# Projective normality and the generation of the ideal of an Enriques surface

```PROJECTIVE NORMALITY AND THE GENERATION OF THE IDEAL
OF AN ENRIQUES SURFACE
arXiv:1308.1267v2 [math.AG] 24 Sep 2013
ANDREAS LEOPOLD KNUTSEN AND ANGELO FELICE LOPEZ*
Abstract. We give necessary and sufficient criteria for a smooth Enriques surface S ⊂ Pr
to be scheme-theoretically an intersection of quadrics. Moreover we prove in many cases
that, when S contains plane cubic curves, the intersection of the quadrics containing S is
the union of S and the 2-planes spanned by the plane cubic curves. We also give a new
(very quick) proof of the projective normality of S if deg S ≥ 12.
1. Introduction
Even though it is a very basic question, it is in general difficult, given a projective variety
X ⊂ Pr , to be able to tell about its equations, as they depend in a nontrivial way on the
geometry and often on the moduli of X. An emblematic example of this is the case of
curves, when the equations are strictly related to its Clifford index and the investigation of
this problem has led to very important results and conjectures, such as Green’s conjecture
[G] and Voisin’s theorem [V1, V2].
In the present paper we deal with the problem of finding the degrees of the equations of
Enriques surfaces S ⊂ Pr . As a matter of fact it is not difficult to see that one can give
an answer as soon as one knows it for a general hyperplane section CP = S ∩ H passing
through a fixed point P ∈ Pr . Now for curves C important results, proved by Green and
Lazarsfeld [GL, Thm. 1], [L, Prop. 2.4.2] (we use also [LS, Thm. 1.3]) come to help: if
deg C ≥ 2g(C) + 1 − 2h1 (OC (1)) − Cliff C then C is projectively normal and if deg C is at
least one more, then C is scheme-theoretically cut out by quadrics unless it has a trisecant
line. It is therefore of crucial importance to know the Clifford index of curves on Enriques
surfaces. Now in [KL3, Thm. 1.1] we proved that Cliff C = gon(C) − 2, where gon(C) is
the gonality of C. A careful study of gon(CP ) leads us then to the main result of this paper
(for the function φ(L) see Def. 2.2):
Theorem 1.1. Let S ⊂ Pr be a smooth linearly normal nondegenerate Enriques surface
and let L = OS (1). Then
(i) S is projectively normal if L2 ≥ 12.
(ii) S is scheme-theoretically cut out by quadrics if and only if φ(L) ≥ 4;
(iii) If L2 ≥ 18 and φ(L) = 3 and L is not of special type (see Def. 2.6), then the
intersection of the quadrics containing S is the union of S and the 2-planes spanned
by cubic curves E ⊂ S such that E 2 = 0, E.L = 3.
We recall that for L to be very ample we need L2 ≥ 10 and φ(L) ≥ 3. Now (ii) and (iii)
improve [GLM2, Thm. 1.3], while (i) gives a very quick new proof of [GLM1, Thm. 1.1]
(except for the case L2 = 10). Note that the proof of (i) consists only of the first five lines
* Research partially supported by the MIUR national project “Geometria delle varietà algebriche” PRIN
2010-2011.
2000 Mathematics Subject Classification : Primary 14J28. Secondary 14H51, 14C20.
1
2
A.L. KNUTSEN AND A.F. LOPEZ
of Section 3, which do not depend on Section 2. Also in case (iii) we have some partial
results when L2 = 16 (see Remark 3.1).
We give some preliminary results in Section 2, among wich a generalization of [KL1, Cor.
1.6]. Then Section 3 is devoted to the proof of Theorem 1.1.
2. Preliminary results
Definition 2.1. We denote by ∼ (resp. ≡) the linear (resp. numerical) equivalence of
divisors or line bundles on a smooth surface. If V ⊆ H 0 (L) is a linear system, we denote
its base scheme by Bs |V |. A nodal curve on an Enriques surface S is a smooth rational
curve contained in S. A nodal cycle is a divisor R > 0 such that, for any 0 < R′ ≤ R we
have (R′ )2 ≤ −2.
Now recall from [CD] the following
Definition 2.2. Let L be a line bundle on an Enriques surface S such that L2 > 0. Set
φ(L) = inf{|F.L| : F ∈ Pic S, F 2 = 0, F 6≡ 0}.
In [KL1] we proved the following result about the variation of the gonality in linear
systems on Enriques surfaces
Proposition 2.3. [KL1, Cor. 1.6] Let |L| be a base-component free complete linear system
on an Enriques surface such that L2 > 0. Let gengon |L| denote the gonality of a general
smooth curve in |L| and mingon |L| denote the minimal gonality of a smooth curve in |L|.
Then
gengon |L| − 2 ≤ mingon |L| ≤ gengon |L|.
q
2
Moreover if equality holds on the left, then φ(L) ≥ ⌈ L2 ⌉.
We will need the ensuing generalization
Proposition 2.4. Let |L| be a base-component free complete linear system on an Enriques
surface such that L2 > 0. Let k := gengon |L| and assume that there is a smooth curve
C0 ∈ |L| with k0 := gon(C0 ) ≥ 2.
If k0 = k − 2 then there exists a line bundle D0 > 0 such that either
(i) D02 = 2, D0 .L = k = 2φ(L) and L2 ≥ 8φ(L) − 8; or
(ii) D02 = 4, φ(D0 ) = 2, k = µ(L) = D0 .L − 2 and L2 ≥ max{4k − 8, 2k + 4}.
If L2 ≥ 4k − 4 and k0 = k − 1 then there exists a line bundle D0 > 0 such that either
(iii) D02 = 2, k ≤ 2φ(L) ≤ D0 .L ≤ k + 1 and L2 ≥ 4D0 .L − 8; or
(iv) D02 = 4, φ(D0 ) = 2, k + 2 ≤ µ(L) + 2 ≤ D0 .L ≤ k + 3 and L2 ≥ max{4D0 .L −
16, 2D0 .L}.
Proof. Let k0 = k − ε with ε = 1 or 2, so that k ≥ 3 and then φ(L) ≥ 2. In particular L is
2
base-point free. When ε = 2 we have k0 ≤ ⌊ L4 ⌋ by [KL1, Lemma 5.1], so that L2 ≥ 4k0 ≥ 8.
This also holds by hypothesis when ε = 1. By [KL1, Rmk 4.9] with b = k0 − 1 we get
L2 ≥ 2k0 − 2 + 2φ(L). Fix an ample divisor H and let A0 be a gk10 on C0 . By [KL1, Prop.
