I am reading Silverman's Arithmetic of Elliptic Curves.I was trying to solve exercise 1.8 but I'm stuck.
1.8. Let $\mathbb{F}_q$ be a finite field with q elements and let $V\subset \mathbb{P}^{n}$ be a variety defined over $\mathbb{F}_q$.
(a) Prove that the $q^{th}$ power map $\phi = [X_{0}^{q} , . . . , X_{n}^{q} ]$ is a morphism.
(b) Prove that $\phi$ is one-to-one and onto.
(c) Prove that $\phi$ is not an isomorphism.
(d)Prove that $V(\mathbb{F}_{q})=$ $\{P\in V:\phi(P)=P\}$
(a)Now a point of $V$ is of the form $[X_{0},...,X_{n}]$ where $X_{i}\in \overline{\mathbb{F}_{q}}$.Then for some $i,k>0$,$X_{i} \in \mathbb{F}_{q^{k_{i}}}$.Then $X_{i}^{q}=0 \iff X_{i}=0$.Therefore,$\phi$ is a morphism since it is regular everywhere.
(b)As before for some $i,k>0$,$X_{i} \in \mathbb{F}_{q^{k_{i}}}$.Then $X_{i}^{q^{k_{i}}}=X_{i}$.Then $X_{i}^{q^{k_{i}-1}}=1$.Take $Y_{i}=X_{i}^{q^{k_{i}-1}}$.This shows $\phi$ is onto.
(d)$V(\mathbb{F_{q}})\subset \{P\in V:\phi(P)=P\}$ is clear.For other inclusion,for some $i,k>0$,let $X_{i} \in \mathbb{F}_{q^{k_{i}}}$; $X_{i}^{q^{k_{i}}}=X_{i}=X_{i}^{q}$,then $k_{i}=1$(?).
I'm not sure how to show injection and part(c).
Part (c): to be precise, the exercise is false as stated; you need to assume $\dim V>0$. Indeed, when $V$ is a point, $\phi$ is an isomorphism.
Now assume $\dim V>0$. Affine locally, the morphism $\phi$ is given by a ring homomorphism $f\colon A\to A:x\mapsto x^p$, where $A$ is a finitely generated $\mathbb F_q$-algebra (in fact, an integral domain). Then, Theorem 8.6A of Hartshorne tells us $\Omega_{A/\mathbb F_q}$ is nonzero. But the differential $df\colon \Omega_{A/\mathbb F_q}\to \Omega_{A/\mathbb F_q}$ of $f$ is zero, since $d(a^p)=pa^{p-1}da=0$. Thus $f$ cannot be an isomorphism.