the diagram

U × Rn ξU

≈- π−1U ⊂ - E

U

p1

?========= U

π

?⊂ - X

π

?

commute and such that ξU induces an R-linear isomorphism ξU,x: x × Rn ≈→ π−1x between the fibresfor each x ∈ U . For example, over the circle S1, the projection map p1:S

1 × R→ S1 is a rank 1 vectorbundle (the trivial rank 1 vector bundle). But also the projection π:M → S1 of the “infinite Mobius”band M to the middle circle is a rank 1 vector bundle, which is not trivial: M is locally of the formU × R, but not globally.

(Just to make sure we are talking about the same thing, M is the quotient of [0, 1] × R obtained byidentifying (0, y) ∼ (1,−y) for all y ∈ R; the middle circle is the image of [0, 1]×0 in this quotient andπ:M → S1 is given by π(x, y) = x)

Two rank n vector bundles π1:E1 → X and π2:E2 → X are isomorphic if there is a homeomorphism

φ:E1≈→ E2 such that the diagram

E1φ

≈- E2

X

π1

?======== X

π2

?

commutes and such that the induced morphism of fibres φx:π−11 x

≈→ π−12 x is R-linear for each x ∈ X .

By definition, any vector bundle π:E → X over X becomes trivial (i.e., isomorphic to Ui × Rn)when restricted to some open cover U = Ui of X . This observation allows for a very simple recipe toobtained all rank n vector bundles (up to isomorphism) that are “trivialised” by U . First suppose thatwe are given a vector bundle π:E → X and a trivialisation

ξi:Ui × Rn ≈→ π−1Ui

Faithfully Flat Descent and Galois Cohomology 9

Then, for each pair (i, j), we have a vector bundle automorphism φij : (Ui ∩ Uj)× Rn ≈→ (Ui ∩ Uj)× Rn

over Ui ∩ Uj given by the composition

φij : (Ui ∩ Uj)× Rn ξi- π−1(Ui ∩ Uj)ξ−1

j- (Ui ∩ Uj)× Rn

where we still write ξi and ξ−1j for their restrictions to (Ui ∩ Uj) × Rn and π−1(Ui ∩ Uj) respectively.

These automorphisms clearly satisfy

1. φii = id for all i;

2. φij = φ−1ji for all pairs i, j;

3. (1-cocycle condition) for every triple i, j, k,

φik = φjk φij

over Ui ∩ Uj ∩ Uk.

The above data φij is called the Cech 1-cocycle of the vector bundle E associated to the trivialisationξi. Now how does a 1-cocycle change when we change the trivialisation? Or, more generally, givenisomorphic vector bundles πE :E → X and πF :F → X with 1-cocycles φij and ψij, what is theirrelation?

To answer that question, let λ:E≈→ F be an isomorphism and let ξi:Ui × Rn ≈→ π−1

E Ui and

χi:Ui×Rn ≈→ π−1F Ui be trivialisations so that φij = ξ−1

j ξi and ψij = χ−1j χi over Ui∩Uj . Then, over

Ui, we have a vector bundle automorphism µi:Ui × Rn → Ui × Rn given by the composition

µi:Ui × Rn ξi- π−1E Ui

λ- π−1F Ui

χ−1i- Ui × Rn

Now a quick computation shows that, for all i, j,

ψij = µj φij µ−1i over Ui ∩ Uj

In case there are automorphisms µi of Ui × Rn such that the above relation holds, we say that the 1-cocycles ψij and φij are cohomologous. As it is easily seen, this establishes an equivalence relationamong the 1-cocycles, and therefore we have showed that to each isomorphism class of rank n vectorbundles that is trivialised by U there is a well-defined 1-cocycle class, which is independent of the chosentrivialisation.

Conversely, given a Cech 1-cocycle φij (that is, automorphisms φij of (Ui ∩ Uj) × Rn satisfyingthe relations 1–3 above), we can construct a vector bundle π:E → X by “glueing” trivial vector bundlesUi × Rn! Just take E to be the quotient

E =

⊔

i(Ui × Rn)

∼

of the disjoint union of the Ui × Rn. Here

v ∼ w for v ∈ Ui × Rn and w ∈ Uj × Rn ⇐⇒ φij(v) = w over Ui ∩ Uj

Now conditions 1–3 above ensure that ∼ is indeed an equivalence relation, so everything is well-defined.Now the projection maps Ui × Rn → Ui induce a global projection map π:E → X , which is a bona fiderank n vector bundle with associated 1-cocycle φij by construction! As easily checked, cohomologous1-cocycles give rise to isomorphic vector bundles so that we have shown that there is a bijection

isomorphism classes of rank nvector bundles trivialised by U

↔

cohomology classes ofU-Cech 1-cocycles

Let us rephrase the above construction in such a way to make it more “algebraisable.” First recallthat given two continuous maps f :A → C and g:B → C, we can form their fibre product A ×C B,defined as the subspace of A×B given by

A×C B = (a, b) ∈ A×B | f(a) = g(b)


We have thus a commutative diagram

A×C B - B

A?

f- C

g

?

where A ×C B → A and A ×C B → B are induced by the projection maps from A × B to A andB, respectively. For instance, if A is a subspace of C and f is the inclusion map, then A ×C B ishomeomorphic to g−1A and the left vertical arrow is just the restriction of g.

Now let X and U be as above and let Y =⊔

i Ui be the disjoint union of the Ui. Consider the maph:Y → X induced by the inclusion maps Ui → X . Then we have homeomorphisms

Y ×X Y =⊔

i,j

Ui ∩ Uj and Y ×X Y ×X Y =⊔

i,j,k

Ui ∩ Uj ∩ Uk

and maps

Y ×X Y ×X Yp12−→−→−→p23

Y ×X Yp1−→−→p2

Yh−→ X

Here p1 and p2 are the projection maps, and p12(y1, y2, y3) = (y1, y2), p13(y1, y2, y3) = (y1, y3) (the“middle one” among the triple arrows, not labelled for lack of space), p23(y1, y2, y3) = (y2, y3). Now togive a collection φij of vector bundle automorphisms of (Ui ∩ Uj)×Rn is the same as to give a singlea vector bundle automorphism over Y ×X Y :

φ:Y ×X Y × Rn ≈→ Y ×X Y × Rn

The 1-cocycle condition is now summarised by a single relation

p∗13φ = p∗23φ p∗12φ

where p∗ijφ denote the pull-backs of φ with respect to pij , that is, the morphism on Y ×X Y ×X Y ×Rn

induced by φ on the (i, j)-components and the identity on the remaining component: for instance,p∗13φ(y1, y2, y3, v) = (y1, y2, y3, w) where φ(y1, y3, v) = (y1, y3, w). Now two automorphisms φ and ψ are

cohomologous if and only if there is an automorphism µ:Y ×Rn ≈→ Y ×Rn such that ψ = p∗2µφp∗1µ−1.

Now we mimic the above algebraically. We work in the more general situation of faithfully flatextensions of commutative rings, whose proof is no more difficult than the case of fields. Recall that acommutative ring extension A ⊂ B is faithfully flat if it is flat, i.e., the functor − ⊗A B is exact, andM ⊗A B = 0 ⇐⇒ M = 0 for any A-module M . In particular, a complex of A-modules

· · · - Mi- Mi−1

- Mi−2- Mi−3

- · · ·

is exact if, and only if, the complex of B-modules

· · · - Mi ⊗A B - Mi−1 ⊗A B - Mi−2 ⊗A B - Mi−3 ⊗A B - · · ·

is exact. In fact, if hi(M•) is the homology of the first complex, then hi(M•) ⊗A B is the homology ofthe second one since B is flat over A, and therefore tensor products commute with quotients. Thereforehi(M•) = 0 ⇐⇒ hi(M•)⊗A B = 0, as required.

It can be shown that B is faithfully flat over A if and only if it is flat and SpecB → SpecA issurjective. In our algebraic setup, Y = SpecB will play the role of the disjoint union of a cover ofX = SpecA. Since we work with rings rather than schemes, arrows will be reversed and fibre productswill be replaced by tensor products. In what follows, we write

p1:B → B ⊗A B

b 7→ b⊗ 1

p2:B → B ⊗A B

b 7→ 1⊗ b


Also, if N is a B-module, let

p∗1N = N ⊗B (B ⊗A B) = N ⊗A B and p∗2N = (B ⊗A B)⊗B N = B ⊗A N

pull-backs of N , namely the B ⊗A B-modules obtained by tensoring N with respect to the first andsecond entries respectively. Then the p∗iN will play the role of trivial bundles Y × Rn. Let

φ: p∗1N≈→ p∗2N

be an isomorphism of B ⊗A B-modules (φ will play the role of the automorphism of Y ×X Y × Rn).Consider the “dual projection maps”

p12:B ⊗A B → B ⊗A B ⊗A B

b1 ⊗ b2 7→ b1 ⊗ b2 ⊗ 1

p13:B ⊗A B → B ⊗A B ⊗A B

b1 ⊗ b2 7→ b1 ⊗ 1⊗ b2p23:B ⊗A B → B ⊗A B ⊗A B

b1 ⊗ b2 7→ 1⊗ b1 ⊗ b2and the pull-backs of N to B ⊗A B ⊗A B, obtained by tensoring the latter ring with N over B withrespect to the first, second and third entries of B ⊗A B ⊗A B, respectively:

N ⊗A B ⊗A B and B ⊗A N ⊗A B and B ⊗A B ⊗A N

Tensoring φ with id over B with respect to the third, second and first entries of B⊗AB⊗AB, we obtainmaps of B ⊗A B ⊗A B-modules

p∗12φ:N ⊗A B ⊗A B → B ⊗A N ⊗A B p∗12φ (n⊗ 1⊗ 1) =∑

i bi ⊗ ni ⊗ 1

p∗13φ:N ⊗A B ⊗A B → B ⊗A B ⊗A N p∗13φ (n⊗ 1⊗ 1) =∑

i bi ⊗ 1⊗ ni

p∗23φ:B ⊗A N ⊗A B → B ⊗A B ⊗A N p∗23φ (1⊗ n⊗ 1) =∑

i 1⊗ bi ⊗ ni

for φ(n ⊗ 1) =∑

i bi ⊗ ni. These maps will play the roles of the pull-backs of the automorphism ofY ×X Y ×X Rn to Y ×X Y ×X Y × Rn. We are now ready to state

Theorem 5.1 (Grothendieck’s Faithfully Flat Descent) Let B ⊃ A be a faithfully flat extensionof commutative rings.

1. For any A-module M , the sequence

0 - Mǫ- M ⊗A B

d0

- M ⊗A B ⊗A Bd1

- M ⊗A B ⊗A B ⊗A Bd2

- · · ·is exact. Here ǫ(m) = m⊗ 1 and

dr(m⊗ b0 ⊗ · · · ⊗ br) =∑

0≤i≤r

(−1)i ·m⊗ b0 ⊗ · · · ⊗ bi−1 ⊗ 1⊗ bi ⊗ · · · ⊗ br

2. Let N be a B-module (or a B-algebra). Then there is a bijection between the set of isomorphismclasses of A-modules (or A-algebras) M such that M⊗AB ∼= N and the set of equivalence classesof B ⊗A B-isomorphisms

φ:N ⊗A B≈→ B ⊗A N

satisfying the 1-cocycle condition p∗13φ = p∗23φ p∗12φ, so that we have a commutative diagram

N ⊗A B ⊗A Bp∗12φ- B ⊗A N ⊗A B

B ⊗A B ⊗A N

p∗23φ

?

p ∗13 φ

-

Two B ⊗A B-isomorphisms φ and ψ are equivalent if and only if there is an automorphism

µ:N≈→ N such that

ψ = p∗2µ φ p∗1µ−1

where p∗1µ = µ⊗ id:N ⊗AB≈→ N ⊗AB and p∗2µ = id⊗µ:B⊗AN

≈→ B⊗AN are the pull-backsof φ with respect to p1 and p2.


Proof 1. First we show that the given complex is homotopic to 0 under the assumption that the mapǫ:M →M ⊗A B has a section s:M ⊗A B →M (that is, s ǫ = id). Define

kr:M ⊗A B⊗(r+1) →M ⊗A B

⊗r

m⊗ b0 ⊗ b1 ⊗ · · · ⊗ br 7→ s(m⊗ b0)⊗ b1 ⊗ · · · ⊗ br

Now check that id = dr−1 kr + kr+1 dr for all r. This shows that the complex is exact. To tackle thegeneral case, it is enough to show that the sequence obtained by base change − ⊗A B is exact since Bis faithfully flat over A, hence it is enough to show that the map ǫB:M ⊗A B → M ⊗A B ⊗A B has asection. Just define s:M ⊗AB⊗AB →M ⊗AB by s(m⊗ b1⊗ b2) = m⊗ b1b2 and check that s ǫ = id.

2. Suppose that M is an A-module such that M ⊗A B = N . Then we may define an isomorphism

φ:N ⊗A B≈→ B ⊗A N of B ⊗A B-modules by

φ: (M ⊗A B)⊗A B≈→ B ⊗A (M ⊗A B)

m⊗ b1 ⊗ b2 7→ b1 ⊗m⊗ b2

which, in our geometric discussion, corresponds to the vector bundle automorphism of Y ×X Y × Rn

that identifies the restrictions of Ui×Rn and Uj ×Rn over Ui ∩Uj . Now a straightforward computationshows that the above φ satisfies the 1-cocycle condition. Observe that by part (1), we may identify Mwith the A-submodule of N given by m⊗ 1 | m ∈M.

Now let M1 and M2 be two A-modules such that M1 ⊗A B = M2 ⊗A B = N , and let φ1 andφ2 are the corresponding isomorphisms N ⊗A B → B ⊗A N . If there is an isomorphism of A-modules

ν:M1≈→M2, then it induces a B-automorphism µ = ν⊗ id of N and we have that φ2 = p∗2µ φ1 p∗1µ−1

sincep∗2µ φ1 p∗1µ−1(m2 ⊗ b1 ⊗ b2) = (id⊗ν ⊗ id) φ1 (ν−1 ⊗ id⊗ id)(m2 ⊗ b1 ⊗ b2)

= (id⊗ν ⊗ id) φ1(ν−1m2 ⊗ b1 ⊗ b2)

= (id⊗ν ⊗ id)(b1 ⊗ ν−1m2 ⊗ b2)= b1 ⊗m2 ⊗ b2 = φ2(m2 ⊗ b1 ⊗ b2)

for all m2 ∈ M2 and b1, b2 ∈ B. Conversely, if φ2 = p∗2µ φ1 p∗1µ−1 ⇐⇒ φ2 p∗1µ = p∗2µ φ1 then

µ:N≈→ N restricts to an isomorphism ν:M1

≈→M2 since

φ2(µ(m1)⊗ 1) = φ2 p∗1µ(m1 ⊗ 1) = p∗2µ φ1(m1 ⊗ 1) = p∗2µ(1⊗m1) = 1⊗ µ(m1),

i.e., µ(m1) ∈M2 for all m1 ∈M1.

Now we have to show that given an isomorphism φ:N ⊗A B≈→ B ⊗A N satisfying the 1-cocycle

condition, there exists an A-module M such that M ⊗A B = N . Secretly, we know that M exists andthat φ is given by the above “swapping formula,” hence it is natural to define

Mdf= m ∈ N | φ(m⊗ 1) = 1⊗m

which, in geometric terms, corresponds to the subset of Y × Rn =⊔

i Ui × Rn of those elements that“agree on the overlaps” Ui ∩ Uj .

The subset M ⊂ N is clearly an A-module. We now show that the natural map

λ:M ⊗A B → N

m⊗ b 7→ bm

is an isomorphism of B-modules. For that, consider the following diagram

M ⊗A B - N ⊗A Bα⊗ id-

β ⊗ id- B ⊗A N ⊗A B

N

λ ≈

? ǫ- B ⊗A N

φ ≈

? e0 -

e1- B ⊗A B ⊗N

≈ p∗23φ

?


where the horizontal maps of the bottom row are the ones of the part (1), namely, e1(b⊗n) = 1⊗ b⊗n,and e0(b⊗ n) = b⊗ 1⊗ n for b ∈ B, n ∈ N , and those of the top are obtained by the faithfully flat basechange −⊗A B of the exact sequence

M - Nα-β- B ⊗A N

that defines the A-module M , where α(n) = φ(n ⊗ 1) and β(n) = 1 ⊗ n. In particular, we have thatM⊗AB is the equaliser of α⊗ id and β⊗ id, while N is the equaliser of e0 and e1. Hence the isomorphismφ will induce the required isomorphism λ once we show that the above diagram commutes. But it iseasy to check that the right square formed by the bottom arrows β⊗ id and e1 commutes, and the sameholds for the left square by the definition of M . The right square formed by the top arrows α ⊗ id ande0 also commutes since p∗23φ α⊗ id(n⊗ b) = p∗23φ p∗12φ(n⊗ 1⊗ b) = p∗13φ(n⊗ 1⊗ b) = e0 φ(n⊗ b).

If N is a B-algebra, we have also to show that the multiplication map m:N ⊗B N → N also“descends,” which can be done by similar computations as the previous ones. Details are left to thereader.

After this little detour in faithfully flat descent, we go back to CSA. Let L ⊃ K be a finite extensionof fields (which is automatically faithfully flat since L is free over K). In order to apply the above to ourstudy, we need to know the automorphism group of Mn(L)⊗K L. Fortunately, it is easy to describe:

Lemma 5.2 (Weak Skolem-Noether) Let K be a field. All automorphisms of Mn(K) are inner,

hence the automorphism group of Mn(K) is the projective linear group PGLn(K)df= GLn(K)/K×.

Proof Any automorphism φ of Mn(K) permutes its minimal non-zero left ideals, i.e., the columnmatrices I1, . . . , In. Conjugating φ by a suitable permutation matrix, we may assume that φ(I1) = I1.In other words, φ restricts to an automorphism of I1 ∼= Kn, and as such it can be represented by aninvertible matrix C ∈ GLn(K), i.e., φ(P ) = CP for all P ∈ I1. Now if M ∈Mn(K) and P ∈ I1, we havethat

φ(MP ) = φ(M)φ(P ) ⇐⇒ CMP = φ(M)CP

Since this holds for all P ∈ I1 ∼= Kn, we have that CM = φ(M)C ⇐⇒ φ(M) = CMC−1. This provesthat φ is inner. Since the centre of GLn(K) is K×, the last statement also follows.

We saw in the last section that any CSA over K is split by some finite Galois extension L ⊃ K,and from now on we work in this setup. Let G = Gal(L/K) and n = [L : K]. By the Lemma ⊗K

Lemma we have that L ⊗K L = Maps(G,L), Mn(L) ⊗K L = Mn(L ⊗K L) = Mn(Maps(G,L)) =Maps(G,Mn(L)) and similarly L⊗KMn(L) = Maps(G,L). Therefore an L⊗KL-isomorphismMn(L)⊗K

L→ L⊗KMn(L) corresponds to a Maps(G,L)-automorphism of Maps(G,Mn(L)), that is, to an elementof Maps(G,PGLn(L)) by the previous lemma. Putting everything together, we obtain a bijection

isomorphism classes of CSAA such that AL

∼= Mn(L)

↔

elements of Maps(G,PGLn(L))satisfying the 1-cocycle condition

We can be more explicit about the 1-cocycle condition. Applying the Lemma⊗KLemma twice, we obtain

(L⊗K L)⊗K L = Maps(G,L)⊗K L = Maps(G,L⊗K L) = Maps(G,Maps(G,L)) = Maps(G×G,L)

given by

L⊗K L⊗K L≈→ Maps(G×G,L)

a⊗ b⊗ c 7→ f(τ, σ) = σ(τ(a) · b) · c for all σ, τ ∈ Gand, similarly, Mn(L) ⊗K L ⊗K L = Maps(G × G,Mn(L)) and so on. Now let f :G → PGLn(L) bean element of Maps(G,PGLn(L)). In terms of the above isomorphisms, we have that the pull-backs off correspond to the Maps(G × G,L)-automorphisms of Maps(G × G,Mn(L)) given by the elements ofMaps(G×G,PGLn(L))

p∗12f(τ, σ) = σf(τ)

p∗23f(τ, σ) = f(σ)

p∗13f(τ, σ) = f(στ)

Hence the 1-cocycle condition now reads f(στ) = f(σ) · σf(τ) for all σ, τ ∈ G.


