PLANAR ARCS
SIMEON BALL AND MICHEL LAVRAUW
Abstract. Let p denote the characteristic of Fq , the finite field with q elements. We
prove that if q is odd then an arc of size q + 2 − t in the projective plane over Fq , which
is not contained in a conic, is contained in the intersection of two curves, which do not
share a common component, and have degree at most t + pblogp tc . This implies that if
√
√
q is odd then an arc of size at least q − q + 3 + max{ q/p, 21 } is contained in a conic.
This is of particular interest in the case that q is an odd square, since there are examples
√
of arcs, not contained in a conic, of size q − q + 1. It has long been conjectured that
if q 6= 9 is an odd square then any larger arc is contained in a conic. We also complete
the classification of arcs in the projective plane over Fq of size q − 1 and q − 2. More
generally, we prove that associated to an arc S of size q +2−t in the projective plane over
Fq , there is a polynomial F (X, Y ), homogeneous of degree t in both X = (X1 , X2 , X3 )
and Y = (Y1 , Y2 , Y3 ), with the property that evaluating F in one of the variables at a
point a of the arc, one obtains the product of the t linear forms whose kernels are the
tangents to S at a. Moreover, modulo the homogeneous polynomials of degree t which
are zero on S, the polynomial F is unique.
1. Introduction.
Let Fq denote the finite field with q elements and let p denote the characteristic of Fq .
Let PGk−1 (Fq ) denote the (k − 1)-dimensional projective space over Fq .
An arc of PGk−1 (Fq ) is a set of points, any k of which span the whole space. An arc is
complete if it cannot be extended to a larger arc. In this article we will be interested in arcs
in PG2 (Fq ), which we will call planar arcs. A planar arc is defined equivalently as a set of
points, no three of which are collinear. One can project any higher dimensional arc to a
planar arc, so the results contained in this article have implications for all low-dimensional
arcs.
Arcs not only play an important role in finite geometry but also forge links to other
branches of mathematics. The matrix whose columns are vector representatives of the
Date: 30 June 2017.
2010 Mathematics Subject Classification. 51E21, 94B05, 05B25.
The first author acknowledges the support of the project MTM2014-54745-P of the Spanish Ministerio
de Economı́a y Competitividad.
1
2
SIMEON BALL AND MICHEL LAVRAUW
points of an arc, generates a linear maximum distance separable code, see [18, Chapter
11] for more details on this. Other areas in which planar arcs play a role include the
representation of matroids, see [20], Del Pezzo surfaces over finite fields, see [3], bent
functions, see [19, Chapter 7] and pro-solvable groups, see [11].
In 1955, Beniamino Segre published the article [21], which contains his now celebrated
theorem that if q is odd then a planar arc of size q + 1 is a conic. He went further in
his 1967 paper [23] and considered planar arcs of size q + 2 − t and proved that the set
of tangents, when viewed as a set of points in the dual plane, is contained in a curve of
small degree d. Specifically, if q is even then d = t and if q is odd then d = 2t.
His starting point, which will also be our starting point, was his lemma of tangents, see
Lemma 11. Lemma 11 is a simplification of the coordinate-free version of Segre’s original
lemma of tangents which first appeared in [2].
Here, we do not apply Segre’s lemma of tangents in the dual setting, nor combine it
with interpolation, as was the approach in [2]. Our aim here is to prove Theorem 13,
which maintains that there is a polynomial F (X, Y ), where X = (X1 , X2 , X3 ) and Y =
(Y1 , Y2 , Y3 ), homogeneous of degree t in both X and Y , which upon evaluation in one
of the variables at a point a of the arc, factorises into t linear forms whose kernels are
precisely the tangents to the arc at a.
This allows us to prove the following theorem.
Theorem 1. Let S be a planar arc of size q + 2 − t not contained in a conic. If q is
odd then S is contained in the intersection of two curves, sharing no common component,
each of degree at most t + pblogp tc .
It is worth remarking that Theorem 1 may hold for q even as well, if we replace the
condition “not contained in a conic” with “not contained in an arc of maximum size”.
