I am reading Lang's book algebra. I don't understand his proof of the first part of theorem 5.1 on page 285.
His definition of norm:
Definition. Let $E$ be a finite extension of $k$. Let $[E:k]_{s}=r$, and let $p^{\mu}=[E:k]_{i}$ if the characteristic is $p>0$, and $1$ otherwise. Let $\sigma_{1},\cdots,\sigma_{r}$ be the distinct embeddings of $E$ in an algebraic closure $k^{a}$ of $k$. If $\alpha\in E$, define its norm from $E$ to $k$ to be $$ N_{E/k}(\alpha)=N_{k}^{E}(\alpha)=\prod_{\nu=1}^{r} \sigma_{\nu}\alpha^{p^{\mu}}. $$
Then he gives the following theorem:
Theorem 5.1. Let $E/k$ be a finite extension. Then the norm $N_{k}^{E}$ is a multiplicative homomorphism of $E^{*}$ into $k^{*}$ and ...
In the proof, he says:
Proof. For the first assertion, we note that $\alpha^{p^{\mu}}$ is separable over $k$ if $p^{\mu}=[E:k]_{i}$. The product $\prod_{\nu=1}^{r}\sigma_{\nu}\alpha^{p^{\mu}}$ is left fixed under any isomorphism into $k^{a}$ because applying such an isomorphism simply permutes the factors. Hence this product must lie in $k$ since $\alpha^{p^{\mu}}$ is separable over $k$. ...
(1) Why is $\alpha^{p^{\mu}}$ separable over $k$?
(2) What isomorphisms is he talking about? I know that if $K/k$ is Galois and if $\beta\in K$ is fixed by all $\sigma\in Gal(K/k)$, then $\beta\in k$. But here we don't have a Galois extension, and the product $\prod_{\nu=1}^{r}\sigma_{\nu}\alpha^{p^{\mu}}$ is only in $k^{a}$.
Thank you for your help!
For (1), see Proposition V.6.1 on page 247. The proposition says that the minimal polynomial of $\alpha$ has roots $\alpha_1,\dots,\alpha_r$ where $r = [k(\alpha) : k]_s$ and each root has the same multiplicity $p^m$ for some $m \ge 0$. Furthermore, $\alpha^{p^m}$ is separable. Since we necessarily have $p^m = [k(\alpha) : k]_i \mid [E : k]_i$, it follows that $\alpha^{p^\mu}$ is also separable.
For (2), note that if $K$ is normal over $k$ (in particular $K = k^a$), then the fixed field of $\operatorname{Aut}(K/k)$ is a purely inseparable extension of $k$. Since $N_{E/k}(\alpha)$ is separable and lies in a purely inseparable extension of $k$ it must be that $N_{E/k}(\alpha) \in k$ (c.f. Proposition V.6.11 on page 251).