Definition 5.3 Let G be a finite group and M be a (not necessarily abelian) G-module, written multi-plicatively. A 1-cocycle is a function f :G→M such that

f(στ) = f(σ) · σ(

f(τ))

for all σ, τ ∈ G

Two 1-cocycles f and g are cohomologous (in symbols f ∼ g) if there exists an element m ∈ M suchthat

f(σ) = m−1 · g(σ) · σ(m) for all σ ∈ GIt is easy to check that ∼ defines an equivalence relation on the pointed set Z1(G,M) of all 1-cocycles (thedistinguished point being the constant function f(σ) = 1). We define the zeroth and first cohomologygroups of G with coefficients in M as the pointed sets

H0(G,M) = MG df= m ∈M | σ(m) = m for all σ ∈ G

H1(G,M) = Z1(G,M)/ ∼

Notice that when M is abelian, the above definitions coincide with the usual cohomology of G withcoefficients in M .

Rewriting the faithful flat descent in terms of the above language we obtain:

Theorem 5.4 (Twisted forms of matrices) Let L ⊃ K be a finite Galois extension of fields withG = Gal(L/K). There is a bijection

isomorphism classes of CSAA such that AL

∼= Mn(L)

↔ H1(G,PGLn(L))

This isomorphism takes (the isomorphism class of) the CSA A to the class of the 1-cocycle

f(σ) = φ−1 σ φ σ−1 ∈ AutL(Mn(L)) = PGLn(L) for σ ∈ G

where φ:Mn(L)≈→ AL is a fixed L-isomorphism. The inverse map associates to each 1-cocycle f ∈

Z1(G,PGLn(L×)) the K-subalgebra of Mn(L) given by

M ∈Mn(L) | f(σ) σ(M) = M for all σ ∈ G

Example 5.5 (Cyclic Algebras Revisited) Let L ⊃ K be a cyclic extension of degree n with G =Gal(L/K) and let σ be a generator ofG. Then an element of H1(G,PGLn(L)) represented by a 1-cocyclef :G→ PGLn(L) is completely determined by the value of f(σ) since f(id ·σ) = f(id)f(σ)⇒ f(id) = 1and

f(σ2) = f(σ) · σf(σ),

f(σ3) = f(σ · σ2) = f(σ) · σf(σ2) = f(σ) · σf(σ) · σ2f(σ),

...

Since f(σn) = f(id) = 1, f(σ) is also subject to

f(σ) · σf(σ) · σ2f(σ) · . . . · σn−1f(σ) = 1 (∗)

For instance, let a ∈ K× and consider “companion matrix” of xn − a:

Cadf=

0 0 · · · 0 a1 0 · · · 0 00 1 · · · 0 0...0 0 · · · 1 0

We may form the 1-cocycle given byf(σ) = Ca mod L×

The Brauer group 15

Since a ∈ K, the condition (∗) is indeed verified:

f(σ) · σf(σ) · . . . · σn−1f(σ) = Cna mod L× = aI mod L× = I mod L×

Here I is the identity matrix. According to the above, the K-algebra corresponding to this 1-cocycle isthe set of matrices M ∈ Mn(L) such that f(σi) σi(M) = M ⇐⇒ Ci

aσi(M)C−i

a = M for all i, whichamounts to σ(M) = C−1

a MCa. Clearly I, Ca, C2a , . . . , C

n−1a satisfy the latter identity. Moreover, since

conjugation by Ca is “almost a cyclic permutation”, it is not difficult to guess and it is immediate toverify that the matrices

Sbdf=

b 0 · · · 00 σ(b) · · · 0

...0 0 · · · σn−1(b)

for b ∈ L

also satisfy that identity as well, that is, SbCa = Caσ(Sb) = CaSσb for all b ∈ L, or equivalently,Sσ−1bCa = CaSb for all b ∈ L. Therefore

Adf= Sb ⊕ SbCa ⊕ SbC

2a ⊕ · · · ⊕ SbC

n−1a

is a K-subalgebra of Mn(L) of the correct dimension n2, hence it must be the algebra defined by the1-cocycle above. But clearly A is isomorphic to our old friend (χ, a), the cyclic algebra given by a andthe character χ(σ) = −1 mod n.

We end this section with a partial generalisation of the long exact sequence of abelian cohomologyto the non-abelian case:

Lemma 5.6 (Not So Long Exact Sequence) Let

1→ A→ B → C → 1

an exact sequence of Z[G]-modules with A ⊂ Z(B). Then the sequence

1 - H0(G,A) - H0(G,B) - H0(G,C)

δ0

- H1(G,A) - H1(G,B) - H1(G,C)

δ1

- H2(G,A)

is an exact sequence of pointed sets. Here, all the maps but the connecting morphisms δi are induced bythe corresponding maps of the short exact sequence. The connecting maps are defined as follows. Givenc ∈ H0(G,C), δ0(c) is the class of the 1-cocycle f(σ) = b−1σ(b) ∈ A, σ ∈ G, where b ∈ B is a pre-imageof c. On the other hand, for any φ ∈ H1(G,C), δ1(φ) is the class of the (abelian) 2-cocycle g:G×G→ Agiven by g(σ, τ) = f(σ) · σ(f(τ)) · f(στ)−1 for σ, τ ∈ G, where f :G → B is such that the compositionwith B → C is a 1-cocycle representing φ.

Proof Straightforward computation shows that the maps δi are well-defined. By an exact sequenceof pointed sets we just mean that the image of the preceding map coincides with the pre-image of thedistinguished point of the next term of the sequence. With this definition, exactness of the above longsequence is a series of boring checks that are better done in private, when no one is looking, so I leaveit to the (patient) reader!

6 The Brauer group

In this section, for each field K we define a very important group, the Brauer group of K, whoseelements classify all division algebras over K. We begin with a

Definition 6.1 Let K be a field. Two CSA A and B over K are Brauer equivalent if they satisfy thetwo following equivalent conditions:

1. A is stably isomorphic to B, that is, there is an isomorphism

Mn(A) = A⊗K Mn(K) ∼= B ⊗K Mm(K) = Mm(B)

for some m and n.

2. the division algebras associated to A and B in Wedderburn’s theorem are isomorphic.


The equivalence of the two conditions above is a consequence of Wedderburn’s theorem. It is easy tosee that Brauer equivalence is an equivalence relation. We write [A] for the equivalence class of A. WhenA and B have the same degree, Brauer equivalence [A] = [B] is the same as the plain old isomorphismA ∼= B of CSA (again by Wedderburn’s theorem).

An important feature of the Brauer equivalence is that it is compatible with tensor products:

[A] = [A′][B] = [B′]

⇒ [A⊗K B] = [A′ ⊗K B′]

Hence the tensor product induces a binary operation

[A] + [B]df= [A⊗K B]

on the set Br(K) of all Brauer equivalence classes of CSA over K. It turns out that Br(K) is anabelian group: the operation is associative and commutative since the tensor product has the sameproperties; the class [K] of the trivial CSA over K is clearly an identity for this operation; and since

A ⊗K Aop ≈→ EndK-mod(A) ⇒ [A] + [Aop] = 0 for every CSA A over K, each element [A] ∈ Br(K) hasan inverse [Aop].

Definition 6.2 Let K be a field. The set Br(K) of all Brauer equivalence classes of CSA over K,equipped with the product induced by the tensor product, is called the Brauer group of K. For anyfield extension L ⊃ K, the restriction map is the group morphism given by

res: Br(K)→ Br(L)

[A] 7→ [AL]

In particular, if L is algebraic over K, we also write Br(L/K) for the kernel of the restriction map, thatis, the subgroup of Br(K) consisting of Brauer classes split by L.

Observe that the elements of Br(K) are exactly the isomorphism classes of division algebras overK. Knowledge of the Brauer group is thus of central importance in the study of CSA. As we shall see,Br(K) is also a very important arithmetic invariant of the field K.

Example 6.3 If K is algebraically closed (separably closed suffices), then Br(K) = 0 since there are nonon-trivial division algebras over K.

Example 6.4 In Br(R), the class [H] of the quaternion algebra H satisfies 2[H] = 0 ⇐⇒ [H] = −[H] =[Hop]. In fact, viewing H as the real subalgebra of M2(C) given by matrices of the form

(

α β−β α

)

α, β ∈ C

we see that the transpose isomorphismM2(C)≈→M2(C)op given byM 7→MT restricts to an isomorphism

H≈→ Hop.

The great thing about the Brauer classes of CSA is that they are much easier to deal with comparedto the isomorphisms classes of CSA, thanks to

Theorem 6.5 (Cohomological Brauer group) Let K be a field and GKdf= Gal(Ksep/K) be its ab-

solute Galois group, where Ksep denotes the separable closure of K. Then there is a natural isomorphism

Br(K)≈- H2(GK ,K

×

sep)

compatible with restriction maps: for any field extension L ⊃ K, one has

Br(L)≈- H2(GL, L

×

sep)

Br(K)

res

6

≈- H2(GK ,K×

sep)

res

6

(Note that L ·Ksep = Lsep so that the restriction map of cohomology groups is well-defined).

The Brauer group 17

Proof Let L ⊃ K be a finite Galois extension with G = Gal(L/K), and let B(n,L) ⊂ Br(K) be thesubset whose elements are the classes of CSA over K of degree n that are split by L. Then we have that

Br(L/K) =⋃

n≥1

B(n,L) and Br(K) =⋃

L finite Galoisover K

Br(L/K)

since every CSA is split by some finite Galois extension. On the other hand, by Wedderburn’s theoremtwo degree n CSA are Brauer equivalent if and only if they are isomorphic, hence by what we showed

in the last section there is a natural bijection B(n,L)≈→ H1(G,PGLn(L)). From the exact sequence of

G-modules1 - L× - GLn(L) - PGLn(L) - 1

we obtain an exact sequence

1 = H1(G,GLn(L)) - H1(G,PGLn(L))δ1

- H2(G,L×) (∗)

where the first term is trivial by Satz 90 Hilberts. Hence, composing B(n,L)≈→ H1(G,PGLn(L)) with

δ1, we obtain a mapιn:B(n,L)→ H2(G,L×)

We have that ιn([A]) = 0 ⇐⇒ A ∼= Mn(K) by the exactness of (∗). Now an explicit computation showsthat for any A1 ∈ B(m,L) and A2 ∈ B(n,L) we have that

ιmn([A1 ⊗K A2]) = ιm([A1]) + ιn([A2]) (∗∗)

In particular, if A2 = Mn(K) then ιmn([A1 ⊗K Mn(K)]) = ιm([A1]), which shows the compatibility ofthe maps ιm with Brauer equivalence, expressed by the following commutative diagram:

B(mn,L)ιmn- H2(G,L×)

B(m,L)∪

6ιm

-

Hence taking the union we obtain a map

ι: Br(L/K)→ H2(GL, L×)

which is group morphism by (∗∗). It is injective by the exactness of (∗). We now show that it isalso surjective. For that it is enough to show that ιn is surjective when n = [L : K]. Consider a2-cocycle g:G × G → L×, and let V be an n-dimensional L-vector space with basis eσ, σ ∈ G. Forσ ∈ G, let f(σ) ∈ PGLn(L) be the image of the matrix corresponding to the V -automorphism given byeτ 7→ g(σ, τ)eστ . Then a computation shows that f is a 1-cocycle and that

g(σ, τ) = f(σ)σ(f(τ))f(στ)−1 for all σ, τ ∈ G

Hence f defines an element of H1(G,PGLn(L)) with δ1([f ]) = [g], proving that ιn is surjective. TheCSA algebra defined by f is the opposite of the usual cross product given by the 2-cocycle g.

Finally, for L′ ⊃ L it is easy to check that

Br(L′/K)ι′- H2(G′, L′×)

Br(L/K)∪

6

ι- H2(G,L×)

inf

∪

6

commutes, where the primes denote the corresponding objects of L′ instead of L. Hence, passing to the

limit, we obtain the desired isomorphism Br(K)≈→ H2(GK ,K

×sep), which is easily shown to be compatible

with restriction maps.


Remark 6.6 A more detailed analysis shows that if K ′ ⊃ K is a separable field extension of degree n,then the connecting maps induce an isomorphism

ker(

H1(GK , PGLn(Ksep))→ H1(GK′ , PGLn(K ′sep))

)

≈→ ker(

H2(GK ,K×

sep)→ H2(GK′ ,K ′sep

×))

between the kernels of the restriction maps, and the right hand side is isomorphic to Br(K ′/K) by thelast theorem.

Corollary 6.7 Let L ⊃ K be a Galois extension with G = Gal(L/K). Then we have an isomorphism

Br(L/K)≈→ H2(G,L×)

Proof Since Lsep = Ksep, (K×sep)GL = L×, G = GK/GL, and H1(GL, L

×sep) is trivial by Hilbert 90, the

inflation-restriction exact sequence reads

0 - H2(G,L×)inf- H2(GK ,K

×

sep)res- H2(GL, L

×

sep)

Hence we can identify H2(G,L×) with the kernel of res: Br(K)→ Br(L), namely with Br(L/K).

Corollary 6.8 The Brauer group is torsion.

Proof With the above notation, we have that H2(G,L×) is killed by |G| by the restriction-corestrictionargument.

Corollary 6.9 If n is prime to charK, the n-torsion of Br(K) is given by

n Br(K)≈→ H2(GK , µn)

where µn denotes the GK -module of all n-th roots of unity in Ksep.

Proof We have an exact sequence of GK-modules (the so-called Kummer sequence)

1 - µn- K×

sep

n- K×

sep- 1

where the superscript n denotes “exponentiation by n.” Since H1(GK ,K×sep) is trivial by Hilbert 90, we

obtain the following fragment of the associated long exact sequence:

1 = H1(GK ,K×

sep) - H2(GK , µn) - H2(GK ,K×

sep)n- H2(GK ,K

×

sep)

Hence we can identify H2(GK , µn) with the n-torsion of Br(K) = H2(GK ,K×sep).

Theorem 6.10 (Cyclic Algebras) Let L ⊃ K be a cyclic extension of degree n with G = Gal(L/K).

1. Let χ ∈ H1(G,Z/n) = Hom(G,Z/n) be a surjective character, viewed as an element ofH1(G,Q/Z) = Hom(G,Q/Z). Consider the connecting map δ:H1(G,Q/Z) → H2(G,Z) as-sociated to the exact sequence of G-modules (with trivial action)

0 - Z - Q - Q/Z - 0

Then for any a ∈ K× = H0(G,L×), we have that the cup product

a ∪ δχ ∈ H2(G,L×) = Br(L/K)

equals the class of the opposite of the cyclic algebra (χ, a).

2. A CSA A over K is Brauer equivalent to a cyclic algebra if and only if there exists a cyclicextension of K splitting A.

The Brauer group 19

3. In Br(K), we have

[(χ, a)] + [(χ, b)] = [(χ, ab)] and [(χ, a)] + [(ψ, a)] = [(χ+ ψ, a)]

for all a, b ∈ K× and χ, ψ ∈ H1(G,Z/n).

Proof 1. Let σ be the generator of G such that χ(σ) = 1. By the explicit formulas of the appendix,a ∪ δχ is represented by the 2-cocycle

f(σi, σj) =

1 if i+ j < na otherwise

i, j = 0, 1, . . . , n− 1

But that is just the coboundary of the element of H1(G,PGLn(L)) given in example 5.5, and the resultfollows.

2. We have already seen that a cyclic algebra (χ, a) is split by the cyclic extension defined by χ.Conversely, suppose that L splits A, that is, [A] ∈ Br(L/K) = H2(G,L×). Let χ ∈ Hom(G,Z/n)be a surjective character. Since G is cyclic, the cup product with δχ induces an isomorphism of Tatecohomology groups

K×

NL/KL×= H0

T (G,L×)∪δχ

≈- H2

T (G,L×) = Br(L/K)

In particular, [A] = a ∪ δχ for some a ∈ K×, hence A is cyclic by the first part.

3. Follows from the bilinearity of the cup product.

Now we come to two of the most important invariants associated to a CSA.

Definition 6.11 Let A be a CSA over a field K.

1. The period or exponent of A (or its class in Br(K)), denoted by perA or expD, is the orderof [A] in Br(K).

2. The (Schur) index of A (or its class in Br(K)), denoted by indA, is the smallest degree [L : K]of a separable splitting field L of A. Equivalently, it is the degree of the unique division algebraD over K which is Brauer equivalent to A.

The last equivalence is a consequence of the next theorem and the fact that a degree n divisionalgebra D is split by a maximal subfield of degree n over K.

Theorem 6.12 If D is a division algebra over K and L ⊃ K is a finite separable field extension splittingD then degD | [L : K].

Proof Let n = [L : K]. By the remark, we have that

[D] ∈ Br(L/K) = ker(

H1(GK , PGLn(Ksep))→ H1(GL, PGLn(Ksep)))

Hence [D] can be represented by a degree n CSA A over K, i.e., A ∼= Mr(D) with degA = n = r ·degD.This shows that degD | n, as required.

Let K be a field and let α ∈ Br(K). From their definitions, it is easy to check that the period andindex have the following properties:

1. For any CSA A over K, indA | degA, with equality if and only if A is a division algebra.

2. If L ⊃ K is any field extension then indαL | indα.

3. ind(α + β) | indα · indβ for any α, β ∈ Br(K).

4. per(α+ β) | perα · per β, with equality if perα and per β are relatively prime.

That they are not unrelated quantities is shown by


Theorem 6.13 (Period-Index) The period always divides the index, and both integers have the sameprime factors.

Proof Let K be a field and α ∈ Br(K). If n = indα then α is split by a separable field extensionL ⊃ K of degree n, i.e., α ∈ Br(L/K). But then nα = 0 by the restriction-corestriction argument (seeappendix). Since perα is the order of α ∈ Br(K), we must have perα | n.

Now let L ⊃ K be a Galois splitting field of α, and suppose that p is a prime number that divides[L : K] but not perα. Let H be the p-Sylow subgroup of G = Gal(L/K) and let M = LH . Then, since[L : M ] = |H | is a power of p, by the restriction-corestriction argument res: Br(M)→ Br(L) is injectiveon the prime-to-p part, hence αM = 0 since αL = 0 and gcd(p, perα) = 1. In other words, M is asplitting field of α, and therefore indα | [M : K] = [G : H ], i.e., indα is not divisible by p. Togetherwith the first part, this shows that perα and indα have the same prime factors.

We can also show that

Lemma 6.14 Let D a division algebra over a field K and let L be an extension of K with [L : K] primeto indD. Then DL is a division algebra over L.

Proof Let n = indD. We know that indDL | n and we have to show that equality holds. Butif M ⊃ L is a field extension splitting DL with [M : L] = indDL, then M also splits D, hencen = indD | [M : K] = indDL · [L : K]. But n is prime to [L : K], hence n | indDL, as required.

Theorem 6.15 (Sylow factors) Let D be a division algebra over a field K and let

indD = pe11 . . . per

r

be the canonical decomposition of indD into powers of distinct primes. Then

D ∼= D1 ⊗K · · · ⊗K Dr

where Di are division algebras over K with indDi = pei

i .

Proof We may write [D] ∈ Br(K) into its primary components, [D] = [D1]+[D2]+· · ·+[Dr] for divisionalgebras Di with per[Di] (and hence indDi) a power of pi. Hence everything follows if we can show thatD1 ⊗K · · · ⊗K Dr is a division algebra. For that, we show that if D1 and D2 are two division algebrasover K with m = indD1 relatively prime with n = indD2 then D1⊗K D2 is also a division algebra, thatis, indD1⊗K D2 = mn. Let L1 and L2 be separable splitting fields of D1 and D2 with [L1 : K] = m and[L2 : K] = n. Then the compositum L1L2 splits D1⊗K D2, hence indD1⊗K D2 | mn = [L1L2 : K]. Onthe other hand, we have that ind(D1⊗KD2)L1 = ind(D2)L1 = indD2 = n by the previous lemma, hencen = ind(D1⊗K D2)L1 | indD1⊗K D2. Similarly, m divides indD1⊗K D2, and since gcd(m,n) = 1, thisproves that mn divides indD1 ⊗K D2.

