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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.

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.

For a point $p\in {\mathbb{A}}_k^1$, let $a$ be the image of $t$ in its residue field $\kappa (p)$. The scheme-theoretic fiber isIf $p\ne (t)$, then $a\ne 0$. The quadratic polynomial $xy-a{z}^2$ is irreducible. Indeed, after setting $z=0$, any hypothetical linear factors must restrict, up to nonzero scalars, to $x$ and $y$; comparing the $xz$ and $yz$ coefficients then forces both $z$ coefficients to vanish, contradicting the nonzero $z^2$ coefficient. Hence its homogeneous coordinate ring is an integral domain, so $Y_p$ is an integral scheme.

At the origin $p=(t)$, the fiber is $V_{+}(xy)$, the union of the two distinct projective lines $x=0$ and $y=0$, and is therefore not irreducible. It is nevertheless a reduced scheme because the ideal $(xy)=(x)\cap (y)$ equals its radical. Every other fiber is integral and hence reduced. Thus the fiber is integral exactly away from the origin, and it is reduced at every point of ${\mathbb{A}}_k^1$.

For a torus character ${\chi }^m$, its principal divisor on a toric variety isHere $t=xy={\chi }^{(1,1)}$, so on the blown-up fanThe scheme-theoretic fiber $X_0=V(f^{\ast }t)$ therefore contains the exceptional divisor $D_{e_1+e_2}$ with multiplicity two. Its defining ideal is not radical along that divisor, so $X_0$ is a nonreduced scheme.