Algebraic curves and maximal arcs

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Testo completo

(1)

A. Aguglia

L. Giuzzi

G. Kor hmáros

Abstra t

Alowerboundontheminimumdegreeoftheplanealgebrai urves

ontainingeverypointinalargepointset

K

oftheDesarguesianplane

PG(2, q)

is obtained. The ase where

K

is a maximal

(k, n)

ar is onsidered ingreaterdepth.

1 Introdu tion

In nite geometry, plane algebrai urves of minimum degree ontaining a

given large pointset

K

in

PG(2, q)

have been a useful tool to investigate

ombinatorialpropertiesof

K

.

When

K

is the whole pointset of

PG(2, q)

,a triviallower bound on the

degree of su h a plane algebrai urve is

q + 1

. G. Tallini pointed out that

thisisattainedonlywhenthe urvesplitsinto

q + 1

distin tlinesof

PG(2, q)

,

all passing through the same point. He also gave a omplete lassi ation

of the absolutely irredu ible urves of degree

q + 2

ontaining all points of

PG(2, q)

;see [23, 24℄ and also[1℄. If

K

isthe omplementary set of alinein

PG(2, q)

,then the bound is

q

; see [13℄.

When

K

onsists of all internal points to a oni

C

in

PG(2, q)

with

q

odd, theabove lowerboundis

q

− 1

. The analogous bound forthe set of the

external pointsto

C

is

q

. These boundswere the mainingredientsfor re ent

ombinatorial hara terisationsof pointsets blo kingallexternallines to

C

;

see [4,12℄.

When

K

is a lassi al unital of

PG(2, q)

, with

q

square, the minimum

degree

d

of an absolutely irredu ible urve

C

through

K

is

d = √q + 1

. For

non lassi al unitals,the best known bound is

d > 2√q

− 4

; see [18℄.

Resear h supported by theItalian MinistryMURST, Strutture geometri he,

(2)

Our purpose is to nd similar bounds for slightly smaller, but still quite

large, pointsets

K

, say

|K| = qt + α

with

t < c

3

q

and

0

≤ α < q

, where

c

isasuitable onstant. Sin eno ombinatorial onditiononthe onguration

of

K

is assumed, we are relying on te hniques and results from algebrai

geometry rather than on the onstru tive methodsused in the papers ited

above.

Themain resultisthat if

q > 8t

3

− 12

2

+ 4t

− 2α + 2

,anyplane algebrai

urve

Γ

ontaining every point of

K

has degree

d

≥ 2t

. The hypothesis on

the magnitude of

t

an be relaxed to

q > 16t

2

− 24t − 2α + 8

whenever

q

is

a prime.

Insome ases,thebound

d

≥ 2t

issharp,asthe followingexampleshows.

Let

K

be the union of

t

disjoint ovals. If

q

odd, or

q

is even and the ovals

are lassi al, then

|K| = qt + t

and

d = 2t

. The latter ase is known to

o ur when

K

is a Denniston maximal ar [10℄ (or one of the maximal ar s

onstru ted by Mathon and others, [25, 26, 16, 20, 14℄) minus the ommon

nu leus of the ovals.

On the otherhand, somerenementof the bound isalsopossible. Let

K

beany maximal ar of size

|K| = qt + t + 1

,that is, a

(qt + t + 1, t + 1)

ar .

Theorem 4.2 shows that if

q

is large enough omparing to

t

, then no plane

algebrai urve of degree

2t

passes through every point of

K

. Therefore, the

minimumdegreeisatleast

2t + 1

forsu h

t

and this boundisattainedwhen

K

is one of the above maximal ar s.

The ase

n = 4

is onsidered in more detail. For

q > 2

6

, the minimum

degreeis

7

andthisisonlyattainedwhen

K

isaDennistonar and the urve

Γ

of minimum degreesplits intothree distin t oni s with the same nu leus

N

∈ K

, together with a linethrough

N

.

2 Some ba kground on plane algebrai urves

over a nite eld

A plane proje tive algebrai urve

is dened over

GF(q)

, but viewed asa

urve over the algebrai losure

GF(q)

of

GF(q)

, if it has anane equation

f (X, Y ) = 0

, where

f (X, Y )

∈ GF(q)[X, Y ]

. The urve

is absolutely

irredu ible if it is irredu ible over the algebrai losure

GF(q)

. Denote by

(3)

irredu ible plane urve

of degree

d

. From the Hasse-Weilbound

N

q

≤ q + 1 + (d − 1)(d − 2)

q.

(1)

This holdstruewhensingularpointsof

lyingin

PG(2, q)

are also ounted;

see [19℄.

