Automorphisms of smooth canonically polarized surfaces in positive characteristic

Advances in Mathematics 310 (2017) 235–289

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Automorphisms of smooth canonically polarized surfaces in positive characteristic ✩ Nikolaos Tziolas 1,2 Department of Mathematics, University of Cyprus, P.O. Box 20537, Nicosia, 1678, Cyprus

a r t i c l e

i n f o

Article history: Received 31 August 2015 Received in revised form 23 January 2017 Accepted 1 February 2017 Communicated by Ravi Vakil This paper is dedicated to my wife Afroditi and my children Victoria and Marko MSC: primary 14J50, 14DJ29, 14J10 secondary 14D23, 14D22

a b s t r a c t This paper investigates the geometry of a smooth canonically polarized surface X defined over an algebraically closed field of characteristic p > 0 in the case when the automorphism scheme of X is not smooth. In particular, it is shown that a 2 smooth canonically polarized surface X with 1 ≤ KX ≤ 2 and non-smooth automorphism scheme tends to be uniruled and simply connected and is the purely inseparable quotient of a ruled or rational surface by a rational vector field. Moreover, restrictions on certain numerical invariants of X are obtained in order for Aut(X) to be smooth. © 2017 Elsevier Inc. All rights reserved.

Keywords: Canonically polarized surfaces Automorphisms Moduli Vector fields Positive characteristic

✩ Part of this paper was written during the author’s stay at the Max Planck Institute for Mathematics in Bonn, from June to August 2016. E-mail addresses: [email protected], [email protected]. 1 Current address: Department of Mathematics, Princeton University, Fine Hall, Washington Road, Princeton, NJ 08544, United States. 2 The author is supported by a Marie Curie International Outgoing Fellowship, grant No. PIOF-GA2013-624345.

http://dx.doi.org/10.1016/j.aim.2017.02.002 0001-8708/© 2017 Elsevier Inc. All rights reserved.

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1. Introduction The objective of this paper is to investigate the geometry of a smooth canonically polarized surface X defined over an algebraically closed field k of characteristic p > 0 such that the automorphism scheme Aut(X) of X is not smooth or equivalently that X has nontrivial global vector fields. This is a situation that appears exclusively in positive characteristic simply because every group scheme in characteristic zero is smooth. The importance of this investigation is twofold. Firstly, the automorphism scheme of a variety is a fundamental invariant of it and is a classical object of study. Secondly, the structure of the automorphism scheme of a canonically polarized surface X is closely related to the moduli of canonically polarized surfaces. In particular, the smoothness of the automorphism scheme is an essential condition in order for the moduli stack of canonically polarized surfaces to be Deligne–Mumford [7], an important property of the stack for the reasons that will be explained in the following paragraph. The classification up to isomorphism of smooth projective algebraic varieties defined over an algebraically closed field k is one of the fundamental problems of algebraic geometry. In order to deal with this problem, the class of varieties is divided into smaller classes where a reasonable answer is expected to exist, a moduli functor is defined and then it is asked whether it is representable. Unfortunately in general the answer is no because the objects that one wants to parametrize have nontrivial automorphisms. There are two alternatives in order to deal with this complication. Coarse moduli spaces and stacks. Coarse moduli spaces are geometric objects whose closed points are in one-to-one correspondence with the parametrized objects but usually they do not support a universal family. Stacks on the other hand are categorical objects which carry information about families. Loosely speaking, to say that a stack is Deligne–Mumford means that there is a family X → S such that for any variety X in the moduli problem, there exist finitely many s ∈ S such that Xs ∼ = X, up to étale base change any other family is obtained from ˆS,s pro-represents it by base change and that for any closed point s ∈ S, the completion O the local deformation functor Def (Xs ). In some sense this family provides a connection between the local moduli functor (which behaves well) and the global one. Early on in the theory of moduli of surfaces of general type in characteristic zero, it was realized that the correct objects to parametrize are not the surfaces of general type themselves but their canonical models. It is well known that in characteristic zero the moduli functor of canonically polarized surfaces with fixed Hilbert polynomial has a separated coarse moduli space which is of finite type over the base field k. Moreover, the moduli stack of stable surfaces is a separated, Deligne–Mumford stack of finite type [18,15]. In positive characteristic the moduli functor of canonically polarized surfaces with a fixed Hilbert polynomial still has a separated coarse moduli space of finite type over the base field [14,15]. However, the moduli stack of smooth canonically polarized surfaces is not Deligne–Mumford. The reason for this failure in positive characteristic is the existence of smooth canonically polarized surfaces with non-smooth automorphism scheme,

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or equivalently of canonically polarized surfaces with nonzero global vector fields. Examples of such surfaces were found in [19,21,33]. From the previous discussion it is clear that the non-smoothness of the automorphism scheme is an essential obstruction for the moduli stack of canonically polarized surfaces to be Deligne–Mumford. As explained earlier, the property of being Deligne–Mumford is very important because it implies the existence of a family which is universal locally in the étale topology. The main objective of this paper is to study the geometry of smooth canonically polarized surfaces with non-smooth automorphism scheme and moreover to find numerical relations (preferably deformation invariant) between the characteristic of the base field and certain numerical invariants of the surface in order for the automorphism scheme to be smooth. This investigation will provide important information on how the moduli functor can be modified in order to obtain proper Deligne–Mumford stacks and how to deal with surfaces with non-smooth automorphism schemes. It will also be interesting to know if surfaces with a non-smooth automorphism scheme form a new component of the moduli space of canonically polarized surfaces. The main result of this paper is the following. Theorem 1.1. Let X be a smooth canonically polarized surface defined over an alge2 ≤ 2. Suppose that Aut(X) braically closed field of characteristic p > 0 such that 1 ≤ KX is not smooth or equivalently that X possesses a nontrivial global vector field. Then: 2 = 2 and p = 3, 5, then X is uniruled. If in addition χ(OX ) ≥ 2, then X is (1) If KX unirational and π1et (X) = {1}. 2 (2) If KX = 1 and p = 7, then X is unirational and π1et (X) = {1}, except possibly if p ∈ {3, 5} and X is a simply connected supersingular Godeaux surface. Moreover, if in addition p = 3, 5, then pg (X) ≤ 1.

In all cases, X is the quotient of a ruled surface by a rational vector field. Corollary 1.2. Let X be a smooth canonically polarized surface defined over an alge2 braically closed field of characteristic p > 0. Suppose that either KX = 2, p ∈ / {3, 5} and 2 X is not uniruled, or KX = 1 and in addition one of the following happens: (1) (2) (3) (4)

pg (X) = 2 and p = 3, 5. χ(OX ) = 3 and p = 3, 5. π1et (X) = {1} and p = 7. X is not unirational and p = 7.

Then Aut(X) is smooth. The conditions of Corollary 1.2 on the Euler characteristic and the étale fundamental group are deformation invariant and therefore they are good conditions for the moduli problem. Hence we get the following.

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Corollary 1.3. Let M1,ns and M1,3 be the moduli stacks of canonically polarized surfaces 2 2 X with KX = 1 and π1et (X) = {1}, or KX = 1 and χ(OX ) = 3, p = 3, 5, respectively. Then M1,3 and M1,ns are Deligne–Mumford stacks. Theorem 1.1 applies in particular to the case of Godeaux surfaces. Considering their significance I find it appropriate to write a statement for this case. Corollary 1.4. Let X be a canonically polarized Godeaux surface defined over an algebraically closed field k of characteristic p > 0, p = 7. Suppose that Aut(X) is not smooth. Then X is unirational and simply connected, except possibly if p ∈ {3, 5} and X is a simply connected supersingular Godeaux surface. Moreover, ⎧ ⎪ ⎪ ⎨Speck τ Pic (X) = Speck or Z/5Z ⎪ ⎪ ⎩(⊕m Z/pni Z) ⊕ N i=1

if p > 7 or

α5

if p = 5 if p = 2, 3,

where Picτ (X) is the torsion subgroup scheme of Pic(X), N is a finite commutative group scheme which is either trivial or is obtained by successive extensions by αp . This corollary follows directly from Theorem 1.1, the discussion about the possible structure of Picτ (X) in Section 2, basic properties about the structure of finite commutative groups schemes and the classification of Godeaux surfaces in characteristic 5 [21,24]. From the characteristics zero and five cases that have been extensively studied [29, 21,24], one might expect that if Aut(X) is not smooth then if p = 2, Picτ (X) is either Z/2Z, Z/4Z or α2 , and if p = 3 it is either Z/3Z or α3 . However, there is no classification in the characteristics 2 and 3 cases yet and since many pathologies appear in these characteristics, one must be careful. Next I would like to discuss the significance and how restrictive or effective the conditions of Theorem 1.1 are. According to Theorem 3.1, non-smoothness of the automorphism scheme happens for small values of K 2 compared to the characteristic of the base field. However, Theorem 3.1 does not give any effective lower bound for the characteristic, relative to K 2 , after which the automorphism scheme will become smooth. From this point of view, the automorphism scheme of surfaces with K 2 = 1 or 2 should not be very complicated. If it is not smooth then these surfaces should be rather pathological. The case when it is expected to be the most complicated and where most pathologies should appear must be the case when p = 2. All these are exhibited by the results of Theorem 1.1. In particular any surface with K 2 ≤ 2 and non-reduced automorphism scheme is uniruled, something that is not possible in characteristic zero. The exclusion in some cases of the characteristics p = 3, 5, 7 in Theorem 1.1 is not because of any fundamental reason. As it can be seen from the proof of the theorem, it

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is due to certain difficulties with numerical calculations and the existence of fibrations with singular general fiber of arithmetic genus 1 or 2 in characteristics 3 and 5. However, these problems are more evident in characteristic 2, a case which is fully studied in Section 7.1. I believe that the method that was used in the characteristic 2 case can in principle be used in order to resolve the problematic characteristic cases. However, at the moment, I am having some technical difficulties and the paper is already too long. One cannot expect that it will be possible to get results similar to those of Theorem 1.1 for any surface with non-reduced automorphism scheme. This is because there are examples in characteristic 2 of surfaces of general type with arbitrary large K 2 and non-reduced automorphism scheme [21,24], many of them not uniruled and not simply connected [33]. However, due to the lack of examples, I do not know if surfaces with K 2 ≤ 2 is the maximal class of surfaces that Theorem 1.1 holds. 2 2 Suppose that X is a canonically polarized surface X with KX ≤ 2. If KX = 2, then by [9, Corollary 1.8], it follows that 0 ≤ χ(OX ) ≤ 4. Moreover, if χ(OX ) ≥ 3, then by Lemma 6.3, π1et (X) = {1} and hence the condition that X is simply connected in Theorem 1.1(2) has value only for χ(OX ) = 2. 2 Suppose that KX = 1. Then pg (X) ≤ 2, 1 ≤ χ(OX ) ≤ 3 and |π1et (X)| ≤ 6 [24, Proposition 1]. In characteristic zero there are examples for all cases and similar examples 2 are expected to exist in any characteristic. Hence many surfaces with KX = 1 are excluded in the theorem which by Corollary 1.2 have smooth automorphism scheme. For example, Godeaux surfaces that are quotients of a smooth quintic in P3k by a free action of Z/5Z, since they are not simply connected. I believe that the biggest disadvantage of this paper is the lack of examples of surfaces with a non-reduced automorphism scheme and low K 2 which will show how effective the results of Theorem 1.1 are. N.I. Shepherd-Barron [33] has constructed an example of a smooth canonically polarized surface X in characteristic 2 with non-reduced automor2 phism scheme and KX = 8. This is the example with the lowest K 2 that I know. Simply connected Godeaux surfaces exist in all characteristics [22,20] but it is not known if there exists any with non-reduced automorphism scheme. The examples constructed in [22] lift to characteristic zero and hence from Lemma 4.2 they have smooth automorphism scheme. This discussion motivates the following problem. Problem 1.5. Are there any simply connected Godeaux surfaces with non-smooth automorphism scheme? Equivalently, are there any with nontrivial global vector fields? If there are, can the torsion part of their Picard scheme be reduced? The main steps in the proof of Theorem 1.1 are the following. In Section 4 it is shown that Aut(X) is not smooth if and only if X admits a nonzero global vector field D such that either Dp = 0 or Dp = D or equivalently that X admits a nontrivial αp or μp action. Corollary 4.2 shows that if X lifts to characteristic zero then Aut(X) is smooth.

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In Section 5 the general strategy for the proof of Theorem 1.1 is explained. Quite generally it is the following. Suppose that Aut(X) is not smooth. Then X has a nontrivial global vector field D such that either Dp = D or Dp = 0. This vector field induces a nontrivial μo or αp action on X. Let π : X → Y be the quotient and g : Y  → Y its minimal resolution. The results are obtained by considering cases with respect to the Kodaira dimension κ(Y  ) of Y  and comparing invariants and the geometry of X and Y  . The basic idea of this method was first used by Rudakov and Shafarevich [30] in order to show that a smooth K3 surface has no global vector fields. In Section 6, Theorem 1.1 is proved in the case p ≥ 3. The case when p = 2 has certain complications which makes it necessary to treat it separately. These complications are explained during the proofs of Theorems 6.1, 6.4 and 6.5. In Section 7 the characteristic 2 case is treated. In addition to the case when p ≥ 3, the geometry of X is investigated in the case when Aut(X) contains μ2 or α2 as subgroup schemes. It is shown that the property that μ2 is a subgroup scheme of Aut(X) is much more restrictive than α2 is a subgroup scheme of Aut(X). Theorem 1.1 is the combination of the results of Theorems 6.1, 6.4, 6.5, and 7.1. Section 8 contains examples of smooth canonically polarized surfaces with smooth automorphism scheme and two examples of canonically polarized surfaces with canonical singularities and non-smooth automorphism scheme. These examples are significant for the following reasons. Singular surfaces should be studied because they naturally appear in the compactification of the moduli problem. The first example is a surface in characteristic 2 with K 2 = 1. This example shows that if there are no restrictions on the singularities then K 2 can be as low as possible and the automorphism scheme not smooth. The second example is a surface in arbitrary characteristic which is smoothable to a smooth canonically polarized surface with smooth automorphism scheme. This shows that the property “smooth automorphism scheme” is not deformation invariant and does not produce a proper moduli stack. Moreover, the smoothing components of the coarse moduli space contain surfaces whose automorphism scheme is not reduced. Therefore such surfaces are an important part of the moduli problem and cannot be ignored. 2. Preliminaries 2.1. Notation–terminology Let X be a scheme of finite type over a field k of characteristic p > 0. X is called a smooth canonically polarized surface if and only if X is a smooth surface and ωX is ample. Derk (X) denotes the space of global k-derivations of X (or equivalently of global vector fields). It is canonically identified with HomX (ΩX , OX ). A nonzero global vector field D on X is called of additive or multiplicative type if and only if Dp = 0 or Dp = D, respectively. The fixed locus of D is the closed subscheme

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of X defined by the ideal sheaf (D(OX )). The divisorial part of the fixed locus of D is called the divisorial part of D. A point P ∈ X is called an isolated singularity of D if and only if the ideal of OX,P generated by D(OX,P ) has an associated prime of height ≥ 2. A prime divisor Z of X is called an integral divisor of D if and only if locally there is a derivation D of X such that D = f D , f ∈ K(X), D (IZ ) ⊂ IZ and D (OX ) ⊂ IZ [30]. Let F be a coherent sheaf on X. By F [n] we denote the double dual (F ⊗n )∗∗ . i (X, Ql ) are independent of l, For any prime number l = p, the cohomology groups Het they are finite dimensional of Ql and are called the l-adic cohomology groups of X. The i i-Betti number bi (X) of X is defined to be the dimension of Het (X, Ql ). It is well known that bi (X) = 0 for any i > 2n, where n = dim X [26, Chapter VI, Theorem 1.1]. The étale Euler characteristic of X is defined by χet (X) =

 i

i (−1)i dimQl Het (X, Ql ) =



(−1)i bi (X).

i

If X is a smooth surface then c2 (X) = χet (X) [26, Chapter V, Theorem 3.12]. Both the Betti numbers and the étale Euler characteristic are invariant under étale equivalence. In particular, if f : X → Y is a purely inseparable morphism of varieties, then f induces an equivalence of the étale sites of X and Y and hence bi (X) = bi (Y ) and χet (X) = χet (Y ) [26]. X is called simply connected if π1et (X) = {1}, where π1et (X) is the étale fundamental group of X. The symbol ≡ denotes numerical equivalence of divisors. The radical ideal of an ideal I of a commutative ring A is denoted by r(I). Let P ∈ X be a normal surface singularity and f : Y → X its minimal resolution. If P ∈ X is canonical, then KY = f ∗ KX . By [17, Theorem 4.22] canonical surface singularities are classified according to the Dynkin diagrams of their minimal resolution and they are called accordingly of type An , Dn , E6 , E7 and E8 . In characteristic zero these are exactly the DuVal singularities and their Dynkin diagrams correspond to explicit equations. However, in positive characteristic I am not aware of a classification with respect to local equations. In this paper, canonical surface singularities will be distinguished according to their Dynkin diagrams. 2 2.2. Surfaces with KX =1

Proposition 2.1 (Proposition 1 [24]). Let X be a minimal surface of general type with 2 KX = 1. Then 1 ≤ χ(OX ) ≤ 3, pg (X) ≤ 2, h1 (OX ) ≤ 1, b1 (X) = 0 and |π1et (X)| ≤ 6. Definition 2.2 ([29]). A numerical Godeaux surface is a minimal surface X of general 2 type such that KX = 1 and χ(OX ) = 1. Numerical Godeaux surfaces are divided in two disjoint classes: classical and nonclassical. A Godeaux surface X is called classical if and only if the torsion part Picτ (X)