3.1](a) we obtain an effective nontrivial decomposition L ∼ N + N ′ with N.N ′ ≤ k0 and
h0 (N ′ ) ≥ 2. Moreover we have (N − N ′ )2 = L2 − 4N.N ′ ≥ 0 and H.(N − N ′ ) ≥ 0, so that
by Riemann-Roch we have that either N ≡ N ′ and L ≡ 2N ′ or L − 2N ′ = N − N ′ ≥ 0.
Consider the set
XL = {D ∈ Pic(S) : h0 (D) ≥ 2, D.(L − D) ≤ k0 ,
(L − 2D)2 ≥ 0 and either L − 2D ≥ 0 or L ≡ 2D}.
PROJECTIVE NORMALITY AND THE IDEAL OF AN ENRIQUES SURFACE.
3
Now XL 6= ∅ since N ′ ∈ XL , whence we can choose an element D0 ∈ XL for which H.D0
is minimal. Note that h0 (D0 ) ≥ 2 implies that D0 .L ≥ 2φ(L). Set G := L − 2D0 so that
we know that either G ≥ 0 or G ≡ 0, whence L.(L − 2D0 ) = L.G ≥ 0. Together with
(L − 2D0 )2 ≥ 0 we get
(1)
L2 ≥ max{4D0 .L − 4D02 , 2D0 .L}.
We first prove that we cannot have D02 ≥ 6. In fact if this is the case, pick any F > 0 such
that F 2 = 0 and F.D0 = φ(D0 ). Then (D0 − F )2 ≥ 2 and D0 − F > 0 by [KL1, Lemma
2.4], so that h2 (D0 − F ) = 0 whence h0 (D0 − F ) ≥ 2 by Riemann-Roch. Moreover
(D0 − F )(L − D0 + F ) = D0 .(L − D0 ) − F.(L − 2D0 ) ≤ k0
since, by [KL2, Lemma 2.1], we have that F.(L − 2D0 ) ≥ 0. For the same reason
(L − 2(D0 − F ))2 = (L − 2D0 )2 + 4F.(L − 2D0 ) ≥ 0.
Finally L − 2(D0 − F ) = L − 2D0 + 2F > 0 both when L − 2D0 ≥ 0 and when L ≡ 2D0 .
Now this contradicts the minimality of D0 . Therefore we have D02 ≤ 4.
We have
(2)
k0 = k − ε ≤ 2φ(L) − ε ≤ D0 .L − ε = D02 + D0 .(L − D0 ) − ε ≤ D02 + k0 − ε
whence D02 ≥ ε ≥ 1, so that D02 ≥ 2.
Suppose D02 = 2. If ε = 2 we have equality in (2), so that D0 .L = k = 2φ(L) and (1) gives
L2 ≥ 8φ(L) − 8, that is (i). If ε = 1 then (2) gives k ≤ 2φ(L) ≤ D0 .L ≤ k + 1 and (1) gives
L2 ≥ 4D0 .L − 8, that is (iii).
Now suppose D02 = 4. We first exclude the case φ(D0 ) = 1. In fact, in the latter case, we
can write D0 ∼ 2F1 + F2 with Fi > 0, Fi2 = 0, i = 1, 2, F1 .F2 = 1 and we get
(3)
3φ(L) ≤ D0 .L = D02 + D0 .(L − D0 ) ≤ 4 + k0 = 4 + k − ε ≤ 4 + 2φ(L) − ε
whence 2 ≤ φ(L) ≤ 4 − ε. If ε = 2 then φ(L) = 2 and we have equality in (3), so
that D0 .L = 6 and the Hodge index theorem applied to D0 and L then gives L2 ≤ 8,
contradicting (1). If ε = 1 then φ(L) ≤ 3 and if equality holds then we have equality in (3),
so that D0 .L = 9 and F1 .L = F2 .L = 3. Now the Hodge index theorem applied to F1 + F2
and L gives L2 ≤ 18, contradicting (1). Therefore φ(L) = 2 and (3) gives 6 ≤ D0 .L ≤ 7
and therefore F1 .L = 2, 2 ≤ F2 .L ≤ 3. Now the Hodge index theorem applied to F1 + F2
and L gives L2 ≤ 8 if F2 .L = 2 and L2 ≤ 12 if F2 .L = 3, both contradicting (1).
Therefore we have proved that, when D02 = 4, we have φ(D0 ) = 2. Since certainly D0 6≡ L,
we have, by [KL1, Lemma 5.1],
k ≤ µ(L) ≤ D0 .L − 2 = 2 + D0 .(L − D0 ) ≤ 2 + k0 = 2 + k − ε
so that, using (1), we deduce that either ε = 2 and k = µ(L) = D0 .L − 2, L2 ≥ max{4k −
8, 2k + 4}, that is (ii), or ε = 1 and k + 2 ≤ µ(L) + 2 ≤ D0 .L ≤ k + 3, L2 ≥ max{4D0 .L −
16, 2D0 .L}, that is (iv).
For the sequel we will also need the following simple
Lemma 2.5. Let |L| be a base-component free complete linear system on an Enriques
surface S such that L2 = 18 and φ(L) = 3. Then there exist E > 0, Ei > 0, 1 ≤ i ≤ 3,
such that E 2 = Ei2 = 0, E.Ei = Ei .Ej = 1 for i 6= j and either
(i) L ∼ 3(E + E1 ); or
(ii) L ∼ 2E + E1 + E2 + E3 .
Moreover, in case (ii) we have that H 1 (E + E1 + KS ) = H 1 (E + Ei + Ej ) = 0.
4
A.L. KNUTSEN AND A.F. LOPEZ
Proof. Let E > 0 be such that E 2 = 0 and E.L = 3. By [KL1, Lemma 2.4] we can write
L ∼ 3E + F with F > 0, F 2 = 0 and E.F = 3. If φ(E + F ) = 1 then F ≡ 3E1 for
some E1 > 0 such that E12 = 0 and we are in case (i). If φ(E + F ) = 2 then we can
write E + F = E1 + E2 + E3 for some Ei > 0 with Ei2 = 0 and E.Ei = Ei .Ej = 1 for
i 6= j and we are in case (ii). To prove that H 1 (E + E1 + KS ) = 0 it is enough, by [KL2,
Cor. 2.5], to see that (E + E1 ).∆ ≥ −1 for every ∆ > 0 such that ∆2 = −2. Now if
(E + E1 ).∆ ≤ −2 then k := −E1 .∆ ≤ −2, whence, by [KL1, Lemma 2.3] we can write
E1 = A + k∆ with A > 0, A2 = 0. As L.∆ > 0 we can find a divisor D ∈ {E, E2 , E3 } such
that D.∆ ≥ 1, giving the contradiction 1 = D.E1 = D.A + kD.∆ ≥ 2. Similarly we can
prove that H 1 (E + Ei + Ej ) = 0.
