Math 254B, UC Berkeley, Spring 2002 (Kedlaya) Profinite Groups and Infinite Galois Theory

Reference. Neukirch, Sections IV.1 and IV.2.

We’ve mostly spoken so far about finite extensions of fields and the corresponding finite Galois groups. However, Galois theory can be made to work perfectly well for infinite extensions, and it’s convenient to do so; it will be more convenient at times to work with the absolute Galois group of field instead of with the Galois groups of individual extensions.

Recall the Galois correspondence for a finite extension: if L/K is Galois and G = Gal(L/K), then the (normal) subgroups H of G correspond to the (Galois) subextensions M of L, the correspondence in each direction being given by

H ↦ Fix H M ↦ Gal(L/M).

To see what we have to be careful about, here’s one example. Let 𝔽_{q} be a finite field; recall that 𝔽_{q} has exactly one finite extension of any degree. Moreover, for each n, Gal(𝔽_{q^{n}}/𝔽_{q}) is cyclic of degree n, generated by the Frobenius map σ which sends x to x^{q}. In particular, σ generates a cyclic subgroup of Gal(\overline{𝔽_{q}}/𝔽_{q}). But this Galois group is much bigger than that! Namely, let {s_{n}}_{n=1}^{∞} be a sequence with s_{n} ∈ ℤ/nℤ, such that if m|n, then s_{m} ≡ s_{n} (mod m). The set of such sequences forms a group ℤ̂ by componentwise addition. This group is much bigger than ℤ, and any element gives an automorphism of \overline{𝔽_{q}}: namely, the automorphism acts on 𝔽_{q^{n}} as σ^{s_{n}}. In fact, Gal(\overline{𝔽_{q}}/𝔽_{q}) ≅ ℤ̂, and it is not true that every subgroup of ℤ̂ corresponds to a subfield of \overline{𝔽_{q}}: the subgroup generated by σ has fixed field 𝔽_{q}, and you don’t recover the subgroup generated by σ by taking automorphisms over the fixed field.

In order to recover the Galois correspondence, we need to impose a little extra structure on Galois groups; namely, we give them a topology.

A profinite group is a topological group which is Hausdorff and compact, and which admits a basis of neighborhoods of the identity consisting of normal subgroups. More explicitly, a profinite group is a group G plus a collection of subgroups of G of finite index designated as open subgroups, such that the intersection of two open subgroups is open, but the intersection of all of the open subgroups is trivial. Profinite groups act a lot like finite groups; some of the ways in which this is true are reflected in the exercises.

Examples of profinite groups include the group ℤ̂ in which the subgroups nℤ̂ are open, and the p-adic integers ℤ_{p} in which the subgroups p^{n}ℤ_{p} are open. More generally, for any local field K, the additive group 𝔬_{K} and the multiplicative group 𝔬_{K}^{∗} are profinite. (The additive and multiplicative groups of K are not profinite, because they’re only locally compact, not compact.) For a nonabelian example, see the exercises.

Warning. A profinite group may have subgroups of finite index that are not open. For example, let G = 1 + t𝔽_{p}[[t]] (under multiplication). Then G is profinite with the subgroups 1 + t^{n}𝔽_{p}[[t]] forming a basis of open subgroups; in particular, it has countably many open subgroups. But G is isomorphic to a countable direct product of copies of ℤ_{p}, with generators 1 + t^{i} for i not divisible by p. Thus it has uncountably many subgroups of finite index, most of which are not open!

If L/K is a Galois extension, but not necessarily finite, we make G = Gal(L/K) into a profinite group by declaring that the open subgroups of G are precisely Gal(L/M) for all finite subextensions M of L.

Theorem 1 (The Galois correspondence). Let L/K be a Galois extension (not necessarily finite). Then there is a correspondence between (Galois) subextensions M of L and (normal) closed subgroups H of Gal(L/K), given by

H ↦ Fix(H) M ↦ Gal(L/M).

For example, the Galois correspondence works for \overline{𝔽_{q}}/𝔽_{q} because the open subgroups of ℤ̂ are precisely nℤ̂ for n ∈ ℕ.

