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Math Help - Krull Topology

  1. #1
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    Krull Topology

    Let K/F be an algebraic (infinite) Galois extension with group G.
    We can define the Krull topology on the group G.
    We wish to show this topology is totally disconnected.
    Let \phi_1,\phi_2 \in X some subset of G.
    Then there exists a subgroup N\subseteq G of the form \text{Gal}(K/E) where E/F is a finite Galois extension with \phi_2\not \in \phi_1 N (this is just a property on how this topology is defined).
    Then, X = (\phi_1 N \cap X) \cup ((G-\phi_1 N) \cap X).
    Thus, X is a union of two non-empty open sets.
    My question is why are those sets open?
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  2. #2
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    Quote Originally Posted by ThePerfectHacker View Post
    Let K/F be an algebraic (infinite) Galois extension with group G.
    We can define the Krull topology on the group G.
    We wish to show this topology is totally disconnected.
    Let \phi_1,\phi_2 \in X some subset of G.
    Then there exists a subgroup N\subseteq G of the form \text{Gal}(K/E) where E/F is a finite Galois extension with \phi_2\not \in \phi_1 N (this is just a property on how this topology is defined).
    Then, X = (\phi_1 N \cap X) \cup ((G-\phi_1 N) \cap X).
    Thus, X is a union of two non-empty open sets.
    My question is why are those sets open?
    the set \{g\text{Gal}(K/E): \ g \in G, \ F \subseteq E \subseteq K \ \text{such \ that} \ [E:F] < \infty \} is a basis for the Krull topology. so \phi_1N is open in G.

    G-\phi_1N is a union of (finitely many) cosets of N and therefore it's also open in G. thus \phi_1N \cap X and (G-\phi_1N) \cap X are

    open in X. (here X is considered with the subspace topology inherited from G.)
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  3. #3
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    Thank you. I did not notice that we were working in the topology of X rather than G.
    (Because it does not need to be open in G).

    This is Mine 13th Post!!!
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    Here is a related question I was thinking about.
    Let \sigma be any automorphism.
    Let H be a subgroup of \text{Gal}(K/F).
    Let C be a closed set such that H\subseteq C.
    If for any open set \sigma N intersects H prove \sigma \in C.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    Here is a related question I was thinking about.
    Let \sigma be any automorphism.
    Let H be a subgroup of \text{Gal}(K/F).
    Let C be a closed set such that H\subseteq C.
    If for any open set \sigma N intersects H prove \sigma \in C.
    so for some \sigma \in \text{Gal}(K/F) we know that \sigma N \cap H \neq \emptyset for all N which are of the form Gal(K/E), for some finite extension E/F. is this what you're assuming? if so, then i don't think we'll need

    H to be a subgroup. that's my first thought, but i guess i need a few minutes to think about it!
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    Quote Originally Posted by NonCommAlg View Post
    so for some \sigma \in \text{Gal}(K/F) we know that \sigma N \cap H \neq \emptyset for all N which are of the form Gal(K/E), for some finite extension E/F. is this what you're assuming? if so, then i don't think we'll need

    H to be a subgroup. that's my first thought, but i guess i need a few minutes to think about it!
    Let H be a subgroup of the Galois group. Let H_0 = \text{Gal}(K/K^H). I wish to prove that \bar H  = H_0 (the topological closure).
    It is sufficient to prove that H_0 is closed and H_0 \subseteq \bar H.
    The closed part is not the problem. It is the inclusion which is giving me a problem.

    Thus, far I know that if N is of form \text{Gal}(K/E) where E/F is finite Galois and \sigma \in H then \sigma N \cap H \not = \emptyset.
    And this is somehow supposed to prove that therefore \sigma \in \bar H.
    That is what I do not see.

    Since \bar H is the intersection of all closed sets which contain H it is sufficient to prove that if C is a closed set which contains H and \sigma N intersects H for all N (those "special" groups) then \sigma \in C and therefore \sigma is in the intersection. This was the theorem I assumed did the trick - and which is why I asked my question above.
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  7. #7
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    Quote Originally Posted by ThePerfectHacker View Post
    Here is a related question I was thinking about.
    Let \sigma be any automorphism.
    Let H be a subgroup of \text{Gal}(K/F).
    Let C be a closed set such that H\subseteq C.
    If for any open set \sigma N intersects H prove \sigma \in C.
    suppose \sigma \notin C. then since C is closed, there exists an open set \alpha N which contains \sigma and \alpha N \cap C = \emptyset. but since \sigma \in \alpha N, we must have \alpha N = \sigma N.

    so \sigma N \cap C = \emptyset, and hence \sigma N \cap H = \emptyset, because H \subseteq C. contradiction! as you see H doesn't need to be a subgroup!
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    I understand what you did but I was thinking.
    Is the theorem "a set is open if and only if it contains all its interior points" still true in general topology?
    One direction is obvious but I do not think the other direction needs to holds.

    The above theorem is of course true for the Krull Topology basically by definition.
    And I know it is true for metric spaces - in fact it can be regarded as a definition of 'open'.
    But I cannot imagine it working in general topology.

    (I never studied topology before this is why I keep on getting these basic questions).
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  9. #9
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    Quote Originally Posted by ThePerfectHacker View Post
    Is the theorem "a set is open if and only if it contains all its interior points" still true in general topology?
    it's very easy to see that the interior of a set in a topological space is the largest open subset of that set. so a set is open if and only if it's equal to the set of its interior points.
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