The Stöhr-Volo h bound depends not only on the degree

d

, but also on

a positive integer, the Frobenius order

ν

of

; see [22℄. This number

ν

is

either

1

or

ε

2

, where

ε

2

is the interse tion number

I(P, ∆

∩ ℓ)

of

with the tangent line

at a general point

P

∈ ∆

. It turns out that

ε

2

is either

2

, or a power, say

p

h

, of the hara teristi

p

of the plane, and is the minimum of

I(Q, ∆

∩ r)

, where

Q

ranges over the nonsingular points of

and

r

is the

tangent to

at

Q

. If

q = p

, then

ε

2

= 2

, and either

ν = 1

, or

ν = 2

and

p = 2

. Withthis notation, the Stöhr-Volo hboundapplied to

is

2N

q

≤ ν(d − 3)d + d(q + 2).

(2)

The followingalgebrai ma hinery anbeusedto ompute

ν

. Let

P = (a, b)

beanonsingularpoint of

su hthat the tangentlineto

at

P

isnot the

verti al line through

P

. The unique bran h (or pla e) entred at

P

has a

lo alparametrisation,also alleda primitivebran h representation,

x = a + t, y = b + ϕ(t)

where

f (x, y) = 0

and

ϕ(t) = b

k

t

k

+ . . .

with

k

≥ 1

is a formalpower series

with oe ients in

GF(q)[[t]]

;see [21℄. Then,

ν

is dened tobethe smallest

integersu h that the determinant

x

− x

q

y

− y

q

1

D

t

(ν)

(y)

=

a

− a

q

+ t

− t

q

b

− b

q

+ ϕ(t)

− ϕ(t)

q

1

D

(ν)

t

(ϕ(t))

does not vanish. Here

D

t

denotes the

ν

th Hasse derivative, that is,

D

t

(ν)

(ϕ(t)) =

k

ν



b

k

t

k−ν

+ . . . .

Theaboveideastillworksifos ulating oni sareusedinpla eoftangent

lines, and, in some ases, the resulting bound improves (2). Before stating

the result, whi h is the Stöhr-Volo h bound for oni s, a further on ept

(4)

respe t to the linear system

Σ

2

of the oni s of the plane is the in reasing sequen e

0, ǫ

1

= 1, ǫ

2

= 2, ǫ

3

, ǫ

4

, ǫ

5

of all interse tion numbers

I(P, ∆

∩ C)

of

with oni s at a general point

P

. The Frobenius

Σ

2

order sequen e is thesubsequen e

ν

0

= 0, ν

1

, ν

2

, ν

3

, ν

4

extra ted in reasinglyfromthe

Σ

2

order sequen e of

,for whi hthe following determinant doesnot vanish:

x

− x

q

x

2

− x

2q

y

− y

q

xy

− x

q

y

q

y

2

− y

2q

1

2x

D

1

)

t

(y) D

1

)

t

(xy) D

1

)

t

(y

2

)

0

1

D

2

)

t

(y) D

2

)

t

(xy) D

2

)

t

(y

2

)

0

0

D

3

)

t

(y) D

3

)

t

(xy) D

3

)

t

(y

2

)

0

0

D

4

)

t

(y) D

4

)

t

(xy) D

4

)

t

(y

2

)

.

Assume that

deg ∆

≥ 3

. The Stöhr-Volo h bound for oni s, that is for

Σ

2

, is

5N

q

≤ [(ν

1

+ . . . + ν

4

)(d

− 3)d + 2d(q + 5)].

(3)

For more onthe Stöhr-Volo h bound see [22℄.

3 Plane algebrai urves of minimum degree

through all the points of a given pointset

Inthis se tion,

K

standsforasetof

qt + α

pointsin

PG(2, q)

,with

0

≤ t ≤ q

and

0

≤ α < q

. Let

Γ

denotea plane algebrai urve of degree

d

ontaining

every point

K

. As already mentioned,

q + 1

is the minimum degree of a

plane algebrai urve ontaining every point of

PG(2, q)

. Thus, sin e we are

lookingforlowerboundson

d

,wewillonlybe on erned withthe asewhere

d

≤ q

.

A straightforward ounting argumentgivesthe following result.

Lemma 3.1. If

d

≤ q

, then

d

≥ t

.

Proof. Sin e

d

≤ q

,thelinear omponentsof

Γ

donot ontainallthepointsof

PG(2, q)

. Chooseapoint

P

∈ PG(2, q)

notinanyoftheselinear omponents.

Ea h of the

q + 1

lines through

P

meets

Γ

at most

d

distin t points. Thus,

(q + 1)d

≥ |K|

,that is

d

qt + α

q + 1

≥ t −

t

q + 1

.

(5)

Our aim is to improve Lemma 3.1. Write

F

for the set of all lines of

PG(2, q)

meeting

K

in at least

1

point. Set

m

0

= min

{|ℓ ∩ K| : ℓ ∈ F}

and

M

0

= max

{|ℓ ∩ K| : ℓ ∈ F}

.

Theorem 3.2. Let

Γ

be an algebrai plane urve over

GF(q)

of minimal

degree

d

whi h passes through all the points of

K

. If

q > 8t

3

− 16t

2

+ 2t + 4

− 2m

0

(2t

2

− 5t + 2) + 2M

0

(2t

− 1),

(4)

then

deg Γ

≥ 2t

. For prime

q

, Condition (4) may be relaxed to

q > 8t

2

− 16t + 8 − 2α + 2M

0

(2t

− 1).

