SSURFACES WITH PRYM-CANONICAL HYPERPLANESECTIONS
MARTINA ANELLI
Abstract.
In this paper, we will explicit some general properties regardingsurfaces with Prym-canonical hyperplane sections and the geometric genus oftheir possible singularities. Moreover, we will construct new examples of thistype of surfaces. Introduction
Let g ≥
3. It is well-known that a
Prym curve is a pair (
C, α ), where C is asmooth genus g = p g ( C ) curve and α is a non-zero 2 − torsion point of Pic ( C ).In the following, we will consider the so called Prym-canonical map , that is therational map φ | ω C ( α ) | : C (cid:57)(cid:57)(cid:75) P g − defined by | ω C ( α ) | . In general, the pair ( C, ω C ( α )) is called Prym-canonical curve .The complete linear system | ω C ( α ) | is base point free unless C is hyperelliptic and α (cid:39) O C ( p − q ), with p and q ramification points of the g . Moreover, it defines anembedding if and only if C does not have a g such that α ∼ O C ( a + b − x − y ),where 2( a + b ) and 2( x + y ) are members of the g (see [3], Lemma 2 . φ | ω C ( α ) | is an embedding, we say that C (cid:39) φ ( C ) ⊂ P g − is a Prym-canonical (embedded)curve . If g <
5, then the Prym-canonical map cannot be an embedding, as observedin [3], so we will work only with g ≥ surface X has Prym-canonical hyperplane sections if it can be bi-rationally realized in some projective space P g − , for g ≥
5, such that a generalhyperplane section C of X is a smooth Prym-canonical (embedded) curve of genus g .In this paper we will analyze the complex projective surfaces with Prym-canonicalhyperplane sections up to birational equivalence. In particular, Section 2 is devotedto studying the first properties regarding surfaces with Prym-canonical hyperplanesections. We will show that these surfaces can be birationally equivalent to ruledsurfaces or to P or to Enriques surfaces. In any case, there is only one effectiveantibicanonical divisor W (cid:48) on X (cid:48) , the minimal resolution of the singularities of X ,and, if π : X (cid:48) → X , then the antibicanonical divisor of X (cid:48) is contracted by π andevery singularity x ∈ X such that π − ( x ) does not meet supp( W (cid:48) ) is a rationaldouble point. We will also show that surfaces with Prym-canonical hyperplanesections birationally equivalent to non-rational ruled surfaces over a base curve ofgenus q > q .At the best of our knowledge, the only known examples of surfaces with Prym-canonical hyperplane sections are the Enriques surfaces and a surface in P of degree10 obtained as image of the blowing up of P in the 10 nodes of an irreduciblerational plane curve of degree 6. In Section 3 we will construct new examples ofthese surfaces. In particular, we will construct four new examples, one birationallyequivalent to an elliptic ruled surface, another birationally equivalent to a ruled a r X i v : . [ m a t h . AG ] F e b MARTINA ANELLI surface over a base curve of genus q ≥
3, again another one birationally equivalentto a rational ruled surface and finally, we will construct a new example of surfacewith Prym-canonical hyperplane sections birationally equivalent to P . Acknowledgements.
The results of this paper are contained in my PhD-thesis.I would like to express my deepest gratitude to my three advisors, Ciro Ciliberto,Concettina Galati and Andreas Leopold Knutsen, for their useful and indispensableadvice and I would also like to acknowledge PhD-funding from the Departmentof Mathematics and Computer Science of the University of Calabria and fundingfrom Research project ”Families of curves: their moduli and their related varieties”(CUP E81—18000100005, P.I. Flaminio Flamini) in the framework of Mission Sus-tainability 2017 - Tor Vergata University of Rome.2.
Preliminary results
We recall that a surface X ⊆ P g − is a surface with Prym canonical hyperplanesections if its general hyperplane section C is a Prym canonical embedded curve.We start with the following remarks. Remarks 2.1. • The generic hyperplane section of X is irreducible and smooth, whence X has at most isolated singularities. • Let π : X (cid:48) → X be the minimal resolution of singularities of X and let C (cid:48) = π ∗ C be the inverse image of a general hyperplane section. For a general C (cid:48) ∈ | π ∗ C | , we have that C (cid:48) ∼ = C and O C (cid:48) ( C (cid:48) ) ∼ = O C (1) ∼ = ω C (cid:48) ( α ), with α a non trivial two-torsion element of Pic ( C (cid:48) ). By the adjunction formulaand because X (cid:48) is smooth, we can say that α = − K X (cid:48) | C (cid:48) , in particular K X (cid:48) · C (cid:48) = 0.From now on, we will assume that C is projectively normal with respect to itsembedding in P g − . For example, if φ | ω C ( α ) | : C (cid:44) → P g − is a Prym-canonicalembedding and the Clifford index Cliff( C ) ≥
3, then C is projectively normal withrespect to the given embedding (see Theorem 1, [10]). In general, there are alsoprojectively normal curves C ⊆ P g − with Cliff( C ) < Theorem 2.1.
Let X be a surface with Prym-canonical hyperplane section C ofgenus g ≥ and let π : X (cid:48) → X be the minimal resolution of its singularities. If C is projectively normal with respect to its embedding in P g − , then: • h ( O X ( n )) = 0 and h ( O X ( n )) = 0 for any n ≥ , in particular h ( O X ) = 0 and h ( O X ) = 0 , whence p a ( X ) = 0 ; • X is projectively normal; • the Kodaira dimension κ ( X (cid:48) ) equals to −∞ or ; • deg( X ) = 2 g − .Proof. • By assumption, C is a projectively normal curve in P g − , thus H ( O P g − ( n )) (cid:16) H ( O C ( n )) , for any n ≥
0. As a consequence the map H ( O X ( n )) → H ( O C ( n )) issurjective for any n ≥ → O X ( n − → O X ( n ) → O C ( n ) → . URFACES WITH PRYM-CANONICAL HYPERPLANE SECTIONS 3
Then, the second part of the long exact sequence associated with (1) is0 → H ( O X ( n − → H ( O X ( n )) → H ( O C ( n )) → (2) → H ( O X ( n − → H ( O X ( n )) → . By Serre’s Theorem, there is a sufficiently large n such that h ( O X ( n )) = 0,for any n ≥ n . From the exact sequence (2) and applying descendinginduction on n , we obtain that h ( O X ( n )) = 0, for any n ≥ H ( O C ( n )) = H ( O C ( n ( K C + α ))). For n = 1, we have that h ( O C ( K C + α )) = h ( O C ( − α )) = 0. Moreover, because deg( K C + α ) =2 g −
2, then deg( n ( K C + α )) > g − n ≥
2, so h ( O C ( n ( K C + α ))) =0 (see [11], Example IV. . . n , from the long exact sequence (2) we can concludethat h ( O X ( n )) = 0, for any n ≥ • To prove that X is projectively normal, it is enough to show that X isnormal and that the map H ( O P g − ( n )) → H ( O X ( n )) is surjective for any n ≥ η : (cid:101) X → X be the normalization of X . We consider the following exactsequence on X :(3) 0 → O X → η ∗ O (cid:101) X → F → , where supp( F ) ⊂ Sing( X ). Since X has isolated singularities, then F ∼ = H ( F ) = ⊕ si =1 ( (cid:102) O i / O i ), where O i is the local ring of x i on X and( η ∗ O (cid:101) X ) x i = (cid:102) O i is the normalization of O i in the function field of X , withSing( X ) = { x , ..., x s } .We know that H ( O X ) = k because X is irreducible. By the proper-ties of pushforward and because (cid:101) X is still irreducible, it is obvious that H ( η ∗ O (cid:101) X ) ∼ = H ( O (cid:101) X ) ∼ = k . Moreover h ( O X ) = 0 by the previous partof this Proposition. For the long exact sequence associated with (3), wehave that h ( F ) = 0. By definition of F , it is true that (cid:102) O i ∼ = O i , for any i = 1 , ..., s . We conclude that X is normal.The surjectivity of H ( O P g − ( n )) → H ( O X ( n )) is trivial for n = 0. Let usconsider the following diagram, where H is a general hyperplane in P g − and C = X ∩ H :0 H ( O P g − ( n − H ( O P g − ( n )) H ( O H ( n )) 00 H ( O X ( n − H ( O X ( n )) H ( O C ( n )) 0 r r r We observe that r is surjective because C is projectively normal and r issurjective by the inductive hypothesis. Then r is also surjective and theclaim is proved. • Consider the following exact sequence:0 → O X (cid:48) ( − C (cid:48) + mK X (cid:48) ) → O X (cid:48) ( mK X (cid:48) ) → O C (cid:48) ( mK X (cid:48) )) → . Since − K X (cid:48) | C (cid:48) ∼ α , for α a non-zero two torsion element of C (cid:48) , then we havethat ( − C (cid:48) + mK X (cid:48) ) · C (cid:48) = − C (cid:48) = 2 − g <
0. Whence h ( O X (cid:48) ( − C (cid:48) + mK X (cid:48) )) = 0 otherwise, if this divisor was effective, it wouldbe a fixed component of | C (cid:48) | that is a linear system without base locus bydefinition. At the same time MARTINA ANELLI h ( O C (cid:48) ( m ( K X (cid:48) ))) = (cid:26) if m odd if m even. Consequently the plurigenus P m ( X (cid:48) ) := h ( O X (cid:48) ( mK X (cid:48) )) ≤ h ( O C (cid:48) ( mK X (cid:48) )) ≤
1. Then the Kodaira dimension κ ( X (cid:48) ) = −∞ or 0. • We have deg( X ) = deg( C ) = C = deg( K C + α ) = 2 g − (cid:3) Remarks 2.2. (1) Since the Kodaira dimension is a birational invariant for smooth varieties,if κ ( X (cid:48) ) = −∞ , then the minimal model X (cid:48)(cid:48) of X (cid:48) is a ruled surface or P (see [11], Theorem V. . κ ( X (cid:48) ) = 0, then X (cid:48) is a minimal Enriques surface. Let us show this.Let us suppose that X (cid:48) is not minimal. Then there is a ( − − curve E (cid:48) on X (cid:48) . Because κ ( X (cid:48) ) = 0, let m > | mK X (cid:48) | contains onlyone effective divisor D (cid:48) . It is obvious that E (cid:48) is a component of D (cid:48) . Now O C (cid:48) ( D (cid:48) ) ∼ = O C (cid:48) ( mK X (cid:48) ) ∼ = O C (cid:48) because − K X (cid:48) | C (cid:48) is a non-zero two torsionelement and m is even as seen in the previous Proposition. Therefore D (cid:48) and consequently E (cid:48) are contracted to a point on X by π , contradicting theminimality of the resolution π .It is true that 12 K X (cid:48) ∼ X (cid:48) is minimal and κ ( X (cid:48) ) = 0 (see [11],Theorem V. . m ≥ mK X (cid:48) ∼ m = { , , , , } (see[6] and [4]). If m = 1, then K X (cid:48) ∼
0, whence O C (cid:48) ( C (cid:48) ) ∼ = O C (cid:48) ( K C (cid:48) − K X (cid:48) ) ∼ = O C (cid:48) ( K C (cid:48) ). This is not possible because C (cid:48) is a Prym-canonical curve. Sowe exclude the cases in which X (cid:48) is a K X (cid:48) was a hyperelliptic surface, it would not contain curves with nega-tive self-intersection, then we would have X = X (cid:48) smooth by Mumford’sTheorem (see [14], Chapter 1). By definition, a hyperelliptic surface is ir-regular, contradicting the first point of Proposition 2.1. In conclusion X (cid:48) isa minimal Enriques surface.We want to determine the possible singularities on X surface with Prym-canonicalhyperplane sections. The following Proposition determines the geometric genus ofthe singularities that occur on X . Proposition 2.3.