Chapter 2

BrauergroupofLocalFields

A local field K is a finite extension of either Qp or Fp((t)). The discrete valuations of Qp and Fp((t))extend uniquely to K, turning it into a complete discretely valued field. A local field arises naturally asthe completion of a global field (i.e. a finite extension of either Q or Fp(t)) with respect to its maximalideals.

The main result of this chapter is the following computation:

Theorem 0.1 (Brauer group of a local field) Let K be a local field. We have an isomorphism

invK :H2(GK ,K×

sep)≈- Q/Z

This isomorphism takes the class of the cyclic algebra (χ, πK) of degree n to the element − 1n mod Z.

Here χ denotes the character given by χ(ΦL/K) = 1, where ΦL/K is the Frobenius automorphism of thedegree n unramified extension L ⊃ K, and πK is a uniformiser of K.

1 Notation

Throughout this chapter, we adopt the following notations and conventions. For any field K we denoteby

Ksepdf= separable closure of K

GKdf= Gal(Ksep/K) = absolute Galois group of K

Kab df= K [GK :GK ]

sep = maximal abelian extension of K

= compositum of all finite abelian extensions of K inside Ksep

GabK = Gal(Kab/K)

Now let K be a local field with residue field k and normalised valuation v:K → Z ∪ ∞. We write

πKdf= uniformiser of K (that is, an element of valuation 1)

OKdf= ring of integers of v = x ∈ K | v(x) ≥ 0

UKdf= group of units of OK = x ∈ OK | v(x) = 0

U(i)K

df= 1 + (πi

K) = closed ball x ∈ K | v(x− 1) ≥ i centred at 1

Knrdf= maximal unramified extension of K in Ksep

= compositum of all finite unramified extensions of K inside Ksep

GnrK

df= Gal(Knr/K) ≈ Gk = Z

ΦKdf= Frobenius automorphism of Knr ⊃ K

where

Zdf= lim←−

n∈N

Z/(n) =∏

p

Zp

and p runs over all prime integers. The last isomorphism follows from the Chinese Remainder Theorem(check!). The Frobenius map ΦK is a topological generator of Gnr

K , corresponding to the element 1 ∈ Z

under the above isomorphism.

22 Brauer group of Local Fields

2 Unramified Cohomology

In this section, L ⊃ K will be a finite unramified extension with G = Gal(L/K), and l ⊃ k will be thecorresponding extension of residue fields. Recall that we have a canonical isomorphism G = Gal(L/K) ≈Gal(l/k), and so G is cyclic.

We now compute H2(G,L×). The starting point is the exact sequence of G-modules

0 - UL- L× v- Z - 0 (†)

where v denotes the normalised valuation of L. We need to compute the cohomology of UL and Z.

To compute the cohomology of Z, we use the exact sequence of G-modules (with trivial G-action)

0→ Z→ Q→ Q/Z→ 0

Since HrT (G,Q) is torsion and multiplication by any non-zero integer is an automorphism of Q, we

conclude that HrT (G,Q) = 0 for all r. Hence we have that Hr

T (G,Z) = Hr−1T (G,Q/Z) for all r.

Next we compute the cohomology of UL.

Theorem 2.1 For all r we have that

HrT (G,UL) = 0

Proof We have exact sequences of G-modules (via the isomorphism G ≈ Gal(l/k))

0→ U(1)L → UL → l× → 0

0→ U(r+1)L → U

(r)L → l+ → 0

The group HrT (G, l+) is trivial for all r since l+ is an induced module by the normal basis theorem

(see appendix). We now show that the group HrT (G, l×) is also trivial for all r. By the periodicity of

cohomology of cyclic groups, it is enough to prove that for r = 0 and r = 1. The case r = 1 is justHilbert 90, while the case r = 0 follows from the surjectivity of the norm map Nl/k : l× → k×: if k = Fq

and b is a generator of the group l× then Nl/k(b) = b1+q+···+qn−1

has order q−1 and hence is a generatorof k×.

Hence, from the long exact sequences associated to the two short ones above, we conclude that

HrT (G,U

(i+1)L ) = Hr

T (G,U(i)L ) for all r and all i ≥ 0. Now to show that Hr

T (G,UL) is trivial, againby periodicity we may assume r > 0. Let f :Gr → UL be an r-cocycle and denote by d the (r − 1)-

th coboundary map. Since Hr(G,U(1)L ) = Hr(G,UL), f differs by a coboundary from an r-cocycle

f1:Gr → U

(1)L , i.e., there exists g0:G

r−1 → UL such that f1 = f ·d(g0)−1. Proceeding in this manner, we

inductively construct functions gi:Gr−1 → U

(i)L such that f · d(g0g1 . . . gi)

−1 is an r-cocycle with values

in U(i+1)L . Then the product g0g1 . . . gi converges to a function g:Gr−1 → UL and we have that dg = f ,

that is, the class [f ] is trivial in Hr(G,UL).

The case r = 0 is of special interest, since it gives a proof of

Corollary 2.2 (Norm groups of unramified extensions) Let L ⊃ K be the unramified extensionof degree n and let π ∈ K be a common uniformiser. Then the norm map NL/K :UL → UK is surjectiveand hence

NL/KL× = πnZ · UK

Back to the computation of H2(G,L×). From (†) and the fact that UL has trivial cohomology, weconclude that the valuation v induces an isomorphism H2(G,L×) = H2(G,Z). On the other hand, we

have another isomorphism given by connecting map δ:H1(G,Q/Z)≈→ H2(G,Z). Putting everything

together, we obtain a canonical isomorphism, called invariant map,

invL/K :H2(G,L×)≈-

1|G|Z

Z

Unramified Cohomology 23

given by the composition

H2(G,L×)v

≈- H2(G,Z) δ

≈H1(G,Q/Z) = Hom(G,Q/Z)

f 7→f(Φ)-1|G|Z

Z

where we write Φ ∈ G for the Frobenius automorphism.

If M ⊃ L is another unramified extension with H = Gal(M/K) then following the isomorphismsabove we obtain a commutative diagram

H2(H,M×)invM/K

≈-

1|H|Z

Z

H2(G,L×)

inf

∪

6

invL/K

≈-

1|G|Z

Z

∪

6

where the left vertical arrow is given by inflation and the right vertical one is the inclusion map. Hencethe invariant maps for the various unramified extensions of K fit together into a single invariant map

invK :H2(GnrK ,K

×

nr)≈- Q/Z

Here GnrK denotes the Galois group of the maximal unramified extension of K.

Theorem 2.3 (Functorial property of the invariant map) Let K ′ ⊃ K be an arbitrary (possiblyramified) finite extension of local fields. We have a commutative diagram

H2(GnrK′ ,K ′×

nr)invK′

≈- Q/Z

H2(GnrK ,K

×

nr)

res

6

invK

≈- Q/Z

[K ′ : K]

6

where the left vertical map is restriction and the right vertical one is multiplication by [K ′ : K]. Hence ifK ′ ⊃ K is an arbitrary finite Galois extension with G = Gal(K ′/K) then H2(G,K ′×) contains a cyclicgroup of order |G|.

Proof First we note that K ′nr = Knr ·K ′ so that the restriction map on H2 is well-defined. Let e and

f be the ramification and inertia degrees of K ′ ⊃ K and denote by v and v′ the normalised valuations

of K and K ′ respectively so that v′|K = e · v. Observe that ΦK′ |Knr = ΦfK . Hence, from the definition

of the invariant map, we obtain a commutative diagram

H2(GnrK′ ,K ′×

nr)v′- H2(Gnr

K′ ,Z) - Homct(GnrK′ ,Q/Z) - Q/Z

H2(GnrK ,K

×

nr)

res

6

v- H2(GnrK ,Z)

e · res6

- Homct(GnrK ,Q/Z)

e · res6

- Q/Z

ef

6

Since ef = [K ′ : K], the first result follows.

To show the second result, assume that K ′ ⊃ K is finite Galois with G = Gal(K ′/K). Using theinflation-restriction sequence and Hilbert 90, we conclude that the inflation maps

inf:H2(GnrK ,K

×

nr) → H2(GK ,K×

sep) and inf:H2(GnrK′ ,K ′×

nr) → H2(GK′ ,K×

sep)


are injective. Identifying H2(GnrK ,K

×nr) and H2(Gnr

K′ ,K ′×nr) with Q/Z via invK and invK′ and using the

result just proven, we obtain a commutative diagram

0 - H2(G,K ′×)inf- H2(GK ,K

×

sep)res - H2(GK′ ,K×

sep)

H2(GnrK ,K

×

nr)

inf

∪

6

res - H2(GnrK′ ,K ′×

nr)

inf

∪

6

Q/Z

invK

w

w

w

w

w

w

w

w

w

[K ′ : K] - Q/Z

invK′

w

w

w

w

w

w

w

w

w

We conclude that H2(G,K ′×) contains a subgroup isomorphic to1

[K′:K]Z

Z, that is, a cyclic group of order

|G| = [K ′ : K].

In order to show that H2(G,K ′×) actually equals1

[K′:K]Z

Z, we shall bound the order of H2(G,K ′×)

from above using a counting argument. Since G is solvable, it will be enough to do that assuming Gcyclic. This is done in the next section.

3 Brauer group of a Local Field

We begin by introducing a very useful tool in the cohomology of cyclic groups that will help simplify ourcounting argument.

Definition 3.1 Let G be a cyclic group and M be a G-module whose Tate cohomology groups are allfinite. We define its Herbrand quotient as

h(G,M) =|H0

T (G,M)||H1

T (G,M)|

The Herbrand quotient plays the same role as the Euler characteristic in Topology. We have twomain computational lemmas:

Lemma 3.2 (Multiplicativity) Let G be a cyclic group and consider an exact sequence of G-modules

0→M ′ →M →M ′′ → 0

If two of the Herbrand quotients h(G,M), h(G,M ′), h(G,M ′′) are defined (i.e. have finite Tate coho-mology groups) then so is the third and

h(G,M) = h(G,M ′) · h(G,M ′′)

Proof From the periodicity of cohomology of cyclic groups, the long exact sequence associated to theabove short one becomes an “exact hexagon”

H0T (G,M ′)

f0- H0

T (G,M)g0- H0

T (G,M ′′)

H1T (G,M ′′)

δ16

g1

H1T (G,M) f

1

H1T (G,M ′)

δ0

?

The first result follows directly from from this hexagon. The second result also follows from this hexagonby elementary counting, as one has

|H0T (G,M ′)| = | ker f0| · | ker g0| |H1

T (G,M ′)| = | ker f1| · | ker g1||H0

T (G,M)| = | ker g0| · | ker δ0| |H1T (G,M)| = | ker g1| · | ker δ1|

|H0T (G,M ′′)| = | ker δ0| · | ker f1| |H1

T (G,M ′′)| = | ker δ1| · | ker f0|

Brauer group of a Local Field 25

Lemma 3.3 (Finite Index Invariance) Let G be a cyclic group, M be a G-module and M ′ be aG-submodule of finite index. Then h(G,M ′) is defined if and only if h(G,M) is defined, in which caseh(G,M ′) = h(G,M).

Proof By the last lemma, it suffices to show that if M is a finite group then h(G,M) = 1. Letσ be a generator of G. Since M is a finite group one has that |H0

T (G,M)| = |MG|/|NG(M)| and|H1

T (G,M)| = | kerNG|/|(σ − 1) ·M |. But |M | = | kerNG| · |NG(M)| and similarly (since MG is thekernel of multiplication by σ − 1) one has that |M | = |MG| · |(σ − 1) ·M |, and the result follows.

Let L ⊃ K be a cyclic extension of local fields with Galois group G of order n. We apply the aboveto the exact sequence of G-modules

0→ UL → L× → Z→ 0

induced by the valuation of L. It is easy to compute h(G,Z) = n and we have

Theorem 3.4 With the above notation and hypotheses, h(G,UL) = 1.

Proof By the previous lemma, it is enough to show that UL contains an induced G-submodule of finiteindex. First assume that charK = 0 and let π be a uniformiser of K. But we have an isomorphism of

G-modules U(i)L∼= mi

L for i sufficiently large, given by the log map. On the other hand, for j sufficiently

large miL ⊃ πjOL

∼= OL. Since UL/U(i)L and mi

L/πjOL are finite, it is enough to show that OL contains

an induced module of finite index. Now let ω1, . . . , ωn be a normal basis of L ⊃ K; multiplying by aconvenient power of π we may assume that ω1, . . . , ωn ∈ OL. Since OL is a finite OK -module, we havethat the induced module M = OKω1 + · · ·+OKωn has finite index in OL.

Now we sketch a proof that works even if charK 6= 0. Let M be as above. Multiplying the ωi by asufficiently large power of π we may assume that M ·M ⊂ πM . Then we may consider the submodule offinite index V = 1+M of UL and the filtration given by V (i) = 1+πiM for i ≥ 0. It is easy to show thatwe have an isomorphism of G-modules V (i)/V (i+1) ∼= M/πM , which has trivial cohomology since thelatter is induced. As in the proof of theorem 2.1, this implies that V itself has trivial cohomology.

Hence we get h(G,L×) = h(G,UL) · h(G,Z) = n. Since H1(G,L×) is trivial by Hilbert 90, weconclude that |H2(G,L×)| = n when G is cyclic. In general, for an arbitrary Galois extension we have

Theorem 3.5 Let L ⊃ K be an arbitrary finite Galois extension of local fields with G = Gal(L/K).Then the group H2(G,L×) is cyclic of order |G|.Proof Since H2(G,L×) contains a cyclic group of order |G| by theorem 2.3, it is enough to show thatthe order of H2(G,L×) divides |G|. For that, we use the special cyclic case above together with the factthat G is solvable. Alternatively, one may use the fact that res:H2(G,L×) → H2(Gp, L

×) defines aninjection on the p-primary components, where Gp is any p-Sylow subgroup of G (see appendix), and wemay work with the solvable group Gp instead of G.

The proof is by induction on |G|. We already know the result for G cyclic. In general, let H ⊳ Gbe a normal subgroup such that H is cyclic and non-trivial. By Hilbert 90, we have an exact inflation-restriction sequence

0→ H2(G/H, (LH)×)→ H2(G,L×)→ H2(H,L×)

Hence the order of H2(G,L×) divides the product of the orders of H2(H,L×) and H2(G/H, (LH)×),which in turn divides |H | · |G/H | = |G| by induction hypothesis, and we are done.

As a corollary, we obtain the following result that allows us to extend the invariant map of lastsection to the whole group H2(GK ,K

×sep). This map plays an important role in the study of division

algebras over a local field.

Theorem 3.6 (Brauer group of a local field) Let K be a local field. We have an isomorphism

invK :H2(GK ,K×

sep)≈- Q/Z

obtained by composing the inflation map inf:H2(GnrK ,K

×nr)

≈→ H2(GK ,K×sep) and the invariant map

invK :H2(GnrK ,K

×nr)

≈→ Q/Z of last section.


Proof By Hilbert 90 and the inflation-restriction sequence, we have an inclusion inf:H2(GnrK ,K

×nr) →

H2(GK ,K×sep), which we now show to be also surjective. Since

H2(GK ,K×

sep) = lim−→K′ finite

Galois over K

H2(

Gal(K ′/K),K ′×)

,

given any γ ∈ H2(GK ,K×sep) we can find a finite Galois extension K ′ such that

γ ∈ H2(

Gal(K ′/K),K ′×)

= ker(

H2(GK ,K×

sep)res- H2(GK′ ,K×

sep))

(by Hilbert 90 and the inflation-restriction sequence the inflation maps in the above limit are injective,so we may view H2

(

Gal(K ′/K),K ′×)

as a subgroup of H2(GK ,K×sep) and we write γ also for the

corresponding element in this subgroup). Now by theorem 2.3 and the above theorem, we have thatinf:H2(Gnr

K ,K×nr) → H2(GK ,K

×sep) allows us to make the identification

ker(

H2(GnrK ,K

×

nr)res- H2(Gnr

K′ ,K ′×nr)

)

= ker(

H2(GK ,K×

sep)res- H2(GK′ ,K×

sep))

(see the second diagram of the proof of theorem 2.3). Hence γ belongs to this kernel and a fortiori to(the inflation of) H2(Gnr

K ,K×nr).

Corollary 3.7 All division algebras over a local field K are cyclic. More precisely, if χ ∈ H1(GK ,Z/n)is the character given by χ(ΦK) = 1, then

invK(χ, π) = − 1

nmod Z

Proof Since Br(K) = Br(Knr/K), every division algebra is split by some finite unramified extensionL ⊃ K. But every such extension is cyclic, and therefore the division algebra is cyclic. For the formulaabove, observe that the class of (χ, πK) in Br(K) is just −πK ∪ δχ, and invL/K(πK ∪ δχ) = 1

n mod Z ascan be checked by an explicit calculation.

4 Local Reciprocity

The computation of the Brauer group of a local field K gives also a very simple and nice description ofthe abelian extensions of K as a byproduct, which will be needed in the next chapter.

4.1 Statements of the main theorem

Theorem 4.1.1 (Local Artin Reciprocity) Let K be a local field. There exists a unique groupmorphism, called local Artin map,

θK :K× → GabK

such that the following holds: for any finite abelian extension L ⊃ K, the map θL/K :K× → Gal(L/K)(also referred to as local Artin map) given by the composition

K× θK- GabK

canonical-- Gal(L/K)

satisfies:

1. θL/K is surjective with kernel given by the norm group NL/KL×. Thus we have an induced

isomorphism (which we still denote by θL/K)

θL/K :K×

NL/K(L×)≈ Gal(L/K)

2. if L ⊃ K is unramified, ΦL/K ∈ Gal(L/K) denotes the corresponding Frobenius map, andv:K× → Z denotes the normalised valuation of K, then for all a ∈ K×

θL/K(a) = Φv(a)L/K

Local Reciprocity 27

Example 4.1.2 Consider the Galois extension M = Q3(√

2,√

3) of Q3 with Galois group Gal(M/Q3) ∼=Z/2× Z/2, generated by automorphisms σ and τ given by

σ(√

2) = −√

2σ(√

3) =√

3

τ(√

2) =√

2τ(√

3) = −√

3

The lattice of subfields is

M = Q3(√

2,√

3)

L0 = Q3(√

2)

τ

L1 = Q3(√

3)

σ

L2 = Q3(√

6)

στ

K = Q3

τ |L1 σ|L2= τ |L2σ|L0

Now we now identify the corresponding norm subgroups in Q×

3 = 3Z × ±1 × U(1)Q3

. Observe that

since the indices of these subgroups divide 4 and U(1)Q3

∼= Z3 is 2-divisible, all of them contain U(1)Q3

.

Since L0 ⊃ Q3 is unramified we know that NL0/Q3(L×

0 ) = 32Z × UQ3 . Moreover −3 ∈ NL1/Q3(L×

1 ) and

−6 ∈ NL2/Q3(L×

2 ), thus 3 ∈ NL2/Q3(L×

2 ) since −2 ∈ U (1)Q3

. Also since M is the compositum of L0 and

L2 we have that NM/Q3(M×) = NL0/Q3

(L×

0 ) ∩NL2/Q3(L×

2 ). Putting everything together, we obtain thefollowing lattice of subgroups of Q×

3 , drawn upside down:

NM/Q3(M×) = 32Z × U (1)

Q3

NL0/Q3(L×

0 ) = 32Z × ±1 × U (1)Q3

NL1/Q3(L×

1 ) = (−3)Z × U (1)Q3

NL2/Q3(L×

2 ) = 3Z × U (1)Q3

Q×

3 = 3Z × ±1 × U (1)Q3

Finally, we have that

θL0/Q3

(

−1 ·NL0/Q3(L×

0 ))

= 1

θL1/Q3

(

−3 ·NL1/Q3(L×

1 ))

= 1

θL2/Q3

(

3 ·NL2/Q3(L×

0 ))

= 1

⇒

θM/Q3

(

−1 ·NM/Q3(M×)

)

= τ

θM/Q3

(

−3 ·NM/Q3(M×)

)

= σ

θM/Q3

(

3 ·NM/Q3(M×)

)

= στ

which completely determines θM/Q3.

