Let $K/F$ be any finite extension and let $\alpha \in K$. Let $L$ be a Galois extension of $F$ containing $K$ and let $H \leq Gal(L/F)$ 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 $Gal(L/F)$ 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 $\prod_{\sigma \in Gal(K/F)} \sigma(\alpha)$.

I want to show that $N_{K/F}(\alpha)\in F$.

I know this must be simple, however I don't see how to show it. First off I think I can show this in one of two ways. Firstly, if I can show that $\sigma(\alpha)\in F$ then clearly $N_{K/F}(\alpha)\in F$ by closure, but I am not sure if $\sigma(\alpha)$ truly is in $F$. The second way is I could show that any $\tau\in Gal(K/F)$ fixes $N_{K/F}(\alpha)$ and therefore $N_{K/F}(\alpha)\in F$, but I am not sure how to show this either. Any input would be greatly appreciated!

  • $\begingroup$ The embeddings $K \to \overline{F}$ are the elements of $Gal(N/F)$ where $N$ is the normal closure of $K/F$. Also (starting with the case $N = F(\beta)$ and generalizing) $F$ is the largest field fixed by every $\sigma \in Gal(N/F)$. Your norm $N_{K/F} = N_{N/F}$ is invariant under those $\sigma$ : $\sigma(N_{K/F}(\alpha)) = N_{K/F}(\alpha)$. Therefore $N_{K/F}(\alpha) \in F$ $\endgroup$ – reuns Apr 21 '17 at 3:19
  • $\begingroup$ You should look aat simple examples, like $\Bbb Q(i)\supset\Bbb Q$. $\endgroup$ – Lubin Apr 21 '17 at 11:53

Let $G = Gal(K/F)$, and suppose $\Omega$ is a set of coset representatives for $H$ in $G$; by your definition, $N_{K/F}(\alpha) = \prod_{\sigma \in \Omega} \sigma(\alpha)$, and this product is independent of the choice of coset representatives. Note that $G$ acts on the coset space $G/H$ by left translation, so in particular, left multiplication by any element $\tau \in G$ yields another set $\tau\Omega$ of coset representatives of $H$ in $G$. Then we see $$\tau(N_{K/F}(\alpha)) = \tau \left(\prod_{\sigma \in \Omega} \sigma(\alpha)\right) = \prod_{\sigma \in \Omega} (\tau\sigma)(\alpha) = \prod_{\rho \in \tau\Omega} \rho(\alpha) = N_{K/F}(\alpha) $$ because as noted above, $\tau\Omega$ is another set of coset representatives of $H$ in $G$. Hence, $N_{K/F}(\alpha)$ is $G$-invariant, as desired.


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