Advances in Mathematics 259 (2014) 89–115

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Advances in Mathematics

www.elsevier.com/locate/aim

Relative Noether inequality on fibered surfaces

Xinyi Yuan a,∗, Tong Zhang b

a Department of Mathematics, University of California, Berkeley, CA 94720, USAb Department of Mathematics, University of Alberta, Edmonton, AB T6G 2G1,Canada

a r t i c l e i n f o a b s t r a c t

Article history:Received 27 September 2013Accepted 18 March 2014Available online 1 April 2014Communicated by Ravi Vakil

Keywords:Noether inequalityAlgebraic surfaceFibered surfaceNef line bundleLinear systemHilbert–Samuel formulaSlope inequalitySeveri inequality

We prove effective upper bounds on the global sectionsof nef line bundles of small generic degree over a fiberedsurface over a field of any characteristic. It can be viewedas a relative version of the classical Noether inequality forsurfaces. As a consequence, we give a new proof of the slopeinequality for fibered surface without using any stabilitymethod. The treatment is essentially different from thoseof Xiao, Cornalba–Harris and Moriwaki. We also study thegeography problem of surfaces in positive characteristicsand show that the Severi inequality is true for surfaces ofgeneral type in positive characteristic whose Albanese map isgenerically finite. Moreover, the geography of surfaces withAlbanese fibrations is studied.

© 2014 Elsevier Inc. All rights reserved.

Contents

1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 901.1. Relative Noether inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 901.2. Idea of the proof: a new filtration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 921.3. New method towards the slope inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 931.4. Geography problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

2. Proof of the relative Noether inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 952.1. The reduction process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

* Corresponding author.E-mail addresses: yxy@math.berkeley.edu (X. Yuan), tzhang5@ualberta.ca (T. Zhang).

http://dx.doi.org/10.1016/j.aim.2014.03.0180001-8708/© 2014 Elsevier Inc. All rights reserved.

90 X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115

2.2. Linear systems on curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 982.3. Hyperelliptic case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1002.4. Non-hyperelliptic case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1022.5. Proof of Theorem 1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

3. Slope inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1033.1. Slope inequality for g(Y ) � 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1043.2. Slope inequality in positive characteristic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1053.3. Slope inequality in characteristic zero . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1063.4. Arakelov inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

4. Irregular surfaces in positive characteristic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1084.1. Proof of Severi inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1084.2. Surfaces with Albanese pencils . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112Appendix A. Base changes of regular varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

1. Introduction

This paper is the algebraic version of the previous work [24] of the authors on linearseries on arithmetic surfaces.

We prove effective upper bounds on the global sections of nef line bundles of smallgeneric degree over a fibered surface over a field of any characteristic. It can be viewedas a relative version of the classical Noether inequality for surfaces.

As a consequence, we give a new proof of the slope inequality for fibered surfacewithout using any stability method. The treatment is essentially different from those ofXiao, Cornalba–Harris and Moriwaki. We also study the geography problem of surfaces inpositive characteristics and show that the Severi inequality is true for surfaces of generaltype in positive characteristic whose Albanese map is generically finite. Moreover, thegeography of surfaces with Albanese fibrations is studied.

We would like to point out that most results in this paper, except the slope inequality,are new in positive characteristic. Nevertheless, we will state our results in full generality,since in characteristic 0 they still hold and Theorems 1.1 and 1.2 have not been statedyet in previous literatures.

1.1. Relative Noether inequality

Let k be an algebraically closed field of any characteristic. Let f : X → Y be a surfacefibration of genus g over k. That is:

(1) X is a smooth projective surface over k;(2) Y is a smooth projective curve over k;(3) f is flat and the general fiber F of f is a geometrically integral curve of arithmetic

genus g := pa(F ).

X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115 91

Note that here we do not assume the general fiber F to be smooth. The generic fiberis regular since X is regular. However, it is not necessarily smooth if char k > 0. See [13]and the appendix of the current paper.

We say f is a hyperelliptic fibration, if the general fiber F is hyperelliptic. Namely,there exists a flat morphism of degree 2 from F onto P1

k. Recall that this is preciselywhat “hyperelliptic” means in the non-smooth case. We refer to [4] for more details onembedded curves and surfaces. If the general fiber F is not hyperelliptic, f is called anon-hyperelliptic fibration. Note that in any case, F is a Cartier divisor on a smoothsurface, and thus, Gorenstein.

The following is the main theorem of this paper.

Theorem 1.1. Let f : X → Y be a surface fibration of genus g � 2 over k, and L be a nefline bundle on X. Denote d = deg(L|F ), where F is a general fiber of f . If 2 � d � 2g−2,then

h0(L) �(

14 + 2 + ε

4d

)L2 + d + 2 − ε

2 .

Here ε = 1 if F is hyperelliptic, d is odd and L|F � KF . Otherwise, ε = 0.

Note that this inequality is sharp. For example, let C be a smooth hyperelliptic curveover k and let f : C × P1

k → P1k be the trivial hyperelliptic fibration. Let L be the

pull-back of a divisor D on C satisfying 2h0(D) = degD+ 1. In this case, we obtain theequality with ε = 1.

The most interesting case of the theorem occurs when L is the relative dualizing sheafωf = ωX/Y = ωX/k ⊗ f∗ω∨

Y/k. We have the following theorem:

Theorem 1.2 (Relative Noether inequality). Let f : X → Y be a surface fibration of genusg � 2 over k. Assume that ωf is nef. Then

h0(ωf ) � g

4g − 4ω2f + g.

The nefness of ωf is closely related to the minimality of the fibration. For any surfacefibration f : X → Y of genus g � 2 over k, we say that f is relatively minimal ifX contains no (−1)-curves in fibers. It is known that if ωf is nef, then f is relativelyminimal. The converse is subtle. In [2,20], it is shown that if f is relatively minimal andthe generic fiber is smooth, then ωf is nef. But it could happen in positive characteristicsthat the generic fiber is not smooth. In this case, up to our knowledge, it is not knownthat ωf is nef in general.

