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Write $a=t_1$, $b=t_2$, $c=t_3$, and $A=\mathbb{C}[a,b,c]/(ac-b^2)$. It is the quadratic cone invariant ringwhere the involution negates both $u$ and $v$. The invariant monomials have even total degree and are generated by $u^2,uv,v^2$; reduction by $ac=b^2$ proves the asserted presentation.

To verify the normal variety property, suppose $q\in \mathrm{Frac}(A)$ is integral over $A$. The same monic equation makes $q$ integral over $\mathbb{C}[u,v]$. This polynomial ring is an integrally closed domain, so $q\in \mathbb{C}[u,v]$. Since $q$ belongs to the fraction field of $A$, it is invariant under the involution, and hence belongs to $A$. Thus $X$ is normal. The gradient of its equation is $(c,-2b,a)$, so the Jacobian criterion shows that the origin $o$ is its only singular point of an algebraic variety. In particular $X$ is not smooth.

Take the prime Weil divisor $D=V_X(a,b)$ and $x=o$. We show that $D$ is not a Cartier divisor at $o$. In the local ringits prime ideal is $I=(a,b)R$. The quotient $I/\mathfrak{m}I$ has dimension of a vector space $2$ over $\mathbb{C}$: $a,b$ generate it, and they are independent since $\mathfrak{m}I\subset {\mathfrak{m}}^2$ while the relation $ac-b^2$ has no linear term. By Nakayama lemma, $I$ cannot be generated by one element.

On a normal variety, the ideal of an effective prime Weil divisor is the divisorial ideal of functions with order at least one along that divisor and order at least zero along every other prime Weil divisor. If $D$ were a Cartier divisor at $o$, a local defining rational function $h$ would identify this ideal with $hR$, because an integrally closed domain is the intersection of its height-one localizations at a prime ideal inside its fraction field. That would make $I$ principal, contradicting the calculation. Therefore $D$ has a nonzero class in the local divisor class group.

Now let $D^{\prime }=D+\mathrm{div}(q)$ be any linearly equivalent Weil divisor. If $o$ were outside its support of a Weil divisor, there would be an open subset containing $o$ on which $D^{\prime }$ is zero. On that neighbourhood $D=-\mathrm{div}(q)$ would be a principal Weil divisor, hence a Cartier divisor, which is impossible. Therefore this point is unavoidable in every representative:The argument does not assume that $D^{\prime }$ is effective. For comparison, $\mathrm{div}(a)=2D$ because on the chart $c\ne 0$, $a=b^2/c$; the obstruction has order two, as in the Divisor class group of an A-type surface singularity.

Let $K$ be the function field of the integral scheme $X$ and choose a nonzero rational section of a line bundle $s\in {\mathcal{L}}_{\eta }$. Choose an open trivializing cover $U_i$ with generators $e_i$ for the invertible sheaf $\mathcal{L}$. On each nonempty $U_i$, write $s=a_i{e}_i$ with $a_i\in K^{\ast }$. If $e_i=u_{ij}{e}_j$ on an overlap, then $a_j=a_i{u}_{ij}$, so $a_i/a_j=u_{ij}^{-1}$ is a regular unit. Consequently the local rational functions $a_i$ define a Cartier divisor $D$.

The divisor line bundle ${\mathcal{O}}_X(D)$ restricts to $a_i^{-1}{\mathcal{O}}_{U_i}$. Define its local isomorphism from $\mathcal{L}$ by $e_i\mapsto a_i^{-1}$. On an overlap,so the maps glue. HenceThe rational section maps to the rational function $1$, which also fixes the sign convention for $D$. Choosing a different nonzero rational section changes $D$ by a principal Cartier divisor, so its linear equivalence of Cartier divisors class is unchanged. No projectivity or Noetherian hypothesis is required.

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.

Choose an ample Cartier divisor $H$. Put initially $A=\delta H$, $A^{\prime }=tH$ for small positive real $\delta ,t$. Since $D$ is nef, the nef-plus-ample ampleness lemma makes $D+tH$ ample. The polynomialhas $P(0,0)=D^n\cdot [X]>0$. Thus the desired strict inequality holds when $\delta ,t$ are sufficiently small and positive.

