# $\mathbb{A}_{\mathbb{F}_p}^1\rightarrow \mathbb{A}_{\mathbb{F}_p}^1$ corresponding to $x\mapsto x^p$ induces an isomorphism on residue fields

I want to prove that the morphism of affine schemes $$\mathbb{A}_{\mathbb{F}_p}^1\rightarrow \mathbb{A}_{\mathbb{F}_p}^1$$ defined by the ring homomorphism $$\varphi :\begin{matrix}\mathbb{F}_p[x]\rightarrow \mathbb{F}_p[x]\\ f(x)\mapsto f(x^p) \end{matrix}$$ is bijective as a map of sets and it induces an isomorphism on residue fields at closed points.
The non zero prime ideals of $$\mathbb{F}_p[x]$$ are of the form $$(f)$$ where $$f$$ is an irreducible polynomial. If $$\mathfrak{p}=(f)$$ is such a prime ideal then $$\varphi^{-1}(\mathfrak{p})=\{g\in\mathbb{F}_p[x],\ f|g^p\}=\{g\in\mathbb{F}_p[x],\ f|g\}=(f)=\mathfrak{p}$$ so the underlying map on topological spaces is just the identity map.
Now let $$\mathfrak{p}=(f)$$ be a closed point of $$\mathbb{A}_{\mathbb{F}_p}^1.$$ The residue field $$\kappa(\mathfrak{p})$$ at $$\mathfrak{p}$$ is $$\mathbb{F}_p[x]_{\mathfrak{p}}/(f\mathbb{F}_p[x]_{\mathfrak{p}})$$ and the induced homomorphism $$\kappa(\mathfrak{p})\rightarrow \kappa(\mathfrak{p})$$ is given by: $$\frac{g}{h}\mapsto \frac{g^p}{h^p}$$. I can't see why this morphism is surjective, any help would be appreciated !

• Field morphisms are always injective, and an injective map between finite sets of the same cardinality must also be surjective. Sep 5 at 19:03
• @ViktorVaughn I don't see why $\kappa(\mathfrak{p})$ is finite ? Sep 5 at 19:04
• @SamiFersi If $f$ has degree $n$, then elements of $\mathbb F_p[x]/\mathfrak p$ can be represented by degree $<n$ polynomials, and there are only finitely many of those. Sep 5 at 19:05
• @KentaS Right ! Thank you ! Sep 5 at 19:07

If $$f$$ has degree $$n$$, then elements of $$\kappa(\mathfrak{p}) = \mathbb{F}_p[𝑥]/\mathfrak{p}$$ can be represented by polynomials of degree $$< n$$, and there are only finitely many of those. Now field morphisms are always injective, and an injective map between finite sets of the same cardinality must also be surjective. Thus $$\varphi$$ is surjective.