There exist several examples where the equality holds. In [23], Xiao has constructed aclass of non-isotrivial hyperelliptic surface fibrations f : X → P1 of genus g over C suchthat deg(f∗ωf ) = g ω2

f . Since f∗ωf is a semipositive vector bundle of rank g over P1,

4g−4

92 X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115

it is actually a direct sum of g line bundles with non-negative degree. In particular,h1(f∗ωf ) = 0. Hence h0(ωf ) = h0(f∗ωf ) = deg(f∗ωf ).

The theorem can be viewed as a relative version of the classical Noether inequalityon algebraic surfaces. Recall that if X is a minimal surface of general type over k, andωX is the canonical bundle of X, then the Noether inequality asserts that

h0(ωX) � 12ω

2X + 2.

See [1] for char k = 0 and [13,12] for char k > 0. If the equality holds, then X has ahyperelliptic pencil (cf. [7]).

1.2. Idea of the proof: a new filtration

The idea of the proof of Theorem 1.1 is to construct a filtration. It comes from [24],where the arithmetic version of this filtration is very natural in the setting of Arakelovgeometry when studying the arithmetic linear series on arithmetic varieties. The newfiltration will serve as a substitute for the Harder–Narasimhan filtration.

Let L be a nef line bundle on X. We can find the smallest integer eL > 0 such thatL− eLF is not nef. Then we take L1 to be the movable part of |L− eLF |. Inductively,we get a filtration

L = L0 � L1 � · · · � Ln � 0.

It corresponds to a family of decompositions:

L = Li + Ni +i−1∑j=0

eLjF, i = 1, . . . , n.

Denote L′i = Li − eLi

F and we can easily compare the following two invariants:

h0(L′i

)− h0(L′

i+1), L′ 2

i − L′ 2i+1.

By taking the summation over i, we can prove Theorem 1.1.In [22], Xiao introduced a filtration to study the geometry of fibered surfaces, namely,

the Harder–Narasimhan filtration. We would like to point out that our constructionis essentially different from Xiao’s method. In fact, Xiao’s method relies heavily onthe (semi)stability of vector bundles on curves. Particularly, it is based on the factthat the pull-back of a semistable vector bundle by a finite map between curves is stillsemistable. This fact holds in characteristic zero, but fails in positive characteristics.For example, the Frobenius pull-back can destabilize a (semi)stable bundle. We refer to[10] for more details. This is one main obstruction for Xiao’s method to be availablein full generality. Another obstruction is that f∗ωf may not be semipositive in positive

X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115 93

characteristics. In [16], Moret-Bailly constructed a semistable family of genus 2 curvesover P1 in characteristic p > 0 such that f∗ωf = OP1(p) ⊕OP1(−1).

On the other hand, our filtration is constructed by some basic properties of linearseries without any stability consideration. Hence it can be applied regardless of thecharacteristic of the base field. It even works very well for arithmetic surfaces in [24],where the arithmetic versions of Theorems 1.1 and 1.2 are proved.

1.3. New method towards the slope inequality

An interesting consequence of our new filtration and relative Noether inequality isthat, it gives a new perspective of the slope inequality. More precisely, we obtain astability-free proof of the slope inequality.

The slope inequality asserts that

ω2f � 4g − 4

gdeg(f∗ωf )

for a relatively minimal fibration of genus g � 2. In fact, all the known proofs are relatedto some sort of stability: geometric invariant theory in [5] (in characteristic 0), stabilityof vector bundle in [22] (in characteristic 0), Chow stability in [17].

Our treatment is completely different from any of the above. We find that in positivecharacteristics,

(relative Noether) + (Frobenius iteration) =⇒ (slope inequality).

Using this idea, we only need to apply our relative Noether inequality to the Frobeniuspull-back of the original fibration and take the limit. The slope inequality in positivecharacteristics follows easily. As a result, we prove the Arakelov inequality in positivecharacteristics, which improves a result of [15].

It turns out that this idea can even be applied to study the similar problem in char-acteristic 0. In this case, the strategy can be expressed in short:

(slope ineq. in pos. char.) + (reduction mod℘) =⇒ (slope ineq. in char. zero).

Namely, in order to get the slope inequality in characteristic 0, it is enough to prove itin positive characteristics.

It should be pointed out that in characteristic 0, our filtration may not be as strongas the Harder–Narasimhan filtration in [22]. Nevertheless, when g(Y ) � 1, our methodcan recover the original slope inequality. When g(Y ) � 2, our filtration might be slightlyweaker. But at least for semistable fibrations, the relative Noether inequality worksperfectly. Therefore, our method is definitely promising for studying the moduli space ofcurves.

94 X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115

1.4. Geography problems

Another related topic is the geography problem for irregular surfaces in positive char-acteristics.

The geography problem asks for the region of points in Z2 which are equal to(c21(X), c2(X)) for certain minimal surface X. We refer to [1] for details over C. However,this problem turns out to be very subtle in positive characteristics. For example, c2 canbe negative for surfaces of general type in positive characteristics, and the positivity ofχ is still unknown. We refer to [13,19] for details.

We study this problem in positive characteristics. For example, we prove the Severiinequality in positive characteristics, i.e.,

ω2X/k � 4χ(OX)

for smooth minimal surface X of general type whose Albanese map is generically finite.In this situation, we know that χ(OX) > 0 (cf. [13,19]). When k = C, it was proved byPardini [18] using a clever covering trick and the slope inequality of Xiao [22].

