Let \(X\) be a ruled surface over a curve \(C\) of genus \(g\), with invariant \(e<0\), and assume that \(\operatorname{char} k=p>0\) and \(g \geqslant 2\).
If \(Y \equiv a C_0+b f\) is an irreducible curve \(\neq C_0, f\), then either
- \(a=1, b \geqslant 0\), or
- \(2 \leqslant a \leqslant p-1, b \geqslant \frac{1}{2} a e\), or
- \(a \geqslant p, b \geqslant \frac{1}{2} a e+1-g\).
If \(a>0\) and \(b>a\left(\frac{1}{2} e+(1 / p)(g-1)\right)\), then any divisor \(D \equiv a C_0+b f\) is ample. On the other hand, if \(D\) is ample, then \(a>0\) and \(b>\frac{1}{2} a e\).
Solution
Recall the numerical formulas for the normalized ruled surface: \[ C_0^2=-e, \qquad C_0\cdot f=1, \qquad f^2=0, \] and \[ K_X\equiv-2C_0+(2g-2-e)f. \]
1. If\[ Y\equiv aC_0+bf \]is an irreducible curve distinct from a fibre, then\[ a=Y\cdot f\ge1. \]
The ruling \[ \pi:X\longrightarrow C \] restricts to a nonconstant finite morphism from \(Y\) to \(C\) unless \(Y\) is contained in a fibre. Its degree is the intersection number with a general fibre: \[ \deg(\pi|_Y)=Y\cdot f=a. \] Since \(Y\) is not a fibre, this degree is positive.
2. If \(a=1\), then \(Y\) is a section and\[ \boxed{b\ge0}. \]
The finite morphism \[ Y\longrightarrow C \] has degree one. Since \(Y\) is integral and \(C\) is nonsingular, hence normal, it is an isomorphism. Thus \(Y\) is a section.
Let \[ 0\longrightarrow N \longrightarrow\mathcal E \longrightarrow L \longrightarrow0 \] be the quotient-line-bundle sequence corresponding to this section. The divisor formula for a section of \(\PP(\mathcal E)\) gives \[ \OO_X(Y) \cong \OO_X(C_0)\tensor\pi^*N^{-1}. \] Numerically, \[ Y\equiv C_0+(-\deg N)f, \] so \[ b=-\deg N. \]
Because \(\mathcal E\) is normalized, every line subbundle of \(\mathcal E\) has degree at most zero: an inclusion of a positive-degree line bundle \(N\) would give a nonzero section of the negative twist \(\mathcal E\tensor N^{-1}\). Hence \[ \deg N\le0, \] and therefore \(b\ge0\).
3. Assume \(a\ge2\). Let\[ \nu:\widetilde Y\longrightarrow Y \]be the normalization and let\[ \varphi=\pi\circ\nu:\widetilde Y\longrightarrow C. \]If \(s\) is the separable degree of \(\varphi\), then\[ \deg\varphi=a=s p^r \]for some \(r\ge0\), and\[ 2g(\widetilde Y)-2\ge s(2g-2). \]
1. In characteristic \(p\), the degree of afinite morphism of nonsingular curves factors as its separable degree timesits inseparable degree, and the latter is a power of \(p\):\[ a=s p^r. \]
Factor \(\varphi\) into its purely inseparable part and its separable part. Over the algebraically closed ground field, a finite purely inseparable map between nonsingular projective curves is a Frobenius power up to Frobenius twist, so it preserves the genus. Riemann–Hurwitz applied to the separable part therefore gives \[ 2g(\widetilde Y)-2 = s(2g-2)+\deg R \ge s(2g-2), \] where \(R\) is the effective ramification divisor.
4. For \(a\ge2\) one has\[ \boxed{ (a-1)(2b-ae) \ge -2(a-s)(g-1).} \]
The arithmetic genus of the integral curve \(Y\) dominates the genus of its normalization: \[ p_a(Y)\ge g(\widetilde Y) \] D-G1AEH. Hence step 3 gives \[ 2p_a(Y)-2\ge s(2g-2). \]
By adjunction T-SRFADJ, \[ 2p_a(Y)-2=Y\cdot(Y+K_X). \] Using the numerical formulas at the start, \[ \begin{aligned} Y\cdot(Y+K_X) &=(aC_0+bf)\cdot \bigl((a-2)C_0+(b+2g-2-e)f\bigr)\\ &=(a-1)(2b-ae)+2a(g-1). \end{aligned} \] Comparing with \(2s(g-1)\) and rearranging gives the claimed inequality.