We must also arrange rationality of the two specified classes; $D$ itself need not be rational. Choose rational ample classes $B$ and $C$ sufficiently near $D+tH$ and $(\delta +t)H$, and defineThen $A^{\prime }$ is close to $tH$ and $A$ is close to $\delta H$, so both are ample real divisors by openness of the ample cone. Also $D+A^{\prime }=B$ and $A+A^{\prime }=C$ are rational and ample. Continuity preserves the strict inequality, giving

Here is the needed algebraic Morse inequality for ample divisors, with its section-count proof. Choose rational Cartier divisor representatives of $B,C$ and a common positive integer $q$ making $qB,qC$ very ample integral Cartier divisors. For the section-count argument rename these scaled representatives $B,C$; undoing this scaling restricts section indices to sufficiently divisible multiples and leaves bigness unchanged. Choose an effective Cartier divisor $G\in |C|$ by taking a defining section that avoids the associated points of ${\mathcal{O}}_X$. Repeated divisor restriction exact sequences giveBecause $C$ is very ample, for each $j$ a section of $jC$ avoiding the finitely many associated points of ${\mathcal{O}}_G$ gives an injection into ${\mathcal{O}}_G(mB)$. Thus every summand is at most $h^0(G,{\mathcal{O}}_G(mB))$. By Serre vanishing and asymptotic Riemann–Roch for the ample $B$,For $n=1$, the restriction term is the constant length of $G$, giving the same formula directly. The positive coefficient proves that $B-C$, hence $D-A$, is big. Scaling back preserves bigness, so

For a general projective scheme, enforce the same strict inequality separately on each positive-dimensional reduced irreducible component $V_j$, using its own dimension $n_j$. At $(\delta ,t)=(0,0)$ every such expression equals the positive number $D^{n_j}\cdot [V_j]$. Finitely many conditions are preserved by one sufficiently small choice and one sufficiently close rational approximation on $N^1(X{)}_{\mathbb{R}}$. The top-dimensional inequalities imply the printed inequality for $[X]$ with its positive generic multiplicities; the section proof on each component makes $D-A$ componentwise big. This avoids inferring bigness on every component from just a positive sum. The displayed inequality is used for $n\ge 1$. For $n=0$, its literal $n-1$ intersection power is undefined; handle this vacuous positivity case separately. Every line bundle is ample, all numerical classes are zero, and the componentwise bigness convention makes the conclusions automatic.

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.

The Nakai–Moishezon criterion says that a Cartier divisor $D$ on a projective scheme is ample exactly whenfor every positive-dimensional integral closed subvariety $Y$. Kleiman's criterion says that the ample cone is the interior of the nef cone; equivalently, the numerical class of $D$ is ample exactly when it is strictly positive on every nonzero element of the closed cone of curves $\overline{\mathrm{NE}}(X)$. Positivity merely on individual curves is insufficient: the closure of the cone is essential. A nef divisor has nonnegative intersection number with every integral curve, and its restriction to every closed subscheme is nef.

For a projective scheme $X$ of dimension of a scheme $n$ and a Cartier divisor $D$, asymptotic Riemann–Roch givesHere Euler characteristic of a coherent sheaf means $\chi ={\sum }_{i=0}^n(-1{)}^i{h}^i$, and $D^n$ is the degree of the top intersection product with the fundamental cycle of $X$, including its component multiplicities. No ampleness assumption on $D$ is needed.

Fix any ample divisor $A$. Choose $q\gg 0$ so that both $qA$ and $(q-1)A$ have nonzero global sections, and choose $F\in |qA|$. Condition (1) and the section subtraction lemma for big divisors yield an integer $j>0$ with $jD-qA\sim E_0\ge 0$. Adding an effective member of $|(q-1)A|$ gives $jD\sim A+E$. Thus (1) implies the stronger condition (2').

If $jD\sim A+E$ with $A$ ample and $E$ effective, multiplication by the section of $qE$ injects $H^0(X,qA)$ into $H^0(X,qjD)$. The positive leading coefficient of the ample Hilbert polynomial supplies condition (1) along the infinite sequence $m=qj$. Condition (2) implies (3). Conversely, if $jD\equiv A+E$, put $P=jD-E$. The Cartier divisor $P$ is numerically equivalent to $A$, hence ample by Kleiman's criterion; the actual equality $jD=P+E$ gives (2).

We have proved (1)$\Rightarrow$(2')$\Rightarrow$(2)$\Rightarrow$(1), and (3)$\Rightarrow$(2)$\Rightarrow$(3). Also (2') implies (3'), while (3') implies (3). Hence all five conditions are equivalent. They characterize a big divisor; the ample-plus-effective formulation is Kodaira's lemma. The inconsistent use of $n$ and $j$ in the printed multiplier is resolved by using $j$ throughout.