It is worth mentioning that both the result and the method are new. The behaviorof the Albanese map in positive characteristics is not as good as in characteristic 0. See[14] or [13, Theorem 8.7] for example, where the Albanese map is purely inseparable.The covering trick in [18] is still available, but due to the failure of the Bertini theoremfor base-point-free line bundle in positive characteristic [11], the usual slope inequalityfor fibrations with smooth general fiber is not enough for here. In particular, Moriwaki’sresult [17] cannot be applied in this case. Also, as we have mentioned before, Xiao’smethod cannot be applied here, either.

Nevertheless, our starting point is again different. We find that

(relative Noether) + (covering trick) =⇒ (Severi inequality).

In this sense, the slope inequality is no longer needed in the proof, even in characteristic 0.In fact, the idea works uniformly in all characteristics.

It is also worth mentioning that our idea works for arbitrary dimension. Just applyingthis idea, the higher-dimensional Severi inequality in characteristic 0 is proved recentlyby the second author in [25]. While, the slope inequality for higher-dimensional fibrationis still far from being known.

Based on the same idea, we also study the case when the Albanese map of X inducesa fibration. We prove that for smooth minimal surface X of general type in positivecharacteristics with non-trivial Albanese map, we have the following results:

(1) ω2X/k � 2χ(OX);

(2) If 2χ(OX) � ω2X/k < 8χ(OX)/3, then the Albanese map of X induces a pencil of

genus 2 curves;

X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115 95

(3) If 8χ(OX)/3 � ω2X/k < 3χ(OX), then the Albanese map of X induces a pencil of

genus 2 curves or hyperelliptic curves of genus 3.

The corresponding results over C were proved by Bombieri [3] for (1) and Horikawa [8]for (2) and (3). In particular, in [8], the slope inequality was essentially used.

Our proof of the result is again by the relative Noether formula. Simply speaking, weuse

(relative Noether) + (base change)

to prove these results. It also works in characteristic 0.

2. Proof of the relative Noether inequality

In this section, we will prove Theorems 1.1 and 1.2. The idea comes from [24], wherea very similar argument was used to prove an effective Hilbert–Samuel formula for arith-metic surfaces.

2.1. The reduction process

We first fix our notations. Let f : X → Y be a surface fibration, and F be a generalfiber. For any nef line bundle L on X, we can find a unique integer c such that

• L− cF is not nef;• L− c′F is nef for any integer c′ < c.

Denote this number c for L by eL. Since L itself is nef, we get eL > 0.We have the following lemma.

Lemma 2.1. Let f : X → Y be a surface fibration, and F be a general fiber. Let L be anef line bundle on X such that LF > 0 and h0(L − eLF ) > 0. Then the linear system|L− eLF | has a fixed part Z > 0. Moreover,

ZF > 0.

Proof. By our assumption, we can write

|L− eLF | = |L1| + Z,

where |L1| is the movable part, and Z is the fixed part. Since L− eLF is not nef, thereexists an irreducible and reduced curve C on X such that

(L− eLF )C < 0,

96 X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115

which implies Z � C > 0. Moreover, LC � 0 since L is nef. Therefore, FC > 0 andZF � FC > 0. �

We have the following general theorem, which will replace the Harder–Narasimhanfiltration.

Theorem 2.2. Let f : X → Y be a surface fibration, and F be a general fiber. Let L

be a nef line bundle on X such that LF > 0 and h0(L) > 0. We can get the followingsequence of triples

{(Li, Zi, ai): i = 0, 1, . . . , n

}where Li is nef, Zi is effective, and ai > 0 for all i such that

• (L0, Z0, a0) = (L, 0, eL) and ai = eLifor i � 0;

• We have

|Li−1 − ai−1F | = |Li| + Zi,

where Li (resp. Zi) is the movable part (resp. fixed part) of the linear system |Li−1−ai−1F | for 0 < i < n;

• We have h0(Li − aiF ) > 0 for i < n and h0(Ln − anF ) = 0;• LF = L0F > L1F > · · · > LnF � 0.

Proof. The triple (Li+1, Zi+1, ai+1) can be obtained by applying Lemma 2.1 to the triple(Li, Zi, ai). The whole process will terminate when h0(Li−aiF ) = 0. It always terminatesbecause by Lemma 2.1, LiF decreases strictly. �

Now, we denote

ri := h0(Li|F ),

di := LiF,

L′i := Li − aiF.

We have the following numerical inequalities:

Proposition 2.3. For any j = 0, 1, . . . n, we have

h0(L0) � h0(L′j

)+

j∑i=0

airi;

L20 � 2a0d0 +

j∑i=1

ai(di−1 + di) − 2d0.

X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115 97

Proof. For any m ∈ Z>0, we have the following short exact sequence:

0 −→ H0(Li+1 −mF ) −→ H0(Li+1 − (m− 1)F)−→ H0(Li+1|F ).

Then it follows that

h0(Li+1 − (m− 1)F)− h0(Li+1 −mF ) � h0(Li+1|F ) = ri+1.

Summing over m = 1, . . . , ai+1, we have

h0(Li+1) − h0(Li+1 − ai+1F ) � ai+1ri+1,

or equivalently,

h0(L′i

)− h0(L′

i+1)� ai+1ri+1.

Note that both L′i + F and L′

i+1 + F are nef, and Zi is effective. We get

L′ 2i − L′ 2

i+1 =(L′i + L′

i+1)(L′i − L′

i+1)

=(L′i + L′

i+1)(ai+1F + Zi+1)

= ai+1(L′i + L′

i+1)F +

[(L′i + F

)+(L′i+1 + F

)− 2F

]Zi+1

� ai+1(di + di+1) − 2Zi+1F

� ai+1(di + di+1) − 2(di − di+1).

For any j = 0, 1, . . . , n− 1, summing over i = 0, 1, . . . , j, we have

h0(L′0)� h0(L′

j

)+

j∑i=1

airi;

L′ 20 � L′ 2

j +j∑

i=1ai(di−1 + di) − 2(d0 − dj).

