A. Aguglia
∗
L. Giuzzi∗
G. Kor hmáros∗
Abstra tAlowerboundontheminimumdegreeoftheplanealgebrai urves
ontainingeverypointinalargepointset
K
oftheDesarguesianplanePG(2, q)
is obtained. The ase whereK
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
inPG(2, q)
have been a useful tool to investigateombinatorialpropertiesof
K
.When
K
is the whole pointset ofPG(2, q)
,a triviallower bound on thedegree of su h a plane algebrai urve is
q + 1
. G. Tallini pointed out thatthisisattainedonlywhenthe urvesplitsinto
q + 1
distin tlinesofPG(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 ofPG(2, q)
;see [23, 24℄ and also[1℄. IfK
isthe omplementary set of alineinPG(2, q)
,then the bound isq
; see [13℄.When
K
onsists of all internal points to a oniC
inPG(2, q)
withq
odd, theabove lowerboundis
q
− 1
. The analogous bound forthe set of theexternal pointsto
C
isq
. These boundswere the mainingredientsfor re entombinatorial hara terisationsof pointsets blo kingallexternallines to
C
;see [4,12℄.
When
K
is a lassi al unital ofPG(2, q)
, withq
square, the minimumdegree
d
of an absolutely irredu ible urveC
throughK
isd = √q + 1
. Fornon lassi al unitals,the best known bound is
d > 2√q
− 4
; see [18℄.∗
Resear h supported by theItalian MinistryMURST, Strutture geometri he,
Our purpose is to nd similar bounds for slightly smaller, but still quite
large, pointsets
K
, say|K| = qt + α
witht < c
3
√
q
and
0
≤ α < q
, wherec
isasuitable onstant. Sin eno ombinatorial onditiononthe onguration
of
K
is assumed, we are relying on te hniques and results from algebraigeometry 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 ofK
has degreed
≥ 2t
. The hypothesis onthe magnitude of
t
an be relaxed toq > 16t
2
− 24t − 2α + 8
whenever
q
isa prime.
Insome ases,thebound
d
≥ 2t
issharp,asthe followingexampleshows.Let
K
be the union oft
disjoint ovals. Ifq
odd, orq
is even and the ovalsare lassi al, then
|K| = qt + t
andd = 2t
. The latter ase is known too ur when
K
is a Denniston maximal ar [10℄ (or one of the maximal ar sonstru 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 tot
, then no planealgebrai urve of degree
2t
passes through every point ofK
. Therefore, theminimumdegreeisatleast
2t + 1
forsu ht
and this boundisattainedwhenK
is one of the above maximal ar s.The ase
n = 4
is onsidered in more detail. Forq > 2
6
, the minimum
degreeis
7
andthisisonlyattainedwhenK
isaDennistonar and the urveΓ
of minimum degreesplits intothree distin t oni s with the same nu leusN
∈ K
, together with a linethroughN
.2 Some ba kground on plane algebrai urves
over a nite eld
A plane proje tive algebrai urve
∆
is dened overGF(q)
, but viewed asaurve over the algebrai losure
GF(q)
ofGF(q)
, if it has anane equationf (X, Y ) = 0
, wheref (X, Y )
∈ GF(q)[X, Y ]
. The urve∆
is absolutelyirredu ible if it is irredu ible over the algebrai losure
GF(q)
. Denote byirredu ible plane urve
∆
of degreed
. From the Hasse-WeilboundN
q
≤ q + 1 + (d − 1)(d − 2)
√
q.
(1)This holdstruewhensingularpointsof
∆
lyinginPG(2, q)
are also ounted;see [19℄.
The Stöhr-Volo h bound depends not only on the degree
d
, but also ona positive integer, the Frobenius order
ν
of∆
; see [22℄. This numberν
iseither
1
orε
2
, whereε
2
is the interse tion numberI(P, ∆
∩ ℓ)
of∆
with the tangent lineℓ
at a general pointP
∈ ∆
. It turns out thatε
2
is either2
, or a power, sayp
h
, of the hara teristi
p
of the plane, and is the minimum ofI(Q, ∆
∩ r)
, whereQ
ranges over the nonsingular points of∆
andr
is thetangent to
∆
atQ
. Ifq = p
, thenε
2
= 2
, and eitherν = 1
, orν = 2
andp = 2
. Withthis notation, the Stöhr-Volo hboundapplied to∆
is2N
q
≤ ν(d − 3)d + d(q + 2).
