On rationality of nonsingular threefolds with a pencil of Del Pezzo surfaces of degree 4

0.2 MB·English

by  Constantin Shramov

On rationality of nonsingular threefolds
with a pencil of Del Pezzo surfaces of degree 4
7
0
0
C.SHRAMOV
2

n
a
J

8 Abstract. We prove a criterion of nonsingularity of a complete
1 intersection of two fiberwise quadrics in PP1(O(d1)⊕...⊕O(d5)).

AsacorollarywederivethefollowingadditiontotheAlexeevtheo-
]
G remonrationalityofstandardDelPezzofibrationsofdegree4over
A P1: we prove that any fibration of this kind with the topological
Euler characteristic χ(X)=−4 is rational.
.
h
t
a 1. Introduction
m
[ Rationality questions of Del Pezzo fibrations of degree d over P1 were

studied by many authors. All the varieties of this type with d > 5 are
1
v rational. Fibrations with d = 1, 2 and 3 were studied, in particular,
1
in [5], [2], [9], [3], [4], [8], and as a result a nearly complete solution of
2
5 rationality problem of nonsingular Del Pezzo fibrations was obtained
1 for d = 1 and (under generality assumptions) for d = 2, 3.
0
In case of degree 4 there is a well-known statement.
7
0
Theorem (Alexeev, see [1]). Let V be a standard Del Pezzo fibration1
/
h of degree 4 over P1. If the topological Euler characteristic χ(V) 6=
t
a 0,−8,−4, then V is nonrational, if χ(V) = 0,−8, then V is rational;
m
finally, if χ(V) = −4, then V if and only if its intermediate Jacobian
:
v is a Jacobian of a curve.
i
X
The main goal of this paper is to prove the following addition to
r
a Alexeev theorem.
Theorem 1.1. Any standard Del Pezzo fibration of degree 4 over P1
with topological Euler characteristic −4 is rational.
This statement will appear as a corollary of the description of in-
tersections of fiberwise quadrics in the scrolls with topological Euler
characteristic equal to −4 (see. Lemma 2.4 after the necessary nota-
tions are introduced), that in turn is a corollary of the nonsingularity
The workwaspartiallysupportedbyRFFI grants04−01−00613and05−01−
00353.
1See section 2 for a definition.
1
criterion for intersections of fiberwise quadrics in a fivefold scroll (The-
orem 2.2).
The author is grateful to I.A.Cheltsov for his nondecreasing interest
to this work and for numerous useful advice, to V.A.Iskovskikh for
attention to the work and to S.S.Galkin for valuable discussions.
2. Definitions, notations and the statements of the main
results
All varieties are assumed to be defined over the field of complex
numbers C.
Definition 2.1 (see, e.g., [1]). A nonsingular threefold X is called a
standard Del Pezzo fibration over P1, if X is endowed with a structure
of Del Pezzo fibration over P1 with normal fibers, and ρ(X) = 2.
Letφ : X → P1 beastandardDelPezzo fibrationofdegreedoverP1.
If d > 3, then X is naturally embedded into the projectivisation PP1(E)
of the bundle E = φ∗ω−1. For example, if d = 4, then X is embedded
X
intosomescrollY = PP1(O(d1)⊕...⊕O(d5))asanonsingular complete
intersection of two divisors, such that the restriction of each of them
on a fiber is a quadric in P4. Hence we may always assume that X is
embedded into Y as a complete intersection of two fiberwise quadrics.
On the other hand, if X is a complete intersection of this kind, and
X is nonsingular with ρ(X) = 2, then X is a standard Del Pezzo
fibration of degree 4. It means that to describe Del Pezzo fibrations of
degree 4 over P1 explicitly the first thing we need is a way to decide
if an intersection of two general members of given linear systems |D |
1
and |D | of fiberwise quadrics on Y is nonsingular (or, equivalently, if
2
there exist such divisors in these linear systems that their intersection
is nonsingular).