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of Pic(X) is reduced, and non-classical if and only if Picτ (X) is not reduced. Non-classical Godeaux’s are further divided into two disjoint classes: singular if Picτ (X) contains μp , and supersingular if Picτ (X) contains αp (since Picτ (X) is a finite commutative group scheme with one dimensional tangent space, these cases are mutually exclusive). Since all group schemes are reduced in characteristic zero, non-classical Godeaux surfaces appear only in positive characteristic p > 0. It is well known that in characteristic zero Picτ (X) can be {1}, Z/2Z, Z/3Z, Z/4Z and Z/5Z [29]. In positive characteristic however, Picτ (X) may have a non-reduced part and one gets either singular or supersingular Godeaux surfaces too. However, non-classical Godeaux surfaces exist only for p ≤ 5 [24]. There is a correspondence between the structure of Picτ (X) and the existence of torsors over X. The correspondence is given by the isomorphism Hom(G, Picτ (X)) ∼ = 1 ∗ Hf l (X, G ) [28, Proposition 6.2.1], where G is any commutative subgroup scheme of Picτ (X), G∗ is its Cartier dual, and Hf1l (X, G∗ ) is the first flat cohomology group of G∗ . According to this, a Godeaux surface is singular if there is a Z/pZ torsor and supersingular if there is an αp torsor over X. Equivalently, if the induced map of the Frobenius on H 1 (OX ) is either bijective or zero. 3. Non-smoothness happens for small p The purpose of this section is to show that non-reducedness of the automorphism scheme of a canonically polarized surface with mild singularities defined over an algebraically closed field of characteristic p > 0 is a property that happens for relatively small values of p compared to certain numerical invariants of X. Moreover, the length of its automorphism scheme is bounded by a number that depends only on the Hilbert polynomial and not on the characteristic of the base field. Theorem 3.1. Let f (x) ∈ Q[x] be a numerical polynomial. Then there are positive integers m and M depending only on f (x) such that for any canonically polarized surface X with canonical singularities defined over an algebraically closed field k of characteristic p > 0 and with Hilbert polynomial f (x), (1) length(Aut(X)) ≤ M . (2) Aut(X) is smooth for all p > m. Proof. Let Ω be the set of all Gorenstein canonically polarized surfaces with fixed Hilbert polynomial f (x) defined over any field of any characteristic. Then this set is bounded [14]. This means that there is a flat morphism f : X → S, where S is of finite type over Z whose geometric fibers are Gorenstein canonically polarized surfaces and such that for any Gorenstein canonically polarized surface X with Hilbert polynomial f (x) defined over an algebraically closed field k, there is a morphism Speck → S such that X ∼ = Speck×S X . Let

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Φ : Aut(X /S) → S be the induced morphism on the relative automorphism scheme. I will show that this map is finite. For this it suffices to show that Φ is proper with finite fibers. It is well known that for any canonically polarized surface X with canonical singularities, Aut(X) is a finite group scheme. Therefore Φ has finite fibers. Properness of Φ follows from the valuative criterion of properness. This is equivalent to the following property. Let R be a discrete valuation ring with function field K and residue field k. Let X → SpecR be a projective flat morphism such that ωX/R is ample. Let XK and Xk be the generic and special fibers. Then any automorphism of XK lifts to an automorphism of X over SpecR. The proof of this statement is identical with the one for characteristic zero [16, Proposition 6] and I omit its proof. Since Φ is a finite morphism, the lengths of its fibers are bounded by some number M that depends only on f (x). Hence if X is any Gorenstein canonically polarized surface defined over an algebraically closed field k of characteristic p > 0, length(Aut(X)) ≤ M . Therefore if p > M , Aut(X) is smooth over k. This follows from the fact that Aut(X) is a finite group scheme defined over a field of characteristic p > 0 and any such group scheme is smooth if p is bigger than its length [27, p. 130]. This concludes the proof of the theorem. 2 The previous theorem shows that the structure of the automorphism scheme is expected to be more complicated in small characteristics. It also suggests that if 2 1 ≤ KX ≤ 2, then the case p = 2 may be the case where most pathologies appear. Theorem 3.1 motivates the following problem. Problem 3.2. Find explicit relations between the characteristic p of the base field k and the coefficients of the Hilbert polynomial of a canonically polarized surface X which imply the smoothness of Aut(X). 4. Some preparatory results for the proof of Theorem 1.1 In this section, unless otherwise specified, X denotes a smooth projective surface defined over an algebraically closed field k of characteristic p > 0. Proposition 4.1. Suppose that X is canonically polarized. Then Aut(X) is not smooth over k if and only if X admits a nontrivial global vector field of either additive or multiplicative type or equivalently X admits a nontrivial αp or μp action. Proof. It is well known that for any canonically polarized Q-Gorenstein surface, Aut(X) is a zero dimensional scheme. Therefore Aut(X) is not smooth if and only if its tangent space at the identity is not trivial. The tangent space of Aut(X) at the identity is HomX (ΩX , OX ) which is the space of global derivations of X. Therefore Aut(X) is not

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smooth if and only if X has a nontrivial global vector field. Then by [30, Lemma 1], X has a nontrivial global vector field D such that either Dp = 0 or Dp = D which by [35, Proposition 3.1] induces a nontrivial αp or μp action, respectively. 2 Lemma 4.2. Suppose that X is canonically polarized and moreover that it lifts to W2 (k), the ring of 2-Witt vectors. Then Aut(X) is smooth. Proof. Since X lifts to characteristic zero or W2 (k), the Kodaira–Nakano vanishing theorem holds for X [10, Corollary 11.3] and hence −1 HomX (ΩX , OX ) = H 0 (TX ) = H 0 (ΩX ⊗ ωX ) = 0,

since ωX is ample. Hence X has no nontrivial global vector fields and therefore by Proposition 4.1, Aut(X) is smooth. 2 This result is a kind of accident since the smoothness of the automorphism scheme is not the consequence of any vanishing theorems but rather the fact that any group scheme is smooth in characteristic zero. This result does not hold for singular surfaces as shown by the Examples 8.3, 8.4 in Section 8. Suppose now that X has a nontrivial global vector field D of either additive or multiplicative type. Let π : X → Y be the quotient of X by the corresponding αp or μp action. It is well known that Y is normal, that the singular points of Y are under the isolated part of the fixed locus of D [1] and their local Picard groups are p-torsion [35, Proposition 4.6]. The next two results describe the size of the singular locus of Y . For more information regarding the structure of the quotient map π and the singularities of Y the reader may refer to [35] where a survey of many known results is included. Proposition 4.3 (Proposition 0.1.13 [6]). Suppose that X has a nontrivial global vector field D of either additive or multiplicative type. Let IZ be the ideal sheaf of the isolated part Z of the fixed locus of D and let Δ be its divisorial part. Then length(OZ ) = KX · Δ + Δ2 + c2 (X). The next proposition shows that KX · Δ decreases after resolving the singularities of D and will be needed for the proof of Theorem 7.1. Proposition 4.4. Suppose that X has a nonzero global vector field D of either additive or multiplicative type and that p = 2. Let π : X → Y be the quotient of X by D. By [13, Proposition 2.6] there exists a commutative diagram X

f

X

π

Y

π g

Y

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such that X  , Y  are smooth and f , g are birational, D lifts to a vector field D on X  and Y  is the quotient of X  by D . Let Δ, Δ be the divisorial parts of D and D , respectively. Then KX  · Δ ≤ KX · Δ,

(4.4.1)

KX  · Δ = 4 (χ(OX  ) − 2χ(OY  )) .

(4.4.2)

Proof. Since f is birational and both X and X  are smooth, f is a composition of blow-ups of points. Moreover, since D lifts to X  , f blows up isolated singular points of D. Hence, in order to prove the first part of the proposition it suffices to assume that f is the blow-up of an isolated singular point of D. Suppose that this is the case and that E is the f -exceptional curve. I will show that Δ = f ∗ Δ + aE, a ≥ 0. Then KX  · Δ = KX · Δ − a ≤ KX · Δ. The proof of the previous claim will be by a direct local calculation of D . In suitable local coordinates at an isolated singular point of D, OX = k[x, y] and D is given by D = h (f ∂/∂x + g∂/∂y), where f , g have no common factor. Locally at the standard open affine covers, the blow-up is given by φ : k[x, y] → k[s, t], φ(x) = s, φ(y) = st. Then it is easy to see that 

1 ∂ ∂ + (tf (s, st) + g(s, st)) D = h(s, st) f (s, st) ∂s s ∂t 

 .

It is now clear that Δ = f ∗ Δ + aE, a ≥ 0. It remains to prove the second part of the proposition. Since Y  is the quotient of X  by the lifting D of D on X  , π  is a finite purely inseparable map of degree 2. Then by the adjunction formula for purely inseparable maps [30, Corollary 1], KX  = (π  )∗ KY  + Δ .

(4.4.3)

Moreover, since Y  is smooth, it is well known that π  : X  → Y  is an αM -torsor [9, Proposition 1.11]. In particular π∗ OX  fits in an exact sequence 0 → OY  → π∗ OX  → M −1 → 0,

(4.4.4)

where M = OY  (C  ) is an invertible sheaf on Y  . If the sequence splits then D2 = D and if it doesn’t split then D2 = 0. Moreover, KX  = (π  )∗ (KY  + C  ). Therefore from (4.4.3) we get that and Δ = (π  )∗ C  . Then from (4.4.4) we get that χ(M −1 ) = χ(π∗ OX  ) − χ(OY  ) = χ(OX  ) − χ(OY  ).

(4.4.5)

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From Riemann–Roch it follows that 1 χ(M −1 ) = χ(OY  ) + ((C  )2 + KY  · C  ) = 2 1 1 χ(OY  ) + C  · (KY  + C  ) = χ(OY  ) + KX  · Δ 2 4

(4.4.6)

Finally from (4.4.5) and (4.4.6) it follows that KX  · Δ = 4 (χ(OX  ) − 2χ(OY  )) .

2

5. Strategy of the proof of Theorem 1.1 Let X be a smooth canonically polarized surface defined over an algebraically closed field of characteristic p > 0 such that Aut(X) is not smooth. Then by Proposition 4.1 X has a nontrivial global vector field D of either additive or multiplicative type which induces a nontrivial αp or μp action. Let π : X → Y be the quotient. Then Y is normal, KY is Q-Cartier and the local class groups of its singular points are p-torsion [35, Proposition 3.5]. Moreover, there is a commutative diagram X

f

π

Z

φ

Y

(5.0.1)

X π

g

Y

such that g : Y  → Y is the minimal resolution of Y , φ : Y  → Z its minimal model, f is birational, D lifts to a vector field D on X  and Y  is the quotient of X  by the corresponding αp or μp action [35, Proposition 3.5]. Let Ei , i = 1, . . . , n, be the f -exceptional curves and Fi = π  (Ei ). Then Fi , i = 1, . . . , n, are exactly the g-exceptional curves. Lemma 5.1. X  has rational singularities and the curves Fi form a configuration of rational (but possibly singular) curves without any loops in Y  . Moreover, c2 (Y  ) = χet (Y ) + n = c2 (X) + n. Proof. Let h : W → X  be the minimal resolution of X  and let σ : W → X be the composition of h and f . Then σ is a composition of smooth blow-ups. Let Ej , j = 1, . . . , m, be the h-exceptional curves. Then the curves Ei and Ej are exactly the σ-exceptional curves, where Ei , 1 ≤ i ≤ n, is the birational transform of Ei in W . Therefore the curves Ei and Ej form a tree of smooth rational curves and moreover the cycle E = i Ei + j Ej is a rational cycle. Since h contracts a subcycle of E, it contracts a rational cycle and therefore X  has rational singularities [2]. Moreover, the curves Ei form a configuration

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of rational curves without any loops. Hence so do the Fi . Since σ is a composition of n + m blow-ups, it follows that c2 (W ) = n + m + c2 (X). Also χet (W ) = χet (X  ) +

 (χet (Mk ) − 1), k

where Mk are the connected components of the reduced exceptional divisor E  = m   j=1 Ej of h. Since Ek are trees of smooth rational curves, it follows that χet (W ) =  χet (X ) + m. Therefore from the above relations it follows that c2 (Y  ) = χet (Y  ) = χet (X  ) = c2 (X) + n = χet (Y ) + n.

2

Let also Bj , j = 1, . . . , m, be the φ-exceptional curves. Taking into consideration that KY has index either 1 or p and g : Y  → Y is the minimal resolution of Y , we get the following adjunction formulas 1 KY  = g ∗ KY − F, p

(5.1.1)

KY  = φ∗ KZ + B, n m where F = i=1 ai Fi , ai ∈ Z≥0 , and B = j=1 bj Bj , bj > 0, j = 1, . . . , m. Moreover, since both Y  and Z are smooth, φ is the composition of m blow-ups. Let Δ be the divisorial part of D. Then by the adjunction formula for purely inseparable maps [30, Corollary 1], [23, Proof of Proposition 4.2] KX = π ∗ KY + (p − 1)Δ.

(5.1.2)

Note that it is possible that Δ = 0. For example, if p = 2 then the homogeneous vector field D = (y + z)∂/∂x + (x + z)∂/∂y + (x + y)∂/∂z of k[x, y, z] gives a vector field on P2 with no divisorial part. As a general strategy, cases with respect to the Kodaira dimension κ(Y  ) of Y  will be considered. Then the classification of surfaces in positive characteristic [5,4,32] will be heavily used in order to get information about Y  and then for X by means of the diagram (5.0.1). Moreover, since π is a purely inseparable map, it induces an equivalence between the étale sites of X and Y . Therefore X and Y have the same algebraic fundamental group, l-adic Betti numbers and étale Euler characteristic. Then by using the fact that g and φ are birational it will be possible to calculate the algebraic fundamental group, l-adic Betti numbers and étale Euler characteristic of X from those of Z. The cases p = 2 and p = 2 will be treated separately. The case p = 2 has certain peculiarities and it requires special attention. In some sense this is expected since 2 is the smallest nonzero characteristic and many special situations appear in this case (as for example the existence of quasi-elliptic fibrations). The difficulties of this case will become evident during the proof of the case p ≥ 3 in Section 6. The case when p ≥ 3 will be treated in section 6 and the case p = 2 will be treated in Section 7.1.

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6. Vector fields on surfaces in characteristic p ≥ 3 The purpose of this section is to prove Theorem 1.1 and Corollary 1.2 in characteristic p ≥ 3. Their statements are a direct consequence of Theorems 6.1, 6.4 and 6.5 of this 2 2 section. I will only do the case where KX = 1 in detail. The proof of the case KX =2 2 is similar. I will sketch its main points, remark on any differences with the case KX =1 but I will leave the details to the reader. The method is exactly the same but certain calculations are lengthier and I see no reason to make an already long paper longer. For the remaining part of this section, fix notation as in Section 5. As was mentioned in Section 5, the divisorial part Δ of D may or may not be zero. These two cases behave quite differently and for this reason they will be considered separately. Case 1: Δ = 0. In this case the following holds. Theorem 6.1. Let X be a smooth canonically polarized surface defined over an alge2 braically closed field of characteristic p ≥ 3. Suppose that KX < p and X admits a nontrivial global vector field D such that Δ = 0. Then X is uniruled. Moreover, if 2 5c21 < c2 , then X is unirational and π1et (X) = {1} (in particular, this happens if KX =1 2 or KX = 2 and χ(OX ) ≥ 2). Proof. From the equation (5.1.2) it follows that 2 KX = KX · π ∗ KY + (p − 1)Δ · KX .

(6.1.1)

Case 1. Suppose that KX ·π ∗ KY < 0. Then κ(Z) = −∞ because if not then |nKZ | = ∅ for n >> 0 and hence |nKY  | = ∅, and also |nKY | = ∅, for n >> 0. But then, since KX is ample, KX · π ∗ KY ≥ 0, which is impossible. Therefore Z is uniruled and hence Y is also uniruled. Let FY : Y(p) → Y be the k-linear Frobenius. Then there is a factorization X δ

Y(p)

π FY

Y

where δ is a finite purely inseparable map of degree p. Since Y uniruled, it follows that Y(p) is uniruled and hence X is purely inseparably uniruled. Moreover, by [30, Section 1, p. 2] the map δ is the quotient of Y(p) by a rational vector field on Y(p) . Case 2. Suppose that KX · π ∗ KY > 0. Then since KX is ample and Δ an effective 2 divisor, it follows from the equation (6.1.1) that KX ≥ p. ∗ Case 3. Suppose that KX · π KY = 0. In this case the equation (6.1.1) only says that 2 KX ≥ p − 1. In particular for p = 2 it provides no information at all. Consider now cases with respect to the Kodaira dimension κ(Z) of Z.

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Case 3.1. Suppose that κ(Z) = −∞. Then Z is uniruled and exactly as in case 1, it follows that X is uniruled as well. 2 Case 3.2. Suppose that κ(Z) = 0. In this case I will show that if p ≥ 3, then KX ≥ p.   If Y = Z, i.e., Y is a minimal surface itself, then KY  ≡ 0 [5,4] and therefore KY ≡ 0. Hence KX ≡ (p − 1)Δ and consequently 2 KX = (p − 1)2 Δ2 ≥ (p − 1)2 ≥ p,

(6.1.2)

if p ≥ 3. Suppose now that Y  is not a minimal surface. Then φ is not trivial and is a composition of blow-ups. Moreover, the usual adjunction formula gives that ∗

KY = φ KZ + 

m 

bi Bi ,

(6.1.3)

i=1

where Bi , i = 1, . . . , m, are the φ-exceptional curves and bi > 0 for all i. Now since m κ(Z) = 0 it follows that KZ ≡ 0 [5,4]. Hence KY  ≡ i=1 bi Bi . Since φ is a compo m sition of blow-ups, it follows that B = i=1 bi Bi contains φ-exceptional curves with  self-intersection −1. However, since Y is the minimal resolution of Y , g does not con m tract any curve with self-intersection −1. Therefore g∗ ( i=1 bi Bi ) = 0. Then from the equations (5.1.1), (6.1.3) it follows that KY ≡ g∗ KY  ≡ g∗

m 

 bi Bi

i=1

Hence KY is numerically equivalent to an effective divisor. Hence KX · π ∗ KY > 0. This 2 together with the equation (6.1.1) shows that if κ(Z) = 0 and p ≥ 3, then KX ≥ p. Case 4. Suppose that κ(Z) = 1 or κ(Z) = 2. I will show that in both cases KY  is linearly equivalent to an effective divisor B with rational coefficients which has at least one irreducible component that is not contracted by g. Then KY is linearly equivalent to a nonzero effective divisor. Hence KX · π ∗ KY > 0 and therefore it follows from the 2 equation (6.1.1) that KX ≥ p. Suppose that κ(Z) = 2. Then nKZ is nef and big for n >> 0 [5,4]. Hence |nKZ | contains an element W ⊂ φ∗ E. This means that φ∗ W is not contained in the g-exceptional set and hence it has at least one irreducible component which is not contracted by g. By the adjunction formula for φ, KY  ≡ φ∗ KZ + B ≡

1 ∗ φ W + B, n

˜ where B ˜ = 1 φ∗ W +B is an effective where B is a φ-exceptional divisor. Hence KY  ≡ B, n divisor such that it has at least one irreducible component which is not contracted by g. Then from the equation (5.1.1) and considering that g∗ φ∗ W = 0, it follows that KY is linearly equivalent to an effective divisor with rational coefficients, as claimed.

