Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 113 / 2

Condition (i) implies (ii) immediately by taking the open set $U=X$.

Every restriction map of the structure sheaf of a scheme is a unital ring homomorphism. Thus $p\cdot 1=0$ in $\Gamma (X,{\mathcal{O}}_X)$ implies the same identity in $\Gamma (U,{\mathcal{O}}_X)$ for every open $U$. This proves (ii)$\Rightarrow$(i), so (i) and (ii) are equivalent.

The unique structure morphism $X\to \mathrm{Spec}\mathbb{Z}$ factors through the closed subscheme $\mathrm{Spec}(\mathbb{Z}/p\mathbb{Z})$ exactly when the ideal sheaf $p{\mathcal{O}}_X$ is zero. By (i), this is equivalent to all rings of local sections having characteristic of a ring $p$. Therefore

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.