Similar questions have been asked in various different settings, but I am not satisfied with the array of answers which have been received. If something truly is a duplicate on the nose, I will be happy to be referred to this question.
Let $q=p^s, A=\mathbb{F}_q [x_1,\dots ,x_n],$ and $\phi: A\to A$ be the map which takes $x_i\mapsto x_{i}^p$. This is a ring map and induces a map of schemes $f: \text{Spec } A\to \text{Spec } A$ given by taking preimage of $\phi$.
I am told that $\phi$ is a bijection. How is this clear? I don't see why injectivity or surjectivity is obvious.
Certainly this map is clearly not an isomorphism; I have no issue regarding this fact. But the bijectivity seems to require something which I do not see.
Notice that $\phi^s$ is just the Frobenius map $x\mapsto x^q$. Thus we have
$$f^s(\mathfrak p)=(\phi^s)^{-1}(\mathfrak p)=\{x\in A\mid \phi^s(x)\in\mathfrak p\}=\{x\in A\mid x^q\in\mathfrak p\}=\mathfrak p.$$
where on the last equality I've used the fact that $\mathfrak p$ is prime. So $f^s$ is the identity map on the level of sets, in particular a bijection, which implies $f$ is a bijection.