Let us fix some notations to formulate the answer to the latter
question. Let Y = F(d1,...,d5) = PP1(O(d1) ⊕ ... ⊕ O(d5)), where
d > ... > d = 0 (the definition and the main properties of Y are
1 5
given, for example, in [10]). Let M be a tautological invertible sheaf
Y
on Y, L be a fiber of the natural projection ϕ : Y → P1 (we’ll often
Y
denotethembyM andLinsteadofM andL ifnoambiguity islikely
Y Y
to arise). Let D ∈ |2M +b L|, i = 1,2, be general divisors, given by
i i
equations f (x,t) = α x x and f (x,t) = β x x respectively,
1 ij i j 2 ij i j
where xi are standardPcoordinates in a fiber ofPϕ, and αij = αij(t0,t1)
(resp. β (t ,t )) are the polynomials in the coordinates on the base
ij 0 1
of degree d + d + b (resp. d + d + b ). We are interested in the
i j 1 i j 2
conditions on d ,b implying that X = D ∩D is a nonsingular variety.
i j 1 2
2
Let Y ,...,Y be negative subscrolls of Y (i.e. Y is a variety given
2 5 i
by the equations x1 = ... = xi−1 = 0). It’s easy to see that the base
locus Bs|D | must coincide with Y , Y , Y or ∅. We’ll always assume
1 3 4 5
that b 6 b (in particular, Bs|D | ⊂ Bs|D |).
1 2 2 1
The following theorem gives criterion for X being nonsingular in
terms of the parameters d , b . Note that our argument is analogous
i j
to one applied to the case of Del Pezzo fibrations of degree 3 in [8,
Lemma 26] and [7, Proposition 31, 32].
Theorem 2.2. Under the assumptions listed above a general variety X
is nonsingular if and only if one (and hence only one) of the following
sets of conditions holds.
1. b > 0.
1
2. b < 0, 2d +b > 0, and one of the following sets of conditions
1 4 1
holds
(a) d +b < 0, b = 0.
1 1 2
(b) d +b > 0, b > 0.
1 1 2
(c) b = −d , b < 0, d +b > 0.
1 1 2 3 2
(d) b = −d , b = −d , d > d .
1 1 2 2 2 3
(e) d +b > 0, d +b > 0, b < 0, d +b > 0.
1 1 2 1 2 3 2
3. 2d + b < 0, 2d + b > 0, and one of the following sets of
4 1 3 1
conditions holds
(a) d > d , d > 0, b = −(d +d ), b = 0.
1 2 4 1 1 4 2
(b) d > d , d = 0, b = −d , b = 0.
1 2 4 1 1 2
(c) d > d +d , d > 0, b = −d , b = −2d , d +b > 0 or
1 2 4 4 1 1 2 4 3 2
d +b = 0.
2 2
(d) d +b < 0, d +d +b > 0, b = 0.
1 1 2 4 1 2
(e) d +b > 0, d +d +b > 0, d +b < 0, d +d +b < 0,
1 1 2 4 1 2 1 3 4 1
b > 0.
2
(f) d > d , d > d + d , d + d > d , b = −d , b < 0,
1 2 1 3 4 2 4 1 1 1 2
2d +b > 0, d +b > 0 or d +b = 0.
4 2 3 2 2 2
(g) d = d +d , d > d , d > 0, b = −d , b < 0, 2d +b > 0,
1 2 4 2 3 4 1 1 2 4 2
d +b > 0 or d +b = 0.
3 2 2 2
(h) d = d +d , d = d +d , d > 0, b = −d , b = −d .
1 2 4 2 3 4 4 1 1 2 2
(i) d +b > 0, d +d +b > 0, d +b < 0, b > 0.
1 1 3 4 1 2 1 2
(j) d + d > d , b = −d , b < 0, 2d + b > 0, d + b > 0
3 4 1 1 1 2 4 2 3 2
d +b = 0.
2 2
(k) d +b > 0, d +d +b > 0, b > 0.
2 1 3 4 1 2
(l) d +b > 0, d +d +b > 0, b < 0, 2d +b > 0, d +b > 0.
2 1 3 4 1 2 4 2 3 2
(m) d = d = d > 0, d = 0, b = b = −d .
