2.8 Rational and real coefficients
To explain Kleiman’s ampleness criterion it’s going to be useful to talk about \(Z_1 X\) and \(\operatorname{\mathrm{Div}}X\) but with rational and real coefficients. So if \(F = \mathbb {Q}\) or \(\mathbf{R}\), we set \(\operatorname{\mathrm{Div}}_F X = \operatorname{\mathrm{Div}}X\otimes F\), and we call an element \(D\in \operatorname{\mathrm{Div}}_F X\) and \(F\)-divisor. Similarly, we set \(Z_1^F X = Z_1 X\otimes F\).
It’s worth pointing out that, since \(\operatorname{\mathrm{Div}}X\) is a free \(\mathbb {Z}\)-module, the inclusions \(\mathbb {Z}\to \mathbb {Q}\to \mathbb {R}\) lead to inclusions \(\operatorname{\mathrm{Div}}X\hookrightarrow \operatorname{\mathrm{Div}_{\mathbb {Q}}}X\hookrightarrow \operatorname{\mathrm{Div}_{\mathbb {R}}}X\). So we can just view \(\operatorname{\mathrm{Div}}X\) as a subgroup of \(\operatorname{\mathrm{Div}_{\mathbb {Q}}}X\) and \(\operatorname{\mathrm{Div}_{\mathbb {Q}}}X\) as a sub-\(\mathbb {Q}\)-vector space of \(\operatorname{\mathrm{Div}_{\mathbb {R}}}X\). We call a divisor \(D\in \operatorname{\mathrm{Div}_{\mathbb {R}}}X\) integral if \(D\in \operatorname{\mathrm{Div}}X\) and rational if \(D\in \operatorname{\mathrm{Div}_{\mathbb {Q}}}X\).
Note that, for \(F=\mathbb {Q}\) or \(F=\mathbb {R}\), if \(D_1,\ldots , D_k\) are \(F\) divisors and \(V\in X_{(k)}\), then, by linearity, we get numbers \(\int _V D_1\cdots D_k \in F\). So, in particular, we get a pairing
and we say that an \(F\)-divisor or an \(F\)-\(1\)-cycle is numerically equivalent to \(0\) if it is in the kernel of the pairing. Equivalently, an \(F\)-divisor \(D\) is numerically equivalent to \(0\) if \(\deg D\cdot C = 0\) for every curve \(C\) on \(X\). It follows that an integral divisor \(D\) is \(\mathbb {Q}\) or \(\mathbb {R}\) numerically equivalent to \(0\) if and only if it is numerically equivalent to \(0\) in \(\operatorname{\mathrm{NS}}X\). And, similarly, a divisor \(D\in \operatorname{\mathrm{Div}_{\mathbb {Q}}}X\) is numerically equivalent to \(0\) if and only if it is numerically equivalent to \(0\) in \(\operatorname{\mathrm{Div}_{\mathbb {R}}}X\).
Set \(\operatorname{\mathrm{Div}}_{F\operatorname{\mathit{num}}} X\) equal to the \(F\)-vector space of all \(F\)-divisors which are numerically equivalent to \(0\). We then write \(N^1_F X = \operatorname{\mathrm{NS}}_F X = \operatorname{\mathrm{Div}}_F X/\operatorname{\mathrm{Div}}_{F\operatorname{\mathit{num}}} X\).
Numerical equivalence of \(\mathbb {Q}\) and \(\mathbb {R}\)-divisors turns out to be easy, but thinking about it at first is a little bit confusing. So I think a lemma about the general algebraic situation helps.
