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Over a perfect field $k$ of characteristic $p$, the scheme-theoretic fiber of the Absolute Frobenius morphism of ${\mathbb{A}}_k^1$ over a rational point $a$ isIts underlying space is one point, but its coordinate ring has a nonzero nilpotent element, so the fibre is a length-$p$ nonreduced scheme.

For a finite surjective map of normal integral schemes with function fields $K\subseteq L$ in characteristic $p$, the condition $L^p\subseteq K$ lets the Absolute Frobenius morphism of the source factor through the map. On an affine chart $A\subseteq B$, send $b$ to $b^p\in A$: it lies in $K$, is integral over $A$, and $A$ is integrally closed. The condition $K\subseteq L^p$ instead lets the map factor after the source Frobenius. Send $a$ to its unique $p$th root in $L$; that root lies in the normal ring $B$ because it satisfies a monic polynomial over $B$. These rules localize and glue; injectivity of Frobenius in the function field supplies uniqueness.

Assume $K\subseteq L^p$. Every $a\in A$ has a unique $p$th root $r(a)$ in $L$: existence is the field inclusion and uniqueness follows from injectivity of the Frobenius endomorphism in a field. Since $r(a{)}^p=a$, this root satisfies the monic polynomial $T^p-a$ over $B$. Because $B$ is an integrally closed domain, this gives $r(a)\in B$. The identitiesand uniqueness show thatis a ring homomorphism.

On $D(a)$ the rule sends a fraction to the fraction of its unique roots. Since $r(a{)}^p=a$, inverting $a$ is equivalent to inverting $r(a)$, so the rule localizes. Also, for a prime $\mathfrak{q}\subset B$, $r(a)\in \mathfrak{q}$ exactly when $a\in \mathfrak{q}$; hence the induced point map agrees with that of $f$. The localized rules consequently glue to a morphism $h:X\to Y$. Its pullback followed by the Frobenius endomorphism sends $a$ to $r(a{)}^p=a$, which provesIf another morphism satisfies this equation, it has the same point map because the Absolute Frobenius morphism fixes points, and its pullback of $a$ must be a $p$th root of $a$ in the normal domain $B$. That root is unique. Thus $h$ is unique.

Assume $L^p\subseteq K$. For $b\in B$, its power $b^p$ belongs to $K$ and is integral over $A$, because $B$ is finite over $A$. Because $A$ is an integrally closed domain, $b^p\in A$. Thusis a ring homomorphism, defining a morphism $g_V:V\to f^{-1}V$. Its composition with $f:f^{-1}V\to V$ is the Absolute Frobenius morphism of $f^{-1}V$, because the composite ring map sends $b$ to $b^p$ in $B$.

These local morphisms glue. Explicitly, over a distinguished open $D(a)\subseteq V$, the ring rule isso it is exactly the localization of ${\gamma }_V$. For the underlying topology, each fibre of $f$ consists of a single point: if primes ${\mathfrak{q}}_1,{\mathfrak{q}}_2\subset B$ contract to the same $\mathfrak{p}$, thenso ${\mathfrak{q}}_1={\mathfrak{q}}_2$. The map $g_V$ sends that point of $V$ back to its unique point above it. Thus localization compatibility on a basis of a topology also gives compatibility on overlaps of arbitrary affine charts. We obtain

For uniqueness, $f$ is surjective on points, so the equation determines the point map of $g$. On each affine chart it then requires $f^{\ast }(g^{\ast }(b))=b^p$. The inclusion $A\hookrightarrow B$ is injective, so this determines $g^{\ast }(b)$ uniquely as $b^p$. Hence the sheaf map and the morphism are unique.

The Absolute Frobenius morphism is the identity on the underlying topological space and acts on the structure sheaf by $a\mapsto a^p$. On an affine scheme it is induced by the ring homomorphism $A\to A$, $a\mapsto a^p$. This is additive in characteristic $p$, and the inverse image of a prime ideal $\mathfrak{p}$ is again $\mathfrak{p}$, since $a^p\in \mathfrak{p}$ exactly when $a\in \mathfrak{p}$. These affine maps agree under localization and hence glue.

A perfect ring of characteristic p is one whose Frobenius endomorphism is bijective. Since the Absolute Frobenius morphism fixes points, its map on the stalk at $x$ is precisely the Frobenius endomorphism of ${\mathcal{O}}_{X,x}$. A morphism of sheaves is an isomorphism exactly when it is an isomorphism on all stalks. ThereforeIn the reverse direction the inverse sheaf map, together with the identity on points, gives the inverse morphism of locally ringed spaces. This is perfection in the Frobenius sense, not the unrelated homological use of the word perfect.

For the function field assertions below, interpret the normal schemes as integral, as is implicit in writing a single field $k(X)$ or $k(Y)$. General normal schemes need not be irreducible; without this interpretation those fields are not defined globally as single fields. Set $K=k(Y)\subseteq L=k(X)$. For an affine open $V=\mathrm{Spec}A$ in $Y$, finiteness gives $f^{-1}V=\mathrm{Spec}B$, with $A\subseteq B$ finite and both rings normal domains. In fact $B$ is the integral closure of $A$ in $L$: any element of $L$ integral over $A$ is also integral over $B$ and therefore belongs to $B$. This local description yields the two forms of Frobenius factorization through a finite normal cover.

On every open set $U$, define $F_U^{\#}(a)=a^p$. In characteristic of a ring $p$, the binomial theorem gives $(a+b{)}^p=a^p+b^p$, so these are ring homomorphisms; they commute with restrictions and hence define a morphism of sheaves of rings.

On an affine chart $\mathrm{Spec}A$, the inverse image of a prime ideal $\mathfrak{p}$ under the Frobenius endomorphism isby primality. The induced continuous map is therefore the identity. These local morphisms agree on overlaps, giving the Absolute Frobenius morphism $F_X:X\to X$. Its action on the underlying space and on every local section was prescribed, so the morphism is unique.

TakeThe global regular functions on projective space give $\Gamma (X,{\mathcal{O}}_X)={\mathbb{F}}_p$, on which the Absolute Frobenius morphism is the identity. On the standard affine line $\mathrm{Spec}{\mathbb{F}}_p[t]$, however, its map on functions is $t\mapsto t^p$, which is not surjective. Thus $F_X$ is not an isomorphism of schemes.

Choose the rational closed point $x$ given by $t=0$. The affine formula for a fibre product of schemes gives its scheme-theoretic fiber:This is the Fibre of absolute Frobenius over a rational point of the affine line: it is a one-point, length-$p$ nonreduced scheme, rather than a reduced point.

A commutative ring of prime characteristic $p$ is perfect in the Frobenius sense when its Frobenius endomorphism is an automorphism. It is necessarily reduced: if $a$ is nilpotent, some $p$-power iterate is zero, and injectivity gives $a=0$. This use of perfect is distinct from the homological notion of a perfect ring. The Absolute Frobenius morphism of a scheme is an isomorphism exactly when all its local rings are perfect in this sense, because it is the identity on points and is an isomorphism on every stalk.