(5)

Proof. We prove that if

d

≤ 2t − 1

, then (4) doesnot hold. For

q

prime we

show that also(5) is not satised. Sin e

Γ

isnot ne essarilyirredu ible, the

followingsetup is required.

The urves

1

, . . . , ∆

l

are the absolutely irredu ible nonlinear ompo-nents of

Γ

dened over

GF(q)

, respe tively of degree

d

i

;

r

1

, . . . , r

k

are the linear omponents of

Γ

over

GF (q)

;

Ξ

1

, . . . , Ξ

s

are the omponents of

Γ

whi h are irredu ible over

GF(q)

but not over

GF(q)

.

Theidea istoestimatethe number ofpointsin

PG(2, q)

that ea hof the

above omponents an have.

Let

N

i

be the number of nonsingular points of

i

lying in

PG(2, q)

. Then, (2) holds for any

i

. Let

ν

(i)

denote the Frobenius order of

i

. If

ν = max

(i)

| 1 ≤ i ≤ l}

and

δ =

P

l

i=1

d

i

,then

2

l

X

i=1

N

i

l

X

i=1

ν

(i)

d

i

(d

i

− 3) + (q + 2)d

i

≤ νδ(δ − 3) + (q + 2)δ.

(6)

For

q

prime,

ν = 1

; see [22℄. Sin e

ν

(i)

= 1

an fail for

q > p

, an upper

bound on

ν

(i)

depending on

d

i

is needed. As

ν

(i)

≤ ε

(i)

2

, a bound on

ε

(i)

2

su es. Sin e

N

i

> 0

maybeassumed,

i

has anonsingular point

P

lying in

PG(2, q)

. If

isthe tangentto

i

at

P

,then

d

i

=

X

Q∈ℓ∩∆

i

I(Q, ℓ

∩ ∆

i

) = I(P, ℓ

∩ ∆

i

) +

X

Q∈ℓ∩∆

i

Q6=P

I(Q, ℓ

∩ ∆

i

)

≥ ε

(i)

2

+ m

0

− 1,

when e

ν

(i)

≤ d

i

− m

0

+ 1

. From (6),

2

l

X

i=1

N

i

(

δ(δ

− 3) + (q + 2)δ

q

≥ 3

prime.

− m

0

+ 1)δ(δ

− 3) + (q + 2)δ

otherwise. (7)

(6)

If

i

has

M

i

singularpoints,fromPlü ker'stheorem

M

i

1

2

(d

i

−1)(d

i

−2)

. Hen e,

2

l

X

i=1

M

i

l

X

i=1

(d

i

− 1)(d

i

− 2) ≤ (δ − 1)(δ − 2).

(8)

The number of points of

K

lying on linear omponents

r

i

is at most

kM

0

≤ dM

0

.

Forevery

Ξ = Ξ

i

, there exists anabsolutely irredu ible urve

Θ

,dened over thealgebrai extension

GF(q

ξ

)

of degree

ξ > 1

of

GF(q)

in

GF(q)

,su h

that the absolutely irredu ible omponents of

Ξ

are

Θ

and its onjugates

Θ

1

, . . . , Θ

ξ−1

. Here, if

Θ

has equation

P a

kj

X

k

Y

j

= 0

and

1

≤ w ≤ ξ − 1

,

then

Θ

w

is the urve of equation

P a

q

w

kj

X

k

Y

j

= 0

. Sin e

Ξ

,

Θ

and the onjugates of

Θ

pass through the same points in

PG(2, q)

, from Bézout's

theorem, see [17, Lemma 2.24℄,

Ξ

has at most

θ

2

points in

PG(2, q)

where

θ = deg Θ

. Notethat

deg Ξ

≥ 2θ

. Let

N

i

denotethe total numberof points (simpleorsingular)of

Ξ

i

lying in

PG(2, q)

. From the above argument,

s

X

i=1

N

i

s

X

i=1

θ

2

i

< (

s

X

i=1

θ

i

)

2

1

4

(

s

X

i=1

deg Ξ

i

)

2

<

1

2

(q + 2)

s

X

i=1

deg Ξ

i

.

(9) As,

qt + α

l

X

i=1

(N

i

+ M

i

) + kM

0

+

s

X

i=1

N

i

,

(10)

from (7), (8), (9) and (10) it follows that

2(qt+α)

(

2d

2

− 6d + 2 + 2dM

0

+ (q + 2)d

q

≥ 3

prime

d

3

− d

2

− 6d + 2 − m

0

d(d

− 3) + 2dM

0

+ (q + 2)d

otherwise.

Sin e

d

≤ 2t − 1

,the mainassertionfollows by straightforward omputation.

