## Galois Theory: Infinite Galois Theory

So far I’ve only discussed finite Galois theory, that is the Galois theory of finite extensions. But there’s also an infinite theory; the extensions it studies are separable and normal, just like finite Galois extensions, but the fundamental theorem needs to be modified to apply. I could transfer it almost in its entirety by using the theory of topological groups, but I think it’s simpler to just make exceptions and note the equivalent topological terminology where necessary.

Infinite Galois theory is wedded to the theory of projective limits of groups, which is based on directed systems.

A set I is a directed system if it comes equipped with a partial order, which I’ll denote <–, such that given any two elements i and j of I, there exists a k in I such that i <– k and j <– k. For example, a total order induces a directed system. For a less trivial example, let I = N (without zero), with a <– b iff a divides b; given any i and j, a suitable k is their least common multiple. Note that if I has a maximal element, then that maximal element is unique.

Now, an inverse system of groups is a family of groups G(i) indexed by the set I, and homomorphisms from G(j) to G(i) whenever i <– j, denoted f(ij). The homomorphisms must be consistent with one another, i.e. given i <– j <– k, every g in G(k) must satisfy $f_{ik}(g) = f_{ij}(f_{jk}(g))$. Typically the homomorphisms are taken to be surjective, but strictly speaking this is not necessary.

For example, if I = N with the normal ordering, letting G(n) = Z/(p^n)Z and f(mn) be the obvious remainder map from Z/(p^n)Z to Z/(p^m)Z yields an inverse system. Alternatively, if I has the divisibility ordering defined above, then G(n) = Z/nZ with the remainder maps is similarly an inverse system.

An inverse system has a limit, defined to be the group that projects on all groups in the system. More precisely, let P be the product of all groups G(i). Note that this is a product rather than a direct sum, the difference being that in a direct sum all but finitely many terms must be the identity element. The inverse or projective limit $\underleftarrow{\lim}G_{i}$ is the subgroup of P consisting of all elements g in P such that for all i <– j in I, $g_{i} = f_{ij}(g_{j})$.

Finally living up to my blog’s name, I’ll also prove the abstract nonsense definition of projective limits. The projective limit comes with obvious projection maps $f_{i}: g \mapsto g_{i}$; if each f(ij) is surjective, then so is each f(i). Furthermore, taken together with those projection maps, it’s universal with the property that $f_{ij}(f_{j}(g)) = f_{i}(g)$. In plain mathematical English, it means that if H is another group with projection maps h(i) such that f(ij)h(j) = h(i), then there exists a unique group homomorphism p from H to the projective limit such that h(i) = f(i)p.

This is likely the first time you encounter universal properties – if it is, consider yourself lucky to have never seen these monsters until now – so I’ll prove it in full instead of appeal to other universal properties. The idea is to explicitly define p in such a way that will yield h(i) = f(i)p.

Every a in H induces an element $h_{i}(a)$ for each i in I. Furthermore, the condition that f(ij)h(j) = h(i) implies that $f_{ij}(h_{j}(a)) = h_{i}(a)$. Therefore, the element of P whose ith entry is $h_{i}(a)$ is in fact in the projective limit, yielding a function p from H to $\underleftarrow{\lim}G_{i}$. This is a group homomorphism, roughly because the ith element of p(ab) is $h_{i}(ab) = h_{i}(a)h_{i}(b)$. Further, that the ith element of p(a) is $h_{i}(a)$ implies immediately that h(i) = f(i)p.

Conversely, this homomorphism is unique because the condition that h(i) = f(i)p forces the ith element of p(a) to be $h_{i}(a)$. This is important because it shows that if H and H’ satisfy the universal property of the projective limit of the same inverse system of groups, then there exists a unique p with h(i) = h‘(i)p and a unique p‘ with h(i)p = h‘(i) so that p and p‘ are isomorphisms that are inverses of each other, and the universal property uniquely defines the projective limit.