Since we still have

L20 − L′ 2

0 = 2a0d0,

L′ 2j + 2dj =

(L′j + F

)2 � 0,

h0(L0) � h0(L′0)

+ a0r0,

the result follows. �We also have the following lemma.

98 X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115

Lemma 2.4. In the above setting, we have

2a0 +n∑

i=1ai − 2 � L2

0d0

.

Proof. Denote b = a1 + · · · + an and Z = Z1 + · · · + Zn. We have the following linearequivalence

L′0 = L′

n + bF + Z.

Since L′0 + F and L′

n + F are both nef, it follows that

(L′

0 + F)2 =

(L′

0 + F)(L′n + F + bF + Z

)�

(L′n + F + bF + Z

)(L′n + F

)+ bd0

�(L′n + F

)2 + b(d0 + dn).

Combine with

L20 −

(L′

0 + F)2 = 2(a0 − 1)d0.

We get

L20 �

(L′n + F

)2 + 2(a0 − 1)d0 + b(d0 + dn) � d0(2a0 + b− 2). �2.2. Linear systems on curves

We need several results on linear systems on algebraic curves.Let C be an integral and Gorenstein curve over k. We say a line bundle L is special if

h0(L) > 0, h1(L) > 0.

We have the following theorem.

Theorem 2.5. Let C be an integral and Gorenstein curve over k, pa(C) � 2. Let L be aline bundle on C such that h0(L) > 0 and deg(L) � 2pa(C) − 2.

(1) [Clifford’s Theorem] If L is special, then

h0(L) � 12 deg(L) + 1.

Moreover, if C is not hyperelliptic, then the equality holds if and only if L = OC orL = ωC/k.

X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115 99

(2) If h1(L) = 0, then

h0(L) � 12 deg(L).

Proof. If L is special, then the theorem is just the generalized version of Clifford’s the-orem in [12]. If h1(L) = 0, by the Riemann–Roch theorem,

h0(L) = deg(L) − pa(C) + 1 � 12 deg(L). �

We also need the following lemma.

Lemma 2.6. Let L be a special line bundle on a hyperelliptic curve C over k such that|L| is base-point-free, then deg(L) is even.

Proof. Denote dL = deg(L). Since L is base-point-free, the complete linear system |L|induces a morphism φL : C → Ph0(L)−1. Thus we can choose D1, D2 ∈ |L| to be thepull-back of two general hyperplane sections such that they have no points in common.Consider the following morphism:

φ = (φ1, φ2) : C → P1 × P1.

Here φ1 is the degree two map from C to P1 by the definition of hyperelliptic curves,and φ2 is the degree dL map from C to P1 induced by D1 and D2. It is easy to see thateither φ is birational from C to φ(C) or degφ = 2.

If degφ = 2, then we are done. Furthermore, we claim that φ cannot be birational toits image.

If φ is birational, denote C ′ = φ(C). By the definition of φ, we can write C ′ ∈|dLF1 + 2F2|, where F1 and F2 are rulings on P1 × P1. Note that L = φ∗OC′(F2). Wehave the following short exact sequence on C ′:

0 −→ OC′(F2) −→ φ∗L −→ E −→ 0,

where E is a skyscraper sheaf. Hence we have a surjection

H1(OC′(F2))� H1(φ∗L).

Since L is special and φ is finite, we get h1(OC′(F2)) � h1(φ∗L) = h1(L) > 0.On the other hand, we have another exact sequence

0 −→ OP1×P1(−C ′ + F2

)−→ OP1×P1(F2) −→ OC′(F2) −→ 0,

which gives

H1(OP1×P1(F2))−→ H1(OC′(F2)

)−→ H2(OP1×P1

(−C ′ + F2

)).

100 X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115

Now by the Serre duality,

h2(OP1×P1(−C ′ + F2

))= h0(OP1×P1

((dL − 2)F1 − F2

))= 0.

Moreover, using the Riemann–Roch formula on P1 × P1, we get

h1(OP1×P1(F2))

= 0,

which forces h1(OC′(F2)) = 0, a contradiction. Thus, φ cannot be birational, and thelemma follows. �

We will divide the proof of Theorem 1.1 into two parts.

2.3. Hyperelliptic case

We first prove Theorem 1.1 when F is hyperelliptic.By Proposition 2.3 for j = n, we get

h0(L0) � h0(L′n

)+

n∑i=0

airi =n∑

i=0airi,

L20 � 2a0d0 +

n∑i=1

ai(di−1 + di) − 2d0.

We claim that if Li|F is not special, neither is Li−1|F . In fact, if Li is not special,then by the Serre duality, h0(ωF − Li|F ) = h1(Li|F ) = 0. As a result, h1(Li−1) =h0(ωF − Li−1|F ) = 0. In particular, Li−1|F is not special.

By the above claim, we only need to deal with the following two different cases:

• Each Li|F is special;• There exists an i0 > 0 such that L0|F , . . . , Li0−1|F are not special and Lk|F , . . . , Ln|F

are special.

First, we assume that each Li|F is special. If d0 is even, applying Clifford’s theorem,we have

ri �12di + 1.

On the other hand, since F is sufficiently general, and the linear system |Li| has nofixed part for i = 1, . . . , n, the line bundle Li|F is base-point-free by construction. ByLemma 2.6, di is even. Thus

X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115 101

di−1 − di � 2

for i = 1, . . . , n. By Lemma 2.4, we have

h0(L0) −14L

20 �

n∑i=0

ai −14

n∑i=1

(di−1 − di)ai + 12d0

� a0 + 12

n∑i=1

ai + 12d0

� 12d0

L20 + 1

2d0 + 1.

If d0 is odd, the only differences are

r0 � 12d0 + 1

2 , d0 − d1 � 1.

Therefore by Lemma 2.4 again,

h0(L0) −14L

20 �

n∑i=0

ai −14

n∑i=1

(di−1 − di)ai + 12d0 −

12a0

� 12a0 + 3

4

n∑i=1

ai + 12d0

� 34d0

L20 + 1

2d0 + 12 .

