Galois extension, algebraic closure

**poincare4223** · December 10th 2008, 05:10 PM

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.]

**ThePerfectHacker** · December 10th 2008, 08:02 PM

Originally Posted by poincare4223

(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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