Surfaces

Report on Meeting 8

We continue to study surfaces that contain an effective divisor C with C.K < 0. (Recall that the existence of such a curve forces all P_{n} = 0 and κ = −1.) We showed last time that if K^{2} > 0, then X is rational. The next step is to consider what happens when K^{2} ≤ 0. A critical role will be played by the following theorem.

Theorem 1. (Tsen) Let π : X → C be a morphism from a smooth projective surface onto a smooth projective curve over an algebraically closed field F. Assume that all the fibres of π have p_{a} = 0. Then the function fields satisfy F (X) ≈ F (C)(t), where t is a transcendental element.

Remark 2. This theorem is not true if you increase the dimesnions by one (so that X is a threefold) or if you use a filed (such as a number field) that is not algebraically closed. The geometric example is due to Artin and Mumford; the arithmetic example, to Colliot-Thélène, Sansuc, and Swinnerton-Dyer.

Lemma 3. If there exists an effective C with C.K < 0 and if K^{2} ≤ 0, then for all postive integers n there are effective divisors D on X such that

(a) |D + K| = ∅, and

(b) dim |D| ≥ n.

Proof: As we saw in meeting 4, the first hypothesis allows us to choose an ample divisor H such that H.K < 0. For all n > 0 and for all m >> 0, we have (nH + mK).H < 0, so that the divisor nH +mK cannot be effective. So, for each n we can choose a non-negative integer m(n) such that

\begin{matrix}|nH + m(n)K| & ≠ ∅ \\ |nH + (m(n) + 1)K| & = ∅. \end{matrix}

Now let D_{n} ∈ |nH + m(n)K| and separate it into pieces D_{n} = A_{n} + B_{n} where both A_{n} and B_{n} are effective, with A_{n} containing the components that intersect K negatively, and B_{n} containing the components that intersect K non-negatively. As we have argued before, if E.K < 0 and E^{2} < 0, then E is exceptional of the first kind. So, by minimality, each component of A_{n} has non-negative self-intersection and thus A_{n}^{2} ≥ 0. Next, note that |K −A_{n}| = ∅, since otherwise K itself would be effective. Using Serre Duality, we conclude that h^{2}(A_{n}) = 0. We have:

dim |A_{n}| = dim H^{0}(O_{X}(A_{n})) − 1 ≥ χ(A_{n}) − 1

= \frac{1}{2}A_{n}.(A_{n} − K) + χ(O_{X}) − 1

≥ −\frac{1}{2}A_{n}.K + χ(O_{X}) − 1

≥ −\frac{1}{2}D_{n}.K + χ(O_{X}) − 1

= −\frac{1}{2}(nH + m(n)K).K + χ(O_{X}) − 1

= −\frac{nH.K}{2} − \frac{m(n)K^{2}}{2} + χ(O_{X}) − 1

≥ \frac{n}{2} + χ(O_{X}) − 1.

Finally, |K + A_{n}| ⊂ |K + D_{n}| = ∅, and the result follows.

Lemma 4. Assume that there exists an effective C with K.C < 0 and that K^{2} ≤ 0. If D is an effective divisor such that |D + K| = ∅, then the natural map on Picard schemes

P ic_{X}^{0} → P ic_{D}^{0}

is surjective.

Proof: By Serre duality, the hypothesis on D forces h^{2}(−D) = 0. Using the canonical exact sequence of a divisor

0 → O_{X}(−D) → O_{X} → O_{D} → 0,

one concludes that H^{1}(O_{X}) → H^{1}(O_{D}) is surjective. However, both Picard schemes are reduced (one because D is a curve, and the other because p_{g} = 0), and these H^{1}’s are the tangent spaces. It follows that the map on group schemes is surjective.

Lemma 5. Assume that there exists an effective C with K.C < 0 and that K^{2} ≤ 0. If D is an effective divisor such that |D + K| = ∅, write D = ∑ n_{i}E_{i} as a sum of its irreducible components. Then

(a) All E_{i} are nonsingular.