Remark 3.3. As

1

≤ m

0

≤ M

0

≤ d

, the proof of Theorem 3.2 shows that Condition (4), and, for

q

prime, Condition (5), may be repla ed by the

somewhatweaker, butmoremanageable, ondition

q > 8t

3

−12

2

+4t

−2α+2

(and

q > 16t

2

− 24t − 2α + 8

for

q

prime).

Remark 3.4. As pointed out in the Introdu tion, Theorem 3.2 is sharp as

(7)

Corollary 3.5. If

Γ

has a omponent not dened over

GF(q)

, then

deg Γ

≥ 2t + 1.

Proof. We use the same arguments as in the proof of Theorem 3.2,

onsid-ering that

δ < d

. In parti ular, we have

2(qt + α) < d

3

− (4 + m

0

)d

2

+ (q + 5m

0

+ 1)d + (4

− 4m

0

− q) + 2dM

0

,

whi h for

d

≤ 2t

proves the assertion.

Remark 3.6. Corollary3.5impliesthata planealgebrai urveof ofdegree

2t

ontaining

K

is always dened over

GF(q)

.

Theorem 3.7. If the urve

Γ

in Theorem 3.2 has no quadrati omponent,

and

q >

750t

3

− 1725t

2

+ 10(10M

0

+ 113)t

− 184 − 40(α + M

0

)

40

,

(11) then

d

5

2

t.

If

q > 5

is prime,then Condition (11) may be relaxed to

q >

125t

2

+ 2(10M

0

− 105)t − 8(α + M

0

− 9)

8

(12)

Proof. Weprovethat if

d

5

2

(t

− 1)

, then(11) (and, for

q

prime, (12))does not hold. It is su ient just a hange in the proof of Theorem 3.2. Let

ν

0

(i)

= 0, . . . , ν

(i)

4

be the Frobenius orders of

i

with respe t to oni s. If

q

is a prime greater than

5

, then

ν

(i)

j

= j

for

0

≤ j ≤ 4

. Otherwise, set

ν

(i)

=

P

4

j=1

ν

(i)

j

. Sin e

ν

(i)

≤ 2 + ε

3

+ ε

4

+ ε

5

and

ε

5

≤ 2d

i

, we have that

ν

(i)

≤ 6d

i

− 1

. From (3),

5

l

X

i=1

N

i

(

10δ(δ

− 3) + (q + 5)2δ

q > 5

prime

.

(6δ

− 1)δ(δ − 3) + (q + 5)2δ,

otherwise. (13)

Using the same argumentas in(10),

s

X

i=1

N

i

1

4

s

X

i=1

deg Ξ

i

!

2

<

2

5

(q + 5)

s

X

i=1

deg Ξ

i

.

(14)

(8)

(qt + α)

(

5

2

d

2

11

2

d + dM

0

+

2

5

qd + 1

q > 5

prime

6

5

d

3

33

10

d

2

+

11

10

d + dM

0

+

2

5

qd + 1

otherwise. (15)

Then,(15)doesnotholdforany

q

. If

q

isprime,also(12)isnotsatised.

4 Algebrai urves passing through the points

of a maximal ar

Remark 3.4 motivates the study of plane algebrai urves passing through

all the points of maximal

(k, n)

ar in

PG(2, q)

.

In this se tion

K

always denotes a maximal

(k, n)

ar . Re all that a

(k, n)

ar

K

of a proje tive plane

π

is a set of

k

points, no

n + 1

ollinear. Barlotti[9℄proved that

k

≤ (n−1)q +n

,forany

(k, n)

ar in

PG(2, q)

;when

equality holds, a

(k, n)

ar is maximal. A purely ombinatorial property

hara terising a

(k, n)

maximal ar

K

is that every line of

PG(2, q)

either

meets

K

in

n

points or is disjoint from it. Trivial examples of maximal

ar s in

PG(2, q)

are the

(q

2

+ q + 1, q + 1)

ar given by all the points of

PG(2, q)

and the

(q

2

, q)

ar s onsisting of the points of an ane subplane

AG(2, q)

of

PG(2, q)

. Ball, Blokhuis and Mazzo a [5℄, [6℄ have shown that

nonontrivial maximalar exists in

PG(2, q)

for

q

odd. On the other hand,

for

q

even, several maximal ar s exists in the Desarguesian plane and many

onstru tions are known; see [10℄, [25℄, [26℄, [16℄ [20℄, [14℄. The ar s arising

from these onstru tions, with the ex eption of those of [25℄, see also [16℄,

all onsist of the union of

n

− 1

disjoint oni s together with their ommon

nu leus

N

. In other words, these ar s are overed by a ompletelyredu ible

urveofdegree

2n

−1

,whose omponentsare

n

−1

oni sand alinethrough

the point

N

.

Remark 4.1. From Corollary3.5, if

Γ

has a omponentdened over

GF(q)

but notover

GF(q)

,anditpassesthrough allthe pointsof

K

, thenitsdegree

d

is atleast

2n

− 1

.

