Profinite groups and infinite Galois theory

Section 3.5 Profinite groups and infinite Galois theory

Reference.

[37], IV.1 and IV.2 (for profinite groups only, not their cohomology); [36], II.4.

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.

Subsection Profinite groups

Recall the Galois correspondence for a finite extension.

Proposition 3.5.1.

Let \(L/K\) be a finite Galois extension of fields and put \(G = \Gal(L/K)\text{.}\) Then the (normal) subgroups \(H\) of \(G\) correspond to the (Galois) subextensions \(M\) of \(L\text{,}\) the correspondence in each direction being given by

\begin{equation*} H \mapsto \Fix{H}, \qquad M \mapsto \Gal(L/M). \end{equation*}

Proof.

We will state a stronger result in Theorem 3.5.7.

To see what we have to be careful about for infinite extensions, consider the following example.

Example 3.5.2.

Let \(\FF_q\) be a finite field; recall that \(\FF_q\) has exactly one finite extension of any degree. Moreover, for each \(n\text{,}\) \(\Gal(\FF_{q^n}/\FF_q)\) is cyclic of degree \(n\text{,}\) generated by the Frobenius map \(\sigma\) which sends \(x\) to \(x^q\text{.}\) In particular, \(\sigma\) generates a cyclic subgroup of \(\Gal(\overline{\FF_q}/\FF_q)\text{.}\) But this Galois group is much bigger than that! Namely, let \(\{s_n\}_{n=1}^\infty\) be a sequence with \(s_n \in \ZZ/n\ZZ\text{,}\) such that if \(m | n\text{,}\) then \(s_m \equiv s_n \pmod{m}\text{.}\) The set of such sequences forms a group \(\widehat{\ZZ}\) by componentwise addition. This group is much bigger than \(\ZZ\text{,}\) and any element gives an automorphism of \(\overline{\FF_q}\text{:}\) namely, the automorphism acts on \(\FF_{q^n}\) as \(\sigma^{s_n}\text{.}\) In fact, \(\Gal(\overline{\FF_q}/\FF_q) \cong \widehat{\ZZ}\text{,}\) and it is not true that every subgroup of \(\widehat{\ZZ}\) corresponds to a subfield of \(\overline{\FF_q}\text{:}\) the subgroup generated by \(\sigma\) has fixed field \(\FF_q\text{,}\) and you don’t recover the subgroup generated by \(\sigma\) 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.

Definition 3.5.3.

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.

Example 3.5.4.

Examples of profinite groups include the group \(\widehat{\ZZ}\) in which the subgroups \(n\widehat{\ZZ}\) are open, and the \(p\)-adic integers \(\ZZ_p\) in which the subgroups \(p^n \ZZ_p\) are open. More generally, for any local field \(K\text{,}\) the additive group \(\gotho_K\) and the multiplicative group \(\gotho_K^*\) 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 Exercise 2.

Remark 3.5.5. Warning.

A profinite group may have subgroups of finite index that are not open. For example, let \(G = 1 + t \FF_p [[ t ]]\) (under multiplication). Then \(G\) is profinite with the subgroups \(1 + t^n \FF_p [[ t ]]\) 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 \(\ZZ_p\text{,}\) with generators \(1 + t^{i}\) for \(i\) not divisible by \(p\text{.}\) Thus it has uncountably many subgroups of finite index, most of which are not open!

By contrast, a theorem of Nikolov and Segal asserts that any finitely generated profinite group (i.e., one which admits a dense finitely generated subgroup) has the property that every subgroup of finite index is open. See [39].

Subsection Infinite Galois groups

Definition 3.5.6.

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 \(L\text{.}\)

Theorem 3.5.7. 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 \(\Gal(L/K)\text{,}\) given by

\begin{equation*} H \mapsto \Fix H, \qquad M \mapsto \Gal(L/M). \end{equation*}

Proof.

See [24], Theorem 8.16.

Example 3.5.8.

The Galois correspondence of Theorem 3.5.7 holds for \(\overline{\FF_q}/\FF_q\) because the open subgroups of \(\widehat{\ZZ}\) are precisely \(n\widehat{\ZZ}\) for all positive integers \(n\text{.}\)

Another way to construct profinite groups uses inverse limits (or projective limits or sometimes just limits).

Definition 3.5.9.

Suppose we are given a partially ordered set \(I\text{,}\) a family \(\{G_i\}_{i \in I}\) of finite groups and a map \(f_{ij}: G_i \to G_j\) for each pair \((i,j) \in I \times I\) such that \(i > j\text{.}\) For simplicity, let’s assume the \(f_{ij}\) are all surjective (this is slightly more restrictive than absolutely necessary, but is always true for Galois groups). Then there is a profinite group \(G\) with open subgroups \(H_i\) for \(i \in I\) such that \(G/H_i \cong G_i\) in a manner compatible with the \(f_{ij}\text{:}\) let \(G\) be the set of families \(\{g_i\}_{i \in I}\text{,}\) where each \(g_i\) is in \(G_i\) and \(f_{ij}(g_i) = g_j\text{.}\)

Example 3.5.10.