Cover $Y$ byThese sets cover because a point of $D_Y$ has $(y,w)\ne (0,0)$. On $U_y$, the relation $x=zw/y$ shows that $D_Y$ has local equation $z$; on $U_w$, the relation $z=xy/w$ gives local equation $x$; and on $U_0$ its local equation is $1$. Thus $D_Y$ is a Cartier divisor.

For the line bundle associated to a divisor ${\mathcal{O}}_Y(D_Y)$, choose local framesThe transition functions areon the corresponding overlaps. Every displayed ratio is a unit in a ring in the corresponding ring of regular functions, and the Čech cocycle condition follows immediately.

The standard affine charts of ${\mathbb{P}}^3$ are copies of ${\mathbb{A}}^3$, and the product charts of ${\mathbb{P}}^1\times {\mathbb{P}}^1$ are copies of ${\mathbb{A}}^2$. Their local rings are localizations of polynomial rings over $k$ and hence are regular local rings. Both schemes are therefore regular.

On a regular integral scheme every Weil divisor is Cartier, so the divisor class group is naturally the Picard group. Pullback of line bundles along the Segre embedding therefore definesThe two groups areThe Segre coordinates are bihomogeneous of bidegree $(1,1)$, so ${\iota }^{\ast }{\mathcal{O}}_{{\mathbb{P}}^3}(1)=\mathcal{O}(1,1)$. In these bases the map is $m\mapsto (m,m)$ and

A Cartier divisor on an integral scheme is an open cover $(U_i)$ together with nonzero rational functions $f_i$ such that every ratio $f_i/f_j$ is a regular unit on $U_i\cap U_j$.

Every prime Weil divisor on ${\mathbb{P}}_k^n$ is cut out by an irreducible homogeneous polynomial because the polynomial ring $k[x_0,\dots ,x_n]$ is a unique factorization domain. Hence any Weil divisor can be represented by a homogeneous rational expressionof some total degree $d$. On the standard chart $U_i=D_{+}(x_i)$, put $f_i=F/x_i^d$. This is a degree-zero rational function, and on $U_i\cap U_j$,is a regular unit. These local equations form a Cartier divisor whose associated Weil divisor is the original one.

Now put $P=X\times {\mathbb{P}}^n$ and let $H=X\times {\mathbb{P}}^{n-1}$ be the hyperplane divisor at infinity. Its complement is $X\times {\mathbb{A}}^n$. Iterating the given invariance under multiplication by ${\mathbb{A}}^1$ givesThe localization sequence for the divisor class group shows that every class on $P$ is a pullback of a class on $X$ plus an integer multiple of $[H]$. Restriction to the generic fiber ${\mathbb{P}}_{K(X)}^n$ kills pullbacks from $X$ and sends $[H]$ to the generator of $\mathrm{Cl}({\mathbb{P}}_{K(X)}^n)\cong \mathbb{Z}$. Therefore the sum is direct, provingThis is the divisor class group of a projective-space bundle with trivial vector bundle.

A Cartier divisor on an integral scheme is given by an open cover $(U_i)$ and rational functions $f_i\in K(X{)}^{\ast }$ such that $f_i/f_j\in {\mathcal{O}}_X^{\ast }(U_i\cap U_j)$. Two are linearly equivalent when their quotient is represented by one global rational function. The Cartier class group is the group of Cartier divisors modulo these principal divisors.

Let ${\mathcal{K}}_X^{\ast }$ be the sheaf of nonzero rational functions. Cartier divisors are the global sections of ${\mathcal{K}}_X^{\ast }/{\mathcal{O}}_X^{\ast }$, and the exact sequencegives a long exact cohomology sequence. On an integral scheme ${\mathcal{K}}_X^{\ast }$ is flasque: every nonempty restriction map is the identity on $K(X{)}^{\ast }$. Hence $H^1(X,{\mathcal{K}}_X^{\ast })=0$, and exactness gives

For a smooth integral variety, every Weil divisor is a Cartier divisor, so the Picard group equals the divisor class group. If $U$ is obtained by removing irreducible Weil divisors $D_j$, the localization sequence for the divisor class group gives the displayed exact sequence. A rational trivialization of a line bundle on $U$ has divisor supported on the removed divisors, describing the kernel concretely.

For a smooth projective algebraic curve of genus $g$, this smooth proper $g$-dimensional scheme represents degree-$d$ line bundles in families. Charts come from nonspecial degree-$g$ bundles: their first sheaf cohomology vanishes, so the Riemann-Roch theorem gives exactly one independent global section, whose zero Cartier divisor varies algebraically. Twisting by fixed line bundles gives charts covering every degree-$d$ class. For degree zero the tensor product gives the Jacobian variety; other components are its torsors.