The following theorem shows that the above hypothesis on the

(9)

Theorem 4.2. For any

n

, there exists

q

0

≤ (2n − 2)

2

su h that if a plane

algebrai urve

Γ

denedover

GF(q)

with

q > q

0

passesthrough allthepoints of a maximal

(k, n)

ar

K

of

PG(2, q)

then its degree

d

isat least

2n

− 1

. If

equality holds then

Γ

has either one linear and

n

− 1

absolutely irredu ible

quadrati omponents or

n

− 2

absolutely irredu ible quadrati omponents

and one ubi omponent.

The proofdepends on the the followinglemma.

Lemma 4.3. Assume that

Γ

is redu ibleand that the number of its

ompo-nents is less than

n

− 1

. Then, the degree

d

of

Γ

satises

d

4

q.

Proof. Weuse the samesetup asinthe proofof Theorem 3.2. This time the

HasseWeil bound is used in pla e of the Stöhr-Volo hbound. The number

of pointsof

i

in

PG(2, q)

is

N

i

+ M

i

; from(1),

M

i

+ N

i

≤ q + 1 + (d

i

− 1)(d

i

− 2)

q.

(16) From (9),

s

X

i=1

N

i

< (q + 1) +

s

X

i=1

(deg Ξ

i

− 1)(deg Ξ

i

− 2)

q.

(17)

If

Γ

has

w

omponents,then

w(q +1)+

s

X

i=1

(deg Ξ

i

−1)(deg Ξ

i

−2)

q +

l

X

j=1

(d

i

−1)(d

i

−2)

q

≥ (n−1)q+n.

(18) Sin e

w

≤ n − 2

, (18) yields

(d

− 1)(d − 2)

q

≥ q + 2;

hen e,

d

4

q

.

Proof of Theorem 4.2. Suppose

Γ

tohavedegree

d < 2n

−1

. ByRemark4.1,

all omponentsof

Γ

aredenedover

GF(q)

. If

Γ

isabsolutelyirredu ible,then

(1)impliesthat

Γ

ontainsatmost

q+(2n

−3)(2n−4)

q+1

points. However,

(10)

When

Γ

has more then one omponent, denote by

t

j

the number of its omponents ofdegree

j

. Let

u

bethe maximum degreeof su h omponents.

Then,

u

≤ 2n − 3

and

d =

u

X

j=1

jt

j

≤ 2n − 2.

From (1),

|Γ ∩ K| ≤ nt

1

+

u

X

j=2

t

j

(q + (j

− 1)(j − 2)

q + 1) =

u

X

j=2

t

j

!

q + c

q + d,

where

c =

u

X

j=2

t

j

(j

− 1)(j − 2),

d = nt

1

+

u

X

j=2

t

j

.

Both

c

and

d

are independent from

q

; therefore,

n

− 1 = lim

q→∞

|Γ ∩ K|

q

=

u

X

j=2

t

j

.

(19) Hen e,

2(n

− 1) = 2

u

X

j=2

t

j

≤ t

1

+

u

X

j=2

jt

j

= d.

Sin e

d

≤ 2n − 2

, by Lemma4.3, for

q > (2n

− 2)

4

the urve

Γ

shouldhave

at least

n

− 1

omponents. This would imply that either

u = 2

,

t

1

= 1

,

t

2

= (n

− 1)

or

u = 3

,

t

1

= 0

,

t

2

= (n

− 2)

,

t

3

= 1

. In parti ular, in both ases

d = 2n

− 1

.

Remark4.4. AsmentionedintheIntrodu tion, ase

d = 2n

−1

inTheorem

4.2 o urs when

K

is a Denniston maximal ar [10℄ (or one of the maximal

ar s onstru ted by Mathon and others, [25, 26, 16, 20, 14℄). This result

may not extend toany of the other known maximal ar s; they are the Thas

maximal

(q

3

− q

2

+ q, q)

ar sin

PG(2, q

2

)

arisingfromthe SuzukiTits ovoid

of

PG(3, q)

;see[25℄. Infa t,

22

istheminimumdegreeofaplane urvewhi h

passes through all points of a Thas' maximal

(456, 8)

ar in

PG(2, 64)

; see

(11)

5 Maximal ar s of degree

4

From Theorem 4.2, for

q > 6

4

, a lower bound on the degree of an algebrai

urve

Γ

passing through all the points of a maximal ar

K

of degree

4

is

7

.

Our aim isto prove inthis ase the followingresult.

Theorem 5.1. Let

K

be a maximal ar of degree

4

and suppose there exists

an algebrai plane urve

Γ

ontaining allthe points of

K

. If

deg Γ = 7

, then

Γ

onsists of three disjoint oni s, all with the same nu leus

N

, and a line through

N

.

Proof. FromTheorem 4.2,the urve

Γ

splitseitherintooneirredu ible ubi

and two irredu ible oni s or into three irredu ible oni s and one line

r

.

These two ases are investigated separately.

