# Galois extension, algebraic closure

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• Dec 10th 2008, 05:10 PM
poincare4223
Galois extension, algebraic closure
Let $\displaystyle K/F$ be any finite extension and let $\displaystyle \alpha \in K$. Let $\displaystyle L$ be a Galois extension of $\displaystyle F$ containing $\displaystyle K$ and let $\displaystyle H \leq \text{Gal}(L/F)$ be the subgroup corresponding to $\displaystyle K$. Define the norm of $\displaystyle \alpha$ from $\displaystyle K$ to $\displaystyle F$ to be $\displaystyle N_{K/F}(\alpha)=\prod_{\sigma}\sigma(\alpha)$, where the product is taken over all the embeddings of $\displaystyle K$ into an algebraic closure of $\displaystyle F$ (so over a set of coset representatives for $\displaystyle H$ in $\displaystyle \text{Gal}(L/F)$ by the Fundamental Theorem of Galois Theory). This is a product of Galois conjugates of $\displaystyle \alpha$. In particular, if $\displaystyle K/F$ is Galois this is $\displaystyle \prod_{\sigma \in \text{Gal}(K/F)}\sigma(\alpha)$

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

[This may be trivial, but I don't know how to show this.]
• Dec 10th 2008, 08:02 PM
ThePerfectHacker
Quote:

Originally Posted by poincare4223
(a) Prove that $\displaystyle N_{K/F}(\alpha \beta)= N_{K/F}(\alpha)N_{K/F}(\beta)$, so that the norm is a multiplicative map from $\displaystyle K$ to $\displaystyle F$.

[This may be trivial, but I don't know how to show this.]

If $\displaystyle \sigma$ is am embedding of $\displaystyle K$ into an algebraic closure of $\displaystyle L$ then $\displaystyle \sigma (K) \subseteq L$ because $\displaystyle L/K$ is normal.
Thus, we can say that $\displaystyle \sigma$ runs through all embeddings of $\displaystyle K$ into $\displaystyle L$.

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