Arcs of maximum size for q even are of size q + 2 and are known as hyperovals. In general,
these are not contained in low degree curves and the subsets of these hyperovals are also
not be contained in low degree curves. Indeed, almost the opposite occurs. In [6], Caullery
and Schmidt prove that if S = {(f (x), x, 1) | x ∈ Fq } ∪ {(1, 0, 0), (0, 1, 0)} is an arc of size
1
k
q + 2 and f has degree less than 12 q 4 , then f (x) = x6 or f (x) = x2 for some positive
integer k. There are many other infinite families of hyperovals known, see [7] for a survey.
PLANAR ARCS
3
In the following sections we will prove various results, which are corollaries to Theorem 1.
We will then go on to prove the aforementioned Lemma 11, Theorem 13 and finally
Theorem 1.
2. The second largest complete arc and the
√
q conjecture.
In 1947, Bose [4] was the first to observe that the largest planar arc has size q + 1 if q is
odd and q + 2 if q is even. The size of the second largest complete arc remains an open
question in general.
√
There are examples of complete arcs of size q − q + 1 in PG2 (Fq ) when q is square,
first discovered by Kestenband, see [17]. These arcs are the intersection of two Hermitian
√
curves of degree q + 1,
√
x1
q+1
√
+ x2
q+1
√
+ x3
q+1
= 0 and
3
X
√
q
xi xj hij = 0,
i,j=1
√
q
where hij = hji and the characteristic polynomial of the matrix H = (hij ) is irreducible
over Fq . It was this construction that motivated us to try and prove that arcs, not
contained in a conic, are contained in the intersection of curves of low degree.
The best bounds on the size of the second largest complete arc are the following.
For q even Segre [23] proved, combining Theorem 12 with the Hasse-Weil theorem, that
√
the second largest complete arc has size at most q − q + 1. The examples above imply
that this bound is tight if q also a square.
For q prime, Voloch [29] proved, by using the Stöhr-Voloch theorem from [24], that the
second largest complete arc has size at most 44
q + 89 .
45
For q non-square, Voloch [30] proved the second largest complete arc has size at most
√
√
q − 14 pq + 29
p − 1 if q is odd and at most q − 2q + 2 if q is even.
16
By, exploiting Theorem 12 (Segre’s theorem from [23]) and bounds on the number of
points on an algebraic curve, Hirschfeld and Korchmáros [15] proved that the second
√
largest complete arc has size at most q − 21 q + 5, provided that the characteristic is at
√
least 5. This was subsequently improved to q − 12 q + 3 by Hirschfeld and Korchmáros
[16], provided that q > 529 and q 6= 36 , 55 .
Theorem 1 leads directly to the following theorem, which improves on the previously
known bounds when q is an odd square, and for small q when q is odd and not a square.
4
SIMEON BALL AND MICHEL LAVRAUW
√
√
Theorem 2. If q is odd then an arc of size at least q − q +3+max{ q/p, 12 } is contained
in a conic.
Proof. Let S be a planar arc of size q + 2 − t which is not contained in a conic. Then, by
Theorem 1, there are two curves sharing no common component, each of degree at most
t + pblogp tc , which contain S. By Bezout’s theorem, |S| 6 (t + pblogp tc )2 .
√
√
Suppose q is not a prime and |S| > q + 2 − t > q − q + q/p + 3. This implies
√
√
√
√
t 6 q − q/p − 1 and hence pblogp tc 6 q/p. Hence, |S| 6 (t + pblogp tc )2 6 ( q − 1)2 , a
contradiction.
√
√
Suppose q is a prime and |S| > q + 2 − t > q − q + 72 . This implies t 6 q − 32 and
√
pblogp tc = 1. Hence, |S| 6 (t + pblogp tc )2 6 ( q − 21 )2 , a contradiction.
We remark that combining Theorem 1 with Bezout’s theorem may give a slightly better
bound in some specific cases, so we state a slightly clumsier theorem, which is precisely
the best that can be deduced.
