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For a smooth projective curve $C$, let $g=h^1(C,{\mathcal{O}}_C)$ be its geometric genus, let $K_C$ be a canonical divisor, and write $\ell (D)=h^0(C,{\mathcal{O}}_C(D))$ for the dimension of a vector space of the Riemann-Roch space. The Riemann-Roch theorem saysfor every divisor on an algebraic curve $D$ on $C$.

Here is the cohomological proof of Riemann-Roch for curves. At a closed point $P$, the local ring is a discrete valuation ring. A local uniformizer shows that increasing the allowed pole order by one gives the short exact sequence of sheavesThe last term is a length-one skyscraper sheaf; its precise identification with $k$ is noncanonical, but its sheaf cohomology has $h^0=1$ and $h^1=0$. All the relevant groups of sheaf cohomology are finite-dimensional because $C$ is a projective variety, and coherent sheaf cohomology above degree one vanishes on a algebraic curve. The long exact sequence in sheaf cohomology therefore givesSince $k$ is an algebraically closed field, every closed point has degree one. Iterating this identity for both positive and negative coefficients of $D$ provesEvery global regular function on a projective variety which is an irreducible variety is constant; hence $h^0(C,{\mathcal{O}}_C)=1$ and $\chi ({\mathcal{O}}_C)=1-g$.

Finally apply Serre duality for a smooth projective curve:Taking dimensions of a vector space and inserting this in the Euler characteristic of a coherent sheaf identity proves the displayed Riemann-Roch theorem. The substantial inputs from sheaf cohomology are finiteness and vanishing for coherent sheaves on a projective curve, and Serre duality; the change of Euler characteristic of a coherent sheaf is proved directly by the point exact sequence. In particular the argument handles negative as well as effective divisors on an algebraic curve.

We prove (c)$\Rightarrow$(a) by induction on dimension. It suffices to work on an integral projective variety $V$. By induction, $D$ is ample on every lower-dimensional integral subvariety, and hence on every lower-dimensional closed subscheme by ampleness on reduced components.

Apply (c) to $V$ itself. The nonzero section of ${\mathcal{O}}_V(mD)$ has a nonempty zero divisor $E$. Because $V$ is integral, this is an effective Cartier divisor and ${\mathcal{O}}_V(E)\cong {\mathcal{O}}_V(mD)$. Its support has dimension less than $\dim V$, so $D{|}_E$, and therefore ${\mathcal{O}}_E(E)$, is ample. Part (ii) makes $E$ semiample, hence some positive multiple of $D$ is basepoint-free.

Let $f:V\to Y\subset {\mathbb{P}}^N$ be the resulting Kodaira map, with ${\mathcal{O}}_V(qD)=f^{\ast }{\mathcal{O}}_Y(1)$. No fibre can have positive dimension: such a projective fibre contains an integral projective curve $C$, on which $D$ has degree zero. But the assumed nonzero section of some ${\mathcal{O}}_C(rD)$ cannot vanish anywhere, since its nonempty effective divisor would have positive degree. This contradicts (c).

Thus $f$ has zero-dimensional fibres. A proper quasi-finite morphism is a finite morphism. The finite pullback of an ample line bundle is ample, so $qD$ and then $D$ are ample. This provesThe fibre argument proves the semiample and curve-positive ampleness criterion. It also explains why testing only existence of a nonzero section, without requiring a zero, would be insufficient: the trivial bundle on a positive-dimensional projective variety has a nowhere-vanishing section.

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.

Take $V=C$ for each integral projective curve in $X$. The assumed positivity gives $D\cdot C>0$, hence certainly $D\cdot C\ge 0$. By the definition of a nef divisor,This is only the curve test. It does not already prove ampleness: positivity against each individual curve can fail to be uniform on limiting classes in the closed cone of curves. The hypotheses in higher dimensions, used in the next sections, are what rule out this failure.

First work on an integral component. Write the big real divisor as $D{\sim }_{\mathbb{R}}B+E$ with $B$ ample real divisor and $E$ effective real Cartier, using Kodaira's lemma. Let $E_1,\dots ,E_k$ be the finitely many integral components of its support. For an integral projective curve $C$ not contained in this support, restriction of each effective Cartier summand to $C$ is effective, soThereforeThis proves negative curves of a big real divisor lie in finitely many divisors.

Now let $A$ be the given ample divisor. Openness of the ample cone gives a $\delta >0$ such that $B-ϵA$ is ample for $0<ϵ\le \delta$. For each of the finitely many $E_i$, the assumption that $D{|}_{E_i}$ is ample similarly gives a ${\delta }_i>0$ such that $(D-ϵA){|}_{E_i}$ is ample for $0<ϵ\le {\delta }_i$. Choose a single positive $ϵ$ smaller than all these bounds.