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Suppose that κ(Z) = 1. Then Z admits an elliptic or quasi-elliptic fibration h : Z → C to a smooth curve C [5,4]. Moreover, for n >> 0, nKZ = h∗ W , where W is an effective divisor on C [5,4]. Moreover, W can be chosen so that h∗ W ⊂ φ∗ E. Now the argument used in the case when κ(Z) = 2 shows that KY  and hence KY are numerically equivalent to a nonzero effective divisor with rational coefficients. Therefore in this case 2 too KX · π ∗ KY > 0 and hence KX ≥ p. Hence it has been shown that if X has a nontrivial global vector field, then either 2 κ(Z) = −∞ and hence X is uniruled, or KX ≥ p. 2 Suppose that 5c1 < c2 . Then according to Lemma 6.3 below, b1 (X) = 0. Then b1 (X  ) = 0 and hence since π  is an étale equivalence, b1 (Z) = b1 (Y  ) = 0. There2 ≥ p or Z is rational and π1et (Z) = {1}. Since the algebraic fore in this case, either KX fundamental group is a birational invariant and also invariant under étale equivalence, it 2 follows from the diagram (5.0.1) that π1et (X) = {1}. Suppose in particular that KX =1 2 or KX = 2 and χ(OX ) ≥ 2. Then from Proposition 2.1 it follows that 1 ≤ χ(OX ) ≤ 3 in the first case, and from [8, Corollary 1.8] 2 ≤ χ(OX ) ≤ 4 in the second case. Then from Lemma 6.3 it follows that b1 (X) = 0 and hence the previous argument shows that X is unirational and simply connected. This concludes the proof of the theorem. 2 Remark 6.2. The proof of Theorem 6.1 shows some of the reasons why the case p = 2 has to be excluded. As will be seen in the remaining part of this section, the existence of quasi-elliptic fibrations in characteristic 2 is another reason. This is to be expected in some sense since p = 2 is the smallest possible characteristic where most pathologies appear. Lemma 6.3. Let X be a smooth surface of general type defined over an algebraically closed field k. Suppose that 5c21 < c2 . Then b1 (X) = 0 and |π1et (X)| ≤

6 36 2 = c − 5c2 . 2χ(OX ) − KX 2 1

Moreover, X does not have an étale cover of degree bigger than 36/(c2 − 5c21 ). The proof of the lemma is a straightforward generalization of [24, Proposition 1] and for this reason its proof is omitted. Case 2: Δ = 0. In this case the following holds. Theorem 6.4. Let X be a smooth canonically polarized surface defined over an algebraically closed field of characteristic p ≥ 3. Suppose that X admits a nontrivial global vector field D with only isolated singularities. Then: 2 (1) If KX = 2 and p = 3, 5, then X is uniruled. If in addition χ(OX ) ≥ 2, then X is unirational and simply connected.

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2 (2) Suppose that KX = 1 and p = 7. Then X is unirational and simply connected except possibly if p ∈ {3, 5} and X is a simply connected supersingular Godeaux surface. 2 2 Proof. I will only do the case KX = 1 in detail. The case KX = 2 is exactly similar and its details are left to the reader. For the remainder of the proof, fix notation as in Section 5. Since Δ = 0, it follows from the equation (5.1.2) that KX = π ∗ KY and therefore 2 KX = pKY2 . Consider now cases with respect to whether Y  is a minimal surface or not. Case 1. Suppose that Y  = Z, i.e., Y  is a minimal surface. 2 Case 1.1. Suppose that κ(Y  ) = 2. In this case I will show that KX ≥ p. From the equation (5.1.1) it follows that 2 KX = pKY2 = pKY2  + KY  · F ≥ p,

since KY2  > 0 and KY  · F ≥ 0. Case 1.2. Suppose that κ(Y  ) = 1. In this case I will show the following. 2 (1) If KX = 1 then p = 3 and X is a simply connected supersingular Godeaux surface. 2 (2) If KX = 2 then p = 3, 5.

Since κ(Y  ) = 1, Y  admits an elliptic or quasi-elliptic (this only for p = 2, 3) fibration ψ : Y  → B [5,4]. In particular KY2  = 0. Let Fi , i = 1, . . . , n, be the g-exceptional curves. Then from the equations (5.1.1) it follows that 1 ai Fi = g ∗ KY , p i=1 n

KY  +

(6.4.1)

where ai ∈ Z≥0 , for all i = 1, . . . , n. Also, since g : Y  → Y is the minimal resolution of Y , over any singular point of Y either ai > 0 for all i, or ai = 0 for all i. Suppose that ai = 0, for all i = 1, . . . , n. Then KY  = g ∗ KY and hence Y has canonical DuVal singularities. Therefore KY2 = KY2  = 0. But this is impossible since 2 KY2 = p1 KX > 0. Therefore not every singular point of Y is canonical DuVal. After a renumbering of the g-exceptional curves we may assume that Fi , 1 ≤ i ≤ m, correspond to non-DuVal singularities and Fi , m < i ≤ n, correspond to DuVal singularities of Y . Then ai > 0, for all i ≤ m and ai = 0, for all i > m. Intersecting the equation (6.4.1) with KY  and taking into consideration that KY2 = 1/p, it follows that m 

ai (Fi · KY  ) = 1.

(6.4.2)

i=1

Then since ai > 0 for all i ≤ m, it follows that, up to a renumbering of the exceptional curves of g, a1 = 1, KY  · F1 = 1 and KY  · Fi = 0, for all i = 2, . . . , n. From this it

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follows that Fi ∼ = P1 and Fi2 = −2, for all i ≥ 2. Also since pa (F1 ) = 1 + 12 (1 + F12 ) and considering that F12 < 0, it follows that either F1 is a singular rational curve of arithmetic genus 1 and F12 = −1, or F1 is a smooth rational curve and F12 = −3. Intersecting now the equation (6.4.1) with F1 it follows that 1 F2 1+ ai (Fi · F1 ) = − 1 . p i=2 p m

(6.4.3)

Now since F12 ∈ {−1, −3} and p ≥ 3, it follows that the only possibility is that p = 3, F12 = −3 and F1 · Fi = 0, for all i = 2, . . . , n. In particular F1 is a smooth rational curve as well. Hence the g-exceptional set consists of an isolated smooth rational curve with self-intersection −3 and chains of −2 curves. Hence Y has one singularity of type 1 3 (1, 1) and perhaps some isolated DuVal singularities. The DuVal singularities must be of type A2 because these are precisely the DuVal singularities whose local class groups are 3-torsion. As was mentioned earlier, Y  admits an elliptic or quasi-elliptic fibration ψ : Y  → B. 2 Since KX = 1, it follows from Lemma 6.3 that b1 (X) = 0 and therefore B = P1 . Then R1 ψ∗ OY  = OP1 (−d) ⊕ T,

(6.4.4)

where d ∈ Z and T is a torsion sheaf on P1 . By using the Grothendieck spectral sequence and Serre duality, we get that χ(OY  ) = 1 + h0 (ωY  ) − h0 (R1 ψ∗ OY  )

(6.4.5)

pg (Y  ) = h0 (ωY  ) = h2 (OY  ) = h1 (R1 ψ∗ OY  ). Since KX = π ∗ KY , it follows that pg (Y  ) ≤ pg (Y ) ≤ pg (X).

(6.4.6)

Moreover, from Proposition 2.1 it follows that pg (X) ≤ 2. Hence 0 ≤ pg (Y  ) ≤ 2. Case 1.2.1. Suppose that pg (Y  ) = 0. Then χ(OY  ) ≤ 1 and hence from the Noether formula it follows that c2 (Y  ) ≤ 12. But also c2 (Y  ) = χet (Y  ) = χet (Y ) + k = χet (X) + k = c2 (X) + k,

(6.4.7)

where k is the number of g-exceptional curves. From Proposition 2.1 it follows that 1 ≤ χ(OX ) ≤ 3 and therefore from the Noether formula it follows that c2 (X) ≥ 11. Then from the equation (6.4.7) and since c2 (Y  ) ≤ 12, it follows that c2 (X) = 11, k = 1 and c2 (Y  ) = 12. Hence Y has exactly one singular point, which is necessarily of type 1/3(1, 1). Let then F be the unique g-exceptional curve. Then F 2 = −3 and hence by adjunction we get that 1 KY  = g ∗ KY − F. 3

(6.4.8)

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Let E be the unique f -exceptional curve. Then π  E = F . Then again by adjunction, KX  = f ∗ KX + aE, a ∈ Z>0 . Then since Δ = 0, it follows that 1 KX  = f ∗ KX + aE = f ∗ π ∗ KY + aE = (π  )∗ KY  + aE + (π  )∗ F. 3

(6.4.9)

Now (π  )∗ F = kE, where k = 1 if E is an integral curve for D and k = 3 if E is not an integral curve. Suppose that k = 1. Then from the equation (6.4.9) and the adjunction formula for purely inseparable maps (5.1.2) it follows that a + 13 = (p − 1)λ = 2λ, where λ is a nonnegative integer. In particular a + 13 ∈ Z, which is impossible. Hence k = 3. Then since (π  )∗ F = 3E it follows that 9E 2 = 3F 2 = −9 and hence E 2 = −1. I will next show that in fact X  → X is just the blow-up of X at a single point. Let h : W → X  be the minimal resolution of X  and let σ = f ◦ h. Let Γi be the h-exceptional curves. Then Γi , E  , where E  is the birational transform of E in W are the σ-exceptional curves. Since X is smooth, there must be a −1 curve among the σ-exceptional curves. But since Γ2i ≤ −2, it follows that (E  )2 = −1. But since also E 2 = −1, this implies that h is an isomorphism and therefore X  is smooth and is just the blow-up of X at a single point. Hence a = 1 and therefore (6.4.9) becomes KX  = (π  )∗ KY  + 2E.

(6.4.10)

Now Y  is the quotient of X  by D . Then from the last equation and from the adjunction formula for inseparable maps (5.1.2) it follows that the divisorial part Δ of D is E. Then by Proposition 4.3, length(OZ ) = KX  · E + E 2 + c2 (X  ) = −2 + c2 (X) + 1 = −2 + 11 + 1 = 10 > 0, where Z is the isolated part of the fixed locus of D . However, since both X  and Y  are smooth, the fixed locus of D does not have an isolated part [1, Theorem 3.3], [35, Proposition 3.5]. Hence we get a contradiction. Hence the case pg (Y  ) = 0 is impossible. Case 1.2.2. Suppose that pg (Y  ) = 1. I will show that in this case, X must be a simply connected supersingular Godeaux surface. From (6.4.5) it follows that h1 (R1 ψ∗ OY  ) = 1. Hence R1 ψ∗ OY  = OP1 (−2) ⊕T , where T is a torsion sheaf on P1 . Suppose that T = 0. Then from equations (6.4.5) it follows that χ(OY  ) ≤ 1. In this case the argument that was used in the case pg (X) = 0 applies giving again a contradiction. Hence T = 0 and therefore χ(OY  ) = 2 and H 1 (OY  ) = 0. Moreover, from the equation (6.4.6) and the discussion immediately after it, it follows that 1 ≤ pg (X) ≤ 2. Suppose that pg (X) = 2. Then from Proposition 2.1 it follows that h1 (OX ) ≤ 1 and hence 2 ≤ χ(OX ) ≤ 3. Then c2 (Y  ) = χet (Y ) + k = c2 (X) + k = 12χ(OX ) − 1 + k,

(6.4.11)

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where k is the number of g-exceptional curves. But since χ(OY  ) = 2, it follows that c2 (Y  ) = 24. Hence 12χ(OX ) − 1 + k = 24. Suppose that χ(OX ) = 3. Then 24 = 35 + k, which is impossible. Suppose that χ(OX ) = 2. Then 24 − 1 + k = 24 and therefore k = 1. Hence there exists only one g-exceptional curve. However, I have shown during the study of the case pg (Y  ) = 0 that this case is also impossible. 2 = 1, from Proposition 2.1 it follows that Suppose that pg (X) = 1. Since KX 1 1 h (OX ) ≤ 1. Suppose that h (OX ) = 0. Then χ(OX ) = 2. Then c2 (X) = 23. Hence, since c2 (Y  ) = 24, it follows from the equation (6.4.11) that g has exactly one exceptional curve and hence Y has exactly one singularity which is necessarily of type 1/3(1, 1). But in the study of the case when pg (Y  ) = 0, I have shown that this case is impossible. Hence the only possibility is that h1 (OX ) = 1 and hence χ(OX ) = 1. Therefore X is a Godeaux surface. I will show that it is a supersingular and simply connected Godeaux. Since h1 (OX ) = 1, X can be either a singular or supersingular Godeaux surface. From the discussion at the end of section 2, X is singular if and only if X has a nontrivial Z/3Z torsor, or equivalently it has nontrivial étale 3-covers. Suppose that this is the case. Étale 1 3-covers are classified by Het (X, Z/3Z). Since π is an étale equivalence, it follows that 1 1 Het (X, Z/3Z) = Het (Y, Z/3Z). But since Y has rational singularities and H 1 (OY  ) = 0, it follows that H 1 (OY ) = 0. But then taking étale cohomology in the exact sequence F −id

0 → (Z/3Z)Y → (Ga )Y → (Ga )Y → 0 1 we find that Het (Y, Z/3Z) = 0, a contradiction. Hence X is a supersingular Godeaux surface. It remains to show that it is simply connected. Since the fundamental group is invariant under étale and birational equivalence, it follows that π1et (X) = π1et (Y ) = π1et (Y  ). Therefore it suffices to show that Y  is simply connected. The first step in order to show this is to study more carefully the structure of the elliptic fibration ψ : Y  → P1 . I will show next that there exists a unique point t0 ∈ P1 such that the fiber Yt0 is a multiple fiber. Moreover, Yt0 = 2C, where C is an indecomposable curve of canonical type. From the adjunction formula for elliptic and quasi-elliptic fibrations [3, Theorem 7.15] we get that

KY  =

k  (mi − 1)Pi ,

(6.4.12)

i=1

where mi Pi = Yti are the multiple fibers of ψ, mi ≥ 2 (note that since T = 0, ψ has no exceptional fibers). By assumption, g has exactly one exceptional curve E such that E 2 = −3. This corresponds to the unique singularity of Y which has index bigger than 1 (the 13 (1, 1) singularity). Then since KY  · E = 1, it follows from the equation (6.4.12) k that i=1 (mi − 1)Pi · E = 1. Taking into consideration that E is not contained in any fiber of ψ (if it did then KY  · E = 0) it follows that Pi · E = 0, for all i = 1, . . . , k. Hence k = 1 and m1 = 1. This means that ψ has exactly one multiple fiber Yt0 and moreover, Yt0 = 2C, where C is an indecomposable curve of canonical type. Then from the equation (6.4.12) it follows that KY  = C and hence Yt ∼ 2KY  , for all t ∈ P1 .

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I will next show that π1et (Y  ) = {1}. In order to do this it suffices to show that Y  does not have nontrivial étale Z/nZ-covers. For n = 3, this is equivalent to the property that Y  does not have torsion line bundles of order n. For n = 3, this is equivalent to the property that μ3 is not a subgroup scheme of Pic(Y  ). Z/3Z étale covers are classified 1 1 by Het (Y  , Z/3Z). However, since H 1 (OY  ) = 0, it follows that Het (Y  , Z/3Z) = 0.  Therefore Y does not have nontrivial étale Z/3Z-covers. Next I will show that Pic(Y  ) is torsion free. Let L be a torsion line bundle on Y  of order n. Then by the Riemann–Roch theorem, χ(L) = χ(OY  ) = 2 > 0. Hence, either H 0 (L) = 0 or H 2 (L) = H 0 (L−1 ⊗ ωY  ) = 0. Since L is torsion, H 0 (L) = 0. Therefore, H 0 (L−1 ⊗ ωY  ) = 0. Then L−1 ⊗ ωY  ∼ = OY  (W ),

(6.4.13)

where W is an effective divisor. I will show that all irreducible components of W are contracted to points by ψ. Suppose on the contrary that W has a component W0 that dominates P1 . Then since ωY  = OY  (C) and 2C = Yt0 ∼ Yt , for any t ∈ P1 , it follows that C 2 = 0 and that C · W0 = 0. But this is impossible since L−1 ⊗ ωY  ∼ = OY  (W ), and L is torsion. Hence every irreducible component of W is contracted by ψ. Now decompose W as W = W 1 + · · · + Wk ,

(6.4.14)

where Wk ⊂ Ytk , tk ∈ P1 and ti = tj , for i = j. I will show that k = 1 and (W1 )red = (Yt1 )red . Suppose that there exists an i such that (Wi )red = (Yti )red . Then it is possible to write (Yti )red = (Wi )red + Z, where Z and Wi do not have any common components. Then since Yti is connected it follows that Z ·Wi = 0. But then from the equation (6.4.14) it follows that 0 = Yt0 · Z = 2W · Z = 2Wi · Z = 0, a contradiction. Hence for all i, (Wi )red = (Yti )red . Since E dominates P1 it follows that E · Wi = 0, for all i. But C · E = 1 and hence W · E = 1. Then the equation (6.4.14) shows that k = 1. Hence W is contained in a single fiber Yt and Wred = (Yt )red . I will now consider cases with respect to the nature of Yt . Suppose that Yt is reduced. Then since Wred = (Yt )red , it follows that W = Yt and hence C ≡ Yt . Therefore 1 = C · E = Yt · E = 2KY  · E = 2, which is impossible. Suppose that Yt is not reduced. If the fiber Yt is a multiple fiber, then since ψ has exactly one multiple fiber, it follows that t = t0 and that W = C or W = 2C. If W = 2C, then from (6.4.13) it follows that C ≡ 0, which is impossible. If W = C, then L ∼ = OY  .