Suppose \(A\) is a commutative ring, \(B\) is a flat commutative \(A\)-algebra, \(F\) is a free \(A\)-module and \(S\subseteq \operatorname{\mathrm{Hom}}(F,A)\). Let \(S_B\) denote the image of \(S\) in \(\operatorname{\mathrm{Hom}}(F\otimes B, B)\) under the natural morphism \(\operatorname{\mathrm{Hom}}(F,A)\to \operatorname{\mathrm{Hom}}(B\otimes F, B)\), and set \(S^{\perp } = \cap _{\lambda \in S} \ker \lambda \), \(S_B^{\perp } = \cap _{\lambda \in S_B} \ker \lambda \). Write \(\phi :B\otimes S^{\perp }\to S_B^{\perp }\) for the \(B\)-module morphism sending \(\sum b_i\otimes f_i\) to \(\sum b_i f_i\). Then
\(\phi \) is an isomorphism.
\(\displaystyle \frac{F}{S^{\perp }}\otimes B = \frac{F\otimes B}{S_B^{\perp }}\).
Set \(G= A^{\oplus S}\) and write \(\Lambda :F\to G\) for the morphism of \(A\)-modules whose projection on the \(\lambda \)-component of \(G\) is \(\lambda :F\to A\). Then we have an exact sequence
Write \(N\) for the image of \(\Lambda \) and \(Q\) for the cokernel of the inclusion \(N\to G\). So we wind up with two exact sequences
with \(i\pi = \Lambda \).
Now, since \(B\) is flat over \(A\), tensoring with \(B\) gives us exact seqences
and \(i_B\pi _B = \Lambda _B: F\otimes B\to G\otimes B\). So we have \(\ker \Lambda _B = S^{\perp }\otimes B\). But \(\ker \Lambda _B = S_B^{\perp }\). And, from this, we get the isomorphism in a. But now this directly implies the isomorphism in b, which just says that \(N\otimes B = (F\otimes B)/S_B^{\perp }\).
We have \(\operatorname{\mathrm{NS}}_{F} X = \operatorname{\mathrm{NS}}X\otimes F\) for \(F=\mathbb {Q}\) or \(\mathbb {R}\).
Set \(B=F\), \(A=\mathbb {Z}\), \(F=\operatorname{\mathrm{Div}}X\), and set \(S\) equal to the set of all maps \(\operatorname{\mathrm{Div}}X\to \mathbb {Z}\) of the form \(D\mapsto \deg D\cdot C\), where \(C\) ranges over all curves in \(X\). Then use Lemma 2.14.
Supose \(D\in \operatorname{\mathrm{Div}}_F X\) with \(F=\mathbb {Q}\) of \(\mathbb {R}\). Then \(D\) is numerically equivalent to zero if and only if there exists integral divisors \(D_1,\ldots , D_k\) which are numberically equivalent to \(0\) and real numbers \(c_1,\ldots , c_k\) such that \(D = \sum c_i D_i\).
If \(D\) is rational, then \(D\) numerically equivalent to \(0\) if and only if there exists a nonzero integer \(r\) such that \(rD\) is integral and numerically equivalent to \(0\).
We say that a \(\mathbb {Q}\)-divisor \(D\) is linearly equivalent to \(0\) if there exists a nonzero integer \(r\) such that \(rD\) is linearly equivalent to \(0\). It might help to keep the following easy lemma in mind.
Suppose \(D\) is an integral divisor. Then \(D\) is linearly equivalent to \(0\) as a \(\mathbb {Q}\)-divisor if and only if \(D\) is torsion as an integral divisor.
Obvious.
If \(F =\mathbb {Q}\) or \(\mathbb {R}\) then, by expanding out using linearity, we get pairings
sending \((D_1,\ldots , D_k, Z)\) with \(Z = \sum a_i V_i\) to \(\sum a_i \deg D_1\ldots D_k\cdot V_i\). Using Theorem 2.5 and Corollary 2.17, we can see that these pairings factor through numerical equivalence to give pairings
Suppose \(D\in \operatorname{\mathrm{Div}}_F X\) for \(F=\mathbb {Q}\) or \(\mathbb {R}\). We say that \(D\) is nef if \(\deg D\cdot C \geq 0\) for all curves \(C\) on \(X\).