Definition 2.6. Let |L| be a base-component free complete linear system on an Enriques
surface S such that L2 = 18 and φ(L) = 3. We say that L is of special type if L ∼
2E + E1 + E2 + E3 with either E2 + E3 − E1 > 0 or E2 + E3 + KS − E1 > 0.
Remark 2.7. Note that line bundles L of special type exist only on nodal Enriques surfaces
(that is that contain a smooth rational curve). In particular they do not exist on the general
Enriques surface.
3. Proof of Theorem 1.1.
Proof. Observe first that, being L very ample, we have φ(L) ≥ 3. Let Cη be a general
hyperplane section of S and set k = gon(Cη ). To see (i) it is enough to prove that Cη is
projectively normal and, as H 1 (L|Cη ) = 0 and deg L|Cη = 2g(Cη ) − 2, by [GL, Thm. 1],
we just need to prove that Cliff(Cη ) ≥ 3. Now gon(Cη ) ≥ 5 by [KL1, Cor. 1.5] whence
Cliff(Cη ) ≥ 3 by [KL3, Thm. 1.1]. This concludes the proof of (i).
To see (ii) note that if φ(L) = 3, then S contains plane cubics, therefore it has trisecant
lines and it cannot be scheme-theoretically cut out by quadrics. This proves the “only if”
part in (ii).
Vice versa suppose now L2 ≥ 18. We will prove (ii) and (iii) together. Let P ∈ Pr be a
point, let |VP | be the linear system cut out on S by hyperplanes passing through P and let
CP be a general element of |VP | . To prove (ii) we will suppose that φ(L) ≥ 4 and P ∈ S,
while to prove (iii) we will suppose that φ(L) = 3, P 6∈ S and P does not belong to the
union of the 2-planes spanned by cubic curves E ⊂ S such that E 2 = 0, E.L = 3.
Observe that, by [GLM2, Prop. 2.1], if φ(L) ≥ 4 then CP has no trisecant lines, while if
φ(L) = 3 then any trisecant line to CP belongs to some 2-plane spanned by a cubic curve
E ⊂ S such that E 2 = 0, E.L = 3. Therefore to see (ii) one easily sees that it is enough
to prove that, if P ∈ S, then CP is scheme-theoretically cut out by quadrics at P (see,
for example, the proof of [GLM2, Thm. 1.3]). Also, as S is linearly normal, any quadric
containing CP lifts to a quadric containing S, whence, to see (iii), it is enough to prove
that the intersection of the quadrics containing CP is CP union all its trisecant lines.
Now, to see (ii), by [L, Prop. 2.4.2] we see that CP is scheme-theoretically cut out by
quadrics as soon as Cliff(CP ) ≥ 4, while, to see (iii), by [LS, Thm. 1.3], we see that the
intersection of the quadrics containing CP is CP union all its trisecant lines as soon as
Cliff(CP ) ≥ 4. Setting k0 = gon(CP ), by [KL3, Thm. 1.1], we therefore need to show, in
both cases, that k0 ≥ 6.
Suppose now that φ(L) ≥ 4. By [KL1, Cor. 1.6 and Cor. 1.5], we are done except
possibly when L2 ≤ 22 and φ(L) = 4. If L2 = 20 or 22, by [KL1, Cor. 1.6 and Cor. 1.5],
we have k = 7, k0 ≥ 5 and if equality holds, by Proposition 2.4(i)-(ii), there exists a divisor
D on S with D 2 = 4, D.L = 9. The case L2 = 22 contradicts the Hodge index theorem,
PROJECTIVE NORMALITY AND THE IDEAL OF AN ENRIQUES SURFACE.
5
while if L2 = 20 we get (L − 2D)2 = 0 and the contradiction 4 = φ(L) ≤ L.(L − 2D) = 2.
If L2 = 18, by [KL1, Cor. 1.6 and Cor. 1.5], we have k = 6, k0 ≥ 4 and if equality
holds, by Proposition 2.4(i)-(ii), there exists a divisor D on S with D 2 = 4, D.L = 8,
contradicting the Hodge index theorem. Hence, to prove (ii), we are left with the cases
(L2 , φ(L), k0 ) = (18, 4, 5), (16, 4, 5) and (16, 4, 4).
Now suppose that φ(L) = 3 so that k = 6, k0 ≥ 4 by [KL1, Cor. 1.6 and Cor. 1.5] and
assume that k0 = 4, 5 with (k0 , L2 ) 6= (5, 18). By Proposition 2.4 there exists a divisor D
on S with either D 2 = 2, D.L = 6 or D 2 = 4, D.L = 8 when k0 = 4 and D 2 = 2, D.L ≤ 7
or D 2 = 4, D.L ≤ 9 when k0 = 5. Then the Hodge index implies that either k0 = 4 and
L2 = 18 or k0 = 5 and 20 ≤ L2 ≤ 24. By [KL1, Prop. 3.1](a) we have a decomposition
L ∼ N + N ′ with N ′ base-component free, 2(N ′ )2 ≤ L.N ′ ≤ (N ′ )2 + k0 ≤ 2k0 and if
φ(N ′ ) = 1 and L.N ′ ≥ (N ′ )2 + k0 − 1 then Bs |N ′ | ∩ CP 6= ∅. If (N ′ )2 = 0 then N ′ ∼ mB
for some m ≥ 1 and a genus one pencil B, giving the contradiction k0 ≥ L.N ′ ≥ 2φ(L) = 6.
If (N ′ )2 = 2 then L.N ′ ≥ 2φ(L) = 6 ≥ k0 + 1 = (N ′ )2 + k0 − 1, whence Bs |N ′ | ∩ CP 6= ∅,
a contradiction because CP is cut out by a general hyperplane passing through P 6∈ S.
Therefore (N ′ )2 = 4 and the Hodge index theorem applied to N ′ and L shows that we are
in the case k0 = 5, L2 = 20 and L.N ′ = 9. But this gives N.N ′ = 5, N 2 = 6, L.N = 11,
whence (N − N ′ )2 = 0 and the contradiction 3 = φ(L) ≤ L.(N − N ′ ) = 2. Hence, to prove
(iii), we are left with the case (L2 , φ(L), k0 ) = (18, 3, 5).