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Suppose that Yt is not a multiple fiber. In this case every irreducible component of  Yred is a smooth rational curve of self-intersection −2 [6, Corollary 5.1.1]. From the relation E · Yt = 2, it follows that E meets at most two irreducible components of Yt . Moreover, 2KY  = Yt . Then from the adjunction formula for g, 1 KY  + E = g ∗ KY , 3 and considering that every irreducible component of Yt is a smooth rational curve of self-intersection −2, it follows that every irreducible component of Yt that does not intersect E must be contracted by g. However, the singular locus of Y consists of a single 1/3(1, 1) singularity and finitely many A2 type canonical singularities. Hence Yt minus the components that meet E is a disjoint union of chains of rational curves of length 2. By taking now into consideration this, the fact that Yt is not multiple or reduced and the classification of reducible fibers of an elliptic fibration [6, p. 288], we see that there are the following two possibilities for Yt : (1) Suppose that E meets exactly one component of Yt . Then Yt is a configuration of ˜6 , rational curves of type E 1

◦

2

◦

•

3

◦

2

◦

◦

2

1

◦

1

where the solid dot corresponds to the curve that meets E and the numbers over each dot indicate the multiplicity with which the corresponding curve appears in Yt . But from the diagram above we see that E meets a curve which has multiplicity 3 in Yt . Hence Yt · E ≥ 3, which is impossible since Yt · E = 2. Hence this case is impossible. (2) Suppose that E meets two component of Yt . Then Yt is a configuration of rational ˜7 , curves of types E 1

◦

2

◦

3

•

4

◦

3

•

2

◦

1

◦

2

◦ where the solid dots correspond to the curves that meet E and the numbers over each dot indicate the multiplicity with which the corresponding curve appears in Yt . But from the above diagram we see that E must meet two curves which appear with multiplicity 2 each one in Yt . Hence Yt · E ≥ 4, which is impossible since Yt · E = 2.

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Hence the only possibility is that W = C and therefore L ∼ = OY  . Hence Pic(Y  ) is torsion free and it does not contain μ3 . Hence Y  is simply connected and therefore so is X. Case 1.2.3. Suppose that pg (Y  ) = 2. I will show that this case is impossible. Suppose that this case happens. I will show that ψ has no multiple fibers and that KY  = Yt , for t ∈ P1 . From the equations (6.4.5) it follows since pg (Y  ) = 2 that h1 (R1 ψ∗ OY  ) = 2 and therefore R1 ψ∗ OY  = OP1 (−3). Then from the adjunction formula for an elliptic or quasi-elliptic fibration [3, Theorem 7.15] it follows that KY  = Yt +



ai P i ,

(6.4.15)

i

where mi Pi = Yti are the multiple fibers of ψ and 0 ≤ ai ≤ mi − 1. But E · KY  = 1. Hence intersecting the previous equation with E it follows that ai = 0, for all i and therefore KY  = Yt , as claimed. Moreover, from this it follows that Yt · E = 1 and hence E is a section of ψ. Hence ψ has no multiple fibers. In particular this shows that T = 0 (since for any t ∈ T , Yt is a multiple fiber). Hence from the equations (6.4.5) it follows that H 1 (OY  ) = 0 and χ(OY  ) = 3. Also from the equation (6.4.6) it follows that pg (X) = 2 and since h1 (OX ) ≤ 1 it follows that 2 ≤ χ(OX ) ≤ 3. Suppose that h1 (OX ) = 1 and hence χ(OX ) = 2. Let F ∗ : H 1 (OX ) → H 1 (OX ) be the map on cohomology induced by the Frobenius. Suppose that F ∗ is not zero. Then X admits an étale Z/3Z cover. But if such a cover existed, then from Lemma 6.3 it would follow that 3≤

6 6 2 = 4 − 1 = 2, 2χ(OX ) − KX

which is clearly impossible. Suppose that F ∗ = 0. Then there exists an α3 -torsor ν : Z → X over X. Then 2 KZ = ν ∗ KX and χ(OZ ) = 3χ(OX ) = 6. Hence KZ2 = 3KX = 3. Then from [24, Proposition 2] it follows that 3 = KZ2 ≥ 2h0 (ωZ ) − 4 = 2(χ(OZ ) − 1 + h1 (OZ )) − 4 ≥ 2χ(OZ ) − 6 = 6, which is impossible. Therefore H 1 (OX ) = 0 and hence χ(OX ) = 3. Then the Noether formula gives that c2 (X) = 35. But also since χ(OY  ) = 3 it follows that c2 (Y  ) = 36. But then from the equation (6.4.11) it follows that g has exactly one exceptional curve and therefore the singular locus of Y consists of one singularity of type 1/3(1, 1). But during the study of the case when pg (Y  ) = 0, I showed that this case is impossible. This concludes the study of the pg (Y  ) = 2. Case 1.3. Suppose that κ(Y  ) = 0. I will show that this case is impossible.

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Indeed. If κ(Y  ) = 0 and Y  was minimal at the same time, then KY  ≡ 0. Hence any g-exceptional curve is a curve with self-intersection −2. Therefore Y has canonical 2 singularities and g is crepant, i.e., KY  = g ∗ KY . Hence KX = pKY2 = pKY2  = 0, which is impossible since KX is ample. Case 1.4. Suppose that κ(Y  ) = −∞. In this case I will show that: 2 = 1, then X is unirational and π1et (X) = {1}. (1) If KX 2 (2) If KX = 2 then X is uniruled. If in addition χ(OX ) ≥ 2, then X is unirational and et π1 (X) = {1}.

Since κ(Y  ) = −∞, it follows that Y  is ruled. Hence arguing as in the proof of Propo2 2 sition 6.1 we get that X is uniruled. In particular, if KX = 1 (or KX = 2 and χ(OX ) ≥ 2) then from Lemma 6.3 it follows that b1 (X) = 0 and hence from the diagram (5.0.1) it follows that b1 (Y  ) = 0 and hence Y  is rational. Therefore X is unirational. Moreover, from the diagram (5.0.1) and the fact that the algebraic fundamental group is invariant under étale equivalence and birational maps, we get that π1et (X) = π1et (X  ) = π1et (Y  ) = {1}. Case 2. Y  is not a minimal surface. Then the map φ : Y  → Z is a composition of m ≥ 1 blow-ups. From the equations (5.1.1) it follows that 1 φ∗ KZ + B + F = g ∗ KY . p Then, KY = g∗ φ∗ KZ + g∗ B.

(6.4.16)

Since Y  is the minimal resolution of Y , g does not contract any curves with selfintersection −1. But since φ is a composition of blow-ups, B has irreducible components of self-intersection −1. Then write B = B  + B  , where B  = i bi Bi such that Bi2 = −1 and B  = j bj Bj such that Bj2 ≤ −2. Then since φ is the composition of m blow-ups it is not hard to see that i bi ≥ m. Hence A = g∗ B = n1 A1 + · · · + ns As , s is an effective divisor such that i=1 ni ≥ m. Consider now cases with respect to the Kodaira dimension κ(Y  ) of Y  . Case 2.1. Suppose that 1 ≤ κ(Y  ) ≤ 2. I will show that this case is impossible if 2 2 KX = 1 and if KX = 2 then p = 3 or 5. In this case it follows from the classification of surfaces [5,4] that nKZ ∼ W , where W is a nontrivial effective divisor whose birational transform in Y  is not contracted by g.

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Hence KX · π ∗ g∗ φ∗ KZ =

1 KX · π ∗ g∗ φ∗ W > 0. n

Then since KX = π ∗ KY and KX is ample, it follows from the equation (6.4.16) that 2 KX = KX · π ∗ KY = KX · π ∗ (g∗ φ∗ KZ ) + KX · π ∗ A ≥ 1 + m. 2 2 Hence since m ≥ 1 it follows that KX ≥ 2. Hence if KX = 1, then it is not possible  2 that 1 ≤ κ(Y ) ≤ 2. If KX = 2 then m = 1, which means that φ is a single blow-up. 2 By following a similar argument as this that will be used in the case when KX = 1 and  2  κ(Y ) = 0 it follows that if KX = 2 and κ(Y ) ≥ 1, then p = 3, 5. The calculations are 2 similar to the case when KX = 1 and the details are left to the reader. 2 = 1 then p = 7 Case 2.2. Suppose that κ(Y  ) = 0. In this case I will show that if KX 2 and if KX = 2 then p = 3. 2 = 1 in detail and leave the other to the reader. As usual I will do the case when KX The method is identical but there is a fairly larger amount of calculations involved. Since κ(Y  ) = 0, it follows that KZ ≡ 0. Therefore from the equation (6.4.16) it follows that KY ≡ g∗ B = A. Then 2 KX = KX · π ∗ KY = KX · π ∗ A = n1 KX · π ∗ A1 + · · · + ns KX · π ∗ As ≥

s 

ni ≥ m,

i=1 2 where m is the number of blow-ups that φ consists of. Hence if KX = 1 then m = 1, and 2 if KX = 2 then m = 1 or 2. 2 = 1 and φ is the blow-up of a single point. Then B ∼ Suppose then that KX = P1 2 2 and KY  = B = −1. From the classification of surfaces [5,4], it is known that c2 (Z) ∈ 2 = 1, it follows from Proposition 2.1 that 1 ≤ χ(OX ) ≤ 3. {0, 12, 24}. Moreover, since KX Now from the diagram (5.0.1) it follows that

c2 (Y  ) = χet (Y  ) = χet (Y ) + k = χet (X) + k = c2 (X) + k, where k is the number of g-exceptional curves. Since φ is a single blow-up, c2 (Y  ) = c2 (Z) + 1 and therefore c2 (Z) = c2 (X) + k − 1.

(6.4.17)

Consider now cases with respect to χ(OX ). Case 2.2.1. Suppose that χ(OX ) = 3. Then c2 (X) = 35 and hence c2 (Z) = 34 + k. But since c2 (Z) ≤ 24, this is impossible. Case 2.2.2. Suppose that χ(OX ) = 2. Then c2 (X) = 23 and hence from (6.4.17) it follows that c2 (Z) = 22 + k. Hence Z is a K3 surface and k = 2, i.e., g has exactly two exceptional curves. Hence F = a1 F1 + a2 F2 , ai ≥ 0. Hence

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1 KY  + (a1 F1 + a2 F2 ) = g ∗ KY , p

(6.4.18)

2 KX = pKY2 = −p + a1 (KY  · F1 ) + a2 KY  · F2 .

(6.4.19)

and therefore

Suppose that Fi2 = −di , di > 0, i = 1, 2. First I will show that di ≥ 2 and that Fi ∼ = P1 , i = 1, 2 (note that in principle, the Fi ’s could be singular rational curves). Intersecting the equation (6.4.18) with F1 and F2 we get that   2pa (F1 ) − 2 + a1 = p 1 + d1   2pa (F2 ) − 2 + a2 = p 1 + d2

a2 (F1 · F2 ) d1

(6.4.20)

a1 (F1 · F2 ) d2

Moreover, F1 · F2 ≤ 1. Indeed. Suppose that F1 · F2 = 0. Then by looking at the map g we get the formula χet (Y  ) = χet (Y ) + χet (F1 + F2 ) − 1. But considering that both F1 and F2 are rational curves, we get that χet (F1 + F2 ) = 4 − (r − 1), where r is the multiplicity of the singular point of F1 + F2 . But since χet (Y  ) = 2 + χet (Y ) it follows that r = 2 and hence F1 · F2 = 1. Suppose that pa (Fi ) ≥ 1, for i = 1, 2. Then ai ≥ p, for i = 1, 2. Moreover, KY  · Fi = 2pa (Fi ) − 2 − Fi2 > 0, since Fi2 < 0. Then from the equation (6.4.19) it follows that 2 KX ≥ p. Hence at most one of pa (Fi ) can be strictly positive. Suppose that pa (F1 ) > 0 and that pa (F2 ) = 0. Hence F2 is a smooth rational curve and hence d2 ≥ 2. If pa (F1 ) ≥ 2, then 2 a1 > p and KY  · F1 ≥ 3 and hence from the equation (6.4.19) it follows that KX ≥ 2p. 2 Hence pa (F1 ) = 1 and KY  · F1 = 1 (if KY  · F1 > 1 then KX ≥ p). Hence F1 is a rational curve of arithmetic genus 1 and moreover from the adjunction formula it follows that F12 = −1. Consider now cases with respect to F1 · F2 . Suppose that F1 · F2 = 1. Then a straightforward calculation shows that KY  + 2F1 + F2 = g ∗ KY . 2 Therefore KY2 = 1 and KX = p. Hence this case is also impossible. Suppose that F1 · F2 = 0. Then an easy calculation shows that KY  = g ∗ KY − F1 . Hence KY2 = 0, which is impossible. Therefore both Fi , i = 1, 2 are smooth rational curves and hence di ≥ 2. Consider next cases with respect to the values of d1 and d2 . Suppose that di ≥ 4, for i = 1, 2. Then from the equations (6.4.20) it follows that ai ≥ p/2, i = 1, 2. Taking into consideration also that KY  · Fi = di − 2 ≥ 2, it follows 2 from the equation (6.4.19) that KX ≥ p. Suppose that at least one of the di is less than 4, say d2 ≤ 3.

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Suppose first that d2 = 2. Then from the equation (6.4.19) it follows that 2 KX = −p + a1 (d1 − 2).

(6.4.21)

I will show that F1 · F2 = 0. Indeed, if F1 · F2 = 0, then Y would have a singularity of type A1 . But by [35, Proposition 3.5] the local class groups of the singularities of Y are p-torsion. Then since the local class group of an A1 singularity is 2-torsion, p must be 2. However, we are assuming that p = 2. Hence F1 ·F2 = 0. Then from the equations (6.4.20) it follows that a2 = 12 a1 (F1 · F2 ) ≥ 12 a1 . Then from the equation (6.4.20) it follows that a1 ≥ p

2(d1 − 2) . 2d1 − 1

(6.4.22)

Then from the equation (6.4.19) it follows that 2 KX ≥ −p + 2p

(d1 − 2)2 . 2d1 − 1

(6.4.23)

It is not difficult to see that (d1 − 2)2 /(2d1 − 1) ≥ 1, if d1 ≥ 5 and therefore in this case 2 KX ≥ p. It remains to examine the cases d1 ≤ 4. Suppose that d1 = 2. Then since d2 = 2, Y has a canonical singularity of type A2 . But then g would be crepant, i.e., KY  = g ∗ KY and hence a1 = a2 = 0. Then from the 2 equation (6.4.19) it follows that KX = −p < 0, which is impossible since KX is ample. Suppose that d1 = 3 and d2 = 2. In this case Y has a singularity of type 1/5(1, 2) and hence p = 5. Moreover, a straightforward calculation shows that a1 = 2 and a1 = 1. Then since KY  · F1 = 1 and KY  · F2 = 0, it follows from the equation (6.4.19) that 2 KX = −5 + 2 = −3 < 0, which is impossible since KX is ample. Suppose that d1 = 4 and d2 = 2. Then Y has a singularity of type 1/7(1, 2) and therefore p = 7. A straightforward calculation shows that a1 = 4 and a2 = 2. Then the 2 equation (6.4.19) shows that KX = 1. Unfortunately I am unable to show that this case is impossible. This is the reason of the assumption p = 7 in 1.1. The last cases that need to be considered are when di ≥ 3, i = 1, 2, and at least one of the di is 3. There are the following possibilities. (1) (2) (3) (4)

F12 F12 F12 F12

= F22 = −3 and F1 · F2 = 1. = F22 = −3 and F1 · F2 = 0. = −3, F22 = −4 and F1 · F2 = 1. = −3, F22 = −4 and F1 · F2 = 0.

All cases are impossible. Indeed. A simple calculation shows that in the first case Y has a singularity of index 8. This is impossible because by [35, Proposition 3.5], Y has singularities of prime index.