Another way to construct profinite groups uses inverse limits. Suppose we are given a partially ordered set I, a family {G_{i}}_{i∈I} of finite groups and a map f_{ij} : G_{i} → G_{j} for each pair (i, j) ∈ I × I such that i > j. For simplicity, let’s assume the f_{ij} are all surjective (this is not strictly necessary). Then there is a profinite group G with open subgroups H_{i} for i ∈ I, such that G/H_{i} ≅ G_{i}, and some other obvious compatibilities hold. Namely, let G be the set of families {g_{i}}_{i∈I}, where each g_{i} is in G_{i}, and f_{ij}(g_{i}) = g_{j}.

For example, this is one of the descriptions we gave last term for ℤ_{p}, as the inverse limit of the groups ℤ/p^{n}ℤ. Similarly, the group ℤ̂ can be viewed as the inverse limit of the groups ℤ/nℤ, with the usual surjections from ℤ/mℤ to ℤ/nℤ if m is a multiple of n (that is, the ones sending 1 to 1). In fact, any profinite group can be reconstructed as the inverse limit of its quotients by open subgroups. (And it’s enough to use just a set of open subgroups which form a basis for the topology, i.e., for ℤ_{p}, you can use p^{2n}ℤ_{p} as the subgroups.)

Rule of thumb. If profinite groups make your head hurt, you can always think instead of inverse systems of finite groups. But that might make your head hurt more!

Cohomology of profinite groups.

One can do group cohomology for groups which are profinite, not just finite, but one has to be a bit careful. These groups only make sense when you carry along the profinite topology. Thus if G is profinite, by a G-module we mean a topological abelian group M with a continuous G-action M × G → M. In particular, we say M is discrete if it has the discrete topology; that implies that the stabilizer of any element of M is open, and that M is the union of M^{H} over all open subgroups H of G. Canonical example: G = Gal(L/K) acting on L^{∗}, even if L is not finite.

The category of discrete G-modules has enough injectives, so you can define cohomology groups for any discrete G-module, and all the usual abstract nonsense will still work. The main point is that you can compute them from their finite quotients.

Proposition 2. The group H^{i}(G, M) is the direct limit of H^{i}(G/H, M^{H}) using the inflation homomorphisms.

That is, if H_{1} ⊆ H_{2} ⊆ H, you have the inflation homomorphism

Inf : H^{i}(G/H_{1}, M^{H_{1}}) → H^{i}(G/H_{2}, M^{H_{2}}),

so the groups H^{i}(G/H, M^{H}) form a direct system, and H^{i}(G, M) is the direct limit of these. (That is, you take the union of all of the H^{i}(G/H, M^{H}), then you identify pairs that become the same somewhere down the line.)

Or if you prefer, you can compute these groups using continuous cochains: use continuous maps G^{i+1} → M that satisfy the same algebraic conditions as do the usual cochains. For example, H^{1}(G, M) classifies continuous crossed homomorphisms modulo principal ones, et cetera.

Warning. The passage from finite to profinite groups is only well-behaved for cohomology. In particular, we will not attempt to define either homology or the Tate groups. (Remember that the formation of the tate groups involves the norm map, i.e., summing over elements of the group.)

Exercises.

1. Prove that every open subgroup of a profinite group contains an open normal subgroup.

2. For any ring R, we denote by GL_{n}(R) the group of n × n matrices over R which are invertible (equivalently, whose determinant is a unit). Prove that GL_{n}(ℤ̂) is a profinite group, and say as much as you can about its open subgroups.

3. Let A be an abelian torsion group. Show that Hom(A, ℚ/ℤ) is a profinite group, if we take the open subgroups to be all subgroups of finite index. This group is called the Pontryagin dual of A.

4. Neukirch exercise IV.2.4: a closed subgroup H of a profinite group G is called a p-Sylow subgroup of G if, for every open normal subgroup N of G, HN/N is a p-Sylow subgroup of G/N. Prove that:

(a) For every prime p, there exists a p-Sylow subgroup of G.

(b) Every subgroup of G, the quotient of which by any open normal subgroup is a p-group, is contained in a p-Sylow subgroup.

(b) Every two p-Sylow subgroups of G are conjugate.

You may use Sylow’s theorem (that (a)-(c) hold for finite groups) without further comment. Warning: Sylow subgroups are usually not open.

5. Neukirch exercise IV.2.4: Compute the p-Sylow subgroups of ℤ̂, of ℤ_{p}^{∗}, and of GL_{2}(ℤ_{p}).