Theorem 3. Let S be a planar arc of size q +2−t. If q is odd and (t+pblogp tc )2 +t 6 q +1
then S is contained in a conic.
Our main motivation to start this work was to prove the following conjecture, which we
√
call the q conjecture, see [14, pp. 236] and also [13, Problem 1.1].
Conjecture 4. If q 6= 9 is an odd square then the second largest planar arc has size
√
q − q + 1.
The examples of Kestenband imply that, if true, the bound in the conjecture is tight.
Combining Segre [23], Theorem 2 and Theorem 1 we get the following theorem.
√
Theorem 5. If there is a counterexample S to Conjecture 4 of size q + 2 − t then q −
√
√
q/p 6 t 6 q and S is contained in the intersection of two curves, sharing no common
component, and each of degree at most t + pblogp tc .
3. Bounds on arcs contained in low degree curves
In 1973, Zirilli [31] constructed arcs of size approximately 12 q, contained in a cubic curve
by exploiting the group structure of an elliptic curve. If the group H of the Fq -rational
points of a non-singular cubic curve has even order, then the coset of a subgroup of H of
PLANAR ARCS
5
index two is a planar arc. This was taken further by Voloch [27] who proved that Zirilli’s
√
construction leads to complete arcs of size 12 (q +1)−, for all || < q, for q odd, (, q) = 1
and q > 175. Further results were obtained by Voloch [28], Szonyi [25] and Giulietti [12].
Theorem 1 leads to some surprising bounds if we assume that S is contained in a curve
of low degree.
Theorem 6. Let S be a planar arc of size q + 2 − t contained in a curve ψ of degree d
and not contained in a conic. If q is odd then (t + pblogp tc )d + t > q + 2.
Proof. If ψ is one of the curves given by Theorem 1 then by Bezout’s theorem, |S| 6
(t + pblogp tc )d which, since |S| = q + 2 − t, gives the bound (t + pblogp tc )d + t > q + 2.
If ψ is not one of the curves given by Theorem 1, then let f and g be the two co-prime
polynomials whose zero-sets are the curves given by Theorem 1. Without losing the coprime property we can assume that f and g have the same degree. Let h be the polynomial
whose zero-set defines the curve ψ. If there is a λ ∈ Fq such that f + λg and h have no
common factor, then Bezout’s theorem implies |S| 6 (t+pblogp tc )d, which gives the desired
bound. Otherwise, for each λ ∈ Fq there is a factor hλ of h that is a factor of f + λg.
Since h has less than q factors, there is a λ and ν for which hλ = hν . But then, this factor
is a factor of both f and g, contradicting the fact that f and g are co-prime.
Theorem 6 has the following corollary in the case q is prime.
Corollary 7. Let S be a planar arc contained in a curve of degree d and not contained
in a conic. If q is prime then |S| 6 d(q + 3)/(d + 1).
Proof. Since q is prime, pblogp tc = 1. The bound follows directly from Theorem 6.
In the worst-case scenario, Theorem 6 has the following corollary.
Corollary 8. Let S be a planar arc contained in a curve of degree d and not contained
in a conic. If q is odd then |S| 6 2d(q + 2)/(2d + 1).
Proof. Observing that pblogp tc 6 t, the bound follows from Theorem 6.
4. Classification of arcs of size q − 1 and q − 2.
It has been known for a long time that any arc of size q is incomplete. This was proved
by Segre [22] (with an amendment by Büke [5]) for q odd, and by Tallini in [26] for q
6
SIMEON BALL AND MICHEL LAVRAUW
even. The next question to answer (“Il primo nuovo quesito da porsi” as Segre writes in
his 1955 paper) is whether there exists a complete arc of size q − 1 in PG2 (Fq ). Since
the standard frame extended with the points (1, 2, 3) and (1, 3, 4) gives a complete arc
of size 6 for q = 7, the answer is affirmative. However, in spite of this small example,
Segre expected the answer to be negative for sufficiently large q: “Rimane tuttavia da
indagare se, com’è da ritenersi probabile, la risposta alla suddetta domanda non diventi
invece negativa quando si supponga q sufficientemente grande”. Segre’s expectations were
correct, there are no complete planar arcs of size q − 1 for q > 13. The classification of
planar arcs of size q − 1 is given in Corollary 9.