If $C$ is contained in some $E_i$, its intersection with $D-ϵA$ is positive by that restriction. OtherwiseIn particular $D-ϵA$ is nef:Only the finitely many exceptional support components are needed for the restriction test; no uniform bound over all subvarieties was assumed.

For a reducible projective scheme, use componentwise bigness on a projective scheme and repeat this argument on each reduced irreducible component. Collect their exceptional supports and take the minimum of all the finitely many positive bounds. Codimension one here is measured in the relevant irreducible component. Every integral curve lies in a component, so the same conclusion holds on $X$. Nilpotent structure does not affect these curve intersection numbers.

For an algebraically closed field $k$, the strong Hilbert Nullstellensatz says that every ideal $J\subseteq k[x_1,\dots ,x_n]$ satisfiesIts weak form says that $V(J)=\varnothing$ exactly when $J=(1)$. Now let $S=\mathbb{C}[x_0,\dots ,x_n]$ and let $I\subseteq S$ be a homogeneous ideal. The affine zero set of $I$ is the affine cone over $V_{+}(I)$ together with the origin. Hence, when the projective zero set is empty, eitheror the affine zero set is exactly the origin. In the latter case the Hilbert Nullstellensatz givesthe irrelevant ideal of projective space. Equivalently, some power $(x_0,\dots ,x_n{)}^N$ is contained in $I$. These are precisely the alternatives in the Projective Nullstellensatz.

Let $V\subseteq {\mathbb{P}}^3$ be a smooth quadric surface. After a projective linear transformation, the Segre embedding identifiesChoose distinct points $p,q\in {\mathbb{P}}^1$. The two fibresare the promised disjoint curves on a smooth quadric surface: each is a smooth projective line, and $C_p\cap C_q=\varnothing$. By the Bézout theorem, any two nonempty curves in the projective plane intersect. An isomorphism of algebraic varieties preserves intersections and takes projective curves to projective curves, so these disjoint curves prove

Next let $W$ be a smooth projective curve and let $f:W⇢Y\subseteq {\mathbb{P}}^N$ be a rational map to a projective variety. At a point $P$ outside its initial domain, writeusing elements of the function field $\mathbb{C}(W)$. The local ring of a smooth algebraic curve at $P$ is a discrete valuation ring. If $m={\min }_i{v}_P(f_i)$, multiplying every coordinate by a function of valuation $-m$ makes all coordinates regular at $P$ and at least one a unit there. They therefore define a morphism near $P$. Repeating this at the finitely many missing points proves the extension of a rational map from a smooth projective curve.

Smoothness is essential. Let $\nu :{\mathbb{P}}^1\to C$ be the normalization of a nodal curve for a rational nodal cubic $C$. The inverse is a rational map $C⇢{\mathbb{P}}^1$ on the smooth locus. If it extended to a morphism $g:C\to {\mathbb{P}}^1$, then $g\circ \nu$ would agree with the identity on a dense Zariski-open subset of ${\mathbb{P}}^1$ and hence everywhere. But the two distinct points $a,b\in {\mathbb{P}}^1$ above the node would satisfya contradiction. This is the failure of rational-map extension on a singular curve.

Finally take the Fermat cubic curveand define the projective hypersurfaceBecause $F$ and $xyz$ have no common irreducible factor, the bihomogeneous polynomial $sF+txyz$ is irreducible, so $Z$ is an algebraic variety. Letbe the second coordinate projection. Every homogeneous cubic in three variables has a projective zero over $\mathbb{C}$, so every fibre is nonempty and $\pi$ is a surjective morphism. Over $p=[1:0]$ the fibre is the Fermat cubic curve, which is smooth and has genusOver $q=[0:1]$ the fibre isthe union of exactly three distinct projective lines, hence exactly three irreducible components. This is the cubic pencil with a triangular member required by the question.

A projective curve is a one-dimensional projective variety.

A projective plane curve is a projective curve embedded in the projective plane, usually given by one homogeneous equation.

A semiample divisor on an integral projective variety is ample if it has positive degree on every integral projective curve. Its Kodaira map cannot contract a positive-dimensional fibre, since such a fibre contains a curve and the pulled-back hyperplane bundle has degree zero there. Thus the morphism is proper and a quasi-finite morphism, hence a finite morphism. The finite pullback of an ample line bundle is ample.

A smooth projective curve is a nonsingular projective variety of dimension one.