Take
$$
X=\mathbb P_{\mathbb F_p}^1.
$$
The <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> $\operatorname{Spec}\mathbb F_p[t]$, 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>:
$$
F_X^{-1}(x)
=\operatorname{Spec}\left(\mathbb F_p[t]\otimes_{\mathbb F_p[t],\,t\mapsto t^p}\mathbb F_p\right)
\cong\boxed{\operatorname{Spec}\mathbb F_p[t]/(t^p)}.
$$
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.

Solution

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

{c}
{title2=$\operatorname{Spec}k[t]/(t-a)^p$}

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 $a$ is
$$
\operatorname{Spec}\bigl(k[t]\otimes_{k[t],\,t\mapsto t^p}k\bigr)
\cong\operatorname{Spec}k[t]/(t-a)^p.
$$
Its underlying space is one point, but its coordinate ring has a nonzero <nilpotent element>, so the fibre is a length-$p$ <nonreduced scheme>.

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