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An ample real divisor is a finite positive real combination of ample Cartier divisors:Equivalently its numerical class lies in the ample cone. This is a numerical condition even when the coefficients are irrational; it does not mean that some integer multiple of $D$ must be an integral divisor.

On an integral projective variety, a big real divisor is a finite positive real combination of big Cartier divisors. Equivalently, by the real form of Kodaira's lemma,For an integral Cartier divisor, bigness means maximal section-growth order $h^0(X,mD)\ge c{m}^{\dim X}$ along sufficiently divisible positive $m$, or Iitaka dimension $\dim X$. The real linear equivalence of divisors formulation permits finite positive combinations of effective Cartier divisors, whose supports are codimension one. The definitions and the ample-plus-effective formulation on integral varieties are discussed in Fujino's notes on big real divisors.

For the paper's assertions on a general projective scheme, use componentwise bigness on a projective scheme: require bigness on every reduced irreducible component. All arguments below can then be carried out on those finitely many integral components; ampleness is also detected there. On a reducible scheme, merely asking for maximal total section growth on one component is insufficient. For example $X={\mathbb{P}}^2⨿{\mathbb{P}}^2$, with $D$ restricting to $\mathcal{O}(1)$ on the first component and $\mathcal{O}(-1)$ on the second, has quadratic total section growth, but negative intersection with every line in the second component. No finite collection of codimension-one subvarieties can contain all those lines. Thus that weaker meaning would make part (ii) false. In dimension zero the positivity statements are vacuous and every line bundle is ample; the compatible bigness convention also regards it as big.

A connected integral projective variety has only constant regular functions, so $h^0(X,{\mathcal{O}}_X)=1$. By Serre duality and $K_X\sim 0$, $h^2(X,{\mathcal{O}}_X)=h^0(X,{\mathcal{O}}_X(K_X))=1$. The assumed first sheaf cohomology vanishes. Therefore the Euler characteristic of a coherent sheaf isThese hypotheses characterize the K3 surface setting used below.

A projective variety is smooth when every local ring is regular, equivalently when its Jacobian has the expected rank at every point. The affine curve $y=x^3$ is smooth because the derivative of $y-x^3$ with respect to $y$ is $1$, while its projective closure $Y{Z}^2=X^3$ is singular at $[0:1:0]$.

The genus is ${\dim }_k{H}^1(C,{\mathcal{O}}_C)$. Intersect the Fermat cubic threefold with the plane $X_3=X_4=0$. This gives the smooth plane cubic $X_0^3+X_1^3+X_2^3=0$, whose genus of a smooth plane curve is $(3-1)(3-2)/2=1$.

An irreducible plane conic is smooth: a singular quadratic form in three variables has rank at most two and factors over the algebraically closed field, contradicting irreducibility. Projection from any point of the conic, or its degree-two complete linear system, then identifies it with the projective line.

Finally, a nonzero ternary quadratic form is determined up to nonzero scalar by its six coefficients. Thus generalized conics are parametrized bijectively by ${\mathbb{P}}^5$.

For an irreducible algebraic variety $X$ and $p\in X$, the local ring ${\mathcal{O}}_{X,p}$ consists of germs of rational functions regular on a neighbourhood of $p$. If $X\subseteq {\mathbb{A}}^n$ has ideal $I(X)$, its Zariski tangent space is

The given variety is the blowup of the affine plane at the origin. On the chart $W\ne 0$, put $W=1$; then $Y=XZ$, so $(X,Z)$ are free affine coordinates. On the chart $Z\ne 0$, put $Z=1$; then $X=WY$, so $(W,Y)$ are free affine coordinates. These two smooth affine-plane charts cover $V$, hence every point of $V$ is smooth.

If $(X,Y)\ne (0,0)$, the equation forcesConsequently $\pi$ restricts to an isomorphism away from the origin, and is therefore birational. Above the origin, however,so $\pi$ is not injective and cannot be an isomorphism of algebraic varieties. Birationality gives

For any morphism $\phi :V\to V^{\prime }$ with $V^{\prime }$ affine, its restriction to the exceptional curve $E\cong {\mathbb{P}}^1$ is constant: every regular function on a projective line is constant, and the affine coordinate functions of $V^{\prime }$ therefore have constant pullbacks. Since $E$ has more than one point, $\phi$ is not injective. If $V$ itself were affine, its identity morphism would contradict this conclusion, so $V$ is not affine.