The group \(\ZZ_p\) can be viewed either as the completion of \(\ZZ\) for the \(p\)-adic absolute value or as the inverse limit of the groups \(\ZZ/p^n\ZZ\text{.}\) Similarly, the group \(\widehat{\ZZ}\) can be viewed as the inverse limit of the groups \(\ZZ/n\ZZ\text{,}\) with the usual surjections from \(\ZZ/m\ZZ\) to \(\ZZ/n\ZZ\) 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 \(\ZZ_p\text{,}\) you can use \(p^{2n}\ZZ_p\) as the subgroups.)

Remark 3.5.11. 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!

Subsection 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.

Definition 3.5.12.

If \(G\) is profinite, by a \(G\)-module we mean a topological abelian group \(M\) with a continuous \(G\)-action \(M \times G \to M\text{.}\) 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 \(G\text{.}\) Canonical example: \(G = \Gal(L/K)\) acting on \(L^*\text{,}\) 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 3.5.13.

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

Proof.

See [36], Proposition II.4.4.

Let us unpack this statement.

Definition 3.5.14.

For \(H_1 \subseteq H_2 \subseteq G\) inclusions of finite index, we have the inflation homomorphism

\begin{equation*} \Inf: H^i(G/H_2, M^{H_2}) \to H^i(G/H_1, M^{H_1}). \end{equation*}

Via these homomorphisms, the groups \(H^i(G/H, M^H)\) form a direct system and Proposition 3.5.13 asserts that \(H^i(G,M)\) is the direct limit (or inductive limit or colimit) of the \(H^i(G/H, M^H)\text{.}\) In concrete terms, you take the disjoint union of \(H^i(G/H, M^H)\) over all \(H\text{,}\) then identify together pairs of elements that become the same somewhere down the line.

Remark 3.5.15.

One can also compute the groups \(H^i(G, M)\) using continuous cochains: this amounts to considering continuous maps \(G^{i+1} \to M\) that satisfy the same algebraic conditions as do the usual cochains. One consequence of this interpretation is that \(H^1(G,M)\) classifies continuous crossed homomorphisms modulo principal ones.

Remark 3.5.16. 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 in the profinite setting. (Remember that the formation of the Tate groups involves the norm map, i.e., summing over elements of the group.)

Exercises Exercises

1.

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

2.

For any ring \(R\text{,}\) we denote by \(\GL_n(R)\) the group of \(n \times n\) matrices over \(R\) which are invertible (equivalently, whose determinant is a unit). Prove that \(\GL_n(\widehat{\ZZ})\) 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 \(\Hom(A, \QQ/\ZZ)\) 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 \(A\text{.}\)

4.

A closed subgroup \(H\) of a profinite group \(G\) is called a Sylow \(p\)-subgroup of \(G\) if, for every open normal subgroup \(N\) of \(G\text{,}\) the image of \(H\) in \(G/N\) (a/k/a \(HN/N\)) is a Sylow \(p\)-subgroup of \(G/N\text{.}\) (It is enough to check this for \(N\) running over a neighborhood basis of the identity.) Using the Sylow theorems for finite groups, prove that:

For every prime \(p\text{,}\) there exists a Sylow \(p\)-subgroup of \(G\text{.}\) (Beware that this subgroup need not be open in \(G\text{.}\))
Every subgroup of \(G\text{,}\) the quotient of which by any open normal subgroup is a \(p\)-group, is contained in a Sylow \(p\)-subgroup.
Every two Sylow \(p\)-subgroups of \(G\) are conjugate.

Hint.

See [37] exercise IV.2.4.

5.

Compute the Sylow \(p\)-subgroups of \(\widehat{\ZZ}\text{,}\) of \(\ZZ_p^*\text{,}\) and of \(\GL_2(\ZZ_p)\text{.}\)

Hint.

See [37], exercise IV.2.4.

6. Artin-Schreier extensions.

Let \(L/K\) be a \(\ZZ/p\ZZ\)-extension of fields of characteristic \(p>0\text{.}\) Prove that \(L = K(\alpha)\) for some \(\alpha\) such that \(\alpha^p - \alpha \in K\text{.}\)

Hint.

Let \(K^{\sep}\) be a separable closure of \(K\) containing \(L\text{,}\) and consider the short exact sequence \(0 \to \FF_p \to K^{\sep} \to K^{\sep} \to 0\) in which the map \(K^{\sep} \to K^{\sep}\) is given by \(x \mapsto x^p - x\text{.}\)

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