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Math Help - Galois Extension

  1. #1
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    Galois Extension

    Let  F be a subfield of  \mathbb{R} . Let  a be an element of  F and let  K=F(\sqrt[n]{a}) where  \sqrt[n]{a} denotes the real  n^{th} root of  a .

    Prove thatif  L is any Galois extension of  F contained in  K then  [L:F]\leq2 .

    I have no idea how to approach this problem.

    thanks everyone!
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  2. #2
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    First, you need to know that the minimal polynomial of \sqrt[n]{a} over F is x^d-b for some d|n,b\in F (not a hard lemma)
    Then if L is Galois over F, we must have \zeta_d\in F in order to throw all conjugates of \sqrt[d]{b} into K. But clearly \zeta_d is not real unless d\leq2.
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  3. #3
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    Quote Originally Posted by FancyMouse View Post
    First, you need to know that the minimal polynomial of \sqrt[n]{a} over F is x^d-b for some d|n,b\in F (not a hard lemma)
    Then if L is Galois over F, we must have \zeta_d\in F in order to throw all conjugates of \sqrt[d]{b} into K. But clearly \zeta_d is not real unless d\leq2.

    Remark: By \zeta_d Fancymouse most probably meant a primitive d-th root of unity in \mathbb{C}

    Tonio
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  4. #4
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    Quote Originally Posted by FancyMouse View Post
    First, you need to know that the minimal polynomial of \sqrt[n]{a} over F is x^d-b for some d|n,b\in F (not a hard lemma)
    Then if L is Galois over F, we must have \zeta_d\in F in order to throw all conjugates of \sqrt[d]{b} into K. But clearly \zeta_d is not real unless d\leq2.
    How is that lemma proven?

    Thanks!
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  5. #5
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    Quote Originally Posted by mathman88 View Post
    How is that lemma proven?

    Thanks!
    Sorry. I mixed up with something else. The ``lemma'' I stated before is not only unnecessary but also wrong. The correct version of the lemma needs the assumption that \zeta_n\in F. The proof of the correct version needs Hilbert Theorem 90 using norms (or traces if n divides the characteristic of F), which is not hard but nontrivial to prove.

    However, we don't even need this. \alpha is a root of x^n-a implies that all roots are of the form \zeta_n^k\sqrt[n]{a} for some k (but not the converse, i.e. not all things of this form are the roots of x^n-a), since the minimal polynomial of \sqrt[n]{a} must divide x^n-a. But use my idea before, that only two of the \zeta_n^k can be real. This implies that \sqrt[n]{a} can only have at most two conjugates, hence the degree is <= 2
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  6. #6
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    Quote Originally Posted by FancyMouse View Post
    But use my idea before, that only two of the \zeta_n^k can be real. This implies that \sqrt[n]{a} can only have at most two conjugates, hence the degree is <= 2
    Sorry, but I don't see why this is always true. I see why only up to 2  \zeta_n^k can be real, but I don't see why this implies that \sqrt[n]{a} can only have at most two conjugates, hence the degree is <= 2.
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