Note that if (L2 , φ(L)) = (18, 4), by [KL1, Prop. 1.4 and Lemma 2.14], there exist,
E > 0, Ei > 0 for i = 1, 2, such that E 2 = Ei2 = 0, E.E1 = E.E2 = 2, E1 .E2 = 1
and L ∼ 2E + E1 + E2 . Moreover we claim that there exists a divisor F > 0 be such that
F 2 = 0, F.L = 5 and P 6∈ Supp(F ). To see the latter, since E1 .L = E2 .L = 5, we can choose
as F one among E1 , E1 + KS , E2 , E2 + KS and we just need to show that their intersection
is empty. This is certainly true and well-known if either E1 or E2 are nef. If not, by [KL1,
Lemma 2.3], for i = 1, 2, there are nodal curves Ri such that, setting li = −Ri .Ei ≥ 1,
there exists an Ai > 0 with A2i = 0 and Ei ∼ Ai + li Ri . Now 5 = Ei .L = Ai .L + li Ri .L ≥ 5
shows that Ai .L = 4, li = 1 and Ai .Ri = 1. Again by [KL1, Lemma 2.3], Ai is nef for
i = 1, 2. Then Supp(Ai ) ∩ Supp(Ai + KS ) = ∅ and it remains to show that R1 .R2 = 0.
It cannot be E.Ai = 0, for then, by [KL2, Lemma 2.1], E ≡ Ai giving the contradiction
2 = E.E1 = E.Ri = Ai .Ri = 1. From 4 = L.Ai = 2E.Ai + E1 .Ai + E2 .Ai ≥ 4 we deduce
that E.Ai = Ej .Ai = 1 for all i, j and R1 .R2 = (E1 −A1 ).(E2 −A2 ) = −1+A1 .A2 . Therefore
it cannot be A1 .A2 = 0, for then R1 .R2 = −1, whence 1 = A1 .E2 = A1 .A2 + A1 .R2 ≥ 1
gives A1 .A2 = 1 and then R1 .R2 = 0.
We now treat together the two cases (L2 , φ(L), k0 ) = (18, 4, 5) or L is as in Lemma 2.5
(i) and k0 = 5. Let A be a g51 on CP and let us apply [KL1, Prop. 3.1].
In case (a) of that Proposition, we have a decomposition L ∼ N + N ′ with N ′ base′
component free, 2(N ′ )2 ≤ L.N ′ ≤ (N ′ )2 + 5 ≤ 10, N|C
≥ A, either N ≥ N ′ or N is
P
base-component free and Bs |N | ⊂ CP . Moreover if φ(N ′ ) = 1 and L.N ′ ≥ (N ′ )2 + 4, then
Bs |N ′ | ∩ CP 6= ∅. Now it cannot be (N ′ )2 = 0 for then, as usual, we get the contradiction
5 ≥ L.N ′ ≥ 2φ(L) ≥ 6. Also we cannot have (N ′ )2 = 2 for then if φ(L) = 4 we get the
contradiction 7 ≥ L.N ′ ≥ 2φ(L) = 8, while if φ(L) = 3, since L.N ′ ≥ 2φ(L) = 6 ≥ (N ′ )2 +4,
we get the contradiction Bs |N ′ | ∩ CP 6= ∅. Therefore (N ′ )2 = 4 and the Hodge index
theorem gives N ′ .L = 9, N.N ′ = 5 and N 2 = 4. Now L.(N − N ′ ) = 0 and (N − N ′ )2 = −2
show that H 0 (N − N ′ ) = H 1 (N − N ′ ) = 0. Then N is base-component free, whence
H 1 (N ) = H 1 (N + KS ) = 0.
′
− Z|. The exact sequence
Let Z ∈ |A| and let Z ′ ∈ |N|C
P
(4)
0 −→ −N −→ N ′ −→ (N ′ )|CP −→ 0
6
A.L. KNUTSEN AND A.F. LOPEZ
′
= Z + Z ′ . Now the exact sequence
shows that there exists D ′ ∈ |N ′ | such that D|C
P
(5)
0 −→ −N −→ JZ ′ /S (N ′ ) −→ A −→ 0
gives h0 (JZ ′ /S (N ′ )) = 2. Let
(6)
|H 0 (JZ ′ /S (N ′ ))| = |V | + G
be the decomposition into moving and fixed part, so that |V | is base-component free and
G ≥ 0. Let M = N ′ − G and pick a general divisor D0 ∈ |H 0 (JZ ′ /S (N ′ ))|, so that there
exist M ′ , M0 ∈ |V | such that D ′ = M ′ + G, D0 = M0 + G and M0 is general in |V |. In
particular M ′ and M0 have no common components. Now
(7)
Z ′ ⊆ D ′ ∩ D0 = (M ′ ∩ M0 ) ∪ G and length(Z ′ ) = N ′ .L − deg Z = (N ′ )2 .
If G > 0, as Z ′ ⊂ CP , (7) gives L.G ≥ 4 − M 2 . From 9 = L.N ′ = L.M + L.G ≥ L.M + 1 we
get L.M ≤ 8 and the Hodge index theorem implies that M 2 ≤ 2. Therefore L.G ≥ 2 and,
as M is base-component free and nontrivial, we have L.M ≥ 2φ(L). When φ(L) = 4 we
get the contradiction 9 = L.M + L.G ≥ 10. When φ(L) = 3 we have L.M ≥ 6, whence, as
L is 3-divisible, L.G = 3, M 2 = 2 and L.M = 6. The Hodge index theorem gives L ≡ 3M ,
whence M.G = 1. Now 4 = (N ′ )2 = M 2 + 2M.G + G2 gives G2 = 0 and G.N ′ = 1, so that
we get the contradiction Bs |N ′ | ∩ CP 6= ∅.
Hence G = 0 and length(Z ′ ) = (N ′ )2 = 4 = M 2 and therefore Z ′ = D ′ ∩ D0 , whence Z ′
is a Cartier divisor on D ′ . Now the exact sequence
(8)
0 −→ OS −→ JZ ′ /S (N ′ ) −→ JZ ′ /D′ (N ′ ) −→ 0
′ (−Z ′ ) = 0, whence, on D ′ we have N ′
′
shows that h0 (JZ ′ /D′ (N ′ )) = 1. But deg N|D
′
|D ′ ∼ Z
and therefore Z ∼ L|D′ − Z ′ ∼ N|D′ + (N ′ )|D′ − Z ′ ∼ N|D′ .
