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Math Help - Galois theory

  1. #1
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    Galois theory

    Please help me with the following question.

    Let $f(x)$ be n-th degree irreducible polynomial in $\mathbb Q[x].$ And let $a_i$ for $i\in [1,n]$ be the roots of $f(x).$ and for each $a_i$ let $d_i(f)$ be the roots of $f(x)$ in $\mathbb Q(a_i)$. Then show that $d_1(f)=...=d_n(f).$

    Now let $d(f):=d_1(f)=...=d_n(f).$ Then show that $d(f)$ is the number of automorphisms for the field $\mathbb Q(a)$ where $a$ is a root of $f(x)$.
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  2. #2
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    Quote Originally Posted by peteryellow View Post
    Let $f(x)$ be n-th degree irreducible polynomial in $\mathbb Q[x].$ And let $a_i$ for $i\in [1,n]$ be the roots of $f(x).$ and for each $a_i$ let $d_i(f)$ be the roots of $f(x)$ in $\mathbb Q(a_i)$. Then show that $d_1(f)=...=d_n(f).$
    I think you want to say d_i(f) is the number of zeros of f(x) in \mathbb{Q}(a_i).
    Is this what you want to show?

    Because otherwise let f(x) = x^4 - 2 with a_i = \zeta^ i \sqrt[4]{2} and consider d_4 (f) and d_1 (f). We have d_4(f) = \{ \sqrt[4]{2} \} and \{ \zeta \sqrt[4]{2} \} \subseteq d_1 (f) which means d_4(f) \not = d_1(f).
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  3. #3
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    yes it is correct.
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  4. #4
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    Since f(x) is irreducible it means we can define an isomorphism \phi_i: \mathbb{Q}(\alpha_i) \to \mathbb{Q}[x]/\left< f(x) \right> by \alpha_i \mapsto x + \left< f(x) \right> and \phi (a) = a for a\in \mathbb{Q}. Now for \alpha_i,\alpha_j it means \phi^{-1}_j\circ\phi_i : \mathbb{Q}(\alpha_i)\to \mathbb{Q}(\alpha_j). Let \theta_{ij} = \phi^{-1}_j \circ \phi_i, from here we see that \theta_{ij} : \mathbb{Q}(\alpha_i)\to \mathbb{Q}(\alpha_j) is an isomorphism with \theta_{ij} (\alpha_i) = \alpha_j and \theta_{ij}(a) = a for a\in \mathbb{Q}.

    Consider \text{Gal}(\mathbb{Q}(\alpha_i)/\mathbb{Q}). The order of this Galois group is equal to the number of roots of f(x) in \mathbb{Q}(\alpha_i) i.e. d_i (f). Therefore, to show that d_i (f) = d_j(f) it sufficies to prove that |\text{Gal}(\mathbb{Q}(\alpha_i)/\mathbb{Q}) | = |\text{Gal}(\mathbb{Q}(\alpha_j)/\mathbb{Q}|. To prove this notice that if \sigma \in \text{Gal}(\mathbb{Q}(\alpha_i)/\mathbb{Q}) then \theta_{ij}\circ\sigma\circ\theta_{ji} \in \text{Gal}(\mathbb{Q}(\alpha_j)/\mathbb{Q}). This means, d_i (f) = | \text{Gal}(\mathbb{Q}(\alpha_i)/\mathbb{Q})| \leq |\text{Gal}(\mathbb{Q}(\alpha_j)/\mathbb{Q})| = d_j(f). By a mirror argument d_j(f) \leq d_i(f). It follows that d_i(f) = d_j(f).
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