4
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.