A pre-base (in fact a base) of the topologycal group G of the infinite Galois extension $F/K$ is given by $\sigma Gal(F/F_i) $ with $F_i/K$ finite Galois subextension. Those opens are uniquely identified by the property $f|_{F_i} =\sigma|_{F_i}$. If we now consider $\sigma Gal(F/F_j) \cap \tau Gal(F/F_i) $ we can get:

  1. The null set if $\tau$ and $\sigma$ doesn't agree on $F_i\cap F_j$

  2. $\gamma Gal(F/F_iF_j) $ otherwise where $\gamma$ is the extension to $F$ of the map on $F_iF_j$ which behave as $\sigma$ on $F_i$ and as $\tau$ on $F_j$

Now I fear there is an error probably in the second point (because my note aren't like this) but I can't find which assumption is wrong, can you help me?


It can be shown that the intersection of two cosets of distinct subgroups is either empty or is a coset of the intersection of the two subgroups. You have correctly identified both of these cases and the intersection of the Galois groups is what you have written: $Gal(F/F_iF_k)$.

It only remains to show what coset we will use in the second case. Let $\phi$ be any element of $\sigma Gal(F/F_i) \cap \tau Gal(F/F_k)$. Then $\phi \in \sigma Gal(F/F_i)$ and $\phi \in \tau Gal(F/F_k)$. We can then write these cosets as $\phi Gal(F/F_i)$ and $\phi Gal(F/F_k)$, so that their intersection is $$\phi Gal(F/F_k) \cap \phi Gal(F/F_k) = \phi ( Gal(F/F_i) \cap Gal(F/F_k))=\phi Gal(F/F_iF_k)$$ This just gives a more general version of your answer, where $\phi$ can be any element in the intersection of the two cosets, rather than the single element you specified. This could save us the trouble of showing that the extension $γ$ you describe exists, once we assume the intersection to be nonempty.


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