With the same assumptions as before, if
Sing( X ) = { x , ..., x s } is the locus of the singular points of X , then: • if X is birationally equivalent to an Enriques surface or P , then X canonly contain rational points as singularities; • if X is birationally equivalent to a ruled surface X (cid:48)(cid:48) over a base curve ofgenus q ≥ , then (cid:80) si =1 p g ( x i ) = q , where p g ( x i ) is the geometric genus ofthe singular point x i .Proof. These results can be obtained using the following exact sequence, which onegets from the Leray spectral sequence for the sheaf O X (cid:48) and the morphism π (see[9], pag. 462):0 → H ( O X ) → H ( O X (cid:48) ) → H ( R π ∗ O X (cid:48) ) → H ( O X ) → ... We know that h ( O X ) = h ( O X ) = 0 by Theorem 2.1, so s (cid:88) i =1 p g ( x i ) = h ( R π ∗ O X (cid:48) ) = h ( O X (cid:48) ) , where p g ( x i ) is the geometric genus of the singular point x i . URFACES WITH PRYM-CANONICAL HYPERPLANE SECTIONS 5 • If X is birationally equivalent to an Enriques surface, then X (cid:48) is a mini-mal Enriques surface by Remark 2.2. So h ( O X (cid:48) ) = 0 by definition and (cid:80) si =1 p g ( x i ) = 0, whence X can only contain rational singularities.If X is birationally equivalent to P , then it is clear that h ( O X (cid:48) ) = 0 and (cid:80) si =1 p g ( x i ) = 0. This proves the first part of this Proposition. • If X is birationally equivalent to a ruled surface X (cid:48)(cid:48) over a base curve ofgenus q ≥
0, then h ( O X (cid:48)(cid:48) ) = q ( X (cid:48)(cid:48) ) = q . Consequently q ( X (cid:48)(cid:48) ) = q ( X (cid:48) ) = q .So (cid:80) si =1 p g ( x i ) = h ( O X (cid:48) ) = q . (cid:3) The following results give other information regarding the singularities that occuron X . First of all, we sketch the proof of a preliminary lemma. Lemma 2.4.
Let X be a surface with Prym-canonical hyperplane section C and let π : X (cid:48) → X be the minimal resolution of singularities of X . Then π ∗ (2 K X (cid:48) ) ∼ .Proof. Let C m ∈ | mC | be smooth satisfing C m ∩ Sing( X ) = ∅ . We put C (cid:48) m = π ∗ ( C m ). We have that − K X (cid:48) · C (cid:48) m ∼ − mK X (cid:48) · C (cid:48) ∼ A ( X (cid:48) ). Since rational equivalence and linear equivalence coincide on a curve,then − K X (cid:48) | C (cid:48) m ∼
0, for any m ≥ π ∗ (2 K X (cid:48) ) | C m ∼
0. Indeed, since π is an isomorphism ina neighbourhood of C (cid:48) m , then π ∗ O C (cid:48) m ∼ = O C m and O C (cid:48) m ∼ = π ∗ ( O C m ). Using theprojection formula (see [11], Exe II. .
1) and the previous results, we obtain that O C m ∼ = π ∗ ( O C (cid:48) m ) ∼ = π ∗ (2 K X (cid:48) ⊗ O C (cid:48) m ) = π ∗ (2 K X (cid:48) ⊗ π ∗ O C m ) ∼ = π ∗ (2 K X (cid:48) ) ⊗ O C m .By a known result of Zariski ([18], Theorem 4), if π ∗ (2 K X (cid:48) ) | C m ∼ m , thenthere exists a divisor D such that D ∼ π ∗ (2 K X (cid:48) ) and D | C m ∼
0. Since C m is veryample on X , then D ∼
0. So π ∗ (2 K X (cid:48) ) ∼
0. This proves the lemma. (cid:3)
Theorem 2.2.
The dimension dim |− K X (cid:48) | = 0 , in particular, if W (cid:48) is the effectiveantibicanonical divisor on X (cid:48) , then either W (cid:48) ∼ or supp( W (cid:48) ) = π − ( x , ..., x r ) for certain singularities x i ∈ X , for i = 1 , .., r .Proof. Since π ∗ (2 K X (cid:48) ) ∼ K X (cid:48) ∼ K X (cid:48) on X (cid:48) with support in π − (Sing( X )). In the lattercase, let 2 K X (cid:48) = (cid:80) m i F i − (cid:80) n j G j be the decomposition in reduced and irreduciblecomponents, with m i , n j ∈ N > and F i (cid:54) = G j , for all i, j . Let F = (cid:80) m i F i and G = (cid:80) n j G j .Suppose F (cid:54) = 0. By Mumford’s Theorem (see [14], Chapter 1), we have that F i < i and, because the intersection form on π − (Sing( X )) is negative definite,also F <
0. So there is an i such that F · F i <
0. Up to renaming the index,we suppose that i = 1. It is obvious that F · G ≥
0. Since F is an irreduciblecomponent, then0 ≤ p a ( F ) = 1 + 12 F · ( F + K X (cid:48) ) = 1 + 12 F + 14 F · F − F · G. The only possibility is F = −
1. Thus F is a ( − − curve, contradicting theminimality of π . Hence F = 0.Thus either 2 K X (cid:48) ∼ π − (Sing( X )). Then dim | − K X (cid:48) | = 0.Let | − K X (cid:48) | = { W (cid:48) } . To conclude the proof, we have only to show that, if x ∈ Sing( X ) is such that π − ( x ) meets supp( W (cid:48) ), then π − ( x ) does not containcurves which are not part of supp( W (cid:48) ). MARTINA ANELLI
Suppose that there is an irreducible curve E ⊂ π − ( x ) which is not part of supp( W (cid:48) ).Since X is normal by Theorem 2.1, Point 2 . , then π − ( x ) is connected, so we canassume that E intersects W (cid:48) and E · W (cid:48) >
0. Then0 ≤ p a ( E ) = 1 + 12 E + 12 E · K X (cid:48) = 1 + 12 E − E · W (cid:48) . Again by Mumford’s Theorem, we have E <
0. So the only possible case for whichthe previous inequality is valid is: E · W (cid:48) = 2, E = − p a ( E ) = 0. Thiscontradicts the minimality of π . (cid:3) Remark 2.5.
By the previous Theorem, we observe that, if X is smooth, then X = X (cid:48) and W (cid:48) ∼
0. Since p a ( X ) = p g ( X ) = 0 by Theorem 2.1, Point 1 . , then X is an Enriques surface by [11], Theorem V. . Lemma 2.6. If W (cid:48) is the unique effective antibicanonical divisor on X (cid:48) , then:a singularity x ∈ X such that π − ( x ) does not meet supp( W (cid:48) ) is a rational doublepoint.Proof. Let x ∈ X be a singularity such that π − ( x ) does not meet supp( W (cid:48) ). Let T be an irreducible component of the connected component π − ( x ), then T · W (cid:48) = 0.So 0 ≤ p a ( T ) = 1 + 12 T + 12 T · K X (cid:48) = 1 + 12 T − T · W (cid:48) = 1 + 12 T . By Mumford’s Theorem T <
0, so the only possible case for which the inequalityabove is valid is: T = − p a ( T ) = 0. Then all the irreducible componentsof π − ( x ) are smooth rational curves with self-intersection −
2. We can call thesecurves E i , for i = 1 , ..., n .We can prove that x must be a rational singularity using [13], Proposition-Definition2 .
1. Let Z = (cid:80) ni =1 a i · E i , for a i ≥ x . First of all, p a ( Z ) = 1 + Z + Z · K X (cid:48) . Now Z · K X (cid:48) = − Z · W (cid:48) = − (cid:80) ni =1 a i E i · W (cid:48) = 0, while Z < Z is contracted by π . So p a ( Z ) < .
1, we have that p a ( E i ) ≤ p a ( Z ) for every E i contained in Z .Since p a ( E i ) = 0 as computed before, then the only possible case is p a ( Z ) = 0, so x is a rational singularity.In conclusion, x ∈ X is a rational double point. Indeed, by the adjunction formula,the self-intersection Z = 2 p a ( Z ) − − Z · K X (cid:48) = 2 p a ( Z ) − − (cid:3) Remark 2.7.
We have already seen that, if X is birationally equivalent to anEnriques surface, then X can only contain rational singularities.Moreover, by the previous Lemma, we conclude that it can only contain rationaldouble points as singularities. 3. Examples
In this chapter, we will construct examples of surfaces with Prym-canonical hyper-plane sections.3.1.
Surfaces with Prym-canonical hyperplane sections birationally equiv-alent to ruled surfaces.
Let us fix some notation about minimal smooth ruledsurfaces in which we will follow [11], Chapter V. X (cid:48)(cid:48) is a minimal smooth ruled surface and p : X (cid:48)(cid:48) → Γ is the natural map on thebase curve Γ of genus q ≥
0, then X (cid:48)(cid:48) = P Γ ( E ), where E is a normalized locally freesheaf of rank 2 on Γ. URFACES WITH PRYM-CANONICAL HYPERPLANE SECTIONS 7
Let ∧ ( E ) = O Γ ( D ), for D ∈ Div(Γ). The integer e = − deg( D ) is an invariant of X (cid:48)(cid:48) .Let C be a section of p such that O X (cid:48)(cid:48) ( C ) = O X (cid:48)(cid:48) (1). Then C = − e . Moreover,we recall that the canonical divisor K X (cid:48)(cid:48) ∼ − C + ( K Γ + D ) · f , where ( K Γ + D ) · f denotes the divisor p ∗ ( K Γ + D ) by abuse of notation, with K Γ + D a divisor on Γ.Finally, if E is decomposable, i.e. E = O Γ ⊕ O Γ ( D ), then we will denote by C afixed section of p disjoint from C . Thus C = e and C ∼ C − D · f .3.1.1. The minimal model is a non-rational ruled surface . We start recallinga simple example of surface with Prym-canonical hyperplane sections.Let X (cid:48)(cid:48) = P Γ ( O Γ ⊕ O Γ ( D )) be a minimal ruled surface over a base curve Γ of genus q ≥
5, with D ∼ − K Γ + α , where α is a not trivial two torsion divisor and K Γ is thecanonical divisor of Γ, and let L (cid:48)(cid:48) = | C | be a linear system on X (cid:48)(cid:48) . By [3], Lemma2 .
