MIT OpenCourseWare http://ocw.mit.edu 18.727 Topics in Algebraic Geometry: Algebraic Surfaces Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.ALGEBRAIC SURFACES, LECTURE 9 LECTURES: ABHINAV KUMAR 1. Castelnuovo’s Criterion for Rationality Theorem 1. Any surface with q = h1(X, OX ) = 0 and p2 = h0(X, ω⊗2) = 0 isX rational. Note. Every rational surface satisfies these: they are birational invariants which vanish for P2 . Reduction 1: Let X be a minimal surface with q = p2 = 0. It is enough to show there is a smooth rational curve C on X with C2 ≥ 0. Proof. First, observe that 2g(C) − 2 = −2 = C (C + K) and χ(OX (C)) = · χ(OX ) + 1 C(C − K). Since p2 = 0, p1 = h0(X, ω) = h2(X, OX ) = 0 and 2 χ(OX ) = 1. Since h2(C) = h0(K − C) ≤ h0(K) = 0, h0(C) ≥ 1 + 21 C(C − K), so h0(C) ≥ 2 + C2 ≥ 2. Choose a pencil inside this system containing C, i.e. a subspace of dimension 2. The pencil has no fixed component (the only possibility is C, but C moves in the pencil): after blowing up finitely many base points, we get a morphism X˜P1 with a fiber isomorphic to C ∼P1 . Therefore, by the = → ˜ ˜Noether-Enriques theorem, X is ruled over P1 and X is rational (as is X). � Reduction 2: Let X be a minimal surface with q = p2 = 0. It is enough to show that ∃ an effective divisor D on X s.t. |K + D| = ∅ and K · D < 0. Proof. This implies that some irreducible component C of D satisfies K C < 0.· Clearly, |K + C| ⊂ |K + D|. Using Riemann-Roch for K + C gives 0 = h0(U + C) + h0(−C) = h0(K + C) + h2(K + C) (1) 1 ≥ 1 + (K + C) C = g(C)2· We thus obtain a smooth, rational curve C on X: −2 = 2g − 2 = C(C + K) and C · K < 0 = ⇒ C2 ≥ −1. Since X is minimal, C2 =� −1, so C2 ≥ 0 as desired. � We now prove our second statement. There are three cases: 12 LECTURES: ABHINAV KUMAR Case 1 (K2 = 0): Riemann-Roch gives h0(−K) = h0(−k) + h0(2K) = h0(−K) + h2(−K) (2) 1 ≥ 1 + 2 K · 2K = 1 + K2 = 1 so |−K = ∅. Take a hyperplane section H of X. Then there is an n ≥ 0 s.t. |H +| �nK| =� ∅ but |H + (n + 1)K| = ∅. Since −K ∼ an effective nonzero divisor, H K < 0 and H (H + nK) is eventually negative and H + nK is not · · effective. Let D ∈ |H + nK|: then |D + K| = ∅ and K · D = K(H + nK) = K H < 0 since −K is effective, H very ample. · Case 2 (K2 < 0): it is enough to find an effective divisor E on X s.t. K E < 0.·Then some component C of E will have K C < 0. The genus formula gives · −2 ≤ 2g − 2 = C(C + K) = ⇒ C2 ≥ −1. C2 = −1 is impossible since X is minimal, so C2 ≥ 0. Now (C + nK) C is negative for n >> 0, so C + nK is· not effective for n >> 0 by the useful lemma. So ∃n s.t. |C + nK| =� ∅ but |C + (n + 1)K| = ∅. Choosing D ∈ |C + nK| gives the desired divisor. We now find the claimed E. Again, let H be a hyperplane section: if K H < 0,·we can take E = H; if K H = 0, we can take K + nH for n >> 0; so assume H · K H > 0. Let γ = −K·> 0 so that (H + γK) K = 0. Also, · K2 · (3) (H + γK)2 > H2 + 2γ(H K) + γ2K = H2 +(K · H)2 > 0· (−K2) So take β rational and slightly larger than γ to get (4) (H + βK) K < (H + γK) K = 0 · · s(since K2 < 0) and (H + βK)2 > 0. Therefore, (H + βK) H > 0. Write β = r .· Then (5) (rH + sK)2 > 0, (rH + sK) K < 0, (rH + sK) H > 0· · by equivalent facts for β. Let D = rH + sK. For m >> 0, by Riemann-Roch we get h0(mD) + h0(K − mD) ≥ 1 mD(mD − K) + 1 → ∞. Moreover, K − mD2 is not effect over for m >> 0 since (K − mD) H = (K H) − m(D H). Thus, · · · mD is effective for large m, and we can take E ∈ |mD|. Case 3 (K2 > 0): Assume that there is no such D as in reduction 2, i.e. K · D ≥ 0 for every effective divisor D s.t. |K + D| = ∅. We will obtain a contradiction. Lemma 1. If X is a minimal surface with p2 = q = 0, K2 > 0 and K D ≥ 0· for every effective divisor D on X s.t. |K + D| = ∅, then (1) Pic (X) is generated by ωX = OX (K), and the anticanonical bundle OX (−K) is ample. In particular, X doesn’t have any nonsingular ra-tional curves.3 ALGEBRAIC SURFACES, LECTURE 9 (2) Every divisor of |−K| is an integral curve of arithmetic genus 1. (3) (K2) ≤ 5, b2 ≥ 5. (Here, b2 = het2´(X, Q�) in general. Proof. First, let us see that every element D of |−K| is an irreducible curve. If not, let C be a component of D s.t. K C < 0 (which we can find, since · K · D = −K2 < 0). If D = C + C�, |K + C| = |−D + C| = |−C�| = ∅ since C� is effective. Also, C K < 0, contradicting the hypothesis. So D is irreducible, · 1and similarly D is not a multiple. Furthermore, pa(D) = 2 D(D + K) + 1 = 1, showing (2). Next, we claim that the only effective divisor s.t. D + K = ∅ is the zero divisor. Assume not, i.e. ∃D > 0 s.t. |K + D| = ∅.|Let x|∈ D: then since h0(−K) ≥ 1 + K2 ≥ 2, there is a C ∈ |−K| passing through x. C is an integral curve, and cannot be a component of D since then (6) |K + D| ⊃ |K + C| = |0| �= ∅ So C D > 0 since they meet at least in x. Then K D = −C D < 0, contradicting · · ·the hypothesis. As an aside, we claim that pn = …