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In the second case, Y has two singularities of type 1/3(1, 1) each. Then a straightforward calculation shows that a1 = a2 = 1 and p = 3. Then from the equation (6.4.19) it 2 follows that KX = −3 + 1 + 1 = −1 < 0, which is impossible. In the third case Y has a singularity of type 1/11(1, 4). Hence p = 11 and a simple calculation shows that a1 = 6 and a2 = 7. Then from the equation (6.4.19) it follows 2 that KX = −11 + 6 + 7 · 2 = 9 > 1. Finally in the fourth case, F2 contracts to a singularity of index 2. But this is again impossible because p = 2. Case 2.2.3. Suppose that χ(OX ) = 1. Then from the Noether formula we get that c2 (X) = 11. Also from the equation (6.4.17) it follows that c2 (Z) = 10 + k, where k is the number of g-exceptional curves. Hence either k = 2 and Z is an Enriques surface or k = 14 and Z is a K3 surface. Suppose that k = 2 and Z is an Enriques surface. This situation is similar to the one of case 2.2.2 above with the only difference that Z is now an Enriques surface instead of a K3 surface. However, the only property of a K3 surface that was used in the argument that was used in the case 2.1.2 is that KZ ≡ 0. Hence it applies in this case too. Therefore this case is impossible unless possibly for p = 7. Suppose that k = 14 and Z is a K3 surface. In this case I will show that p ∈ {3, 5} and X is a simply connected supersingular Godeaux surface. Since Z is a K3 surface, it follows that π1et (Z) = {1} [5,4]. Then from the diagram (5.0.1) it follows that π1et (X) = {1}. Moreover, since pg (Y  ) = 0, it follows that pg (Y ) = h0 (ωY ) > 0. Then pg (X) > 0 and since χ(OX ) = 1 it follows from Proposition 2.1 that h1 (OX ) = 1. Therefore X is either a singular or a supersingular Godeaux surface and therefore p ≤ 5 [24]. Suppose that X was a singular Godeaux surface. Then X admits a nontrivial étale p-cover. But since π1et (X) = {1}, there is no such cover. Hence X is a simply connected supersingular Godeaux surface. This concludes the proof of Theorem 6.4. 2 The last result of this section gives restrictions on the geometric genus of a smooth canonically polarized surface that has nontrivial global vector fields. Theorem 6.5. Let X be a smooth canonically polarized surface defined over an alge2 braically closed field of characteristic p ∈ / {2, 3, 5} such that KX = 1. Suppose that X admits a nontrivial global vector field D. Then pg (X) ≤ 1. 2 = 1, Remark 6.6. By Proposition 2.1, if X is a canonically polarized surface with KX 1 then pg (X) ≤ 2 and 1 ≤ χ(OX ) ≤ 3. Considering that χ(OX ) = 1 + pg (X) − h (OX ), it follows that if χ(OX ) = 3 then pg (X) = 2 and if χ(OX ) = 2 then either pg (X) = 2 and h1 (OX ) = 1 or pg (X) = 2 and h1 (OX ) = 0. The previous theorem says that if either χ(OX ) = 2 and pg (X) = 2 or χ(OX ) = 3, then X does not have nontrivial global vector fields if p > 5.

In order to prove Theorem 6.5 the following results will be needed.

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Lemma 6.7 (with assumptions as in Theorem 6.5). Suppose that D has a divisorial part and moreover suppose that |KX | contains an integral curve. Let C ∈ |KX | be a member of the canonical system which is not an integral curve of D. Let π : X → Y be the quotient of X by the αp or μp action induced by D and let C˜ = π(C). Assume that Y has rational singularities. Then C˜ is a Cartier divisor in Y and C˜ ∼ = C. Proof. Let C0 ∈ |KX | be an integral curve and Δ the divisorial part of D. Then since ˜ Therefore since C ∼ C0 , it follows that C is not an integral curve, π∗ C = C. C˜ = π∗ C ∼ π∗ C0 = pC˜0 Hence since the local Picard groups of Y are p-torsion it follows that C˜ is Cartier. Then ˜ in Y . Then from since Y has rational singularities the adjunction formula holds for C the adjunction formula and (5.1.2) it follows that ˜ − 2 = C˜ 2 + KY · C˜ = p + KX · C − (p − 1)Δ · C = p + 1 − (p − 1)Δ · C. 2pa (C) (6.7.1) ˜ ≥ pa (C) = 2 with Since π∗ C = C˜ it follows that C → C˜ is birational and hence pa (C) equality holding only if the map is in fact an isomorphism. Then from (6.7.1) it follows ˜ = pa (C) = 2. Hence C˜ ∼ that Δ · C = 1 and hence pa (C) = C as claimed. 2 Proposition 6.8. Let X be a normal projective variety of finite type over an algebraically closed field of characteristic p > 0 and let D be a nontrivial global vector field on X such that either Dp = 0 or Dp = D. Let π : X → Y be the quotient of X by the αp or μp action induced by D. Let L be a rank one reflexive sheaf on Y and M = (π ∗ L)[1] . Then D induces a k-linear map D∗ : H 0 (X, M ) → H 0 (X, M ) with the following properties: (1) Let s ∈ H 0 (X, M ) be a section and Z(s) its divisor of zeros. Then the divisor of zeros Z(D∗ (s)) of D∗ (s) is the divisor of X defined by the ideal sheaf r(D(IZs )), where IZ(s) is the ideal sheaf of OX defining Z(s). (2) Ker(D∗ ) = H 0 (Y, L) (considering H 0 (Y, L) as a subspace of H 0 (X, M ) via the map π ∗ ). (3) If Dp = 0 then D∗ is nilpotent and if Dp = D then D∗ is a diagonalizable map whose eigenvalues are in the set {0, 1, . . . , p − 1}. Corollary 6.9. Let D be a nontrivial global vector field of either additive or multiplicative type on a smooth projective variety X defined over an algebraically closed field k of characteristic p > 0. Suppose that pg (X) = 0 and that D does not have a divisorial part. Then |KX | contains an integral curve of D.

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Proof. Let π : X → Y be the quotient of X by the αp or μp action induced by D. Then since D does not have a divisorial part it follows that KX = π ∗ KY . Now the corollary follows immediately from Proposition 6.8 since any eigenvector of D∗ corresponds to an integral curve of D. 2 Proof of Proposition 6.8. The action can be defined on any codimension two open subset of X and hence by removing the isolated singular points of D we can assume that Y is smooth and hence L is Cartier. Then L is defined by giving an affine cover Vi of Y and elements aij ∈ OV∗ ij , where Vij = Vi ∩ Vj , satisfying the usual cocycle condition. D Let Ui = π −1 (Vi ). Then the Ui form an affine open cover of X and OVi = OU = {b ∈ i ∗ OUi | Db = 0}. Then M = π L is defined by the affine cover Ui of X and the images ∗ ∗ aij ∈ OU of aij in OU , where Uij = Ui ∩ Uj . Let s ∈ H 0 (X, M ) be a section. This ij ij is defined by giving local sections si ∈ OUi such that si |Uij = aij sj |Uij . Then since aij ∈ OV∗ ij , it follows that D(aij ) = 0 and hence D(si )|Uij = aij D(sj )|Uij . Therefore the sections D(si ) ∈ OUi form a global section D∗ (s) of M . Let Z(s) be the divisor of zeros of s. Then from the above construction it is clear that the divisor of zeros of D∗ (s) is defined by the ideal sheaf of OX generated by D(IZ(s) ), where IZ(s) is the ideal sheaf defining Z(s). Let now s ∈ Ker(D∗ ). As explained above, s is defined by giving local sections si ∈ OUi such that si |Uij = aij sj |Uij . Then since s ∈ Ker(D∗ ) it follows that D(si ) = 0, for all i. Therefore si ∈ OVi and hence the si define a section t of L whose pullback in X is s. Suppose that Dp = 0. Then (D∗ )p = 0 and hence D∗ is nilpotent. If on the other hand Dp = D, then (D∗ )p = D∗ . This implies that the k-linear map D∗ is diagonalizable and its eigenvalues are in the set {0, 1, . . . , p − 1}. 2 Proposition 6.10. Let X be a smooth canonically polarized surface defined over an alge2 braically closed field of characteristic p = 2 such that KX = 1 and pg (X) = 2. Suppose that X has a nontrivial global vector field D of either additive or multiplicative type. Then the linear system |KX | has a unique base point P ∈ X. Moreover, every member of |KX | is smooth at P and P is an isolated fixed point of D. Proof. First I will show that |KX | has a unique base point P and that every member of 2 |KX | is smooth at P . Let C be a member of |KX |. Then KX · C = KX = 1. Therefore, since KX is ample, it follows that C is irreducible and reduced. Since dim |KX | = 2 and 2 KX = 1, it follows that |KX | has base points. Let P1 , . . . , Pn be its base points and let C1 , C2 be two different members of |KX |. Since C1 , C2 are integral curves, it follows 2 that C1 · C2 ≥ n. However, since C1 · C2 = KX = 1 it follows that n = 1 and therefore |KX | has a unique base point P . Next I will show that every member of |KX | is smooth at P . Let f : X  → X be the blow-up of X at P and let E be the f -exceptional curve. Then for any two different members C1 and C2 of |KX |, f ∗ Ci = Ci + mi E, where Ci are the birational transforms

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of Ci in X  , i = 1, 2. Hence 1 = f ∗ C1 · f ∗ C2 = C1 · f ∗ C2 + m1 E · f ∗ C2 = C1 · C2 + m1 m2 . Hence C1 · C2 = 0 and m1 = m2 = 1. Therefore C1 and C2 are smooth at P as claimed. Next I will show that P ∈ X is an isolated fixed point of D. In order to prove this consider cases with respect to the divisorial part Δ of D. Suppose that Δ = 0. Then KX = π ∗ KY . Then from Proposition 6.8 it follows that for any C ∈ |KX | the ideal generated by D(IC ) defines another member D∗ (C) ∈ |KX |. Let now C1 and C2 be two general members of |KX |. Then C1 ∩ C2 = {P } and therefore mP = IC1 + IC2 . Then D(mp ) = D(IC1 + IC2 ) = D(IC1 ) + D(IC2 ) ⊂ r(D(IC1 )) + r(D(IC2 )) = ID∗ (C1 ) + ID∗ (C2 ) = mp , since D∗ (C1 ) and D∗ (C2 ) are members of |KX | and P is the unique base point of |KX |. Therefore P is a fixed point of D and since Δ = 0, it is an isolated fixed point. Suppose now that Δ = 0 and that P is not an isolated fixed point of D. In this case I will show that KY is Cartier and in fact Y has canonical DuVal singularities. 2 = 1 and X First notice that from the proof of Proposition 6.1 it follows that if KX  has nontrivial global vector fields, then κ(Y ) = −∞ and hence the minimal resolution of Y is a rational surface. 2 ≥ p. If Δ = 0 then from the equation (5.1.2) it follows that if pg (Y ) = 0, then KX Hence we can assume that pg (Y ) = 0. Then since the minimal resolution g : Y  → Y of Y is a rational surface, it follows from the Leray spectral sequence that R1g∗ OY  = 0. Hence Y has rational singularities. Next observe that no member of |KX | is an integral curve of D. Indeed. If on the ˜ where C˜ = contrary a member C ∈ |KX | was an integral curve of D, then C = π ∗ C, ∗ ˜ But now as in the case where Δ = 0, one can use π(C). Therefore KX = π (OY (C)). Proposition 6.8 to show that D acts on |KX | and conclude similarly that P is an isolated fixed point of D, a contradiction since we are assuming that P is not an isolated singular point of D. The singularities of Y are the images of the isolated singular points of D. Let P ∈ X be an isolated fixed point of D, Q = π(P ) ∈ Y and let C0 ∈ |KX | such that P ∈ C0 . Let C˜0 = π(C0 ). Then π∗ C0 = C˜0 and therefore the map C0 → C˜0 is birational. I will show that it is in fact an isomorphism and that C˜0 is a Cartier divisor in Y . Indeed, let ˜ where C˜ = π(C). However, since C is C ∈ |KX | be a general member. Then C˜0 ∼ C, general, and P is not an isolated singular point of D, C does not contain any isolated singular point of D and therefore C˜ is in the smooth part of Y and hence it is Cartier. Therefore C˜0 is Cartier as well. Hence since Y has rational singularities the adjunction formula for C˜0 holds. Hence considering that π ∗ C˜0 = pC0 and the equation (5.1.2), it follows that

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2pa (C˜0 ) − 2 = C˜02 + KY · C˜0 = pC02 + π ∗ KY · C0 = 2 − (p − 1)KX · Δ = p + 1 − (p − 1)KX · Δ. p + KX

Now since C0 → C˜0 is birational it easily follows that pa (C˜0 ) ≥ pa (C0 ) = 2 with equality only in the case of isomorphism. Then from the above relation we get that KX · Δ = 1 and hence pa (C˜0 ) = 2. Hence C˜0 ∼ = C0 . Hence C˜0 has embedding dimension 2 at Q and hence Y has a hypersurface singularity at Q. Therefore KY is Cartier. Moreover, since Y has rational singularities, its singularities must be DuVal canonical singularities [2], as claimed. Next I will show that Δ is not a member of |KX |. Since Δ · KX = 1 it follows that Δ is reduced and irreducible. Suppose that Δ ∈ |KX |. Then from the adjunction 2 formula (5.1.2), it follows that π ∗ KY = (2 − p)KX and therefore KY2 = (p−2) ∈ / Z, if p p = 2. But this is impossible since it has already been shown that KY is Cartier. Suppose now that there is an isolated singular point Q of D on Δ. Since Δ ∈ / |KX |, there exists a curve C0 ∈ |KX | such that Q ∈ C0 . Moreover, since C0 · Δ = 1, the same arguments that have been used in order to show that every member of |KX | is smooth at the base point of |KX |, show that Δ and C0 are smooth at Q. Then from the adjunction formula (5.1.2) it follows that (p − 1)π∗ Δ = π∗ KX − pKY = C˜0 − pKY . Considering that pπ∗ Δ and C˜0 are Cartier divisors, it follows that π∗ Δ is also Cartier. Also note that we can assume Δ is not an integral curve of D. If it was then OX (Δ) = ˜ and therefore KX = π ∗ (OY (KY + Δ)). ˜ Then by Proposition 6.8, D acts on π ∗ OY (Δ) |KX | and the same arguments that were used in the case where Δ = 0 show that P is an isolated fixed point of D. Therefore assume that Δ is not an integral curve of D and ˜ is Cartier. Then the adjunction formula gives that ˜ = π(Δ) and hence Δ hence π∗ Δ = Δ ˜ − 2 = KY · Δ ˜ +Δ ˜ 2 = π ∗ KY · Δ + pΔ2 = 2pa (Δ) KX · Δ − (p − 1)Δ2 + pΔ2 = KX · Δ + Δ2 = 2pa (Δ) − 2. ˜ Considering now that the map Δ → Δ ˜ is birational, it follows Therefore pa (Δ) = pa (Δ). ˜ ˜ is Cartier that it is an isomorphism. Hence if R = π(Q), Δ is smooth at R. But since Δ in Y it follows that R ∈ Y is a smooth point. This is a contradiction since Q ∈ X was assumed to be an isolated singular point of D and hence R ∈ Y must be singular. Therefore there are no isolated singular points of D on Δ. Hence if Q ∈ X is an isolated singular point of D, then Q ∈ / Δ. Let C0 ∈ |KX | be such that Q ∈ C0 . Let f : X  → X be the blow-up of X at Q and let E be the f -exceptional curve. Then since Q ∈ X is an isolated singular point of D, D lifts to a vector field D on X  inducing an action of either αp or μp . Let π  : X  → Y  be the quotient. Then there exists a commutative diagram:

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X

f

Y

(6.10.1)

X

π

267

π g

Y

where g is birational. Then F = π(E) is the g-exceptional curve. Let C˜0 = π(C0 ). Notice that Q ∈ C0 must be a singular point of multiplicity m ≥ 2 since otherwise, since C˜0 ∼ = C0 is Cartier in Y , it would follow that Y is smooth at R = π(Q). Let Cˆ0 be the birational transform of C˜0 in Y  . Then there is an m ≥ 1 such that g ∗ C˜0 = Cˆ0 + m F . Then standard calculations in the above diagram and considering that π ∗ C˜0 = pC0 give that m (π  )∗ F = pmE.

(6.10.2)

Let now a ∈ Q such that KY  = g ∗ KY + aF . Since Y has DuVal canonical singularities, it follows that a is a nonnegative integer. Then straightforward calculations in the diagram (6.10.1) give that KX  = (π  )∗ KY + (p − 1)f ∗ Δ + E − a(π  )∗ F ,

(6.10.3)

where Δ = f ∗ Δ ∼ / Δ. = Δ since Q ∈ Consider now cases with respect to whether E is an integral curve of D or not. Suppose that E is an integral curve of D . Then (π  )∗ F = E. Then from (6.10.2) it follows that m = pm. Also the equation (6.10.3) gives that KX  = (π  )∗ KY + (p − 1)f ∗ Δ + (1 − a)E, where Δ = f ∗ Δ ∼ = Δ. But the general adjunction formula for purely inseparable maps says that (p − 1)f ∗ Δ + (1 − a)E = (p − 1)Δ , where Δ is the divisorial part of D . Hence 1 − a = (p − 1)λ, where λ ≥ 0. Now since a is a nonnegative integer, if p = 2 this is possible only if a = 1 and λ = 0. Then we have the following formulas KY  = g ∗ KY + F g ∗ C˜0 = Cˆ0 + mpF. Since C˜0 is Cartier and all the local Picard groups of Y  are p-torsion, the second formula implies that Cˆ0 is Cartier. Then since (π  )∗ F = E, it follows that F 2 = −1/p and therefore Cˆ02 = C˜02 + m2 p2 F 2 = pC02 − m2 p = p − m2 p, since π ∗ C˜0 = pC0 . Also from the above equations it follows that F · Cˆ0 = m and that KY  · Cˆ0 = KY · C˜0 + m = π ∗ KY · C0 + m = 1 − (p − 1)Δ · C0 + m = 2 − p + m,