1, if Γ is non-hyperelliptic and it does not admit g , then a general hyperplanesection of X (cid:48)(cid:48) is Prym-canonically embedded. The images of fibres of X (cid:48)(cid:48) by i L (cid:48)(cid:48) are lines since C (cid:48)(cid:48) · f = C · f = 1. It is not difficult to prove that X has only onesingularity, so the map i L (cid:48)(cid:48) : X (cid:48)(cid:48) (cid:57)(cid:57)(cid:75) P q − defined by the linear system L (cid:48)(cid:48) is suchthat X = i L (cid:48)(cid:48) ( X (cid:48)(cid:48) ) is a cone on a Prym-canonical embedded curve and if a generalhyperplane section C of X is projectively normal, then the geometric genus of theonly singularity is q (see Proposition 2.3). Remark 3.1.
Let us consider L (cid:48)(cid:48) ⊆ | aC + ∆ · f | , for a ≥ ∈ Pic(Γ). If X (cid:48)(cid:48) = P Γ ( O Γ ⊕O Γ ( D )) is a minimal ruled surface over a base curve Γ of genus q ≥ X (cid:48)(cid:48) by the map i L (cid:48)(cid:48) associated with L (cid:48)(cid:48) are not lines since C (cid:48)(cid:48) · f = ( aC + ∆ · f ) · f = a >
1, for C (cid:48)(cid:48) a general curve in L (cid:48)(cid:48) . Furthermore thereis not another family of rational curves on X (cid:48)(cid:48) mapped into lines by i L (cid:48)(cid:48) because thegenus of the base curve Γ is q >
0. Indeed, by the Riemann-Hurwitz formula (see[11], Corollary
IV. . q >
0. Hence we never obtain X = i L (cid:48)(cid:48) ( X (cid:48)(cid:48) )as a cone.Now we focus our attention on surfaces with Prym-canonical hyperplane sectionsbirationally equivalent to ruled surfaces X (cid:48)(cid:48) over a non-hyperelliptic base curve ofgenus q ≥ Proposition/Example 3.1.
Let X (cid:48)(cid:48) = P Γ ( O Γ ⊕O Γ ( D )) be a minimal ruled surfaceover a non-hyperelliptic smooth base curve Γ of genus q ≥ , for D ∈ Div(Γ) .Let L (cid:48)(cid:48) = | aC | be a linear system with a ≥ . If D ∼ − K Γ + α , for α a non-zero two-torsion divisor, then the image X = i L (cid:48)(cid:48) ( X (cid:48)(cid:48) ) ⊆ P a ( q − of the morphismassociated with L (cid:48)(cid:48) has Prym-canonical hyperplane sections and only one singularity.In particular, if a general hyperplane section C of X is projectively normal, thenthe geometric genus of the only singularity x is p g ( x ) = q .Proof. It is easy to prove that the linear system L (cid:48)(cid:48) = | aC | is base-point free using[8], Proposition 36, for any a ∈ N ≥ . Moreover, since ( C (cid:48)(cid:48) ) >
0, for C (cid:48)(cid:48) a generalelement of L (cid:48)(cid:48) , then, by Bertini’s Theorem, C (cid:48)(cid:48) is smooth and irreducible.CLAIM 1 : We have that − K X (cid:48)(cid:48) is not effective, while − K X (cid:48)(cid:48) is. Proof.
We know that h ( O X (cid:48)(cid:48) ( − K X (cid:48)(cid:48) )) = h ( O X (cid:48)(cid:48) (4 C )) = h ( O Γ )+ h ( O Γ ( D ))+ h ( O Γ (2 D )) + h ( O Γ (3 D )) + h ( O Γ (4 D )) by [8], Lemma 35. Because deg( D ) =2 − q <
0, then h ( O X (cid:48)(cid:48) ( − K X (cid:48)(cid:48) )) = 1 and − K X (cid:48)(cid:48) is effective.On the other hand, we have that h ( O X (cid:48)(cid:48) ( − K X (cid:48)(cid:48) )) = h ( O X (cid:48)(cid:48) (2 C − ( K Γ + D ) · f )) = h ( O Γ ( − K Γ − D )) + h ( O Γ ( − K Γ )) + h ( O Γ ( − K Γ + D )) by [8], Lemma MARTINA ANELLI
35. Since deg( D − K Γ ) = 4 − q < − K Γ ) <
0, then h ( − K X (cid:48)(cid:48) ) = h ( O Γ ( − K Γ − D )) = h ( O Γ ( − α )) = 0, so − K X (cid:48)(cid:48) is not effective. (cid:3) CLAIM 2 : We prove that O C (cid:48)(cid:48) ( − K X (cid:48)(cid:48) ) (cid:29) O C (cid:48)(cid:48) while O C (cid:48)(cid:48) ( − K X (cid:48)(cid:48) ) ∼ = O C (cid:48)(cid:48) , where C (cid:48)(cid:48) ∈ L (cid:48)(cid:48) is a general curve. Proof.
Since − K X (cid:48)(cid:48) is effective as seen in Claim 1 and C (cid:48)(cid:48) · ( − K X (cid:48)(cid:48) ) =( aC − aD · f ) · (4 C ) = 4 aC − a deg( D ) = 0, then O C (cid:48)(cid:48) ( − K X (cid:48)(cid:48) ) ∼ = O C (cid:48)(cid:48) .Clearly also C (cid:48)(cid:48) · ( − K X (cid:48)(cid:48) ) = 0. Since h ( O X (cid:48)(cid:48) ( − K X (cid:48)(cid:48) )) = 0 as seen in Claim 1, ifwe prove that h ( O X (cid:48)(cid:48) ( − K X (cid:48)(cid:48) − C (cid:48)(cid:48) )) = 0, then, from the exact sequence0 → O X (cid:48)(cid:48) ( − K X (cid:48)(cid:48) − C (cid:48)(cid:48) ) → O X (cid:48)(cid:48) ( − K X (cid:48)(cid:48) ) → O C (cid:48)(cid:48) ( − K X (cid:48)(cid:48) ) → , we have that h ( O C (cid:48)(cid:48) ( − K X (cid:48)(cid:48) )) = 0, which implies that O C (cid:48)(cid:48) ( − K X (cid:48)(cid:48) ) (cid:29) O C (cid:48)(cid:48) .Thus, by Serre Duality, we have that h ( O X (cid:48)(cid:48) ( − K X (cid:48)(cid:48) − C (cid:48)(cid:48) )) = h ( O X (cid:48)(cid:48) (2 K X (cid:48)(cid:48) + C (cid:48)(cid:48) )). If we prove that K X (cid:48)(cid:48) + C (cid:48)(cid:48) is ample, then, by the Kodaira vanishing Theorem(see [11], Remark III. . h ( O X (cid:48)(cid:48) (2 K X (cid:48)(cid:48) + C (cid:48)(cid:48) )) = 0 and the claimis proved.By [11], Proposition V. .
20, if a >
2, then K X (cid:48)(cid:48) + C (cid:48)(cid:48) is ample and the claim issatisfied, instead, if a = 2, we have that K X (cid:48)(cid:48) + C (cid:48)(cid:48) is not ample. About that, let ussuppose that O C (cid:48)(cid:48) ( − K X (cid:48)(cid:48) ) ∼ = O C (cid:48)(cid:48) . Then the image of X (cid:48)(cid:48) by i L (cid:48)(cid:48) is a surface withcanonical hyperplane sections. By [7], Corollary 5 . X (cid:48)(cid:48) contains only one effectiveanticanonical divisor. This contradicts Claim 1, then this claim is also satisfied for a = 2 . (cid:3) We know that h ( O Γ ) = 1. Using the Riemann-Roch Theorem, we also have that h ( O Γ ( − D )) = h ( O Γ ( α )) + 2 q − − q = (2 q −
2) + 1 − q and h ( O Γ ( − mD )) = m (2 q −
2) + 1 − q , for any m ∈ N > . Hence, using [8], Lemma 35, we obtain that h ( O X (cid:48)(cid:48) ( L (cid:48)(cid:48) )) = (1 + ... + a )(2 q −
2) + a (1 − q ) + 1 == a ( a + 1)( q −
1) + a (1 − q ) + 1 = a ( q −
1) + 1 . So X = i L (cid:48)(cid:48) ( X (cid:48)(cid:48) ) is contained in P a ( q − , for a ( q − ≥ · L (cid:48)(cid:48) defines a birational morphism i L (cid:48)(cid:48) , in par-ticular an isomorphism outside C . So i L (cid:48)(cid:48) | C (cid:48)(cid:48) is a Prym-canonical embedding, for C (cid:48)(cid:48) ∈ L (cid:48)(cid:48) a general divisor. Proof.
We can prove that L (cid:48)(cid:48) defines a birational map, in particular an isomorphismoutside C , using [8], Theorem 38. Indeed, this happens if − aD is very ample andif | − aD + D | , | − aD + ( a − D | = | − D | and | − aD + aD | are base-point free.The last case is trivial. By [11], Corollary IV. .
2, since deg( − aD ) = a (2 q − ≥ q − q + (2 q − ≥ q + 1, then − aD is very ample. Again by [11], Corollary IV. .
2, if a ≥
3, since deg( − aD + D ) = a (2 q −
2) + (2 − q ) = ( a − q − ≥ q − q + (2 q − ≥ q , then | − aD + D | is base-point free.It remains to show that | − D | is base-point free. Since D ∼ − K Γ + α , if | − D | hasbase points, then Γ must be hyperelliptic (see [3], Lemma 2 . L (cid:48)(cid:48) defines an isomorphism outside C , called i L (cid:48)(cid:48) .A general C (cid:48)(cid:48) ∼ aC in L (cid:48)(cid:48) is disjoint from C by definition, then i L (cid:48)(cid:48) | C (cid:48)(cid:48) : C (cid:48)(cid:48) → P a ( q − − is an embedding. By the adjunction formula, we have that L (cid:48)(cid:48) | C (cid:48)(cid:48) ∼ = K C (cid:48)(cid:48) − K X (cid:48)(cid:48) | C (cid:48)(cid:48) URFACES WITH PRYM-CANONICAL HYPERPLANE SECTIONS 9 but in Claim 2 we have already proved that − K X (cid:48)(cid:48) | C (cid:48)(cid:48) is a non trivial two-torsiondivisor, so i L (cid:48)(cid:48) | C (cid:48)(cid:48) is a Prym-canonical embedding. (cid:3) We observe that, since i L (cid:48)(cid:48) | C (cid:48)(cid:48) : C (cid:48)(cid:48) → P g − by definition of Prym-canonical map,then g = g ( C (cid:48)(cid:48) ) = a ( q −
1) + 1 . The image x ∈ X of − K X (cid:48)(cid:48) ∼ C by i L (cid:48)(cid:48) is a singular point. There are notother possible singularities because L (cid:48)(cid:48) is an isomorphism outside C . We havefound examples of surfaces in P a ( q − with Prym-canonical hyperplane sectionsbirationally equivalent to non-rational ruled surfaces, for a ≥ q ≥ C of X is projectively normal, then, by Proposition2.3, the singularity x has geometric genus equal to q . (cid:3) We can construct an example of surface with Prym-canonical hyperplane sectionsbirationally equivalent to an elliptic ruled surface X (cid:48)(cid:48) . Example 3.2.