(b) If n_{i} ≥ 2, then either (i) E_{i} is rational, or (ii) E_{i} is elliptic with E_{i}^{2} = 0 and nontrivial normal bundle, or (iii) E_{i}^{2} < 0.

(c) The E_{i} are connected together without loops.

Proof: The natural map P ic_{D}^{0} → P ic_{E_{i}}^{0} is surjective. Moreover, by the previous result, P ic_{D}^{0} is a quotient of the abelian variety P ic_{X}^{0}. So, P ic_{D}^{0} and each P ic_{E_{i}}^{0} are abelian varieties. In particular, each E_{i} must be nonsingular, proving (a). If (c) were false, then P ic_{D}^{0} would contain factors of G_{m} corresponding to the loops. To check (b), notice that the hypothesis shows that P ic_{D}^{0} → P ic_{2E_{i}}^{0} is surjective, hence P ic_{2E_{i}}^{0} is an abelian variety, and hence P ic_{2E_{i}}^{0} = P ic_{E_{i}}^{0}. However, there are exact sequences

0 → O_{E_{i}}(−E_{i}^{2}) → O_{2E_{i}} → O_{E_{i}} → 0.

Using the fact that the first cohomology group is the tangent space to the Picard scheme, one concludes from the long exact sequence that H^{1}(O_{E_{i}}(−E_{i}^{2})) = 0. If E_{i}^{2} ≥ 0, then the degree of the sheaf on E_{i} is negative, and so one can only get H^{1} to vanish in one of the enumerated cases.

Lemma 6. Assume that there exists an effective C with K.C < 0 and that K^{2} ≤ 0. If P ∈ X is an arbitrary closed point, then X contains a nonsingular rational curve passing through P.

Proof: Suppose not. Then there are only finitely many rational curves on X of any given degree. Moreover, on any surface there are only fintely many nonsingular curves of each degree satisfying one of the conditions

(i) E^{2} < 0, or (ii) E is elliptic, E^{2} = 0, and the normal bundle of E is nontrivial. Taking all these curves together (for all degrees) we still only get a countable set of curves, which cannot exhaust X. Choose q distinct points P_{1}, …, P_{q} lying on none of these curves. Use the previous lemma to choose an effective divisor D on X such that |D + K| = ∅ and dim |D| ≥ 3q. The dimension is chosen large enough that we can choose a divisor C ∈ |D| with a double point at each P_{j}. Write C = ∑ n_{i}E_{i} as a sum of irreducible components. If any n_{i} ≥ 2, then the corresponding component E_{i} must lie in the countable set of curves that was described earlier, and so it cannot pass through any P_{j}. So, each P_{j} lies on components having multiplicity one. Since each E_{i} is nonsingular, each P_{j} must lie on 2 components. Since the E_{i} are connected to form a tree, we must have at least q + 1 distinct components containing some P_{j}. Moreover, by construction, each component is nonrational. Therefore

q = dim(P ic_{X}^{0}) ≥ dim(P ic_{D}^{0}) ≥ ∑ dim(P ic_{E_{i}}^{0}) ≥ q + 1.

We have arrived at a contradiction, so our assumption that there not enough rational curves must have been false.

Proposition 7. Assume that there exists an effective C with K.C < 0 and that K^{2} ≤ 0. Then X is a ruled surface.

Proof: First, suppose q = 0. Together with the previous result, this forces the existence of a positive dimensional linear system of nonsingular rational curves on X. Take a pencil of such curves, yielding a map X → P^{1} whose fibres have p_{a} = 0. Using Tsen’s theorem, we see that the function field of X is F (X) = F (P^{1})(t) = F (s, t), which is a rational function field in two variables. Hence, X is rational and is therefore birational to a ruled surface.

On the other hand, suppose q > 0. Consider the Albanese mapping π : X → A, where A is an abelian variety. By the universal property of the Albanese (and by the fact that Alb(P^{1}) = 0), the Albanese mapping must collapse all nonsingular rational curves to points. It thus follows from the previous lemma that the image of π is a curve C, and that the fibres are rational curves. The result now follows from Tsen’s theorem.