4.2 Tate-Nakayama theorem

Let G be a finite group. In this section, we prove a purely group theoretic result giving an isomorphism

Gab ≈ CG

NG(C)

where C is a G-module satisfying some conditions and NG:C → CG denotes the norm map of C (seeappendix). The methods of this section are inspired in computations of Algebraic Topology, with thestarting point being the well-known relation H1(X,Z) = π1(X)ab between the first singular homologygroup of a topological space X and the maximal abelian quotient π1(X)ab of the fundamental group ofX . Since Galois extensions are algebraic analogs of covering spaces in Topology, this turns out to be aquite natural point of view (if, of course, you’ve studied Algebraic Topology before, but don’t worry ifyou haven’t, the proofs below are purely algebraic, but it’s a good idea to eventually look at the sourceof inspiration for them).


Theorem 4.2.1 (Twin number vanishing criterion) Let G be a finite group and M be a G-module.If there are two consecutive numbers i and i+ 1 for which

HiT (H,M) = Hi+1

T (H,M) = 0 for all subgroups H ≤ G

then HrT (G,M) = 0 for all r ∈ Z.

Proof First observe that by dimension shifting (see appendix) it is enough to show the “weaker”conclusion Hr

T (G,M) = 0 for all r ≥ 1. In fact, writing an exact sequence of G-modules

0→ N → P →M → 0

for some induced G-module P (which is thus also induced as an H-module for all H ≤ G) we obtain an

isomorphism Hj+1T (H,N) = Hj

T (H,M) for all H ≤ G and j ∈ Z. Hence if we know that HrT (H,M) = 0

for all r ≥ 1 then we know that HrT (H,N) = 0 for all r ≥ 2, and applying the “weak twin number

criterion” to N in place of M we conclude that H1T (H,N) = 0 as well, that is H0

T (H,M) = 0, so that theconclusion of the weak criterion holds also for r ≥ 0. Proceeding inductively in this manner, we extendthe result to all r ∈ Z.

A similar proof using dimension shifting (check!) also allows us to assume that i = 1 (or any otherfixed number we deem convenient). Hence from now on we assume that H1(H,M) = H2(H,M) = 0 forall H ≤ G and prove that these conditions imply that Hr(G,M) = 0 for all r ≥ 1.

Since Hr(G,M) is a torsion abelian group, it is enough to show that its p-primary componentvanishes for all prime numbers p. Let Gp be a p-Sylow subgroup of G. A restriction-corestrictionargument shows that res:Hr(G,M)→ Hr(Gp,M) is injective on p-primary components (see appendix),thus it is enough to show that Hr(Gp,M) = 0.

Hence we may assume that G is solvable and proceed by induction on the order of G. If G iscyclic, the theorem follows from the periodicity of cohomology. Now let H ⊳ G be a proper normalsubgroup such that G/H is cyclic. By induction Hr(H,M) = 0 for all r ≥ 1 and hence we have an exactinflation-restriction sequence

0→ Hr(G/H,MH)→ Hr(G,M)→ Hr(H,M) = 0

for all r ≥ 1. Therefore Hr(G/H,MH) = Hr(G,M) for all r ≥ 1 and in particular H1(G,M) =H2(G,M) = 0 implies that H1(G/H,MH) = H2(G/H,MH) = 0. But G/H is cyclic, so periodicityyields Hr(G/H,MH) = 0 for all r ≥ 1, and hence Hr(G,M) = 0 for all r ≥ 1 too.

Now we can prove the main result of this section.

Theorem 4.2.2 (Tate-Nakayama) Let G be a finite group and let C be a G-module such that for allsubgroups H ≤ G1. H1(H,C) = 0

2. H2(H,C) is cyclic of order |H |Let γ be a generator of H2(G,C). Then for all r ∈ Z the cup product with γ gives an isomorphism

HrT (G,Z)

∪γ

≈- Hr+2

T (G,C)

Observe that if γ ∈ H2(G,C) is a generator then res(γ) ∈ H2(H,C) is also a generator sincecor res(γ) = [G : H ] · γ has order |H |, hence res(γ) must have order |H | as well in view of 2.

Proof The key idea of the proof is to apply a “double dimension shifting” (wow!) and for that weconstruct a cohomologically trivial G-module C(γ) fitting into an exact sequence

0→ C → C(γ)→ IG → 0 (∗)

Here IG = 〈σ − 1 | σ ∈ G〉 is the kernel of the augmentation map Z[G] → Z (see appendix), so that wehave an exact sequence of G-modules

0→ IG → Z[G]→ Z→ 0 (∗∗)


Once C(γ) is constructed, the proof of the theorem follows easily: from (∗∗), using the fact that Z[G]has trivial cohomology, we conclude that the connecting map δa:Hr

T (G,Z) ≈ Hr+1T (G, IG) is an isomor-

phism; similarly, from (∗), we have that the connecting map δc:Hr+1T (G, IG) ≈ Hr+2

T (G,C) is also an

isomorphism. The composition δc δa gives the desired isomorphism HrT (G,Z) ≈ Hr+2

T (G,C), whichequals to the cup product with γ, as we later show.

Let c be a 2-cocycle representing γ. Since we wish H2(G,C(γ)) to vanish, the idea is to constructC(γ) so that c becomes a coboundary in C(γ). Take C(γ) to be the direct sum of C with the free abeliangroup with basis xσ , σ ∈ G, σ 6= 1:

C(γ)df= C ⊕

⊕

σ∈Gσ 6=1

Zxσ

We extend the G-action from C to C(γ) in such a way that c becomes the coboundary of σ 7→ xσ:

σ · xτ = xστ − xσ + c(σ, τ)

where we interpret “x1” to be c(1, 1). The 2-cocycle relation then guarantees that 1 · xτ = xτ and(ρσ) · xτ = ρ · (σ · xτ ) hold for all ρ, σ, τ ∈ G (check!), turning C(γ) into a G-module containing C as aG-submodule. Finally the map C(γ) → IG in (∗) is given by xσ 7→ σ − 1 for σ 6= 1, and is identicallyzero on C. Another easy check shows that the latter map preserves the G-action.

In order to show that C(γ) is cohomologically trivial we apply the twin number vanishing criterion:we need to verify that H1(H,C(γ)) = H2(H,C(γ)) = 0 for all subgroups H ≤ G. Here the hypotheses1 and 2 of the theorem come into play. First observe that from (∗∗) and the explicit description of theconnecting map in terms of the standard resolution (see appendix) we obtain

(i) H1(H, IG) = H0T (H,Z) is cyclic of order |H | with generator given by the class [f ] of the 1-cocycle

f(σ) = σ − 1, σ ∈ H .

(ii) H2(H, IG) = H1(H,Z) = Hom(H,Z) = 0

From (∗) we have an exact sequence

0 = H1(H,C) - H1(H,C(γ)) - H1(H, IG)

δ- H2(H,C) - H2(H,C(γ)) - H2(H, IG) = 0

Hence everything falls through if we can show that the connecting map δ is an isomorphism. At least weknow that both H1(H, IG) and H2(H,C) are cyclic groups of order |H |, so it is enough to show that δ issurjective. We show by explicit computation that δ([f ]) = res(γ) where [f ] ∈ H1(H, IG) is as in (i). First

we lift f to the function f :H → C(γ) given by f(σ) = xσ. Now (df)(σ, τ) = σ · xτ − xστ + xσ = c(σ, τ)

for all σ, τ ∈ G, where d denotes the coboundary map, hence δ([f ]) = [df ] = [c] = res(γ), as required.

Finally, we verify that the composition δcδa of the two connecting maps δa:HrT (G,Z)

≈→ Hr+1T (G, IG)

and δc:Hr+1T (G, IG)

≈→ Hr+2T (G,C) is indeed the cup product with γ. Denote by µ ∈ H0

T (G,Z) = Z/|G|the generator 1 mod |G| and by φ = [f ] = δa(µ) ∈ H1(G, IG) where f is as in (i). By the compatibilityof cup products with the connecting maps (see appendix) we obtain, for all α ∈ Hr

T (G,Z),

α ∪ γ = α ∪ δc(φ) = (−1)r · δc(α ∪ φ) = (−1)r · δc(

α ∪ δa(µ))

= (−1)r · δc(

(−1)r · δa(α ∪ µ))

= (−1)r · δc(

(−1)r · δa(α))

= δc δa(α)

since − ∪ µ is the identity on HrT (G,Z).

Recall that (see appendix) H−2T (G,Z) = H−1

T (G, IG) = IG/I2G and that Gab = IG/I

2G via the

isomorphism σ · [G : G] 7→ (σ − 1) · I2G. As a consequence we obtain

Corollary 4.2.3 (“Abstract” Reciprocity) With the notation and hypotheses of the Tate-Nakayamatheorem, we have an isomorphism

CG

NG(C)= H0

T (G,C) ≈ H−2T (G,Z) = Gab


4.3 Reciprocity law

We wish to apply the Tate-Nakayama’s theorem to G = Gal(L/K) and C = L× where L ⊃ K is a finiteGalois extension of local fields. Hence we need to verify that, for all subgroups H ≤ G,

1. H1(H,L×) = 0;

2. H2(H,L×) is cyclic of order |H |.The first condition is just Hilbert Satz 90 (see appendix), and the second one follows from the computationof Br(K). We can now make the following

Definition 4.3.1 For any finite abelian extension L ⊃ K of local fields, we define the local reciprocitymap θL/K :K× → Gal(L/K) as the composition of the natural projection map K×

։ K×/NL/KL× with

the inverse of the Tate-Nakayama isomorphism

Gal(L/K)≈→ K×

NL/KL×

given by the cup product with the unique element γ ∈ H2(Gal(L/K), L×) such that invL/K(γ) =1

[L:K] mod Z. Such element γ is called a fundamental class.

To prove that the above isomorphism satisfies the two properties of theorem 4.1.1, we describe θL/K

in terms of characters. Let L ⊃ K be as above with G = Gal(L/K). Recall that we have two exactsequences of G-modules

0→ Z→ Q→ Q/Z→ 0 (∗)

0→ IG → Z[G]→ Z→ 0 (∗∗)

where Q and Z[G] have trivial cohomology, hence the connecting maps

δ:H1(G,Q/Z)≈→ H2(G,Z)

δa:H−2T (G,Z)

≈→ H−1T (G, IG)

are isomorphisms. Besides we have a natural isomorphism Gab = H−1T (G, IG) given by

Gab ≈→ H−1T (G, IG) =

IGI2G

σ · [G : G] 7→ (σ − 1) · I2G

Theorem 4.3.2 (Local Reciprocity Revisited) With the above notation, for any a ∈ K× and anycharacter χ ∈ H1(G,Q/Z) = Hom(G,Q/Z) one has

χ(

θL/K(a))

= invK(a ∪ δχ)

where a is the image of a in H0T (G,L×) = K×/NL/KL

×. In other words, the cup product gives a perfectpairing

H0T (G,L×)⊗H2

T (G,Z)∪- H2

T (G,L×) ⊂inf- H2

T (GK ,K×

sep)

K×

NL/KL×⊗Hom(G,Q/Z)

id⊗δ ≈6

-1|G|Z

Z

≈?

⊂ - Q/Z

≈ invK

?


Proof Let γ ∈ H2(G,L×) be the fundamental class and n = |G|. Given a ∈ K× and χ ∈ H1(G,Q/Z),

write σdf= θL/K(a) ∈ G and χ(σ) = i

n mod Z with 0 ≤ i < n. Since invK(γ) = 1n mod Z, in order to

show that χ(σ) = invK(a ∪ δχ) we have to show that a ∪ δχ = i · γ in H2(G,L×). However by thevery definition of θL/K we have that σ ∪ γ = a, where σ ∈ H−2

T (G,Z) denotes the image of σ under the

isomorphism H−2T (G,Z) = Gab = G. Hence

a ∪ δχ = σ ∪ γ ∪ δχ = σ ∪ δχ ∪ γ

and we are left to show that σ ∪ δχ = i mod n in H0T (G,Z). Note that σ ∪ δχ = δ(σ ∪ χ) and by the

lemma below σ ∪ χ ∈ H−1T (G,Q/Z) is represented by χ(σ) = i

n mod Z ∈ Q/Z. On the other hand, the

connecting map δ:H−1T (G,Q/Z)

≈→ H0T (G,Z) is induced by the norm map (see appendix), which in our

case is just multiplication by n. Hence δ(σ ∪ χ) = i mod n, as required.

Finally, the above identity implies that the pairing

H0T (G,L×)⊗H1

T (G,Q/Z)id⊗δ

≈- H0

T (G,L×)⊗H2T (G,Z)

∪- H2T (G,L×)

is perfect. In fact, if χ ∈ H1T (G,Q/Z) is such that a ∪ δχ = 0 ⇐⇒ χ(θL/K(a)) = 0 for all a ∈ K× then

χ = 0 since θL/K :K×։ Gal(L/K) is surjective. On the other hand, if a ∈ K× is such that a ∪ δχ =

0 ⇐⇒ χ(θL/K(a)) = 0 for all χ ∈ H1T (G,Q/Z) then clearly θL/K(a) = 0 ⇐⇒ a ∈ ker θL/K = NL/KL

×

and thus a = 0, showing that the left kernel of this pairing is also trivial.

Lemma 4.3.3 Let G be a finite group and A and B be G-modules. Let f :G→ B be a 1-cocycle.

1. for a ∈ ker(ANG- A), the cup product of [a] ∈ H−1

T (G,A) and [f ] ∈ H1T (G,B) is given by

[a] ∪ [f ] = −[

∑

σ∈G

σ(a)⊗ f(σ)]

∈ H0T (G,A⊗B)

Here brackets denote the corresponding cohomology classes.

2. for σ ∈ G denote by σ ∈ H−2T (G,Z) the image of σ under the isomorphism Gab ≈ H−2

T (G,Z).Then

σ ∪ [f ] = [f(σ)] ∈ H−1T (G,B)

Proof We use dimension shifting. Write an exact sequence of G-modules

0→ B → B′ → B′′ → 0

with B′ induced, and such that it is split as a sequence of abelian groups, so that

0→ A⊗B → A⊗B′ → A⊗B′′ → 0

is still exact (check the appendix for more details). Then the connecting map δ:H0T (G,B′′)

≈→ H1T (G,B)

is an isomorphism and thus we may write [f ] = δ[b′′] for some b′′ ∈ B′′G and

[a] ∪ [f ] = [a] ∪ δ[b′′] = −δ[a⊗ b′′]

On the other hand, the connecting map δ:H−1T (G,A ⊗ B′′) → H0

T (G,A ⊗ B) is induced by the norm.There exists a pre-image b′ ∈ B′ of b′′ such that f(σ) = σ(b′) − b′ ∈ B for all σ ∈ G. Since a ⊗ b′ is apre-image of a⊗ b′′ under A⊗B′ → A⊗B′′, we conclude that δ[a⊗ b′′] is represented by

NG(a⊗ b′) =∑

σ∈G

σa⊗ σ(b′) =∑

σ∈G

σa⊗ f(σ) +∑

σ∈G

σa⊗ b′

=∑

σ∈G

σa⊗ f(σ) +NG(a)⊗ b′ =∑

σ∈G

σa⊗ f(σ)

To show 2, observe that tensoring the augmentation sequence (∗∗) with B we obtain an exactsequence

0→ IG ⊗B → Z[G]⊗B → Z⊗B = B → 0


since Z is a free Z-module and hence TorZ1 (B,Z) = 0. Since Z[G]⊗B is induced (see appendix), we have

that the connecting map δa:H−1T (G,B)

≈→ H0T (G, IG ⊗ B) is an isomorphism, hence it suffices to show

that δa(σ ∪ [f ]) = δa[f(σ)]. Since δa(σ ∪ [f ]) = δa(σ)∪ [f ] and δa(σ) = [σ− 1] ∈ H−1T (G, IG), applying 1

we obtain

δa(σ ∪ [f ]) = [σ − 1] ∪ [f ] = −[

∑

τ∈G

τ(σ − 1)⊗ f(τ)]

=[

∑

τ∈G

τ ⊗ f(τ) −∑

τ∈G

τσ ⊗ f(τ)]

=[

∑

τ∈G

τσ ⊗ f(τσ) −∑

τ∈G

τσ ⊗ f(τ)]

=[

∑

τ∈G

τσ ⊗ τf(σ)]

∈ H0T (G, IG ⊗B)

On the other hand, since δa:H−1T (G,B)

≈→ H0T (G, IG⊗B) is induced by the norm and 1⊗f(σ) ∈ Z[G]⊗B

is a pre-image of f(σ) ∈ B = Z⊗B we have that

δa[f(σ)] =[

NG

(

1⊗ f(σ))]

=[

∑

τ∈G

τ ⊗ τf(σ)]

∈ H0T (G, IG ⊗B)

Hence the two classes δa(σ ∪ [f ]) and δa[f(σ)] are equal in H0T (G, IG ⊗ B) since

∑

τ∈G

τσ ⊗ τf(σ) −∑

τ∈G

τ ⊗ τf(σ) =∑

τ∈G

τ(σ − 1)⊗ τf(σ) = NG

(

(σ − 1)⊗ f(σ))

∈ NG(IG ⊗B)

Using the description of the reciprocity map given in theorem 4.3.2, it is easy to show that for an

unramified extension of local fields L ⊃ K one has θL/K(a) = Φv(a)L/K , where v denotes the normalised

valuation of K and ΦL/K is the Frobenius automorphism of L ⊃ K. In fact, let n = [L : K] and

G = Gal(L/K) and consider the character χ:G → Q/Z given by χ(ΦL/K) = 1n mod Z. Then a ∪ δχ is

represented by the 2-cocycle c:G×G→ L× given by

c(ΦiL/K ,Φ

jL/K) =

a if i+ j ≥ 00 otherwise

0 ≤ i, j < n

Now since the invariant map is given by the composition

H2(G,L×)v

≈- H2(G,Z) δ

≈H1(G,Q/Z) = Hom(G,Q/Z) ≈ Q/Z

and v([c]) = δ[f ] where f :G→ Q/Z is the 1-cocycle given by f(ΦL/K) = v(a)n mod Z, we conclude that

χ(

θL/K(a))

= invK(a ∪ δχ) = invK([c]) =v(a)

nmod Z = χ(Φ

v(a)L/K)

showing that θL/K(a) = Φv(a)L/K since χ is injective.

We can also show the compatibility of the various maps θL/K in the sense that given finite extensionsM ⊃ L ⊃ K with M and L abelian over K we have a commutative diagram

K×θM/K- Gal(M/K)

Gal(L/K)

can.

??

θL/K

-

This will prove that the maps θL/K fit together into a single map θK :K× → GabK , which will then satisfy

all the properties required in theorem 4.1.1.

Let a ∈ K×, let χ ∈ Hom(Gal(L/K),Q/Z) and denote by χ′ = inf(χ) ∈ Hom(Gal(M/K),Q/Z).Then inf(a ∪ δχ) = a ∪ δχ′ and thus

χ(θL/K(a)) = invK(a ∪ δχ) = invK(a ∪ δχ′) = χ′(θM/K(a)) = χ(θM/K(a)|L)

Since this holds for all χ ∈ Hom(Gal(L/K),Q/Z), we conclude that θL/K(a) = θM/K(a)|L, as was to beshown.

With the above, we finish the proof of the local reciprocity theorem. Hurray!

Chapter 3

BrauergroupofGlobalFields

By a global field we mean either a function field or a number field, that is, a finite extension of Fp(t)or a finite extension of Q, respectively. The main result of this chapter is the following computation:

Theorem 0.1 (Brauer group of global fields) Let K be a global field. One has an exact sequence

0 - Br(K)

∑

vresv-

⊕

v

Br(Kv)

∑

vinvKv- Q/Z - 0

Here Kv denotes the completion of K with respect to the place v of K, resv: Br(K) → Br(Kv) andinvKv : Br(Kv) → Q/Z are the usual restriction and invariant maps, and the direct sum runs over allplaces of K, including the archimedean ones.

The proof of the above theorem is the main step of the proof of the global reciprocity law. In termsof central simple algebras, it describes the local-global or Haße principle for division algebras: adivision algebra D over a global field K is completely determined by its restrictions D ⊗K Kv to thelocal fields Kv. In particular, D is trivial if and only if D ⊗K Kv is trivial for all v.

1 The geometric case

In the next section, we will present a proof that works over a general global field K. However thegeometric case, that is, the case when K is a finite extension of Fp(t), and therefore the function field ofa non-singular projective curve C, is much easier to deal with, and serves as inspiration for the generalmethod. We thus give a sketch of the proof of this special case first, assuming a few facts from the theoryof algebraic curves.

