Group and Galois Cohomology — Chapter 2 Galois cohomology

Chapter 2
Galois cohomology

2.1. Profinite groups

Definition 2.1.1. §

A topological group $G$ is a group endowed with a topology with respect to which both the multiplication map $G \times G \to G$ and the inversion map $G \to G$ that takes an element to its inverse are continuous.

Examples 2.1.2. §

a.: The groups $ℝ$ , $ℂ$ , $ℝ^{\times}$ , and $ℂ^{\times}$ are continuous with respect to the topologies defined by their absolute values.
b.: Any group can be made a topological group by endowing it with the discrete topology.

Remark 2.1.3. §

We may consider the category of topological groups, in which the maps are continuous homomorphisms between topological groups.

Definition 2.1.4. §

A homomorphism $ϕ : G \to G^{'}$ between topological groups $G$ and $G^{'}$ is a topological isomorphism if it is both an isomorphism and a homeomorphism.

The following lemma is almost immediate, since elements of a group are invertible.

Lemma 2.1.5. §

Let $G$ be a topological group and $g \in G$ . Then the map $m_{g} : G \to G$ with $m_{g} (a) = 𝑔𝑎$ for all $a \in G$ is a homeomorphism.

We also have the following.

Lemma 2.1.6. §

A group homomorphism $ϕ : G \to G^{'}$ between topological groups is continuous if and only if, for each open neighborhood $U$ of $1$ in $G^{'}$ with $1 \in U$ , the set $ϕ^{- 1} (U)$ contains an open neighborhood of $1$ .

Proof.

We consider the non-obvious direction. Let $V$ be an open set in $G^{'}$ , and suppose that $g \in G$ is such that $h = ϕ (g) \in V$ . Then $h^{- 1} V$ is open in $G^{'}$ as well, by Lemma 2.1.5. By assumption, there exists an open neighborhood $W$ of $1$ in $G$ contained in $ϕ^{- 1} (h^{- 1} V)$ , and so $𝑔𝑊$ is an open neighborhood of $g$ in $G$ such that $ϕ (𝑔𝑊) \subseteq V$ . Hence, $ϕ$ is continuous. □

Lemma 2.1.7. §

Let $G$ be a topological group.

a.: Any open subgroup of $G$ is closed.
b.: Any closed subgroup of finite index in $G$ is open.

Proof.

If $H$ is an open (resp., closed) subgroup of $G$ , then its cosets are open (resp., closed) as well. Moreover, $G - H$ is the union of the nontrivial cosets of $H$ . Therefore, $G - H$ is open if $G$ is open and closed if $G$ is closed of finite index, so that there are only finitely many cosets of $H$ . □

Lemma 2.1.8. §

Every open subgroup of a compact group $G$ is of finite index in $G$ .

Proof.

Let $H$ be a open subgroup of $G$ . Note that $G$ is the union of its distinct $H$ -cosets, which are open and disjoint. Since $G$ is compact, there can therefore only be finitely many cosets, which is to say that $H$ is of finite index in $G$ . □

We leave it to the reader to verify the following.

Lemma 2.1.9. §

a.: A subgroup of a topological group is a topological group with respect to the subspace topology.
b.: The quotient of a topological group $G$ by a normal subgroup $N$ is a topological group with respect to the quotient topology, and it is Hausdorff if $N$ is closed.
c.: A direct product of topological groups is a topological group with respect to the product topology.

Recall the definitions of a directed set, inverse system, and the inverse limit.

Definition 2.1.10. §

A directed set $I = (I, \geq)$ is a partially ordered set such that for every $i, j \in I$ , there exists $k \in I$ with $k \geq i$ and $k \geq j$ .

Definition 2.1.11. §

Let $I$ be a directed set. An inverse system $(G_{i}, ϕ_{i, j})$ of groups over the indexing set $I$ is a set

{G_{i} ∣ i \in I}

of groups and a set

{ϕ_{i, j} : G_{i} \to G_{j} ∣ i, j \in I, i \geq j}

of group homomrphisms.

Definition 2.1.12. §

An inverse limit

G = \underset{i}{lim_{\leftarrow}} G_{i}

of an inverse system of groups $(G_{i}, ϕ_{i, j})$ over a directed indexing set $I$ is a pair $G = (G, {π_{i} ∣ i \in I})$ consisting of a group $G$ and homomorphisms $π_{i} : G \to G_{i}$ such that $ϕ_{i, j} \circ π_{i} = π_{j}$ for all $i, j \in I$ with $i \geq j$ that satisfy the following universal property: Given a group $G^{'}$ and maps $π_{i}^{'} : G^{'} \to G_{i}$ for $i \in I$ such that $ϕ_{i, j} \circ π_{i}^{'} = π_{j}^{'}$ for all $i \geq j$ , there exists a unique map $ψ : G^{'} \to G$ such that $π_{i}^{'} = π_{i} \circ ψ$ for all $i \in I$ .

By the universal property, any two inverse limits of an inverse system of groups are canonically isomorphic (via compatible maps).

Remark 2.1.13. §

We may make the latter definition more generally with any category $𝒞$ replacing the category of groups. The groups are replaced with objects in $𝒞$ and the group homomorphisms with morphisms in $𝒞$ . Moreover, we may view the system of groups as a covariant functor to the category $𝒞$ from the category that has the elements of $I$ as its objects and morphisms $i \to j$ for each $i, j \in I$ with $i \geq j$ .

We may give a direct construction of an inverse limit of an inverse system of groups as follows. The proof is left to the reader.

Proposition 2.1.14. §

Let $(G_{i}, ϕ_{i, j})$ be an inverse system of groups over an indexing set $I$ . Then the an inverse limit of the system is given explicitly by the group

G = {{(g_{i})}_{i} \in \prod_{i \in I} G_{i} ∣ ϕ_{i, j} (g_{i}) = g_{j}}

and the maps $π_{i} : G \to G_{i}$ for $i \in I$ that are the compositions of the $G \to \prod_{i \in I} G_{i} \to G_{i}$ of inclusion followed by projection.

We may endow an inverse limit of groups with a topology as follows.

Definition 2.1.15. §

Let $(G_{i}, ϕ_{i, j})$ be an inverse system of topological groups over an indexing set $I$ , with continuous maps. Then the inverse limit topology on the inverse limit $G$ of Proposition 2.1.14 is the subspace topology for the product topology on $\prod_{i \in I} G_{i}$ .

Lemma 2.1.16. §

The inverse limit of an inverse system $(G_{i}, ϕ_{i, j})$ of topological groups (over a directed indexing set $I$ ) is a topological group under the inverse limit topology.

Proof.

The maps

\begin{array}{rcl} \prod_{i \in I} G_{i} \times \prod_{i \in I} G_{i} \to \prod_{i \in I} G_{i} & and & \prod_{i \in I} G_{i} \to \prod_{i \in I} G_{i} \end{array}

given by componentwise multiplication and inversion are clearly continuous, and this continuity is preserved under the subspace topology on the inverse limit. □

Remark 2.1.17. §

In fact, the inverse limit of an inverse system of topological groups and continuous maps, when endowed with the product topology, is an inverse limit in the category of topological groups.

When we wish to view it as a topological group, we typically endow a finite group with the discrete topology.

Definition 2.1.18. §

A profinite group is an inverse limit of a system of finite groups, endowed with the inverse limit topology for the discrete topology on the finite groups.

Recall the following definition.

Definition 2.1.19. §

A topological space is totally disconnected if and only if every point is a connected component.

We leave the following as difficult exercises.

Proposition 2.1.20. §

A compact Hausdorff space is totally disconnected if and only if it has a basis of open neighborhoods that are also closed.

Proposition 2.1.21. §

A compact Hausdorff group that is totally disconnected has a basis of neighborhoods of $1$ consisting of open normal subgroups (of finite index).

We may now give a topological characterization of profinite groups.

Theorem 2.1.22. §

A profinite topological group $G$ is compact, Hausdorff, and totally disconnected.

Proof.

First, suppose that $G$ is profinite, equal to an inverse limit of a system $(G_{i}, ϕ_{i, j})$ of finite groups over an indexing set $I$ . The direct product $\prod_{i \in I} G_{i}$ of finite (discrete) groups $G_{i}$ is compact Hausdorff (compactness being Tychonoff’s theorem). As a subset of the direct product, $G$ is Hausdorff, and to see it is compact, we show that $G$ is closed. Suppose that

{(g_{i})}_{i} \in \prod_{i \in I} G_{i}

with ${(g_{i})}_{i} \notin G$ , and choose $i, j \in I$ with $i > j$ and $ϕ_{i, j} (g_{i}) \neq g_{j}$ . The open subset

{{(h_{k})}_{k} \in \prod_{k \in I} G_{k} ∣ h_{i} = g_{i}, h_{j} = g_{j}}

of the direct product contains ${(g_{i})}_{i}$ and has trivial intersection with $G$ . In that the complement of $G$ is open, $G$ itself is closed. Finally, note that any open set $\prod_{i \in I} U_{i}$ with each $U_{i}$ open in $G_{i}$ (i.e., an arbitrary subset) and $U_{i} = G_{i}$ for all but finitely many $i$ is also closed. That is, its complement is the intersection

⋂_{j \in I} ((G_{j} - U_{j}) \times \prod_{i \in I - {j}} U_{i})

of open sets, which is actually equal to the finite intersection over $j \in I$ with $U_{i} \neq G_{i}$ . It is therefore open, and by Proposition 2.1.20, the group $G$ is totally disconnected. □

Remark 2.1.23. §

We leave it to the reader to check that the converse to Theorem 2.1.22 also holds. They key is found in the proof of part a of the following proposition.

