Why does Frobenius automorphism satisfy $x^n -1$? If we think about the Frobenius automorphism as map from $F_{p^n}$ to itself defined $ x \mapsto x^p$, and we have that $x^{p^n} = x$, for all $x \in F_{p^n}$, why do we say that the Frobenius automorphism satisfies $x^n -1$?
I have seen this claim a couple of times now in my Algebra class but I don't understand. If we plug in the automorphism we get
\begin{equation}
(x^p)^n = x^{pn}.
\end{equation}
Is the statement implying that $x^{pn}$ should be equal to $1$?
 A: As indicated in the comments, this really comes down to how you interpret the question. There is a standard way to do so, which is not the interpretation you had.
First of all, for extra clarity, I will refer to elements of the field as $a \in \mathbb F_{p^n}$ and reserve $x$ for the formal variable in the polynomial ring. The Frobenius map will then be defined as $\phi(a) = a^p$.
Now, to evaluate the polynomial $x^n - 1$ at $\phi$ we need to ask ourselves what does $\phi^n$ mean? And for that matter, what does $1$ mean? And $-$? That is, evaluating a polynomial at $\phi$ is meaningless unless we specify a ring in which $\phi$ lives. Now, your interpretation seems to be that $\phi$ is an element of the ring $R$ of functions $\mathbb F_{p^n} \longrightarrow \mathbb F_{p^n}$ whose operations are componentwise. So $fg$ is defined as $a \mapsto f(a)g(a)$, $f+g$ is $a \mapsto f(a) + g(a)$, and $0$, $1$ are interpreted as constant functions. Then $\phi$ is an element of the ring $R$, so we can evaluate the polynomial $x^n - 1$ at $\phi \in R$ as $a \mapsto \phi(a)^n - 1 = a^{pn} - 1$. As you correctly noted, there is no reason for this to be zero, so when $\phi$ is thought of as an element of the ring $R$ it does not satisfy the polynomial $x^n - 1$.
$R$ is not the right ring to work in in this context. For instance, $\phi$ has much more structure than just being a function $\mathbb F_{p^n} \longrightarrow \mathbb F_{p^n}$ - it preserves addition. That is, it is a group homomorphism. We say specifically that it is an "endomorphism" of $\mathbb F_{p^n}$. We write the set of all endomorphisms of $\mathbb F_{p^n}$ (i.e. all additive group homomorphisms $\mathbb F_{p^n} \longrightarrow \mathbb F_{p^n}$) as $End(\mathbb F_{p^n})$. Now, this is a subset of $R$ but not a subring, as it is not closed under componentwise multiplication. In fact, the constant function $1 \in R$ is certainly not an element of $End(\mathbb F_{p^n})$. But this can be given the structure of a ring in a very natural way. We'll define addition to be componentwise again, and $0$ will still be the constant function $a \mapsto 0$. As for multiplication, we will define the product $f g$ as their composition $f \circ g$. In that case, the multiplicative identity of $End(\mathbb F_{p^n})$ is none other than the identity function.
I won't show that these operations make $End(\mathbb F_{p^n})$ into a ring, but they doand this ring is the correct context for this question. What does $\phi^n$ mean when interpreted in $End(\mathbb F_{p^n})$? It is the $n$-fold composition $\phi \circ \phi \circ \dots \circ \phi$. For instance, $\phi^2(a) = \phi(\phi(a)) = (a^{p})^p = a^{p^2}$. Similarly, $\phi^3(a) = a^{p^3}$ and you can show $\phi^n(a) = a^{p^n}$. So finally, we can evaluate $x^n - 1$ at $\phi \in End(\mathbb F_{p^n})$ as $a \mapsto (\phi^n - id)(a) = \phi^n(a) - a$. As just discussed, $\phi^n(a) = a^{p^n}$ so this equals $a^{p^n} - a$. And in $\mathbb F_{p^n}$ this is always $0$. Hence, $\phi^n - id$ is the zero function in $End(\mathbb F_{p^n})$ so $\phi$ satisfies the polynomial $x^n - 1$ when evaluated in this ring.