Now let us deal with the other case. From Theorem 2.5, we know

ri �12di, i = 0, . . . , i0 − 1,

and

ri �12di + 1, i = i0, . . . , n.

Similarly, we have

di−1 − di � 1, i = 0, . . . , i0,

and

di−1 − di � 2, i = i0 + 1, . . . , n.

102 X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115

Thus we have

h0(L0) −14L

20 �

n∑i=i0+1

ai −14

n∑i=i0+1

(di−1 − di)ai + 12d0

+(ri0 −

12di0 −

14(di0−1 − di0)

)ai0

� 12

n∑i=i0+1

ai + 12d0 +

(ri0 −

12di0 −

14(di0−1 − di0)

)ai0 .

If di0−1 − di0 � 2, we have

h0(L0) −14L

20 � 1

2

n∑i=i0

ai + 12d0.

If di0−1 − di0 = 1, we know that ri0 � ri0−1 − 1. So

ri0 � ri0−1 − 1 � 12di0−1 − 1 = 1

2di0 + 12 .

Hence we still have

h0(L0) −14L

20 <

12

n∑i=i0

ai + 12d0.

By Lemma 2.4 again,

h0(L0) �(

14 + 1

2d0

)L2

0 + 12d0 + 1 − a0 −

12

i0−1∑i=1

ai

�(

14 + 1

2d0

)L2

0 + 12d0.

2.4. Non-hyperelliptic case

In this case, for i = 1, . . . , n− 1, we have the following stronger Clifford’s theorem:

ri �12di + 1

2 .

For i = 0 or i = n, it also holds if Li|F is neither trivial nor ωF/k.

X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115 103

First, we assume Ln|F is not trivial. Using the strong bound and Lemma 2.4, we get

h0(L0) −14L

20 � a0 + 1

2

n∑i=1

ai −14

n∑i=1

ai(di−1 − di) + 12d0

� a0 + 12

n∑i=1

ai + 12d0

� 12d0

L20 + 1

2d0 + 1.

If Ln|F is trivial, then rn = 1 and dn = 0. Note that dn−1 − dn > 2 since F is nothyperelliptic. It gives

h0(L0) −14L

20 � a0 + 1

2

n−1∑i=1

ai + an − 14

n∑i=1

ai(di−1 − di) + 12d0

< a0 + 12

n∑i=1

ai + 12d0.

By Lemma 2.4 again, it yields

h0(L0) �(

14 + 1

2d0

)L2

0 + 12d0 + 1.

It ends the proof of Theorem 1.1.

2.5. Proof of Theorem 1.2

Now, Theorem 1.2 becomes straightforward. Let f : X → Y be a surface fibrationof genus g � 2. Then the relative dualizing sheaf ωf = ωX ⊗ f∗ω∨

Y is a line bundleand ωfF = 2g − 2 is even. By our assumption, we know that ωf is nef. Just applyingTheorem 1.1, we can directly get Theorem 1.2.

3. Slope inequality

In this section, we prove the following slope inequality.

Theorem 3.1 (Slope inequality). Let f : X → Y be a semistable fibration of genus g � 2over k. Then

ω2f � 4g − 4

gdeg(f∗ωf ).

104 X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115

Note that in our definition of semistability, a rational curve in a geometric fiber isrequired to intersect other curves at two or more points. By a result of Tate [21], thegeneric fiber of f is smooth under the assumption of the theorem, since it is alreadyregular. See Corollary A.2 of Appendix A of our paper. Hence, ωf is nef.

As mentioned in the introduction, the case of characteristic zero was proved by [5]and [22], and the case of positive characteristic was proved by [17]. Here we derive thetheorem as a consequence of Theorem 1.2.

By the Riemann–Roch theorem, Theorem 1.2 implies the slope inequality with an“error term”. To get rid of the “error term”, we consider a sequence of base changes ofY and take the limit. The argument is straightforward if g(Y ) � 1. In general, the basechange trick only works in positive characteristics. To get the result in characteristic 0,we use reduction modulo ℘.

3.1. Slope inequality for g(Y ) � 1

We first give the following consequence of the relative Noether inequality.

Lemma 3.2. Let f : X → Y be a surface fibration of genus g � 2 such that ωf is nef.Then

deg(f∗ωf ) � g

4g − 4ω2f + gb.

Here b = g(Y ).

Proof. By Theorem 1.2, we have

h0(ωf ) � g

4g − 4ω2f + g.

On the other hand, using the Riemann–Roch theorem on Y , we get

h0(ωf ) = h0(f∗ωf ) � deg(f∗ωf ) + g(1 − b).

It follows that

deg(f∗ωf ) � g

4g − 4ω2f + gb. �

In the case g(Y ) = 0, the lemma gives the same inequality as Theorem 3.1. In the caseg(Y ) = 1, the lemma implies Theorem 3.1 with an extra argument. For convenience, weinclude the following result.

X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115 105

Lemma 3.3. Let f : X → Y be a semistable fibration over k. Let Y ′ → Y be a finite mor-phism of degree m. Let X ′ be the minimal desingularization of the base change X ×Y Y ′,and denote by f ′ : X ′ → Y ′ the composition morphism. Then

deg(f ′∗ωf ′

)= m deg(f∗ωf ), ω2

f ′ = mω2f .

Proof. The results follow from the basic fact that ωf ′ is the pull-back of ωf via themorphism X ′ → X. �

Now we can deduce Theorem 3.1 in the case g(Y ) = 1. Note that Y is an elliptic curveover k (by fixing an identity point). Let μ : Y → Y be the multiplication by a positiveinteger n indivisible by char k. Denote X ′ = X ×μ Y , which is already smooth since μ isétale. Applying Lemma 3.2 to the base change f ′ : X ′ → Y , we have

deg(f ′∗ωf ′

)� g

4g − 4ω2f ′ + g.