Proposition 2.1.24. §

Let $G$ be a profinite group, and let $𝒰$ be the set of all open normal subgroups of $G$ . Then the following canonical homomorphisms are homeomorphisms:

a.: $G \to {lim_{\leftarrow}}_{N \in 𝒰} G ∕ N$ ,
b.: $H \to {lim_{\leftarrow}}_{N \in 𝒰} H ∕ (H \cap N)$ , for $H$ a closed subgroup of $G$ , and
c.: $G ∕ K \to {lim_{\leftarrow}}_{N \in 𝒰} G ∕ 𝑁𝐾$ , for $K$ a closed normal subgroup of $G$ .

Proof.

We prove part $a$ . The continuous map $ϕ$ from $G$ to the inverse limit $Q$ of its quotients has closed image, and $ϕ$ is injective since $𝒰$ is a basis of $1$ in $G$ as in Proposition 2.1.21. Suppose that ${(g_{N} N)}_{N \in 𝒰}$ is not in the image of $ϕ$ , which is exactly to say that the intersection of the closed sets $g_{N} N$ is empty. Since $G$ is compact this implies that some finite subset of the ${g_{N} N ∣ N \in 𝒰}$ is empty, and letting $M$ be the intersection of the $N$ in this subset, we see that $g_{M} M = \emptyset$ , which is a contradiction. In other words, $ϕ$ is surjective. □

The following is a consequence of Proposition 2.1.24a. We leave the proof to the reader.

Corollary 2.1.25. §

Let $G$ be a profinite group and $𝒱$ a set of open normal subgroups of $G$ that forms a basis of open neighborhoods of $1$ . Then the homomorphism

G \to \underset{N \in 𝒱}{lim_{\leftarrow}} G ∕ N

is a homeomorphism.

The following lemma will be useful later.

Lemma 2.1.26. §

The closed subgroups of a profinite group are exactly those that may be written as intersections of open subgroups.

Proof.

In a topological group, an open subgroup is also closed, an arbitrary intersection of closed sets is closed, and an arbitrary intersection of subgroups is a subgroup, so an intersection of open subgroups is a closed subgroup. Let $𝒰$ denote the set of open subgroups of a profinite group $G$ . Let $H$ be a closed subgroup of $G$ . It follows from Proposition 2.1.24b and the second isomorphism theorem that the set of subgroups of the form $𝑁𝐻$ with $N$ open normal in $G$ has intersection $H$ . Note that each $𝑁𝐻$ is open as a union of open subgroups, so it is open. □

We may also speak of pro- $p$ groups.

Definition 2.1.27. §

A pro- $p$ group, for a prime $p$ , is an inverse limit of a system of finite $p$ -groups.

We may also speak of profinite and pro- $p$ completions of groups.

Definition 2.1.28. §

Let $G$ be a group.

a.: The profinite completion $\hat{G}$ of $G$ is the inverse limit of its finite quotients $G ∕ N$ , for $N$ a normal subgroup of finite index in $G$ , together with the natural quotient maps $G ∕ N \to G ∕ N^{'}$ for $N \leq N^{'}$ .
b.: The pro- $p$ completion $G^{(p)}$ of $G$ , for a prime $p$ , is the inverse limit of the finite quotients of $G$ of $p$ -power order, i.e., of the $G ∕ N$ for $N ⊴ G$ with $[G : N]$ a power of $p$ , together with the natural quotient maps.

Remark 2.1.29. §

A group $G$ is endowed with a canonical homomorphism to its profinite completion $\hat{G}$ by the universal property of the inverse limit.

Remark 2.1.30. §

We may also speak of topological rings and fields, where multiplication, addition, and the additive inverse map are continuous, and in the case of a topological field, the multiplicative inverse map on the multiplicative group is continuous as well. We may speak of profinite rings as inverse limits by quotients by two-sided ideals of finite index (or for pro- $p$ rings, of $p$ -power index).

The next proposition shows that $ℤ_{p}$ is the pro- $p$ completion of $ℤ$ .

Proposition 2.1.31. §

Let $p$ be a prime. We have an isomorphism of rings

ψ : ℤ_{p} \to_{}^{\sim} \underset{k \geq 1}{lim_{\leftarrow}} ℤ ∕ p^{k} ℤ, \sum_{i = 0}^{\infty} a_{i} p^{i} \mapsto (\sum_{i = 0}^{k - 1} a_{i} p^{i})_{k},

where the maps $ℤ ∕ p^{k + 1} ℤ \to ℤ ∕ p^{k} ℤ$ in the system are the natural quotient maps. Moreover, $ψ$ is a homeomorphism.

Proof.

The canonical quotient map $ψ_{k} : ℤ_{p} \to ℤ ∕ p^{k} ℤ$ is the $k$ th coordinate of $ψ$ , which is then a ring homomorphism by the universal property of the inverse limit. The kernel $ψ$ is the intersection of the kernels of the maps $ψ_{k}$ , which is exactly

⋂_{k} p^{k} ℤ_{p} = 0 .

Moreover, any sequence of partial sums modulo increasing powers of $p$ has a limit in $ℤ_{p}$ , which maps to the sequence under $ψ$ . The open neighborhood $p^{n} ℤ_{p}$ of $0$ in the $p$ -adic topology is sent to the intersection

(\prod_{k = 1}^{n} {0} \times \prod_{k = n + 1}^{\infty} ℤ_{p} ∕ p^{k} ℤ_{p}) \cap (\underset{k \geq 1}{lim_{\leftarrow}} ℤ ∕ p^{k} ℤ),

which is open in the product topology. On the other hand, the inverse image of a basis open neighborhood

(\prod_{k = 1}^{n} U_{k} \times \prod_{k = n + 1}^{\infty} ℤ_{p} ∕ p^{k} ℤ_{p}) \cap (\underset{k \geq 1}{lim_{\leftarrow}} ℤ ∕ p^{k} ℤ)

with $0 \in U_{k}$ for all $1 \leq k \leq n$ under $ψ$ clearly contains $p^{n} ℤ_{p}$ . It then follows from Lemma 2.1.6 that $ψ$ is a homeomorphism. □

Definition 2.1.32. §

The Prüfer ring $\hat{ℤ}$ is the profinite completion of $ℤ$ . That is, we have

ℤ≅ \underset{n \geq 1}{lim_{\leftarrow}} ℤ ∕ 𝑛ℤ

with respect to the quotient maps $ℤ ∕ 𝑛ℤ \to ℤ ∕ 𝑚ℤ$ for $m ∣ n$ .

Since $ℤ ∕ 𝑛ℤ$ may be written as a direct product of the $ℤ ∕ p^{k} ℤ$ for primes $p$ with $p^{k}$ exactly dividing $n$ , we have the following.

Lemma 2.1.33. §

We have an isomorphism of topological rings

\hat{ℤ} ≅ \prod_{p prime} ℤ_{p} .

Example 2.1.34. §

The free profinite (or pro- $p$ ) group on a generating set $S$ is the profinite (resp., pro- $p$ ) completion of the free group on $S$ .

Remark 2.1.35. §

As with free groups, closed subgroups of free profinite (or pro- $p$ ) groups are free profinite (or pro- $p$ ) groups. Moreover, every profinite (resp., pro- $p$ ) group is a topological quotient of the free group on a set of its generators, so we may present such groups via generators and relations much as before.

Definition 2.1.36. §

A subset $S$ of a topological group $G$ is said to be a topological generating set of $G$ if $G$ is the closure of the subgroup generated by $S$ .

Definition 2.1.37. §

We say that a topological group is (topologically) finitely generated if it has a finite set of topological generators.

Remark 2.1.38. §

If $G$ is a free profinite (or pro- $p$ ) group on a set $S$ , then it is topologically generated by $S$ .

We leave a proof of the following to the reader.

Lemma 2.1.39. §

Let $G$ be a topological group, and let $H$ be a (normal) subgroup. Then the closure $\bar{H}$ of $H$ is also a (normal) subgroup of $G$ .

