Numerical equivalence and the Neron-Severi space of a surface

Assume the Axiom of Choice, inherited from the intersection product and the Euler-characteristic suppliers (The Axiom of Choice). Let $k$ be a field and let $X$ be an integral regular projective surface over $k$ (Intersection numbers of Cartier divisors on a smooth projective surface).

Numerical equivalence. Two invertible ${\mathcal{O}}_X$-modules $\mathcal{L},\mathcal{M}$ (Invertible sheaves) are numerically equivalent, written $\mathcal{L}\equiv \mathcal{M}$, if $\mathcal{L}\cdot \mathcal{N}=\mathcal{M}\cdot \mathcal{N}\text{for every invertible }{\mathcal{O}}_X\text{-module }\mathcal{N}\mathrm{.}$ By $\mathbb{Z}$-bilinearity and symmetry of the intersection product (The surface intersection product is symmetric and bilinear) this is equivalent to $(\mathcal{L}\otimes {\mathcal{M}}^{\vee })\cdot \mathcal{N}=0$ for every invertible $\mathcal{N}$. An invertible sheaf is numerically trivial if it is numerically equivalent to ${\mathcal{O}}_X$.

For Cartier divisors $C,D$ on $X$ (Cartier divisor) write $C\equiv D$ when ${\mathcal{O}}_X(C)\equiv {\mathcal{O}}_X(D)$. Because $D\mapsto [{\mathcal{O}}_X(D)]$ induces an isomorphism $\mathrm{CaDiv}(X)\mathrm{/}{\mathrm{Prin}}_C(X)\stackrel{\sim }{\to }\mathrm{Pic}(X)$ and every invertible sheaf is ${\mathcal{O}}_X(C)$ for a Cartier divisor $C$ (On an integral scheme, Cartier divisors modulo principal divisors compute the Picard group, Rational sections of line bundles are Cartier divisors), this is equivalent to $C\cdot E=D\cdot E$ for every Cartier divisor $E$; by linear-equivalence invariance and the fact that every Cartier divisor is linearly equivalent to a difference of effective Cartier divisors (Intersection numbers of Cartier divisors on a smooth projective surface, The surface intersection product is symmetric and bilinear), it suffices to test $E$ running over effective Cartier divisors. For an effective Cartier divisor $E$ the number $C\cdot E$ is the degree of ${\mathcal{O}}_X(C){\mathrm{\mid }}_E$ (Intersection with a curve is the degree of the restriction), which is the formulation used in the sources.

Basic facts.

1. Numerical equivalence is an equivalence relation: the defining equalities of integers are reflexive, symmetric and transitive. It is compatible with tensor products and duality, and the numerically trivial classes form a subgroup of the Picard group $\mathrm{Pic}(X)$ (Picard group of a scheme): if $\mathcal{L}\equiv {\mathcal{L}}^{\mathrm{\prime }}$ and $\mathcal{M}\equiv {\mathcal{M}}^{\mathrm{\prime }}$ then $\mathcal{L}\otimes \mathcal{M}\equiv {\mathcal{L}}^{\mathrm{\prime }}\otimes {\mathcal{M}}^{\mathrm{\prime }}$ because $\cdot$ is bilinear, and ${\mathcal{L}}^{\vee }\equiv {\mathcal{L}}^{\mathrm{\prime }\vee }$ because bilinearity gives ${\mathcal{L}}^{\vee }\cdot \mathcal{N}={\mathcal{L}}^{\mathrm{\prime }\vee }\cdot \mathcal{N}$ (Dual of a line bundle is its tensor inverse).
2. Linearly equivalent Cartier divisors are numerically equivalent (Intersection numbers of Cartier divisors on a smooth projective surface): the intersection product depends only on the isomorphism classes of the associated invertible sheaves.
3. Consequently the intersection pairing descends to a well-defined symmetric $\mathbb{Z}$-bilinear pairing on the quotient ${\mathrm{N}}^1(X):=\mathrm{Pic}(X)\mathrm{/}\{\text{numerically trivial classes}\},$ and $\mathcal{L}\cdot \mathcal{M}$ depends only on the classes of $\mathcal{L}$ and $\mathcal{M}$: if $\mathcal{L}\equiv {\mathcal{L}}^{\mathrm{\prime }}$ and $\mathcal{M}\equiv {\mathcal{M}}^{\mathrm{\prime }}$ then $\mathcal{L}\cdot \mathcal{M}={\mathcal{L}}^{\mathrm{\prime }}\cdot {\mathcal{M}}^{\mathrm{\prime }}$ by testing $\mathcal{L}\equiv {\mathcal{L}}^{\mathrm{\prime }}$ against $\mathcal{M}$ and $\mathcal{M}\equiv {\mathcal{M}}^{\mathrm{\prime }}$ against ${\mathcal{L}}^{\mathrm{\prime }}$.

Real Neron-Severi space. Set ${\mathrm{N}}_{\mathbb{R}}^1(X):={\mathrm{N}}^1(X){\otimes }_{\mathbb{Z}}\mathbb{R}$, the extension of scalars along $\mathbb{Z}\to \mathbb{R}$ (Restriction of scalars and extension of scalars $S{\otimes }_{R}M$ along a ring homomorphism $R\to S$). It is a real vector space; no finiteness theorem is used in this definition. The pairing extends uniquely to a symmetric $\mathbb{R}$-bilinear form ${\mathrm{N}}_{\mathbb{R}}^1(X)\times {\mathrm{N}}_{\mathbb{R}}^1(X)⟶\mathbb{R},(x,y)⟼x\cdot y,$ the intersection form. For an invertible sheaf $H$ we write $[H]\in {\mathrm{N}}_{\mathbb{R}}^1(X)$ for its class.

The form remains nondegenerate after extending scalars. Indeed, ${\mathrm{N}}^1(X)$ is torsion-free: if $nv=0$ with $n\mathrm{\ne }0$, then $n(v\cdot w)=0$ for every integral class $w$, so $v$ is numerically trivial and $v=0$. Put $V_{\mathbb{Q}}={\mathrm{N}}^1(X)\otimes \mathbb{Q}$. Every rational class has an integral multiple, so the pairing on $V_{\mathbb{Q}}$ is nondegenerate by the same vanishing criterion. For any finite-dimensional rational subspace $W\subseteq V_{\mathbb{Q}}$, the functionals $v\mapsto v\cdot w$, with $w\in V_{\mathbb{Q}}$, span $W^{\ast }$: otherwise their common kernel in $W$ would contain a nonzero vector, contradicting nondegeneracy. Choose a basis of these functionals. Its evaluation matrix on a rational basis of $W$ is invertible over $\mathbb{Q}$, hence over $\mathbb{R}$. Every real class is a finite real combination of vectors in a rational basis of some such $W$. If it pairs to zero with every class, this invertible evaluation matrix forces all its coefficients to be zero. In particular $[H]=0$ exactly when $H$ is numerically trivial.