(2)The followingalgebrai ma hinery anbeusedto ompute
ν
. LetP = (a, b)
beanonsingularpoint of
∆
su hthat the tangentlineto∆
atP
isnot theverti al line through
P
. The unique bran h (or pla e) entred atP
has alo alparametrisation,also alleda primitivebran h representation,
x = a + t, y = b + ϕ(t)
where
f (x, y) = 0
andϕ(t) = b
k
t
k
+ . . .
with
k
≥ 1
is a formalpower serieswith oe ients in
GF(q)[[t]]
;see [21℄. Then,ν
is dened tobethe smallestintegersu 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 theν
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
respe t to the linear system
Σ
2
of the oni s of the plane is the in reasing sequen e0, ǫ
1
= 1, ǫ
2
= 2, ǫ
3
, ǫ
4
, ǫ
5
of all interse tion numbersI(P, ∆
∩ C)
of∆
with oni s at a general pointP
. The FrobeniusΣ
2
order sequen e is thesubsequen eν
0
= 0, ν
1
, ν
2
, ν
3
, ν
4
extra ted in reasinglyfromtheΣ
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
(ν
1
)
t
(y) D
(ν
1
)
t
(xy) D
(ν
1
)
t
(y
2
)
0
1
D
(ν
2
)
t
(y) D
(ν
2
)
t
(xy) D
(ν
2
)
t
(y
2
)
0
0
D
(ν
3
)
t
(y) D
(ν
3
)
t
(xy) D
(ν
3
)
t
(y
2
)
0
0
D
(ν
4
)
t
(y) D
(ν
4
)
t
(xy) D
(ν
4
)
t
(y
2
)
.
Assume that
deg ∆
≥ 3
. The Stöhr-Volo h bound for oni s, that is forΣ
2
, is5N
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
standsforasetofqt + α
pointsinPG(2, q)
,with0
≤ t ≤ q
and
0
≤ α < q
. LetΓ
denotea plane algebrai urve of degreed
ontainingevery point
K
. As already mentioned,q + 1
is the minimum degree of aplane algebrai urve ontaining every point of
PG(2, q)
. Thus, sin e we arelookingforlowerboundson
d
,wewillonlybe on erned withthe asewhered
≤ q
.A straightforward ounting argumentgivesthe following result.
Lemma 3.1. If
d
≤ q
, thend
≥ t
.Proof. Sin e
d
≤ q
,thelinear omponentsofΓ
donot ontainallthepointsofPG(2, q)
. ChooseapointP
∈ PG(2, q)
notinanyoftheselinear omponents.Ea h of the
q + 1
lines throughP
meetsΓ
at mostd
distin t points. Thus,(q + 1)d
≥ |K|
,that isd
≥
qt + α
q + 1
≥ t −
t
q + 1
.
Our aim is to improve Lemma 3.1. Write
F
for the set of all lines ofPG(2, q)
meetingK
in at least1
point. Setm
0
= min
{|ℓ ∩ K| : ℓ ∈ F}
andM
0
= max
{|ℓ ∩ K| : ℓ ∈ F}
.Theorem 3.2. Let
Γ
be an algebrai plane urve overGF(q)
of minimaldegree
d
whi h passes through all the points ofK
. Ifq > 8t
3
− 16t
2
+ 2t + 4
− 2m
0
(2t
2
− 5t + 2) + 2M
0
(2t
− 1),
(4)then
deg Γ
≥ 2t
. For primeq
, Condition (4) may be relaxed toq > 8t
2
− 16t + 8 − 2α + 2M
0
(2t
− 1).
(5)Proof. We prove that if
d
≤ 2t − 1
, then (4) doesnot hold. Forq
prime weshow that also(5) is not satised. Sin e
Γ
isnot ne essarilyirredu ible, thefollowingsetup is required.
The urves
∆
1
, . . . , ∆
l
are the absolutely irredu ible nonlinear ompo-nents ofΓ
dened overGF(q)
, respe tively of degreed
i
;r
1
, . . . , r
k
are the linear omponents ofΓ
overGF (q)
;Ξ
1
, . . . , Ξ
s
are the omponents ofΓ
whi h are irredu ible overGF(q)
but not overGF(q)
.Theidea istoestimatethe number ofpointsin
PG(2, q)
that ea hof theabove omponents an have.
