Given a field extension $E/k$ why do the norm and trace functions of $E/k$ take values in $k$

I don't quite follow one piece of argument Lang uses about the norm and trace functions.

Given a finite field extension $E/k$, he defines the norm of $E/k$ s.t. $N(a)=\Pi_{i=1}^{r} \sigma_i(a^{p^u})$ where $r$ is the degree of separability and $p^u$ is the degree of inseparability of $E/k$. He then argues that $N(a)$ is invariant under the action of any embedding $\sigma_i$ of $E$ over $k$ and hence that $N(a)$ must lie in $k$ since $a^{p^u}$ is separable over $k$.

I don't understand how the separability of $a^{p^u}$ over $k$ comes into play here and why it forces $N(a)$ to lie in $k$.

EDIT: I can see that $N(a)$ is the product of the roots of the minimal polynomial of $a$ over $k$ and hence is a coefficient in that polynomial and hence lies in $k$. Similarly, the trace would be the sum of the roots. Is this right?

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That an element $\alpha \in E$ is separable over $k$ means that $k(\alpha)$ has $[k(\alpha) : k]$ distinct embeddings over $k$ into an algebraic closure. Since there is only one possible embedding... – Dylan Moreland Oct 11 '11 at 4:12
@Dylan: Thanks. So, since $a^{p^u}$ is separable, so is $N(a)$ and hence $k(N(a))/k$ has both separable and inseparable degree equal to 1 and hence the degree of the extension is $1$. Is there anything wrong with my argument? – Venchi Da Oct 11 '11 at 4:22