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Thread: Galois Extension

  1. #1
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    Galois Extension

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

    Prove thatif $\displaystyle L $ is any Galois extension of $\displaystyle F $ contained in $\displaystyle K $ then $\displaystyle [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 $\displaystyle \sqrt[n]{a}$ over F is $\displaystyle x^d-b$ for some $\displaystyle d|n,b\in F$ (not a hard lemma)
    Then if L is Galois over F, we must have $\displaystyle \zeta_d\in F$ in order to throw all conjugates of $\displaystyle \sqrt[d]{b}$ into K. But clearly $\displaystyle \zeta_d$ is not real unless $\displaystyle d\leq2$.
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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 $\displaystyle \sqrt[n]{a}$ over F is $\displaystyle x^d-b$ for some $\displaystyle d|n,b\in F$ (not a hard lemma)
    Then if L is Galois over F, we must have $\displaystyle \zeta_d\in F$ in order to throw all conjugates of $\displaystyle \sqrt[d]{b}$ into K. But clearly $\displaystyle \zeta_d$ is not real unless $\displaystyle d\leq2$.

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

    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 $\displaystyle \sqrt[n]{a}$ over F is $\displaystyle x^d-b$ for some $\displaystyle d|n,b\in F$ (not a hard lemma)
    Then if L is Galois over F, we must have $\displaystyle \zeta_d\in F$ in order to throw all conjugates of $\displaystyle \sqrt[d]{b}$ into K. But clearly $\displaystyle \zeta_d$ is not real unless $\displaystyle d\leq2$.
    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 $\displaystyle \zeta_n\in F$. 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. $\displaystyle \alpha$ is a root of x^n-a implies that all roots are of the form $\displaystyle \zeta_n^k\sqrt[n]{a}$ 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 $\displaystyle \sqrt[n]{a}$ must divide x^n-a. But use my idea before, that only two of the $\displaystyle \zeta_n^k$ can be real. This implies that $\displaystyle \sqrt[n]{a}$ 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 $\displaystyle \zeta_n^k$ can be real. This implies that $\displaystyle \sqrt[n]{a}$ 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 $\displaystyle \zeta_n^k$ can be real, but I don't see why this implies that $\displaystyle \sqrt[n]{a}$ can only have at most two conjugates, hence the degree is <= 2.
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