Let
N
i
be the number of nonsingular points of∆
i
lying inPG(2, q)
. Then, (2) holds for any∆
i
. Letν
(i)
denote the Frobenius order of
∆
i
. Ifν = max
{ν
(i)
| 1 ≤ i ≤ l}
andδ =
P
l
i=1
d
i
,then2
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 eν
(i)
= 1
an fail for
q > p
, an upperbound on
ν
(i)
depending ond
i
is needed. Asν
(i)
≤ ε
(i)
2
, a bound onε
(i)
2
su es. Sin e
N
i
> 0
maybeassumed,∆
i
has anonsingular pointP
lying inPG(2, q)
. Ifℓ
isthe tangentto∆
i
atP
,thend
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 eν
(i)
≤ d
i
− m
0
+ 1
. From (6),2
l
X
i=1
N
i
≤
(
δ(δ
− 3) + (q + 2)δ
q
≥ 3
prime.(δ
− m
0
+ 1)δ(δ
− 3) + (q + 2)δ
otherwise. (7)If
∆
i
hasM
i
singularpoints,fromPlü ker'stheoremM
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 omponentsr
i
is at mostkM
0
≤ dM
0
.Forevery
Ξ = Ξ
i
, there exists anabsolutely irredu ible urveΘ
,dened over thealgebrai extensionGF(q
ξ
)
of degree
ξ > 1
ofGF(q)
inGF(q)
,su hthat the absolutely irredu ible omponents of
Ξ
areΘ
and its onjugatesΘ
1
, . . . , Θ
ξ−1
. Here, ifΘ
has equationP a
kj
X
k
Y
j
= 0
and
1
≤ w ≤ ξ − 1
,then
Θ
w
is the urve of equationP a
q
w
kj
X
k
Y
j
= 0
. Sin eΞ
,Θ
and the onjugates ofΘ
pass through the same points inPG(2, q)
, from Bézout'stheorem, see [17, Lemma 2.24℄,
Ξ
has at mostθ
2
points in
PG(2, q)
whereθ = deg Θ
. Notethatdeg Ξ
≥ 2θ
. LetN
′
i
denotethe total numberof points (simpleorsingular)ofΞ
i
lying inPG(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
primed
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, forq
prime, Condition (5), may be repla ed by thesomewhatweaker, 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
Corollary 3.5. If
Γ
has a omponent not dened overGF(q)
, thendeg 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 have2(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
ontainingK
is always dened overGF(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) thend
≥
5
2
t.
If
q > 5
is prime,then Condition (11) may be relaxed toq >
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, forq
prime, (12))does not hold. It is su ient just a hange in the proof of Theorem 3.2. Letν
0
(i)
= 0, . . . , ν
(i)
4
be the Frobenius orders of∆
i
with respe t to oni s. Ifq
is a prime greater than5
, thenν
(i)
j
= j
for0
≤ j ≤ 4
. Otherwise, setν
(i)
=
P
4
j=1
ν
(i)
j
. Sin eν
(i)
≤ 2 + ε
3
+ ε
4
+ ε
5
andε
5
≤ 2d
i
, we have thatν
(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)(qt + α)
≤
(
5
2
d
2
−
11
2
d + dM
0
+
2
5
qd + 1
q > 5
prime6
5
d
3
−
33
10
d
2
+
11
10
d + dM
0
+
2
5
qd + 1
otherwise. (15)Then,(15)doesnotholdforany
q
. Ifq
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 inPG(2, q)
.In this se tion
K
always denotes a maximal(k, n)
ar . Re all that a(k, n)
arK
of a proje tive planeπ
is a set ofk
points, non + 1
ollinear. Barlotti[9℄proved thatk
≤ (n−1)q +n
,forany(k, n)
ar inPG(2, q)
;whenequality holds, a
(k, n)
ar is maximal. A purely ombinatorial propertyhara terising a
(k, n)
maximal arK
is that every line ofPG(2, q)
eithermeets
K
inn
points or is disjoint from it. Trivial examples of maximalar 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)
ofPG(2, q)
. Ball, Blokhuis and Mazzo a [5℄, [6℄ have shown thatnonontrivial maximalar exists in
PG(2, q)
forq
odd. On the other hand,for
q
even, several maximal ar s exists in the Desarguesian plane and manyonstru 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 ommonnu leus
N
. In other words, these ar s are overed by a ompletelyredu ibleurveofdegree
2n
−1
,whose omponentsaren
−1
oni sand alinethroughthe point
N
.Remark 4.1. From Corollary3.5, if
Γ
has a omponentdened overGF(q)
but notover
GF(q)
,anditpassesthrough allthe pointsofK
, thenitsdegreed
is atleast2n
− 1
.The following theorem shows that the above hypothesis on the
Theorem 4.2. For any
n
, there existsq
0
≤ (2n − 2)
2
su h that if a plane
algebrai urve
Γ
denedoverGF(q)
withq > q
0
passesthrough allthepoints of a maximal(k, n)
arK
ofPG(2, q)
then its degreed
isat least2n
− 1
. Ifequality holds then
Γ
has either one linear andn
− 1
absolutely irredu iblequadrati omponents or
n
− 2
absolutely irredu ible quadrati omponentsand one ubi omponent.