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since KX · Δ = 1. Since Cˆ0 is Cartier in Y  and Y  has rational singularities (since Y has), the adjunction formula for Cˆ0 holds in Y  . From this we find that 2pa (Cˆ0 ) − 2 = Cˆ02 + KY  · Cˆ0 = p − m2 p + 2 − p + m = 2 + m(1 − mp) and therefore 1 pa (Cˆ0 ) = 2 + m(1 − mp). 2 But this relation is impossible for p ≥ 3 and m ≥ 2. Hence E cannot be an integral curve of D . Suppose now that E is not an integral curve of D . Hence (π  )∗ F = pE. Then the equation (6.10.3) becomes KX  = (π  )∗ KY + (p − 1)f ∗ Δ + (1 − pa)E. Then the adjunction formula for purely inseparable maps (5.1.2) shows again that 1 − pa = (p − 1)λ, for some λ ≥ 0. But this is impossible unless a = 0 and p = 2. Hence this case is impossible too. This concludes the proof that the base point of |KX | is an isolated fixed point of D. 2 Proof of Theorem 6.5. Suppose that X has a nontrivial global vector field D. According to Proposition 4.1 we may assume that D is either of additive or multiplicative type. 2 Moreover, from Proposition 2.1 it follows that if KX = 1 then pg (X) ≤ 2. Therefore in order to prove the proposition it suffices to show that the case pg (X) = 2 is impossible. Suppose then that pg (X) = 2. Now since the linear system |KX | is 2-dimensional it follows from Propositions 6.10, 6.8, that |KX | has a unique base point P ∈ X which is also an isolated fixed point of D. In addition every member C ∈ |KX | is a reduced and irreducible curve which is smooth at P . Consider now cases with respect to whether the general member of |KX | is an integral curve of D or not. Case 1. Suppose that the general member of |KX | is an integral curve of D. In this case I will show that p ∈ {2, 3, 5}. Suppose that the general member C of |KX | is smooth. Then D restricts to a vector field on C. However, since C is smooth of arithmetic genus 2 and such curves do not have global vector fields, the restriction of D on C must be zero and hence C is in the divisorial part of D. However, since there are infinitely many such C, this implies that D = 0. Suppose that every member of |KX | is singular. Let then f : X  → X be the blow-up of the base point P ∈ X of |KX | and let E be the f -exceptional curve. Let C ∈ |KX | be any member and C  be its birational transform in X  . Then since by Proposition 6.10 C is smooth at P , it follows that C  ∈ |KX  − 2E| and (C  )2 = 0. Then the linear system

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|KX  − 2E| defines a fibration h : X  → P1 whose fibers are the birational transforms of the fibers of |KX |. The general fiber of h is a singular normal integral curve Ck(t) defined over the function field k(t) of P1 . Since it is not smooth, there exists a purely inseparable extension k(t) ⊂ K of k(t) such that CK = Ck(t) ⊗k(t) K is not normal. Let C˜K be its normalization. Then by [31,34], the difference pa (Ck(t) ) − pa (C˜K ) is divisible by (p − 1)/2. But pa (Ck(t) ) = 2 and therefore, since pa (C˜K ) ≤ 1, p ∈ {2, 3, 5}. Case 2. The general member of |KX | is not an integral curve of D. In this case we may also assume that the general member of |KX | is also smooth. Otherwise the arguments of Case 1 show that p ∈ {2, 3, 5}. Next I will show that the divisorial part Δ of D is not zero. In order to do this one has to consider separately the cases when D is of additive or multiplicative type. I will only do the additive case. The multiplicative case is treated with similar arguments. Suppose that Dp = 0 and that Δ = 0. Since Δ = 0, it follows that KX = π ∗ KY . Then by Proposition 6.8 it follows that D acts on |KX |. In particular it induces a k-linear map D∗ : H 0 (X, OX (KX )) → H 0 (X, OX (KX )) such that Ker(D∗ ) = H 0 (Y, OY (KY )). Then by Proposition 6.8 D∗ is nilpotent. Therefore since dimk H 0 (X, OX (KX )) = 2, it follows that (D∗ )2 = 0 and hence Im(D∗ ) ⊂ Ker(D∗ ). Note that necessarily pg (Y ) = dimk H 0 (Y, O(KY )) > 0. If pg (Y ) = 0 then D∗ would be an isomorphism. However this is impossible since D∗ is nilpotent. Hence 1 ≤ pg (Y ) ≤ 2. Suppose that pg (Y ) = 2. Then H 0 (X, OX (KX )) = H 0 (Y, OY (KY )) and therefore every member of |KX | is the pullback of a member of |KY |. Hence every member of |KX | is an integral curve of D, which is impossible since we are assuming that this is not the case. Suppose that pg (Y ) = 1. Then dimk Ker(D∗ ) = 1. Hence, since H 0 (X, OX (KX )) is two dimensional, it follows that dimk Im(D∗ ) = 1 and therefore Ker(D∗ ) = Im(D∗ ). But this says that for any C ∈ |KX |, D(IC ) ⊂ IC0 . Then since mp = IC + IC0 D(mp ) = D(IC ) + D(IC0 ) ⊂ IC0 . Let C ∈ |KX | be general. Then C0 · C = 1. This implies that C0 and C intersect with multiplicity 1 at the base point P of |KX |, which by Lemma 6.10 is an isolated fixed point of D. Hence the local equations of C0 and C at P generate the maximal ideal mP of OX,P . Hence we may choose local coordinates x and y at P such that C is given by x = 0 and C0 by y = 0. Then D = f (x, y)

∂ ∂ + yg(x, y) . ∂x ∂y

Since D(mp ) ⊂ IC0 it follows that D(x) ∈ (y) and hence f (x, y) = yf1 (x, y). But this now implies that C0 is in the divisorial part of D and therefore Δ = 0, as claimed. Then

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2 from the proof of Proposition 6.1 it follows that if KX = 1 and X has nontrivial global  vector fields, then if p = 2, κ(Y ) = −∞. Next I will show first that in fact every member of |KX | is not an integral curve of D. Suppose that there exists a curve C0 ∈ |KX | which is an integral curve of D. Then by Lemma 6.7 it follows that for general C ∈ |KX |, π(C) = C˜ ∼ = C and C˜ is Cartier in Y . ˜ Since C is smooth at P , it follows that C is smooth at Q = π(P ), and since it is Cartier, it follows that Y is smooth at Q. This is impossible since by assumption P ∈ X is an isolated fixed point of D and therefore Q is a singular point of Y . Therefore no member of |KX | is an integral curve. Let f1 : X1 → X be the blow-up of X at P and let E be the f1 -exceptional curve. Since P is a fixed point of D, D lifts to a vector field D1 on X1 . Let ν : X1 → Y1 be the quotient of X1 by the αp or μp action on X1 induced by D1 . Let F = ν(E). Then there exists a commutative diagram

X1

f1

ν

Y1

(6.10.4)

X π

g1

Y

where g1 is birational and F is the g1 -exceptional curve. Claim. Y1 is smooth along F . Indeed. For any C ∈ |KX |, let C  denote its birational transform in X1 , Cˆ = ν(C  ) and ˆ C˜ = π(C). Next I will show that for any C ∈ |KX |, Cˆ is Cartier in Y1 and C ∼ = C ∼ = C. We know that C is an integral curve of arithmetic genus 2 and moreover it is smooth at P . Hence it easily follows that KX1 · C  = 2 and (C  )2 = 0. Let Δ be the divisorial part of D . This is certainly nonzero since Δ = 0. Then since C  is not an integral curve of D it follows that KY1 · Cˆ = ν ∗ KY1 · C  = KX1 · C  − (p − 1)Δ · C  = 2 − (p − 1)Δ · C  . Let now C be a general member of |KX |. Then no isolated singular point of D is on C  and therefore Cˆ is in the smooth part of Y1 . Moreover, since (C  )2 = 0, it follows that Cˆ 2 = 0. Since Cˆ is in the smooth part of Y1 it is Cartier and the adjunction formula ˆ From this it follows that holds for general C. ˆ − 2 = KY · Cˆ + Cˆ 2 = 2 − (p − 1)Δ · C  . 2pa (C) 1 ˆ = 2 = pa (C  ). It now easily This relation is possible only if Δ · C  = 0 and hence pa (C) ν ˆ since it is birational, is in fact an isomorphism. Note also that the follows that C  → C,   relation Δ · C = 0 implies that Δ ∈ |KX | and E is not in the divisorial part of D .

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Let now C0 ∈ |KX | be any member. Since it is not an integral curve it follows that ˆ where C ∈ |KX | is a general curve. Then since Cˆ is Cartier, ν∗ C0 = Cˆ0 . Hence Cˆ0 ∼ C, ˆ ˆ so is C0 . Hence C is Cartier in Y1 for any C ∈ |KX |. Considering now that all C ∈ |KX | are smooth at P it follows that C  is smooth at C  ∩ E and therefore Cˆ is smooth at Cˆ ∩ F . Hence since all Cˆ are Cartier, it follows that Y1 is smooth along F . Next I will show that ν ∗ F = pE, or equivalently that E is not an integral curve of D1 . Suppose otherwise that E is an integral curve of D1 . Then ν ∗ F = E. Then E 2 = pF 2 and hence F 2 = −1/p ∈ / Z. However this is impossible since Y1 is smooth along F and hence F is a Cartier divisor in Y1 . Hence E is not an integral curve of D and therefore ν ∗ F = pE. Hence F 2 = −p. This implies that F contracts to a rational singularity in Y of type p1 (1, 1). Hence Q = π(P ) is a rational singularity. Now let m > 0 such that ˜ Then g1∗ C˜ = Cˆ + mF . Then from the diagram (6.10.4) it follows that f1∗ π ∗ C˜ = ν ∗ g1∗ C. ∗˜ ∗  ∗ ∗ˆ  since π C = pC, f C = C + E, ν F = pE and ν C = pC it follows that 1

pC  + pmE = pC  + pE. Therefore m = 1 and hence g1∗ C˜ = Cˆ + F . Since Cˆ and F are Cartier divisors it follows that g1∗ C˜ is Cartier and hence, since Q ∈ Y is a rational singularity, C˜ is Cartier as well. The arguments now used in the proof of Proposition 6.7 show that C˜ ∼ = C. But since C is smooth at P , then C˜ is smooth at Q. Then since C˜ is Cartier in Y this implies that Y is smooth at Q, which is impossible since P is an isolated fixed point of D and hence Y is singular at Q. 2 7. Vector fields on surfaces in characteristic 2 The purpose of this section is to study smooth canonically polarized surfaces defined over an algebraically closed field of characteristic 2 with nontrivial global vector fields. Even though 2 is the smallest nonzero characteristic where many unusual situations appear (like the existence of quasi-elliptic fibrations) and in the view of Theorem 3.1 this is the characteristic that it is most likely that the automorphism scheme of a surface with K 2 = 1, 2 is not reduced, there are certain advantages over the higher characteristic case, essentially because purely inseparable maps of degree 2 between normal varieties are torsors in codimension 2. The following theorem is the main result of this section and is slightly stronger than Theorem 1.1. It provides additional information about the geometry of X in the case when μ2 is a subgroup scheme of Aut(X). Theorem 7.1. Let X be a smooth canonically polarized surface defined over an algebraically closed field of characteristic 2 such that Aut(X) is not smooth. Then: 2 = 2 then X is uniruled. If in addition χ(OX ) ≥ 2, then X is unirational and (1) If KX et π1 (X) = {1}.

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2 (2) If KX = 1 then π1et (X) = {1}, pg (X) ≤ 1 and X is unirational.

Moreover, suppose that μ2 is a subgroup scheme of Aut(X). Then: 2 (1) If KX = 2 then X is uniruled and 0 ≤ χ(OX ) ≤ 1. 2 = 1 then X is a simply connected supersingular Godeaux surface. (2) If KX

In all cases X is the quotient of a ruled or rational surface (maybe singular) by a rational vector field. Remarks 7.2. (1) If Aut(X) is not smooth then it contains either μ2 or α2 . Theorem 7.1 shows that the condition “μ2 is a subgroup scheme of Aut(X)” is more restrictive than “α2 is a subgroup scheme of Aut(X)”. Equivalently, it is more rare that a surface has vector fields of multiplicative type than of additive type. (2) The proof of the statement of the theorem in the case when μ2 is a subgroup scheme of Aut(X) strongly uses the fact that the quotient map π : X → Y is a torsor in codimension 2, something that is not true in general for higher characteristics. This is the reason that I am unable at the time to generalize it in all characteristics. Proof of Theorem 7.1. Fix notation as in Section 5. As was explained there, if Aut(X) is not smooth then X has a nontrivial global vector field D such that either D2 = 0 or D2 = D inducing a nontrivial α2 or μ2 -action on X. Let π : X → Y be the quotient of X by this action. Then by [35, Proposition 3.6], [13, Proposition 2.6], Y has Gorenstein singularities and there are commutative diagrams X 

f 

π 

Y 

X

X



Y

X

π

π g

f

Z

φ

Y

π g

(7.2.1)

Y

with the following properties: (1) f  is a resolution of the isolated singularities of D through successive blow-ups of its isolated singular points. X  and Y  are smooth, D lifts to a vector field D in X  with only divisorial singularities and Y  is the quotient of X  by the corresponding α2 or μ2 -action. However, Y  may not be the minimal resolution of Y . (2) Y  is the minimal resolution of Y and Z its minimal model. However, X  may now be singular. In any case, as was explained in section 5, X  has rational singularities and the reduced f and g-exceptional divisors are configurations of rational curves without any loops (the g-exceptional curves may be singular).

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(3) By [9, Proposition 1.11], π is a torsor over a codimension 2 open subset of Y and since Y  is smooth, π  is a torsor. In particular X  has hypersurface singularities and hence KX  is Cartier. (4) Suppose that D2 = D, i.e., that X has a nontrivial μ2 action. Then the two diagrams (7.2.1) are the same. This means that Y  = Y  , X  = X  and π  = π  . Since g and φ are birational morphisms and Y is Gorenstein, there are adjunction formulas KY  = g ∗ KY − F

(7.2.2)

KY  = φ∗ KZ + B where F , B are effective g and φ-exceptional divisors, respectively. Since p = 2, the adjunction formula (5.1.2) becomes KX = π ∗ KY + Δ.

(7.2.3)

Moreover, since π and π  are torsors, it follows from [9, Proposition 1.7], [35, Theorem 5.4] that there are exact sequences 0 → OY → E → L−1 → 0

(7.2.4)

0 → OY  → E  → M −1 → 0 where E = π∗ OX , E  = π∗ OX  , L = O(C) is a reflexive sheaf on Y and M = OY  (C  ) is an invertible sheaf on Y  . In the multiplicative case the sequences are split exact. Moreover, again from [9, Proposition 1.7], KX = π ∗ (KY + C)

(7.2.5)

KX  = (π  )∗ (KY  + C  )

(7.2.6)

From this and (7.2.3) it follows that π ∗ C = Δ. Also, since KX is ample and π a finite morphism, it follows that KY + C is ample too. Finally, from (7.2.5) and the fact that π is finite of degree 2 it follows that 2 KX = 2(KY + C)2 = 2KY · (KY + C) + 2C · (KY + C).

(7.2.7)

2 Claim. If Δ = 0, then KX ≥ 4. Hence, unlike the case when p ≥ 3, the possibility that Δ = 0 does not create any problems here.

I proceed to prove the claim. Suppose that Δ = 0. By its construction, π factors through the geometric Frobenius F : X → X (2) . In fact there is a commutative diagram

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(7.2.8)

Y π

ν F

X

X (2)

Since X (2) is smooth and Y is normal, then ν is a torsor over X (2) [9, Proposition 1.11]. Therefore, KY = ν ∗ (KX (2) + W (2) ) where W (2) is a divisor on X (2) . Recall that the geometric Frobenius is constructed from the next commutative diagram X F

Fab

X (2)

pr1

X

pr2

Spec(k)

π Fab

Spec(k)

where Fab is the absolute Frobenius. Since k is algebraically closed, pr1 is an isomorphism. Hence W (2) = pr1∗ W , where W is a divisor on X. Then ∗ KX = π ∗ KY = π ∗ ν ∗ (KX (2) + W (2) ) = F ∗ (KX (2) + W (2) ) = Fab (KX + W ) = 2KX + 2W. 2 Therefore KX = −2W and hence KX = 4W 2 ≥ 4, as claimed. Hence in the following we can assume the Δ = 0. As in the case of Theorems 6.1, 6.4 and 6.5, the proof of the Theorem 7.1 will be in several steps, according to the Kodaira dimension κ(Y  ) of Y  . 2 ≥ 3. Case 1. Suppose κ(Y ) = 2 or κ(Y ) = 1. In this case I will show that KX ∗ Since π C = Δ, it follows that 2C ∼ π∗ Δ. Hence 2C is equivalent to an effective divisor in Y . Therefore, since KY + C is ample, it follows that (KY + C) · C > 0. Hence since 2C is Cartier, it follows that 2(KY + C) · C ≥ 1. From the equations (7.2.2) it follows that

KY = g∗ φ∗ KZ + g∗ B.

