2
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.