1 2 3 4 1 2 1
4. 2d + b < 0, 2d + b > 0, and one of the following sets of
3 1 2 1
conditions holds
3
(a) d +d = d +d , d > d > 0, b = −(d +d ), b = 0.
1 4 2 3 3 4 1 1 4 2
(b) d = d > d = d > 0, b = −(d +d ), b = 0.
1 2 3 4 1 1 4 2
(c) d = d +d , d > d > 0, b = −d , b = −2d , d +b > 0
1 2 3 3 4 1 1 2 4 3 2
or d +b = 0.
2 2
(d) d = d +d , d > d = 0, b = −d , b = 0.
1 2 3 3 4 1 1 2
(e) d = d , d = d = 0, b = −d , b = 0.
1 2 3 4 1 1 2
(f) d > 0, d = d , d = 2d , d = 3d , b = −3d , b = −2d .
4 3 4 2 4 1 4 1 4 2 4
Remark 2.3. It’s easy to check that for each set of conditions in Theo-
rem 2.2 there is a variety X such that those conditions hold for X.
Theorem 2.2 is proved in section 4. Its awkwardness is partially
compensated by the following corollary (see section 5).
Lemma 2.4. Let X be a standard Del Pezzo fibration of degree 4 over
P1, and χ(X) = −4. Then X is isomorphic to a complete intersection
of two fiberwise quadrics in a scroll, and there are only the following
possibilities for parameters d , b .
i j
(X ) d = d = d = d = 0, b = 0, b = 1 (case 1 of Theorem 2.2).
1 1 2 3 4 1 2
(X ) d = 2, d = d = d = 1, b = −2, b = −1 (case 2c of
2 1 2 3 4 1 2
Theorem 2.2).
Finally, Theorem 1.1 is an implication of Lemma 2.4 provided that
the varieties X and X are rational (this is checked in section 5).
1 2
3. Preliminaries
Lemma 3.1. The variety X is nonsingular if and only if the following
conditions hold:
(∗) the intersection of Bs|D |\Bs|D | with D ∩SingD is empty,
1 2 2 1
(∗∗) in any point of Bs|D | the vectors grad (f ) and grad (f ) are
2 x 1 x 2
not proportional.
Proof. If X is nonsingular, than the conditions (∗) and (∗∗) apparently
hold. Assume that the conditions (∗) and (∗∗) hold. In the points
of D ∩ (D \ Bs|D |) the variety D is nonsingular by (∗), and the
1 2 2 1
divisor D is movable, so X is nonsingular outside Bs|D | by Bertini
2 2
theorem. In the points of Bs|D | the variety X is nonsingular if and
2
only if the vectors grad(f ) = (grad (f ),grad (f )) and grad(f ) =
1 x 1 t 1 2
(grad (f ),grad (f )) are not proportional in any point of Bs|D |. Let
x 2 t 2 2
D′, i = 1,2, be divisors given by equations ∂fi = 0. Then |D′| = |2M+
i ∂t0 i
′
(b − 1)L|, and Bs|D | ⊃ Bs|D |. Hence grad (f ) and grad (f ) are
i i i t 1 t 2
zero onBs|D |, andthe nonproportionalitycondition may berewritten
2
as (∗∗). (cid:3)
4
To check that X is nonsingular we shall use the conditions (∗) and
(∗∗) everywhere below.
Let’s fix some notations. Let M denote the 2×3-matrix
4
α x +α x α x +α x α x +α x
M = 14 4 15 5 24 4 25 5 34 4 35 5 ,
4 (cid:18) β x +β x β x +β x β x +β x (cid:19)
14 4 15 5 24 4 25 5 34 4 35 5
and M — the 2×4-matrix
5
α α α α
M = 15 25 35 45 .
5 (cid:18) β β β β (cid:19)
15 25 35 45
Let m(l), 1 6 i < j 6 3, l = 4,5 denote a 2 × 2-minor of the matrix
ij
M , containing its i’th and j’th columns.
l
Consider the following conditions:
(4)
(∗∗) the polynomials m have no common zeros on Y .
4 ij 4
(5)
(∗∗) the polynomials m have no common zeros on Y .