(7.2.9)

From the classification of surfaces [5,4] it follows that since κ(Y  ) ≥ 1, then Z is either a surface of general type or has an elliptic or quasi-elliptic fibration over a curve B. In both cases, nKZ ∼ W , for n >> 1, where W is an effective divisor in Z. Moreover, W can be chosen so that φ∗ W is not contained in the exceptional locus of g. Therefore g∗ φ∗ KZ ≡ n1 g∗ φ∗ W = 0. Now considering that KY + C is ample and KY is Cartier, it

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follows from the equation (7.2.9) that KY ·(KY +C) ≥ 1. Hence from the equation (7.2.7) 2 ≥ 3, as claimed. it follows that KX 2 ≤ 2, then Case 2. Suppose that κ(Y  ) = 0. In this case I will show that if 1 ≤ KX et 2 X is unirational and π1 (X) = {1}. Moreover, if KX = 1, then X is a simply connected supersingular Godeaux surface. 2 = 1. The proof for the case I will only prove the statement for the case when KX 2 KX = 2 is similar with minor differences and it is left to the reader. Suppose that Y  in the diagram (7.2.1) is not minimal. Then B = 0. Moreover, since  Y is the minimal resolution of Y , g does not contract any −1 curves. However, since φ is not trivial and both Y  and Z are smooth, B contains −1 curves. Therefore g∗ B = 0. Since Z is a minimal surface of Kodaira dimension zero it follows that 12KZ = 0. Hence from the equations (7.2.2) it follows that KY  = φ∗ KZ + B ≡ B and therefore KY ≡ g∗ B = B  , where B  is a nonzero effective Cartier divisor. Hence since KY + C is ample it follows that KY · (KY + C) = B  · (KY + C) ≥ 1 Moreover, for the same reasons as in Case 1, 2C · (KY + C) ≥ 1. Therefore from (7.2.7) 2 ≥ 3. we get that KX 2 = 1. Then from Proposition 2.1 it Suppose now that Y  is minimal. Suppose that KX follows that 1 ≤ χ(OX ) ≤ 3 and b1 (X) = 0. Also from [9, Corollary 1.8] it follows that if 2 KX = 2 then 0 ≤ χ(OX ) ≤ 4. Hence in order to show the claim it suffices to show that the cases χ(OX ) ∈ {2, 3} are impossible, that π1et (X) = {1} and that X is unirational and not singular. Since b1 (X) = 0 it follows that b1 (Y  ) = b1 (Y ) = b1 (X) = 0. Hence Y  is either an Enriques or a K3 surface and hence c2 (Y  ) = 12, if Y  is Enriques, and 24 if it is K3. Now from the diagram (7.2.1) it follows that c2 (Y  ) = χet (Y  ) = χet (X  ) = c2 (X) + k,

(7.2.10)

where k is the number of f -exceptional curves. Suppose that χ(OX ) = 3. Then from the Noether formula we get that c2 (X) = 35 and from (7.2.10) that c2 (Y  ) = 35 + k > 24. Hence this case is impossible. Suppose that χ(OX ) = 2. Then from the Noether formula we get that c2 (X) = 23. Then c2 (Y  ) = c2 (X) +k = 23 +k. Hence the only possibility is that Y  is a K3 and k = 1. Then I claim that Y  has exactly one singular point which must be canonical of type A1 . Y has canonical singularities since KY  = 0. Let E and F be the f and g-exceptional curves, respectively. Both are smooth rational curves. Then since KY  = 0, it follows 2 that F 2 = −2 and so Y has exactly one A1 singular point (if KX = 2 then k = 2 and Y has canonical singularities whose minimal resolution has two exceptional curves. Then by [35, Proposition 3.6], Y has exactly two A1 singular points). Let Δ be the divisorial

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part of D and Δ the divisorial part of D , the lifting of D on X  . Then since Y  is K3, KX = Δ and KX  = Δ . Moreover, Δ = (π  )∗ M , where M is as in the equations (7.2.4). From (7.2.4) it follows that χ(M −1 ) = χ(E  ) − χ(OY  ) = χ(OX  ) − χ(OY  ) = 2 − 2 = 0. Now Riemann–Roch gives that 1 1 0 = χ(M −1 ) = χ(OY  ) + (M 2 + M · KY  ) = 2 + M 2 . 2 2 2  2 Hence M 2 = −4 and therefore, since Δ = (π  )∗ M , KX = −8. Since X is  = (Δ ) smooth and KX  Cartier, there is a positive a ∈ Z such that

KX  = f ∗ KX + aE.

(7.2.11)

Now E may or may not be an integral curve for D . If it is an integral curve, then (π  )∗ F = E and hence E 2 = −4. Then from (7.2.11) we get that 2 2 2 2 −8 = KX  = KX − 4a = 1 − 4a ,

(7.2.12)

which is impossible. Suppose that E is not an integral curve of D . Then 2E = (π  )∗ F and hence E 2 = −1. Then it is easy to see that X  is just the blow-up of X at a single 2 2 point. Hence KX  = KX − 1 = 0. But this is a contradiction since we have already found 2 that KX  = −8. Hence the case χ(OX ) = 2 is impossible too. Hence χ(OX ) = 1 and therefore X is a Godeaux surface. Next I will show that Y  is a K3 surface with a special configuration [33] of rank 13. Arguing similarly as before we find that c2 (Y  ) = c2 (X) + k = 11 + k,

(7.2.13)

where k is the number of f -exceptional curves. If Y  was Enriques, then c2 (Y  ) = 12 and 2 hence k = 1. Repeat the previous argument. Exactly as before we get that KX  = −4.  ∗ 2 Suppose that (π ) F = E. Then E = −4 and hence from (7.2.12) we get that −4 = 1 − 4a2 , which is impossible. If on the other hand (π  )∗ F = 2E, then E 2 = −1 and hence again from (7.2.12) we get that −4 = 1 − a2 and hence a2 = 5, which is again impossible since a ∈ Z. Hence Y  is a K3 surface and therefore c2 (Y  ) = 24. Then from (7.2.13) it follows that k = 13 and hence g has exactly 13 exceptional curves. Moreover, Y has canonical singularities and therefore by [35] they must be of type either A1 or D2n . Hence by [33], Y  has a special configuration of rank 13. Then Y  is unirational [33, Corollary 3.2] and hence X is unirational as well. Moreover, from the diagram (7.2.1) we get that π1et (X) = π1et (X  ) = π1et (Y  ) = π1et (Y  ) = {1}.

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Finally, I will show that X is a supersingular Godeaux surface. This means that pg (X) = h1 (OX ) = 1 and the induced map F ∗ of the Frobenius on H 1 (OX ) is zero. Since ωY ∼ = OY we get from duality for finite morphisms [11, Chapter III, Ex. 7.2] that ωX = π ! OY . Hence H 0 (ωX ) = HomY (π∗ OX , OY ). But this is nonzero. Indeed, by [35, Corollary 4.3] it follows that if D2 = D, then π∗ OX = OY ⊕ L−1 and hence in this case clearly H 0 (ωX ) = 0. If on the other hand D2 = 0, then the map φ : π∗ OX → OY defined by φ(a) = Da is nonzero and OY -linear and again this shows that H 0 (ωX ) = 0. Then since χ(OX ) = 1, it follows that pg (X) = h1 (OX ) = 1. Since h1 (OX ) = 1, X is either a singular or supersingular Godeaux surface. By the discussion at the end of Section 2, X is a singular Godeaux surface if and only if there exists a Z/2Z-torsor over X, i.e., if X has a nontrivial étale 2-cover. However this is impossible since π1et (X) = {1}. Therefore X is supersingular. Case 3. Suppose that κ(Y  ) = −∞. In this case I will show the following. 2 (1) If KX = 2, then X is uniruled. Moreover, if χ(OX ) ≥ 2, then X is unirational and et π1 (X) = {1}. 2 (2) If KX = 1 then X is unirational, simply connected and pg (X) ≤ 1. In particular 1 ≤ χ(OX ) ≤ 2. 2 2 (3) If D2 = D, then KX ≥ 2 and 0 ≤ χ(OX ) ≤ 1. In particular the case KX = 1 is impossible.

Proof of Case 3.1. Only the statement that if χ(OX ) ≥ 2, then X is unirational and π1et (X) = {1} needs some justification. From Lemma 6.3 it follows that if χ(OX ) ≥ 2, then b1 (X) = 0. Hence from the diagram (7.2.1) it follows that b1 (Z) = b1 (Y  ) = b1 (Y ) = b1 (X) = 0. Therefore Z is rational and simply connected. Hence from the diagram (7.2.1) it follows that Y is rational and therefore from the diagram (7.2.8) it follows that X (2) and hence X is unirational and simply connected. 2 Proof of Case 3.2. Suppose that KX = 1. Fix notation as in the diagram (7.2.1). Since  k(Y ) = −∞, then Z is ruled over a smooth curve B. From Lemma 6.3 it follows that b1 (X) = 0 and hence B ∼ = P1 . Hence X is unirational and for the same reasons as in et the previous cases, π1 (X) = π1et (Z) = {1}. Hence X is simply connected. It remains to show that pg (X) ≤ 1. By Proposition 2.1 it follows that pg (X) ≤ 2. Hence to show the claim it suffices to show that the case pg (X) = 2 is impossible. 2 Suppose that pg (X) = 2. For the same reasons as in Case 1, if Δ = 0 then KX ≥ 4. So we may assume that Δ = 0. Next I will study some properties of the linear system |KX |.

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Claim. (1) The linear system |KX | has a unique base point P . Moreover, every member of |KX | is smooth at P . ˜ ∈ |KY + C| such that (2) For any W ∈ |KX | there exists a smooth rational curve W ∗ ˜ π W = W. (3) Q = π(P ) is the unique singular point of Y . Moreover, it is of type A1 . 2 Note that Y is certainly singular since (KY + C)2 = 1/2KX = 1/2. Let W ∈ |KX | be any member. Then since KX is ample and W · KX = 1 it follows that W is reduced and irreducible. Moreover, for any two distinct C1 , C2 ∈ |KX |, C1 · C2 = 1 and their point of intersection is the base point P of |KX |. The proof of the first part of the claim is exactly the same as in the case when p ≥ 3 in Proposition 6.5 and for this reason it is omitted. Next I will show the second part of the claim. Let W ∈ |KX |. I will show that there ˜ ∈ |KY + C| such that π ∗ W ˜ = W . In the notation of the diagram (7.2.1), exists W 0 0  H (ωX  ) = H (ωX ). Let then W ∈ |KX  | be a lifting of W in X  . Then I will show ¯ ∈ |KY  + M | such that (π  )∗ W ¯ = W  . Then W ˜ = g∗ W ¯ ∈ |KY + C| that there exists W ∗ ˜ and π W = W . This is because all the previous equations hold over a codimension 2 open subset of Y (its smooth part) and so everywhere. Now from the equation (7.2.5) it follows that

H 0 (ωX  ) = H 0 ((π  )∗ (ωY  ⊗ M )) = H 0 (ωY  ⊗ M ⊗ π∗ OX  ) Then from the equation (7.2.4) we get the exact sequence 0 → ωY  ⊗ M → π∗ OX  ⊗ ωY  ⊗ M → ωY  → 0. This gives an exact sequence in cohomology 0 → H 0 (ωY  ⊗ M ) → H 0 (π∗ OX  ⊗ ωY  ⊗ M ) → H 0 (ωY  ) → · · · Now since Y  is rational, it follows that H 0 (ωY  ) = 0. Hence H 0 (ωY  ⊗ M ) = H 0 (π∗ OX  ⊗ ωY  ⊗ M ) ¯ ∈ |KY  + M | such that (π  )∗ W ¯ = W  . This concludes the and therefore there exists W proof of the second part of the claim. ˜ ∼ ˜ ∈ |KY + C|, there exists an exact sequence I will next show that W = P1 . Since W 0 → OY (−KY − C) → OY → OW ˜ → 0. This gives an exact sequence in cohomology 2 · · · → H 1 (OY (−KY − C)) → H 1 (OY ) → H 1 (OW ˜ ) → H (OY (−KY − C)) → · · ·

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I will show that H 1 (OY ) = H 2 (OY (−KY − C)) = 0. Therefore, H 1 (OW ˜ ) = 0 and hence 1 ∼ ˜ W = P . From the Leray spectral sequence we get the exact sequence 0 → H 1 (OY ) → H 1 (OY  ) → H 0 (R1 g∗ OY  ) → H 2 (OY ) Since Y  is rational it follows that H i (OY  ) = 0, i = 1, 2. Therefore H 1 (OY ) = 0. Next I will show that H 2 (OY (−KY − C)) = 0. By Serre duality for Cohen–Macaulay sheaves [17, Theorem 5.71], we get that H 2 (OY (−KY − C)) = H 0 (OY (2KY + C)). Suppose that H 0 (OY (2KY + C)) = 0. Then there exists a nonzero effective divisor Z ∈ |2KY + C|. Then KY + C = Z − KY . Therefore from (7.2.5) we get that 2 KX = 2(KY + C)2 = 2KY · (KY + C) + 2C · (KY + C) =

(7.2.14)

2Z · (KY + C) − 2KY · (KY + C). Since KY + C is ample and 2Z, 2C are Cartier, 2Z · (KY + C) ≥ 1 and 2C · (KY + C) ≥ 1 (since 2C is equivalent to π∗ Δ, which is effective). Suppose that KY · (KY + C) < 0. Then from the second equality of (7.2.14) it follows 2 that KX ≥ 2. Suppose that KY · (KY + C) > 0. Then from the first equality of (7.2.14) 2 it follows that KX ≥ 3. Suppose now that KY · (KY + C) = 0. Then π ∗ KY · KX = 0 and hence from the 2 adjunction KX = π ∗ KY + Δ we get that KX · Δ = KX = 1. Fix notation as in the diagram (7.2.1). Then from Proposition 4.4 we get that KX  · Δ = 4 (χ(OX  ) − 2χ(OY  )) . Since Y  is rational, χ(OY  ) = 1 and hence KX  · Δ = 4 (χ(OX  ) − 2)

(7.2.15)

If pg (X) = 2, then either χ(OX ) = 3 or χ(OX ) = 2. Suppose that χ(OX ) = 3. Then from (7.2.15) it follows that KX  · Δ = 4 > 1 = KX · Δ, which is impossible by Proposition 4.4. Suppose that χ(OX ) = 2. In this case, KX  · Δ = 0. Claim 7.3. Y has exactly one singular point which must be of type A1 . Indeed. From Proposition 4.4, KX · Δ decreases from X to X  . In order to show the claim I will study how exactly KX · Δ decreases. Since f is a composition of blow-ups of isolated singular points of D, it suffices to examine what happens after a single blow-up. So let f1 : X1 → X be the blow-up of an isolated singular point of D. Let D1 be the

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lifting of D on X1 , Δ1 its divisorial part and Y1 the quotient of X1 by D1 . Then there exists a commutative diagram X1

f1

(7.3.1)

X

π1

π g1

Y1

Y

Let E be the f1 -exceptional curve and F = π1 (E) the g-exceptional curve. Then, since KY is Cartier, there is a ∈ Z such that KY1 = g1∗ KY + aF. Moreover, KX = π ∗ KY + Δ. Hence from (7.3.1) we that KX1 = f1∗ KX + E = f1∗ π ∗ KY + f1∗ Δ + E = π1∗ KY1 + f1∗ Δ + (1 − ka)E where k ∈ {1, 2} is such that π1∗ F = kE. If E is an integral curve for D , then k = 1. Otherwise k = 2. Therefore, Δ1 = f1∗ Δ + (1 − ka)E.

(7.3.2)

From the proof of Proposition 4.4 it follows that 1 − ka ≥ 0. If 1 − ka > 1, then KX1 · Δ1 = KX · Δ − (1 − ka) ≤ KX · Δ − 2. But considering that KX · Δ = 1 and KX  · Δ = 0, this cannot happen. Hence 1 − ka ∈ {0, 1}. Suppose that 1 − ka = 1. Then a = 0 and KX1 · Δ1 = KX · Δ − 1, hence KX · Δ drops by one. Again considering that KX · Δ = 1 and KX  · Δ = 0, this case can happen only once. In all other cases, KX · Δ stays the same and hence 1 − ka = 0, i.e., a = 1/k (and hence k = 1 since a ∈ Z). In the notation then of the diagram (7.2.1), one can write KY  = h∗ KY +



ai Fi ,

i

such that ai ∈ Z and there is exactly one i such that ai = 0 and aj > 0 for all j = i. Hence Y has canonical singularities. Now consider the commutative diagram Y φ

Y 

g h

Y

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where g : Y  → Y is the minimal resolution. Then since h has exactly one crepant divisor, g has exactly one crepant divisor too. Therefore since Y has canonical singularities, it follows that KZ = g ∗ KY and g contracts exactly one smooth rational curve of selfintersection −2. Hence Y has exactly 1 singular point which must be of type A1 , as claimed. Then c2 (Y  ) = χet (Y ) + 1 = c2 (X) + 1. If χ(OX ) = 2, then c2 (X) = 23. Therefore, c2 (Y  ) = 24. However, since Y  is rational, χ(OY  ) = 1 and hence from the Noether formula KY2 = KY2  = 12 − 24 = −12. Moreover, since KX = π ∗ KY + Δ it follows that 2 −24 = 2KY2 = (KX − Δ)2 = KX + Δ2 − 2KX · Δ = −1 + Δ2

and hence Δ2 = −23. However, since KX · Δ = 1 and KX is ample, it follows that Δ is irreducible and reduced. But then from the genus formula 2pa (Δ) − 2 = KX · Δ + Δ2 = 1 − 23 = −22.

(7.3.3)

But this is impossible. Therefore KY · (KY + C) = 0 and hence H 0 (OY (2KY + C)) = 0, ˜ ∼ ˜ is smooth. as claimed. Hence W = P1 . In particular, W Next I will show that Q = π(P ), where P is the unique base point of |KX |, is the only singular point of Y . Indeed. Since P is the base point of |KX |, Q is the only base point of |KY + C|. Let W1 , W2 ∈ |KX | be two distinct members. Then as was shown earlier, there ˜ 2 ∈ |2KY + 2C|, and ˜ i ∈ |KY + C| such that Wi = π ∗ W ˜ i , i = 1, 2. Since W ˜1 + W are W ˜1 + W ˜ 2 is Cartier. Moreover, the local class groups of Y are 2-torsion, it follows that W ˜ ˜ since both W1 and W2 are smooth, it follows that Y is smooth everywhere except Q (as explained earlier Y is singular and therefore cannot be smooth at Q). Hence Y has exactly one singular point. Next I will show that Q ∈ Y is an A1 point. Let f1 : X1 → X be the blow-up of P . Let D1 be the lifting of D on X1 , Δ1 its divisorial part and Y1 the quotient of X1 by D1 . Then there exists a commutative diagram X1

f1

π1

Y1

(7.3.4)

X π

g1

Y

Let E be the f1 -exceptional curve and F = π1 (E) the g1 -exceptional curve. Then there is a ∈ Z such that KY1 = g1∗ KY + aF . I will show that Y1 is smooth, a = 0 and F 2 = −2.