Definition 2.1.40. §

The Frattini subgroup $Φ (G)$ of a pro- $p$ group $G$ , where $p$ is a prime, is smallest closed normal subgroup containing the commutator subgroup $[G, G]$ and the $p$ th powers in $G$ .

The following lemma is a consequence of the well-known case of finite $p$ -groups.

Lemma 2.1.41. §

Let $G$ be a pro- $p$ group for a prime $p$ . Then $Φ (G)$ is normal in $G$ , and a subset $S$ of $G$ generates $G$ if and only if its image in $G ∕ Φ (G)$ generates $G ∕ Φ (G)$ .

Remark 2.1.42. §

In the case that $G$ is an abelian pro- $p$ group, the Frattini subgroup $Φ (G)$ in Lemma 2.1.41 is $G^{p}$ .

Finally, we state without proof the structure theorem for (topologically) finitely generated abelian pro- $p$ groups. In fact, this is an immediate consequence of the structure theorem for finitely generated modules over a PID.

Theorem 2.1.43. §

Let $A$ be a topologically finitely generated abelian pro- $p$ group. Then there exist $r, k \geq 0$ and $n_{1} \geq n_{2} \geq \dots \geq n_{k} \geq 1$ such that we have an isomorphism

𝐴≅ ℤ_{p}^{r} \oplus ℤ ∕ p^{n_{1}} ℤ \oplus \dots \oplus ℤ ∕ p^{n_{k}} ℤ

of topological groups.

2.2. Cohomology of profinite groups

In this section, $G$ will denote a topological group.

Definition 2.2.1. §

A topological $G$ -module $A$ is an abelian topological group such that the map $G \times A \to A$ defining the action of $G$ on $A$ is continuous.

Definition 2.2.2. §

A $G$ -module $A$ is discrete if it is a topological $G$ -module for the discrete topology on $A$ .

Proposition 2.2.3. §

Let $G$ be a profinite group, and let $A$ be a $G$ -module. The following are equivalent:

i.: $A$ is discrete,
ii.: $A = ⋃_{N \in 𝒰} A^{N}$ , where $𝒰$ is the set of open normal subgroups of $G$ , and
iii.: the stabilizer of each $a \in A$ is open in $G$ .

Proof.

Let $π : G \times A \to A$ be the map defining the $G$ -action on $A$ . For $a \in A$ , let $G_{a}$ denote the stabilizer of $a$ . If $A$ is discrete, then $π^{- 1} (a) \cap (G \times {a})$ is open and equal to $G_{a} \times {a}$ , so $G_{a}$ is open as well. Thus, (i) implies (iii). Conversely, suppose that (iii) holds. To see (i), it suffices to check that for any $a, b \in A$ , then set $X_{a, b} = {g \in G | 𝑔𝑎 = b}$ is open. If $X_{a, b}$ is nonempty, then for any $g \in X_{a, b}$ , we clearly have $X_{a, b} = G_{b} g$ , which is open by the continuity of the multiplication on $G$ . Thus, (iii) implies (i).

If $G_{a}$ is open for a given $a \in A$ , then as $𝒰$ is a base of open neighborhoods of $1$ in $G$ , there exists $N \in 𝒰$ with $N \subseteq G_{a}$ . In other words, $a \in A^{N}$ . Thus (iii) implies (ii). Conversely, suppose that (ii) holds. Take $a \in A$ and let $N \in 𝒰$ be such that $a \in A^{N}$ . Since $N$ has finite index in $G$ , the stabilizer $G_{a}$ is a finite union of $N$ -cosets, so $G_{a}$ is open as well. Thus (ii) implies (iii). □

Remark 2.2.4. §

Note that our notion of a discrete $G$ -module $A$ says only that the $G$ -action on $A$ is continuous with respect to the discrete topology, so $A$ can be thought of as a topological module when endowed with said topology. It is possible that the discrete topology is not the unique topology that makes $A$ a topological $G$ -module. For instance, $ℤ ∕ 2 ℤ$ acts on $ℝ$ by $x \mapsto - x$ , and this is continuous with respect to both the discrete and the usual topology on $ℝ$ .

Examples 2.2.5. §

a.: Every trivial $G$ -module is a discrete $G$ -module.
b.: If $G$ is finite (with the discrete topology), then every $G$ -module is discrete.
c.: If $G$ is profinite, then every finite $G$ -module is necessarily discrete.
d.: The action of $ℂ^{\times}$ on $ℂ$ by left multiplication gives $ℂ$ the structure of a $ℂ^{\times}$ -module that is not discrete.
e.: The action of ${\hat{ℤ}}^{\times}$ on the group of roots of unity in $ℂ$ by $u \cdot ζ = ζ^{u}$ , for $u \in {\hat{ℤ}}^{\times}$ and $ζ$ a root of unity, is discrete. Here, $ζ^{u}$ is $ζ$ raised to the power of any integer that is congruent to $u$ modulo the order of $ζ$ .

Definition 2.2.6. §

We say that a topological $G$ -module $A$ is discrete if its topology is the discrete topology.

Definition 2.2.7. §

For a topological $G$ -module $A$ and $i \in ℤ$ , the group of continuous $i$ -cochains of $G$ with $A$ -coefficients is

C_{cts}^{i} (G, A) = {f : G^{i} \to A ∣ f continuous} .

Lemma 2.2.8. §

Let $A$ be a topological $G$ -module. The usual differential $d_{A}^{i}$ on $C^{i} (G, A)$ restricts to a map $d_{A}^{i} : C_{cts}^{i} (G, A) \to C_{cts}^{i + 1} (G, A)$ . Thus, $(C_{cts}^{\cdot} (G, A), d_{A}^{\cdot})$ is a cochain complex.

Proof.

Set $X = G^{i + 1}$ . Since $f \in C_{cts}^{i} (G, A)$ and the multiplication maps $G \times G \to G$ and $G \times A \to A$ are continuous, so are the $i + 2$ maps $X \to A$ taking $(g_{1}, \dots, g_{i + 1})$ to $g_{1} f (g_{2}, \dots, g_{i + 1})$ , to $f (g_{1}, \dots, g_{j} g_{j + 1}, \dots, g_{i})$ for some $1 \leq j \leq i$ , and to $f (g_{1}, \dots, g_{i})$ . The alternating sum defining $d_{A}^{i} (f)$ from these $i + 2$ maps is the composition of the diagonal map $X \to X^{i + 2}$ , the direct product $X^{i + 2} \to A^{i + 2}$ of the maps in question, and the alternating sum map $A^{i + 2} \to A$ . Since all of these maps are continuous, so is $d_{A}^{i} (f)$ . □

Remark 2.2.9. §

In general, $C^{\cdot} (G, \cdot)$ is a left exact functor from the category of topological $G$ -modules with continuous $G$ -module homomorphisms to the category of abelian groups. However, it need not be exact.

Proposition 2.2.10. §

Let $G$ be a topological group. If

0 \to A \to_{}^{ι} B \to_{}^{π} C \to 0

is an exact sequence of discrete $G$ -modules, then endowing $A$ , $B$ , and $C$ with the discrete topology, the sequence

0 \to C_{cts}^{i} (G, A) \to_{}^{ι^{i}} C_{cts}^{i} (G, B) \to_{}^{π^{i}} C_{cts}^{i} (G, C) \to 0

is exact for each $i$ .

Proof.

We need only show right-exactness. Choose a set-theoretic splitting of $s : C \to B$ of $π$ . In that $B$ and $C$ are discrete, $s$ is necessarily continuous. For any continuous $f : G^{i} \to C$ , the map $s \circ f : G^{i} \to B$ is therefore continuous, and $π^{i} (s \circ f) = f$ . □

Definition 2.2.11. §

Let $G$ be a profinite group and $A$ a discrete $G$ -module. The $i$ th profinite cohomology group of $G$ with coefficients in $A$ is $H^{i} (G, A) = H^{i} (C_{cts}^{\cdot} (G, A))$ , where $A$ is endowed with the discrete topology.

Notation 2.2.12. §

If $f : A \to B$ is a $G$ -module homomorphism between discrete $G$ -modules $A$ and $B$ , where $G$ is profinite, then the induced maps on cohomology are denoted $f^{*} : H^{i} (G, A) \to H^{i} (G, B)$ .

As a corollary of Proposition 2.2.10, any short exact sequence of discrete $G$ -modules gives rise to a long exact sequence of profinite cohomology groups.

Theorem 2.2.13. §

Suppose that

0 \to A \to_{}^{ι} B \to_{}^{π} C \to 0

is a short exact sequence of discrete $G$ -modules. Then there is a long exact sequence of abelian groups

0 \to H^{0} (G, A) \to_{}^{ι^{*}} H^{0} (G, B) \to_{}^{π^{*}} H^{0} (G, C) \to_{}^{δ^{0}} H^{1} (G, A) \to \dots .

Moreover, this construction is natural in the short exact sequence in the sense of Theorem 1.2.13.

