Frobenius mophism, Exercise 7.3 R of Ravi Vakil's book on algebraic geometry I am a geometry person and got stuck in Exercise 7.3.R which is about Frobenius morphism.
Suppose $p$ is a prime and $r \in \mathbb{Z}^+$. Let $q=p^r$ and $k=\mathbb{F}_q$. Define $\phi :k[x_1,\dots,x_n] \rightarrow k[x_1,\dots,x_n]$ by $\phi(x_i)=x_i^p$ for each $i$, and let $F:\mathbb{A}_k^n \rightarrow \mathbb{A}_k^n$ be the map of schemes corresponding to $\phi$.
(a) Show that $F^r$ is the identity on the level of sets, but is not the identity morphism.
(b) Show that $F$ is a bijection, but is not an isomophism of schemes.
(c) If $K=\bar{F}_p$, show that the morphism $F:\mathbb{A}_K^n \rightarrow \mathbb{A}_K^n$ of $K$-schemes corresponding to $x_i \rightarrow x_i^p$ is a bijection, but no power of $F$ is the identity on the level of sets!
For (a) and (b), I only know the solution when $n=1$ since $k[x]$ is a PID which makes it very easy. But I got stuck when $n>1$.
 A: For $c)$, notice that surjections are stable under base change, i.e. $F: \mathbb A_K^n \to \mathbb A_K^n$ is a surjection as the base change of of $F: \mathbb A_k^n \to \mathbb A_k^n$. For injectivity, let $F(\mathfrak p) = F(\mathfrak q)$ for two primes in $K[x_1, \dotsc, x_n]$. In general we have $\phi(\phi^{-1}(A))=A \cap \operatorname{image}(\phi)$ for any subset $A$, i.e. we deduce 
$$\mathfrak p \cap K[x_1^p, \dotsc, x_n^p] = \mathfrak q \cap K[x_1^p, \dotsc, x_n^p].$$
From this, you can easily deduce $\mathfrak p = \mathfrak q$: Let $f \in \mathfrak p$. Then $f^p \in \mathfrak p \cap K[x_1^p, \dotsc, x_n^p] = \mathfrak q \cap K[x_1^p, \dotsc, x_n^p] \subset \mathfrak q$, hence $f \in \mathfrak q$, since it is a prime ideal.
Thus we have shown hat $F$ is a bijection on the level of sets. But no power is the identity map:
Let $r > 0$ and $a \in K \setminus \mathbb F_{p^r}$. $(\phi^r)^{-1}((x_1-a))=(x_1-a^{p^r})$, i.e. the prime ideal $(x_1-a)$ is mapped by $F^r$ to $(x_1-a^{p^r})$, which is not the same, since $a \neq a^{p^r}$. Hence $F^r$ is not the identity map.
A: Thanks to the hints by Hoot, I try to answer (a) and (b) by the following,
We have $\phi^r(f(x_i))=f(x_i^{q})=f(x_i)^q$, so it is obvious that for every prime ideal $P \subset k[x_1,\cdots,x_n]$, $\phi^r(P) \subset P$, so $P \subset (\phi^r)^{-1}(P)$. Now we show that $(\phi^r)^{-1}(P) \subset P$, which will show that $F^r$ is identity.
For $f\in (\phi^r)^{-1}(P)$, which means $\phi^r(f(x_i))=f(x_i^{q})=f(x_i)^q \in P$, so $f \in P$ since $P$ is a prime ideal.
$F^r$ is not the identity morphism since $\phi^r$ is not identity homomophism.
(b) follows from (a).