We have therefore proved that there exists a decomposition L ∼ N + N ′ with N ′ basecomponent free, N.N ′ = 5, H 1 (N ) = H 1 (N + KS ) = H 0 (N − N ′ ) = H 1 (N − N ′ ) = 0,
(N ′ )2 = N 2 = 4, a divisor D ′ ∈ |N ′ | and a divisor Z ∈ |N|D′ | such that Z ⊂ CP . Let X be
the family of such pairs (D ′ , Z) and consider the incidence subvariety of |L| × X:
J = {(C, (D ′ , Z)) : C ∈ |L|, (D ′ , Z) ∈ X, Z ⊂ C}
together with its two projections

/ |L| × X
J ●
●
⑥
●
●●π2
π1 ⑥⑥⑥
●●
⑥
●●
⑥
●#
~⑥⑥
|L|
X.
Of course we have proved that CP ∈ Im π1 . One easily checks that dim X = dim |N ′ | +
dim |N|D′ | = 4 and the dimension of a general fiber of π2 is at most h0 (JZ/S (L)) − 1. The
exact sequence
(9)
0 −→ N −→ JZ/S (L) −→ JZ/D′ (L) −→ 0
now shows that h0 (JZ/S (L)) = h0 (N ) + h0 ((N ′ )|D′ ) = 5, so that dim J ≤ 8. On the
other hand let C ∈ Im π1 be general, so that there exist D ′ ∈ |N ′ | and Z ∈ |N|D′ | such that
′ −Z
Z ⊂ C. Therefore Z ⊂ D ′ ∩C and we can find an effective divisor Z ′ such that Z ′ ∼ D|C
′ . Now
on C. Note that Z is Cartier on D ′ , whence, on D ′ , we find Z ′ ∼ L|D′ − Z ∼ N|D
′
0
′
0
′
(8) gives h (JZ ′ /S (N )) = 1 + h (OD ) ≥ 2 and from the exact sequence
0 −→ −N −→ JZ ′ /S (N ′ ) −→ JZ ′ /C (N ′ ) −→ 0
PROJECTIVE NORMALITY AND THE IDEAL OF AN ENRIQUES SURFACE.
7
we get H 0 (JZ ′ /S (N ′ )) ∼
= H 0 (JZ ′ /C (N ′ )) and h0 (JZ ′ /C (N ′ )) ≥ 2. Pick a general divisor
Z ′′ ∈ |JZ ′ /C (N ′ )|, so that there is a divisor D ′′ ∈ |N ′ | such that Z ′ ⊂ D ′′ , D ′′ ∩ C = Z ′′ + Z ′
and D ′′ is general in |H 0 (JZ ′ /S (N ′ ))|. Exactly as we did above we deduce that Z ′ = D ′ ∩D ′′ ,
whence Z ′ is a Cartier divisor on D ′′ . From the exact sequence
0 −→ OS −→ JZ ′ /S (N ′ ) −→ JZ ′ /D′′ (N ′ ) −→ 0
′
′
′ ∼ N′
′′
we get h0 (JZ ′ /D′′ (N ′ )) ≥ 1. Since deg N|D
′′ = deg Z we find Z
|D ′′ on D and
therefore Z ′′ ∼ N|D′′ . This shows that the at least one-dimensional family of pairs (Z ′′ , D ′′ )
belong to X and therefore any fiber of π1 has dimension at least one, so that dim Im π1 ≤ 7.
As dim |L| = 9 and the possible decompositions L ∼ N + N ′ as above are finitely many, we
have shown that this case cannot occur.
Therefore we must be in case (b) of [KL1, Prop. 3.1]. Let F > 0 be such that F 2 =
0, F.L = 5 and P 6∈ Supp(F ) in case φ(L) = 4 or F = F1 or E in case φ(L) = 3.
Since h1 (A) = 6 > F.L, we find as usual that h0 (E(CP , A)(−F )) > 0, whence we have a
decomposition L ∼ N + N ′ with N ≥ F , N ′ = N + R + KS base-component free, R a
′
≥ A and Bs |N ′ | ⊂ CP . It cannot be (N ′ )2 = 0 for then
nodal cycle, N.N ′ = 5, N|C
P
5 = N.N ′ + (N ′ )2 = L.N ′ ≥ 2φ(L) ≥ 6. Also we cannot have φ(N ′ ) = 1 (and, in particular,
(N ′ )2 = 2) for then we have the contradiction ∅ =
6 Bs |N ′ | ⊂ CP . Hence N ′ is base-point
′
2
2
free and (N ) ≥ 4. From N + 5 = L.N ≥ L.F we see that N 2 ≥ 0 when φ(L) = 4 and
N 2 ≥ −2 when φ(L) = 3. But if N 2 = −2 we get L.N = 3 ≥ L.F ≥ 3 so that N = F
giving the contradiction F 2 = −2. Therefore N 2 ≥ 0 and from 18 = L2 = N 2 + (N ′ )2 + 10
we deduce the three possibilities (N 2 , (N ′ )2 ) = (4, 4), (2, 6) or (0, 8). In the first case we get
the contradiction L.R = L.N ′ − L.N = 0. If (N 2 , (N ′ )2 ) = (2, 6) we have L.N = 7 which
is not divisible by 3, whence φ(L) = 4, giving the contradiction 7 = L.N ≥ 2φ(L) = 8. If
(N 2 , (N ′ )2 ) = (0, 8) we have L.N = 5, whence again φ(L) = 4 and 5 = L.N ≥ L.F = 5, so
that N = F .
Hence the proof in the case φ(L) = 3 and L as in Lemma 2.5(i) is completed.
When φ(L) = 4 we note that H 1 (N ) = H 1 (N + KS ) = 0 since L.N = 5 < 2φ(L) (see
for example [KLM, Lemma 2.5]). We also record that, since N − N ′ = −R + KS , we have
H 0 (N − N ′ ) = 0 and h1 (N − N ′ ) = 1.
′
− Z|. As above there exists D ′ ∈ |N ′ | such that
Let Z ∈ |A| and let Z ′ ∈ |N|C
P
′
= Z + Z ′ and again (5) gives h0 (JZ ′ /S (N ′ )) = 2. With the same notation as in (6)
D|C
P
we now deduce by (7) that, if G > 0, then L.G ≥ 8 − M 2 . From 13 = L.N ′ = L.M + L.G ≥
L.M + 1 we get L.M ≤ 12 and the Hodge index theorem implies that M 2 ≤ 8. If equality
holds we get 2L ≡ 3M and the contradiction 10 = 2L.E1 = 3M.E1 . If M 2 = 6 then
L.G ≥ 2 and L.M ≥ 3φ(L) = 12, giving the contradiction L.M + L.G ≥ 14. If M 2 = 2, 0
then L.G ≥ 6, 8 and L.M ≥ 2φ(L) = 8, giving the same contradiction. If M 2 = 4 then
L.G ≥ 4, L.M = 13 − L.G ≤ 9 and the Hodge index theorem gives L.M = 9 and L.G = 4.
