Frobenius map of polynomial ring Let $k$ be the finite field with $p^r$ elements for $p$ a prime.  define $$\varphi:  k[x_1,x_2,...,x_n]\rightarrow k[x_1,x_2,...,x_n],  x_i\mapsto x_i^p$$ ($\varphi$ is identity on $k$).  how to prove $\mathfrak{p}\mapsto \varphi^{-1}(\mathfrak{p})$  is a bijection of the prime ideals of $k[x_1,x_2,...,x_n]$?
 A: $\varphi$ is the homomorphism $k[x] \to k[x^p], f(x) \mapsto f(x^p)$, 
$I$ a prime ideal of $k[x]$,


*

*For $f(x) \in I$ then $(f(x))^p = f^p(x^p) \in I \cap k[x^p]$ 
(where $f^p(x)$ means applying the Frobenius to the coefficients of $f$)
thus $f^p(x) \in \varphi^{-1}(I)$.

*Since $k$ is a finite field $f(x) \mapsto f^p(x)$ is an automorphism of $k[x]$.

*If $f^p(x) \in \varphi^{-1}(I)$ then $f^p(x^p) \in I$, since $f^p(x^p)=(f(x))^p$ and $I$ is prime it means $f(x) \in I$ 
Thus $I \mapsto \varphi^{-1}(I \cap k[x^p])$ is the natural bijective map on prime ideals obtained from the automorphism $f(x) \mapsto f^p(x)$.
A: First of all, the map $\varphi$ you define is not an homomorphism, unless $k$ is the finite field with $p$-elements, since you defined $\varphi$ to be the identity on $k$, and it must be the $p$-power map, which is an automorphism of $k$. This means that $\varphi(f)=f^p$, since the $p$-power map commutes with the sum in characteristic $p$. (The other option, if $q=|k|=p^r$, is to consider $\varphi$ as the map which is the identity on $k$ and sends $\varphi(x_i)=x_i^q$, or, equivalently, it is the $q$-power map, since $x^q=x$ for all $x\in k$; this homomorphism is just $r$-times the previous one, so we can proved the result for the previous $\varphi$ and we get the result for the new one.)
So I will show the result for that $\varphi$
To show the assertion, first note that the fact that $\varphi^{-1}(\mathfrak{p})$ is prime if $\mathfrak{p}$ is prime is a general result valid for any ring homomorphism. So we only need to show if gives a bijection. 
Now, observe that $\varphi$ is an injective homomorphism of rings. Since $\varphi_{|k}:k\to k$ is a bijection (here is where we use $k$ is a finite field), the image of $\varphi$ is equal to the subring $$k[x_1^p,\dots,x_n^p]\subseteq k[x_1,\dots,x_n].$$
Hence $\varphi$ gives an isomorphism between $k[x_1^p,\dots,x_n^p]$ and $ k[x_1,\dots,x_n]$, and the assertion is equivalent to show that the map $$\Phi: \{\text{ Prime ideals of } k[x_1,\dots,x_n]\} \to \{\text{ Prime ideals of } k[x_1^p,\dots,x_n^p]\} $$
$$ \mathfrak{p} \mapsto \mathfrak{p}\cap k[x_1^p,\dots,x_n^p]$$
gives a bijection. 
The following observation is proved by reuns in his answer.

Key Observation: For any element $f$ of $k[x_1^p,\dots,x_n^p]$ there exists a unique $g\in k[x_1,\dots,x_n]$ such that $f=g^p$. 

This implies that $$\mathfrak{p}\cap k[x_1^p,\dots,x_n^p]=\{g(x_1,\dots,x_n)^p\ | \ g(x_1,\dots,x_n)\in \mathfrak{p}\}.$$
That the RHS is inside the LFS is clear. Now, taking $f\in \mathfrak{p}\cap k[x_1^p,\dots,x_n^p]$, by the observation $f=g^p$, with $g\in k[x_1,\dots,x_n]$. But $g^p\in \mathfrak{p}$, that is prime, so $g\in \mathfrak{p}$. 
This result shows easily that $\Phi$ is injective, since the $p$-power map is injective in $k[x_1,\dots,x_n]$. 
To show surjectivity, take a prime ideal $\mathfrak{q}$ in $k[x_1^p,\dots,x_n^p]$, and let $I=k[x_1,\dots,x_n]\mathfrak{q}$ be the ideal generated by $\mathfrak{q}$ in the bigger ring. If we show that $I$ is a prime ideal, then we showed surjectivity of $\Phi$, since $I\cap k[x_1^p,\dots,x_n^p]=\mathfrak{q}$. But observe that $$I=\{f\in k[x_1,\dots,x_n]\ | \ f^p\in \mathfrak{q}\}.$$
If $f\in  I$, it is clear that $f^p\in \mathfrak{q}$. But the RHS is clearly an ideal, being just the preimage by $\varphi$, and it clearly contains $\mathfrak{q}$, so it contains $I$. Now, if $f$ and $g\in k[x_1,\dots,x_n]$ verify that $fg\in I$, then $(fg)^p=f^pg^p\in I\cap k[x_1^p,\dots,x_n^p]=\mathfrak{q}$, so $f^p\in \mathfrak{q}$ or $g^p\in \mathfrak{q}$, so $f\in I$ or $g\in I$. 
Note that it is not true in general that if $R$ is a subring of a commutative ring $S$, the map $\mathfrak{p}\mapsto \mathfrak{p}\cap R$ from prime ideals of $S$ to prime ideals of $R$ is injective or surjective (for example, if $R=\mathbb{Z}$ and $S=\mathbb{Z}[\sqrt{-1}]$).