Let Γ be an elliptic curve and let X (cid:48)(cid:48) = P Γ ( O Γ ⊕ O Γ ( D )) be aminimal ruled surface with base curve Γ, for D ∈ Div(Γ). If Q i , for i = 1 , ,
3, isa general point on Γ and α, β ∈ Γ are two points such that α − β is a two torsionelement of Γ, then we assume that D = − Q − Q − Q − α + β . So e = − deg( D ) = 3.We consider the linear system | C | = | C + 3( Q + Q + Q + α − β ) · f | . It is easyto prove that this linear system is base-point free using [8], Proposition 36, so, byBertini’s Theorem, its general element L is smooth and, since L = 9 C = 9 e > f i := Q i · f , for i = 1 , ,
3. For any fibre f , we have that L · f = 3, so wecan fix the 9 points Z := { x , , x , , x , , x , , x , , x , , x , , x , , x , } of intersection between L and f i , for i = 1 , ,
3. Since 3 C is disjoint from C , wecan assume that Z ∩ C = ∅ . So we can consider the linear system L (cid:48)(cid:48) ⊂ | C | on X (cid:48)(cid:48) with Z as base locus. In particular, we suppose that every curve C (cid:48)(cid:48) ∈ L (cid:48)(cid:48) simply passes through the 9 points, so L is an element of L (cid:48)(cid:48) . By [8], Lemma 35, wehave that h ( O X (cid:48)(cid:48) (3 C + 3( Q + Q + Q + α − β ) · f )) = 19, so dim L (cid:48)(cid:48) ≥ − L ∈ L (cid:48)(cid:48) and smoothness is an open condition, then the general element C (cid:48)(cid:48) of L (cid:48)(cid:48) is smooth.We know that − K X (cid:48)(cid:48) ∼ C − D · f and h ( O X (cid:48)(cid:48) ( − K X (cid:48)(cid:48) )) = 4 by [8], Lemma 35.We can show that h ( O X (cid:48)(cid:48) ( − K X (cid:48)(cid:48) − I Z )) = 0. Indeed, if we suppose that there isan effective divisor T ∈ | − K X (cid:48)(cid:48) − I Z | , then T · C = 2 C − deg( D ) = 2( −
3) + 3 < C is a fixed component of T . Moreover T · f = 2 but T contains 3 points of thefibres f i , so it also contains f , f , f . Then | − K X (cid:48)(cid:48) − I Z | = | C − D · f − I Z | = C + f + f + f + | C + ( α − β ) · f | , so dim | − K X (cid:48)(cid:48) − I Z | = dim | C + ( α − β ) · f | . By [8], Lemma 35, we have that h ( O X (cid:48)(cid:48) ( C + ( α − β ) · f )) = h ( O Γ ( α − β )) + h ( O Γ ( α − β − Q − Q − Q − α + β )) = 0. Hence T effective does not exist.It is not difficult to prove that h ( O X (cid:48)(cid:48) ( − K X (cid:48)(cid:48) )) = 10 and, since − K X (cid:48)(cid:48) ∼ C +2( Q + Q + Q ) · f and 4 C + 2( Q + Q + Q ) · f contains Z with multiplicity 2,then also h ( O X (cid:48)(cid:48) ( − K X (cid:48)(cid:48) ) ⊗ I Z ) > φ : X (cid:48) → X (cid:48)(cid:48) be the blowing up of X (cid:48)(cid:48) along the 9 points defining Z . Let E i,j ∈ X (cid:48) be the exceptional divisor of x i,j , for i, j ∈ { , , } , and let (cid:101) f i be the strict transform of f i , for i = 1 , ,
3. With abuse of notation, we call C := φ ∗ ( C ) (weremark that φ ∗ ( C ) is the strict transform of C since Z ∩ C = ∅ ), φ ∗ ( α · f ) := f α and φ ∗ ( β · f ) := f β . Let L (cid:48) be such that L (cid:48)(cid:48) = φ ∗ L (cid:48) . Then the strict transform C (cid:48) ∈ L (cid:48) of a general C (cid:48)(cid:48) ∈ L (cid:48)(cid:48) is of the form C (cid:48) = φ ∗ ( C (cid:48)(cid:48) ) − E , − ... − E , ∼ C + 3 (cid:88) i =1 (cid:101) f i + 3 f α − f β ++2( E , + ... + E , ) . Instead, using [11], Proposition V. .
3, we obtain that − K X (cid:48) = φ ∗ ( − K X (cid:48)(cid:48) ) − E , − ... − E , ∼ C + (cid:88) i =1 (cid:101) f i + f α − f β and − K X (cid:48) ∼ C + 2 (cid:101) f + 2 (cid:101) f + 2 (cid:101) f . It is clear that h ( O X (cid:48) ( − K X (cid:48) )) = h ( O X (cid:48)(cid:48) ( − K X (cid:48)(cid:48) −I Z )) = 0 while h ( O X (cid:48) ( − K X (cid:48) )) = h ( O X (cid:48)(cid:48) ( − K X (cid:48)(cid:48) − I Z )) > C (cid:48) of X (cid:48) is aPrym-canonical embedded curve.CLAIM 1 : We have that O C (cid:48) ( − K X (cid:48) ) (cid:29) O C (cid:48) and O C (cid:48) ( − K X (cid:48) ) ∼ = O C (cid:48) , where C (cid:48) ∈ L (cid:48) is a general curve. In particular, − K X (cid:48) | C (cid:48) is a non-zero two torsion divisor. Proof.
The intersection C (cid:48) · ( − K X (cid:48) ) = 6 C + 6 (cid:88) i =1 C · (cid:101) f i + 3 (cid:88) i =1 ( C · (cid:101) f i + (cid:101) f i ) + 2[ (cid:101) f · ( E , + E , + E , )+ (cid:101) f · ( E , + E , + E , )+ (cid:101) f · ( E , + E , + E , )] = 6( − − C (cid:48) · ( − K X (cid:48) ) = 0.Since − K X (cid:48) is effective, then the antibicanonical divisor of X (cid:48) is contracted by i L (cid:48) ,in particular O C (cid:48) ( − K X (cid:48) ) ∼ = O C (cid:48) .On the contrary, we have that h ( O X (cid:48) ( − K X (cid:48) )) = 0, so − K X (cid:48) is not effective. Asseen in Claim 2 of Proposition 3.1, if we proved that h ( O X (cid:48) ( − K X (cid:48) − C (cid:48) )) = 0,then O C (cid:48) ( − K X (cid:48) ) (cid:29) O C (cid:48) .Thus, by Serre Duality, it is clear that h ( O X (cid:48) ( − K X (cid:48) − C (cid:48) )) = h ( O X (cid:48) (2 K X (cid:48) + C (cid:48) )).If we prove that K X (cid:48) + C (cid:48) is big and nef, then, by the Kawamata-Viehweg vanishingTheorem (see [12] and [17]), the first cohomology h ( O X (cid:48) (2 K X (cid:48) + C (cid:48) )) = 0.In our case, since 2( α − β ) ∼
0, then 2 f α − f β ∼ K X (cid:48) + C (cid:48) ∼ C + 2 (cid:101) f + 2 (cid:101) f + 2 (cid:101) f + 2 E , + ... + 2 E , . Since K X (cid:48) + C (cid:48) is written as sum of irreducible and effective curves, then, to provethat K X (cid:48) + C (cid:48) is nef, it is enough to prove that ( K X (cid:48) + C (cid:48) ) · δ ≥
0, for any itsirreducible component δ . With some simple computations we obtain that this istrue and because we have strictly positive intersections between K X (cid:48) + C (cid:48) and itscomponents, then K X (cid:48) + C (cid:48) is also big. So the claim is satisfied. (cid:3) URFACES WITH PRYM-CANONICAL HYPERPLANE SECTIONS 11
It is not difficult to compute that C (cid:48) = 18, so, by the adjunction formula, thegenus g ( C (cid:48) ) = 1 + ( C (cid:48) + K X (cid:48) · C (cid:48) ) = 10. We also observe that C (cid:48) is smoothbecause it is the strict transform of a general element C (cid:48)(cid:48) of L (cid:48)(cid:48) , that is smooth.Since − K X (cid:48) | C (cid:48) is a non-zero two torsion divisor as seen in Claim 1, we have that L (cid:48) | C (cid:48) = | K C (cid:48) − K X (cid:48) | C (cid:48) | defines a Prym-canonical map φ L (cid:48) | C (cid:48) : C (cid:48) (cid:57)(cid:57)(cid:75) P . CLAIM 2 : The rational map φ L (cid:48) | C (cid:48) : C (cid:48) (cid:57)(cid:57)(cid:75) P is an embedding, for any generalcurve C (cid:48) ∈ L (cid:48) . Proof.
First of all, we know that, if L (cid:48) | C (cid:48) has base points, then C (cid:48) is hyperellipticby [3], Lemma 2 .
1. Moreover, since C (cid:48) · (cid:101) f = 3, where (cid:101) f is the pullback of a generalfibre f of X (cid:48)(cid:48) , then C (cid:48) is also a covering 3 : 1 of the elliptic curve Γ. This is notpossible by Castelnuovo-Severi inequality otherwise we would have 10 = g ( C (cid:48) ) ≤ · · · L (cid:48) | C (cid:48) is base-point free.Thanks to [3], Corollary 2 .