Note that it’s possible to define similar inverse limits of every algebraic structure – rings, modules, algebras, and so on – as well as of topological spaces. Also note that if I has a maximal element then the projective limit is just that element, making the case of interest the one with no maximal element. Finally, duplicating an index – i.e. splitting i into i1 and i2, with i(j) <– k iff i <– k and k <– i(j) iff k <– i, and with i1 and i2 either comparable or incomparable – doesn’t change the projective limit.

As an example, the projective limit of G(n) = Z/(p^n)Z is $\mathbb{Z}_{p}$, the additive group of the p-adic integers; it’s certainly true that every sequence of residue classes modulo p^n compatible under taking remainders induces a p-adic integer and vice versa.

More interestingly, the projective limit of G(n) = Z/nZ is the product of Z(p) over all primes p. This is because by the universal property, $\underleftarrow{\lim}(G_{i} \times H_{i}) = \underleftarrow{\lim}G_{i} \times \underleftarrow{\lim}H_{i}$, and Z/(p1^a1)(p2^a2)…(p(k)^a(k))Z = Z/(p1^a1)Z * Z/(p2^a2)Z * … * Z/(p(k)^a(k))Z.

Back to Galois theory now. If L/K is an infinite Galois extension (i.e. separable and normal, as in the finite case), then the Galois groups of the intermediate Galois extensions form an inverse system, roughly because if F/E/K is a tower of Galois extensions, then Gal(F/K) projects onto Gal(E/K). If you deal with sequences better than with abstract inverse systems, then let L be the splitting field of {f1, f2, …} and look at the splitting field of f1, then this of f1f2, then this of f1f2f3…

A K-automorphism of L is defined by its action on each a in L. But L/K is assumed to be algebraic, so each a in L is contained in a finite extension of K, which has a Galois closure, say F. As F/K is normal, every K-automorphism of any extension of F will map F to itself; this was proved together with the part of the fundamental theorem concerning normal subgroups of the Galois group and normal extensions. Therefore, a K-automorphism of L is completely determined by the automorphisms it induces on each finite Galois extension F/K.

Finally, to see that Gal(L/K) is the inverse limit of Gal(F/K) over all intermediate F, note that given a tower of Galois extensions L/F/E/K, the map from Gal(L/K) to Gal(E/K) is fairly obviously the same as the map from Gal(L/K) to Gal(F/K) composed with the one from Gal(F/K) to Gal(E/K).

Here is where the standard Galois correspondence breaks down. To see why, let K = F(p) for any p, and L be the union of F(p^(q^n)) over all n; then Gal(L/K) is Z(q). The group Z(q) is uncountable and so has uncountably many subgroups, but L/K only has one proper intermediate extension for each integer n.

It’s necessary to view the projective limit as not just a group but also the inverse system it’s based on and the homomorphisms to each of the groups in the system. In this view, a subgroup must come from the groups in the system. More precisely, a sub-projective limit of G(i) arises as the projective limit of a collection of subgroups H(i) such that f(ij)(H(j)) = H(i). That will correspond to a field generated by the fixed fields of all H(i)’s.

Conversely, let F be an intermediate extension of L/K. Then F intersects every finite Galois extension in some finite intermediate extension, yielding a corresponding sub-projective limit. These two correspondences are inverses of each other by the fundamental theorem of (finite) Galois theory. Further, the sub-projective limit defined by H(i) is a normal subgroup iff each H(i) is normal in G(i), so the second part of the fundamental theorem holds as well.

Obviously, L/F is finite iff the sub-projective limit is a finite group, and F/K is finite iff the sub-projective limit has finite index in the projective limit. In the latter case, F is contained in some finite Galois extension, so it arises from a single H(i); then for all j with i <– j, H(j) is the preimage of H(i) in G(j), and for all k <– j, H(k) = f(kj)(H(j)). Conversely, a sub-projective limit that arises from a single H(i) this way has the property every g in the inverse limit that maps into H(i) in G(i) maps into the preimage of H(i) in each G(j) with i <– j, so that it’s in the sub-projective limit; the sub-projective limit is then the inverse image of H(i) in the projective limit, so its index in the projective limit is [G(i):H(i)], which is finite.

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