5 ij 5
(They are useful because the condition (∗∗) is equivalent to (∗∗) if
4
Bs|D | = Y , and to (∗∗) if Bs|D | = Y .)
2 4 5 2 5
Lemma 3.2. Under the assumptions made above the condition (∗∗)
4
holds if and only if b = −d = −(d +d ), b = −d = −(d +d ).
1 1 2 4 2 2 3 4
Proof. Clearly, to satisfy (∗∗) we need the last column of M to be
4 4
nonzero, hence it is necessary d +d +b > 0. Moreover, if β = 0 or
3 4 2 25
α = 0, then m(4) are zero on Y , hence (∗∗) implies d +b > 0 and
15 ij 5 4 2 2
d +b > 0. Finally, if α = 0, then (∗∗) also doesn’t hold, so (∗∗)
1 1 24 4 4
implies d +d +b > 0.
2 4 1
Now assume that the inequalities listed above hold. Consider
the surface Y ∼= F(d ,0), the divisors C ∈ |2M + (d + d +
4 4 ij Y4 i j
(4)
b + b )L | on Y , given by the equations m = 0, the divisor
1 2 Y4 4 ij
A ∈ |M + (d + b )L |, given by the equation α x + α x = 0,
Y4 1 1 Y4 14 4 15 5
and the divisor B ∈ |M + (d + b )L |, given by the equation
Y4 1 2 Y4
β x + β x = 0. The divisors C , A and B are movable, C in-
14 4 15 5 12 12
tersects C and A intersects B transversally. Hence, C ∩C consists
13 12 13
of C C = 4(d +d +b +b )+2(d +d ) points, and A∩B consists
12 13 1 4 1 2 2 3
of AB = 2d +d +b +b points.
1 4 1 2
If C C = 0, then the condition (∗∗) holds; the equality
12 13 4
0 = C C = 4(d +b )+2(d +b )+2(d +d +b )+2d
12 13 1 1 2 2 3 4 2 4
is equivalent (under the above assumptions) to b = b = −d = −d =
1 2 1 2
(4) (4)
−d , d = 0. Let C C 6= 0. Note that the vanishing of m and m
3 4 12 13 12 13
(4)
implies the vanishing of m provided that the first column of M is
23 4
nonzero. Hence if C C 6= 0, then for the condition (∗∗) to hold it is
12 13 4
5
necessaryandsufficient thatC ∩C ⊂ A∩B. SinceA∩B ⊂ C ∩C ,
12 13 12 13
the latter is equivalent to
0 = (4(d +d +b +b )+2(d +d ))−(2d +d +b +b ) =
1 4 1 2 2 3 1 4 1 2
= 2(d +d +d )+3(d +b +b ) =
1 2 3 4 1 2
= 2(d +b )+(d +b )+(d +d +b )+2(d +d +b ).
1 1 2 2 2 4 1 3 4 2
Under the above assumptions this equality is equivalent to b = −d =
1 1
−(d + d ), b = −d = −(d + d ). To get the final answer note
2 4 2 2 3 4
that these conditions are weaker than those that appeared in the case
C C = 0. (cid:3)
12 13
Lemma 3.3. Under the assumptions made above the condition (∗∗)
5
holds if and only if either the conditions d + b > 0 and d +b > 0
1 1 3 2
hold, or the conditions d + b = 0, d + b = 0 hold together with at
1 1 2 2
least one of the conditions d +b = 0 and d +b < 0.
1 2 2 1
Proof. If d +b < 0, then the first row of M is zero, so d +b > 0 is
1 1 5 1 1
necessaryfor(∗∗) . Ifd +b > 0,then(∗∗) holdseitherwhenthereare
5 1 1 5
(5) (5) (5)
two nonzero minorsm (we may assume that these arem andm ),
ij 12 13
i.e. when deg(β ) = d + b > 0; or when deg(m(5)) = deg(α β −
25 3 2 12 15 25
α β ) = 0, i.e. when deg(α ) = d +b = 0, deg(β ) = d +b = 0
25 15 15 1 1 25 2 2
and one of the following conditions holds: deg(β ) = d + b = 0 or
15 1 2
deg(α ) = d +b < 0. (cid:3)
25 2 1
The following statement will simplify the calculations in section 4.4.