Here is a quick review of the facts we shall need. Given a non-singular projective curve C over afield k, we can view it as a collage of two Dedekind domains, namely the integral closures of k[t] andk[ 1t ] in its function field K = k(C). For technical reasons, we assume that k is algebraically closed in Kfrom now on. Each closed point P ∈ C0 (not necessarily defined over k) corresponds to a maximal idealmP of one of these two Dedekind domains, and hence P defines a discrete valuation vP :K → Z ∪ ∞:vP (h) equals the exponent of mP in the factorisation of the principal fractional ideal (h). In geometricterms, vP (h) measures the order of the zero or pole of the function h at the point P .

A divisor in C is just an element of the free abelian group generated by the closed points of C:

Div(C) =⊕

P∈C0

ZP

Roughly, Div(C) corresponds to the multiplicative group of non-zero fractional ideals of a Dedekinddomain, written additively. The degree degD of a divisor D =

∑

P nPP ∈ Div(C), nP ∈ Z, is definedas a weighted sum of the finitely many non-zero coefficients nP of D: degD =

∑

P nP fP ∈ Z, wherefP = [k(P ) : k] is the residue degree of P , that is, the residue degree of the discrete valuation vP overits restriction to k[t] or k[ 1t ].

Each element h ∈ K× defines a principal divisor

div(h)df=

∑

P∈C0

vP (h)P

since vP (h) = 0 for all but finitely many P . A principal divisor corresponds to a principal fractionalideal in a Dedekind domain. An explicit computation shows that deg div(h) = 0 for any principal divisor

34 Brauer group of Global Fields

div(h) (do it first for P1k and then use the formula

∑

i eifi = n for extension of primes in Dedekinddomains). It is also easy to check that div(h) = 0 ⇐⇒ h ∈ k× (only constant functions have no zerosor poles).

The Picard group Pic(C) of C is defined as the quotient of Div(C) by the subgroup of principaldivisors. It corresponds to a globalised form of the usual class group of a Dedekind domain. We havethus an exact sequence

0 - k× - K× div- Div(C) - Pic(C) - 0

Since the degree of a principal divisor is zero, the degree of an element of Pic(C) is well-defined.The kernel of the degree map is denoted by Pic0(C). If C has k-rational points (i.e. points with residuedegree 1, or points “defined over k”), the degree map is surjective and we have an exact sequence

0 - Pic0(C) - Pic(C)deg- Z - 0

End of the review. From now on, let C be a non-singular projective curve defined over a finite fieldk, and let K = k(C) be its function field, which is a global field. Let us assume for simplicity thatC has a k-rational point (even though the proof can be harnessed to eliminate this hypothesis). Thecomputation of Br(K) is based on the following theorem:

Theorem 1.1 Let C be a non-singular projective curve defined over a finite field k. Write C = C×Spec k

Spec ksep for the curve over ksep obtained by base change from C, and K = K · ksep for the function field

of C. Then

1. Pic0(C) is a torsion abelian group;

2. (Lang) Hi(Gk,Pic0(C)) = 0 for i > 0;

3. (Tsen) Br(K) = 0.

Proof Omitted. The proof of (1) follows from the fact that there is a natural bijection Pic0(C) =JC(ksep) where JC is the Jacobian variety of C (see Milne’s notes “Abelian varieties”, for instance),and JC(ksep) is the union of the finite groups JC(l) where l runs over all finite extensions of k. On the

other hand, (2) holds more generally for any connected algebraic group instead of Pic0 by Lang’s articleAlgebraic groups over finite fields, Amer. J. Math. 78 (1956), 555–563. For a proof of Tsen’s theorem,see Shatz’s book for instance.

From Lang’s theorem and the long exact sequence associated to the short one

0 - Pic0(C) - Pic(C)deg- Z - 0

we obtain an isomorphism deg:Hi(Gk,Pic(C))≈→ Hi(Gk,Z) for i > 0 (actually, these groups are 0 for

i > 2 since Gk∼= Z has cohomological dimension 1). Finally, from the exact sequence of Gk-modules

0 - K×

/k×

sep- Div(C) - Pic(C) - 0

and the above computations, we obtain an exact sequence

H1(Gk,Div(C))deg- H1(Gk,Z)

- H2(Gk,K×

/k×

sep) - H2(Gk,Div(C)) - H2(Gk,Z)

- H3(Gk,K×/k×

sep) = 0

(∗)

Here the last term vanishes since Gk has cohomological dimension 1. Since we are assuming that C hasa k-rational point, we have that the map deg is surjective. On the other hand, we already know fromthe computations of last chapter that H2(Gk,Z) = Q/Z. We also have that

H2(Gk,Div(C)) = H2(Gk,⊕

P∈C0

Ind ZP ) =⊕

P∈C0

H2(Gk,Z)

The general case 35

and each factor H2(Gk,Z) = Q/Z can be identified with Br(KP ), the Brauer group of the completionof K with respect to the valuation vP . Finally, using once more the fact that Gk has cohomologicaldimension 1, from the exact sequence

0 - k×

sep- K

× - K×

/k×

sep- 0

we obtain an isomorphism

H2(Gk,K×

) = H2(Gk,K×

/k×

sep)

But the first group is Br(K/K), which equals Br(K) by Tsen’s theorem.

Putting everything together, we obtain from (∗) the exact sequence

0 - Br(K) -⊕

P∈C0

Br(KP ) - Q/Z - 0

Following the isomorphisms, it is easy to check that the second map is just the restriction map while thethird is the sum of the invariant maps.

2 The general case

Now we turn to the general case. Let K be an arbitrary global field. One of the key steps in thecomputation of Br(K) of the geometric case was Lang’s theorem, which shows that H1(Gk,Pic0(C)) =H1(Gk, JC(ksep)) = 0. For number fields, we do not have an “arithmetic Jacobian” at our disposal.Instead, we shall define another object, the idele class group, which will play an analogous role to theJacobian (actually JC × Z) in the geometric case.

In what follows, K will denote an arbitrary global field, and for any place v of K (archimedean ornot), we write Kv for the completion of K with respect to v. For K a number field, an archimedeanplace v corresponds to an embedding σ:K → Kalg, which defines an absolute value

‖a‖v df=

|σ(a)| if σ is a real embedding|σ(a)|2 if σ is a complex embedding

a ∈ K

and Kv is isomorphic to R and C, respectively. On the other hand, when v is non-archimedean, that is,v:K ։ Z ∪ ∞ is a (surjective) valuation, we define its normalised absolute value by

‖a‖v df= N(v)−v(a) a ∈ K

where N(v) is the cardinality of the residue field of v (which is a finite field). This definition is made sothat the product formula holds:

∏

v

‖a‖v = 1 a ∈ K×

Here v runs over all places of v, both archimedean or not. For non-archimedean places we also write

Ov = x ∈ Kv | v(x) ≥ 0Uv = O×

v = x ∈ Kv | v(x) = 0

for its ring of integers and corresponding unit group, respectively.

Now let L ⊃ K be a finite Galois extension of global fields with G = Gal(L/K). Recall that G actstransitively on the set of places w1, . . . , wg of L that divide v, via σ(wi) = wi σ−1. For w | v, denote by

Gw = σ ∈ G | σw = w

the decomposition group of w. Since L is dense in Lw, each σ ∈ G extends uniquely to a continuous

isomorphism σ:Lw≈→ Lσw. In particular, if σ ∈ Gw, σ extends to an automorphism of Lw, and this

gives a natural map Gw≈→ Gal(Lw/Kv), which is an isomorphism.

It will be convenient to write Lv for the completion of L with respect to any place w dividing v.They will all be isomorphic to each other by the above, and the corresponding decomposition groupswill be conjugate in G. In particular, if G is abelian, the decomposition groups will be all equal and wewill be able to identify Gal(Lv/Kv) with a subgroup of G = Gal(L/K). In particular, if Lv ⊃ Kv isunramified, we will be able to view the Frobenius automorphism Φv of Lv as an element of G.


2.1 Ideles

We begin with a

Definition 2.1.1 The idele group IK of K is the multiplicative subgroup of∏

v K×v given by

IK = (av) ∈∏

v

K×

v | av ∈ Uv for all but finitely many v

where v runs over all places (both archimedean and non-archimedean) ofK. We turn IK into a topologicalgroup as follows: as S runs over all finite sets of places of K containing all archimedean ones, the subsetsof the form

∏

v∈S

Wv ×∏

v/∈S

Uv ⊂ IK , where Wv is an open subset of K×

v ,

define a basis of neighbourhoods of 1.

The idele group plays the role of the divisor group Div(C) of the geometric case. Note howeverthat it is actually larger when K = k(C) is the function field of a curve C, since in that case one has asurjective map

IK ։ Div(C)

(aP )P∈C0 7→∑

P∈C0

vP (aP )P

with a nontrivial kernel. Yet IK is not too large: it can be shown that IK is a locally compact topologicalgroup, and we will manage to compute its cohomology without too much effort.

The embeddings K → Kv allow us to view K× as a subgroup of IK via the diagonal map a 7→(. . . , a, a, a, . . .). An element of the image of K× → IK is called a principal idele; principal ideles willplay the role of principal divisors of the geometric case.

Definition 2.1.2 The idele class group CK of K is defined to be the quotient

CK =IK

K×

The idele class group will play the role of the Jacobian (actually JC ×Z) of the geometric case. It isnot a compact group nor a countable union of compact ones (as one would expect from a generalisationof a projective group variety), but it is “almost compact,” which will do for our purposes.

Example 2.1.3 The group IQ consists of the tuples (a∞, a2, a3, a5, . . .) ∈ R××Q×

2 ×Q×

3 ×Q×

5 ×· · · withap ∈ Up for all but finitely many primes p.

Given an idele α = (a∞, a2, a3, . . .) ∈ IQ, let

r = (signa∞)∏

p

pvp(ap) ∈ Q×

where signa∞ = ±1 depending on whether a∞ is positive or negative. This is is well-defined sincevp(ap) = 0 for all but finitely many primes p. We have αr−1 ∈ R>0 ×U2 ×U3× · · ·. On the other hand,if r ∈ Q× ∩ (R>0 × U2 × U3 × · · ·), then r = 1. Hence

CQ = IQ/Q× = R>0 × U2 × U3 × · · ·

Each Up is a compact group, but R>0 is not (the archimedean place is the “culprit” for the non-compactness of CQ).

Now let L ⊃ K be a finite Galois extension of number fields with G = Gal(L/K). There is aninjection IK → IL which maps the v-th component av to (av, . . . , av) ∈ L×

w1× · · · × L×

wgwhere the wi

denote the places of L dividing v. Hence we can view IK as a subgroup of IL. Moreover, the G-action onthe wi induces a G-action on IL, defined by σ(aw) = (bw), where bσw = σaw for σ ∈ G and (aw) ∈ IL.Pictorially,

(a1, . . . , ag) ∈ L×

w1× · · · × L×

wg

↓ σ(σa1, . . . , σag) ∈ L×

σw1× · · · × L×

σwg

The general case 37

Hence IL is a G-module, with “induced action” on each component over v. This fact will be quite usefullater since it will allow us to apply Shapiro’s lemma during the cohomology computations.

If (aw) ∈ IL is fixed by all σ ∈ G, taking σ ∈ Gw = Gal(Lw/Kv) we have that aw ∈ K×v for each

w | v, and hence IGL = IK . Therefore we may define a norm map

NL/K : IL → IK

α 7→∏

σ∈G

σ(α)

We can also define a norm map between idele class groups thanks to

Lemma 2.1.4 (Norm compatibility) The diagram

L× ⊂ - IL

K×

NL/K

?⊂ - IK

NL/K

?

commutes. Hence NL/K :CL → CK is well-defined.

Proof Let wdf= w1, w2, . . . , wg be the places dividing a fixed place v of K, and let σi ∈ G df

= Gal(L/K)be such that σiwi = w. Since τ ∈ G | τwi = w = Gwσi, we have that the w-th component of the normof b ∈ L×, viewed as an element of IL, is just

∏

1≤i≤g

∏

τ∈Gwσi

τ(b) =∏

τ∈G

τ(b) = NL/K(b)

Definition 2.1.5 For any global field K and any finite set S of places of K containing all archimedeanones, define the following subgroup of IK :

ISK

df= (av) ∈ IK | av ∈ Uv for v /∈ S =

∏

v∈S

K×

v ×∏

v/∈S

Uv

The next lemma is just an idelic version of the finiteness of the class number.

Lemma 2.1.6 (Idelic Class Group Finiteness) Let K be a global field. There exists a finite set Sof places of K such that

IK = K×ISK

Hence CK = ISK/(I

SK ∩K×).

Proof Take S to be the set of all archimedean places together with enough non-archimedean ones togenerate the whole class group of K. Then, given any (av) ∈ IK , consider the corresponding fractional

ideal a =∏

v pv(av)v , where now v runs over all non-archimedean places of K, and pv denotes the prime

ideal of OK corresponding to v. By the choice of S, there exists f ∈ K× and nv ∈ Z such that

(f) · a =∏

v∈Sv non-arch.

pnvv

But that amounts to saying that f · av ∈ Uv for v /∈ S, that is, f · (av) ∈ ISK .

2.2 Global reciprocity

In this section, we want to give a description of the abelian extensions of a global field in terms of theidele class group.


Definition 2.2.1 Let L ⊃ K be a finite Galois extension of global fields with abelian G = Gal(L/K).The global Artin reciprocity map θL/K is defined as the product of the local reciprocity maps:

θL/K : IK → G

(av) 7→∏

v

θLv/Kv(av)

Here, if v is archimedean and Lv ⊃ Kv is isomorphic to C ⊃ R, we define θLv/Kv(av) to be

the conjugation map if av < 0 and the identity map otherwise, so that we have a trivial reciprocityisomorphism (the “sign map”)

θC/R:R×

NC/RC×=

R×

R>0

≈→ Gal(C/R)

Also, observe that for all but finitely many v, the extension Lv ⊃ Kv is unramified and av ∈ Uv, henceθLv/Kv

(av) = 1. Hence the above product does make sense.

We know from the local reciprocity theorem that NL/KIL is contained in the kernel of θL/K . Weactually want to show that the principal ideles are also killed by θL/K and that

Theorem 2.2.2 (Global Artin Reciprocity) Let L ⊃ K be a finite Galois extension of global fieldswith abelian G = Gal(L/K). The global Artin map induces an isomorphism

θL/K :CK

NL/KCL

≈→ G

The relation between the Brauer group and the Artin reciprocity map is given by the following

Lemma 2.2.3 Let L ⊃ K be a finite abelian Galois extension of global fields with G = Gal(L/K).Consider the following statements:

1. the sequence

0 - Br(L/K) -⊕

v

Br(Lv/Kv)

∑

vinvv- Q/Z - 0

is a complex.

2. K× ⊂ ker θL/K

Then 1⇒ 2 unconditionally, and 2⇒ 1 if G is cyclic.

Proof Let χ ∈ H1(G,Q/Z) = Hom(G,Q/Z) and denote by δχ ∈ H2(G,Z) the image of χ under theisomorphism given by the connecting map associated to the short exact sequence

0 - Z - Q - Q/Z - 0

Consider the following diagram:

K× - IK

∏

vθv-

⊕

v

Gv

∏

- G

Br(L/K)

∪δχ?

-⊕

v

Br(Lv/Kv)

∪δχ? ∑

vinvv- Q/Z

∑

v χ

?====== Q/Z

χ

?

Cup product with δχ gives a morphism K×/NL/KL× = H0

T (G,L×) → H2(G,L×) = Br(L/K), which

is an isomorphism if G is cyclic and χ generates H1(G,Q/Z) (the “periodicity isomorphism”, see theappendix). Similarly, still denoting by χ its restriction to Lv ⊃ Kv, cup product with δχ gives morphismsK×

v → Br(Lv/Kv), which induce the second vertical map of the diagram: given (av) ∈ IK , for all but

The general case 39

finitely many v, av is a unit and the extension Lv ⊃ Kv is unramified, hence av ∈ UKv = NLv/KvULv

and av ∪ δχ = 0 (see corollary II.2.2 and theorem II.4.1.1).

Now the whole diagram commutes: the left square, by the compatibility of the restriction maps withthe cup product; the middle square, by the formula χ(θv(av)) = invv(av ∪ δχ) of the local reciprocitylaw (see theorem II.4.3.2); and finally, the right square, by the definition of group morphism.

Now assume (1) holds. To show that θL/K(a) = 1 for a ∈ K×, we have to show that χ(θL/K(a)) = 0

for all χ ∈ H1(G,Q/Z). But from the commutativity of the diagram, χ(θL/K(a)) =∑

v invv(a ∪ δχ),which vanishes since the bottom row is a complex by assumption. Conversely, assume that G is cyclicand (2) holds. Let χ be a generator of H1(G,Q/Z). Then the leftmost vertical arrow is surjective, henceany element of Br(L/K) can be written as a ∪ δχ for some a ∈ K×. But then again from the diagramwe conclude that

∑

v invv(a ∪ δχ) = χ(θL/K(a)) = 0, and hence the bottom row is a complex.

In other words, the computation of the Brauer group of a global field is essentially equivalent toproving Artin’s reciprocity law! As a first step toward this common goal, let us prove

Theorem 2.2.4 (Reciprocity Law for Cyclotomic Fields) Let K be a number field and let L ⊃ Kbe a subextension of the cyclotomic extension K(ζn) ⊃ K, where ζn denotes a primitive n-th root ofunity. Then statement (2) of the last lemma holds.

Proof First we show that it is enough to prove the theorem for K = Q and L = Q(ζn). This followsfrom the functorial properties of the local reciprocity map, which imply that the following diagramcommutes for any abelian extension L ⊃ Q (use the fact that the local reciprocity map is given by thecup product with the fundamental class and that the cup product commutes with corestriction):

IK

θKL/K- Gal(KL/K)

IQ

NK/Q

? θL/Q- Gal(L/Q)?

∩

Since the right vertical arrow is injective, if we can show that Q× ⊂ ker θL/Q then θKL/K(a)|L =θK/Q(NL/Q(a)) = 1⇒ θKL/K(a) = 1 for all a ∈ K×.

Let Qmc df=

⋃

n≥1 Q(ζn) be the maximal cyclotomic extension of Q. We have that

Gal(Qmc/Q) = lim←−n∈N

Gal(Q(ζn)/Q) = lim←−n∈N

(Z/n)× = Z×

We now construct a continuous map ψ: IQ → Gal(Qmc/Q), which we secretly know to coincide withthe inverse limit θQ: IQ → Gal(Qmc/Q) of the reciprocity maps θQ(ζn)/Q: IQ → Gal(Q(ζn)/Q) (they fittogether by the functorial properties of the local reciprocity maps). By construction, we will have thatQ× ⊂ kerψ, and then we will have to show that ψ = θQ.

The map ψ: IQ → Gal(Qmc/Q) will be the inverse limit of ψn: IQ → Gal(Q(ζn)/Q), defined asfollows. Let S be a finite set of places containing ∞ and the ramified ones (i.e., those dividing n). For(av) ∈ IQ, we use the approximation theorem to find f ∈ Q× such that

‖f − av‖v < min‖nav‖v, ‖av‖v ⇒ ‖1− f−1av‖v < ‖n‖v for each v ∈ S

Set

ψn(av)df=

∏

v/∈S

Φv(f−1av)v =

∏

v/∈S

θQv(ζn)/Qv(f−1av)

where Φv denotes the Frobenius automorphism of Qv(ζn), namely the element of Gal(Qv(ζn)/Qv) ⊂Gal(Q(ζn)/Q) given by Φv(ζn) = ζp

n, where p is the prime corresponding to the place v. In other words,ψn(av) is the automorphism given by ζn 7→ ζr

n, where r =∏

v/∈S ‖f−1av‖−1v mod n.

The product formula∏

v

‖a‖v = 1 a ∈ Q×


ensures that the definition of ψn does not depend on the choice of S or f . For instance, if g ∈ Q× isanother element such that ‖g−av‖v < min‖nav‖v, ‖av‖v for all v ∈ S, then since ‖f‖v = ‖g‖v = ‖av‖vfor all v ∈ S we have that

∏

v/∈S

‖f−1av‖−1v ≡

∏

v/∈S

‖g−1av‖−1v (mod n) ⇐⇒

∏

v∈S

‖f−1av‖v ≡∏

v∈S

‖g−1av‖v (mod n)

holds. It is also easy to check that ψm(av) = ψn(av)|Q(ζm) for m | n, so that we may define ψ(av) =lim←−ψn(av), and another routine check shows that the resulting map ψ: IQ → Gal(Qmc/Q) is continuous.

