MIT OpenCourseWare http://ocw.mit.edu18.727 Topics in Algebraic Geometry: Algebraic SurfacesSpring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.ALGEBRAIC SURFACES, LECTURE 10 LECTURES: ABHINAV KUMAR Recall that we had left to show that there are no surfaces in characteristic p > 0 satisfying (1) Pic (X) is generated by ωX = OX (K), and the anticanonical bundle is ample. In particular, X doesn’t have any nonsingular rational curves. (2) Every divisor of |−K| is an integral curve of arithmetic genus 1. (3) (K2) ≤ 5, b2 ≥ 5. Lemma 1. Let X be as above. Then ∃ a nonsingular curve D ∈ |−K|. Proof. Suppose not. Since pa(D) = 1, we may assume that every D in |−K has exactly one singular point, a node or a cusp. Then dim |K K2 1. Let L ⊂ |−K| be a one-dimensional |−linear ≥ ≥subsystem. The fiber |s of φ1 L : X ��� Pare exactly the curves in L, because L has no fixed components (all the elements of L are integral). Let Y be the set of all singular points on curves in L, and let x be a base point of L (if any). We claim that x ∈/ Y . This is because if x ∈ Y , then for D ∈ L a curve singular ˜at x, π : X → X the blowup of X at x, then ˜D is a nonsingular rational curve ˜an ˜d a fiber of φ = φ ◦ π : X → P1 . After further (at most K2) blowups, we get an X� s.t. φ� : X� → P1 is a morphism and one fiber of φ�is a smooth, rational curve. So X�is geometrically ruled over P1 , implying that X� is rational and so X is rational. This, however, is impossible by the classification of rational surfaces (Pic (X) is never ZωX ), giving the desired contradiction. ˜˜1 So blowup the base points of φ so that φ : X P is a quasi-elliptic fibration →(by Tate, all singular points are cusps in characteristic 2 or 3, e.g. y2 = x3 + t). All the fibers are integral rational curves with one singular point, and the set of singularities Y˜of φ˜is a nonsingular irreducible curve with φ˜|Y : Y˜→P1 a bijective, purely inseparable morphism. Thus, Y˜= P1 = Y = P1 ,∼⇒∼contradicting the fact that there are no smooth rational curves on X. � Lemma 2. Let X be as above. Then H2(X, TX ) = 0 for TX = (Ω1 )∨ theX/ktangent sheaf. 12 LECTURES: ABHINAV KUMAR Proof. ∃ a nonsingular elliptic curve D ∈ |−K| by the above lemma. The short exact sequence(1) 0 → OX ((n − 1)D) ⊗ TX → OX (nD) ⊗ TX → OX (nD) ⊗ TX ⊗ OD → 0 gives the long exact sequence in cohomology H1(X, O T2X (nD) ⊗ X ⊗ OD) → H (X, OX ((n − 1)D) ⊗ TX ) (2) → H2(D, OX (nD) ⊗ TX ) = 0 for large n by Serre vanishing (since D is ample). By reverse induction, it is enough to show that H1(X, OX (nD) ⊗ TX ⊗ OD) = 0 for n ≥ 1. This will show H2((3) O2X ((n − 1)D) ⊗ TX ) = 0 =⇒ H (OX ((n − 2)D) ⊗ TX ) = 0 =⇒ H2(OX ⊗ TX ) = 0 Dualizing the conormal exact sequence 0 → I/I2 → ΩX ⊗ OD → ΩD → 0, we get(4) 0 → OD ∼= TD → TX ⊗ OD → N = OX (D) ⊗ OD → 0where we used that TD = ωD∨ ∼= OD since D is an elliptic curve. Taking coho-mology gives (5) H1(O1X (nD) ⊗ OD) → H (OX (nD) ⊗ TX ⊗ OD) → H1(OX (nD) ⊗ N) D2 > 0, so we get H1(OX (nD) ⊗ OD) = 0 = H1(OX (nD) ⊗ N) as desired, since OX (nD) ⊗ OD and OX (nD) ⊗ N are sheaves of large positive degree on D for n >> 0. � We now return to the proof of the proposition for p = char(k) > 0. Let A = W (k) be the Witt vectors of k. A is a complete DVR of characteristic 0, with maximal ideal m and residue field A/m = k. The idea is to lift X to characteristic 0 and use the result already proved. Note that H2(X, TX ) = 0: in addition, X is projective and H2(X, OX ) = 0, so by SGA1, Theorem III, 7.3, there is a smooth projective morphism f : U → V = Spec (A) which closed fiber isomorphic to X.Let X� be the general fiber. Then X� is a nonsingular projective surface over the fraction field K� of A (which is unfortunately not algebraically closed). The fibers of f are 2-dimensional, so Rif∗OU = 0 for i ≥ 3. The base change theorem gives (6) (R2f∗OU ) ⊗A A/m → H2(f−1(m), OU /mOU ) = H2(X, OX ) = 0 is an isomorphism. By Nakayama’s lemma, we get that R2f∗OU = 0, and sim-ilarly for R1 . Thus, H1(X�, OX� ) = H2(X�, OX� ) = 0. See Mumford’s Abelian Varieties or Chapters on Algebraic Surfaces for more details. Now, let K�� be an algebraic closure of K� and Ki� the family of finite extensions of K� inside K��. Let X�� = X� ×K� k��, Xi = X� ×K� Ki�. Let A�� be the integralclosure of A inside K��, and A i = A��∩Ki��. Let m��be a maximal ideal of A�� lying over m and B�� = A��m its localization. Similarly, let mi = m�� ∩ Ai, Bi = (Ai)��miand set n�� = m��B��, ni = miBi. Since K = A/m is algebraically closed, we see, B��/n�� = Bi/ni = K. Now let Vi = Spec Bi, Ui = U ⊗A Bi, fi = f ⊗A Bi : Ui → Vi. Xi Ui U X (7) fif Spec K i�Vi Spec A Spec K Since Bi/ni = k, the closed fiber of fi is canonically isomorphic to X. The generic fiber of fi is isomorphic to Xi and since Ki ⊃ K is finite, Bi is a DVR and Viis an inductive system. By EGA and general nonsense, lim Pic Xi → Pic { }X�� is an isomorphism. −→ Lemma 3. There is a group isomorphism b : Pic Xi Pic X defined by thefollowing: for L an →invertible OXi -module, ∃ an invertible OU∼i -module Li s.t. Li|Xi = L, and we set b([L]) = [Li|X ].Proof. Omitted. � Proof of theorem: So we get a canonical isomorphism between Pic X and Pic X�� which takes ωX to ωX�� . Since Pic X = ZωX , we see that Pic X�� = ZωX�� and 1 ������ ��3 ALGEBRAIC SURFACES, LECTURE 10 �� �� ���� �� ��ω−is ample, giving us (1) for X��. Also, ωX��usingX��ωX��= ωX�ωX�= ωX ωX· · · flatness and the definition of ( ) by χ(), so ωX�� ωX�� ≤ 5 and b2(X��) ≥ 5 by· · Noether’s formula. Since q(X�) = 0, q(X��) = 0. But X�� is over an …