Surfaces

Report on Meeting 7

As in the last meeting, we will assume that P_{2} = q = 0 and that K^{2} > 0. We saw that these hypotheses imply that p_{g} = 0, and that 10 = K^{2} + b_{2}, so 1 ≤ K^{2} ≤ 9. Also, recall the following result:

Proposition 1. If | − K| contains a reducible linear subsystem δ, then X is a rational surface.

Remark 2. The idea is to use this result to prove Castelnuovo’s Criterion. Unfortunately, we need to proceed differently depending on the value of K^{2}.

Proposition 3. Assume K^{2} ≥ 4 and P_{2} = q = 0. Then X is rational.

Proof: Fix a rational point P ∈ X, and consider the linear system δ of all divisors in | − K| that have a singular point at P. Then dim(δ) ≥ dim | − K| − 3 ≥ K^{2} − 3 > 0. (Note: the dimensiojn decreases by three because imposing a singular point on a curve on a surface is three conditions: one for containing the point, and one each for the vanishing of two partial derivatives.) If δ is reducible, then X is rational by the previous result. So, assume that δ is irreducible. Since p_{a}(−K) = 1, it follows that the general member of δ is a rational curve (with a singularity that causes it to have arithmetic genus 1). Blow up the point P. The total transform of a divisor D ∈ δ has the form C + 2E, where C is effective. This produces a positive dimensional linear system of genus zero curves on the blowup. So, X is rational.

Proposition 4. Assume K^{2} = 3 and P_{2} = q = 0. Then X is rational.

Proof: Without loss of generality, we may assume that | − K| is an irreducible linear system. Letting r denote the dimension of this linear system, we see that it defines a rational map X → Y where Y is a variety of dimension at most 2 in a projective space P^{r}. By irreducibility, |−K| is not composite with a pencil, so the image has dimension exactly equal to 2. Since the degree of the map is (−K)^{2} = 3, which is a prime, the map must either be an isomorhism on function fields or a (purely inseparable) extension of degree 3. In the latter case, Y must have degree 1, and so is a plane. This is impossible, since the image does not lie in a proper linear subspace. So, the map is birational to a surface of degree 3 in projective space. By repeatedly projecting into smaller projective spaces, we see that X is birational to a surface of degree 2 or 3 in P^{3}. All surfaces of degree 2 are rational. The only nonrational cubic surfaces are cones over elliptic curves, which have arithmetic genus p_{a} > 0. However, our surface has p_{a} = p_{g} − q = 0, hence must be rational.

Proposition 5. Assume K^{2} = 2 and P_{2} = q = 0. Then X is rational.

Proof: We may assume that | − K| is irreducible. We claim that dim | − K| = K^{2}. Since we already know that dim | − K| ≥ K^{2}, we will proceed by assuming that dim | − K| > K^{2} and derive a contradiction. Let C be a general (irreducible) member of the linear system | − K|. As we have seen, p_{a}(C) = 1. If C is nonsingular, then it is an elliptic curve, and every effective divisor D on C satisfies l(D) = deg(D). However, restricting the linear system | − K| to C yields a linear system of degree C.(−K) = (−K)^{2} and of dimension greater than K^{2}. Thus C must contain a singular point P. But then the divisors in | − K| that pass through the point Q would cut out a linear system of degree ≤ K^{2} − 2 and of dimension ≥ K^{2} − 1, which is impossible. So, dim | − K| = K^{2}.

If | − K| has base points, then (after removing the fixed part of the divisor at the base points) restricting | − K| to an irreducible member C of the family would again cut out a linear system of degree < K^{2} and of dimension = K^{2} − 1. As before, this forces C to be a singular rational curve. Once again, take the linear subsystem of anticanonical divisors passing through the singularity (which is not a base point) to conclude that | − K| has no base points.

Now we know that | − K| defines a morphism φ : X → P^{2}. Since the degree of this map is K^{2} = 2, the extension of functions fields F (X)/F (x, y) is also of degree 2. Let us make life simple and assume the characteristic of the ground field is not 2. Then the field extension is separable, and X must be birationally equivalent to an affine surface Y with equation z^{2} = f(x, y), where the polynomial f has no multiple factors. Now let Δ be the curve in the affine plane defined by f(x, y) = 0; this is the branch locus of the cover. Let L be any line in P^{2}, and let C ∈ |−K| on Y be the corresponding curve. A singular point on C corresponds to a multiple intersection of Δ with L. It follows that the general member of the anticanonical class is nonsingular, and hence is an elliptic curve. But the quadratic map C → L has exactly 4 branch points, so the degree of Δ (and of f) must also equal 4. Now, every degree 4 curve in the plane has either a bitangent or a line that hits it at one point with multiplicity 4. The corresponding curve C in the anticanonical linear system either has two singular points or a tacnode, and thus cannot be an irreducible rational curve with p_{a} = 1. To finish the proof, we assert:

If X is a minimal surface with p_{a} = P_{2} = 0 and K^{2} > 0, and if X contains a divisor C that is not linearly equivalent to an integral multiple of K, then X is rational. To see this, we can assume that p_{a}(C) = 0. If C^{2} ≥ 0, we will see later that X is birationally equivalent to a ruled surface, which will force it to be rational since p_{a} = 0. So, we may assume that if D is any irreducible curve with p_{a}(D) = 0, then D^{2} < 0. By minimality, this means in particular that D^{2} ≤ −2. So

D.(−K) = D^{2} − 2p_{a}(D) + 2 = D^{2} + 2 ≤ 0.

If −K.D < 0, then D is a fixed component of | − K|. Then | − K| is reducible and X is rational. So, we must have D.K = 0 and D^{2} = −2. Then there must exist an effective divisor in |−K| that contains D as a component. But every component of this divisor must have arithmetic genus 0 (and hence self-intersection -2 and must intersect the anticanonical divisor trivially). But that contradicts the fact that K^{2} > 0.

Proposition 6. Assume K^{2} = 1 and P_{2} = q = 0. Then X is rational.

Proof: We may assume that | − K| is irreducible. We have dim | − K| = K^{2} = 1, so that | − K| defines a rational map X → P^{1}, with a single base point P that is nonsingular on each member of the pencil, and the tangents through P are distinct for distinct members of the pencil. One can also conclude that all the systems | − nK| are irreducible for n ≥ 1, of dimension n(n + 1)/2. In particular, l(−2K) = 4. Letting t be a parameter for the pencil, we have

\begin{aligned}Γ(X, O_{X}(−K)) & =< 1, t > \\ Γ(X, O_{X}(−2K)) & =< 1, t, t^{2}, x > . \end{aligned}

Since | − 2K| is irreducible, it is not composite with a pencil. It follows that x and t are algebraically independent over the ground field F, and that F (X) is an algebraic extension of the rational function field F (t, x). The image W of the rational map (1, t, t^{2}, x) is the rational cone z_{0}z_{2} = z_{1}^{2}, and its plane sections correspond naturally to the cycles in |−2K|. Since W has degree 2 and | − K| is degree 4, it follows that the function field F (X) is a quadratic extension of the function field of W (which is itself rational). One can proceed as before to give a simple proof of the rationality of X in the case that the characteristic differs from 2.

Remark 7. We have shown (more or less) that if there exists an effective divisor C with C.K < 0 and if K^{2} > 0, then X is rational. The next step is to consider what happens when such a curve exists, but K^{2} ≤ 0. That will have to wait until next time.