2, we know that, if C (cid:48) is not bielliptic, then L (cid:48) | C (cid:48) is anembedding. Because C (cid:48) is a triple cover of Γ as observed before, then C (cid:48) cannotbe bielliptic again by Castelnuovo-Severi inequality otherwise we would have 10 ≤ · · · (cid:3) At this point, since L (cid:48) | C (cid:48) is base-point free, it is clear that L (cid:48) is also base-point free.Since the restriction L (cid:48) | C (cid:48) defines an embedding for each generic curve C (cid:48) ∈ L (cid:48) ,then φ L (cid:48) is a birational map, generically 1 : 1.Before we have showed that dim( L (cid:48)(cid:48) ) = dim( L (cid:48) ) ≥
9. From the exact sequence0 → O X (cid:48) ( C (cid:48) − C (cid:48) ) → O X (cid:48) ( C (cid:48) ) → O C (cid:48) ( C (cid:48) ) → , we conclude that h ( O X (cid:48) ( C (cid:48) )) ≤
10 since O X (cid:48) ( C (cid:48) − C (cid:48) ) ∼ = O X (cid:48) and h ( O C (cid:48) ( C (cid:48) )) = 9.So we have that h ( O X (cid:48) ( C (cid:48) )) = 10.Then X (cid:48) has hyperplane sections that are Prym-canonically embedded and, in par-ticular, we have found a new surface X = φ L (cid:48) ( X (cid:48) ) ⊂ P with Prym-canonicalhyperplane sections. Since the antibicanonical divisor of X (cid:48) is connected, then itsimage x ∈ X by φ L (cid:48) is a singular point. There are other possible rational doublesingularities on X whose exceptional divisors on X (cid:48) do not intersect − K X (cid:48) .If a general hyperplane section C of X is projectively normal, then, by Proposition2.3, the geometric genus p g ( x ) is equal to 1. Remark 3.3.
We can compute how many moduli the couple ( X (cid:48)(cid:48) , L (cid:48)(cid:48) ) of previousexample depends on.The choice of the elliptic curve Γ depends on one parameter. In addition we fix adivisor D = − Q − Q − Q − α + β of degree −
3, where Q i is a general point onΓ, for i = 1 , ,
3, and α − β is a non-zero two torsion element of Γ.We know that there are only three non-zero two torsion points on Γ. Instead weobserve that | Q + Q + Q | is a linear system of dimension 2, so the choice of O Γ ( D ) depends on 3 − X (cid:48)(cid:48) that means that,if φ : Γ → Γ is an automorphism, then X (cid:48)(cid:48) = P Γ ( O Γ ⊕ O Γ ( D )) ∼ = P Γ ( O Γ ⊕ O Γ ( φ ∗ ( D ))). The group Aut(Γ) has dimension 1.To construct the surface with Prym-canonical hyperplane sections of the previousexample, we also fix a linear system L (cid:48)(cid:48) ⊂ | C | = | C − D · f | with 9 simple base points. The linear system | C | depends on the parameters fixed before. Insteadthe 9 simple base points are the points of intersection between a general element L ∈ | C | and the three fibres f i := Q i · f , for i = 1 , ,
3. The choice of the effectivedivisor in a linear system of the type | Q + Q + Q | that defines the three fibres f , f , f depends on 2 parameters. In addition, as seen in the previous example,the nine points { x , , x , , x , , x , , x , , x , , x , , x , , x , } impose independentconditions on the linear system L (cid:48)(cid:48) , so they depend on 9 parameters.The choice of the pair ( X (cid:48)(cid:48) , L (cid:48)(cid:48) ) depends on 1 + 1 − P )) = 15. We can consider X (cid:48)(cid:48) as the blowing up ofthe vertex of the cone C X (cid:48)(cid:48) on a plane cubic of P . If C Γ is the base curve of C X (cid:48)(cid:48) , there are ∞ plane cubics isomorphic to C Γ . Since we can choose the vertexamong all the possible points of P obtaining always isomorphic cones, then thereare ∞ (8+3) = ∞ isomorphic cones of the type of C X (cid:48)(cid:48) in P .Thus there are ∞ automorphism of P that fix X (cid:48)(cid:48) so, in conclusion, the couple( X (cid:48)(cid:48) , L (cid:48)(cid:48) ) depends on 12 − The minimal model is a rational ruled surface . We construct an ex-ample of surface with Prym-canonical hyperplane sections birationally equivalentto a rational ruled surface.
Example 3.4.
Let Γ be a rational smooth curve and let X (cid:48)(cid:48) = P Γ ( O Γ ⊕ O Γ ( D )) bea minimal ruled surface with base curve Γ, for D ∈ Div(Γ). We assume that e = 4,so deg( D ) = −
4. Hence X (cid:48)(cid:48) is a Hirzebruch surface F .We know that − K X (cid:48)(cid:48) ∼ C − ( K Γ + D ) · f , where deg( − K Γ − D ) = 2 + 4 = 6. Weput − K X (cid:48)(cid:48) = 2 C + 2 (cid:88) i =1 F i , where F , F and F are distinct and fixed fibres. We also set W (cid:48)(cid:48) = 4 C + 3 F + 3 (cid:88) i =1 F i ∈ | − K X (cid:48)(cid:48) | , where F is a generic fibre distinct from F i , for i = 1 , , | C | = | C − D · f | on X (cid:48)(cid:48) . Every element in | C | intersects every fibre of X (cid:48)(cid:48) in 4 points since 4 C · f = 4. In addition, we know that h ( O X (cid:48)(cid:48) (4 C )) = 45 by [8], Lemma 35.CLAIM 1 : There is a smooth curve L ∈ | C | such that L is tangent to F , F , F and F respectively in two points. Proof.
We can assume that X (cid:48)(cid:48) = F is the blowing up of the vertex of a cone in P on a rational normal curve of P . It is clear that C is the exceptional divisorassociated with the vertex.Let E = F + F + F + F be the curve intersection between the cone and a hyperplane H of P passing through the vertex of the cone. With abuse of notation, the totaltransform of E on X (cid:48)(cid:48) is E = C + F + F + F + F . URFACES WITH PRYM-CANONICAL HYPERPLANE SECTIONS 13
It is obvious that the linear systems | C | and | C | are base-point free. Then theyrespectively contain a general quadric Q and a general cubic C , that are smooth byBertini’s Theorem.It is clear that 2 Q intersects every fibre in two points with multiplicity 2. We call { x j , x ,j , x ,j , x ,j } , for j = 1 ,
2, the intersections points between 2 Q and F + F + F + F .Let us consider a pencil P generated by 2 Q and E + C . By Bertini’s Theorem, itscurves may have singular points only on the base locus of the pencil. At this pointwe observe that 2 Q · C ∼ C ) · (3 C ) = 12 · C is a rational normalcurve of degree 4. These 48 points are base points for the pencil, different from { x j , x ,j , x ,j , x ,j } , with j = 1 ,
2, for the generality of C . Now E + C has only 16singular points since E has only 4 singular points on C , while C is smooth anddisjoint from C for its generality and E · C ∼ ( C + 4 F ) · C = 12. Since Q isalso disjoint from C and it is general, then these 16 points are different from the48 base points. Then E + C is smooth in the 48 base points. The same is true for2 Q . Hence also a general divisor L in the pencil P is smooth in the 48 base points.Instead 2 Q | E = { x , x , x , , x , , x , , x , , x , , x , } . These are other 8 base pointsfor P . Since 2 Q passes through { x j , x ,j , x ,j , x ,j } , for j = 1 ,
2, with multiplic-ity 2 and since E + C simply passes through the eight points ( E contains thefibres F, F , F , F and C does not contain these 8 points), then a general curve L is smooth in these 8 points and, in particular, it is tangent to F, F , F , F in { x , x , x , , x , , x , , x , , x , , x , } .Finally, we observe that 2 Q ∼ C ) = 4 C and similarly, we have C ∼ C and E ∼ C + 4 F , so E + C ∼ C . Then we have found a smooth curve L ∈ | C | tangent to F in x and x and tangent to F i in x i, and x i, , for i = 1 , , (cid:3) In the following figure, we analyze what happens blowing up all the intersectionpoints between L and W (cid:48)(cid:48) , also infinitely near. We observe that, since L ∼ C and4 C is disjoint from C , then L does not intersect C . With abuse of notation, wewill call the strict transforms of C , F i and L with the same names.In the figure, we only focus on F , it is the same for F , F and F . • STEP 1 We blow up the intersection points x i, and x i, on X (cid:48)(cid:48) ; • STEP 2 In X (cid:48)(cid:48) := Bl x , ,x , ,x , ,x , ,x , ,x , ( X (cid:48)(cid:48) ), the curve L simply passesthrough the infinitely near base points y i, and y i, , for i = 1 , ,
3. We alsoblow up these six points; • STEP 3 Again L intersects the exceptional divisors E i, and E i, respec-tively in z i, and z i, on X (cid:48)(cid:48) := Bl y , ,y , ,y , ,y , ,y , ,y , ( X (cid:48)(cid:48) ), for i = 1 , , Y = Bl z , ,z , ,z , ,z , ,z , ,z , ( X (cid:48)(cid:48) ) blowing up these other sixpoints. With the same techniques as before, we also blow up { x , x , y , y , z , z } ∈ L ∩ F (as seen in the previous figure, they are infinitely near points). We define X (cid:48) := Bl z ,z ( Bl y ,y ( Bl x ,x ( Y ))) . In X (cid:48) , there are no intersection points between L and − K X (cid:48) .After observing how the blowing up works, we consider L (cid:48)(cid:48) ⊂ | C | as the linearsystem of the curves of | C | simply passing through Z := { x , x , y , y , z , z , x i, , x i, , y i, , y i, , z i, , z i, } , f or i = 1 , , . Then dim L (cid:48)(cid:48) ≥ − · L is an element of L (cid:48)(cid:48) and sincesmoothness is an open condition, then the general element C (cid:48)(cid:48) of L (cid:48)(cid:48) is smooth.With the same notation as before, we can obtain that − K Y = 2 C + (cid:88) i =1 (2 F i + J i, + J i, + 2 E i, + 2 E i, + B i, + B i, )while W (cid:48)(cid:48) Y = 4 C + 3 F + (cid:88) i =1 (3 F i + J i, + J i, + 2 E i, + 2 E i, ) ∈ | − K Y | . URFACES WITH PRYM-CANONICAL HYPERPLANE SECTIONS 15
At this point, we use the fact that, if M is an effective divisor and N i are irreducibledivisors, if M · N <
0, ( M − N ) · N <
0, and so on, the (cid:80) N i is a partial fixedpart of | M | . Thus, inductively, one can verify that all 2 C + (cid:80) i =1 (2 F i + J i, + J i, +2 E i, + 2 E i, + B i, + B i, ) is a fixed component of | − K Y | , so this is the only oneeffective curve in its linear system. In addition, the part 4 C + (cid:80) i =1 (3 F i + J i, + J i, + 2 E i, + 2 E i, ) is the fixed part of | − K Y | while its variable part is 3 F .Similarly to before, we can compute that − K X (cid:48) ∼ C + (cid:88) i =1 (2 F i + J i, + J i, + 2 E i, + 2 E i, + B i, + B i, )+ − J − J − E − E − B − B . This is clearly not effective. Instead W (cid:48) = 4 C +3 F + J + J +2 E +2 E + (cid:88) i =1 (3 F i + J i, + J i, +2 E i, +2 E i, ) ∈ |− K X (cid:48) | is effective. In addition, all 4 C + 3 F + J + J + 2 E + 2 E + (cid:80) i =1 F i + J i, + J i, + 2 E i, + 2 E i, is a fixed component of | − K X (cid:48) | and consequently it is the onlyone effective divisor in its linear system.Since all the divisors on Γ of the same degree are linearly equivalent, then we observethat − D · f ∼ F + 4 (cid:80) i =1 F i , so we can assume that C (cid:48)(cid:48) ∼ C + 4 F + 4 (cid:88) i =1 F i , where C (cid:48)(cid:48) is a general element in L (cid:48)(cid:48) . Then, its strict transform C (cid:48) on X (cid:48) is linearlyequivalent to C (cid:48) ∼ C + 4 F + (cid:88) i =1 (3 J j + 6 E j + 5 B j )++ (cid:88) i =1 (4 F i + 3 J i, + 3 J i, + 6 E i, + 6 E i, + 5 B i, + 5 B i, ) . If φ : X (cid:48) → X (cid:48)(cid:48) is the blowing up of X (cid:48)(cid:48) along the points of Z , let L (cid:48) be such that L (cid:48)(cid:48) = φ ∗ L (cid:48) , with C (cid:48) a general element.Step by step, we can prove that a general hyperplane section C (cid:48) of X (cid:48) is a Prym-canonical embedded curve.CLAIM 2 : We have that O C (cid:48) ( − K X (cid:48) ) (cid:29) O C (cid:48) while O C (cid:48) ( − K X (cid:48) ) ∼ = O C (cid:48) . In partic-ular, − K X (cid:48) | C (cid:48) is a non-zero two torsion divisor. Proof.
