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Math Help - Galois extension, algebraic closure

  1. #1
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    Galois extension, algebraic closure

    Let K/F be any finite extension and let \alpha \in K. Let L be a Galois extension of F containing K and let H \leq \text{Gal}(L/F) be the subgroup corresponding to K. Define the norm of \alpha from K to F to be N_{K/F}(\alpha)=\prod_{\sigma}\sigma(\alpha), where the product is taken over all the embeddings of K into an algebraic closure of F (so over a set of coset representatives for H in \text{Gal}(L/F) by the Fundamental Theorem of Galois Theory). This is a product of Galois conjugates of \alpha. In particular, if K/F is Galois this is \prod_{\sigma \in \text{Gal}(K/F)}\sigma(\alpha)

    (a) Prove that N_{K/F}(\alpha \beta)= N_{K/F}(\alpha)N_{K/F}(\beta), so that the norm is a multiplicative map from K to F.

    [This may be trivial, but I don't know how to show this.]
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  2. #2
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    Quote Originally Posted by poincare4223 View Post
    (a) Prove that N_{K/F}(\alpha \beta)= N_{K/F}(\alpha)N_{K/F}(\beta), so that the norm is a multiplicative map from K to F.

    [This may be trivial, but I don't know how to show this.]
    If \sigma is am embedding of K into an algebraic closure of L then \sigma (K) \subseteq L because L/K is normal.
    Thus, we can say that \sigma runs through all embeddings of K into L.

    Remember that, \sigma (\alpha \beta) = \sigma (\alpha) \sigma (\beta).
    This means, N_{K/F}(\alpha \beta) = \prod_{\sigma} \sigma(\alpha \beta) = \prod_{\sigma} \sigma (\alpha) \sigma (\beta) = \left( \prod_{\sigma} \sigma(\alpha) \right) \left( \prod_{\sigma} \sigma(\beta) \right) = N(\alpha)(\beta).
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