Prove that the Frobenius map $f_p : \mathbb{P}^n \rightarrow \mathbb{P}^n$ is a regular map

Question

I wish to prove that the Frobenius map $f_p : \mathbb{P}^n \rightarrow \mathbb{P}^n$, defined by $[x_0, \dots, x_n] \mapsto [x_0^p, \dots, x_n^p]$, is a regular map but not an isomorphism (for $p \geq 2$, of course).

Answer 1

Let $k$ be the base field, of characteristic $p>0$, and let $k'$ denote the same field $k$, with the $k$-algebra structure given by the Frobenius endomorphism $b\mapsto b^p$ (in other words: if $c\in k$ and $c'\in k'$, then the new product is defined by $c.c'=c^pc'$).

First of all, $f_p$ is not an isomorphism because it is not linear: $\textrm{Aut }\mathbb P^n=\textrm{PGL}_n(k)$.

To see that it is a regular map, I see two possible routes:

1. An abstract one: by definition, the Frobenius map $X\to X^{(p)}$ of an algebraic variety $X$ arises from the universal property of fiber product in the category of $k$-schemes: there is a certain fiber square (remember that $X^{(p)}$ is a pullback by definition) and the object $X$ appears "outside", with two arrows compatible with that square. I can be more precise on request, but I do not know what definition you know of $X^{(p)}$.
2. More concretely, $\mathbb P^n\to (\mathbb P^n)^{(p)}$ can be defined by glueing together the following morphisms: if $U_i=D_+(x_i)\cong\textrm{Spec }k[y_1,\dots,y_n]$ is an open subset of $\mathbb P^n$ (one of the standard open covering, so that $y_j=x_j/x_i$), then there is a morphism $$\mathcal O(U_i)^{(p)}:=\mathcal O(U_i)\otimes_kk'\to \mathcal O(U_i)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f\otimes \lambda\mapsto \lambda f^p.$$ This is a morphism of $k$-algebras. Then, $f_p$ is defined by the glueing of the corresponding morphisms $$\textrm{Spec }\mathcal O(U_i)=U_i\to \textrm{Spec }\mathcal O(U_i)^{(p)}.$$