Remark 2.2.14. §

If $G$ is a profinite group and $A$ is a discrete $G$ -module, then $H^{i} (G, A)$ in the sense of Definition 2.2.11 need not be the same as $H^{i} (G, A)$ in the sense of (abstract) group cohomology. They do, however, agree in the case that $G$ is finite, since in that case $G$ is a discrete group, and every cochain $G^{i} \to A$ is continuous. Whenever $G$ is a profinite group and $A$ is discrete, we take $H^{i} (G, A)$ to be the profinite cohomology group.

Example 2.2.15. §

For a pro- $p$ group $G$ , the first cohomology group $H^{1} (G, 𝔽_{p})$ consists of the continuous homomorphisms from $G$ to $𝔽_{p}$ . It is then canonically isomorphic to the $𝔽_{p}$ -dual of $G ∕ Φ (G)$ , with $Φ (G)$ the Frattini subgroup. It follows from Lemma 2.1.41 that the $𝔽_{p}$ -dimension of $H^{1} (G, 𝔽_{p})$ is equal to the order of the smallest (topological) generating set of $G$ .

The following proposition shows that profinite cohomology groups are direct limits of usual cohomology groups of finite groups under inflation maps.

Proposition 2.2.16. §

Let $G$ be a profinite group, and let $𝒰$ be the set of open normal subgroups of $G$ . For each discrete $G$ -module $A$ , we have an isomorphism

H^{i} (G, A) ≅ \underset{N \in 𝒰}{lim_{\to}} H^{i} (G ∕ N, A^{N}),

where the direct limit is taken with respect to inflation maps, and these isomorphisms are natural in $A$ .

Proof.

It suffices to check that we have natural isomorphisms

C_{cts}^{i} (G, A) ≅ \underset{N \in 𝒰}{lim_{\to}} C^{i} (G ∕ N, A^{N})

commuting with connecting homomorphisms. We verify the isomorphism, which then clearly has the other properties. Let $f : G^{i} \to A$ be continuous. Since $G$ is compact and $A$ is discrete, the image of $f$ is finite. For each $a \in im f$ , let $M_{a} \in 𝒰$ be such that $a \in A^{M_{a}}$ . Then $M = ⋂_{a \in im f} M_{a} \in 𝒰$ , and $im f \subset A^{M}$ .

We next check that $f$ factors through ${(G ∕ H)}^{i}$ for an open subgroup $H \in 𝒰$ . For this, note that the continuity of $f$ forces it to be constant on an open neighborhood of any $x \in G^{i}$ , and inside such a neighborhood is a neighborhood of the form $x \prod_{j = 1}^{i} H_{j} (x)$ with $H_{j}$ an open normal subgroup of $G$ . Take $H (x) = ⋂_{j = 1}^{i} H_{j} (x)$ , which again is an open normal subgroup, so $f$ is constant on $𝑥𝐻 {(x)}^{i}$ . Now $G^{i}$ is covered by the $𝑥𝐻 {(x)}^{i}$ for $x \in G^{i}$ . Compactness of $G^{i}$ tells us that is a finite subcover corresponding to some $x_{1}, \dots, x_{n} \in G^{i}$ . The intersection $H = ⋂_{k = 1}^{n} H (x_{k})$ is then such that $f$ factors through ${(G ∕ H)}^{i}$ , since for any $y \in G^{i}$ , we have $y \in x_{k} H {(x_{k})}^{i}$ for some $k$ , and therefore $f$ is constant on $𝑦𝐻 \subseteq x_{i} H {(x_{i})}^{i}$ . Thus $f$ factors through ${(G ∕ H)}^{i}$ .

We have shown that $f$ is the inflation of a map ${(G ∕ H)}^{i} \to A^{M}$ . If we take $N = H \cap M$ , then $f$ factors through a map ${(G ∕ N)}^{i} \to A^{N}$ , proving the result. □

The notion of a compatible pair passes to profinite group cohomology if we merely suppose that our map of profinite groups is continuous.

Definition 2.2.17. §

Let $G$ and $G^{'}$ be profinite groups, $A$ a discrete $G$ -module and $A^{'}$ a $G^{'}$ -module. We say that a pair $(ρ, λ)$ with $ρ : G^{'} \to G$ a continuous group homomorphism and $λ : A \to A^{'}$ a group homomorphism is compatible if

λ (ρ (g^{'}) a) = g^{'} λ (a)

for all $a \in A$ and $g^{'} \in G^{'}$ .

Consequently, we have inflation, restriction, and conjugation maps as in Definition 1.8.6 and Proposition 1.8.12 so long as the subgroup is taken to be closed, which ensures that it is a profinite group. By Proposition 2.2.16 and exactness of the direct limit, it is easy to see that these maps are just direct limits of the analogous maps for usual group cohomology under inflation, as holds for any map on profinite cohomology induced by a compatible pair. In fact, we also have corestriction, defined simply as the direct limit of corestriction maps at finite level. Moreover, the inflation-restriction sequence is still exact, and this works for any closed normal subgroup. We state the higher degree version of this result for later use.

Proposition 2.2.18. §

Let $G$ be a profinite group, let $N$ be a closed normal subgroup of $G$ , and let $A$ be a discrete $G$ -module. Let $i \geq 1$ , and suppose that $H^{j} (N, A) = 0$ for all $j \leq i - 1$ . Then the sequence

0 \to H^{i} (G ∕ N, A^{N}) \to_{}^{Inf} H^{i} (G, A) \to_{}^{Res} H^{i} (N, A)

is exact.

2.3. Galois theory of infinite extensions

Recall that an algebraic extension of fields $L ∕ K$ is Galois if it is normal, so that every polynomial in $K [x]$ that has a root in $L$ splits completely, and separable, so that no irreducible polynomial in $K [x]$ has a double root in $L$ . The Galois group $Gal (L ∕ K)$ of such an extension is the group of automorphisms of $L$ that fix $K$ .

In the setting of finite Galois extensions $L ∕ K$ , the subfields $E$ of $L$ containing $K$ are in one-to-one correspondence with the subgroups $H$ of $Gal (L ∕ K)$ . In fact, the maps $E \mapsto Gal (L ∕ E)$ and $H \mapsto L^{H}$ give inverse bijections between these sets. This is not so in the setting of infinite Galois extensions, where there are rather more subgroups than there are subfields. To fix this issue, we place a topology on $Gal (L ∕ K)$ and consider only the closed subgroups under this topology. The above-described correspondences then work exactly as before.

Proposition 2.3.1. §

Let $L ∕ K$ be a Galois extension of fields. Let $E$ denote the set of finite Galois extensions of $K$ contained in $L$ , ordered by inclusion. This is a directed set. Let $ρ$ be the map

ρ : Gal (L ∕ K) \to \underset{E \in E}{lim_{\leftarrow}} Gal (E ∕ K)

defined by the universal property of the inverse limit, with the maps $Gal (E^{'} ∕ K) \to Gal (E ∕ K)$ for $E, E^{'} \in E$ with $E \subseteq E^{'}$ and the maps $Gal (L ∕ K) \to Gal (E ∕ K)$ for $E \in E$ being restriction maps. Then $ρ$ is an isomorphism.

Proof.

Let $σ \in Gal (L ∕ K)$ . If $σ |_{E} = 1$ for all $E \in E$ , then since

L = ⋃_{E \in E} E,

we have that $σ = 1$ . On the other hand, if elements $σ_{E} \in Gal (E ∕ K)$ for each $E \in E$ are compatible under restriction, then define $σ \in Gal (L ∕ K)$ by $σ (α) = σ_{E} (α)$ if $α \in E$ . Then, if $α \in E^{'}$ for some $E^{'} \in E$ as well, then

σ_{E^{'}} (α) = σ_{E \cap E^{'}} (α) = σ_{E} (α),

noting that $E \cap E^{'} \in E$ . Therefore, $σ$ is well-defined, and so $ρ$ is bijective. □

Proposition 2.3.1 gives us an obvious topology to place on the Galois group of a Galois extension.

Definition 2.3.2. §

Let $L ∕ K$ be a Galois extension of fields. The Krull topology on $Gal (L ∕ K)$ is the unique topology under which the set of $Gal (L ∕ E)$ for $E ∕ K$ finite Galois with $E \subseteq L$ forms a basis of open neighborhoods of $1$ .

Remark 2.3.3. §

The Krull topology agrees with the inverse limit topology induced by the isomorphism of Proposition 2.3.1, since

1 \to Gal (L ∕ E) \to Gal (L ∕ K) \to Gal (E ∕ K) \to 1

is exact. Therefore, if $L ∕ K$ is Galois, then $Gal (L ∕ K)$ is a topological group under the Krull topology.

Lemma 2.3.4. §

Let $L ∕ K$ be a Galois extension of fields. The open subgroups in $Gal (L ∕ K)$ are exactly those subgroups of the form $Gal (L ∕ E)$ with $E$ an intermediate field in $L ∕ K$ of finite degree over $K$ .

