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Thread: Confusion about Normal Field Extension

  1. #1
    Senior Member slevvio's Avatar
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    Confusion about Normal Field Extension

    Hello I was a little bit confused by something and would appreciate any help.

    Let $\displaystyle \mathbb{F}$ be a field, and let $\displaystyle \mathbb{K}:\mathbb{F}$ be a finite, normal field extension. So it is the splitting field of some polynomial $\displaystyle f\in\mathbb{F}[x]$.

    Let $\displaystyle \theta:K \rightarrow \overline{\mathbb{F}}$ be an $\displaystyle \mathbb{F}$-homomorphism (i.e. it fixes the ground field) into the algebraic closure of $\displaystyle \mathbb{F}$.

    I am trying to show that $\displaystyle \theta(\mathbb{K}) = \mathbb{K}$.

    So $\displaystyle \mathbb{K}=\mathbb{F}(\alpha_1,\ldots,\alpha_s)$ where $\displaystyle \alpha_i$ are the roots of the polynomial $\displaystyle f$. It follows then that $\displaystyle \theta$ permutes these root since it is an injective field homomorphism, but why does this then automatically mean $\displaystyle \theta (\mathbb{K}) = \mathbb{K}$ ? Is it true that $\displaystyle \mathbb{F}(\theta(\alpha_1),\ldots,\theta(\alpha_s )\)=\theta(\mathbb{F}(\alpha_1,\ldots,\alpha_s))?$ Thanks for any help
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  2. #2
    MHF Contributor

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    what does a typical element of $\displaystyle \mathbb{F}(\alpha_1,\dots ,\alpha_s)$ look like? what happens when you apply $\displaystyle \theta$ to such an element?
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  3. #3
    Senior Member slevvio's Avatar
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    I think it looks like some multivariate polynomial function of $\displaystyle \alpha_1,\ldots,\alpha_n$,

    so if i apply theta i just replace each $\displaystyle \alpha_i$ by $\displaystyle \theta(\alpha_i)$)?

    But why is $\displaystyle \theta(\mathbb{K})$ generated by $\displaystyle \{ \theta(\alpha_1),\ldots\theta(\alpha_n)\}$?
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  4. #4
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    because θ just permutes the roots, and fixes F. a permutation is a bijection, and θ maps the set {$\displaystyle \alpha_1,\dots ,\alpha_s$} to itself.
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  5. #5
    Senior Member slevvio's Avatar
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    ok thanks
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