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Thread: 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 $\displaystyle d_i(f)$ is the number of zeros of $\displaystyle f(x)$ in $\displaystyle \mathbb{Q}(a_i)$.
    Is this what you want to show?

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

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