Proof.

First, let $E$ be an intermediate field in $L ∕ K$ of finite degree. Let $E^{'}$ be the Galois closure of $E$ in $L$ , which is of finite degree over $K$ . Then $Gal (L ∕ E^{'})$ is an open normal subgroup under the Krull topology, contained in $Gal (L ∕ E)$ . Since $Gal (L ∕ E)$ is then a union of left $Gal (L ∕ E^{'})$ -cosets, which are open, we have that $Gal (L ∕ E)$ is open.

Conversely, let $H$ be an open subgroup in $Gal (L ∕ K)$ . Then $H$ contains $Gal (L ∕ E)$ for some finite Galois extension $E ∕ K$ in $L$ . Any $α \in L^{H}$ , where $L^{H}$ is the fixed field of $H$ in $L$ , is contained in $M^{Gal (L ∕ E)}$ , where $M$ is the Galois closure of $E (α)$ . Since the restriction map $Gal (L ∕ E) \to Gal (M ∕ E)$ is surjective, we then have $α \in M^{Gal (M ∕ E)}$ . But $M ∕ K$ is finite, so $M^{Gal (M ∕ E)} = E$ by the fundamental theorem of Galois theory. Thus $L^{H} \subseteq E$ .

Let $\bar{H}$ be the image of $H$ under the restriction map $π : Gal (L ∕ K) \to Gal (E ∕ K)$ . As $Gal (L ∕ E) \leq H$ , we have that $π^{- 1} (\bar{H}) = H$ . We remark that $\bar{H} = Gal (E ∕ L^{H})$ , since $\bar{H} = Gal (E ∕ E^{\bar{H}})$ by the fundamental theorem of Galois theory for finite extensions and $L^{H} = E^{H} = E^{\bar{H}}$ . But $π^{- 1} (\bar{H})$ is then $Gal (L ∕ L^{H})$ as well. □

From this, we may derive the following.

Lemma 2.3.5. §

Let $L ∕ K$ be a Galois extension of fields. The closed subgroups of $Gal (L ∕ K)$ are exactly those of the form $Gal (L ∕ E)$ for some intermediate field $E$ in the extension $L ∕ K$ .

Proof.

Under the Krull topology on $Gal (L ∕ K)$ , the open subgroups are those of the form $Gal (L ∕ E)$ with $E ∕ K$ finite. By Lemma 2.1.26, we have therefore that the closed subgroups are those that are intersections of $Gal (L ∕ E)$ over a set $S$ of finite degree over $K$ intermediate fields $E$ . Any such intersection necessarily fixes the compositum $E^{'} = \prod_{E \in S} E$ , while if an element of $Gal (L ∕ K)$ fixes $E^{'}$ , then it fixes every $E \in S$ , so lies in the intersection. That is, any closed subgroup has the form

Gal (L ∕ E^{'}) = ⋂_{E \in S} Gal (L ∕ E) .

□

Theorem 2.3.6 (Fundamental theorem of Galois theory). §

Let $L ∕ K$ be a Galois extension. Then there are inverse one-to-one, inclusion reversing correspondences

Infinite Galois correspondence. A full diagram description follows.

Diagram description: Infinite Galois correspondence

The two maps give mutually inverse, inclusion-reversing correspondences between intermediate fields and closed subgroups.

Objects, listed by row and column:

Row 1, from left to right: column 1: left brace intermediate extensions inL / K right brace; column 2: left brace closed subgroups of Gal (L / K) right brace.

Arrows and lines:

An arrow from left brace intermediate extensions inL / K right brace to left brace closed subgroups of Gal (L / K) right brace, labelled psi.
An arrow from left brace closed subgroups of Gal (L / K) right brace to left brace intermediate extensions inL / K right brace, labelled theta.

given by $ψ (E) = Gal (L ∕ E)$ for any intermediate extension $E$ in $L ∕ K$ and $𝜃 (H) = L^{H}$ for any closed subgroup $H$ of $Gal (L ∕ K)$ . These correspondences restrict to bijections between the normal extensions of $K$ in $L$ and the closed normal subgroups of $Gal (L ∕ K)$ , as well as to bijections between the finite degree (normal) extensions of $K$ in $L$ and the open (normal) subgroups of $Gal (L ∕ K)$ . Moreover, if $E$ is normal over $K$ $($ resp., $H ⊴ Gal (L ∕ K)$ is closed), then restriction induces a topological isomorphism

Gal (L ∕ K) ∕ Gal (L ∕ E) \to_{}^{\sim} Gal (E ∕ K)

$($ resp., $Gal (L ∕ K) ∕ H \to_{}^{\sim} Gal (L^{H} ∕ K))$ .

Proof.

We will derive this from the fundamental theorem of Galois theory for finite Galois extensions. Let $E$ be an intermediate extension in $L ∕ K$ . Then $E \subseteq L^{Gal (L ∕ E)}$ by definition. Let $x \in L^{Gal (L ∕ E)}$ . The Galois closure $M$ of $E (x)$ in $L$ is of finite degree over $E$ . But every element of $Gal (M ∕ E)$ extends to an element of $Gal (L ∕ E)$ , which fixes $x$ . So $x \in M^{Gal (M ∕ E)}$ , which equals $E$ by fundamental theorem of Galois theory for finite Galois extensions. Since $x$ was arbitrary, we have $E = L^{Gal (L ∕ E)}$ . In other words, $𝜃 (ψ (E)) = E$ .

Let $H$ be a closed subgroup of $Gal (L ∕ K)$ . In Lemma 2.3.5, we saw that $H = Gal (L ∕ E)$ for some intermediate $E$ in $L ∕ K$ . Since $E = L^{Gal (L ∕ E)} = L^{H}$ from what we have shown, we have that $H = Gal (L ∕ L^{H})$ . Therefore, $ψ (𝜃 (H)) = H$ . It follows that we have the desired inclusion-reserving one-to-one correspondences. The other claims are then easily checked and are left to the reader. □

Definition 2.3.7. §

A separable closure of a field $L$ is any field that contains all roots of all separable polynomials in $L$ .

Notation 2.3.8. §

We typically denote a separable closure of $L$ by $L^{sep}$ .

Remark 2.3.9. §

If one fixes an algebraically closed field $Ω$ containing $L$ , then there is a unique separable closure of $L$ in $Ω$ , being the subfield generated by the roots of all separable polynomials in $L [x]$ .

Definition 2.3.10. §

The absolute Galois group of a field $K$ is the Galois group

G_{K} = Gal (K^{sep} ∕ K),

where $K^{sep}$ is a separable closure of $K$ .

Remark 2.3.11. §

The absolute Galois group, despite the word “the”, is not unique, but rather depends on the choice of separable closure. An isomorphism of separable closures gives rise to a canonical isomorphism of absolute Galois groups, however.

Example 2.3.12. §

Let $q$ be a power of a prime number. Then there is a unique topological isomorphism $G_{𝔽_{q}} \to_{}^{\sim} \hat{ℤ}$ sending the Frobenius automorphism $φ_{q} : x \mapsto x^{q}$ to $1$ . To see this, note that $Gal (𝔽_{q^{n}} ∕ 𝔽_{q}) \to ℤ ∕ 𝑛ℤ$ given by sending $φ_{q}$ to $1$ is an isomorphism, and these give rise to compatible isomorphisms in the inverse limit

G_{𝔽_{q}} \to_{}^{\sim} \underset{n}{lim_{\leftarrow}} Gal (𝔽_{q^{n}} ∕ 𝔽_{q}) \to_{}^{\sim} \underset{n}{lim_{\leftarrow}} ℤ ∕ 𝑛ℤ \to_{}^{\sim} \hat{ℤ} .

Example 2.3.13. §

Let $ℚ (μ_{p^{\infty}})$ denote the field given by adjoining all $p$ -power roots of unity to $ℚ$ . Then

Gal (ℚ (μ_{p^{\infty}}) ∕ ℚ) ≅ \underset{n}{lim_{\leftarrow}} Gal (ℚ (μ_{p^{n}}) ∕ ℚ) ≅ \underset{n}{lim_{\leftarrow}} {(ℤ ∕ p^{n} ℤ)}^{\times} ≅ ℤ_{p}^{\times}

the middle isomorphisms arising from the $p^{n}$ th cyclotomic characters.

Terminology 2.3.14. §

The isomorphism $Gal (ℚ (μ_{p^{\infty}}) ∕ ℚ) \to ℤ_{p}^{\times}$ of Example 2.3.13 called the $p$ -adic cyclotomic character.

Since the compositum of two abelian extensions of a field inside a fixed algebraic closure is abelian, the following makes sense.

Notation 2.3.15. §

Let $K$ be a field. The maximal abelian extension of $K$ inside an algebraic closure of $K$ is denoted $K^{ab}$ .