In particular, we get that Q× ⊂ kerψ for free since for a principal idele a ∈ Q× we may choosef = a. In other words, ψ is defined so as to coincide with the product of the local reciprocity mapsat the unramified places, and so that Q× ⊂ kerψ: this allow us to replace an arbitrary idele by anequivalent one which is “very close to 1” at the ramified places, so that the corresponding local reciprocitymaps also vanish there. Besides, for any finite subextension L ⊃ Q of Qmc ⊃ Q we also have thatNL/Q IL ⊂ kerψ|L (which is no surprise if we believe ψ equals the reciprocity map θQ). In fact, forβ ∈ IL, by the approximation theorem we may find b ∈ L× such that bβ is arbitrarily close to 1 at thearchimedean and ramified places of L. Since the norm map is continuous, NL/Q(bβ) = NL/Q(b) ·NL/Q(β)will also be arbitrarily close to 1 at the archimedean and ramified places of Q, and since we already knowthat NL/Q(b) ∈ Q× belongs to kerψ, in order to prove that NL/Q(β) ∈ kerψ|L it suffices to show that

NL/Q(bβ) ∈ kerψ|L. But now ψ(

NL/Q(bβ))

|L coincides with the product of the local maps at theunramified local extensions, and the result follows from the corresponding property of the local maps.

For each v, let ιv: Q×v → IK be v-th inclusion map given by ιv(a) = (. . . , 1, 1, a, 1, . . .) which equals

the identity map at the v-th component. Let v be a fixed place of Q and Qmcv

df=

⋃

n≥1 Qv(ζn) be the

maximal cyclotomic extension of Qv. We now show that the composition ψvdf= ψιv induces a continuous

map ψv: Q×v → Gal(Qmc

v /Qv) which coincides with the local reciprocity map θv. This will prove thatψ = θQ, finishing the proof of the theorem.

We now show that the image of ψv indeed lies in Gal(Qmcv /Qv) ⊂ Gal(Qmc/Q). We have to show

that, for any finite subextension L ⊃ Q of Qmc ⊃ Q with G = Gal(L/Q), ψv(a)|L belongs to thedecomposition group Gv = Gal(Lv/Qv) of v for all a ∈ Qv. Let M = LGv ; then Mv = Qv, and henceιv(a) ∈ NM/QIM . Thus ψv(a)|M is trivial, that is, ψv(a)|L ∈ Gv, as required.

Next we show that ψv = θv. First assume that v 6=∞. Let Qnrv ⊂ Qmc

v be the maximal unramifiedextension of Qv, and for a uniformiser π ∈ Q×

v of v, denote by Qπv the fixed field of θv(π). Then

Qmcv = Qnr

v · Qπv . Since the uniformisers π generate Q×

v , to prove that ψv = θv it suffices to show thatψv(π) and θv(π) agree on Qnr

v and Qπv . But both equal the Frobenius automorphism Φv on Qnr

v , sowe are left to show that ψv(π) is trivial on Qπ

v , or equivalently, on any finite subextension M ⊃ Qv ofQπ

v ⊃ Qv. Let L ⊂ Qmc be the fixed field of Gal(Qmcv /Qv) ⊂ Gal(Qmc/Q) so that M = Lv. We have that

π ∈ NLv/Qv(Lv)× since θv(π) is trivial on Lv, therefore ιv(π) ∈ NL/QIL, showing that ψv(π) is trivial on

M = Lv as well.

Now if v = ∞, θv(a) is the conjugation map if a < 0 and equals the identity otherwise. SinceR>0 = NC/RC× is contained in kerψv as well, ψv(a)2 = id for all a ∈ R×, and so ψv(a) must be theconjugation map if it is not trivial. Hence we just have to check that ψv is not trivial, and for that welook at its restriction to Q(i), which ramifies only at 2 and ∞ over Q. Therefore

1 = ψ(−7) = ψ∞(−7) · ψ2(−7) · ψ7(−7)

We have that −7 is a 2-adic norm so ψ2(−7) = 1. We have thus to show that ψ7(−7) is the conjugationmap. But ψ7(−7) is the Frobenius map i 7→ i7 = −i, so we are done.

2.3 The Brauer group of a global field

The computation of the Brauer group Br(K) of a global field and the proof of the reciprocity law is aconsequence of the following computation, which we shall dub “class field theory axioms.”

Theorem 2.3.1 (Class Field Theory Axioms) Let L ⊃ K be a Galois extension of global fields withabelian G = Gal(L/K). Then

1. (Cohomology of Ideles) For all i ≥ 0,

HiT (G, IL) =

⊕

v

HiT (Gv, (L

v)×)

The general case 41

2. (Cohomology of CL) Suppose further that G is cyclic of prime order p, and that µp ⊂ K×. Then

|H0T (G,CL)| = |G| and |H1

T (G,CL)| = 1

We postpone the proof of this theorem to the next section. We now derive some important conse-quences from it. Note that the vanishing of H1(G,CL) is the analogue of Lang’s theorem for a generalglobal field.

Corollary 2.3.2 Let L ⊃ K be a Galois extension of global fields with G = Gal(L/K). Then

1. H0T (G,CL) is finite, and its order divides |G|;

2. H1(G,CL) = 0;

3. H2(G,CL) is finite, and its order divides |G|.

Proof Let us first show that H1(G,CL) vanishes. First observe that the class field theory axiomH1(G,CL) = 1 holds for all cyclic extensions L ⊃ K of prime degree p without the extra hypothesisµp ⊂ K×. In fact, since the degree [K(ζp) : K] divides p−1, it is prime to p. But H1(G,CL) is p-torsion(it is killed by |G|), hence res:H1(G,CL) → H1(Gal(L(ζp)/K(ζp)), CL(ζp)) is injective by the restriction-

corestriction argument (see appendix). By the class field theory axiom, H1(Gal(L(ζp)/K(ζp)), CL(ζp)) =

1, and therefore H1(G,CL) = 1 as well.

For the general case, since H1(G,CL) is torsion, it is enough to show that its p-primary part vanishesfor all primes p. If Sp ⊂ G is a p-Sylow subgroup, then by the restriction-corestriction argument, wehave that res:H1(G,CL) → H1(Sp, CL) is injective on p-primary parts, hence it is enough to considerthe case where G = Sp is a p-group, hence solvable. Now we proceed by induction on |G|, the base case|G| = p already solved. Let H ⊳ G be a non-trivial normal subgroup, and denote by M = LH the fixedfield of L by H . By induction hypothesis, H1(H,CL) = 1, and hence from the short exact sequence

1 - L× - IL- CL

- 1

we have that

1 - M× - IM- CH

L- H1(H,CL) = 1

is exact and thus CHL = CM . By the inflation-restriction sequence (see appendix), we thus obtain an

exact sequence

0 - H1(G/H,CM )inf- H1(G,CL)

res- H1(H,CL)

We conclude that the middle term vanishes since the left and right terms vanish by induction hypothesis.

The proof of (1) proceeds along similar lines. Since H0T (G,CL) is torsion, it is enough to show that

its p-primary part is finite and divides |Sp| where Sp ⊂ G denotes once more a p-Sylow subgroup ofG. Restriction-corestriction allows us to assume that G = Sp, and shows that we may dispense withthe roots of unity condition of the class field theory axiom |H0

T (G,CL)| = |G| for G cyclic. The onlydifference is that induction on the order of the p-group G utilises the following observation instead ofthe inflation-restriction sequence: if H ⊳ G then since

[CK : NL/KCL] = [CK : NM/KCM ] · [NM/KCM : NL/KCL]

and NM/K :CM/NL/MCL ։ NM/KCM/NL/KCL is surjective, we obtain that |H0T (G,CL)| divides

|H0T (G/H,CM )| · |H0

T (H,CL)|.We also leave the details of the proof of (2) to the reader. We just observe that the base case |G| = p

of the induction now follows from |H2(G,CL)| = |H0T (G,CL)| = |G| by the periodicity of cohomology of

cyclic groups (see appendix), and that the inflation-restriction sequence now reads

0 - H2(G/H,CM )inf- H2(G,CL)

res- H2(H,CL)

since we have already shown that H1(H,CL) = 1.


Theorem 2.3.3 Let L ) K be a solvable extension of global fields with G = Gal(L/K). Then

1. there exist infinitely many prime ideals p in OK that do not split completely in OL;

2. if G is abelian, the Frobenius automorphisms Φv ∈ Gal(Lv/Kv) ⊂ G generate G, as v runs overall valuations that do not belong to a fixed finite set S of places of K containing all archimedeanand ramified ones.

Proof Replacing L ⊃ K by some cyclic subextension of prime order, we may assume that |H0T (G,CL)| ≥

|G| by the class field theory axioms. Suppose that the set S of primes that do not split completely in Lis finite. Given α = (av) ∈ IK , using the approximation theorem (c.f. the proof of lemma 2.1.6) we mayfind f ∈ K× such that f · av ∈ NLv/Kv

(Lv)× for all v ∈ S (norm groups of local fields are open of finiteindex by local class field theory). But then f · α ∈ NL/KIL since Lv = Kv for all v /∈ S by assumption.

This proves that K× ·NL/KIL = IK . But since IK/(K× ·NL/KIL) = CK/NL/KCL = H0

T (G,CL), thiscannot happen, proving (1).

To obtain (2), observe that the prime corresponding to a valuation v of K splits completely in L ifand only if Φv = 1. Hence, applying (1) to the fixed field M of all Φv, v /∈ S, we conclude that M = K,that is, the Φv, v /∈ S, generate G, as required.

Remark 2.3.4 Observe that the above proof only relies on the fact that |H0T (G,CL)| ≥ |G|. This will

be used later in the proof of the so-called second inequality.

Theorem 2.3.5 (Haße Principle for CSA) Let L ⊃ K be a Galois extension of global fields withG = Gal(L/K). Then

0 - Br(L/K) -⊕

v

Br(Lv/Kv) - H2(G,CL) - 0

is complex that is exact at the first two terms. If G is cyclic, it is also exact at the last term. In particular,taking the limit over all L, we have that

0 - Br(K) -⊕

v

Br(Kv)

is exact, that is, an element α ∈ Br(K) is trivial if and only if it is locally trivial.

Proof Let G = Gal(L/K). From the short exact sequence of G-modules

1 - L× - IL- CL

- 1

we obtain an exact sequence

0 = H1(G,CL) - H2(G,L×) - H2(G, IL) - H2(G,CL) - H3(G,L×)

which can be rewritten as

0 - Br(L/K) -⊕

v

Br(Lv/Kv) - H2(G,CL) - H3(G,L×)

using the class field axioms. Moreover, if G is cyclic, by Hilbert 90 and the periodicity of cohomology ofG, H3(G,L×) = H1(G,L×) = 0 as well, which finishes the proof.

Corollary 2.3.6 (Haße Norm Theorem) Let L ⊃ K be a cyclic extension of global fields. Then anelement of K× is a norm if and only if it is a local norm, that is, a ∈ K× belongs to NL/KL

× if andonly if a ∈ NLv/Kv

(Lv)× for all v.

Proof Just use the previous theorem together with the commutative diagram:

K×

NL/KL×= H0

T (G,L×)∪δχ≈

- H2(G,L×) = Br(L/K)

K×v

NLv/Kv(Lv)×

= H0T (G, (Lv)×)

?∪δχ≈- H2(G, (Lv)×) = Br(Lv/Kv)

?

Here χ ∈ H1(G,Q/Z) is a generator of this group, and the rows are isomorphisms by the periodicity ofthe cohomology of the cyclic group G.

The general case 43

Now we can show

Lemma 2.3.7 Let K be a number field. Any α ∈ Br(K) is split by a cyclic extension L of K containedin some cyclotomic extension K(ζ) of K.

Proof From the commutative diagram with exact top row

0 - Br(Lv/Kv) - Br(Kv)res - Br(Lv)

Q/Z

inv ≈? [Lv : Kv]- Q/Z

≈ inv

?

together with the Haße principle for CSA, we reduce the problem to proving the following claim: givenan integer n and a finite set of places S of K, there is a cyclic extension L ⊃ K contained in a cyclotomicextension of K such that n divides the local degrees [Lv : Kv] for all v ∈ S.

Clearly, it suffices to prove the claim for K = Q. Let p be a prime. Since Gal(Q(ζpr )/Q) = (Z/pr)×,Q(ζpr ) ⊃ Q contains a cyclic subextension F (pr) ⊃ Q of degree pr−2. Now if l is another prime, wehave that [Ql(ζpr ) : Ql] equals φ(lr) = lr−1(l− 1) if p = l (totally ramified case), and equals the smallestpositive integer f such that pr | lf − 1 if p 6= l (totally unramified case). In either case, [F (pr)l : Ql] is apower of p that tends to infinity as r grows.

Now let n = pe11 . . . pet

t be the factorisation of n into powers of distinct primes pi. Choose r largeenough so that pei

i | [F (pri )l : Ql] for all i = 1, 2, . . . , t and l ∈ S. We may take L = F (pr

1) . . . F (prt ),

which is a cyclic extension of Q with the required properties.

Observe that the above theorem shows that Br(Qmc) = 0 for the maximal cyclotomic extension Qmc

of Q, much in the same way that Tsen’s theorem asserts that the Brauer group of the function field ofa curve over Fsep

p is trivial. The field Qmc (more precisely, a cyclic quotient of it) plays the same role asthe Fsep

p in the geometric case.

We now can finish the proof of the Reciprocity Law:

Proof (of the Reciprocity Law) Let L ⊃ K be the cyclic extension contained in a cyclotomic extensionof K, as the previous theorem. Then by the explicit computation of theorem 2.2.4 we know thatK× ⊂ ker θL/K . Hence, by lemma 2.2.3, the sequence

0 - Br(L/K) -⊕

v

Br(Lv/Kv)

∑

vinvv- Q/Z (∗)

is a complex. But the previous theorem says that Br(K) is the union of all Br(L/K) as L runs over allcyclic extension contained in a cyclotomic extension of K, hence we obtain a complex

0 - Br(K) -⊕

v

Br(Kv)

∑

vinvv- Q/Z - 0 (∗∗)

This, in turn, implies that (∗) is a complex for all L. Finally, we conclude that K× ⊂ ker θL/K for all Lby lemma 2.2.3.

Now suppose that G is abelian, and observe that

θL/K : IK ։ G

is surjective since the Frobenius maps Φv generate G (theorem 2.3.3). But ker θL/K ⊃ K× · NL/KIL,and since the subgroup K× ·NL/KIL has index at most |G| in IK by corollary 2.3.2, it must equal thekernel, and θL/K induces an isomorphism

θL/K :IK

K× ·NL/KIL=

CK

NL/KCL

≈→ G


Finally, to show that (∗∗) is exact, it is enough to show that

0 - Br(L/K) -⊕

v

Br(Lv/Kv)

∑

vinvv-

1[L:K]Z

Z- 0 (†)

is exact for the cyclic extension of the beginning of the proof. Observe that we already know that (†) isa complex and that it is exact at the first (Haße principle) and last term since the Frobenius maps Φv

generate G, hence invv(Φv ∪ δπv) = 1fv

mod Z generate the last group: here fv = [Lv : Kv] is the order

of Φv and πv is a uniformiser of v.

Comparison of the exact sequence in theorem 2.3.5 with the last sequence yields a surjective map

H2(G,CL) ։

1[L:K]

Z

Z, and we have to show that this map is an isomorphism. For that, just observe that

H2(G,CL) is cyclic of order |G| = [L : K] since by the global Artin reciprocity we have isomorphisms

G θL/K

≈H0

T (G,CL)∪δχ

≈- H2(G,CL)

where χ ∈ H1(G,Q/Z) is a generator and cup product with δχ ∈ H2(G,Z) is the periodicity isomor-phism.

3 Proof of the Class Field Theory Axioms

We are left to compute the cohomology groups of IL and CL. The first computation is fairly straight-forward, but verifying that |H0

T (G,CL)| = |G| and |H1T (G,CL)| = 1 is quite delicate, and the proof

is broken into two parts, traditionally called “first and second inequalities:” the first inequality is thecomputation of the Herbrand quotient h(G,CL) = |H0

T (G,CL)|/|H1T (G,CL)| = |G|, which already shows

that |H0T (G,CL)| ≥ |G|, while the second inequality is the opposite one.

3.1 Cohomology of IL

Recall that for any set S of valuations of K, we write

ISK =

∏

v∈S

K×

v ×∏

v/∈S

Uv

Theorem 3.1.1 (Cohomology of Ideles) Let L ⊃ K be a finite Galois extension of global fields withabelian G = Gal(L/K). Then

Hp(G, IL) =⊕

v

Hp(Gv, (Lv)×)

In particular, H1(G, IL) = 1 and H2(G, IL) =⊕

v

1nv

Z

Zwhere nv = [Lv : Kv].

Proof Let S be a finite set of primes containing all archimedean and ramified ones. Using the fact thatcohomology commutes with products,

Hp(G, ISL) =

∏

w|vv∈S

Hp(G,L×

w)×∏

w|vv /∈S

Hp(G,Uw)

By Shapiro’s lemma,

Hp(G, ISL) =

∏

v∈S

Hp(Gv, (Lv)×)×

∏

v/∈S

Hp(Gv, Uv)

Since every v /∈ S is unramified, Hp(Gv, Uv) = 0, and therefore

Hp(G, ISL) =

∏

v∈S

Hp(Gv, (Lv)×)

Taking the injective limit over all finite sets S yields the result.

Proof of the Class Field Theory Axioms 45

From now on we concentrate on the number field case, which is the difficult one. We leave to thereader the task of adapting the proofs to the function field case.

3.2 First inequality

Let L ⊃ K be a cyclic extension of number fields with Galois group G = Gal(L/K). Our goal is tocompute the Herbrand quotient h(G,CL).

Theorem 3.2.1 (First Inequality) Let L ⊃ K be a Galois extension of number fields with cyclicG = Gal(L/K). Then

h(G,CL) = [L : K]

In particular, |H0T (G,CL)| ≥ [L : K].

Proof We invoke lemma 2.1.6 and choose a finite set T of valuations of L which

1. includes all archimedean and ramified primes;

2. IL = L×ITL ;

3. is G-stable, i.e., w ∈ T ⇒ σ(w) ∈ T for all σ ∈ G.

and let S be the set of places of K dividing the places of T . We have

h(G,CL) = h(

G,ITL

ITL ∩ L×

)

=h(G, IT

L)

h(G, ITL ∩ L×)

We already know that

h(G, ITL) =

|H0T (G, IT

L)||H1

T (G, ITL)| =

∏

v∈S

|H0T (Gv, L

v)||H1

T (Gv, Lv)| =∏

v∈S

nv

where nv = [Lv : Kv]. On the other hand, LT df= IS

L ∩ L× is just the group of T -units of L. Denote by

Vdf= Hom(T,R) =

⊕

w∈T

Rew

the |T |-dimensional real vector space equipped with G-action

(σf)(w) = f(σ−1w), f ∈ V and σ ∈ G,w ∈ T

Here ew denotes the dual of the standard basis of R|T |, that is, ew(w) = 1 and ew(w′) = 0 if w′ 6= w. ByDirichlet’s unity theorem, the map of G-modules

log u:LT → V

u 7→∑

w∈T

log ‖u‖w · ew

has a finite kernel µ ⊂ L× and a G-lattice M0 of rank |T | − 1 as image, contained in the G-invarianthyperplane

∑

w∈T xwew |∑

w∈T xw = 0 by the product formula. We have thus exact sequences ofG-modules

0 - µ - LT log- M0 - 0

0 - M0 - M - Z - 0

where M = M0 ⊕ Zv and v = (1, 1, . . . , 1) ∈ V . Hence

h(G,LT ) =h(G,M)h(G,Z)

h(G,µ)= h(G,M) · |G|

To compute h(G,M), we may replace M with any G-lattice N such that M ⊗Z Q ∼= N ⊗Z Q (as G-

modules) since Herbrand quotients of a finite G-modules are trivial. Take Ndf= Hom(T,Z) ⊂ V . We

have that M ⊗Z R ∼= N ⊗Z R⇒M ⊗Z Q ∼= N ⊗Z Q and

h(G,M) = h(G,N) = h(G,∏

v∈S

IndGGv

Z) =∏

v∈S

h(Gv,Z) =∏

v∈S

nv

Putting everything together, we obtain the desired result.