√
Segre’s bound |S| 6 q − q + 1 for q even, proves the non-existence of complete arcs
of size q − 1 for q even. For q odd the situation is quite different. The combination
of computer aided calculations (see for example the articles by Coolsaet and Sticker [9,
10] and Coolsaet [8]) with theoretical upper bounds (see Section 2) were not enough to
complete the classification of complete arcs of size q − 1.
According to [14, Theorem 10.33], there are 15 values of q, all odd and satisfying q > 31
(q = 31 has since been ruled out by Coolsaet [8]), for which it is not yet determined
whether an arc of size q − 1 is complete or not. Theorem 2 allows us to complete the
classification of planar arcs of size q − 1.
Corollary 9. The only complete planar arcs of size q − 1 occur for q = 7 (there are 2
projectively distinct arcs of size 6), for q = 9 (there is a unique arc of size 8), q = 11
(there is a unique arc of size 10, see Example 3) and q = 13 (there is a unique arc of size
12, see Example 2).
Proof. Combining Theorem 3 with [14, Table 9.4].
Theorem 3 also allows us to complete the classification of planar arcs of size q − 2.
Corollary 10. The only complete planar arcs of size q − 2 occur for q = 8 (there are 3
projectively distinct arcs of size 6), for q = 9 (there is a unique arc of size 7) and q = 11
(there are 3 projectively distinct arcs of size 9).
Proof. Combining Theorem 3 with [14, Table 9.4], [9] and [10].
As Segre would say “Il primo nuovo quesito da porsi” is whether there are any more arcs
of size q − 3 to be found. The computational results of Coolsaet and Sticker [9, 10] and
Coolsaet [8] rule out any examples being found for 19 6 q 6 31. Theorem 3 rules out any
PLANAR ARCS
7
new examples being found for q > 41. Voloch’s bound from [30] rules out q = 32. We
can deduce from [14, Table 9.4] that there are examples of complete arcs of size q − 3 for
q = 9, 13, 16 and 17. The only remaining case is now q = 37. Furthermore, Theorem 1
implies that if there is an arc of size 34 in PG2 (F37 ) then it is contained in the intersection
of two sextic curves, which do not share a common component.
5. The tangent functions.
Throughout, S will be an arc of PG2 (Fq ) of size q + 2 − t.
Each point of S is incident with precisely t tangents, where a tangent is a line of PG2 (Fq )
which is incident with exactly one point of S.
We define a homogeneous polynomial of degree t in three variables,
fa (X) =
t
Y
αi (X),
i=1
where αi (X) are linear forms whose kernels are the t tangents to the arc S incident with a
point a of S. This defines fa (X) up to scalar factor. Let e1 be an arbitrary fixed element
of S and scale fa (X) so that
fa (e1 ) = (−1)t+1 fe1 (a).
The following is Segre’s lemma of tangents, originally from [23]. It appears again in a
coordinate-free version in [2], and here it is further simplified by scaling.
Lemma 11. For all x, y ∈ S,
fx (y) = (−1)t+1 fy (x).
Proof. Let B = {e1 , x, y}. Since S is an arc, it follows that B is a basis.
For a ∈ S, let fa∗ (X) be the polynomial we obtain from fa (X) when we change the basis
from the canonical basis to the basis B. Then
t
t
t
Y
Y
Y
∗
∗
∗
fx (X) =
(bi1 X1 +bi3 X3 ), fy (X) =
(ci1 X1 +ci2 X2 ), and fe1 (X) =
(di2 X2 +di3 X3 ),
i=1
i=1
i=1
for some bij , cij , dij ∈ Fq .
