Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 89 / 2

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.

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.

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.