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Math Help - Degree of a finite extension of towered fields.

  1. #1
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    Degree of a finite extension of towered fields.

    I have a question that goes like this: Suppose that F,K,L are all fields and L contains K which contains F. If L is a finite extension of F show that L is a finite extension of K and K is a finite extension of F.

    One idea I had was letting A[i]B[j] be a bases of L over F and showing we can have A[i] as a bases of L over K and B[j] as a bases of K over F but I got stuck in the algebra and think I am making heavy work of it.
    Is it enough to just say that as L is a finite extension over F both F and K must be field extensions and so as L contains them they must be finite dimensional? So K must be finite dimensional over F and L must be finite dimensional over F?
    Thanks
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Prove the following:

    (i) If \{a_1,\ldots,a_r\}\subset L is linearly independent over K , which implies r\leq (L:K)

    and

    \{b_1,\ldots,b_s\}\subset K is linearly independent over F , which implies  s\leq (K:F)

    then,

    the rs elements a_ib_j of L are linearly independent over F .

    (ii) If r= (L:K), s= (K:F) those rs elements of L generate L as a vector space over F .

    This will lead to:

    (L:F)=(L:K)(K:F)

    Fernando Revilla


    Edited: Sorry, I read too quickly and my hints were for a different problem, although related with it.
    Last edited by FernandoRevilla; January 20th 2011 at 06:12 AM.
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  3. #3
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    Yes this is sort of the reverse of the problem I am dealing with, I am familiar with the proof of the above but am still stuck with the original problem
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  4. #4
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    Quote Originally Posted by FernandoRevilla View Post
    Prove the following:

    (i) If \{a_1,\ldots,a_r\}\subset L is linearly independent over K , which implies r\leq (L:K)

    and

    \{b_1,\ldots,b_s\}\subset K is linearly independent over F , which implies s\leq (K:F)

    then,

    the rs elements a_ib_j of L are linearly independent over F .

    (ii) Those rs elements of L generate L as a vector space over F .

    This will lead to:

    (L:F)=(L:K)(K:F)

    Fernando Revilla


    Edited: Sorry, I read too quickly and my hints were for a different problem, although related with it.
    Do you really need that much?

    You can just say [L:K] is finite because the basis for L as an F-vector space will also span L as a K-vector space (it won't necessarily be a basis, but that doesn't matter).

    Also, [K:F] is finite because K is an F-subspace of the finite dimensional F-vector space L.
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by topspin1617 View Post
    Do you really need that much?
    Of course we don't. The proof I provided (see Lessons in Modern Algebra , Dubreil-Jacotin) allows to construct a basis of L over F in terms of basis of L over K and K over F .


    Fernando Revilla
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by kevinlightman View Post
    Yes this is sort of the reverse of the problem I am dealing with, I am familiar with the proof of the above but am still stuck with the original problem
    Hint :

    Following the outline (i) of my first post, analyze what would happen if (L:K)=+\infty or (K:F)=+\infty


    Fernando Revilla
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