It is easy to compute that C (cid:48) · W (cid:48) = 0. Since W (cid:48) is effective, then it iscontracted by the map defined by L (cid:48) , in particular O C (cid:48) ( − K X (cid:48) ) ∼ = O C (cid:48) .It is clear that also C (cid:48) · ( − K X (cid:48) ) = 0 but this time we have that h ( O X (cid:48) ( − K X (cid:48) )) = 0.As seen in Claim 2 of Proposition 3.1, it is sufficient to show that h ( O X (cid:48) ( − K X (cid:48) − C (cid:48) )) = 0 to prove that O C (cid:48) ( − K X (cid:48) ) (cid:29) O C (cid:48) .Using Serre Duality and the Kawamata-Viehweg vanishing Theorem (see [12] and[17]), if we prove that K X (cid:48) + C (cid:48) is big and nef, then the claim is satisfied. Now K X (cid:48) + C (cid:48) = 2 C + (cid:88) i =1 (2 F i + 2 J i, + 2 J i, + 4 E i, + 4 E i, + 4 B i, + 4 B i, )++4 F + 4 J + 4 J + 8 E + 8 E + 8 B + 8 B . Since K X (cid:48) + C (cid:48) is written as sum of irreducible and effective curves, then, to provethat K X (cid:48) + C (cid:48) is nef, it is enough to prove that ( K X (cid:48) + C (cid:48) ) · δ ≥
0, for any itsirreducible component δ . It is possible to compute that this is true. Because wehave strictly positive intersections between K X (cid:48) + C (cid:48) and its components, then K X (cid:48) + C (cid:48) is also big. So the claim is satisfied. (cid:3) It is not difficult to compute that C (cid:48) = 40, so, by the adjunction formula, thegenus g ( C (cid:48) ) = 1 + ( C (cid:48) ) = 21. We also observe that C (cid:48) is smooth because it is thestrict transform of a general element C (cid:48)(cid:48) of L (cid:48)(cid:48) , that is smooth. Since − K X (cid:48) | C (cid:48) is anon-zero two torsion divisor as seen in Claim 2, we have that L (cid:48) | C (cid:48) = | K C (cid:48) − K X (cid:48) | C (cid:48) | defines a Prym-canonical map φ L (cid:48) | C (cid:48) : C (cid:48) (cid:57)(cid:57)(cid:75) P . We have already observed that dim( L (cid:48)(cid:48) ) = dim( L (cid:48) ) ≥
20. From the exact sequence0 → O X (cid:48) ( C (cid:48) − C (cid:48) ) → O X (cid:48) ( C (cid:48) ) → O C (cid:48) ( C (cid:48) ) → , we conclude that h ( O X (cid:48) ( C (cid:48) )) ≤
20 since O X (cid:48) ( C (cid:48) − C (cid:48) ) ∼ = O X (cid:48) and h ( O C (cid:48) ( C (cid:48) )) =19. So we have that h ( O X (cid:48) ( C (cid:48) )) = 21 and φ L (cid:48) ( X (cid:48) ) ⊆ P . CLAIM 3 : The rational map φ L (cid:48) | C (cid:48) : C (cid:48) (cid:57)(cid:57)(cid:75) P is an embedding, for any generalcurve C (cid:48) ∈ L (cid:48) . Proof.
First of all, we know that, if the Prym-canonical system L (cid:48) | C (cid:48) has basepoints, then C (cid:48) is hyperelliptic by [3], Lemma 2 . C (cid:48) is hyperelliptic. Since we know that C (cid:48) ∈ X (cid:48) is isomorphicto C (cid:48)(cid:48) ∼ C on X (cid:48)(cid:48) , then C (cid:48)(cid:48) is also hyperelliptic.The self-intersection C (cid:48)(cid:48) = (4 C ) = 64. Furthermore C (cid:48)(cid:48) is nef since C (cid:48)(cid:48) ∼ C +4 F +4 (cid:80) i =1 F i and (4 C +4 F +4 (cid:80) i =1 F i ) · C = 0 and (4 C +4 F +4 (cid:80) i =1 F i ) · F =(4 C + 4 F + 4 (cid:80) i =1 F i ) · F i = 4. Moreover, by [11], Proposition IV. .
2, we havethat | K C (cid:48)(cid:48) | is not very ample, precisely it does not separate any pair of points p and q such that p + q is a member of the g on C (cid:48)(cid:48) . By the adjunction formula, we alsohave that | K X (cid:48)(cid:48) + C (cid:48)(cid:48) | does not separate such p and q .By [15], Theorem 1 . , there exists an effective divisor E on X (cid:48)(cid:48) passing through p and q such that C (cid:48)(cid:48) · E <
4. Since C (cid:48)(cid:48) ∼ C , then C (cid:48)(cid:48) · E = 0. This is not possibleand thus E cannot exist. We exclude the case C (cid:48)(cid:48) hyperelliptic and hence L (cid:48) | C (cid:48) isbase-point free.Furthermore, we can prove that L (cid:48) | C (cid:48) defines a birational map. Indeed, if this didnot happen, we would have C (cid:48) bielliptic and the image of X (cid:48) via the map associatedwith L (cid:48) would be a surface in P with elliptic sections (see [3], Corollary 2 . >
9, then the surface image in P could not be a Del Pezzo surface but it wouldbe an elliptic cone. Anyway X (cid:48) is a rational surface, so it cannot cover an ellipticcone. Then L (cid:48) | C (cid:48) defines a birational map.More precisely, we can also show that L (cid:48) | C (cid:48) defines an embedding, for any general C (cid:48) ∈ L (cid:48) . By [3], Lemma 2 .
1, we know that L (cid:48) | C (cid:48) does not separate p and q (possiblyinfinitely near) if and only if C (cid:48) has a g and − K X (cid:48) | C (cid:48) ∼ O C (cid:48) ( p + q − x − y ), where2( p + q ) and 2( x + y ) are members of the g .We know that C (cid:48) ∼ = C (cid:48)(cid:48) and C (cid:48)(cid:48) ∼ C has a g defined by the fibres of the ruledsurface X (cid:48)(cid:48) . This is the only one. Indeed, if C (cid:48) had two g , then there would be amap ψ : C (cid:48) → P × P . If ψ was a birational map, the image curve would be of thetype (4 ,
4) on P × P . Then its geometric genus would be at most (4 − −
1) = 9.Since C (cid:48) has genus 21, this case is excluded. Thus ψ would be a map 2 : 1 on acurve D . The image curve D would be a curve of type (2 ,
2) on P × P , so its URFACES WITH PRYM-CANONICAL HYPERPLANE SECTIONS 17 geometric genus would be g ( D ) ≤
1. Since C (cid:48) is non-hyperelliptic as seen before,then g ( D ) = 1 and C (cid:48) is bielliptic. Then C (cid:48) admits a singular correspondence. ByCorollary 2 . | C (cid:48) | is not birational,in particular it is 2 : 1 on a surface with elliptic sections. We have already excludedthis possibility, so C (cid:48)(cid:48) has only one g .It is clear that C (cid:48)(cid:48) ∩ F ∼ C (cid:48)(cid:48) ∩ F ∼ C (cid:48)(cid:48) ∩ F ∼ C (cid:48)(cid:48) ∩ F and we know that C (cid:48)(cid:48) istangent in two points to this four fibres. So we have four pairs of points ( p, q ) ∈ F ,( x, y ) ∈ F , ( z, w ) ∈ F and ( a, b ) ∈ F such that 2( p + q ) ∼ x + y ) and so onfor all the possible cases. Since C (cid:48) is the strict transform of C (cid:48)(cid:48) , it has the samecharacteristics of C (cid:48)(cid:48) and, after the blowing up, the four pairs of points that satisfythis property are the intersection points between C (cid:48) and B i,j and C (cid:48) and B j , for i = 1 , , j = 1 ,
2. Now, with abuse of notation and using the expression of − K X (cid:48) seen before, we have that − K X (cid:48) | C (cid:48) = (cid:88) i =1 ( B i, + B i, ) | C (cid:48) − (3 B + 3 B ) | C (cid:48) == x + y + w + z + a + b − p − q ∼ x + y − w − z + a + b − p − q. At this point, we observe that x + y − w − z + a + b − p − q (cid:28) x + y − p − q otherwise, if a + b − w − z ∼ , then C (cid:48) would have a g . Hence L (cid:48) | C (cid:48) separate eachpair of points and it defines an embedding. (cid:3) At this point, since L (cid:48) | C (cid:48) is base-point free, it is clear that L (cid:48) is also base-point free.Since the restriction L (cid:48) | C (cid:48) defines an embedding for each generic curve C (cid:48) ∈ L (cid:48) ,then φ L (cid:48) is a birational map, generically 1 : 1.Then X (cid:48) has hyperplane sections that are Prym-canonically embedded. In partic-ular, φ L (cid:48) ( X (cid:48) ) is a surface with Prym-canonical hyperplane sections.We have found a new surface X = φ L (cid:48) ( X (cid:48) ) ⊂ P with Prym-canonical hyperplanesections. Since W (cid:48) is connected, then the image x ∈ X of W (cid:48) is a rational singularpoint (see Proposition 2.3). There are other possible rational double singularitieson X whose exceptional divisors on X (cid:48) do not intersect − K X (cid:48) .3.2. More surfaces with Prym-canonical hyperplane sections birationallyequivalent to P . We construct a new example of such surface whose minimalmodel is X (cid:48)(cid:48) = P . Example 3.5.