Over $\mathbb{C}$, $V$ is not compact in the Euclidean topology: the sequence $(n,0,[1:0])$ has no convergent subsequence. Every complex projective variety is Euclidean compact. Since compactness is preserved by homeomorphisms, $V$ cannot be homeomorphic to a projective variety.

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.

Two irreducible varieties $X$ and $Y$ are birational if they contain nonempty Zariski-open subsets $U\subseteq X$ and $V\subseteq Y$ that are isomorphic. Equivalently, their function fields are isomorphic:For an irreducible variety,An isomorphism of function fields preserves transcendence degree, so $\dim X=\dim Y$. This is the dimension from the function field.

Now let $K/\mathbb{C}$ be a finitely generated field extension. Choose field generators $a_1,\dots ,a_m$ and setThen $A$ is a finitely generated integral $\mathbb{C}$-algebra and $\mathrm{Frac}(A)=K$. The affine irreducible varietyhas function field $K$. Embed $X_0$ as a closed affine subvariety of some affine space ${\mathbb{A}}^N$, identify that affine space with the standard open chart of projective space ${\mathbb{P}}^N$, and take the projective closure $X$ of $X_0$. The closure of an irreducible topological space is irreducible, and $X_0$ is dense and open in $X$. Thus $X$ is a projective variety andThis constructs the projective model of a finitely generated field.

For the curveputIn its function field, the defining equation givesHence $\mathbb{C}(X)=\mathbb{C}(t)$, so $X$ is rational. Equivalently, this is the normalization of y squared equals x times x minus one squared, whose smooth projective normalization is ${\mathbb{P}}^1$ and has geometric genus zero.

A smooth projective curve in the projective plane of degree $d$ has, by the genus of a smooth plane curve formula,Its genus can be zero only for $d=1$ or $d=2$. Conversely, a projective line and every smooth conic over $\mathbb{C}$ are rational, so each is birational to $X$. Therefore the required degrees are exactly

Finally, in ${\mathbb{A}}_{x,y,z}^3$ takeThe polynomial $F$ is irreducible: as a quadratic in $y$, it could factor only if $x(x-1{)}^2$ were a square polynomial, which it is not. Thus $S$ is an irreducible affine hypersurface of dimension two. Its function field isso $S$ is birational to ${\mathbb{A}}^2$.

The partial derivatives areSolving $F=F_x=F_y=0$ gives $x=1$, $y=0$, with $z$ arbitrary. By the Jacobian criterion,This is an irreducible subvariety of dimension one, giving the requested singular cylinder over a nodal curve.

For a smooth projective variety, its $m$th plurigenus is $P_m(X)=h^0(X,{\omega }_X^{\otimes m})$, for $m\ge 1$. These dimensions describe pluricanonical linear systems. Blowup invariance of plurigenera proves their invariance under a point blowup on a smooth surface.

Suppose a nonconstant morphism of varieties $f:X\to Z$ has fiber $Y$ over $z_0$, where $X$ is normal, connected and projective. Choose an affine neighborhood $U$ of $z_0$ and a regular function on $U$ nonconstant on $f(X)\cap U$. Its pullback is a nonconstant rational function on $X$, so its pole Weil divisor is nonzero. A pole-free rational function on a normal variety is regular, and global regular functions on a connected projective variety are constant. The pole divisor avoids $f^{-1}(U)$, hence avoids $Y$. No projectivity assumption on $Z$ is needed.

A projective completion of an affine variety $X$ is a projective variety containing $X$ as a dense open subvariety. Homogenizing affine defining equations and taking their projective zero set gives the projective closure, although it may add singularities at infinity.

A projective curve is a one-dimensional projective variety.

A finite-type morphism of schemes is quasi-finite if its fibres are finite discrete schemes. A proper quasi-finite morphism is a finite morphism. In particular, a proper finite-type morphism of projective varieties with no positive-dimensional fibres is finite; a positive-dimensional projective fibre would contain an integral curve.

The Segre embedding sendsand realizes ${\mathbb{P}}^m\times {\mathbb{P}}^n$ as a projective variety.

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.