With respect to the basis B, the line joining e1 and (s1 , s2 , s3 ) ∈ S \ B is the kernel of the
linear form X2 − (s2 /s3 )X3 . Since these lines are distinct from the tangent lines incident
8
SIMEON BALL AND MICHEL LAVRAUW
with e1 we have
t
Y
Y
di3 Y −s2
=
b = −1.
d
s3
i=1 i2 s∈S\B
b∈Fq \{0}
Q
Q
Observing that ti=1 di3 = fe∗1 (y) and ti=1 di2 = fe∗1 (x), this gives
Y
Y
fe∗1 (y)
s2 = (−1)t+1 fe∗1 (x)
s3 .
s∈S\B
s∈S\B
Similarly,
fx∗ (e1 )
Y
s3 = (−1)t+1 fx∗ (y)
s∈S\B
Y
s1
s∈S\B
and
fy∗ (x)
Y
s1 = (−1)t+1 fy∗ (e1 )
s∈S\B
Y
s2 .
s∈S\B
Combining these three equations we obtain,
fe∗1 (y)fx∗ (e1 )fy∗ (x) = (−1)t+1 fe∗1 (x)fx∗ (y)fy∗ (e1 ).
Since fa and fa∗ define the same function on the points of PGk−1 (Fq ), we also have that
fx∗ (e1 ) = (−1)t+1 fe∗1 (x) and fy∗ (e1 ) = (−1)t+1 fe∗1 (y), and hence fx∗ (y) = (−1)t+1 fy∗ (x), and
therefore fx (y) = (−1)t+1 fy (x).
In 1967 Segre [23] proved that the set of tangents, viewed as points in the dual plane, lie
on an algebraic curve of degree t, when q is even, and an algebraic curve of degree 2t,
when q is odd. His main tool in the proof of this theorem was Lemma 11. However, to
prove this result for q odd, he only uses the fact that
fx (y)2 = fy (x)2 ,
whereas Lemma 11 maintains that the stronger statement
fx (y) = (−1)t+1 fy (x)
holds.
In terms of the polynomials fa (X), Segre’s theorem is the following theorem. Although
we will not need this to derive the main results of this article we include a proof here,
since it is quite short.
Theorem 12. Let m ∈ {1, 2} such that m − 1 = q modulo 2. If S is a planar arc of
size q + 2 − t, where |S| > mt + 2, then there is a homogeneous polynomial in three
variables φ(Z), of degree mt, which gives a polynomial G(X, Y ) under the substitution
PLANAR ARCS
9
Z1 = X2 Y3 − Y2 X3 , Z2 = X1 Y3 − Y1 X3 , Z3 = X2 Y1 − Y2 X1 , with the property that for all
a∈S
G(X, a) = fa (X)m .
Proof. Order the set S arbitrarily and let E be a subset of S of size mt + 2. Define
X
Y det(X, Y, u)
G(X, Y ) =
.
fa (b)m
det(a,
b,
u)
a<b
u∈E\{a,b}
where the sum runs over subsets {a, b} of E.
Observe that det(X, Y, u) = u1 Z1 + u2 Z2 + u3 Z3 , where Z1 = X2 Y3 − Y2 X3 , Z2 = X3 Y1 −
Y3 X1 and Z3 = X1 Y2 − X2 Y1 , so G can be obtained from a homogeneous polynomial in
Z of degree mt under this change of variables.
Then, for y ∈ E, the only non-zero terms in G(X, y) are obtained for a = y and b = y.
Lemma 11 implies
X
Y det(X, y, u)
.
G(X, y) =
fa (y)m
det(a, y, u)
a∈E\y
u∈E\{a,y}
With respect to a basis containing y, the polynomials G(X, y) and fy (X)m are homogeneous polynomials in two variables of degree mt. Their values at the mt + 1 points
x ∈ E \ {y} are the same, so we conclude that G(X, y) = fy (X)m .
If y 6∈ E then we still have that with respect to a basis containing y, the polynomial
G(X, y) is a homogeneous polynomial in two variables of degree mt. For x ∈ E,
G(x, y) = G(y, x) = fx (y)m = fy (x)m ,
the last equality following from Lemma 11, and so again we conclude that G(X, y) =
fy (X)m .
For an arc S of PG2 (Fq ) of size q + 2 − t, let Φ[X] denote the subspace of the vector space
of homogeneous polynomials of degree t in X = (X1 , X2 , X3 ) which are zero on S.