Lemma 3.4. Let Bs|D | = Y , d +d +b > 0. Then for the condition
1 3 2 3 1
(∗) to hold it is necessary that 2(d +d +d +d )+4b +b = 0. If
1 2 3 4 1 2
the latter holds and either d + d + b > 0, b > 0, or d + b > 0,
1 4 1 2 1 1
2d +b > 0, then this condition is also sufficient.
4 2
Proof. If Bs|D | = Y , then all the components of grad (f ) with the
1 3 x 1
possible exception of ∂f1 = α x +α x + α x and ∂f1 = α x +
∂x1 13 3 14 4 15 5 ∂x2 23 3
α x +α x , are zero on Y . Since d +d +b > 0, the polynomials
24 4 25 5 3 2 3 1
α and α are nonzero.
23 13
Let A and A be the divisors given by the equations α x +α x +
1 2 13 3 14 4
α x = 0 andα x +α x +α x = 0 respectively. Since Sing(D ) =
15 5 23 3 24 4 25 5 1
A ∩ A ∩ Y , the condition (∗) means that D ∩ A ∩ A ∩ Y = ∅.
1 2 3 2 1 2 3
In particular, A A D | = 0 must hold, i.e., as |A | = |M + (d +
1 2 2 Y3 1 Y3 1
b )L |, |A | = |M + (d + b )L |, and Y ∼= F(d ,d ,0), the latter
1 Y3 2 Y3 2 1 Y3 3 3 4
condition may be rewritten as 2(d +d +d +d )+4b +b = 0.
1 2 3 4 1 2
If d +d +b > 0, then A andA have no common components, and
1 4 1 1 2
their intersection is an effective curve on Y , hence if Y 6⊂ Bs|D | the
3 5 2
6
condition A A D | = 0 implies D ∩A ∩A ∩Y = ∅. If d +b > 0,
1 2 2 Y3 2 1 2 3 1 1
then A and A have no common components, and Y 6⊂ A , so that if
1 2 5 1
Y 6⊂ Bs|D | the condition A A D | = 0 implies D ∩A ∩A ∩Y =
4 2 1 2 2 Y3 2 1 2 3
∅. (cid:3)
4. Proof of Theorem 2.2
4.1. Case Bs|D | = ∅. This case occurs if and only if b > 0. The
1 1
conditions (∗) and (∗∗) hold automatically (case 1 of Theorem 2.2).
4.2. Case Bs|D | = Y . This case occurs if and only if b < 0, 2d +
1 5 1 4
b > 0. All the components of grad (f ) with the possible exception of
1 x 1
∂f1 = α , ∂f1 = α , ∂f1 = α and ∂f1 = α are zero on Y .
∂x1 15 ∂x2 25 ∂x3 35 ∂x4 45 5
Let us consider several possibilities.
4.2.1. Case d +b < 0. We have grad (f ) = 0 on Y . The condition
1 1 x 1 5
(∗∗) implies Bs|D | = ∅, and (∗) implies that D ∩Y = ∅, i.e. b = 0
2 2 5 2
(case 2a of Theorem 2.2). This condition is apparently sufficient for
(∗) and (∗∗).
4.2.2. Case d +b = 0. Under this assumption we have deg(α ) = 0.
1 1 15
If Bs|D | = ∅ (i.e. b > 0), then (∗) and (∗∗) hold automatically
2 2
(case 2b of Theorem 2.2). If Bs|D | = Y (b < 0), it is necessary
2 5 2
and sufficient to check the condition (∗∗) on Y . Since any pair of
5
polynomials α , β has no common zeros on Y , the condition (∗∗)
ij kl 5
is equivalent to (∗∗) . Hence by Lemma 3.3 the condition (∗∗) holds
5
either if d +b > 0 (case 2c of Theorem 2.2), or if d +b = 0 and one
3 2 2 2
of the following conditions holds: d +b = 0 or d +b < 0; it’s easy to
1 2 2 1
see that under the assumptions made above d + b = 0 is equivalent
1 2
to b = b , and d +b < 0 is equivalent to b < b , hence one of these
1 2 2 1 1 2
two holds automatically — this gives case 2d of Theorem 2.2.