Let

C

denote any of the above oni s of nu leus

N

. We showthat every

point of

C

= C

∪ {N}

lying in

PG(2, q)

is ontained in

K

: in fa t, if there

were a point

P

∈ C

\ K

, then there would be at least

q

4

lines through

P

external to

K

. All these lines would meet

C

in

q

4

distin tpoints, whi h, in turn, wouldnot beon

K

. Hen e,

Γ

wouldhaveless than

3q + 4

pointsonthe

ar

K

, a ontradi tion.

Nowassumethat

Γ

splitsintoa ubi

D

andtwo oni s

C

i

,with

i = 1, 2

. Denote by

N

i

the nu leus of

C

i

and set

X = C

1

∪ C

2

∪ {N

1

, N

2

}

. Sin e

|X | ≤ 2q + 4

, there exists a point

P

∈ K \ X

. Obviously,

P

∈ D

. Every line through

P

meets

K

infourpoints;thus,there isnoline

through

P

meeting

both

C

1

and

C

2

in

2

points; otherwise,

|ℓ ∩ X ∩ K| = 4

and

|ℓ ∩ K| ≥ 5

, a ontradi tion. Hen e, there are at least

q

− 1

lines through

P

meeting

D

in another point

P

. There are at most

5

bise ants to the irredu ible ubi

urve

D

through any given point

P

∈ D

, namely the tangent in

P

to

D

and,possibly,fourothertangentsindierentpointsto

D

passingthrough

P

.

Hen e,thereare

q

−6

linesthrough

P

meeting

D

inthreepoints. Ifthiswere

the ase,

D

would onsist ofatleast

2(q

− 6) + 6

points,whi his impossible.

Therefore, we may assume that

Γ

splits into three oni s, say

C

1

,

C

2

,

C

3

, with nu lei

N

1

,

N

2

,

N

3

, and a line

r

.

Re all that, asseen above, the nu lei of allthe oni sbelong to

K

. Now

we show that at least one nu leus, say

N

1

, lies on the line

r

. Sin e

Γ

is a urve ontaining

K

of minimum degree with respe t to this property, there

is at least a point

P

on

r

∩ K

not on

C

1

. Ea h line through

P

isa

4

se ant to

K

hen e, it meets

X

in an odd number of points. If

P

6= N

1

, then the numberoflines through

P

meeting

X

inanodd numberof pointsisatmost

(12)

A

N

1

N

2

N

3

C

1

C

2

C

3

Figure 1: Case 1 inTheorem 5.1 PSfrag repla ements

C

i

C

j

C

s

N

i

= N

j

N

s

Figure 2: Case 2 inTheorem 5.1

15

,whi hislessthan

q +1

for

q

≥ 2

4

,a ontradi tion. A tually,allthenu lei

N

i

lie on

r

. In fa t, suppose that

N

j

∈ r

/

for

j

∈ {2, 3}

. Then,

N

j

∈ C

s

, with

s

6= 1, j

and the line

N

1

N

j

joining

N

1

and

N

j

is tangentto

C

1

and

C

j

. Consequently,

N

1

N

j

meets

C

s

in another point dierent from

N

j

that is, it is a

5

se ant to

K

, againa ontradi tion.

We are left with three ases, namely:

(1)

N

1

6= N

2

6= N

3

,

C

i

∩ C

j

=

{A}

for any

i

6= j

and

A

∈ N

1

N

2

.

(2)

N

i

= N

j

, N

j

6= N

s

,

N

i

∈ C

s

,

C

i

∩ C

j

=

, with

i, j, s

∈ {1, 2, 3}

.

(3)

N

1

= N

2

= N

3

and

C

i

∩ C

j

=

for

i

6= j

.

We are going toshowthat ases (1)and (2)do not a tually o ur.

PSfrag repla ements

C

1

C

2

C

3

N

1

= N

2

= N

3

(13)

Lemma 5.2. Let

C

1

,

C

2

be two oni s with a ommon point

A

but dierent nu lei

N

1

,

N

2

. If

A

∈ N

1

N

2

thenthere is a line

s

with

A

∈ s

and

s

6= N

1

N

2

su h that for any point on

s

\ (C

1

∪ C

2

)

there passes a line

with

|ℓ ∩ (C

1

∪ C

2

∪ {N

1

, N

2

})| ≥ 3.

Proof. Let

(X, Y, Z)

denotehomogeneous oordinates of points of the plane

PG(2, q)

. Chooseareferen esystem su h that

A = O = (0, 0, 1)

andthe line joining

N

1

and

N

2

is the

X

axis. We may suppose

C

i

to have equation

α

i

X

2

+ XY + β

i

Y

2

+ λ

i

Y Z = 0,

where

α

i

, β

i

, λ

i

∈ GF(q)

and

i = 1, 2

.

Sin e both

C

i

are nondegenerate oni s, we have

α

i

6= 0

and

λ

i

6= 0

. Furthermore,

λ

1

6= λ

2

as the nu lei

N

1

and

N

2

are distin t.