The proofdepends on the the followinglemma.
Lemma 4.3. Assume that
Γ
is redu ibleand that the number of itsompo-nents is less than
n
− 1
. Then, the degreed
ofΓ
satisesd
≥
√
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
inPG(2, q)
isN
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
hasw
omponents,thenw(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 ew
≤ n − 2
, (18) yields(d
− 1)(d − 2)
√
q
≥ q + 2;
hen e,d
≥
4
√
q
.Proof of Theorem 4.2. Suppose
Γ
tohavedegreed < 2n
−1
. ByRemark4.1,all omponentsof
Γ
aredenedoverGF(q)
. IfΓ
isabsolutelyirredu ible,then(1)impliesthat
Γ
ontainsatmostq+(2n
−3)(2n−4)
√
q+1
points. However,
When
Γ
has more then one omponent, denote byt
j
the number of its omponents ofdegreej
. Letu
bethe maximum degreeof su h omponents.Then,
u
≤ 2n − 3
andd =
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,
wherec =
u
X
j=2
t
j
(j
− 1)(j − 2),
d = nt
1
+
u
X
j=2
t
j
.
Both
c
andd
are independent fromq
; 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, forq > (2n
− 2)
4
the urve
Γ
shouldhaveat least
n
− 1
omponents. This would imply that eitheru = 2
,t
1
= 1
,t
2
= (n
− 1)
oru = 3
,t
1
= 0
,t
2
= (n
− 2)
,t
3
= 1
. In parti ular, in both asesd = 2n
− 1
.Remark4.4. AsmentionedintheIntrodu tion, ase
d = 2n
−1
inTheorem4.2 o urs when
K
is a Denniston maximal ar [10℄ (or one of the maximalar 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 hpasses through all points of a Thas' maximal
(456, 8)
ar inPG(2, 64)
; see5 Maximal ar s of degree
4
From Theorem 4.2, forq > 6
4
, a lower bound on the degree of an algebrai
urve
Γ
passing through all the points of a maximal arK
of degree4
is7
.Our aim isto prove inthis ase the followingresult.