4.2.3. Case d +b > 0. Ifd +b < 0, thenD hasisolatedsingularities
1 1 2 1 1
on Y , and by (∗) it is necessary that Bs|D | = ∅, i.e. b > 0. The
5 2 2
condition b > 0 is apparently sufficient for (∗) and (∗∗) (regardless
2
to d + b < 0). This is case 2b of Theorem 2.2. If d + b > 0 and
2 1 2 1
b < 0, then Bs|D | = Y , and (∗∗) is equivalent to (∗∗) , i.e. (since
2 2 5 5
d +b 6= 0)byLemma 3.3thecondition(∗∗)isequivalent tod +b > 0
1 1 3 2
(case 2e of Theorem 2.2).
4.3. Case Bs|D | = Y . This case occurs if and only if 2d + b < 0,
1 4 4 1
2d +b > 0. Under this assumption all the components of grad (f )
3 1 x 1
with the possible exception of ∂f1 = α x +α x , ∂f1 = α x +α x
∂x1 14 4 15 5 ∂x2 24 4 25 5
and ∂f1 = α x +α x are zero on Y .
∂x3 34 4 35 5 4
7
Let us consider several possibilities.
4.3.1. Case d + d + b < 0. We have Sing(D ) = Y ; in particular,
1 4 1 1 4
Sing(D ) contains a line l in a fiber of ϕ. Since D ∈ |2M +b L|, we
1 2 2
have l∩D 6= ∅, and the condition (∗) doesn’t hold.
2
4.3.2. Case d + d + b > 0, d + b < 0, d + d + b < 0. We
1 4 1 1 1 2 4 1
have α = α = α = α = α = 0, and if deg(α ) > 0, then
15 24 25 34 35 14
the divisor D is singular along some line in a fiber of ϕ, so (∗) implies
1
deg(α ) = 0, i.e. d +d +b = 0. Furthermore, asD is singular along
14 1 4 1 1
Y , the condition (∗) implies b = 0. These conditions are apparently
5 2
sufficient for (∗) and (∗∗) (case 3a of Theorem 2.2).
4.3.3. Case d + b > 0, d + d + b < 0. We have α = α =
1 1 2 4 1 24 25
α = α = 0, and Sing(D ) = C, where the curve C is given by the
34 35 1
equation α x + α x = 0 on Y . Since C 6= ∅, the condition (∗)
14 4 15 5 4
implies Bs|D | ⊂ Y , i.e. 2d +b > 0. The condition (∗) also implies
2 5 4 2
0 = D C = 2(d +d +b )+b = 2(d +b )+2d +b ,
2 1 4 1 2 1 1 4 2
i.e. d +b = 0, 2d +b = 0. Since C 6⊃ Y , this is sufficient for (∗). If
1 1 4 2 5
d = 0, then (∗∗) holds automatically (case 3b of Theorem 2.2), and if
4
d > 0, then Bs|D | = Y , and by Lemma 3.3 the condition (∗∗) holds
4 2 5
either if d +b > 0, or if d +b = 0 (case 3c of Theorem 2.2).
3 2 2 2
4.3.4. Case d +b < 0, d +d +b > 0. We have α = α = α = 0.
1 1 2 4 1 15 25 35
Since the polynomials α and α have no common zeros, Sing(D ) =
14 24 1
Y . The condition (∗) implies b = 0. The latter is apparently sufficient
5 2
for (∗) and (∗∗) (case 3d of Theorem 2.2).
4.3.5. Case d +b > 0, d +d +b > 0, d +b < 0, d +d +b < 0. We
1 1 2 4 1 2 1 3 4 1
have α = α = α = 0, and D has at most isolated singularities. If
25 34 35 1
d +b > 0, then D has some singular points on Y , hence (∗) implies
1 1 1 5
b > 0. The latter is apparently sufficient for (∗) and (∗∗) regardless
2
to the assumption d +b > 0 (case 3e of Theorem 2.2).