Let X (cid:48)(cid:48) = P be such that − K X (cid:48)(cid:48) is an irreducible sextic with10 nodes { x , ..., x } . Let L (cid:48)(cid:48) be a linear system of curves of degree 18 with basepoints { x , ..., x } ∈ X (cid:48)(cid:48) of multiplicity respectively r i = 4, for i = 1 , ,
3, and r i = 6, for i = 4 , ...,
10. Let X (cid:48) = Bl { x ,...,x } ( P ) be the blowing up of X (cid:48)(cid:48) alongthe base points of L (cid:48)(cid:48) and let L (cid:48) be the strict transform of L (cid:48)(cid:48) . We observe that theanticanonical divisor − K X (cid:48) is not effective.Let C (cid:48) ∼ l − (cid:88) i =1 E i − (cid:88) i =4 E i be a general curve in L (cid:48) , where E i is the exceptional divisor associated with x i , for i = 1 , ...,
10. It is obvious thatdeg( C (cid:48) | C (cid:48) ) = 18 − · − ·
36 = 24 . We have that h ( O X (cid:48) ( C (cid:48) )) ≥ (cid:0) (cid:1) − · − · = 13, so φ L (cid:48) ( X (cid:48) ) = X ⊂ P r , for r ≥ − K X (cid:48) ∼ J = 6 l − (cid:80) i =1 E i is effective and C (cid:48) · ( − K X (cid:48) ) = 0 by construction,then O C (cid:48) ( − K X (cid:48) ) ∼ = O C (cid:48) . So L (cid:48) contracts J in a single point since J is irreducible.Moreover, since J is also rational, then φ L (cid:48) ( J ) is a rational singularity of multiplicity4 because the fundamental cycle Z = J is such that Z = J = − L (cid:48) ) = 12, so φ L (cid:48) ( X (cid:48) ) = X ⊆ P . Proof.
We can consider the following exact sequence, already tensored with O X (cid:48) ( C (cid:48) ):(4) 0 → O X (cid:48) ( C (cid:48) − J ) → O X (cid:48) ( C (cid:48) ) → O J ( C (cid:48) ) → . Since O C (cid:48) ( − K X (cid:48) ) ∼ = O C (cid:48) and J ∈ | − K X (cid:48) | , then O J ( C (cid:48) ) ∼ = O J ∼ = O P since J isrational. Thus we can rewrite (4) as(5) 0 → O X (cid:48) (12 l − (cid:88) i =1 E i − (cid:88) i =4 E i ) → O X (cid:48) (18 l − (cid:88) i =1 E i − (cid:88) i =4 E i ) → O P → . Similarly we obtain that0 → O X (cid:48) (6 l − (cid:88) i =4 E i ) → O X (cid:48) (12 l − (cid:88) i =1 E i − (cid:88) i =4 E i ) → (6) → O J (12 l − (cid:88) i =1 E i − (cid:88) i =4 E i ) → . It is possible to choose a quintuple of points among the 10 nodes { x , ..., x } of J such that three of these points are not aligned, then an irreducible conic passingthrough this quintuple of points exists. Up to renaming the nodes of J , we supposethat a conic passing through { x , ..., x } exists.So let us consider the following exact sequences:(7)0 → O X (cid:48) (4 l − (cid:88) i =4 E i − (cid:88) i =9 E i ) → O X (cid:48) (6 l − (cid:88) i =4 E i ) → O l − (cid:80) i =4 E i (6 l − (cid:88) i =4 E i ) → → O X (cid:48) (3 l − (cid:88) i =4 E i ) → O X (cid:48) (4 l − (cid:88) i =4 E i − (cid:88) i =9 E i ) → (8) → O l − E − E (4 l − (cid:88) i =4 E i − (cid:88) i =9 E i ) → . It obvious that h ( O X (cid:48) (3 l − (cid:80) i =4 E i )) = (cid:0) (cid:1) − X (cid:48) and it has the excepted dimension, then h ( O X (cid:48) (3 l − (cid:80) i =4 E i )) = 0.Because l − E − E is rational and ( l − E − E ) · (4 l − (cid:80) i =4 E i − (cid:80) i =9 E i ) = 0,then h ( O l − E − E (4 l − (cid:80) i =4 E i − (cid:80) i =9 E i )) = 1 and h ( O l − E − E (4 l − (cid:80) i =4 E i − (cid:80) i =9 E i )) = 0.From the exact sequence (8), we conclude that h ( O X (cid:48) (4 l − (cid:80) i =4 E i − (cid:80) i =9 E i )) =3 + 1 = 4 and h ( O X (cid:48) (4 l − (cid:80) i =4 E i − (cid:80) i =9 E i )) = 0. URFACES WITH PRYM-CANONICAL HYPERPLANE SECTIONS 19
Using the Riemann-Roch Theorem, since J and 2 l − (cid:80) i =4 E i are rational, we obtainthat h ( O J (12 l − (cid:88) i =1 E i − (cid:88) i =4 E i )) = 4 + 1 = 5and h ( O l − (cid:80) i =4 E i (6 l − (cid:88) i =4 E i )) = 2 + 1 = 3 . From the exact sequence (7), we conclude that h ( O X (cid:48) (6 l − (cid:80) i =4 E i )) = 7 and h ( O X (cid:48) (6 l − (cid:80) i =4 E i )) = 0. Again, from the exact sequence (6), we have that h ( O X (cid:48) (12 l − (cid:80) i =1 E i − (cid:80) i =4 E i )) = 12 and h ( O X (cid:48) (12 l − (cid:80) i =1 E i − (cid:80) i =4 E i )) =0. Finally, from the exact sequence (5), we obtain that h ( O X (cid:48) (18 l − (cid:88) i =1 E i − (cid:88) i =4 E i )) = 12 + 1 = 13 . Then the claim is proved. (cid:3)
Step by step we want to show that a general hyperplane section of X (cid:48) is a Prym-canonical embedded curve.CLAIM 2 : We prove that O C (cid:48) ( − K X (cid:48) ) (cid:29) O C (cid:48) . Proof.
If we show that h ( O X (cid:48) ( − K X (cid:48) − C (cid:48) )) = 0, then, using the long exact sequenceassociated with0 → O X (cid:48) ( − K X (cid:48) − C (cid:48) ) → O X (cid:48) ( − K X (cid:48) ) → O C (cid:48) ( − K X (cid:48) ) → − K X (cid:48) is not effective, we have that h ( O C (cid:48) ( − K X (cid:48) )) = 0, thus O C (cid:48) ( − K X (cid:48) ) (cid:29) O C (cid:48) .Since − K X (cid:48) − C (cid:48) ∼ − l + 3 (cid:88) i =1 E i + 5 (cid:88) i =4 E i , then it is not effective and h ( O X (cid:48) ( − K X (cid:48) − C (cid:48) )) = 0. By Serre Duality, we havethat h ( O X (cid:48) ( − K X (cid:48) − C (cid:48) )) = h ( O X (cid:48) (2 K X (cid:48) + C (cid:48) )) = h ( O X (cid:48) (12 l − (cid:80) i =1 E i − (cid:80) i =4 E i )) = 12 as proved in Claim 1.Using the Riemann-Roch Theorem, we conclude that − h ( O X (cid:48) ( − K X (cid:48) − C (cid:48) )) ++ h ( O X (cid:48) ( − K X (cid:48) − C (cid:48) )) = − h ( O X (cid:48) ( − K X (cid:48) − C (cid:48) )) + 12 = ( − l + 3 (cid:80) i =1 E i +5 (cid:80) i =4 E i ) · ( − l +2 (cid:80) i =1 E i +4 (cid:80) i =4 E i )+1 − h ( O X (cid:48) ( − K X (cid:48) − C (cid:48) )) = 0and the claim is proved. (cid:3) CLAIM 3 : There are irreducible curves of degree 18 with exactly 3 quadruplepoints and 7 points of multiplicity six in the ten nodes of J . Proof.
We observe that curves of the type J + D , with D ∈ | l − (cid:80) i =1 E i − (cid:80) i =4 E i | and J fixed part, are contained in L (cid:48) = | l − (cid:80) i =1 E i − (cid:80) i =4 E i | .As proved in Claim 1, we have that dim | l − (cid:80) i =1 E i − (cid:80) i =4 E i | = 12 whiledim | l − (cid:80) i =1 E i − (cid:80) i =4 E i | = 11, so the reducible curves J + D do not fillup all the linear system of the curves of degree 18. As consequence of Bertini’sTheorem (see [2], pag. 1), the generic curve of L (cid:48) is irreducible (indeed the curvesof the linear system L (cid:48) with fixed part J define a sublinear system and moreoverthe sublinear system is not composed by a pencil, even more so the linear system L (cid:48) ). Also curves of the type 2 J + F , with F ∈ | l − (cid:80) i =4 E i | and 2 J fixed part, arecontained in L (cid:48) . Since these special curves of L (cid:48) have exactly quadruple points inthree of the 10 nodes of J and points of multiplicity 6 in seven of the 10 nodes of J , then the generic curves of the linear system L (cid:48) have the same property. Thusirreducible curves of degree 18 with exactly quadruple points in three of the 10nodes of J and points of multiplicity 6 in the remaining nodes of J exist. (cid:3) Therefore the arithmetic genus, that is equal to the geometric genus of C , is g ( C (cid:48) ) = 17 · − · − ·
52 = 13by the Pl¨ucker Formula.It remains to show that L (cid:48) defines an embedding outside the contracted curve J .CLAIM 4 : The linear system L (cid:48) is base-point free. Proof.