We also prove, for later, that H 1 (L − M ) = H 0 (L − 2M ) = H 1 (L − 2M ) = 0. Suppose first
that h1 (L − M ) > 0. Then, by Riemann-Roch, as (L − M )2 = 4, we get h0 (L − M ) ≥ 4 and
we can write |L − M | = |B| + Γ with |B| base-component free, Γ ≥ 0 the fixed component
and h0 (B) ≥ 4. Also L.B ≤ L.(L − M ) = 9. But then either B 2 = 0 and B ∼ hQ for a
genus one pencil Q and h ≥ 3, giving the contradiction 9 ≥ L.B ≥ 6φ(L) = 24 or B 2 > 0 so
that H 1 (B) = 0 and Riemann-Roch gives B 2 ≥ 6, contradicting the Hodge index theorem.
This proves that H 1 (L − M ) = 0. Now (L − 2M )2 = −2 and L.(L − 2M ) = 0 easily gives
H 0 (L − 2M ) = H 1 (L − 2M ) = 0.
Suppose first that G = 0.
8
A.L. KNUTSEN AND A.F. LOPEZ
By (7) we have Z ′ = D ′ ∩ D0 , whence Z ′ is a Cartier divisor on D ′ . Now (8) gives
′
′ (−Z ′ ) = 0, whence, on D ′ we have N ′
h0 (JZ ′ /D′ (N ′ )) = 1. But deg N|D
′
|D ′ ∼ Z and
therefore Z ∼ L|D′ − Z ′ ∼ N|D′ + (N ′ )|D′ − Z ′ ∼ N|D′ .
We have therefore proved that there exists a decomposition L ∼ N + N ′ with N ′ basepoint free, N.N ′ = 5, (N 2 , (N ′ )2 ) = (0, 8), H 1 (N ) = H 1 (N + KS ) = 0, H 0 (N − N ′ ) = 0,
h1 (N − N ′ ) = 1, a divisor D ′ ∈ |N ′ | and a divisor Z ∈ |N|D′ | such that Z ⊂ CP . Let X be
the family of such pairs (D ′ , Z) and consider the incidence subvariety of |VP | × X:
(10)
J = {(C, (D ′ , Z)) : C ∈ |VP |, (D ′ , Z) ∈ X, Z ⊂ C}
together with its two projections
|VP |
④
π1 ④④④
④④
}④④
/ |VP | × X
❍❍ π2
❍❍
❍❍
❍❍
\$

J ❍
❍
X.
Of course, as CP ∈ Im π1 , we have proved that π1 is surjective. One easily checks that X
is irreducible and dim X = dim |N ′ | + dim |N|D′ | = 5. As Z moves on CP , we see that a
general element (D ′ , Z) ∈ X is such that P 6∈ Z, and the dimension of a general fiber of π2
is at most dim{T ∈ |VP | : Z ⊂ T } = h0 (JZ∪{P }/S (L)) − 1. Since P 6∈ Supp(F ), then either
P 6∈ Supp(D ′ ) and we find an effective divisor D ′ + F ∼ L such that P 6∈ Supp(D ′ + F ) and
Z ⊂ Supp(D ′ + F ) or P ∈ Supp(D ′ ) and the natural map H 0 (JZ/D′ (L)) → H 0 (L|{P } ) is
′ is base-point free and from (9) we get that H 0 (J
surjective since JZ/D′ (L) ∼
= N|D
′
Z/S (L)) →
0
0
0
H (JZ/D′ (L)) is surjective. Therefore h (JZ∪{P }/S (L))−1 = h (JZ/S (L))−2 and (9) gives
′ ) = 5, so that the dimension of a general fiber of π is at
h0 (JZ/S (L)) = h0 (N ) + h0 (N|D
′
2
most 3 and dim J ≤ 8. On the other hand, exactly as in case (a) above, any fiber of π1 has
dimension at least one, so that dim Im π1 ≤ 7, a contradiction.
Suppose now that G > 0, so that, as we have seen above, M 2 = 4, L.M = 9 and
L.G = 4. Recall that Z ′ ⊂ D ′ ∩ CP with D ′ = M ′ + G. Let ZG = G ∩ Z ′ ⊆ G ∩ CP
be the scheme-theoretic intersection and let ZM = Z ′ − ZG , seen as effective divisors
on CP . Then ZM ∩ G = ∅ and (7) shows that ZM ⊆ M ′ ∩ M0 . On the other hand
M 2 ≥ length(ZM ) = length(Z ′ ) − length(ZG ) ≥ length(Z ′ ) − L.G = N ′ .L − 9 = 4 = M 2
and therefore ZM = M ′ ∩ M0 , ZG = G ∩ CP and, in particular, ZM is a Cartier divisor on
M ′ . Moreover on CP we have Z + ZM + ZG = Z + Z ′ = D ′ ∩ CP = M ′ ∩ CP + G ∩ CP , so
that Z + ZM = M ′ ∩ CP . Also note that, on M ′ , we have Z ∼ L|M ′ − ZM ∼ (L − M )|M ′ .
We have therefore proved that there exists a decomposition L ∼ M + G + N with N > 0,
M base-component free, M 2 = 4, N 2 = 0, H 1 (L − M ) = H 0 (L − 2M ) = H 1 (L − 2M ) = 0,
a divisor M ′ ∈ |M | and a divisor Z ∈ |(L − M )|M ′ | such that Z ⊂ CP . Let X be the family
of such pairs (M ′ , Z) and consider the incidence subvariety of |VP | × X:
J = {(C, (M ′ , Z)) : C ∈ |VP |, (M ′ , Z) ∈ X, Z ⊂ C}
together with its two projections
④④
④④
④
}④
④
π1
|VP |
/ |VP | × X
❍❍ π2
❍❍
❍❍
❍❍
\$

J ❍
❍
④
X.
Of course, as CP ∈ Im π1 , we have proved that π1 is surjective. One easily checks that X
is irreducible and dim X = dim |M | + dim |(L − M )|M ′ | = 4. Now, exactly as before, the
PROJECTIVE NORMALITY AND THE IDEAL OF AN ENRIQUES SURFACE.
