2.9 Ampleness of \(\mathbb {Q}\)-divisors
Suppose \(D\) is a \(\mathbb {Q}\)-divisor in \(\operatorname{\mathrm{Div}}X\). The following are equivalent.
\(D = \sum c_i A_i\) with \(c_i\) positive and \(A_i\) integral and ample.
There exists a positive integer \(r\) such that \(rD\) is an ample divisor in \(\operatorname{\mathrm{Div}}X\).
\(D\) satisfies Nakai-Moishezon. In other words, for every closed \(k\)-dimensional subvariety \(V\) of \(X\), \(\int _V D^k {\gt} 0\).
(a)\(\Rightarrow \)(b): Clear denominators. Then use Proposition 2.10.
(b)\(\Rightarrow \)(a): Obvious.
(b)\(\Rightarrow \)(c): Clear denominators with \(r\), and then pull \(r^k\) out of the integral sign.
(c)\(\Rightarrow \)(b): Find a positive integer \(r\) such that \(rD\) is integral. Then, by the usual Nakai-Moishezon, \(rD\) is ample.
We say that \(D\) is \(\mathbb {Q}\)-ample if the equivalent conditions of Propostion 2.22 hold.
Suppose \(D\in \operatorname{\mathrm{Div}}X\). Then \(D\) is ample as a \(\mathbb {Q}\)-divisor if and only if it is ample as a \(\mathbb {Q}\)-divisor.
If \(A\) is an ample \(\mathbb {Q}\)-divisor then any \(\mathbb {Q}\)-divisor numerically equivalent to \(A\) is also ample.
Suppose \(D_1\) and \(D_2\) are two ample \(\mathbb {Q}\)-divisors and \(a_1\) and \(a_2\) are two nonnegative rational numbers which are not both \(0\). Then \(a_1 D_1 + a_2 D_2\) is ample.
Obvious by clearing denominators.
Suppose \(H, E_1, \ldots , E_k\in \operatorname{\mathrm{Div}}_{\mathbb {Q}} X\) are \(\mathbf{Q}\)-divisors with \(H\) \(\mathbb {Q}\)-ample. Then there exists a positive rational number \(\epsilon \) such that, for all \(x_1,\ldots , x_n\in \mathbb {Q}\) with \(|x_i|{\lt}\epsilon \) for all \(i\), the divisor \(H+x_1E_1 + \cdots x_k E_k\) is \(\mathbb {Q}\)-ample.
After clearing denominators, we can assume that \(H\) and all the \(E_i\) are integral. Then we can find a positive integer \(n\) such that \(nH+E_i\) and \(nH-E_i\) are all ample (for \(i=1,\cdots , k\)). Then, if we set \(\epsilon = 1/n\), we get that \(H+\epsilon E_i\) and \(H-\epsilon E_i\) are both ample. But then, for \(x\in (-\epsilon , \epsilon )\), there exists \(s\in [0,1]\) such that \(k^{-1}H+xE_i = s(k^{-1}H-\epsilon E_i) + (1-s) (k^{-1}H+\epsilon E_i)\). So it follows from Corollary 2.26 that \(k^{-1}H + x_iE_i\) is ample. But then, using Corollary 2.26 again, it follows that \(H+x_1 E_1 + \cdots x_k E_k\) is ample.
The next Proposition is a special case (for surfaces) of a more general theorem of Kleiman’s.
Suppose \(\dim X = 2\) and \(D\) is a nef \(\mathbb {Q}\)-divisor. Then \(D^2\geq 0\).
To get a contradiction, suppose that \(D\) is a nef \(\mathbb {Q}\)-divisor with \(D^2 {\lt} 0\). Pick an ample \(\mathbb {Q}\)-divisor \(H\), and define polynomials \(P\) and \(Q\) with rational coefficients by settings \(P(t) = (D+tH)^2\) and \(Q(t) = (D\cdot H)t + D^2\). Then \(Q(t) = bt + c\) with \(b= D\cdot H\) and \(c = D^2\), while \(P(t) = at^2 + 2bt + c\) with \(a= H^2\). Since \(H\) is ample and \(D\) is nef, we have \(a{\gt}0\) and \(b\geq 0\). But, by assumption, we have \(c {\lt} 0\).
It follows that \(P\) has a positive real root \(\displaystyle t_0 = \frac{-b +\sqrt{b^2 - ac}}{a}\). For each rational number \(s\), set \(D(s) = D+sH\).
If \(s\) is a rational number with \(s{\gt}t_0\) and \(C\) is a curve, then \(D(s)\cdot C = D\cdot C + s H\cdot C {\gt} 0\). And \(D(s)^2 = P(s) {\gt} 0\). So \(D(s)\) is ample by Nakai-Moishezon. Since \(D\) is nef, it follows that \(Q(s) = D^2 + sD\cdot H = D\cdot D(s) \geq 0\) for all rational \(s{\gt}t_0\). But then, by continuity, \(Q(t_0) \geq 0\).
However, \(P(t) = at^2 + bt + Q(t)\). So, as \(a\) and \(t_0\) are positive and \(b\geq 0\), it follows that \(P(t_0) {\gt} 0\). This contradicts our assumption that \(t_0\) is a root of \(P\).