We will prove the following theorem which, for q odd, uses the full force of Lemma 11.
Theorem 13. Let S be a planar arc of size q + 2 − t. There is a polynomial F (X, Y ),
which is a homogeneous polynomial of degree t in both X and Y , such that
F (X, Y ) = (−1)t+1 F (Y, X)
and with the property that for all a ∈ S,
F (X, a) = fa (X)
(mod Φ[X]).
10
SIMEON BALL AND MICHEL LAVRAUW
Moreover, modulo (Φ[X], Φ[Y ]) the polynomial F is unique.
The remainder of the article is then dedicated to proving Theorem 1.
6. Some examples.
Before we go on and prove Theorem 13, we first consider some examples, which not only
give an idea of what Theorem 13 is saying, but also give a first indication of how useful
it might be.
Example 1. If t = 1 then clearly Φ[X] = {0} and Theorem 13 implies that the polynomial
F (X, Y ) is a symmetric bilinear form.
If F (X, X) ≡ 0 then F (X, Y ) is also an alternating bilinear form, so q is even in this case.
The polynomial F (X, Y ) is equal to det(X, Y, z) for some point z, which is incident with
all the tangents and is called the nucleus. The arc S can be extended to an arc of size
q + 2 by adjoining the nucleus.
If F (X, X) 6≡ 0 then Theorem 13 implies that the points of S are the zeros of a conic,
since fx (x) = 0 for all x ∈ S. Thus we have that if q is odd then an arc of size q + 1 is a
conic, which is Segre’s theorem from [21].
Example 2. The planar arc of 12 points in PG2 (F13 ),
S = {(3, 4, 1), (−3, 4, 1), (3, −4, 1), (−3, −4, 1), (4, 3, 1), (4, −3, 1), (−4, 3, 1), (−4, −3, 1),
(1, 1, 1), (1, −1, 1), (−1, 1, 1), (−1, −1, 1)}
is an arc with t = 3 and it is not contained in a curve of degree 3. Consequently,
Theorem 13 implies that there is a unique polynomial F (X, Y ) of degree three in both X
and Y with the property that F (X, a) = fa (X) for all a ∈ S. It is given by
F (x, y) = 5(x22 x3 y12 y3 + y22 y3 x21 x3 + x2 x23 y12 y2 + x21 x2 y2 y32 + x1 x23 y1 y22 + x1 x22 y1 y32 )
+6x1 x2 x3 y1 y2 y3 + x31 y13 + x32 y23 + x33 y33 .
Example 3. The planar arc of 10 points in PG2 (F11 ),
S = {(1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 8, 7), (1, 5, 2), (1, 3, 8), (1, 4, 1), (1, 7, 6), (1, 9, 4), (1, 10, 5)}
is an arc with t = 3 and it is not contained in a curve of degree 3. Consequently,
Theorem 13 implies that there is a unique polynomial F (X, Y ) of degree three in both X
PLANAR ARCS
11
and Y with the property that F (X, a) = fa (X) for all a ∈ S. In this case F has 59 terms
and is given by
F (x, y) = 9x33 y23 + 7x33 y1 y22 + 4x33 y12 y2 + 2x33 y13 + 5x2 x23 y2 y32 + 4x2 x23 y1 y32 + x2 x23 y1 y2 y3
+9x2 x23 y1 y22 + 2x2 x23 y12 y3 + 8x2 x23 y12 y2 + 5x22 x3 y22 y3 + 9x22 x3 y1 y32 + 4x22 x3 y1 y2 y3 + 8x22 x3 y1 y22
+5x22 x3 y12 y3 + x22 x3 y12 y2 + 8x22 x3 y13 + 9x32 y33 + 3x32 y12 y3 + x32 y13 + 4x1 x23 y2 y32 + 9x1 x23 y22 y3
+5x1 x23 y1 y32 +10x1 x23 y1 y2 y3 +8x1 x23 y1 y22 +9x1 x23 y12 y2 +x1 x2 x3 y2 y32 +4x1 x2 x3 y22 y3 +10x1 x2 x3 y1 y32
+7x1 x2 x3 y1 y2 y3 + 8x1 x2 x3 y1 y22 + 4x1 x2 x3 y12 y3 + 3x1 x2 x3 y12 y2 + 7x1 x22 y33 + 9x1 x22 y2 y32
+8x1 x22 y22 y3 + 8x1 x22 y1 y32 + 8x1 x22 y1 y2 y3 + 8x1 x22 y1 y22 + 10x1 x22 y12 y3 + x1 x22 y12 y2
+2x21 x3 y2 y32 + 5x21 x3 y22 y3 + 3x21 x3 y23 + 4x21 x3 y1 y2 y3 + 10x21 x3 y1 y22 + 5x21 x3 y12 y3
+3x21 x3 y12 y2 + 4x21 x2 y33 + 8x21 x2 y2 y32 + x21 x2 y22 y3 + 9x21 x2 y1 y32 + 3x21 x2 y1 y2 y3
+x21 x2 y1 y22 + 3x21 x2 y12 y3 + 8x21 x2 y12 y2 + 2x31 y33 + 8x31 y22 y3 + x31 y23 .