3.3 Second inequality

Now we finish the proof of the class field theory axioms by showing

Theorem 3.3.1 (Second Inequality) Let L ⊃ K be a Galois extension of number fields with cyclicG = Gal(L/K) of prime order n. Suppose that µn ⊂ K×. Then

|H0T (G,CL)| ≤ n

The key idea of the proof is to use Kummer theory in place of the reciprocity law, and a variant ofHaße’s norm theorem dealing with n-th powers instead of norms, given by the first lemma below.

By Kummer theory, L = K( n√a) for some a ∈ K×. The second inequality follows from

Theorem 3.3.2 Let L ⊃ K be as above, and let S be a sufficiently large finite set of primes of K suchthat

• all archimedean primes are in S;

• all primes that ramify in L are in S;

• all primes dividing a or n are in S;

• IK = K×ISK

Suppose that there is a set T of |S| − 1 primes of K, disjoint from S, satisfying

1. Lv = Kv for all v ∈ T .

2. (local power) KS∪T ∩D(S, T ) = (KS∪T )n, where

D(S, T ) =∏

v∈S

(K×

v )n ×∏

v∈T

K×

v ×∏

v/∈S∪T

Uv

and KS∪T denotes the group of (S ∪ T )-units.

Then |H0T (G,CL)| divides n.

Proof Since D(S, T ) ⊂ NL/KIL (use corollary II.2.2), it is enough to prove that | IK

K×D(S,T ) | divides n.

Using the relation

[AC : BC] = [A : B] · [A ∩C : B ∩ C]

for subgroups A,B,C of an abelian group, and condition 2, we have that

∣

∣

∣

∣

IK

K×D(S, T )

∣

∣

∣

∣

=

∣

∣

∣

∣

K×IS∪TK

K×D(S, T )

∣

∣

∣

∣

=

∣

∣

∣

IS∪TK

D(S,T )

∣

∣

∣

∣

∣

∣

K×∩IS∪TK

K×∩D(S,T )

∣

∣

∣

=

∏

v∈S [K×v : (K×

v )n]

[KS∪T : (KS∪T )n]

But by Kummer theory and the structure theorem of the multiplicative group of a local field, we havethat [K×

v : (K×v )n] = n2/‖n‖v (which also works for archimedean v in view of the assumption µn ⊂ K×).

On the other hand, since µn ⊂ K×, Dirichlet’s unit theorem implies that [KS∪T : (KS∪T )n] = n|S|+|T | =n2|S|−1. Hence

∣

∣

∣

∣

IK

K×D(S, T )

∣

∣

∣

∣

=

∏

v∈Sn2

‖n‖v

n2|S|−1=

n2|S|

n2|S|−1= n

using the product formula.

We now construct T . First we make the following reduction:

Lemma 3.3.3 Let L ⊃ K and S be as above. Suppose T is a set of places of K, disjoint from S, suchthat the map

KS։

∏

v∈T

Uv/Unv

is surjective. Then KS∪T ∩D(S, T ) = (KS∪T )n.


Proof Observe that x ∈ K is an n-th power if and only if the field M = K( n√x) is equal to K, and by

the first inequality (or rather theorem 2.3.3) this happens if and only if

H0T (Gal(M/K), CM ) = 1 ⇐⇒ IK = K×NM/KIM

Hence we have to show that if x ∈ KS∪T ∩D(S, T ) then IK = K×NM/KIM , so we need to find manynorms. Notice that

NM/KIM ⊃ E df=

∏

v∈S

K×

v ×∏

v∈T

Unv ×

∏

v/∈S∪T

Uv

since x ∈ D(S, T ), hence x ∈ (K×v )n, and x ∈ KS∪T , hence M is unramified outside S ∪ T and

norm maps are surjective by corollary II.2.2. But now given α ∈ ISK there exists f ∈ KS such that

fα ∈ E ⊂ NM/KIM by assumption, and therefore IK = K×ISK = K×NM/KIM , as required.

The next lemma finishes the proof of the second inequality.

Lemma 3.3.4 With the above notation and hypotheses, there is a set T of |S|−1 primes in K such that

1. Lv = Kv for all v ∈ T .

2. the map KS։

∏

v∈T Uv/Unv is surjective.

Proof Let s = |S|, and H be the kernel of the map in (2). We claim that it suffices to choose T so

that L = K( n√H) (by the Kummer correspondence). In fact, (1) is clear, while in order to show (2), it

suffices to show that KS/H and∏

v∈T Uv/Unv have the same cardinality. But

|KS/H | = [KS : (KS)n]

[H : (KS)n]=

ns

[L : K]= ns−1

by Dirichlet’s unit theorem and Kummer theory, while

∏

v∈T

|Uv/Unv | =

∏

v∈T

n/‖n‖v = ns−1

by the structure theorem of units of local fields and the product formula.

Now choose representatives a1 = a, a2, . . . , as ∈ KS of a basis of the s-dimensional Z/n-vector spaceKS/(KS)n. Let

M = K(n√KS) and Mi = K( n

√a1, . . . , n

√ai−1, n

√ai+1 . . . , n

√as)

Then by Kummer theory we have that L =⋂

2≤i≤s Mi, and each M ⊃ Mi is a cyclic extension ofdegree n. Hence by theorem 2.3.3 we may choose places wi of Mi such that their restrictions vi toK are all distinct and lie outside S, and such that wi does not split completely in M . Hence Mi isthe decomposition field of the unique place of M lying over wi, and therefore L ⊃ K is the maximalsubextension of M ⊃ K where all places vi split completely. Let T = v2, . . . , vs. Now H = (L×)n∩KS

since, for x ∈ KS , we have

x ∈ (L×)n ⇐⇒ K( n√x) ⊂ L

⇐⇒ Kvi(n√x) = Kvi for i = 2, . . . , s (since each vi splits completely)

⇐⇒ x ∈ Unvi

for i = 2, . . . , s

⇐⇒ x ∈ H

and by Kummer theory we have that L = K( n√H), as required.

Chapter 4

Appendix: Galoiscohomology

0.1 Definitions

In this subsection, G will denote a finite group.

Definition 0.1.1 A G-module M is just a left Z[G]-module, where Z[G] is the group ring of G withinteger coefficients. In other words, M is an abelian group together with a G-action, that is, a mapG×M →M sending (σ,m) to an element σ ·m ∈M such that, for all m,m′ ∈M , and σ, σ′ ∈ G,

1. 1 ·m = m

2. σ · (m+m′) = σ ·m+ σ ·m′

3. (σσ′) ·m = σ · (σ′ ·m)

A morphism of G-modules f :M → N is a map of left Z[G]-modules, i.e., a group morphism suchthat f(σm) = σf(m) for all σ ∈ G and m ∈M .

If M is a G-module, we write MG for the subgroup of G-fixed points:

MG df= m ∈M | σ ·m = m for all σ ∈ G

Example 0.1.2 Examples of G-modules arising in nature are

1. M = Z, M = Q, or M = Q/Z where G operates trivially: σm = m for all σ ∈ G and m ∈M ;

2. M = Z[G] or M = IG, where G operates by left multiplication. Here IG is the so-calledaugmentation ideal of Z[G], defined as the kernel of the augmentation map ǫ: Z[G] → Z

given by∑

σ∈G

nσσ 7→∑

σ∈G

nσ

Observe that IG is a free Z-module with basis σ − 1, σ ∈ G, σ 6= 1. If Z is given the trivialaction as above, then the augmentation map is a morphism of G-modules;

3. if L ⊃ K is a finite Galois extension with G = Gal(L/K), then M = L+, M = L× and M = µL

are all examples of G-modules where G operates via the Galois action. Here L+ and L× are theadditive and multiplicative groups of L and µL ⊂ L× is the subgroup of roots of 1 contained inL.

4. for any G-module M we have a morphism of G-modules NG:M →MG given by

NG(m)df=

∑

σ∈G

σm, m ∈M

called norm map. In the previous example, when M = L× the norm map NG coincides withthe usual norm of fields NL/K :L→ K. When M = L+ then it becomes the trace TL/K :L→ K.

From now on, unless otherwise stated the abelian groups M of the example will always be given theG-actions above. With this convention, for any G-module M we have an isomorphisms of abelian groups

HomZ[G](Z,M) = MG

and

Z⊗Z[G] M =M

IG ·M

50 Appendix: Galois cohomology

since Z[G]/IG = Z (induced by augmentation).

Now we show how to build new modules from old ones. Let M and N be two G-modules. Then theset

HomG(M,N)

of all morphisms of G-modules between M and N can be made into a G-module by “conjugation”

(σf)(m)df= σf(σ−1m) for f ∈ HomG(M,N) and m ∈M . Similarly, the tensor product of M and N over

Z

M ⊗N

can be made into a G-module by “diagonal action” σ(m ⊗ n)df= σ(m) ⊗ σ(n) for m ∈ M and n ∈ N .

The next definition shows how to “lift” modules from subgroups:

Definition 0.1.3 Let H ≤ G be a subgroup and N be an H-module. The induced module from N isthe G-module

IndGH(N)

df= HomH(G,N)

of all H-linear functions f :G → N (i.e. f(h · g) = h · f(g) for all h ∈ H and g ∈ G) and where the

G-action is given by (σf)(g) = f(gσ) for all σ, g ∈ G. When H = 1 we simply write IndG(N). A

G-module M is called induced if M = IndG(N) for some abelian group N .

Another way to “lift” an H-module is via the “base change” Z[G] ⊗Z[H] N , where the G-action isgiven by multiplication on the left component (in the tensor product, the left component is viewed as a

right Z[H ]-module via m · σ df= σ−1 ·m for all σ ∈ H and m ∈ Z[G]). Base change is related to induced

modules via the isomorphism of G-modules

IndG(N) ∼= Z[G]⊗N

given by φ 7→ ∑

g∈G g ⊗ φ(g−1). Here N is any abelian group with trivial G-action. Observe that an

induced G-module M = Z[G] ⊗N is also induced as an H-module: if Hσ1, . . . , Hσn are right cosets ofH then M = Z[H ]⊗N ′ where N ′ =

⊕

1≤i≤n σiN .

Example 0.1.4 Let L ⊃ K be a finite Galois extension with G = Gal(L/K). Then L+ is an inducedG-module. In fact, by the normal basis theorem there exists ω ∈ L such that σ(ω) | σ ∈ G is a basisof L over K. Then we have a G-isomorphism K[G] ∼= L+ given by

∑

σ∈G aσσ 7→∑

σ∈G aσσ(ω). Since

K[G] = Z[G]⊗K is induced by the above, so is L+.

The following lemma will be useful in arguments involving dimension shifting (which we will seelater).

Lemma 0.1.5 Let M be an arbitrary G-module and denote by M0 the underlying abelian group. Thenthere is an injective map of G-modules

M → IndG(M0)

sending m to the function φm:G → M0 given by φm(σ) = σ · m. There is also a surjective map ofG-modules

IndG(M0) ։ M

given by φ 7→∑

σ∈G σφ(σ−1).

Definition 0.1.6 Let M be a G-module and i ≥ 0. The i-th cohomology group of M is defined as

Hi(G,M)df= Exti

Z[G](Z,M)

while the i-th homology group of M is defined as

Hi(G,M)df= Tor

Z[G]i (Z,M)

We briefly recall the definitions of Ext and Tor below, but we warn you that except for low degreesi = 0, 1 the computations of these groups are not done directly from their definitions but rather via theirfunctorial properties and some “vanishing theorems” that tell you sufficient conditions under which thesegroups are trivial.


A G-module I is called injective if the functor HomG(−, I) is exact. A G-module P is calledprojective if the functor HomG(P,−). For instance, free Z[G]-modules are projective. It can be shownthat any G-module M can be embedded into an injective module and it can also be written as a quotientof a projective module. Hence we can inductively construct an injective resolution of M , namely anexact sequence

0→M → I0 → I1 → I2 → · · ·

where the G-modules Ii are all injective, and similarly one may construct a projective resolution

· · · → P 2 → P 1 → P 0 →M → 0

Now we can define ExtiZ[G](Z,M) as follows. Choose an injective resolution of M as above and apply

the G-fixed point functor HomG(Z,−) = (−)G to it. We obtain a complex

0→MG → (I0)G → (I1)G → (I2)G → · · ·

Then

ExtiZ[G](Z,M)df=

ker(

(Ii)G → (Ii+1)G)

im(

(Ii−1)G → (Ii)G)

(Here we interpret I−1 = 0). Now an easy but rather tedious computation shows that the groups thusobtained are independent of the choice of the injective resolution of M . Alternatively, one can choose aprojective resolution of Z

· · · → P 2 → P 1 → P 0 → Z→ 0

and apply the functor HomG(−,M) to it, defining

ExtiZ[G](Z,M)

df=

ker(

HomG(P i,M)→ HomG(P i+1,M))

im(

HomG(P i−1,M)→ HomG(P i,M))

(Here we interpret P−1 = 0) Again, this is independent of the chosen projective resolution of Z, and itcan be shown that either procedure, via projective resolutions of Z or injective resolutions of M , yieldisomorphic groups.

For TorZ[G]i (Z,M), the procedure is similar, except that Hom is replaced by the tensor product and

the we use projective resolutions for both entries (in the tensor product M ⊗Z[G]N we view M as a right

Z[G]-module via m · σ df= σ−1 ·m for σ ∈ G and m ∈ M). For instance, if P• → M → 0 is a projective

resolution of M , applying the functor Z⊗Z[G] − = −/IG · − we obtain

TorZ[G]i (Z,M) =

ker(

Pi/IG · Pi → Pi−1/IG · Pi−1

)

im(

Pi+1/IG · Pi+1 → Pi/IG · Pi

)

Example 0.1.7 (Cyclic groups) Let G be a cyclic group of order n and let σ be a generator of G.Then we have a projective resolution of Z

· · · I- Z[G]N- Z[G]

I- Z[G]N- Z[G]

I- Z[G]ǫ- Z - 0

where ǫ is the augmentation map, and I and N denote multiplication by σ − 1 and 1 + σ + · · ·+ σn−1,respectively. Hence we obtain

Hi(G,M) =

MG if i = 0

kerNG

IG ·Mif i is odd

MG

NG(M)if i > 0 is even


where NG:M → MG is the norm map and IG is the augmentation ideal of Z[G]. Using the sameprojective resolution of Z, for homology we obtain

Hi(G,M) =

M

IG ·Mif i = 0

MG

NG(M)if i is odd

kerNG

IG ·Mif i > 0 is even

The main functorial property of Tor and Ext, and thus of the homology and cohomology groups,are their long exact sequences. For any short exact sequence of G-modules

0→ A→ B → C → 0

one has long exact sequences

0 - H0(G,A) = AG - H0(G,B) = BG - H0(G,C) = CG

δ0

- H1(G,A) - H1(G,B) - H1(G,C)

δ1

- H2(G,A) - H2(G,B) - H2(G,C)δ2

- · · ·

and

· · · δ3- H2(G,A) - H2(G,B) - H2(G,C)

δ2- H1(G,A) - H1(G,B) - H1(G,C)

δ1- H0(G,A) =A

IG ·A- H0(G,B) =

B

IG · B- H0(G,C) =

C

IG · C- 0

for cohomology and homology respectively. The maps δi and δi are called connecting morphisms.The other maps are the natural ones induced by the maps A→ B and B → C.

Think of the short exact sequence as a way to “decompose” B into simpler modules, a submoduleA and a quotient module C. If we know the homology/cohomology of A and C then the long exactsequence allows us to find out the homology/cohomology of B.

Example 0.1.8 Let ǫ be the augmentation map. From the exact sequence of G-modules

0 - IG - Z[G]ǫ- Z - 0

and the fact that Z[G] is free (and thus has trivial homology), we conclude that the connecting morphismsgive isomorphisms Hp(G,Z) = Hp−1(G, IG) for all p ≥ 1. In particular, for p = 1 we have that

H1(G,Z) = H0(G, IG) =IGI2G

= Gab

where Gab df= G/[G : G] is the maximal abelian quotient of G. The isomorphism Gab ≈ IG/I

2G is given

by σ · [G : G] 7→ (σ − 1) · I2G.

The main vanishing theorem is Shapiro’s lemma.

Lemma 0.1.9 (Shapiro) Let H ⊂ G be a subgroup and N be an H-module. Then for all i ≥ 0 wehave isomorphisms

Hi(H,N) = Hi(G, IndGH(N)) and Hi(H,N) = Hi(G, IndG

H(N))

In particular, any induced G-module has trivial cohomology and homology.

Proof (Sketch) For any any G-module M and any H-module N we have a canonical isomorphism

HomG(M, IndGH N) = HomH(M,N) and the functor IndG

H(−) is exact. From these two properties it

follows that IndGH(−) preserves injectives, hence given an injective resolution 0 → N → I• of N , we

obtain an injective resolution 0 → IndGH(N) → IndG

H(I•) of IndGH(N). The result follows by applying

HomG(Z,−) and using the fact that HomG(Z, IndGH N) = HomH(Z, N). The proof for homology is

similar.


Corollary 0.1.10 If L ⊃ K is a finite Galois extension with G = Gal(L/K) then H0(G,L+) = K+

and Hp(G,L+) = 0 for p ≥ 1.

Definition 0.1.11 The p-th Tate cohomology group is defined by

HpT (G,M) =

Hp(G,M) if p ≥ 1

MG

NGMif p = 0

kerNG

IG ·Mif p = −1

H−p−1(G,M) if p ≤ −2

The importance of the Tate cohomology groups is that it allows us to splice the long exact sequencesof homology and cohomology into a single very long one. Given a short exact sequence of G-modules

0→ A→ B → C → 0

we have a commutative diagram with exact rows

H1(G,C) - H0(G,A) - H0(G,B) - H0(G,C) - 0

0 - H0(G,A)

NG

?- H0(G,B)

NG

?- H0(G,C)

NG

?- H1(G,A)

where the vertical arrows are induce by the norm maps. Hence we obtain an exact sequence

· · · → H−1T (G,A)→ H−1

T (G,B)→ H−1T (G,C)

→ H0T (G,A)→ H0

T (G,B)→ H0T (G,C)

→ H1T (G,A)→ H1

T (G,B)→ H1T (G,C)→ · · ·

Example 0.1.12 (Periodicity of Tate cohomology for cyclic groups) If G is cyclic we have that

HpT (G,M) =

MG

NGMif p is even

kerNG

IG ·Mif p is odd

0.2 Explicit Resolutions

Here we show how to define homology and cohomology via an explicit projective resolution of Z:

Definition 0.2.1 The standard resolution of Z is the projective resolution

· · · d3- C2d2- C1

d1- C0d0- Z - 0

where Cp = Z[Gp+1] is the free Z-module with basis (σ0, σ1, . . . , σp) ∈ Gp+1 = G× · · · ×G (p+ 1 times)

with “diagonal” G-action s · (σ0, . . . , σp) = (s · σ0, . . . , s · σp), and where dp+1: Cp+1 → Cp is given by

d(σ0, σ1, . . . , σp+1) =∑

0≤k≤p+1

(−1)k(σ0, σ1, . . . , σk−1, σk+1, . . . , σp+1)

Applying HomG(−,M) to the standard resolution, we obtain a complex of abelian groups

0 - C0(G,M)d0

- C1(G,M)d1

- C2(G,M)d2

- · · ·

where

Cp(G,M)df=

functions f :Gp+1 →M∣

∣

∣

f(s · σ0, . . . , s · σp) = s · f(σ0, . . . , σp) for alls ∈ G and (σ0, . . . , σp) ∈ Gp+1


and dp: Cp(G,M)→ Cp+1(G,M) is given by

(dpf)(σ0, . . . , σp+1) =∑

0≤k≤p+1

(−1)kf(σ0, . . . , σk−1, σk+1, . . . , σp+1)

We have an isomorphism between Cp(G,M) and the abelian group Cp(G,M) of all functions from Gp

to M : it takes f ∈ Cp(M) to the function

(σ1, σ2, . . . , σp) 7→ f(1, σ1, σ1σ2, σ1σ2σ3, . . . , σ1σ2 · · ·σp)

The above complex is therefore isomorphic to the following “inhomogeneous” one

0 - C0(G,M)d0

- C1(G,M)d1

- C2(G,M)d2

- · · ·where dp:Cp(G,M)→ Cp+1(G,M) is given by

(dpf)(σ1, . . . , σp+1) = σ1 · f(σ2, . . . , σp+1)

+∑

1≤k≤p

(−1)kf(σ1, . . . , σk−1, σk · σk+1, σk+2, . . . , σp+1)

+ (−1)p+1f(σ1, . . . , σp)

Hence we obtain an explicit formula

Hp(G,M) =kerdp

im dp+1

An element of ker dp is called a p-cocycle while an element of im dp−1 is called a p-coboundary. Let

0→ A→ B → C → 0

be a short exact sequence of G-modules. In terms of cocycles and coboundaries, one has a very explicitdescription of the connecting morphism

δ:Hp(G,C)→ Hp+1(G,A)

as follows: given a p-cocycle f :Gp → C representing an element ϕ = [f ] ∈ Hp(G,C), we may lift it

to a function f :Gp → B. Then dpf is in the image of the map Cp+1(G,A) → Cp+1(G,B) induced by

A→ B, so we may view it as a p-cocycle dpf :Gp+1 → A, and δ(ϕ) = [dpf ] ∈ Hp+1(G,A).