By Lemma 3.3, it becomes

n2 deg(f∗ωf ) � n2g

4g − 4ω2f + g,

i.e.,

deg(f∗ωf ) � g

4g − 4ω2f + g

n2 .

Setting n → ∞, the theorem follows.In the case g(Y ) � 2, if we directly use a separable base change π : Y ′ → Y , then

g(Y ′) also increases. The new idea is to take a reduction to positive characteristics, anduse pull-back via the Frobenius map in positive characteristics.

3.2. Slope inequality in positive characteristic

It is easy to prove Theorem 3.1 for char k = p > 0. Let π : Y ′ → Y be the k-linearFrobenius morphism of degree pn, and let X ′ be the minimal desingularization of X×Y Y

′.Applying Lemma 3.2 to f ′ : X ′ → Y , we have

deg(f ′∗ωf ′

)� g

4g − 4ω2f ′ + gb.

By Lemma 3.3, it becomes

deg(f∗ωf ) � g

4g − 4ω2f + gb

pn.

The theorem is obtained by setting n → ∞.

106 X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115

Using the same idea, we can get a more general result for line bundles of small degreein positive characteristics, which is similar to Theorem 3.1.

Theorem 3.4. Let f : X → Y be a surface fibration of genus g > 0 over a field k of positivecharacteristic, and L be a nef line bundle on X. Assume that 2 � d = deg(L|F ) � 2g−2,where F is a general fiber of f . Then

L2 � 4dd + 2 + ε

deg(f∗L).

Here, ε = 1 if F is hyperelliptic, d is odd and L|F � KF . Otherwise, ε = 0.

Proof. The proof is the same as the proof of the slope inequality in positive characteristic.We just sketch here.

First, by Theorem 1.1, we have

h0(L) �(

14 + 2 + ε

4d

)L2 + d + 2 − ε

2 .

The Riemann–Roch theorem on Y gives us

h0(L) = h0(f∗L) � deg(f∗L) + r(1 − b),

where b = g(Y ) and r = h0(L|F ). Combine them together and we get

deg(f∗L) �(

14 + 2 + ε

4d

)L2 + d + 2 − ε

2 + r(b− 1).

Now we apply the Frobenius base change iteration as above. Finally, we eliminate theconstant term. �3.3. Slope inequality in characteristic zero

Now we can prove the slope inequality for char k = 0. We first introduce the followinglemma.

Lemma 3.5. Let X , Y, Z be integral schemes, and f : X → Y and g : Y → Z beproper and flat morphisms of relative dimensions one. Assume that f is a local completeintersection and g is smooth. Then the numbers

deg fz∗(ωXz/Yz) and ω2

Xz/Yz

are independent of z ∈ Z. Here fz : Xz → Yz denotes the fiber of f : X → Y over z, andωXz/Yz

denotes the relative dualizing sheaf of Xz over Yz.

X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115 107

Proof. The invariance of deg fz∗(ωXz/Yz) is an interpretation of the determinant line

bundle. Recall that for any line bundle L on X , the determinant line bundle λf (L) is aline bundle on Y such that, for any y ∈ Y, there is a canonical isomorphism

λf (L) � detH∗(Xy,Ly) = detH0(Xy,Ly) ⊗ detH1(Xy,Ly)∨.

The construction is functorial.In the setting of the lemma, consider the determinant line bundle M = λf (ωX/Y).

Restricted to Yz, we have

M|Yz= (det fz∗ωXz/Yz

) ⊗(detR1fz∗ωXz/Yz

)∨ = det fz∗ωXz/Yz.

Here we used the canonical isomorphism R1fz∗ωXz/Yz= OYz

following the duality the-orem. Therefore, we simply have

deg fz∗ωXz/Yz= deg(M|Yz

).

It is independent of z.The invariance of ω2

Xz/Yzfollows from the definition of the Deligne pairing introduced

in [6]. In fact, the Deligne pairing N = 〈ωX/Y , ωX/Y〉 is a line bundle on Y such that

ω2Xz/Yz

= deg(N|Yz), ∀z ∈ Z.

It is also independent of z. �Remark 3.6. The proof does not work for arbitrary line bundles since it uses the dualitytheorem.

Now we are ready to prove Theorem 3.1 in characteristic zero by that in positivecharacteristics.

We first introduce a basic fact. Let f : X → Y be as in the theorem. Let X◦ → Y

be the (unique) stable model of the generic fiber of X → Y . Then X is the minimaldesingularization of X◦, and ωX/Y is just the pull-back of ωX◦/Y via X → X◦. Hence,the slope inequality is true for X → Y if and only if it is true for X◦ → Y .

Go back to the proof of Theorem 3.1. By the Lefschetz principle, we can assume thatk is finitely generated over Q. Let Z be an integral scheme of finite type over Z withfunction field k, and let Y → Z be a projective and flat morphism of integral schemesextending Y → Spec k. By the stable reduction theorem, replacing Y by a modificationif necessary, we can assume that the generic fiber of X → Y has a stable model X → Yover Y. Note that a generic desingularization can make the generic fiber Y of Y smooth.The replacement amounts to replacing Y by a finite (possibly ramified) cover and X◦

by the base change by this cover, which does not change the slope inequality. By the

108 X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115

uniqueness of the stable model, the preimage of Y in X is exactly X◦. Replacing Z by anopen subscheme if necessary, we can assume that Y → Z is smooth (and still projective).

In one word, we have obtained an extension X → Y → Z of X → Y → Spec(k)satisfying the conditions of the lemma. Now choose a closed point ℘ ∈ Z. Then theresidue field of ℘ has a positive characteristic. By Lemma 3.5, the slope inequality forX℘ → Y℘ → ℘ implies that for X → Y → Spec k.

3.4. Arakelov inequality

As a direct consequence of the slope inequality, we have the following Arakelov in-equality in positive characteristic.