1 1
If d + b = 0, b < 0 and d + d + b > 0, then D has some
1 1 2 2 4 1 1
singular points on Y \Y , and by (∗) it is necessary that Bs|D | =6 Y ,
4 5 2 4
i.e. 2d +b > 0, hence Bs|D | = Y . By Lemma 3.3 in this case the
4 2 2 5
condition (∗∗) holds if and only if either d + b > 0, or d + b = 0
3 2 2 2
(case 3f of Theorem 2.2).
If d +b = 0, b < 0 and d +d +b = 0, then D is nonsingular.
1 1 2 2 4 1 1
If under these assumptions Bs|D | = Y , i.e. 2d + b > 0, then by
2 5 4 2
Lemma 3.3 the condition (∗∗) holds if and only if either d +b > 0, or
3 2
d +b = 0 (case 3g of Theorem 2.2). If Bs|D | = Y , i.e. 2d +b < 0,
2 2 2 4 4 2
8
then (∗∗) is equivalent to (∗∗) . By Lemma 3.2 this is equivalent to
4
b = −d = −(d +d ),b = −d = −(d +d )(case3hofTheorem2.2).
1 1 2 4 2 2 3 4
4.3.6. Case d + b > 0, d + d + b > 0, d + b < 0. We have
1 1 3 4 1 2 1
α = α = 0. If d +b > 0, then D has (isolated) singularities on
25 35 1 1 1
Y , hence (∗) implies b > 0. The latter is sufficient for (∗) and (∗∗)
5 2
regardless to the assumption d +b > 0 (case 3i of Theorem 2.2).
1 1
If d + b = 0, b < 0, then D is nonsingular. If Bs|D | = Y ,
1 1 2 1 2 5
i.e. 2d +b > 0, then by Lemma 3.3 the condition (∗∗) holds if and
4 2
only if either d + b > 0, or d + b = 0 (case 3j of Theorem 2.2).
3 2 2 2
If Bs|D | = Y , then (∗∗) is equivalent to (∗∗) . By Lemma 3.2 the
2 4 4
latter implies d = d +d , contradicting the assumptions d +b < 0,
2 3 4 2 1
d +d +b > 0.
3 4 1
4.3.7. Case d +d +b > 0, d +b > 0. The divisor D is nonsingular.
3 4 1 2 1 1
If b > 0, then (∗∗) holds automatically (case 3k of Theorem 2.2). If
2
b < 0, 2d +b > 0 (i.e. Bs|D | = Y ), then, since the latter implies
2 4 2 2 5
b > b , by Lemma 3.3 the condition (∗∗) holds only if d + b >
2 1 3 2
0 (case 3l of Theorem 2.2). If 2d + b < 0 (i.e. Bs|D | = Y ),
4 2 2 4
the condition (∗∗) is equivalent to (∗∗) . By Lemma 3.2 the latter is
4
equivalent to b = −d = −(d + d ), b = −d = −(d + d ). Under
1 1 2 4 2 2 3 4
the current assumptions this implies b = b , d = d = d > 0, d = 0
1 2 1 2 3 4
(case 3m of Theorem 2.2).
4.4. Case Bs|D | = Y . This case occurs if and only if 2d + b < 0,
1 3 3 1
2d +b > 0. Under this assumption all the components of grad (f )
2 1 x 1
with the possible exception of ∂f1 = α x +α x +α x and ∂f1 =
∂x1 13 3 14 4 15 5 ∂x2
α x +α x +α x are zero on Y .
23 3 24 4 25 5 3
Let us consider several possibilities.
4.4.1. Case d + d + b < 0. We have α = α = α = 0, and
2 3 1 23 24 25
Sing(D ) contains a line in a general fiber of ϕ, hence (∗) doesn’t hold.
1
4.4.2. Case d +d +b > 0, d +d +b < 0, d +b < 0. We have
2 3 1 2 4 1 1 1
α = α = α = 0. If deg(α ) > 0, then Sing(D ) contains a curve
15 24 25 23 1
in a fiber of ϕ over a zero of the polynomial α . Hence (∗) implies
23
d + d + b = 0. In this case Sing(D ) ⊂ Y . If deg(α ) 6= 0 (in
2 3 1 1 4 14
particular, if this degree is negative, i.e. the polynomial α is zero),
14
then Sing(D ) contains a line in a fiber of ϕ, hence it is necessary that
1
d +d +b = 0. Inthiscase Sing(D ) = Y , andfor(∗) and(∗∗)tohold
1 4 1 1 5
the equality b = 0 is necessary and sufficient (case 4a of Theorem 2.2).