Denote by

T

(x)

the tra e of

GF(q

2

)

over

GF(q)

; namely

T

(x) = x + x

q

.

If

T

((α

1

λ

2

+ α

2

λ

1

)(β

1

λ

2

+ β

2

λ

1

)) = 0

, then the two oni s

C

1

and

C

2

have more than one point in ommon, whi h is impossible. Hen e,

T

1

λ

2

+

α

2

λ

1

)(β

1

λ

2

+ β

2

λ

1

)) = 1

; inparti ular,

β

1

λ

2

6= β

2

λ

1

. A generi point

P

i

m

of

C

i

\ {A}

has homogeneous oordinates

P

m

i

=



λ

i

m

α

i

m

2

+ m + β

i

,

λ

i

α

i

m

2

+ m + β

i

, 1



,

with

m

∈ GF(q) \ {0}

. Consider now a point

P

ε

= (0, ǫ, 1)

on the

Y

axis,

with

ε

∈ GF(q) \ {0,

λ

1

β

1

,

λ

2

β

2

}

. The points

P

1

m

,

P

2

t

and

P

ε

are ollinearif and only if

λ

1

m

α

1

m

2

+ m + β

1

λ

1

α

1

m

2

+ m + β

1

1

λ

2

t

α

2

t

2

+ t + β

2

λ

2

α

2

t

2

+ t + β

2

1

0

ε

1

= 0,

that is

α

1

λ

2

εm

2

t+ α

2

λ

1

ǫmt

2

+ ε(λ

2

+ λ

1

)mt+ (ελ

1

β

2

+ λ

1

λ

2

)m+ (ǫλ

2

β

1

+ λ

1

λ

2

)t = 0.

(20)

Equation (20) may be regarded as the ane equation of a ubi urve

D

in the indeterminate

m

and

t

. Observe that

(0, 0, 1)

∈ D

. The only points

at innity of

D

are

Y

= (0, 1, 0)

,

X

= (0, 1, 0)

and

B = (

λ

1

α

2

λ

2

α

1

, 1, 0)

(14)

Therefore,

D

doesnot splitintothree onjugate omplexlines. Thus, by [17,

Theorem 11.34℄ and [17, Theorem 11.46℄, there is at least one ane point

T = (m, t, 1)

on

D

dierent from

(0, 0, 1)

.

Sin e

T

((α

1

λ

2

+ α

2

λ

1

)(β

1

λ

2

+ β

2

λ

1

)) = 1

, the line

m = t

isa

1

se antto

D

in

(0, 0, 1)

;hen e, the point

T

isnot onthis line.

This impliesthat, for any given

ε

∈ GF(q) \ {0}

, there existat least two

distin t values

m, t

∈ GF(q) \ {0}

satisfying (20). Hen e,

P

1

m

,

P

2

t

and

P

ε

are ollinear and the line

P

1

m

P

t

2

meets

C

1

∪ C

2

∪ {N

1

, N

2

}

in at least three points.

From Lemma

5.2

,in ase (1)the set

C

1

∪ C

2

∪ {N

1

, N

2

}

annotbe om-pleted to a maximal ar just by adding a third oni

C

3

, together with its nu leus

N

3

, sin e, in this ase, there would be at least a

5

se ant to

C

1

∪ C

2

∪ C

3

∪ {N

1

, N

2

, N

3

}

. Hen e, ase (1) is ruled out.

Lemma 5.3. Given any two disjoint oni s

C

1

,

C

2

with the same nu leus

N

, there isa unique degree

4

maximal ar ontaining

X = C

1

∪ C

2

.

Proof. Thereis aline

r

in

PG(2, q)

externalto

X

,sin e,otherwise,

X

would

bea

2

blo kingsetwithlessthan

2q+

2q+1

points,whi hisa ontradi tion; see [8℄.

Choose a referen e system su h that

N = O = (0, 0, 1)

and

r

is the line

at innity

Z = 0

. The oni s

C

i

, for

i = 1, 2

,have equation:

α

i

X

2

+ XY + β

i

Y

2

+ λ

i

Z

2

= 0,

(21)

where

α

i

, β

i

, λ

i

∈ GF(q)

and

T

i

β

i

) = 1

.

Sin e both

C

i

are nondegenerate,

λ

i

6= 0

. We rst show that,

λ

1

6= λ

2

, as

C

1

and

C

2

are disjoint. We argue by ontradi tion. If it were

λ

1

= λ

2

, then we ould assume

α

1

6= α

2

; in fa t, if

α

1

= α

2

and

λ

1

= λ

2

, the linear system generated by

C

1

and

C

2

would ontain the line

Y = 0

; thus their interse tionwould not be empty. Letnow

γ =

r β

1

− β

2

α

1

− α

2

;

hen e,

α

1

γ

2

+ γ + β

1

= α

2

γ

2

+ γ + β

2

,

and the line

X = γY

would meetthe two oni s inthe same point

P =



γ

λ

1

α

1

γ

2

+γ+β

1



1

2

,

λ

1

α

1

γ

2

+γ+β

1



1

2

, 1



,

(15)

ontradi ting

C

1

∩ C

2

=

∅.