Similarly, for homology one obtains an inhomogeneous complex

· · · d3- C2(G,M)d2- C1(G,M)

d1- C0(G,M) - 0

where Cp(G,M) = Cp(G,M) and dp:Cp(G,M)→ Cp(G,M) is given now by

(dpf)(σ1, . . . , σp−1) =∑

σ∈G

σ−1 · f(σ, σ1, . . . , σp−1)

+∑

1≤k≤p−1

(−1)k∑

σ∈G

f(σ1, . . . , σk−1, σk · σ, σ−1, σk+1, . . . , σp−1)

+ (−1)p+1∑

σ∈G

f(σ1, . . . , σp−1, σ)

Example 0.2.2 We have that a 1-cocycle is a function f :G→ M such that f(στ) = σf(τ) + f(σ). Itis a coboundary if and only if it has the form f(σ) = σm −m for some m ∈ M . In particular, if theG-action on M is trivial, then all 1-coboundaries are trivial and a 1-cocycle is just a group morphism:H1(G,M) = Hom(G,M) in this case.

Example 0.2.3 Let G be a cyclic group of order n generated by σ. Then one has an exact sequence ofG-modules

0→ Z→ Q→ Q/Z→ 0

As we shall see later, Hi(G,Q) = 0 for all i ≥ 1 and hence the connecting maps induce isomorphisms

δ:Hi(G,Q/Z)≈→ Hi+1(G,Z) for all i ≥ 0. Then H1(G,Q/Z) = Hom(G,Q/Z) is a group of order n,

generated by a morphism f :G → Q/Z given by f(σ) = 1n mod Z, and hence H2(G,Z) is also cyclic of

order n, generated by the class of the 2-cocycle

δf(σi, σj) =

1 if i+ j ≥ n0 otherwise

for 0 ≤ i, j < n

which can be computed as described above using the lift f :G→ Q of f given by f(σi) = in for 0 ≤ i < n.

Now we use the explicit characterisation of cohomology in terms of cocycles to prove a very importantvanishing theorem in Galois cohomology:


Theorem 0.2.4 (Hilbert’s Satz 90) Let L ⊃ K be a finite Galois extension with Galois group G =Gal(L/K). Then

H1(G,L×) = 0

Proof Let f :G→ L× be a 1-cocycle, and consider the element

mdf=

∑

σ∈G

f(σ) · σ(a)

where a ∈ L× is chosen so that m 6= 0, which is possible by Dedekind’s independency of characters.Then for every τ ∈ G we have that

m =∑

σ∈G

f(τσ) · τσ(a) =∑

σ∈G

τ(

f(σ))

· f(τ) · τσ(a) = τ(m) · f(τ)

That is, f(τ)−1 = τ(m)/m, showing that f is a coboundary.

0.3 Dimension shifting; Inflation, Restriction, Corestriction

We have seen how changing modules alters the cohomology groups via the long exact sequence. Now weshow how changing the group alters the cohomology. Let f :G′ → G be a group morphism and A be aG-module. We may view A as a G′-module as well via

σ′ · a df= f(σ′) · a for a ∈ A, σ′ ∈ G′

We denote this G′-module by f∗A. Clearly AG is a subgroup of (f∗A)G′

, hence the inclusion mapdefines a functorial map H0(G,A)→ H0(G′, f∗A). By general properties of derived functors, this mapin degree 0 extends uniquely for all p ≥ 0 to a functorial map Hp(G,A)→ Hp(G′, f∗A), compatible withthe connecting maps. We recall the proof of this fact since it relies on a recurrent technique in grouphomology/cohomology known as dimension shifting.

First write an exact sequence of G-modules

0→ A→ I → B → 0

with I injective (cohomologically trivial would do, for instance an induced module). Since Hp(G, I) = 0

for all p > 0 we have that the connecting maps give isomorphisms δ:Hp−1(G,B)≈→ Hp(G,A) for all

p > 1. Now, for p > 1, if we already know the map Hp−1(G,−)→ Hp−1(G′, f∗−) in dimension p− 1 wemay define it in dimension p via the composition

Hp(G,A)δ−1

≈- Hp−1(G,B) - Hp−1(G′, f∗B)

∂- Hp(G′, f∗A)

where ∂ denotes the connecting map with respect to the exact sequence of G′-modules

0→ f∗A→ f∗I → f∗B → 0

For the “base case” p = 1 the procedure is similar but we need to use the commutative diagram instead

0 - AG - IG - BG - H1(G,A) - 0

0 - (f∗A)G′

?

∩

- (f∗I)G′

?

∩

- (f∗B)G′

?

∩

- H1(G′, f∗A)

?- · · ·

Similar procedures work for homology and also for Tate cohomology groups. One may consider themore general case where A′ is a G′-module and g:A→ A′ is a group morphism which is compatible withf in the sense that g(f(σ′) · a) = σ′ · g(a) for all a ∈ A and σ′ ∈ G′. Then the degree 0 map induced byg extends to a map Hp

T (G,A)→ HpT (G′, A′) for all p.


Definition 0.3.1 Let H ≤ G be a subgroup and M be a G-module. We define the restriction map

res:HpT (G,M)→ Hp

T (H,M)

to be the map induced by the inclusion map MG →MH in degree 0 (take f :H → G to be the inclusionmap).

Now suppose that H ⊳ G is normal. We define the inflation

inf:HpT (G/H,MH)→ Hp

T (G,M)

to be the map induced by the identity (MH)G/H = MG in degree 0 (take f :G ։ G/H to be the quotientmap and g:MH →M to be the inclusion).

For p ≥ 0, the restriction and inflation maps have a very simple description in terms of the standardresolution. Given a p-cocycle f :Gp → M , res([f ]) is represented by the restriction f :Hp → M of f toHp. On the other hand, for a p-cocycle f : (G/H)p →MH we have that inf([f ]) is given by the p-cocycle

f :Gp →M given by the composition

Gp։ (G/H)p f- MH ⊂ - M

where the unlabelled maps are the natural ones.

Theorem 0.3.2 (Inflation-restriction sequence) Let M be a G-module, and H be a normal subgroupof G. Suppose that Hp(H,M) = 0 for p = 1, . . . , q − 1. Then

0 - Hq(G/H,MH)inf- Hq(G,M)

res- Hq(H,M)

is exact.

Proof The result follows easily for p = 1 by direct computation with cocycles. The general case followsby dimension shifting. Suppose that q ≥ 2 and consider an exact sequence ofG-modules (see lemma 0.1.5)

0→M → IndG(M0)→ N → 0 (∗)

Note that the middle term is induced also as an H-module. Since H1(H,M) = 0 by hypothesis we havethat the sequence of G/H-modules

0→MH →(

IndG(M0))H → NH → 0 (∗∗)

is still exact. The middle term(

IndG(M0))H

= IndG/H(M0) is induced as a G/H-module, and we havea commutative diagram

0 - Hq(G/H,MH)inf- Hq(G,M)

res- Hq(H,M)

0 - Hq−1(G/H,NH)

≈6

inf- Hq−1(G,N)

≈6

res- Hq−1(H,N)

≈6

where the vertical arrows are the connecting maps associated to (∗) and (∗∗), which are isomorphismssince the middle terms of (∗) and (∗∗) are cohomologically trivial. Since Hp(H,N) = Hp+1(H,M) = 0for p = 1, 2, . . . , q− 2 we have that the bottom row is exact by induction on q. Hence the top row is alsoexact.


Remark 0.3.3 For the expert (but then you shouldn’t be reading this appendix!): the previous resultfollows directly from the Hochschild-Serre spectral sequence

Hp(G/H,Hq(H,M))⇒ Hp+q(G,M)

Remark 0.3.4 Using the inflation map one can define cohomology for profinite groups as well. A groupis profinite if it is the projective limit of finite groups. For instance, for any field k, its absolute Galois

group Gkdf= Gal(ksep/k) is profinite. If G = lim←−

i∈I

Gi, giving the discrete topology to the finite groups

Gi, G is made into a topological space as well, namely the projective limit of the topological spaces Gi.Since the product of discrete groups is compact by Tychonoff’s theorem and G is a closed subgroup of∏

i∈I Gi we have that G is compact. Now let M a continuous G-module, namely a G-module for whichthe action G×M →M is continuous. This means that for every m ∈M the orbit G ·m is finite. Thenwe can define the p-th cohomology group as

Hp(G,M)df= lim←−

H

Hp(G/H,MH)

where H runs over all open normal subgroups of G and the transition maps are given by inflation.

Let us go back to the finite case. Let H ≤ G be a subgroup and M be a G-module. We define thenorm map NG/H :MH →MG via

NG/H(m)df=

∑

σ∈S

σ(m)

where S is a set of left cosets representatives of G/H . Clearly this does not depend on the choice ofS and the above sum is G-invariant. Now dimension shifting allows us to extend this map to all otherdimensions using an exact sequence of G-modules (see lemma 0.1.5)

0→M → IndG(M0)→ N → 0

and the fact that the middle term is also induced as an H-module.

Definition 0.3.5 The map

cor:HpT (H,M)→ Hp

T (G,M)

induced by the norm map NG/H :MH →MG in degree 0 is called corestriction.

Theorem 0.3.6 (Restriction-Corestriction) Let H ≤ G be a subgroup and M be a G-module. Thenthe composition

HpT (G,M)

res- HpT (H,M)

cor- HpT (G,M)

equals multiplication by [G : H ]. In particular, HpT (G,M) is killed by |G|.

Proof For p = 0 we have that the above composition is

MG

NGM- MH

NHM

NG/H- MG

NGM

where the first map is the natural one. This composition is clearly multiplication by [G : H ]. The generalcase now follows easily by dimension shifting.

Example 0.3.7 Let Gp be any p-Sylow subgroup of G. Then res:Hp(G,M)→ Hp(Gp,M) is injectiveon the p-primary components of these groups. In fact, cor res is multiplication by [G : Gp], which isprime to p, and thus is an automorphism on these p-primary components.

Example 0.3.8 We have that HpT (G,Q) = 0 for all p. In fact, multiplication by |G| is an automorphism

of Q. Since Hp(G,−) is a functor, this implies that multiplication by |G| is also an automorphism ofHp

T (G,Q) which must then be zero by the theorem.


0.4 Cup product

Definition 0.4.1 There exists a unique family of maps

HpT (G,A) ⊗Hq

T (G,B)∪- Hp+q

T (G,A⊗B)

(tensor products over Z) called cup products, which are characterised by the following properties:

1. the cup product is a morphism of bifunctors (a.k.a natural transformations) in the pair (A,B);

2. for p = q = 0 the cup product is induced by the natural map AG ⊗BG → (A⊗B)G;

3. the cup product is compatible with connection morphisms: if

0→ A′ → A→ A′′ → 0

is an exact sequence of G-modules and B is a G-module such that

0→ A′ ⊗B → A⊗B → A′′ ⊗B → 0

is exact then the diagram

HpT (G,A′′)⊗Hq

T (G,B)∪- Hp+q

T (G,A′′ ⊗B)

Hp+1T (G,A′)⊗Hq

T (G,B)

δp ⊗ id

? ∪- Hp+q+1T (G,A′ ⊗B)

δp+q

?

commutes. On the other hand, if

0→ B′ → B → B′′ → 0

is an exact sequence of G-modules and A is a G-module such that

0→ A⊗B′ → A⊗B → A⊗B′′ → 0

is exact then the diagram

HpT (G,A) ⊗Hq

T (G,B′′)∪- Hp+q

T (G,A⊗B′′)

HpT (G,A) ⊗Hq+1

T (G,B′)

id⊗δq

? ∪- Hp+q+1T (G,A ⊗B′)

(−1)p · δp+q

?

commutes.

Uniqueness of this family is easily proven by dimension shifting, while existence is given by theexplicit formulas in cocycles. First we extend the standard resolution by setting C−p = C∗

p−1 for p ≥ 1,

where C∗p−1 is the dual of Cp−1 = Z[Gp], namely the free Z-module with basis given by functions

(σ∗1 , . . . , σ

∗p), which send (σ1, . . . , σp) to 1 ∈ Z and every other basis element of Cp−1 to 0 ∈ Z. The

boundary map d−p: C−p → C−p−1 is given by

d−p(σ∗1 , . . . , σ

∗p) =

∑

s∈G

∑

0≤i≤q

(−1)i(σ∗1 , . . . , σ

∗i , s

∗, σ∗i+1, . . . , σ

∗p)

and d0: C0 → C−1 is given by d0(σ0) =∑

s∈G(s∗). One may then compute the Tate cohomology groupsby applying HomG(−,M) to this sequence and computing the cohomology of the resulting complex.

Now define φp,q: Cp+q → Cp ⊗ Cq as follows:


• For p, q ≥ 0φp,q(σ0, . . . , σp+q) = (σ0, . . . , σp)⊗ (σp, . . . , σp+q)

• For p, q ≥ 1φ−p,−q(σ

∗1 , . . . , σ

∗p+q) = (σ∗

1 , . . . , σ∗p)⊗ (σ∗

p+1, . . . , σ∗p+q)

• For p ≥ 0, q ≥ 1

φp,−p−q(σ∗1 , . . . , σ

∗q ) =

∑

(σ1, s1, . . . , sp)⊗ (s∗p, . . . , s∗1, σ

∗1 , . . . , σ

∗q )

φ−p−q,p(σ∗1 , . . . , σ

∗q ) =

∑

(σ∗1 , . . . , σ

∗q , s

∗1, . . . , s

∗p)⊗ (sp, . . . , s1, σq)

φp+q,−q(σ0, . . . , σp) =∑

(σ0, . . . , σp, s1, . . . , sq)⊗ (s∗q , . . . , s∗1)

φ−q,p+q(σ0, . . . , σp) =∑

(s∗1, . . . , s∗q)⊗ (sq, . . . , s1, σ0, . . . , σp)

where the sum runs over all (s1, . . . , sp) ∈ Gp.

Then a straightforward but long check shows that (ǫ ⊗ ǫ) φ0,0 = ǫ (here ǫ: C0 → Z denotes theaugmentation map) and that

φp,q d = (d⊗ 1) φp+1,q + (−1)p(1⊗ d) φp,q+1

for all p, q ∈ Z (we omitted the indices of the coboundary maps d for notational clarity). Now given

f ∈ HomG(Cp, A) and g ∈ HomG(Cq , B) we define f ∪ g ∈ HomG(Cp+q , A⊗B) by

f ∪ g = (f ⊗ g) φp,q

and now it is easy to check that if f and g are cocycles, so is f ∪ g, and that its class depends only onthe classes of f and g. A lengthy but easy check shows that the pairing ∪ so defined has the desiredproperties of the cup product.

The following lemma can be easily proved by dimension shifting, and is left as an exercise for thereader since I’m running out of time and energy:

Lemma 0.4.2 Let H ≤ G be a subgroup. We have

1. (a∪b)∪c = a∪(b∪c) in Hp+q+rT (G,A⊗B⊗C) for a ∈ Hp

T (G,A), b ∈ HqT (G,B), c ∈ Hr

T (G,C);

2. a ∪ b = (−1)pq · b∪ a under the isomorphism A⊗B = B ⊗A for a ∈ HpT (G,A), b ∈ Hq

T (G,B);

3. res(a ∪ b) = res a ∪ res b ∈ Hp+qT (H,A⊗B) for a ∈ Hp

T (G,A) and b ∈ HqT (G,B);

4. cor(

a ∪ res b)

= cora ∪ b for a ∈ Hp(H,A) and b ∈ Hq(G,B).

Our last theorem shows that how the cup product enters in the periodicity of the cohomology ofcyclic groups:

Theorem 0.4.3 Let G be a cyclic group of order n and σ be a generator. Consider the 2-cocyclef :G×G→ Z given by

f(σi, σj) =

1 if i+ j ≥ n0 otherwise

for 0 ≤ i, j < n

Then for any G-module M the cup product − ∪ [f ] gives an isomorphism

HpT (G,M)

∪[f ]

≈- Hp+2

T (G,M)

Proof We have exact sequences of G-modules

0 - IG - Z[G]ǫ- Z - 0

0 - Zµ- Z[G]

σ−1- IG - 0

where ǫ is the augmentation map, µ denotes multiplication by 1+σ+ · · ·+σn−1 and Z[G]→ IG is given

by multiplication by σ − 1. The connecting maps ∂:HpT (G,Z)

≈→ Hp+1T (G, IG) and δ:Hp+1

T (G, IG)≈→

Hp+2T (G,Z) are isomorphisms and by explicit computations with cocycles we have that [f ] = δ ∂(φ)

where φ = 1 mod n ∈ H0T (G,Z) = Z/n.

Since all terms of the above sequences are free Z-modules they stay exact after tensoring with M .Therefore for any α ∈ Hp

T (G,M) we have that

α ∪ [f ] = α ∪ δ ∂(φ) = δ ∂(α ∪ φ) = δ ∂(α)

since − ∪ φ is the identity, as can be easily checked by dimension shifting for example. Now since

Z[G]⊗M is induced the connecting maps ∂:HpT (G,M)

≈→ Hp+1T (G,M ⊗ IG) and δ:Hp+1

T (G,M ⊗ IG)≈→

Hp+2T (G,M) are isomorphisms, hence so is δ ∂ and we are done.

Chapter 5

Bibliography

1. E. Artin and J. Tate, Class field Theory, Addison-Wesley.

2. Z. I. Borevich, I. R. Shafarevich, Number Theory, Academic Press.

3. J. W. S. Cassels, Local Fields, London Mathematical Society Student Text 3, Cambridge Uni-versity Press.

4. J. W. S. Cassels and A. Frohlich, Algebraic Number Theory, Academic Press.

5. P. Gille and T. Szamuely, Central Simple Algebras and Galois Cohomology, Cambridge Univer-sity Press.

6. J. S. Milne, Algebraic Number Theory, course notes, available at http://www.jmilne.org/math/

7. J. S. Milne, Class Field Theory, course notes, available at http://www.jmilne.org/math/

8. J. S. Milne, Abelian Varieties, course notes, available at http://www.jmilne.org/math/

9. J. Neukirch, Algebraic Number Theory, Grundlehren der mathematischen Wissenschaften 322,Springer-Verlag.

10. J. Neukirch, A. Schmidt, and K. Wingberg, Cohomology of Number Fields, Grundlehren derMathematischen Wissenschaften 323, Springer-Verlag.

11. R. S. Pierce, Associative Algebras, Graduate Texts in Mathematics 88, Springer-Verlag.

12. W. Scharlau, Quadratic and Hermitian forms, Grundlehren der Mathematischen Wissenschaften,vol. 270, Springer-Verlag.

13. J-P. Serre, A Course in Arithmetic, Graduate Texts in Mathematics 7, Springer-Verlag

14. J-P. Serre, Local Fields, Graduate Texts in Mathematics 67, Springer-Verlag

15. S. S. Shatz, Profinite groups, Arithmetic, and Geometry, Annals of Mathematical Studies 67,Princeton University Press.

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