Example 4. The 24 points which form the zero set of the polynomial
x31 x2 + x32 x3 + x33 x1
is a planar arc of PG2 (F29 ) with t = 7. The dimension of Φ, the space of homogeneous
polynomials of degree t whose zero set contains the 24 points of the arc, is 13.
In this case F is unique modulo (Φ[X], Φ[Y ]) and an example with 136 terms can be found
in the Appendix.
This example comes close to the upper bound of 25 given by Corollary 7 for the size of a
planar arc in PG2 (F29 ) contained in a quartic curve.
7. Proof of Theorem 13.
Proof. (of Theorem 13) Let V = Ψ[X] ⊕ Φ[X] be the vector space of all homogeneous
polynomials of degree t in three variables, where Φ[X] is the subspace of polynomials
vanishing on the arc S. Consider the subspace
Ω = {(g(a))a∈S | g ∈ V }
|S|
of Fq .
|S|
Let λa denote the a coordinate of a vector λ ∈ Fq .
For each λ ∈ Ω⊥ and each x ∈ S we have
X
λa fa (x) = 0,
a∈S
12
SIMEON BALL AND MICHEL LAVRAUW
since, by Lemma 11, fa (x) = (−1)t+1 fx (a) for all a, x ∈ S, and fx (a) is a homogeneous
polynomial in a of degree t, by definition.
Define the subspace Π to be
Π = {λ ∈ Ω⊥ |
X
λa fa (X) ≡ 0}.
a∈S
|S|
Let U and W be subspaces of Fq such that Ω⊥ = Π ⊕ W and Π⊥ = Ω ⊕ U . Observe that
dim U = dim W and let m = dim U .
For each i, j ∈ {0, . . . , t}, with i + j 6 t, we define a function fij from S to Fq , where for
each a ∈ S the value of fij (a) is the coefficient of X1i X2j X3t−i−j of the polynomial fa (X),
i.e.
t
X
fa (X) =
fij (a)X1i X2j X3t−i−j .
i+j=0
For each λ ∈ Π,
X
λa fij (a) = 0,
a∈S
so the vector (fij (a))a∈S ∈ Π⊥ . We can write
fij = pij + hij ,
for some functions pij and hij , where (pij (a))a∈S ∈ Ω and (hij (a))a∈S ∈ U . Observe that
the function pij is the evaluation of a homogeneous polynomial pij [Y ] ∈ Ψ[Y ] of degree t.
Let u1 , . . . , um be functions from S to Fq such that {((u1 (a))a∈S , . . . , ((um (a))a∈S } is a
basis for U . Then
m
X
qijk uk ,
fij = pij +
k=1
for some qijk ∈ Fq .
For λ ∈ W \ {0},
X
λa fa (X),
a∈S
is a non-zero polynomial vanishing at the points of S, so is an element of Φ[X].
