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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 !

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    $\begingroup$ Field morphisms are always injective, and an injective map between finite sets of the same cardinality must also be surjective. $\endgroup$ Sep 5 at 19:03
  • $\begingroup$ @ViktorVaughn I don't see why $\kappa(\mathfrak{p})$ is finite ? $\endgroup$
    – Sami Fersi
    Sep 5 at 19:04
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    $\begingroup$ @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. $\endgroup$
    – Kenta S
    Sep 5 at 19:05
  • $\begingroup$ @KentaS Right ! Thank you ! $\endgroup$
    – Sami Fersi
    Sep 5 at 19:07
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Combining comments into an answer:

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.

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