9
dimension of a general fiber of π2 is at most h0 (JZ∪{P }/S (L)) − 1 = h0 (JZ/S (L)) − 2 = 3,
so that dim J ≤ 7 and dim Im π1 ≤ 7, a contradiction.
We now deal with the case (L2 , φ(L), k0 ) = (18, 3, 5) and L is as in Lemma 2.5 (ii) and
not of special type (see Def. 2.6). Let A be a g51 on CP and let E := E(C, A) be the
vector bundle as defined in [KL1]. Consider the decomposition L + KS ∼ D1 + D2 with
D1 = E + E1 and D2 = E + E2 + E3 + KS . Since D12 + D22 = 8, we get by [KL1, Lemma
3.2], that h0 (E(−D0 )) > 0 with D0 = D1 or D0 = D2 and (L − D0 )|CP ≥ A. Saturating we
get an exact sequence
(11)
0 −→ N −→ E −→ N ′ ⊗ JX −→ 0
with N, N ′ ∈ Pic S, X a zero-dimensional scheme, L ∼ N + N ′ and N ≥ D0 . Moreover N ′
is base component free and nontrivial, Bs |N ′ | ⊂ CP ∪ X and (N ′ )|CP ≥ A.
Since N ′ ≤ L − D0 and h1 (L − D0 ) = 0 we get (N ′ )2 ≤ (L − D0 )2 ≤ 6. Also, taking c2
in (11), we find that N.N ′ ≤ N.N ′ + length(X) = 5.
Now if (N ′ )2 = 0 we get the contradiction 6 = 2φ(L) ≤ L.N ′ = N.N ′ ≤ 5. If (N ′ )2 = 2
then |N ′ | has two base points, whence, as CP is general, Bs |N ′ | ∩ CP = ∅ and therefore
length(X) ≥ 2 and N.N ′ ≤ 3, giving the contradiction 6 = 2φ(L) ≤ L.N ′ = 2 + N.N ′ ≤ 5.
If (N ′ )2 = 4 we have L.N ′ = 4 + N.N ′ ≤ 9 and then L.N ′ = 9 by the Hodge index theorem.
Therefore N.N ′ = 5 and X = ∅. As above it cannot be φ(N ′ ) = 1 for then |N ′ | has
two base points which must lie on CP and therefore φ(N ′ ) ≥ 2 giving the contradiction
9 = L.N ′ ≥ 5φ(N ′ ) ≥ 10. Hence (N ′ )2 = 6 and then D0 = E + E2 + E3 + KS , so that
N ′ ≤ E + E2 + E3 . But 11 = L.(E + E2 + E3 ) ≥ L.N ′ ≥ 11 by the Hodeg index theorem
′
− Z|.
and therefore N ′ = E + E2 + E3 and N = E + E1 . Let Z ∈ |A| and let Z ′ ∈ |N|C
P
′
′
′
′
As above, (4) shows that there exists D ∈ |N | such that D|CP = Z + Z and (5) gives
h0 (JZ ′ /S (N ′ )) = 2. With notation as in (6) we get N ′ = M + G with |M | base-component
free nontrivial and G ≥ 0. We now claim that G = 0. If G > 0 we get by (7) that
L.G ≥ 6 − M 2 . From 11 = L.N ′ = L.M + L.G we get L.M ≤ 10 and the Hodge
index theorem implies that M 2 ≤ 4. If M 2 = 0 we have L.G ≥ 6 and the contradiction
5 ≥ L.M ≥ 2φ(L) = 6. If M 2 = 2 we can write M = F1 + F2 with Fi > 0, Fi2 = 0 and
F1 .F2 = 1. Now L.G ≥ 4, whence 7 ≥ L.M = L.F1 + L.F2 . Since E is (numerically) the
only F > 0 such that F 2 = 0, L.F = 3 and the Ej ’s are (numerically) the only F > 0
such that F 2 = 0, L.F = 4, we deduce that F1 ≡ E, F2 ≡ Ej for some j such that
1 ≤ j ≤ 3. Then M ≡ E + Ej and G ≡ E2 + E3 − Ej . It cannot be j = 1, for this
case is excluded by the hypothesis that L not of special type, whence we can assume that
j = 2, M ≡ E + E2 and G ≡ E3 . Recall that Z ′ ⊂ D ′ ∩ CP with D ′ = M ′ + G. Let
ZG = G ∩ Z ′ ⊆ G ∩ CP be the scheme-theoretic intersection and let ZM = Z ′ − ZG , seen
as effective divisors on CP . Then ZM ∩ G = ∅ and (7) shows that ZM ⊆ M ′ ∩ M0 . On the
other hand M 2 ≥ length(ZM ) = length(Z ′ ) − length(ZG ) ≥ length(Z ′ ) − L.G = 2 = M 2
and therefore ZM = M ′ ∩ M0 , ZG = G ∩ CP and, in particular, ZM is a Cartier divisor on
M ′ . Moreover on CP we have Z + ZM + ZG = Z + Z ′ = D ′ ∩ CP = M ′ ∩ CP + G ∩ CP ,
so that Z + ZM = M ′ ∩ CP . But now M|CP ∼ A + ZM and h0 (M|CP ) = h0 (M ) = 2 and
then ZM = Bs |M | ⊂ CP , a contradiction. Therefore M 2 = 4, L.G ≥ 2 and L.M ≤ 9. Now
either φ(M ) = 1, but then we can write M = 2F1 + F2 with Fi > 0, Fi2 = 0, F1 .F2 = 1
and we get the contradiction 9 ≥ L.M = 2L.F1 + L.F2 ≥ 10 or φ(M ) ≥ 2, but then
M.N ≥ 2φ(M ) ≥ 4, M.N ′ ≥ 3φ(M ) ≥ 6 so that M.G ≥ 2, giving the contradiction
9 ≥ L.M = M 2 + M.G + M.N ≥ 10. This gives that G = 0.
We have therefore proved that there exists a decomposition L ∼ N + N ′ with N ′ basecomponent free, (N ′ )2 = 6, N 2 = 2, H 1 (N ) = 0 and H 1 (N −N ′ ) = H 1 (E2 +E3 +KS −E1 ) =
10
A.L. KNUTSEN AND A.F. LOPEZ
0, a divisor D ′ ∈ |N ′ | and a divisor Z ∈ |N|D′ | such that Z ⊂ CP . Let X be the family
of such pairs (D ′ , Z) and consider the incidence subvariety in (10) together with its two
projections. Of course we have proved that CP ∈ Im π1 , that is π1 is surjective. One easily
checks that dim X = dim |N ′ | + dim |N|D′ | = 4 and, using (9), the dimension of a general
fiber of π2 is at most 4. On the other hand, exactly as above, any fiber of π1 has dimension
at least one, so that dim Im π1 ≤ 7, a contradiction.
