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

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

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

    So \mathbb{K}=\mathbb{F}(\alpha_1,\ldots,\alpha_s) where \alpha_i are the roots of the polynomial f. It follows then that \theta permutes these root since it is an injective field homomorphism, but why does this then automatically mean \theta (\mathbb{K}) = \mathbb{K} ? Is it true that \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 \mathbb{F}(\alpha_1,\dots ,\alpha_s) look like? what happens when you apply \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 \alpha_1,\ldots,\alpha_n,

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

    But why is \theta(\mathbb{K}) generated by \{ \theta(\alpha_1),\ldots\theta(\alpha_n)\}?
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  4. #4
    MHF Contributor

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    because θ just permutes the roots, and fixes F. a permutation is a bijection, and θ maps the set { \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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