Theorem 5.1. Let
K
be a maximal ar of degree4
and suppose there existsan algebrai plane urve
Γ
ontaining allthe points ofK
. Ifdeg Γ = 7
, thenΓ
onsists of three disjoint oni s, all with the same nu leusN
, and a line throughN
.Proof. FromTheorem 4.2,the urve
Γ
splitseitherintooneirredu ible ubiand 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 leusN
. We showthat everypoint of
C
′
= C
∪ {N}
lying in
PG(2, q)
is ontained inK
: in fa t, if therewere a point
P
∈ C
′
\ K
, then there would be at least
q
4
lines throughP
external toK
. All these lines would meetC
′
in
q
4
distin tpoints, whi h, in turn, wouldnot beonK
. Hen e,Γ
wouldhaveless than3q + 4
pointsonthear
K
, a ontradi tion.Nowassumethat
Γ
splitsintoa ubiD
andtwo oni sC
i
,withi = 1, 2
. Denote byN
i
the nu leus ofC
i
and setX = C
1
∪ C
2
∪ {N
1
, N
2
}
. Sin e|X | ≤ 2q + 4
, there exists a pointP
∈ K \ X
. Obviously,P
∈ D
. Every line throughP
meetsK
infourpoints;thus,there isnolineℓ
throughP
meetingboth
C
1
andC
2
in2
points; otherwise,|ℓ ∩ X ∩ K| = 4
and|ℓ ∩ K| ≥ 5
, a ontradi tion. Hen e, there are at leastq
− 1
lines throughP
meetingD
in another point
P
′
. There are at most
5
bise ants to the irredu ible ubiurve
D
through any given pointP
∈ D
, namely the tangent inP
toD
and,possibly,fourothertangentsindierentpointsto
D
passingthroughP
.Hen e,thereare
q
−6
linesthroughP
meetingD
inthreepoints. Ifthiswerethe ase,
D
would onsist ofatleast2(q
− 6) + 6
points,whi his impossible.Therefore, we may assume that
Γ
splits into three oni s, sayC
1
,C
2
,C
3
, with nu leiN
1
,N
2
,N
3
, and a liner
.Re all that, asseen above, the nu lei of allthe oni sbelong to
K
. Nowwe show that at least one nu leus, say
N
1
, lies on the liner
. Sin eΓ
is a urve ontainingK
of minimum degree with respe t to this property, thereis at least a point
P
onr
∩ K
not onC
1
. Ea h line throughP
isa4
se ant toK
hen e, it meetsX
in an odd number of points. IfP
6= N
1
, then the numberoflines throughP
meetingX
inanodd numberof pointsisatmostA
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 hislessthanq +1
forq
≥ 2
4
,a ontradi tion. A tually,allthenu lei
N
i
lie onr
. In fa t, suppose thatN
j
∈ r
/
forj
∈ {2, 3}
. Then,N
j
∈ C
s
, withs
6= 1, j
and the lineN
1
N
j
joiningN
1
andN
j
is tangenttoC
1
andC
j
. Consequently,N
1
N
j
meetsC
s
in another point dierent fromN
j
that is, it is a5
se ant toK
, 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 anyi
6= j
andA
∈ N
1
N
2
.(2)
N
i
= N
j
, N
j
6= N
s
,N
i
∈ C
s
,C
i
∩ C
j
=
∅
, withi, j, s
∈ {1, 2, 3}
.(3)
N
1
= N
2
= N
3
andC
i
∩ C
j
=
∅
fori
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
Lemma 5.2. Let
C
1
,C
2
be two oni s with a ommon pointA
but dierent nu leiN
1
,N
2
. IfA
∈ N
1
N
2
thenthere is a lines
withA
∈ s
ands
6= N
1
N
2
su h that for any point ons
\ (C
1
∪ C
2
)
there passes a lineℓ
with|ℓ ∩ (C
1
∪ C
2
∪ {N
1
, N
2
})| ≥ 3.
Proof. Let
(X, Y, Z)
denotehomogeneous oordinates of points of the planePG(2, q)
. Chooseareferen esystem su h thatA = O = (0, 0, 1)
andthe line joiningN
1
andN
2
is theX
axis. We may supposeC
i
to have equationα
i
X
2
+ XY + β
i
Y
2
+ λ
i
Y Z = 0,
where
α
i
, β
i
, λ
i
∈ GF(q)
andi = 1, 2
.Sin e both
C
i
are nondegenerate oni s, we haveα
i
6= 0
andλ
i
6= 0
. Furthermore,λ
1
6= λ
2
as the nu leiN
1
andN
2
are distin t.Denote by
T
(x)
the tra e ofGF(q
2
)
over
GF(q)
; namelyT
(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 sC
1
andC
2
have more than one point in ommon, whi h is impossible. Hen e,T
(α
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 pointP
i
m
ofC
i
\ {A}
has homogeneous oordinatesP
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 pointP
ε
= (0, ǫ, 1)
on theY
axis,with
ε
∈ GF(q) \ {0,
λ
1
β
1
,
λ
2
β
2
}
. The pointsP
1
m
,P