Theorem 3.7 (Arakelov inequality). Let f be a non-isotrivial semistable fibration of genusg � 2 over the field k of positive characteristic. Let S ⊂ Y be the singular locus overwhich f degenerates. Then

deg(f∗ωf ) < g2 degΩY (S).

Proof. Let f be a non-isotrivial semistable fibration of genus g � 2 over k. Let S ⊂ Y bethe singular locus over which f degenerates. We have the following Szpiro’s inequality(cf. [20, Prop. 4.2]):

ω2f < (4g − 4)g degΩY (S).

Therefore,

deg(f∗ωf ) � g

4g − 4ω2f < g2 degΩY (S). �

4. Irregular surfaces in positive characteristic

In this section, we always assume that X is a smooth minimal surface of general typeover a field k of positive characteristic.

4.1. Proof of Severi inequality

We have the following theorem.

Theorem 4.1. Suppose the Albanese map of X is generically finite. Then

ω2X � 4χ(OX).

Before the proof, we first show a lemma.

X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115 109

Lemma 4.2. Let X be a smooth surface of general type over k. Let B be a big and nef linebundle on X such that |2B| is base point free. Then for any nef line bundle L � ωX +2Bsatisfying d = 2LB � 2, we have

h0(L) �(

14 + 1

2d

)L2 + d + 2

2 .

Proof. By Bertini’s theorem in [11], we can choose two general member B1, B2 ∈ |2B|such that B1, B2 are both geometrically integral and intersect each other transversally.Now, let σ : X̃ → X be the blow-up of X along B1 ∩B2 and we get a fibration from X̃

to P1 with the general fiber B̃ being the proper transform of Bi. Apply Theorem 1.1 tothis fibration and note that d is even. It follows that

h0(L) = h0(σ∗L)�

(14 + 1

2d

)(σ∗L

)2 + d + 22 =

(14 + 1

2d

)L2 + d + 2

2 . �Proof of Theorem 4.1. Let X be a minimal surface of general type over k with maximalAlbanese dimension. Denote AlbX : X → A to be the Albanese map, where A = Alb(X)is an abelian variety of dimension m over k. We only have m � h1,0(X) (cf. [9]), butsince X is of maximal Albanese dimension, we have m � 2 in any case.

Let H ′ be a very ample line bundle on A, and L be the pull-back of H = 2H ′ on X.Set

α = L2, β = ωXL.

Since X is of general type, α and β are both strictly positive.Let μ : A → A be the multiplication by n, where n > 1 is an integer. If char k = p > 0,

we assume n and p to be coprime. We have the following base change:

X ′ τ

ν

X

AlbX

Aμ

A

where X ′ = X ×μ A. We have

ω2X′ = n2mω2

X , χ(OX′) = n2mχ(OX).

We also have the following numerically equivalence on A:

μ∗H ∼num n2H,

which yields

τ∗L ∼num n2L′.

110 X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115

Here L′ = ν∗H. It follows that

L′ 2 = n2m−4α, ωX′nL′ = n2m−2β.

Now, apply Lemma 4.2 to X ′. It follows that

χ(OX′) � h0(ωX′n) �

(14 + 1

2n2m−2β

)ω2X′

n+ n2m−2β + 2

2 .

Hence we have

(14 + 1

2n2m−2β

)n2mω2

X + n2m−2β + 22 � n2mχ(OX).

Therefore, the results follows by letting n → ∞. �4.2. Surfaces with Albanese pencils

Here we treat the following more general case.

Theorem 4.3. Suppose χ(OX) > 0 and the Albanese map of X is non-trivial. Then thefollowing are true:

(1) ω2X � 2χ(OX);

(2) If 2χ(OX) � ω2X < 8χ(OX)/3, then the Albanese map of X induces a pencil of

genus 2 curves;(3) If 8χ(OX)/3 � ω2

X < 3χ(OX), then the Albanese map of X induces a pencil ofgenus 2 curves or hyperelliptic curves of genus 3.

Before the proof, we recall that over complex numbers, we have h1,0 = h0,1 = 12b1 and

that X has non-trivial Albanese map if and only if one (and thus all) of these numbersis non-zero. In positive characteristics, we only have h0,1 � 1

2b1 and h1,0 � 12b1, and a

non-trivial Albanese map is equivalent to b1 �= 0.

Proof of Theorem 4.3. We can assume ω2X < 3χ(OX), else there is nothing to show. Then

from Theorem 4.1, we know that the Albanese map of X induces a pencil. Passing tothe Stein factorization, we denote this pencil by f : X → Y , where Y is a smooth curvesatisfying b = g(Y ) > 0, the general fiber F of f is geometrically integral of arithmeticgenus g = pa(F ) � 2.

Let A = Jac(Y ) and μ : A → A be the multiplication by a positive integer n coprimeto char k. Let X ′ = X ×μ A. We have the following diagram:

X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115 111

X ′ τ

ν

X

f

Y ′ μY

Aμ

A

Apply Theorem 1.1 to ωX′ . We can get

χ(OX′) � h0(ωX′) � g

4g − 4ω2X′ + g.

It yields

n2bχ(OX) � n2bg

4g − 4ω2X + g.

By letting n → ∞, it follows that

ω2X � 4g − 4

gχ(OX).

Therefore, from our assumptions, (1) and (2) are proved and g � 3. We devote the restpart to proving (3) by steps.

Step 1. Fix a very ample line divisor H on A. Write D = f∗H and D′ = ν∗H. Weclaim that ωX′ � D′ when n is sufficiently large. It implies that ν factors through thecanonical map φωX′ of X ′.

In fact, we have the following exact sequence

0 −→ H0(ωX′ −D′) −→ H0(ωX′) −→ H0(ωX′ |D′).

From the étale cover, we know that

h0(ωX′) � χ(OX′) = n2bχ(OX).