2
9
4.4.3. Case d +d +b > 0, d +b < 0. We have α = α = 0, and
2 4 1 1 1 15 25
Y ⊂ Sing(D ). Hence (∗) implies b = 0. By Lemma 3.4 the condition
5 1 2
(∗) is equivalent to
0 = 2(d +d +d +d )+4b +b = 2(d +d +b )+2(d +d +b ),
1 2 3 4 1 2 2 4 1 1 3 1
i.e. d +d +b = d +d +b = 0. Since b = 0, the condition (∗∗)
2 4 1 1 3 1 2
holds automatically (case 4b of Theorem 2.2).
4.4.4. Case d + d + b > 0, d + d + b < 0, d + b > 0. We
2 3 1 2 4 1 1 1
have α = α = 0. If d + b > 0, then Sing(D ) ∩ Y 6= ∅, and
24 25 1 1 1 5
the inequality b > 0 must hold. By Lemma 3.4 the condition (∗) is
2
equivalent to
0 = 2(d +d +d +d )+4b +b = 2(d +b )+2(d +d +b )+2d +b > 0,
1 2 3 4 1 2 1 1 2 3 1 4 2
a contradiction. Hence it is necessary that d +b = 0. Since d +d +
1 1 2 3
b > 0 implies that Sing(D ) contains a line in a fiber of ϕ, the equality
1 1
d +d +b = 0 is also necessary.
2 3 1
By Lemma 3.4 the condition (∗) implies
0 = 2(d +d +d +d )+4b +b = 2(d +b )+2(d +d +b )+2d +b ,
1 2 3 4 1 2 1 1 2 3 1 4 2
that is equivalent to 2d +b = 0, i.e. (again by Lemma 3.4) under the
4 2
assumptions made above (∗) is equivalent to 2d + b = 0. If d > 0,
4 2 4
then Bs|D | = Y , and (∗∗) is equivalent to (∗∗) , and as d +b < 0
2 5 5 2 1
and d + b = 0 the latter condition holds if and only if d + b > 0
1 1 3 2
or d +b = 0 by Lemma 3.3 (case 4c of Theorem 2.2). If d = 0, i.e.
2 2 2
Bs|D | = ∅, we have b = 0, which apparently implies (∗∗) (case 4d
2 2
of Theorem 2.2).
4.4.5. Case d + d + b > 0, d + b > 0. Since Sing(D ) 6= ∅, we
2 4 1 1 1 1
have Bs|D | =6 Y , i.e. 2d +b > 0. By Lemma 3.4 the condition (∗)
2 3 3 2
implies
0 = 2(d +d +d +d )+4b +b = 2(d +b )+2(d +d +b )+2d +b ,
1 2 3 4 1 2 1 1 2 4 1 3 2
i.e. d +d +b = d +b = 2d +b = 0. If under these assumptions
2 4 1 1 1 3 2
d > d , then Bs|D | = Y , and by Lemma 3.2 the condition (∗∗)
3 4 2 4
implies b = −(d +d ), i.e. d = d , a contradiction. Hence we have
2 3 4 3 4
d = d , and by Lemma 3.4 the assumptions made above are sufficient
3 4
for (∗).
The condition 2d +b = 0 means that Bs|D | ⊂ Y . If Bs|D | = ∅,
4 2 2 5 2
then b > 0 implies b = d = d = 0; in this case the condition
2 2 3 4
(∗∗) holds automatically (case 4e of Theorem 2.2). If Bs|D | = Y ,
2 5
i.e. d > 0, then (∗∗) is equivalent to (∗∗) . Since d + b < 0 and
4 5 2 1
d +b < 0,thecondition(∗∗) isequivalenttod +b = 0byLemma3.3
3 2 5 2 2
(case 4f of Theorem 2.2).
10