Remark 2.3.16. §

The abelianization $G_{K}^{ab}$ of the absolute Galois group $G_{K}$ of a field $K$ canonically isomorphic to $Gal (K^{ab} ∕ K)$ via the map induced by restriction on $G_{K}$ .

2.4. Galois cohomology

Definition 2.4.1. §

Let $L ∕ K$ be a Galois extension of fields, and let $A$ be a discrete $Gal (L ∕ K)$ -module with respect to the Krull topology on $Gal (L ∕ K)$ . For $i \geq 0$ , the $i$ th Galois cohomology group of $L ∕ K$ with coefficients in $A$ is the profinite cohomology group $H^{i} (Gal (L ∕ K), A)$ .

Example 2.4.2. §

Let $L ∕ K$ be a Galois extension with Galois group $G$ . Then the additive and multiplicative groups of $L$ are discrete $G$ -modules. That is, $L$ is the union of the finite Galois subextensions $E$ of $K$ in $L$ , and $E = L^{Gal (L ∕ E)}$ by the fundamental theorem of infinite Galois theory.

Hilbert’s Theorem 90 admits the following generalization to Galois cohomology.

Theorem 2.4.3. §

Let $L ∕ K$ be a Galois extension of fields. Then $H^{1} (Gal (L ∕ K), L^{\times}) = 0$ .

Proof.

Let $E$ denote the set of finite Galois extensions of $K$ in $L$ . Then

H^{1} (Gal (L ∕ K), L^{\times}) = \underset{E \in E}{lim_{\to}} H^{1} (Gal (E ∕ K), E^{\times}),

which reduces us to the case that $L ∕ K$ is finite Galois. Let $G = Gal (L ∕ K)$ , and let $f : G \to L^{\times}$ be a $1$ -cocycle. We may view the elements $σ \in G$ as abelian characters $L^{\times} \to L^{\times}$ . As distinct characters of $L^{\times}$ , these characters form a linearly independent set. The sum $\sum_{σ \in G} f (σ) σ$ is therefore a nonzero map $L^{\times} \to L$ . Let $α \in L^{\times}$ be such that $z = \sum_{σ \in G} f (σ) σ (α) \neq 0$ . For any $τ \in G$ , we have

\begin{matrix} τ^{- 1} (z) = \sum_{σ \in G} τ^{- 1} (f (σ)) \cdot τ^{- 1} σ (α) = \sum_{σ \in G} τ^{- 1} (f (𝜏𝜎)) σ (α) \\ = \sum_{σ \in G} τ^{- 1} (f (τ) \cdot 𝜏𝑓 (σ)) σ (α) = τ^{- 1} (f (τ)) \sum_{σ \in G} f (σ) σ (α) = τ^{- 1} (f (τ)) z . \end{matrix}

Thus,

f (τ) = \frac{z}{τ (z)},

so $f$ is the $1$ -coboundary of $z^{- 1}$ . □

This has the usual statement of Hilbert’s Theorem 90 as a corollary.

Corollary 2.4.4. §

Let $L ∕ K$ be a finite cyclic extension of fields, and let $N_{L ∕ K} : L^{\times} \to K^{\times}$ be the norm map. Then

\ker N_{L ∕ K} = {α \in L^{\times} ∣ α = \frac{σ (β)}{β} for some β \in L^{\times}},

where $σ$ is a generator of $Gal (L ∕ K)$ .

Proof.

Since $σ$ generates $G = Gal (L ∕ K)$ , the element $σ - 1 \in ℤ [G]$ generates $I_{G}$ , and so the statement at hand is $\ker N_{L ∕ K} = I_{G} L^{\times}$ , which is to say ${\hat{H}}^{- 1} (G, L^{\times}) = 0$ . Since $G$ is cyclic, we have ${\hat{H}}^{- 1} (G, L^{\times}) ≅ H^{1} (G, L^{\times})$ . Thus, the result follows from Theorem 2.4.3. □

For the additive group, we have the following much stronger generalization of the additive version of Hilbert’s Theorem 90.

Theorem 2.4.5. §

Let $L ∕ K$ be a Galois extension of fields. Then $H^{i} (Gal (L ∕ K), L) = 0$ for all $i \geq 1$ .

Proof.

As in the proof of Theorem 2.4.3, this reduces quickly to the case that $L ∕ K$ is finite, which we therefore suppose. As a $K [G]$ -module, $L$ is free on a single generator by the normal basis theorem, and therefore it is isomorphic to

Z [G] \otimes_{ℤ} K ≅ {Ind}^{G} (K) ≅ {CoInd}^{G} (K) .

So, the result follows from the acyclicity of coinduced modules. □

Notation 2.4.6. §

For a field $K$ , we let $K^{sep}$ denote a fixed separable closure and $G_{K}$ denote its absolute Galois group.

Definition 2.4.7. §

The Brauer group $Br (K)$ of a field $K$ is $H^{2} (G_{K}, {(K^{sep})}^{\times})$ .

We have the following inflation-restriction theorem for Brauer groups.

Proposition 2.4.8. §

For any Galois extension $L ∕ K$ , there is an exact sequence

0 \to H^{2} (Gal (L ∕ K), L^{\times}) \to_{}^{Inf} Br (K) \to_{}^{Res} Br (L)

of abelian groups.

Proof.

Let $K^{sep}$ be a separable closure of $K$ containing $L$ . Note that ${({(K^{sep})}^{\times})}^{G_{L}} = L^{\times}$ by the fundamental theorem of Galois theory, and we have $H^{1} (G_{L}, {(K^{sep})}^{\times}) = 0$ by Theorem 2.4.3. The sequence is then just the inflation-restriction sequence of Proposition 2.2.18 for $i = 2$ , $G = G_{K}$ , $N = G_{L}$ , and $A = {(K^{sep})}^{\times}$ . □

Example 2.4.9. §

Consider the finite field $𝔽_{q}$ for a prime power $q$ . For $n \geq 1$ , we know that $𝔽_{q^{n}} ∕ 𝔽_{q}$ is cyclic of degree $n$ , so we have an isomorphism

H^{2} (Gal (𝔽_{q^{n}} ∕ 𝔽_{q}), 𝔽_{q^{n}}^{\times}) ≅ {\hat{H}}^{0} (Gal (𝔽_{q^{n}} ∕ 𝔽_{q}), 𝔽_{q^{n}}^{\times}) ≅ 𝔽_{q}^{\times} ∕ N_{𝔽_{q^{n}} ∕ 𝔽_{q}} 𝔽_{q^{n}}^{\times} .

Now, the norm of any primitive $(q^{n} - 1)$ th root of unity $ξ$ is

N_{𝔽_{q^{n}} ∕ 𝔽_{q}} (ξ) = \prod_{i = 0}^{n - 1} ξ^{q^{i}} = ξ^{\frac{q^{n} - 1}{q - 1}},

which is a primitive $(q - 1)$ th root of unity. In other words, the norm map is surjective, so

Br (𝔽_{q}) = \underset{n}{lim_{\to}} H^{2} (Gal (𝔽_{q^{n}} ∕ 𝔽_{q}), 𝔽_{q^{n}}^{\times}) = 0 .

2.5. Kummer theory

Notation 2.5.1. §

For a field $K$ of characteristic not dividing $n \geq 1$ , we use $μ_{n}$ to denote the group of $n$ th roots of unity in $K^{sep}$ .

Notation 2.5.2. §

For an abelian group $A$ and $n \geq 1$ , let $A [n]$ denote the elements of exponent dividing $n$ in $A$ .

Example 2.5.3. §

We have $K^{sep} [n] = μ_{n}$ for any $n \geq 1$ not divisible by $char (K)$ .

Proposition 2.5.4. §

Let $K$ be a field of characteristic not dividing $n \geq 1$ , and let $μ_{n}$ be the group of roots of unity in a separable closure $K^{sep}$ of $K$ . Let $G_{K} = Gal (K^{sep} ∕ K)$ be the absolute Galois group. Then there are canonical isomorphisms

\begin{array}{rcl} K^{\times} ∕ K^{\times n} \to_{}^{\sim} H^{1} (G_{K}, μ_{n}) & and & H^{2} (G_{K}, μ_{n}) \to_{}^{\sim} {Br}_{K} [n] . \end{array}

Proof.

Since $K^{sep}$ is separably closed, we have an exact sequence

1 \to μ_{n} \to {(K^{sep})}^{\times} \to_{}^{n} {(K^{sep})}^{\times} \to 1

(2.5.1)

of discrete $G_{K}$ -modules. By Hilbert’s Theorem 90, the long exact sequence attached to (2.5.1) breaks into exact sequences

\begin{array}{rcl} K^{\times} \to_{}^{n} K^{\times} \to H^{1} (G_{K}, μ_{n}) \to 0 & and & 0 \to H^{2} (G_{K}, μ_{n}) \to Br (K) \to_{}^{n} Br (K) . \end{array}

□

Terminology 2.5.5. §

The sequence in (2.5.1) is often called a Kummer sequence.