5. If\[ 2\le a\le p-1, \]then\[ \boxed{b\ge\frac12ae}. \]
Since \(a<p\), no nontrivial power of \(p\) divides the inseparable degree in \[ a=s p^r. \] Thus \(r=0\) and \(s=a\). Step 4 becomes \[ (a-1)(2b-ae)\ge0. \] Because \(a-1>0\), \[ 2b-ae\ge0, \] which is the desired inequality.
6. If\[ a\ge p, \]then\[ \boxed{b\ge\frac12ae+1-g}. \]
The separable degree satisfies \[ 1\le s\le a. \] From step 4, \[ 2b-ae \ge -2\frac{a-s}{a-1}(g-1). \] Since \[ 0\le\frac{a-s}{a-1}\le1, \] one obtains \[ 2b-ae\ge-2(g-1). \] Equivalently, \[ b\ge\frac12ae+1-g. \] Together with steps 2 and 5, this proves part (a).
7. Now let\[ D\equiv aC_0+bf \]satisfy\[ a>0, \qquad b>a\left(\frac e2+\frac{g-1}{p}\right). \]Then\[ D^2>0, \qquad D\cdot f>0, \qquad D\cdot C_0>0. \]
First, \[ D\cdot f=a>0. \] Also \[ D^2=-a^2e+2ab = a(2b-ae). \] The hypothesis gives \[ 2b-ae>\frac{2a(g-1)}p>0, \] so \(D^2>0\).
Finally, \[ D\cdot C_0=b-ae. \] Since \(e<0\), \[ b-ae > -\frac12ae+\frac{a(g-1)}p >0. \]
8. Under the hypotheses of step 7,\[ D\cdot Y>0 \]for every irreducible curve\[ Y\ne C_0,f. \]
Write \[ Y\equiv cC_0+df. \] Part (a), already proved, gives exactly three possibilities. In all cases \[ D\cdot Y=-ace+ad+bc. \]
If \(c=1\), then \(d\ge0\), so \[ D\cdot Y = -ae+ad+b \ge -ae+b. \] The hypothesis implies \(b>ae/2\), and \(e<0\), hence \[ -ae+b>-\frac12ae>0. \]
If \(2\le c\le p-1\), then \[ d\ge\frac12ce. \] Therefore \[ \begin{aligned} D\cdot Y &\ge -ace+\frac12ace+bc\\ &=c\left(b-\frac12ae\right)\\ &>0. \end{aligned} \]
Finally, if \(c\ge p\), then \[ d\ge\frac12ce+1-g. \] Hence \[ \begin{aligned} D\cdot Y &\ge -ace+a\left(\frac12ce+1-g\right)+bc\\ &= c\left(b-\frac12ae\right)-a(g-1). \end{aligned} \] The hypothesis says \[ b-\frac12ae>\frac{a(g-1)}p, \] so \[ D\cdot Y > a(g-1)\left(\frac cp-1\right) \ge0. \] The inequality is strict, so \(D\cdot Y>0\) also when \(c=p\).
9. The divisor \(D\) in step 7 is ample.
Step 7 gives \(D^2>0\) and positive intersection with \(C_0\) and \(f\). Step 8 gives positive intersection with every other irreducible curve on \(X\). The Nakai–Moishezon criterion T-SRFNAKAI therefore gives \[ \boxed{D\text{ is ample}.} \] This proves the forward implication in part (b).
10. Conversely, if\[ D\equiv aC_0+bf \]is ample, then\[ \boxed{a>0, \qquad b>\frac12ae}. \]
Ampleness and Nakai–Moishezon give \[ D\cdot f>0. \] Since \[ D\cdot f=a, \] one has \(a>0\).
Again by Nakai–Moishezon, \[ D^2>0. \] But \[ D^2=a(2b-ae). \] Since \(a>0\), this forces \[ 2b-ae>0, \] or \[ b>\frac12ae. \] This proves the converse implication in part (b).
11. Q.E.D.
Steps <1>1–<1>6 prove part (a), and steps <1>7–<1>10 prove part (b).