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Math Help - Normalizer subgroup corresponding to a closed intermediate field

  1. #1
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    Normalizer subgroup corresponding to a closed intermediate field

    Hello, this was another problem that I had difficulty with:

    Let F be a field extension of K, and let L be a closed intermediate field. Let G=Aut_kF. Show that the normalizer of L' in G is the set of elements of G which map L onto itself.

    L' is the corresponding subgroup (in G) of L.

    I'm not sure. I wanted to use FTGT (Fundamental Theoreom of Galois Theory), but this problem doesn't state that the extension is finite.
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  2. #2
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    Re: Normalizer subgroup corresponding to a closed intermediate field

    ok, so G = Aut(F/K), and thus L' = Aut(F/L). we want to show that NG(L') = {g in G: g(L) = L} (this is a looser requirement than g(a) = a, for all a in L).

    so suppose g is (any fixed element) in NG(L'), and h is any (arbitrary) element of L'. this means g-1hg(a) = a, for all a in L.

    now suppose g(a) = b (at the moment, all we know is b is some element of F).

    then g(g-1hg(a)) = g(a) = b, so:

    hg(a) = b, and thus h(b) = b, for all h in L'. this means b is in the fixed field of h, so b is in L. hence g(L) is contained in L, and since g is bijective, g(L) = L.

    on the other hand, suppose g in G maps L to L. we want to show that g normalizes any h in L'.

    so let g(a) = b, where a,b are in L, and let h be any element of Aut(F/L). then:

    g-1hg(a) = g-1(h(g(a))) = g-1(h(b)) = g-1(b) = a, so

    g-1hg is in L', thus g normalizes any h in L'.

    **************

    it may seem mysterious that the closure of L was not mentioned explicitly, here. so where did we use it? when we said that h(b) = b implies b is in L. since L is closed, the fixed field of L' is precisely L, no more, no less (that is: given any u in F, but not in L, some element h of L' has: h(u) ≠ u. so if b = g(a) is not in L, there is SOME h in L' with h(b) ≠ b. but the a in L we picked was arbitrary, as was h).

    note that F itself may not have this property, since F might not be galois over K. in other words, the fixed field of G could be larger than K (it has to be at least K).
    Thanks from Bingk
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  3. #3
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    Re: Normalizer subgroup corresponding to a closed intermediate field

    Thank you for that. I was actually thinking along similar lines. I knew that the normalizer contained L' and g(L)=L for g in L'. But I didn't know what to do with the other automorphisms in the normalizer, so I gave up on that track. Thanks again!
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