This concludes the proof for the case L2 = 18.
We now treat the cases L2 = 16, φ(L) = 4 and k0 = 4, 5. Let A be a gk10 on CP .
If k0 = 4 we are in case (a) of [KL1, Prop. 3.1], so that there is a decomposition
′
L ∼ N + N ′ with N ′ base-component free, 2(N ′ )2 ≤ L.N ′ ≤ (N ′ )2 + 4 ≤ 8 and N|C
≥ A.
P
′
′
′
2
′
Moreover if φ(N ) = 1 and L.N ≥ (N ) + 3, then Bs |N | ∩ CP 6= ∅. Now it cannot be
(N ′ )2 = 0 for this gives the contradiction 4 ≥ L.N ′ ≥ 2φ(L) ≥ 8. Also we cannot have
(N ′ )2 = 2 for then 6 ≥ L.N ′ ≥ 2φ(L) ≥ 8. Therefore (N ′ )2 = 4 and the Hodge index
theorem gives L ≡ 2N ′ , N ≡ N ′ and H 1 (N ) = H 1 (N + KS ) = 0.
′
−Z|. As above, (4) shows that there exists D ′ ∈ |N ′ | such
Let Z ∈ |A| and let Z ′ ∈ |N|C
P
′
that D|C
= Z + Z ′ and (5) gives h0 (JZ ′ /S (N ′ )) = 2. With notation as in (6) we get N ′ =
P
M +G with |M | base-component free nontrivial and G ≥ 0. From 8 = L.N ′ = L.M +L.G ≥
2φ(L) + L.G ≥ 8 we get that G = 0 and (7) gives length(Z ′ ) = (N ′ )2 = 4 = M 2 . Therefore
Z ′ = D ′ ∩ D0 , whence Z ′ is a Cartier divisor on D ′ . Now (8) gives h0 (JZ ′ /D′ (N ′ )) = 1.
′ (−Z ′ ) = 0, whence, on D ′ we have N ′
′
But deg N|D
′
|D ′ ∼ Z and therefore Z ∼ N|D ′ .
We have therefore proved that there exists a decomposition L ∼ N + N ′ with N ′ basecomponent free, N ≡ N ′ , H 1 (N ) = H 1 (N + KS ) = 0, (N ′ )2 = 4, a divisor D ′ ∈ |N ′ | and a
divisor Z ∈ |N|D′ | such that Z ⊂ CP . Let X be the family of such pairs (D ′ , Z) and consider
the incidence subvariety in (10) together with its two projections. Of course we have proved
that CP ∈ Im π1 , that is π1 is surjective. One easily checks that dim X = dim |N ′ | +
dim |N|D′ | ≤ 4 and the dimension of a general fiber of π2 is at most h0 (JZ∪{P }/S (L)) − 1.
Now either P 6∈ Supp(D ′ ) and, as N is base-point free, we find an effective divisor D ′ +N1 ∼
L, N1 ∈ |N |, such that P 6∈ Supp(D ′ + N1 ) and Z ⊂ Supp(D ′ + N1 ) or P ∈ Supp(D ′ ) and
′
the natural map H 0 (JZ/D′ (L)) → H 0 (L|{P } ) is surjective since JZ/D′ (L) ∼
= N|D
′ is base0
0
point free and from (9) we get that H (JZ/S (L)) → H (JZ/D′ (L)) is surjective. Therefore
′ ) = 5,
h0 (JZ∪{P }/S (L)) = h0 (JZ/S (L)) − 1 and (9) gives h0 (JZ/S (L)) = h0 (N ) + h0 (N|D
′
so that the dimension of a general fiber of π2 is at most 3 and dim J ≤ 7. On the other
hand, exactly as above, any fiber of π1 has dimension at least one, so that dim Im π1 ≤ 6,
This proves that the case L2 = 16, φ(L) = 4 and k0 = 4 does not occur.
Assume now L2 = 16, φ(L) = 4 and k0 = 5. Recall that, by [KL1, Prop. 1.4], there exist
E > 0, E1 > 0 such that E 2 = E12 = 0, E.E1 = 2 and L ≡ 2(E + E1 ). Set D1 = E + E1
and D2 = L + KS − D1 , so that h1 (D1 + KS ) = 0 and [KL1, Lemma 3.2] gives a divisor
D0 ≡ E + E1 such that h0 (E(−D0 )) > 0, where E = E(CP , A).
As in the proof of [KL3, Lemma 2.4] we have an exact sequence
0 −→ N −→ E −→ N ′ ⊗ JW −→ 0
with N, N ′ ∈ Pic S, W a zero-dimensional scheme, N ≥ D0 and L ∼ N + N ′ . Moreover N ′
′
is base-component free and nontrivial, N|C
≥ A and N.N ′ ≤ N.N ′ + length(W ) = 5.
P
Since N ′ ≤ D0 and both are nef, we get (N ′ )2 ≤ D02 = 4. If (N ′ )2 = 0, 2 we get
the contradiction 8 = 2φ(L) ≤ L.N ′ = (N ′ )2 + N.N ′ ≤ 7. Therefore (N ′ )2 = 4 and
PROJECTIVE NORMALITY AND THE IDEAL OF AN ENRIQUES SURFACE.
11
L.N ′ = (N ′ )2 + N.N ′ ≤ 9, whence, as L is 2-divisible, L.N ′ ≤ 8. Now the Hodge index
theorem gives L ≡ 2N ′ and this case can be excluded exactly as the previous one.
This concludes the proof for the case L2 = 16, φ(L) = 4.
Remark 3.1. With our methods, we could also extend Theorem 1.1(iii) when (L2 , φ(L)) =
(16, 3) or (18, 3). Here, for the general hyperplane section Cη of S, we have that gon(Cη ) = 6
and, as above, if CP is cut out on S by a general hyperplane section of passing through a
point P ∈ Pr − S, one easily excludes the case gon(CP ) = 4, but we do not know if it can
be gon(CP ) = 5. If this can be excluded then (iii) holds for S also in these cases.
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Andreas Leopold Knutsen, Department of Mathematics, University of Bergen, Postboks 7800,
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Angelo Felice Lopez, Dipartimento di Matematica e Fisica, Università di Roma Tre, Largo
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