We also see that

α

1

λ

2

6= α

2

λ

1

and

β

1

λ

2

6= β

2

λ

1

, sin e, otherwise, the points

P = (

q

λ

1

α

1

, 0, 1) = (

q

λ

2

α

2

, 0, 1),

Q = (0,

q

λ

1

β

1

, 1) = (0,

q

λ

2

β

2

, 1)

would lie onboth oni s.

Letnow

C

3

be the oni with equation

α

1

λ

2

+ α

2

λ

1

λ

1

+ λ

2

X

2

+ XY +

β

1

λ

2

+ β

2

λ

1

λ

1

+ λ

2

Y

2

+ (λ

1

+ λ

2

)Z

2

= 0.

Set

ν =

1

λ

2

+ α

2

λ

1

)(β

1

λ

2

+ β

2

λ

1

)

λ

2

1

+ λ

2

2

.

The ollineation H of

PG(2, q)

given by the matrix

a

−1

0 0

0

a 0

b

c 1

,

where

a =

q

α

1

λ

2

2

λ

1

λ

1

2

, b =

q

1+α

1

/a

2

λ

1

and

c = a

q

β

1

2

λ

1

2

,

maps the oni s

C

i

,

i = 1, 2

, to

C

i

2

: X

2

+ XY + νY

2

+ λ

i

Z

2

= 0,

and

C

3

to

C

3

2

: X

2

+ XY + νY

2

+ (λ

1

+ λ

2

)Z

2

= 0.

If it were

T

(ν) = 0

, then

C

1

2

and

C

2

2

would share some pointin ommon on the lineat innity. Hen e,

T

(ν) = 1

. In parti ular,

C

1

2

,

C

2

2

and

C

3

2

together with their ommonnu leus

O

, forma degree

4

maximalar

K

of Denniston

type; see [10℄, [2℄and [20, Theorem 2.5℄.

It remains to show the uniqueness of

K

. We rst observe that

X

is a

(0, 2, 4)

set with respe t to lines of the plane. No point lying ona

4

se ant to

X

an be added to

X

to get a degree

4

maximal ar .

Take

P /

∈ X

and denoteby

u

i

with

i = 0, 2, 4

,the numberof

i

se antsto

X

through

P

, that is the number of lines meeting

X

in

i

points. The lines through

P

whi h areexternalto

X

are alsoexternal

K

and the onverse also

holds. Therefore, when

P

6∈ K

,wehave

u

0

=

1

4

q

. From

(16)

2u

2

+ 4u

4

= 2q + 2,

also

u

4

=

1

4

q

. Hen e, no point

P

6∈ K

may be added to

X

to obtain a maximal ar of degree

4

.

Finally,Lemma5.3 shows that ase (2) doesnot o ur.

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Segre on regular points with respe t to an ellipse of an ane Galois

plane, Ann. Mat.Pura Appl. 72 (1997),87102.

[2℄ V. Abatangelo, B. Larato; A hara terization of Denniston's maximal

ar s, Geom. Dedi ata30 (1989), 197203.

[3℄ A. Aguglia, L. Giuzzi; An algorithm for onstru ting some maximal

ar s in

PG(2, q

2

)

, Results Math., toappear.

[4℄ A. Aguglia,G.Kor hmáros;Blo king sets of externallines to a oni in

PG(2, q)

,

q

odd , Combinatori a 26 (2006),no. 4,379394.

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[6℄ S. Ball, A. Blokhuis; An easier proof of the maximal ar s onje ture,

Pro . Amer. Math. So . 126 (1998),no. 11,33773380.

[7℄ S.Ball, A.Blokhuis; The lassi ation of maximal ar s in small

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[8℄ S. Ball, A. Blokhuis; On the size of a double blo king set in

PG(2, q)

,

Finite FieldsAppl. 2 (1996),no. 2, 125137.

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(k, n)

ar hi di un piano libero nito, Boll. Un. Mat.

Ital. (3) 11 (1956), 553556.

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(17)

hara teristi , Manus riptaMath. 59 (1987), no. 4, 457-469.

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PG(2, q)

,

q

even, European J. Combin 28 (2007), no. 1, 3642.

[13℄ V.D. GoppaGeometry and Codes, Kluwer(1988).

[14℄ N.Hamilton,R.Mathon;Moremaximalar s in Desarguesianproje tive

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Authors' addresses: Angela AGUGLIA Dipartimentodi Matemati a Polite ni odi Bari ViaOrabona 4 70125Bari (Italy) Email: a.agugliapoliba.it Lu a GIUZZI Dipartimentodi Matemati a Polite ni odi Bari Via Orabona 4 70125 Bari(Italy) Email: l.giuzzipoliba.it GáborKORCHMÁROS Dipartimentodi Matemati a

Università della Basili ata

Contrada Ma hiaRomana

85100Potenza(Italy).

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