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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.

No. Let $k={\mathbb{F}}_2$ be the field with two elements, put $A=k[\epsilon ]/({\epsilon }^2)$, and take the closed immersioninduced by the ring homomorphism $q:A\twoheadrightarrow k$ with $q(\epsilon )=0$.

For every field $K$, a ring homomorphism $\alpha :A\to K$ must send $\epsilon$ to zero: $\alpha (\epsilon {)}^2=0$, and a field has no nonzero nilpotent elements. Consequently $\alpha$ factors uniquely through $q$. Passing from ring homomorphisms to morphisms of schemes between affine schemes gives a bijectionThis concerns all morphisms of schemes, without requiring them to be over $k$: if $K$ does not have characteristic two, both sets are empty; otherwise each contains exactly one element.

Nevertheless, $f$ is not an isomorphism of schemes. Its map on global sections is $q$, whose kernel of a ring homomorphism contains the nonzero element $\epsilon$, so $q$ is not a ring isomorphism. The target is the nonreduced scheme defined by the dual numbers, whereas the source is a reduced scheme. This illustrates why field-valued points do not detect nilpotents.

The quotient. Evaluation $\mathbb{Z}[t]\to \mathbb{Z}[i]$, $t\mapsto i$, is a surjective ring homomorphism whose kernel of a ring homomorphism is $(t^2+1)$. Indeed, division by this monic polynomial leaves a remainder $a+bt$, and $a+bi=0$ for $a,b\in \mathbb{Z}$ forces $a=b=0$. Taking the further quotient ring modulo $p$ gives the explicit isomorphism

All fibers, including the generic fiber. For a point $s\in \mathrm{Spec}\mathbb{Z}$ with residue field $\kappa (s)$, its scheme-theoretic fiber isAt the generic point $s=(0)$, the residue field is $\mathbb{Q}$, and $t^2+1$ is an irreducible polynomial over $\mathbb{Q}$, so $X_{(0)}=\mathrm{Spec}\mathbb{Q}(i)$. For a closed point $(p)$ the residue field is ${\mathbb{F}}_p$. The permitted criterion for a square root of $-1$ gives three cases. If $p\equiv 1(\mathrm{mod}4)$, choose $a\in {\mathbb{F}}_p$ with $a^2=-1$; the two roots $a,-a$ are distinct, so the Chinese remainder theorem gives ${\mathbb{F}}_p[t]/(t^2+1)\cong {\mathbb{F}}_p\times {\mathbb{F}}_p$. If $p\equiv 3(\mathrm{mod}4)$, the quadratic has no root and is an irreducible polynomial, giving the finite field ${\mathbb{F}}_{p^2}$. Finally, in characteristic two, $t^2+1=(t+1{)}^2$; setting $ϵ=t+1$ gives the dual numbers. ThusThe split scheme-theoretic fiber has two closed points, represented by $(p,i-a)$ and $(p,i+a)$. The inert scheme-theoretic fiber has one closed point, represented by $(p)$. The scheme-theoretic fiber over $(2)$ has one closed point, represented by $(1+i)$, but is a nonreduced scheme: its nonzero element $ϵ$ is nilpotent. These descriptions retain the residue fields and the nilpotent structure, which a count of points alone would lose.

The topology. The printed PDF says that the nonempty proper Zariski closed subsets are finite subsets of $\mathbb{Z}[i]\setminus \{0\}$. Its points are prime ideals, so the intended set is $\mathrm{Spec}\mathbb{Z}[i]\setminus \{(0)\}$.

Put $R=\mathbb{Z}[i]$. Since $R$ is a principal ideal domain, every nonzero prime ideal is a maximal ideal. Each such prime ideal contains a nonzero Gaussian integer $z$, and therefore contains the positive integer $z\overline{z}$. Its contraction to $\mathbb{Z}$ is consequently a nonzero prime ideal $(p)$, so the scheme-theoretic fibers just computed list all the closed points of $X$. The sole remaining point is the generic point $(0)$.

Every proper Zariski closed subset is $V((a))$ with $a\ne 0$: if its defining ideal were zero, it would be all of $X$. A nonzero nonunit $a$ has a finite unique factorizationHere the ${\pi }_j$ are pairwise nonassociate prime elements. Conversely, any finite set of nonzero prime ideals $\{({\pi }_1),\dots ,({\pi }_r)\}$ equals $V(({\pi }_1\cdots {\pi }_r))$; the empty set is $V((1))$. ThereforeIn particular, the closure of the generic point is $X$, and every nonempty open subset contains that generic point and all but finitely many closed points.

A nonidentity automorphism. Complex conjugationis a ring automorphism fixing $\mathbb{Z}$. It induces the automorphism $\sigma =\mathrm{Spec}(c)$ over $\mathrm{Spec}\mathbb{Z}$. Since $c(i)=-i\ne i$, this is not the identity morphism of schemes. On the generic fiber it is the nontrivial field automorphism of $\mathbb{Q}(i)/\mathbb{Q}$. On a split scheme-theoretic fiber it swaps the two closed points, since $i-a$ maps to $-(i+a)$. On an inert scheme-theoretic fiber it fixes the only point but induces the nonidentity finite-field Frobenius automorphism $z\mapsto z^p$ on its residue field, because $i^p=-i$. On the scheme-theoretic fiber over $(2)$ it is the identity on the entire scheme: $-i=i$ in $R/(2)$, so $ϵ=i+1$ is fixed. Thus

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.