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¯ i be the birational transforms of Wi in Y1 , i = 1, 2. Then I claim that Let W ˜i = W ¯ i + 1 F, g1∗ W 2

(7.3.5)

˜i = W ¯ i + mi F . Then clearly W  = π ∗ W ¯ for i = 1, 2. Indeed. Suppose that g ∗ W 1 i , where i  Wi are the birational transforms of Wi in X1 . Then, since Wi are smooth at P , ˜ i = π∗ g∗ W ˜ i = W  + mi π ∗ F. Wi + E = f1∗ Wi = f1∗ π ∗ W 1 i 1 Therefore, mi π1∗ F = E. This implies that m1 = m2 . Moreover, if π1∗ F = E, then m1 = m2 = 1. If on the other hand, π1∗ F = 2E, then m1 = m2 = 1/2. Suppose that π ∗ F = E. Then F 2 = −1/2. Also, ¯ i + F. ˜i = W g1∗ W

(7.3.6)

˜ 2 , it follows that W ¯ 2 + F and hence W ¯ 2 . Therefore, ˜1 ∼ W ¯1 + F ∼ W ¯1 ∼ W Since W ∼ ¯ ¯ ¯ ¯ ¯ i ), OY1 (W1 ) = OY1 (W1 ). Considering now that W1 ∩ W2 = ∅ it follows that both OY1 (W ¯ 1 and W ¯ 2 are Cartier. In addition, W ¯ 2 = 0, i = 1, 2. i = 1, 2, are invertible and hence W i But then from (7.3.6) it follows that ¯ i · F = −F 2 = 1/2 W ¯ i are Cartier. Hence π ∗ F = 2E and m1 = m2 = 1/2 which is impossible since W 1 and (7.3.5) holds. In particular, F 2 = −2. Next I will show that Y1 is smooth. From (7.3.5) it follows that ˜1 + W ˜ 2) = W ¯1 + W ¯ 2 + F. g1∗ (W ˜ 1 +W ˜ 2 ∈ |2KY +2C|, W ˜ 1 +W ˜ 2 is Cartier and hence W ¯ 1 +W ¯ 2 +F is Cartier as well. Since W ˜1 ∼ W ˜ 1 +W ˜ 2 , and are both Cartier, it follows that g ∗ (2W ˜ 1 ) ∼ g ∗ (W ˜ 1 +W ˜ 2 ). Also, since 2W 1 Hence ¯1 + F ∼ W ¯1 + W ¯2 + F 2W ¯1 ∼ W ¯ 2 . Then since the local Picard groups of the singularities of Y1 are and therefore W ¯ 1 is Cartier and hence F is also Cartier. Hence W ¯1 + W ¯ 2 is 2-torsion it follows that 2W ¯ ¯ ¯ also Cartier. However, W1 ∩ W2 = ∅. Hence both Fi are Cartier. Moreover, from (7.3.5) ¯ i · F = 1. From this, since both F and W ¯ i are Cartier and W ¯ i is smooth, it follows that W ¯ i . But since the W cover Y , it follows that F is smooth at the intersection point with W ¯ the W cover Y1 , and hence it follows that F is smooth (note that the map E → F is only birational and in principle F could be singular). Therefore, since F is Cartier and smooth, it follows that Y1 is smooth. Therefore since g1 contracts a −2 curve, Y has exactly one singular point which is of type A1 . Moreover, the diagram (7.3.4) is its resolution.

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Next I claim that KX · Δ + Δ2 + c2 (X) = 1.

(7.3.7)

Indeed. With notation as in (7.3.4), since Y1 is smooth, D1 has no isolated fixed points. Hence if Δ1 is its divisorial part, then from Proposition 4.3 KX1 · Δ1 + Δ21 + c2 (X1 ) = 0. But from (7.3.2) we get that Δ1 = f1∗ Δ + E. Now from this, the fact that c2 (X1 ) = c2 (X) + 1, the previous equation and some straightforward calculations we get (7.3.7). If pg (X) = 2 then χ(OX ) ∈ {2, 3}. 2 = 1, from the Noether formula we get that Suppose that χ(OX ) = 2. Then since KX c2 (X) = 23. Hence c2 (Y1 ) = χet (Y ) + 1 = c2 (X) + 1 = 24. Then since Y1 is rational, from the Noether formula for Y1 it follows that KY21 = 12χ(OY1 ) − 24 = 12 − 24 = −12 Hence since f1 is crepant, KY2 = KY21 = −12. From the adjunction formula KY = π ∗ KY + Δ we get that −24 = 2KY2 = (KX − Δ)2 = 1 + Δ2 − 2KX · Δ, and hence Δ2 − 2KX · Δ = −25.

(7.3.8)

However, since c2 (X) = 23, (7.3.7) gives also that KX · Δ + Δ2 = −22.

(7.3.9)

Now from (7.3.8) and (7.3.9) it follows that KX · Δ = 1 and Δ2 = −23. But now for the exactly the same reasons as in (7.3.3), this is impossible. Hence the case χ(OX ) = 2 is impossible. Suppose that χ(OX ) = 3. Arguing similarly as before we get that KX · Δ = 5

(7.3.10)

Δ = −39. 2

I will show that these relations are impossible by examining all possible cases for the structure of Δ as a cycle. Suppose that Δ=

k  i=1

ni Δi ,

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where Δi are distinct prime divisors. Since KX is ample and KX · Δ = 5, it follows that k ≤ 5 and there are the following possibilities. (1) Δ is reduced. Then Δ has at most 5 irreducible components. (2) Δ is not reduced. Then there are the following possibilities: (a) Δ = 2Δ1 + Δ2 + Δ3 + Δ4 , and KX · Δ1 = 1, for all i. (b) Δ = 2Δ1 + Δ2 + Δ3 , and KX · Δ1 = KX · Δ2 = 1, KX · Δ3 = 2. (c) Δ = 2Δ1 + Δ2 , and KX · Δ1 = 1, KX · Δ2 = 3, or KX · Δ1 = 2, KX · Δ2 = 1. (d) Δ = 2Δ1 + 2Δ2 + Δ3 , and KX · Δi = 1, for all i. (e) Δ = 4Δ1 + Δ2 , and KX · Δ1 = KX · Δ2 = 1. (f) Δ = 5Δ1 , KX · Δ1 = 1. (g) Δ = 3Δ1 + Δ2 + Δ3 , and KX · Δi = 1, for all i. (h) Δ = 3Δ1 + 2Δ2 , and KX · Δi = 1, for all i. The proof that the equations (7.3.10) are impossible will be by studying each one of the above cases separately. In all cases except (2)(h), there are no solutions to (7.3.10) under the restriction coming from the genus formula 2pa (Δi ) − 2 = KX · Δi + Δ2i . Next I will work the cases (1) and (2)(h). The cases (2)(a) to (2)(g) are treated in exactly the same way as (1) but (2)(h) needs some deeper geometric argument since in this case there is a numerical solution to (7.3.10). k Suppose then that Δ = i=1 Δi , i ≤ 5 is reduced. Then KX · Δ + Δ 2 =

k  i=1 k  i=1



(KX · Δi + Δ2i ) + 2

Δi · Δ j ≥

1≤i
(KX · Δi + Δ2i ) =

k  (2pa (Δi ) − 2) ≥ −2k ≥ −10, i=1

which is impossible since from (7.3.10), KX · Δ + Δ2 = −34. Suppose now that Δ = 3Δ1 + 2Δ2 , KX · Δ1 = KX · Δ2 = 1. Then from the genus formula it follows that KX · Δi + Δ2i ≥ −2 and hence Δ2i ≥ −3. Then it is easy to see that the only solutions to (7.3.10) are Δ2i = −3, i = 1, 2, and Δ1 · Δ2 = 0. In particular pa (Δi ) = 0 and hence Δi = P1 , i = 1, 2. ˜i , i = 1, 2, be the images of Δi in Y , with reduced structure. From the previous Let Δ discussion, X  is the blow-up of the unique isolated fixed point P of D and Y  the minimal resolution of the singularity Q = π(P ) ∈ Y . Moreover, Q ∈ Y is an A1 singular point and in particular KY is Cartier. Then KX = π ∗ KY + 3Δ1 + 2Δ2 . From this it follows that π ∗ KY · Δ1 = 10

(7.3.11)

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π ∗ KY · Δ2 = 7 From the projection formula we get that KY · π∗ Δ1 = 10 KY · π∗ Δ2 = 7. ˜ i or 2Δ ˜ i , depending on whether Δi is an integral Moreover, π∗ Δi is equal to either Δ curve for D or not. From the second equation and since KY is Cartier, it follows that ˜ 2 = 2Δ2 and ˜ 2 and hence Δ2 is not an integral curve for D. Hence π ∗ Δ π ∗ Δ2 = Δ 2 ˜ ˜2 ˜ ˜ ˜ 2 , then Δ therefore, Δ2 = −6 and KY · Δ2 = 7. I will now show that Q ∈ Δ2 . If Q ∈ /Δ is in the smooth part of Y . Hence from the genus formula again it follows that ˜ 2 ) − 2 = KY · Δ ˜2 + Δ ˜ 2 = 7 − 6 = 1, 2pa (Δ 2 ˜ 2 and therefore P ∈ Δ2 . Let C ∈ |KX | be any member. which is impossible. Hence Q ∈ Δ Then C · Δ2 = 1 and since P ∈ Δ2 , C  · Δ2 = 0, where C  and Δ2 are the birational transforms of C and Δ2 in X  . But the birational transforms C  form a base point free two dimensional linear system on X  and hence they define a morphism h : X  → P1 . This map is surjective since X  is projective and its fibers are the birational transforms C  of C ∈ |KX |. But then, since Δ2 · C  = 0, Δ2 must be contained in a fiber and since the fibers are irreducible and reduced curves, Δ2 is a fiber of h. But this implies that Δ2 ∈ |KX |. But this is impossible since it has genus zero and all members of |KX | have genus 2. Therefore the case (2)(h) is impossible too. 2 2 Proof of Case 3.3. Suppose that D2 = D. Suppose that either KX = 1 or KX ≥ 2 and χ(OX ) ≥ 2. In order to prove the claim it suffices to show that these cases are impossible. From Lemma 6.3 and its proof it follows that in both of the above cases b1 (X) = 0. Since κ(Z) = −∞, Z is ruled over a curve B. Since b1 (Z) = b1 (X) = 0, it follows that B∼ = P1k and hence both Y  , Y are rational and therefore X and X  are unirational.

Claim. X lifts to W2 (k), the ring of second order Witt vectors over k. Suppose that the claim is true. Since X lifts to W2 (k), then by Lemma 4.2 X has no nontrivial global vector fields and therefore Aut(X) is smooth. It remains to prove the claim. Recall that Y  is the quotient of X  by the μ2 action on X  induced by the lifting D of D on X  . Since both X  and Y  are smooth, it follows from [9, Proposition 1.11] that π  : X  → Y  is a torsor. In particular,

X  = SpecY  OY  ⊕ M −1 where M is an invertible sheaf on Y  and the ring structure on OY  ⊕ M −1 is induced by a section s of M ⊗2 . It is well known [12, Theorem 10.2] that H 2 (TY  ) is an obstruction space for deformations of X over local Artin rings and H 1 (OY  ), H 2 (OY  ) are

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obstruction spaces for deformations of line bundles and sections of line bundles on Y  [12, Theorem 6.4]. Since Y  is rational, all these spaces are zero. Therefore, Y  , M and the section s ∈ H 0 (M ⊗2 ) all lift compatibly to W2 (k). Let Y2 , M2 and s2 be the liftings of Y  , M and s, respectively. Then

X2 = Spec OY2 ⊕ M2−1 is a lifting of X  over W2 (k). Next I will show that X lifts over W2 (k) too. Let X2 be the ringed space (X, f∗ OX2 ). Then X2 is a deformation of X over W2 (k). Indeed. From the construction of the second order Witt vectors W2 (k) there exists an exact sequence [10] σ

0 → k → W2 (k) → k → 0, where σ(x) = x · p. Then tensoring with OX2 we get the exact sequence σ

0 → OX  → OX2 → OX  → 0 Applying f∗ we get the exact sequence 0 → f∗ OX  → f∗ OX2 → f∗ OX  → R1 f∗ OX  Considering that f∗ OX  ∼ = OX and that R1 f∗ OX = 0, we get the following exact sequence p

0 → OX → f∗ OX2 → OX → 0. Now tensoring with k over W2 (k) we get that OX2 ⊗W2 (k) k ∼ = OX . Finally from the infinitesimal criterion of flatness, X2 is flat over W2 (k) and hence X2 is a deformation of X over W2 (k), as claimed. This concludes the proof of Case 3.3 and Theorem 7.1. 2 8. Examples Example 8.1. A Godeaux surface X with π1et (X) = {1} has no nonzero global vector fields. This follows from Theorem 1.1. In particular Godeaux surfaces π1et (X) = Z/5Z do not have nonzero global vector fields. It is known [20] that a general Godeaux surface with π1et (X) = Z/5Z in any characteristic p = 5 is the quotient of a smooth quintic in P3 by a free action of Z/5Z. Then the fact that X has no global vector fields follows also from the fact that smooth hypersurfaces have no global vector fields. However, this proof only works for general X while Theorem 1.1 gives the result for any X. Example 8.2. Smooth canonically polarized surfaces which lift to characteristic zero or at least to W2 (k) have no vector fields. This follows from Lemma 4.2. In particular:

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Smooth complete intersections X = Xd1 ,...,dn−2 ⊂ Pn such that d1 + · · · + dn−2 ≥ n + 2 lift to W2 (k) and hence they have no nonzero global vector fields. The case of hypersurfaces in P3k of degree ≥ 5 was studied in [25] where it was shown by direct calculations using standard exact sequences of P3k that H 0 (TX ) = 0. Any Godeaux surface X with Picτ (X) = Z/5Z in characteristic 5 does not have nonzero global vector fields. Such a surface is simply connected and by [20] it is the quotient of a singular quintic hypersurface in P3k by a nontrivial μ5 action. The whole construction explained in [20] lifts to characteristic zero and therefore X has no nonzero global vector fields. This was proved in [20] only for general X. Next I give some examples of singular canonically polarized surfaces with canonical singularities, low K 2 and non-smooth automorphism scheme. The study of singular surfaces with canonical singularities is important because they appear naturally in the moduli problem of canonically polarized surfaces and in particular its compactification. 2 = 1 Example 8.3. This is an example of a canonically polarized surface X with KX defined over an algebraically closed field of characteristic 2 which has DuVal singularities such that Aut(X) is not smooth. Moreover, this surface lifts to characteristic zero and unlike smooth surfaces that lift to characteristic zero, it has vector fields. Let k be an algebraically closed field of characteristic 2 and Y be the weighted projective space Pk (2, 1, 1). Let

π : X = Spec (OY ⊕ OY (−5)) → Y be the 2-cyclic cover defined by OY (−5) and a general section s of OY (5)[2] = OY (10). 2 Then X has canonical DuVal singularities, KX is ample and KX = 1. Moreover, X is liftable to characteristic zero and has global vector fields of multiplicative type. From the theory of weighted projective spaces it follows that H 0 (L[2] ) = H 0 (OY (10)) = k[x0 , x1 , x2 ](10) , the space of homogeneous polynomials of degree 10 with weights, w(x0 ) = 2, w(xi ) = 1, i = 1, 2, 3. Let s = x50 + f (x1 , x2 ) ∈ k[x0 , x1 , x2 ](10) . Then it is not difficult to check by using the local description of the cyclic cover π that X has canonical DuVal singularities. 2 Next I will show that KX is ample and KX = 1. Indeed, by the construction of X as a 2-cyclic cover over Y , ωX = π ∗ (ωY ⊗ OY (5))∗∗ = π ∗ OY (1)∗∗ . From this it follows that ωX = π ∗ OY (2) and hence KX is ample. Moreover, [2]

2 KX =

1 1 2 [2] 1 1 1 1 [2] c1 (ωX ) = 2c21 (ωY ⊗ OY (10)) = c21 (OY (2)) = 4c21 (OY (1)) = · 4 · = 1. 4 4 2 2 2 2

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Finally since Y , OY (5) and s lift to characteristic zero, the construction of X as a 2-cyclic cover lifts also to characteristic zero. However, by its construction as a 2-cyclic cover, X has nontrivial global vector fields of multiplicative type. Also, since X has canonical singularities, its automorphism scheme Aut(X) is zero dimensional, and hence since X has vector fields, it is not smooth. Example 8.4. This is an example of a surface X defined over an algebraically closed field of characteristic p > 0 which has canonical singularities of type An such that 2 KX = (2p + 1)(2p − 3), X has nonzero global vector fields and moreover there is a flat morphism f : X → C, where C is a curve of finite type over k such that X = f −1 (s) for some s ∈ C and whose general fiber has no vector fields. Therefore the property smooth automorphism scheme is not deformation invariant and cannot be used to construct proper moduli stacks in positive characteristic. Let k be an algebraically closed field of characteristic p > 0 and X ⊂ P3k be the hypersurface given by xy(x2p−1 + y 2p−1 + z 2p−1 ) + zw2p = 0. It is not difficult to check that its singularities are locally isomorphic to xy + zf (x, y, z) = 0 and therefore they are canonical of type An . Moreover, the equation ∂ of X is invariant under the graded derivation D = z ∂w of k[x, y, z, w], which therefore induces a nonzero global vector field on X. X is smoothable by Xt given by



(1 − t) xy(x2p−1 + y 2p−1 + z 2p−1 ) + zw2p + t x2p+1 + y 2p+1 + z 2p+1 + w2p+1 = 0, t ∈ k. For t = 0, Xt is a smooth surface in P3k of degree 2p + 1 ≥ 5 and hence by Example 8.2 it has no global vector fields. Therefore Aut(Xt ) is smooth for t = 0. Finally I would like to mention that all existing examples in characteristic p = 2 are uniruled. However, N.I. Shepherd-Barron [33, Theorem 5.3] has constructed in characteristic 2 a smooth non-uniruled canonically polarized surface with nontrivial global vector fields. At the moment of this writing it is not known if non-uniruled examples exist in characteristic p = 2. Problem 8.5. Do there exist non-uniruled smooth canonically polarized surfaces with nontrivial global vector fields in characteristic p > 2? References [1] A. Aramova, L. Avramov, Singularities of quotients by vector fields in characteristic p > 0, Math. Ann. 273 (1986) 629–645. [2] M. Artin, On isolated rational singularities on surfaces, Amer. J. Math. 88 (1) (1962) 129–136. [3] L. Badescu, Algebraic Surfaces, Universitext, Springer, 2001. [4] E. Bombieri, D. Mumford, Enriques classification of surfaces in char.p, III, Invent. Math. 35 (1976) 197–232.

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