2
t
andP
ε
are ollinearif and only ifλ
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 isα
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
andt
. Observe that(0, 0, 1)
∈ D
. The only pointsat innity of
D
areY
∞
= (0, 1, 0)
,X
∞
= (0, 1, 0)
andB = (
λ
1
α
2
λ
2
α
1
, 1, 0)
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)
onD
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 linem = t
isa1
se anttoD
in(0, 0, 1)
;hen e, the pointT
isnot onthis line.This impliesthat, for any given
ε
∈ GF(q) \ {0}
, there existat least twodistin t values
m, t
∈ GF(q) \ {0}
satisfying (20). Hen e,P
1
m
,P
2
t
andP
ε
are ollinear and the lineP
1
m
P
t
2
meetsC
1
∪ C
2
∪ {N
1
, N
2
}
in at least three points.From Lemma
5.2
,in ase (1)the setC
1
∪ C
2
∪ {N
1
, N
2
}
annotbe om-pleted to a maximal ar just by adding a third oniC
3
, together with its nu leusN
3
, sin e, in this ase, there would be at least a5
se ant toC
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 leusN
, there isa unique degree4
maximal ar ontainingX = C
1
∪ C
2
.Proof. Thereis aline
r
inPG(2, q)
externaltoX
,sin e,otherwise,X
wouldbea
2
blo kingsetwithlessthan2q+
√
2q+1
points,whi hisa ontradi tion; see [8℄.Choose a referen e system su h that
N = O = (0, 0, 1)
andr
is the lineat innity
Z = 0
. The oni sC
i
, fori = 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)
andT
(α
i
β
i
) = 1
.Sin e both
C
i
are nondegenerate,λ
i
6= 0
. We rst show that,λ
1
6= λ
2
, asC
1
andC
2
are disjoint. We argue by ontradi tion. If it wereλ
1
= λ
2
, then we ould assumeα
1
6= α
2
; in fa t, ifα
1
= α
2
andλ
1
= λ
2
, the linear system generated byC
1
andC
2
would ontain the lineY = 0
; thus their interse tionwould not be empty. Letnowγ =
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 pointP =
γ
λ
1
α
1
γ
2
+γ+β
1
1
2
,
λ
1
α
1
γ
2
+γ+β
1
1
2
, 1
,
ontradi ting
C
1
∩ C
2
=
∅.
We also see that
α
1
λ
2
6= α
2
λ
1
andβ
1
λ
2
6= β
2
λ
1
, sin e, otherwise, the pointsP = (
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 equationα
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.
Setν =
(α
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
,
wherea =
q
α
1
λ
2
+α
2
λ
1
λ
1
+λ
2
, b =
q
1+α
1
/a
2
λ
1
andc = a
q
β
1
+β
2
λ
1
+λ
2
,
maps the oni s
C
i
,i = 1, 2
, toC
i
2
: X
2
+ XY + νY
2
+ λ
i
Z
2
= 0,
andC
3
toC
3
2
: X
2
+ XY + νY
2
+ (λ
1
+ λ
2
)Z
2
= 0.
If it wereT
(ν) = 0
, thenC
1
2
andC
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
andC
3
2
together with their ommonnu leusO
, forma degree4
maximalarK
of Dennistontype; see [10℄, [2℄and [20, Theorem 2.5℄.
It remains to show the uniqueness of
K
. We rst observe thatX
is a(0, 2, 4)
set with respe t to lines of the plane. No point lying ona4
se ant toX
an be added toX
to get a degree4
maximal ar .Take
P /
∈ X
and denotebyu
i
withi = 0, 2, 4
,the numberofi
se antstoX
throughP
, that is the number of lines meetingX
ini
points. The lines throughP
whi h areexternaltoX
are alsoexternalK
and the onverse alsoholds. Therefore, when
P
6∈ K
,wehaveu
0
=
1
4
q
. From2u
2
+ 4u
4
= 2q + 2,
also
u
4
=
1
4
q
. Hen e, no pointP
6∈ K
may be added toX
to obtain a maximal ar of degree4
.Finally,Lemma5.3 shows that ase (2) doesnot o ur.
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ar s in
PG(2, q
2
)
, Results Math., toappear.
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PG(2, q)
,q
odd , Combinatori a 26 (2006),no. 4,379394.[5℄ S.Ball,A.Blokhuis, F.Mazzo a;Maximal ar s inDesarguesian planes
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Rend. Mat. e Appl. (5) 20 (1961),431479.[24℄ G. Tallini; Le ipersuper ie irridu ibili d'ordine minimo he invadono
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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).