Definition 2.5.6. §

Let $K$ be a field of characteristic not dividing $n \geq 1$ , let $a \in K^{\times}$ , and choose an $n$ th root $α \in {(K^{sep})}^{\times}$ of $a$ . The Kummer cocycle $χ_{a} : G_{K} \to μ_{n}$ attached to $a$ (or more precisely, $α$ ) is the $1$ -cocycle defined on $σ \in G_{K}$ by

χ_{a} (σ) = \frac{σ (α)}{α} .

Remarks 2.5.7. §

We maintain the notation of Definition 2.5.6.

a.: If $μ_{n} \subset K$ , then $χ_{a}$ is independent of the choice of $α$ and is in fact a group homomorphism, since $G_{K}$ acts trivially on $μ_{n}$ . In this case, we refer to $χ_{a}$ as the Kummer character attached to $a$ .
b.: The class of $χ_{a}$ in $H^{1} (G_{K}, μ_{n})$ is independent of the choice of $α$ , as the difference between two such choices is the $1$ -coboundary of an $n$ th root of unity.

Lemma 2.5.8. §

Let $K$ be a field of characteristic not dividing $n \geq 1$ . Then the isomorphism $K^{\times} ∕ K^{\times n} \to_{}^{\sim} H^{1} (G_{K}, μ_{n})$ of Proposition 2.5.4 takes the image of $a \in K^{\times}$ to $χ_{a}$ .

Proof.

The connecting homomorphism yielding the map is the snake lemma map in the diagram

The Kummer connecting homomorphism. A full diagram description follows.

Diagram description: The Kummer connecting homomorphism

This is the snake-lemma diagram used to compute the Kummer connecting homomorphism. The horizontal power maps are labelled n, and the vertical maps are d superscript 0.

Objects, listed by row and column:

Row 1, from left to right: column 1: 1; column 2: mu subscript (n); column 3: (K superscript (sep)) superscript (times); column 4: (K superscript (sep)) superscript (times); column 5: 1.
Row 2, from left to right: column 1: 0; column 2: Z superscript (1)(G subscript (K), mu subscript (n)); column 3: Z superscript (1)(G subscript (K),(K superscript (sep)) superscript (times)); column 4: Z superscript (1)(G subscript (K),(K superscript (sep)) superscript (times)); column 5: 0.

Arrows and lines:

An arrow from 1 (row 1, column 1) to mu subscript (n), without a label.
An arrow from mu subscript (n) to (K superscript (sep)) superscript (times) (row 1, column 3), without a label.
An arrow from mu subscript (n) to Z superscript (1)(G subscript (K), mu subscript (n)), labelled d superscript (0).
An arrow from (K superscript (sep)) superscript (times) (row 1, column 3) to (K superscript (sep)) superscript (times) (row 1, column 4), labelled n.
An arrow from (K superscript (sep)) superscript (times) (row 1, column 3) to Z superscript (1)(G subscript (K),(K superscript (sep)) superscript (times)) (row 2, column 3), labelled d superscript (0).
An arrow from (K superscript (sep)) superscript (times) (row 1, column 4) to 1 (row 1, column 5), without a label.
An arrow from (K superscript (sep)) superscript (times) (row 1, column 4) to Z superscript (1)(G subscript (K),(K superscript (sep)) superscript (times)) (row 2, column 4), labelled d superscript (0).
An arrow from 0 (row 2, column 1) to Z superscript (1)(G subscript (K), mu subscript (n)), without a label.
An arrow from Z superscript (1)(G subscript (K), mu subscript (n)) to Z superscript (1)(G subscript (K),(K superscript (sep)) superscript (times)) (row 2, column 3), without a label.
An arrow from Z superscript (1)(G subscript (K),(K superscript (sep)) superscript (times)) (row 2, column 3) to Z superscript (1)(G subscript (K),(K superscript (sep)) superscript (times)) (row 2, column 4), labelled n.
An arrow from Z superscript (1)(G subscript (K),(K superscript (sep)) superscript (times)) (row 2, column 4) to 0 (row 2, column 5), without a label.

so given on $a \in K^{\times}$ by picking $α \in {(K^{sep})}^{\times}$ with $α^{n} = a$ , taking $d^{0} (α) \in Z^{1} (G_{K}, K^{\times})$ and noting that it takes values in $μ_{n}$ . Since $d^{0} (α) = χ_{a}$ by definition, we are done. □

Terminology 2.5.9. §

The isomorphism $K^{\times} ∕ K^{\times n} \to_{}^{\sim} H^{1} (G_{K}, μ_{n})$ of Lemma 2.5.8 is called the Kummer isomorphism.

Proposition 2.5.10. §

Let $L ∕ K$ be a Galois extension of fields of characteristic not dividing $n \geq 1$ , and suppose that $μ_{n}$ is contained in $L$ . Then the Kummer isomorphism restricts to an isomorphism

(K^{\times} \cap L^{\times n}) ∕ K^{\times n} \to_{}^{\sim} H^{1} (Gal (L ∕ K), μ_{n}) .

Proof.

This is a simple consequence of the inflation-restriction sequence combined with the Kummer isomorphisms for $K$ and $L$ . These yield a left exact sequence

0 \to H^{1} (Gal (L ∕ K), μ_{n}) \to K^{\times} ∕ K^{\times n} \to L^{\times} ∕ L^{\times n}

that provides the isomorphism. □

Proposition 2.5.11. §

Let $K$ be a field of characteristic not dividing $n \geq 1$ , and suppose that $K$ contains the $n$ th roots of unity. Let $L ∕ K$ be a cyclic extension of degree $n$ . Then $L = K (\sqrt[n]{a})$ for some $a \in K^{\times}$ .

Proof.

Let $ζ$ be a primitive $n$ th root of unity in $K$ . Note that $N_{L ∕ K} (ζ) = ζ^{n} = 1$ , so Hilbert’s Theorem 90 tells us that there exists $α \in L$ and a generator $σ$ of $Gal (L ∕ K)$ with $\frac{σ (α)}{α} = ζ$ . Note that

N_{L ∕ K} (α) = \prod_{i = 1}^{n} σ^{i} α = \prod_{i = 1}^{n} ζ^{i} α = ζ^{n (n - 1) ∕ 2} α^{n} = {(- 1)}^{n - 1} α^{n},

so setting $a = - N_{L ∕ K} (- α)$ , we have $α^{n} = a$ . Since $α$ has $n$ distinct conjugates in $L$ , we have that $L = K (α)$ . □

Notation 2.5.12. §

Let $Δ$ be a subset of a field $K$ , and let $n \geq 1$ be such that $K$ contains the $n$ th roots of unity in $\bar{K}$ . Then the field $K (\sqrt[n]{Δ})$ is the field given by adjoining an $n$ th root of each element of $Δ$ to $K$ .

Theorem 2.5.13 (Kummer duality). §

Let $K$ be a field of characteristic not dividing $n \geq 1$ , and suppose that $K$ contains the $n$ th roots of unity. Let $L$ be an abelian extension of $K$ of exponent dividing $n$ , and set $Δ = L^{\times n} \cap K^{\times}$ . Then $L = K (\sqrt[n]{Δ})$ , and there is a perfect bimultiplicative pairing

⟨, ⟩ : Gal (L ∕ K) \times Δ ∕ K^{\times n} \to μ_{n}

given by $⟨ σ, a ⟩ = χ_{a} (σ)$ for $σ \in Gal (L ∕ K)$ and $a \in Δ$ .

Proof.

Since $μ_{n} \subset K$ , Proposition 2.5.10 tells us that the map taking $a \in Δ$ to its Kummer cocycle $χ_{a}$ yields

Δ ∕ K^{\times n} ≅ Hom (Gal (L ∕ K), μ_{n}) .

This isomorphism gives rise to the bimultiplicative pairing $⟨, ⟩$ , and it implies that any $a \in Δ ∕ K^{\times n}$ of order $d$ dividing $n$ pairs with some element of $Gal (L ∕ K)$ to a $d$ th root of unity. It remains to show that the pairing also induces an isomorphism

Hom (Δ ∕ K^{\times n}, μ_{n}) ≅ Gal (L ∕ K) .

Clearly, $K (\sqrt[n]{Δ})$ is contained in $L$ . On the other hand, $L ∕ K$ is a compositum of cyclic extensions of exponent dividing $n$ , we have by Proposition 2.5.11 that $L = K (\sqrt[n]{Γ})$ for some subset $Γ$ of $K^{\times}$ , which then can be taken to be $Δ$ . So, let $σ \in Gal (L ∕ K)$ be of order $d$ dividing $n$ . Since $L = K (\sqrt[n]{Δ})$ , we have that there exists $a \in Δ$ such that $σ (α) ∕ α$ for $α \in L$ with $α^{n} = a$ is a primitive $d$ th root of unity times $α$ . Hence, the pairing is perfect. □

Remark 2.5.14. §

One may replace $Δ$ in Theorem 2.5.13 by any $Γ \subseteq Δ$ with $Δ = Γ K^{\times n}$ . Then $Δ ∕ K^{\times n}$ should be replaced by the isomorphic $Γ ∕ (Γ \cap K^{\times n})$ .

