# Infinite Galois theory and $\mathbb{N}^{ \mathbb{N}}$.

I recently tried exploring a little bit of infinite Galois theory, although not in depth as i just wanted to build a few basic elements of the theory before going back to the classical finite theory.

Up until now i have defined the Krull Topology on a Galois extension E/K and given necessary and sufficient conditions on subsets of a group to be the neighbourhood system of the neutral element of that topological group. I have noticed that if E/K is a finite extension then we get the discrete topology on G.

I am not sure where to start with the following questions :

• Let H be a subgroup of G with $K^H = K$. Show that for all Galois sub-extension L of E, finite over K, any K-automorphism of L is the restriction of an automorphism belonging to H.

• By considering G a subset of $\mathbb{N}^{ \mathbb{N}}$ reformulate this last result as " H is dense in G".

• Show that the topology of $G = Gal(E/K)$ is induced by the product topology of the discrete topologies on the factors of $\mathbb{N}^{ \mathbb{N}}$.

• Deduce from this that the topological group G is compact and totally disconnected.

If i am not mistaken, these questions enable us to study G without using the language of profinite groups. But how exactly, in simple terms, do I consider G as a subset of the sequence space $\mathbb{N}^{ \mathbb{N}}$ ?

• In your first bullet, I suppose that $H$ is being supposed not closed? In your second bullet, what is your recipe for mapping $G$ into $\Bbb N^{\Bbb N}$? Since the latter is only a semigroup, not a group, were you constraining your mapping so it would go into a subgroup of the semigroup? – Lubin Mar 27 '18 at 23:56
• H is not closed and thus one cannot build any galois correspondence with it. Yes apparently the author wants to embed G into a monoid, so i guess we are looking for an induced homeomorphism rather than a group homomorphism. The topological structure of a profinite group seems to ressemble N^N: each copy of N could be a discrete finite group endowed with the discrete topology and then the cartesian product may give us a product topology... Perhaps it is possible to study G without using category theory after all. Not that i dislike that field but i haven't had any courses on it yet. – Psylex Mar 28 '18 at 10:43