Let X (cid:48) = Bl x ( X (cid:48) ), where x is a point of a general C (cid:48) ∈ L (cid:48) . If L (cid:48) = | l − (cid:80) i =1 E i − (cid:80) i =4 E i − E | , for E the exceptional divisor associated with x , then L (cid:48) is base point free if and only ifdim( L (cid:48) ) = dim( L (cid:48) ) − , for any point x ∈ C (cid:48) , for a general C (cid:48) ∈ L (cid:48) .We have already proved that dim( L (cid:48) ) = 12, instead dim( L (cid:48) ) ≥ (cid:0) (cid:1) − · − · − − x ∈ C (cid:48) and C (cid:48) and J are disjoint by assumptions,then O J ( C (cid:48) ) ∼ = O P , where C (cid:48) is a general curve in L (cid:48) and J is the strict transformof J on X (cid:48) .Similarly to the exact sequences (5), (6), we have the following:(9)0 → O X (cid:48) (12 l − (cid:88) i =1 E i − (cid:88) i =4 E i − E ) → O X (cid:48) (18 l − (cid:88) i =1 E i − (cid:88) i =4 E i − E ) → O P → → O X (cid:48) (6 l − (cid:88) i =4 E i − E ) → O X (cid:48) (12 l − (cid:88) i =1 E i − (cid:88) i =4 E i − E ) → (10) → O J (12 l − (cid:88) i =1 E i − (cid:88) i =4 E i − E ) → . As in Claim 1, we suppose that an irreducible conic passing through { x , ..., x } exists. • If x ∈ l − (cid:80) i =4 E i , we can consider the exact sequence0 → O X (cid:48) (4 l − (cid:88) i =4 E i − (cid:88) i =9 E i ) → O X (cid:48) (6 l − (cid:88) i =4 E i − E ) → (11) → O l − (cid:80) i =4 E i − E (6 l − (cid:88) i =4 E i − E ) → . From the exact sequence (8), we know that h ( O X (cid:48) (4 l − (cid:80) i =4 E i − (cid:80) i =9 E i )) =4 and h ( O X (cid:48) (4 l − (cid:80) i =4 E i − (cid:80) i =9 E i )) = 0.Since h ( O l − (cid:80) i =4 E i − E (6 l − (cid:80) i =4 E i − E )) = 2 by Riemann-Roch’sTheorem, then h ( O X (cid:48) (6 l − (cid:80) i =4 E i − E )) = 6 and h ( O X (cid:48) (6 l − (cid:80) i =4 E i − E )) = 0 from the exact sequence (11). URFACES WITH PRYM-CANONICAL HYPERPLANE SECTIONS 21 • If x / ∈ l − (cid:80) i =4 E i , then we consider the following0 → O X (cid:48) (4 l − (cid:88) i =4 E i − E − E − E ) → O X (cid:48) (6 l − (cid:88) i =4 E i − E ) → (12) → O l − (cid:80) i =4 E i (6 l − (cid:88) i =4 E i − E ) → . (cid:7) If x ∈ l − E − E , we consider0 → O X (cid:48) (3 l − (cid:88) i =4 E i ) → O X (cid:48) (4 l − (cid:88) i =4 E i − (cid:88) i =9 E i − E ) → (13) → O l − E − E − E (4 l − (cid:88) i =4 E i − (cid:88) i =9 E i − E ) → . It is obvious that h ( O X (cid:48) (3 l − (cid:80) i =4 E i )) = 3 and h ( O l − E − E − E (4 l − (cid:80) i =4 E i − (cid:80) i =9 E i − E )) = 0. From the exact sequence (13), weobtain that h ( O X (cid:48) (4 l − (cid:80) i =4 E i − (cid:80) i =9 E i − E )) = 3. (cid:7) If x / ∈ l − E − E , we can consider the following exact sequences0 → O X (cid:48) (3 l − (cid:88) i =4 E i ) → O X (cid:48) (4 l − (cid:88) i =4 E i − (cid:88) i =9 E i − E ) → (14) → O l − E − E (4 l − (cid:88) i =4 E i − (cid:88) i =9 E i − E ) → → O X (cid:48) ( l − (cid:88) i =9 E i ) → O X (cid:48) (3 l − (cid:88) i =4 E i ) → (15) → O l − (cid:80) i =4 E i (3 l − (cid:88) i =4 E i ) → . By assumption we have that h ( O X (cid:48) ( l − (cid:80) i =9 E i )) = 0. Because h ( O l − (cid:80) i =4 E i (3 l − (cid:80) i =4 E i )) = 2, then, from the exact sequence(15) we conclude that h ( O X (cid:48) (3 l − (cid:80) i =4 E i )) ≤
2. Since h ( O X (cid:48) (3 l − (cid:80) i =4 E i )) ≥ (cid:0) (cid:1) − h ( O X (cid:48) (4 l − (cid:80) i =4 E i − (cid:80) i =9 E i − E )) ≥ (cid:0) (cid:1) − − − h ( O l − E − E (4 l − (cid:80) i =4 E i − (cid:80) i =9 E i − E )) = 1, then,from the exact sequence (14), we have that h ( O X (cid:48) (4 l − (cid:80) i =4 E i − (cid:80) i =9 E i − E )) ≤
3, so equality holds.
In both previous cases, we have found h ( O X (cid:48) (4 l − (cid:80) i =4 E i − (cid:80) i =9 E i − E )) = 3. Since it is the expected dimension, then h ( O X (cid:48) (4 l − (cid:80) i =4 E i − (cid:80) i =9 E i − E )) = 0.Using the Riemann-Roch Theorem, we have that h ( O l − (cid:80) i =4 E i (6 l − (cid:80) i =4 E i − E )) = 3. So we obtain that h ( O X (cid:48) (6 l − (cid:80) i =4 E i − E )) = 6 and h ( O X (cid:48) (6 l − (cid:80) i =4 E i − E )) = 0 from theexact sequence (12).In all two cases we have that h ( O X (cid:48) (6 l − (cid:80) i =4 E i − E )) = 6.With the same techniques as before, from the exact sequence (10), we have that h ( O X (cid:48) (12 l − (cid:80) i =1 E i − (cid:80) i =4 E i − E )) = 11 and h ( O X (cid:48) (12 l − (cid:80) i =1 E i − (cid:80) i =4 E i − E )) = 0 and finally, from the exact sequence (9), we obtain that h ( O X (cid:48) (18 l − (cid:80) i =1 E i − (cid:80) i =4 E i − E )) = 12. Then we can conclude that L (cid:48) is base-point free. (cid:3) By Bertini’s Theorem, since L (cid:48) is base-point free, then the generic C (cid:48) ∈ L (cid:48) is smooth.CLAIM 5 : The linear system L (cid:48) defines an embedding outside the contracted curve J . Proof.
It is sufficient to show that dim L (cid:48) = dim( L (cid:48) ) −
2, where either L (cid:48) = | l − (cid:80) i =1 E i − (cid:80) i =4 E i − E − E | , for E and E the exceptional di-visors associated with any two distinct points x and x not belonging to J , or L (cid:48) = | l − (cid:80) i =1 E i − (cid:80) i =4 E i − E − E | , for E and E the exceptionaldivisors associated with any two points x and x infinitely near not belonging to J .We have already proved that dim( L (cid:48) ) = 12. Moreover we have that dim( L (cid:48) ) ≥ (cid:0) (cid:1) − · − · − − C (cid:48) is a general curve in L (cid:48) and J is the stricttransform of J on X (cid:48) , then C (cid:48) · J = 0 since we choose x and x not belonging to J .We will show that L (cid:48) defines an embedding outside the contracted curve J assuming x and x distinct. The proof is similar for x and x infinitely near.We can consider the following exact sequences:0 → O X (cid:48) (12 l − (cid:88) i =1 E i − (cid:88) i =4 E i − E − E ) → (16) → O X (cid:48) (18 l − (cid:88) i =1 E i − (cid:88) i =4 E i − E − E ) → O P → → O X (cid:48) (6 l − (cid:88) i =4 E i − E − E ) → O X (cid:48) (12 l − (cid:88) i =1 E i − (cid:88) i =4 E i − E − E ) → (17) → O J (12 l − (cid:88) i =1 E i − (cid:88) i =4 E i − E − E ) → → O X (cid:48) (2 (cid:88) i =1 E i − E − E ) → O X (cid:48) (6 l − (cid:88) i =4 E i − E − E ) → URFACES WITH PRYM-CANONICAL HYPERPLANE SECTIONS 23 (18) → O J (6 l − (cid:88) i =4 E i − E − E ) → . It is clear that h ( O X (cid:48) (2 (cid:80) i =1 E i − E − E )) = 0. Again, by Serre Duality, wehave that h ( O X (cid:48) (2 (cid:80) i =1 E i − E − E )) = 0. Using the Riemann-Roch Theorem,we obtain that − h (2 (cid:80) i =1 E i − E − E )) = (2 (cid:80) i =1 E i − E − E ) · (3 l + (cid:80) i =1 E i − (cid:80) i =4 E i − E − E ) + 1 = − J · (6 l − (cid:80) i =4 E i − E − E ) = 8, then h ( O J (6 l − (cid:80) i =4 E i − E − E )) = 9.Moreover h ( O X (cid:48) (6 l − (cid:80) i =4 E i − E − E )) ≥ (cid:0) (cid:1) − · − h ( O X (cid:48) (6 l − (cid:80) i =4 E i − E − E )) = 0.We observe that curves of the type (3 l − (cid:80) i =4 E i ) + F , with F ∈ | l − (cid:80) i =4 E i | and 3 l − (cid:80) i =4 E i fixed part, are contained in | l − (cid:80) i =4 E i − E − E | .We have that dim | l − (cid:80) i =4 E i − E − E | ≥ | l − (cid:80) i =4 E i | = (cid:0) (cid:1) − · − l − (cid:80) i =4 E i ) + F do not fill up allthe linear system of the curves of degree 6. As consequence of Bertini’s Theorem(see [2], pag. 1), the generic curve D of | l − (cid:80) i =4 E i − E − E | is irreducible(indeed the curves of the linear system | l − (cid:80) i =4 E i − E − E | with fixed part3 l − (cid:80) i =4 E i define a sublinear system and moreover the sublinear system is notcomposed by a pencil, even more so the linear system | l − (cid:80) i =4 E i − E − E | ).Let us consider the exact sequence(19) 0 → O X (cid:48) → O X (cid:48) ( D ) → O D ( D ) → . Since D = 36 − − p a ( D ) = · − h ( O D ( D )) = 0 (see [11],Example IV. . . h ( O X (cid:48) ) = 0 by definition, then h ( O X (cid:48) (6 l − (cid:80) i =4 E i − E − E )) = 0 from the exact sequence (19). Con-sequently h ( O X (cid:48) (6 l − (cid:80) i =4 E i − E − E )) = 5 from the exact sequence (18).Since J is rational, then h ( O J (12 l − (cid:80) i =1 E i − (cid:80) i =4 E i − E − E )) = 5, so,from the exact sequence (17), we have that h ( O X (cid:48) (12 l − (cid:80) i =1 E i − (cid:80) i =4 E i − E − E )) = 10 and h ( O X (cid:48) (12 l − (cid:80) i =1 E i − (cid:80) i =4 E i − E − E )) = 0 . Finally, from the exact sequence (16), we obtain that h ( O X (cid:48) (18 l − (cid:80) i =1 E i − (cid:80) i =4 E i − E − E )) = 11. The claim is proved. (cid:3) We have found a new example of rational surface X ⊂ P of degree deg( X ) = C (cid:48) deg φ L (cid:48) = 24 with Prym-canonical hyperplane sections and only one singularity, aquartic rational singularity. References [1]
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