Remark 2.5.15. §

The pairing of Proposition 2.5.13 is perfect with respect to the Krull topology on $Gal (L ∕ K)$ and the discrete topology on $Δ$ .

Terminology 2.5.16. §

We say that $Gal (L ∕ K)$ and $Δ ∕ K^{\times n}$ in Proposition 2.5.13 are Kummer dual to each other.

Corollary 2.5.17. §

Let $K$ be a field of characteristic not dividing $n \geq 1$ , and suppose that $K$ contains the $n$ th roots of unity. The Galois group of the maximal abelian extension of $K$ of exponent $n$ is Kummer dual to $K^{\times} ∕ K^{\times n}$ .

Remark 2.5.18. §

Suppose that $K$ contains $μ_{n}$ , where $n$ is not divisible by the residue characteristic of $K$ . Let $L ∕ K$ be abelian of exponent $n$ and $G = Gal (L ∕ K)$ . Write $L = K (\sqrt[n]{Δ})$ for some $Δ \leq K^{\times}$ . Then $L_{d} = K (\sqrt[d]{Δ})$ is the maximal subextension of exponent dividing $d$ , and $G_{d} = Gal (L_{d} ∕ K) ≅ G ∕ G^{d}$ . Moreover, we have a commutative diagram of pairings

Compatibility of Kummer pairings for different exponents. A full diagram description follows.

Diagram description: Compatibility of Kummer pairings for different exponents

The structural squares and triangles displayed here commute.

Objects, listed by row and column:

Row 1, from left to right: column 1: G times capital Delta / ( capital Delta intersection K superscript (times n)); column 2: mu subscript (n).
Row 2, from left to right: column 1: G subscript (d) times capital Delta / ( capital Delta intersection K superscript (times d)); column 2: mu subscript (d).

Arrows and lines:

An arrow from G times capital Delta / ( capital Delta intersection K superscript (times n)) to mu subscript (n), labelled left angle bracket , right angle bracket.
An arrow from G times capital Delta / ( capital Delta intersection K superscript (times n)) to G subscript (d) times capital Delta / ( capital Delta intersection K superscript (times d)), without a label.
An arrow from mu subscript (n) to mu subscript (d), without a label.
An arrow from G subscript (d) times capital Delta / ( capital Delta intersection K superscript (times d)) to mu subscript (d), labelled left angle bracket , right angle bracket subscript (d).

where the left vertical map is the direct product of the restriction (or the quotient map) with the map induced by the identity and the map $μ_{n} \to μ_{d}$ is the $n ∕ d$ -power map. That is, we have

{⟨ σ, a ⟩}_{d} = \frac{σ (\sqrt[d]{a})}{\sqrt[d]{a}} = (\frac{σ (\sqrt[n]{a})}{\sqrt[n]{a}})^{n ∕ d} = {⟨ σ, a ⟩}^{n ∕ d},

where $σ$ denotes both an element of $G$ and its image in $G_{d}$ and $a$ denotes the image of an element of $Δ$ . In particular, the composition

G \to Hom (Δ, μ_{n}) \to Hom (Δ, μ_{d})

in which the first map is given by $⟨, ⟩$ and the second by the $(n ∕ d)$ -power map agrees with the map $G \to Hom (Δ, μ_{d})$ given by ${⟨, ⟩}_{d}$ .

Remark 2.5.19. §

By Kummer duality, if $K$ has characteristic $0$ and contains all roots of unity, then

G_{K}^{ab} ≅ \underset{n}{lim_{\leftarrow}} G_{K}^{ab} ∕ {(G_{K}^{ab})}^{n} ≅ \underset{n}{lim_{\leftarrow}} Hom (K^{\times}, μ_{n}) ≅ \underset{n}{lim_{\leftarrow}} {Hom}_{cts} (\hat{K^{\times}}, μ_{n}) ≅ {Hom}_{cts} (\hat{K^{\times}}, \underset{n}{lim_{\leftarrow}} μ_{n}),

where $\hat{K^{\times}}$ denotes the profinite completion of $K^{\times}$ .

Example 2.5.20. §

Let $K$ be a field of characteristic not $p$ containing the group $μ_{p^{\infty}}$ of all $p$ -power roots of unity, and let $a \in K^{\times}$ with $a \notin K^{\times p}$ . Then the field $L = K (\sqrt[p^{\infty}]{a})$ given by adjoining all $p$ -power roots of $a$ to $K$ is the union of the fields $L_{n} = K (\sqrt[p^{n}]{a})$ , each of which has degree $p^{n}$ over $K$ by Theorem 2.5.13 since $a$ has order $p^{n}$ in $K^{\times} ∕ K^{\times p^{n}}$ . Let $Δ = ⟨ a ⟩$ . Then

Gal (L ∕ K) ≅ \underset{}{lim_{\leftarrow}} Gal (L_{n} ∕ K) ≅ \underset{n}{lim_{\leftarrow}} Hom (Δ, μ_{p^{n}}) ≅ \underset{n}{lim_{\leftarrow}} μ_{p^{n}} ≅ ℤ_{p},

since a homomorphism from $Δ$ is determined by where it sends $a$ .

Definition 2.5.21. §

Let $K$ be a field of characteristic $p$ . The Tate module $ℤ_{p} (1)$ is the topological $G_{K}$ -module that is $ℤ_{p}$ as a topological group together with the action of the $G_{K}$ given by

σ \cdot a = χ (σ) a

for $a \in ℤ_{p}$ , where $χ : G_{K} \to ℤ_{p}^{\times}$ is the $p$ -adic cyclotomic character.

Remark 2.5.22. §

Let $K$ be a field of characteristic not $p$ , set $G = Gal (K (μ_{p^{\infty}}) ∕ K)$ . The group

T_{p} = \underset{n}{lim_{\leftarrow}} μ_{p^{n}}

is a Galois module also referred to as the Tate module, the action of $G$ given by multiplication by the $p$ -adic cyclotomic character $χ : G \to ℤ_{p}^{\times}$ (which factors through $G$ ) in the sense that

σ ({(ξ_{n})}_{n}) = {(ξ_{n}^{χ (σ)})}_{n}

for all ${(ξ_{n})}_{n} \in T_{p}$ . The group $T_{p}$ is noncanonically topologically isomorphic to the Tate module $ℤ_{p} (1)$ , with the isomorphism given by choice of a compatible sequence ${(ζ_{p^{n}})}_{n}$ of primitive $p^{n}$ th roots of unity, which is taken to $1$ .

Example 2.5.23. §

Let $K$ be a field of characteristic not $p$ and $a \in K^{\times} - K^{\times p}$ . Set $L = K (μ_{p^{\infty}})$ and $M = L (\sqrt[p^{\infty}]{a})$ . By Example 2.5.20, we know that $Gal (M ∕ L) ≅ ℤ_{p}$ as topological groups. But note that $M ∕ K$ is Galois. In fact, take $σ \in Gal (L ∕ K)$ and lift it to an embedding $δ$ of $M$ into a separable closure of $M$ . Then

\tilde{σ} {(\sqrt[p^{n}]{a})}^{p^{n}} = a,

so $\tilde{σ} (\sqrt[p^{n}]{a}) = ξ \sqrt[p^{n}]{a}$ for some $p^{n}$ th root of unity $ξ$ , which is in $M$ by definition. To determine the Galois group, take $τ \in Gal (M ∕ L)$ , and let $ζ$ be the $p^{n}$ th root of unity such that $τ (\sqrt[p^{n}]{a}) = ζ \sqrt[p^{n}]{a}$ . For $n \geq 1$ , we then have

\tilde{σ} τ {\tilde{σ}}^{- 1} (\sqrt[p^{n}]{a}) = \tilde{σ} (τ ({\tilde{σ}}^{- 1} (ξ^{- 1}) \sqrt[p^{n}]{a})) = \tilde{σ} (σ^{- 1} (ξ^{- 1}) ζ \sqrt[p^{n}]{a}) = σ (ζ) \sqrt[p^{n}]{a} = ζ^{χ (σ)} \sqrt[p^{n}]{a},

where $χ$ is the $p$ -adic cyclotomic character. In other words, we have

Gal (M ∕ L) ≅ Gal (M ∕ L) ⋊ Gal (L ∕ K) ≅ ℤ_{p} ⋊ ℤ_{p}^{\times},

where through the conjugation action of $Gal (L ∕ K)$ on $Gal (M ∕ L)$ , the latter module is isomorphic to the Tate module $ℤ_{p} (1)$ .