Let $K$ be a finite normal extension of a field $F$. Prove that for every $\alpha \in K$, the norm of $\alpha$ over $F$, given by $$\prod_{\sigma \in G(K/F)}\sigma(\alpha)$$ is an element of $F$.
I know the norm of $\alpha$ is fixed by G(K/F), but having no idea how to show $\alpha \in F$. Could you please give me some hints?
For the finite field extension $K/F$ we claim that the norm of $\alpha$ is given by $\det (m_{\alpha})\in F$, where $m_{\alpha}: K \rightarrow K$ is the $F$-linear map of multiplication by $\alpha$. The proof that this equals $\prod_{\sigma \in G(K/F)}\sigma(\alpha)$ is standard for algebraic number theory, see for example K. Conrad's article: However, most of the time it is also assumed in the beginning that the extension is separable. Then we can of course say immediately that $\alpha$ is fixed by the Galois group $G=G(K/F)$ of $K/F$ and hence $\alpha\in K^G=F$.