Also, from the proof of Theorem 4.1, we see that

ωX′D′ = n2b−2ωXD.

By adjunction, ωX′ |D′ is a special divisor on D′. Hence by Clifford’s theorem,

h0(ωX′ |D′) � n2b−2ωXD.

In particular, when n is sufficiently large, h0(ωX′) > h0(ωX′ |D′). It follows that

h0(ωX′ −D′) > 0.

112 X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115

Step 2. We can further assume that

ω2X′ < 3χ(OX′) − 10.

In fact, by our assumption,

ω2X � 3χ(OX) − 1.

When n > 3, we will have

ω2X′ = n2bω2

X � n2b(3χ(OX) − 1)

= 3χ(OX′) − n2b < 3χ(OX′) − 10.

Step 3. We claim that ν is the only possible hyperelliptic pencil on X ′. Otherwise,X ′ has another hyperelliptic pencil which is not ν. Denote the general member in thispencil by G. Then ν(G) = Y ′. On the other hand, since G is hyperelliptic, φωX′ (G) = P1,where φωX′ is the canonical map of X ′. This is impossible, because by Step 1, ν factorsthrough φωX′ .

Step 4. Now we are ready to prove (3). Assume that F is not hyperelliptic, other-wise there is nothing to prove. Denote by F ′ a general fiber of ν. Note that under thisassumption, F ′ is not hyperelliptic, either. By Step 3, X ′ has no hyperelliptic pencil.

If dimφωX′ (X ′) = 1, since ν factors through φωX′ , we can write

ωX′ ∼alg aF ′ + Z,

where F ′ is a general fiber of ν, Z is the fixed part of |ωX′ |, a � h0(ωX′). Also note thatpa(F ′) = 3. Hence

ω2X′ � aωX′F ′ � 4h0(ωX′) � 4χ(OX′) − 4.

If dimφωX′ (X ′) = 2, since X ′ has no hyperelliptic pencil, X ′ cannot be a double coverof ruled surfaces. By the Castelnuovo inequality in [13],

ω2X′ � 3h0(ωX′) − 7 � 3χ(OX′) − 10.

In any case, it contradicts Step 2. It finishes the proof. �Acknowledgments

The authors would like to thank Yanhong Yang for the communication explaining thesubtlety of the stability in positive characteristics. They are grateful to Xi Chen for alot of valuable help and suggestions, and also Huayi Chen, Jun Lu, Sheng-Li Tan, HangXue, Kang Zuo for their interests on this paper. Special thanks goes to Christian Liedtke

X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115 113

for his careful reading of the earlier version of this paper, his valuable comments and hisunpublished preprint [14].

The first author is supported by the grant DMS-1161516 of the National ScienceFoundation. The second author is supported by a discovery grant of the Natural Sciencesand Engineering Research Council of Canada.

Appendix A. Base changes of regular varieties

For a variety X over a field k, it is well known that being smooth implies beingregular, and vice versa if k is perfect. However, the inverse direction fails if k is notperfect. Similarly, there is a regular X such that the base change Xk̄ to the algebraicclosure k̄ is not regular.

For example, let k = Fp(t) for a prime p > 2, and let

X = Spec k[x, y]/(x2 − yp + t

).

Then X is regular at the maximal ideal (x) = (x, yp − t), but Xk̄ is not regular at themaximal ideal (x, y − t1/p) (of its affine ring).

The following result is essentially attributed to Tate [21] by many authors. We includea proof here since we could not find it in [21].

Theorem A.1. Let k be any field, X be a regular and geometrically integral variety over k,and X ′

k̄be the normalization of Xk̄. Then the natural morphism X ′

k̄→ Xk̄ is a homeo-

morphism between the underlying topological spaces.

Proof. First, Xk′ is regular for any finite separable extension k′ of k. In fact, the mapXk′ → X is étale, since it is a base change of the map Spec k′ → Spec k. Hence, Xk′ isregular by the assumption that X is regular.

Second, Xks is regular for the separable closure ks of k. Let P be any closed pointof Xks . Then there is a subfield k′ of ks finite over k, such that under the mapπ : Xks → Xk′ , the preimage π−1(π(P )) exactly consists of the single point P . Thenthe maximal ideal of the local ring OXk′ ,π(P ) generates the maximal ideal of OXks ,P .Hence the regularity of OXk′ ,π(P ) implies that of OXks ,P .

If k is of characteristic 0, then k̄ = ks and X ′k̄

= Xk̄ is regular (and thus smooth).Assume that k is of characteristic p > 0 in the following.

Replacing k by ks if necessary, we can assume that k is separably closed. We canalso assume that X = SpecR for an affine ring R over k. Denote by X ′

k̄= SpecR′

k̄the

normalization of Xk̄ = SpecRk̄. We have R ⊂ Rk̄ ⊂ R′k̄.

Third, we claim that (R′k̄)q ⊂ Rk̄ for some integer power q = pr of p. Let x ∈ R′

k̄be

any element. It suffices to prove xq ∈ Rk̄ for some power q since R′k̄

is finite over Rk̄.By definition, x satisfies a monic polynomial f(t) ∈ Rk̄[t]. Since k̄ is purely inseparableover k, there is a power q of p such that f(t)q ∈ R[t] and that xq lies in the fraction

114 X. Yuan, T. Zhang / Advances in Mathematics 259 (2014) 89–115

field of R. The key is that xq satisfies the monic polynomial g(t) ∈ R[t] defined byg(tq) = f(t)q. We conclude that xq ∈ R ⊂ Rk̄ since R is normal.

Finally, the map SpecR′k̄→ SpecRk̄ is a homeomorphism by (R′

k̄)q ⊂ Rk̄ ⊂ R′

k̄. �

As a consequence, we have the following basic result.

Corollary A.2. Let X and k be as in the theorem. If furthermore dimX = 1, then Xk̄

does not contain any node.

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