On Lang's Algebra theorem 5.4 states
The group of automorphisms of $\mathbb{F}_q$ where $q=p^n$ is cyclic of degree $n$, generated by the Frobenius isomorphism.
On the last part of the theorem Lang says that since any automorphism of $\mathbb{F}_q$ must fix the field $\mathbb{F}_p$, then the number of such automorphisms must be the degree of separability $[\mathbb{F}_q : \mathbb{F}_p]_s$ which is less or equal than the degree of the field extension (theorem 4.1 ibidem). It is not clear to me why the number of such automorphisms is degree of separability $[\mathbb{F}_q : \mathbb{F}_p]_s$ when Lang defined it as following:
given a field extension $E/k$ and an embedding $\sigma: k \rightarrow L$ where $L$ is an algebraically closed field, the degree of separability is the number of extensions of $\sigma$ to all $E$.
Using this definition the degree of separability of $\mathbb{F}_q$ over $\mathbb{F}_q$ is the number of extensions of $\sigma : \mathbb{F}_p\rightarrow L$ to all $\mathbb{F}_q$, but since there are no algebraically closed field that are finite I can't hope to have the situation $\sigma : \mathbb{F}_p\rightarrow \mathbb{F}_q$ that extends to $\sigma : \mathbb{F}_q\rightarrow \mathbb{F}_q$ since $\mathbb{F}_q$ is not algebraically closed. I feel like this should be trivial but I really can't see a way around it. Maybe this has to do with the fact that these extensions are Galois and therefore separable? Any help is appreciated.
Here a proof. Let $q= p^n.$ We know that as a group, $F_q^*$ is a cyclic group. let $a$ be a generator of this cyclic group (a primitive element). As the subgroup generated by $a$ is $F_q^*$, $a$ generates $F_q$, and it minimal polynomial is of degree $n$. This polynomial divides $X^q-X= \Pi _{x\in F_q}(X-x)$ and is therefore split in $F_q$, with all distinct roots (separable). (The extension is therefore Galois.) Let us check that the distinct roots of $P$ are $a, \Phi(a), \Phi^2(a),...\Phi^{n-1}(a)$. if not two of those guys are the same, so that for a certain $k<n$ $\Phi ^k(a)=a$, and $\Phi ^k= Id$. But the set of fixed point of $\phi ^k$ is $F_{p^k}$ so is not $F_q$ unless $k=n$. It follows that the order of $\Phi$ is $n$, and that this automophism generated the Galois group.