Let K be a char 0 field with algebraic closure ˉK and absolute Galois group G. Let U be an ultrafilter on N and F=ˉKN/U be the ultrapower of ˉK on U, which has a natural action of G.

What can be said of the field FG of G-invariant points in F ? In particular, can we describe in some way the field extension FG/K ?

Edit : the afore-mentionned field extension is quite large. It would be more interesting and more natural to look at FG/(KN/U) (the ultrapower of K over U, which can easily be seen as a subfield of FG).

**Answer**

This is quite trivial but not enough for a comment: FG is not always KN/U.

Let x1,x2,… be elements of K such that no product of finitely many of them is a perfect square in K. For instance, if K=Q we can take them to be the primes.

Our element of ˉKN/U will come from the following sequence of elements of ˉK: If n is written in base 2 as 2e1+2e2+⋯+2ek, the nth element of the sequence is √xe1xe2…xek.

For each finitely generated subgroup of G, infinitely many elements of the sequence are invariant under it. This is because any finite codimension subspace of Fn2 stil has infinitely many elements, by induction on the codimension.

Thus, the set of all subsets of N containing the elements of the sequence invariant under any finitely generated subgroup is a filter, which can be extended to an ultrafilter.

Since for each element of G, the subsequence invariant under that element is in the ultrafilter, the associated element of F is G-invariant. It lives in an algebraic extension of G.

Conversely, if G is topologically finitely generated, then FG is KN/U, because being if the set invariant under each generator is in the ultrafilter, the set invariant under all the generators is in the ultrafilter, and that’s also the set of elements of K.

**Attribution***Source : Link , Question Author : Cyrille Corpet , Answer Author : Will Sawin*