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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.

The spectrum of the dual number ring $k[ϵ]/({ϵ}^2)$ is a nonreduced scheme of length two supported at one point. Its nonzero nilpotent $ϵ$ retains first-order information that the underlying singleton topological space forgets. It differs from the disjoint union of two reduced $k$-points.

The Nakai–Moishezon criterion says that a Cartier divisor $D$ on a projective scheme is ample exactly whenIn particular the test includes each positive-dimensional irreducible component. Here and below, “proper” means proper over the ground field, not necessarily a strict subset of $X$. Interpreting it as a strict subset would make the later criteria false even for an integral projective curve, whose strict closed subvarieties have dimension zero.

The intersection product of Cartier divisors with cycles depends only on their numerical equivalence of divisors classes. One way to see this is to intersect all but one factor first, obtaining a one-cycle; replacing the remaining factor by a numerically equivalent divisor does not change its pairing with that cycle. Multilinearity then handles replacement of every factor. Consequently $D\equiv D^{\prime }$ makes all the displayed numbers equal. ThusThe criterion applies to the integral reduced subvarieties of a possibly nonreduced scheme; ampleness on reduced components explains why the nilpotent structure does not change this condition.

Homogenization gives $F=X^4+X^2{Y}^2+Y^2{Z}^2+Z^4$, withIf the characteristic is not $2$, a singular point with $Y=0$ would force $X=Z=0$, impossible. If $Y\ne 0$, then $X^2+Z^2=0$. If either $X$ or $Z$ vanishes, both vanish, giving $[0:1:0]$. Otherwise the other two derivative equations require $Y^2=-2X^2=-2Z^2$, whence $X^2=Z^2$ and $2X^2=0$, again impossible. ThusIn the chart $Y=1$, the quadratic term is $X^2+Z^2$, the product of two distinct linear factors over $k$, so this is an ordinary double point.

In characteristic $2$,All first derivatives of this equation vanish. The equation defines a nonreduced scheme, singular everywhere; it no longer satisfies the irreducible/reduced hypersurface premise. If “curve” means the associated reduced variety, it is the union of two distinct lines, and only their intersection $\boxed{[1:0:1]}$ is singular. These two interpretations must not be conflated.

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 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.