Since (pij (a))a∈S ∈ Ω ⊆ W ⊥ , we have
!
m
t
X
X
X
X
λa fa (X) =
λa uk (a) vk (X), where vk (X) =
qijk X1i X2j X3t−i−j .
a∈S
k=1
a∈S
i+j=0
PLANAR ARCS
Suppose that λ, λ0 ∈ W and that
X
λa fa (X) =
X
a∈S
0
13
λ0a fa (X).
a∈S
0
Then λ − λ ∈ Π and so λ = λ . This implies that as λ ranges over the q m elements in
P
W , we obtain q m different polynomials a∈S λa fa (X) in the subspace spanned by the m
polynomials v1 , . . . , vm . Hence,
X
{
λa fa (X) | λ ∈ W } = hv1 (X), . . . , vm (X)i,
a∈S
and in particular v1 , . . . , vm ∈ Φ[X].
Let gij (X) ∈ Ψ[X] be the polynomial such that X1i X2j X3t−i−j = gij (X) modulo Φ[X].
Define a homogeneous polynomial of degree t in X and Y as
F (X, Y ) =
t
X
pij (Y )gij (X),
i+j=0
Then, for all a ∈ S,
F (X, a) =
t
X
pij (a)gij (X) =
i+j=0
=
t
X
t
X
fij (a)gij (X) −
i+j=0
fij (a)X1i X2j X3t−i−j
i+j=0
−
m X
t
X
m X
t
X
qijk uk (a)gij (X)
k=1 i+j=0
qijk uk (a)X1i X2j X3t−i−j
(mod Φ[X])
k=1 i+j=0
= fa (X) −
m
X
uk (a)vk (X) = fa (X)
(mod Φ[X]).
k=1
It only remains to prove the uniqueness claim and the quasi-symmetric claim.
To prove the uniqueness claim, consider
H(X, Y ) = F (X, Y ) − F 0 (X, Y )
(mod Φ[X], Φ[Y ])
as a polynomial in X written out over a basis for Ψ[X], whose coefficients are polynomials
in Ψ[Y ]. For all a ∈ S, we have that H(X, a) = 0. If there is a coefficient f (Y ) of H(X, Y )
which is not zero, then S is contained in the zero set of f , so f ∈ Φ[Y ]. However, f ∈ Ψ[Y ],
which implies f = 0. Hence H = 0 and so F ≡ F 0 modulo (Φ[X], Φ[Y ]).
To prove the quasi-symmetric claim, observe that each term of F (X, Y ) is the product of
a polynomial in Ψ[X] with a polynomial in Ψ[Y ]. Then consider
F (X, Y ) − (−1)t+1 F (Y, X)
14
SIMEON BALL AND MICHEL LAVRAUW
again as a polynomial in X written out over a basis for Ψ[X], whose coefficients are
polynomials in Ψ[Y ]. Suppose there is a coefficient f (Y ) for which there is an a ∈ S with
the property that f (a) 6= 0. Then, by Lemma 11 and the fact that F (x, a) = fa (x) for
any a, x ∈ S, the polynomial
F (X, a) − (−1)t+1 F (a, X) ∈ Ψ[X]
vanishes on S, and is therefore zero. So f (Y ) is zero when evaluated at a point of S.
Hence, f ∈ Φ[Y ]. However, f ∈ Ψ[Y ], which implies f = 0. Therefore, every coefficient
of F (X, Y ) − (−1)t+1 F (Y, X) is identically zero, which completes the proof.
8. Arcs in odd characteristic are contained in low degree curves.
In this section we investigate some consequences of Theorem 13.
Let S be a planar arc of size q+2−t and let F (X, Y ) be a polynomial given by Theorem 13,
i.e. a representative from the equivalence class modulo (Φ[X], Φ[Y ]).
For each i, j, k ∈ {0, . . . , t − 1} where i + j + k 6 t − 1, define ρijk (Y ) to be the coefficient
of X1i X2j X3k in
F (X + Y, Y ) − F (X, Y ).
Lemma 14. For all i, j, k ∈ {0, . . . , t − 1} where i + j + k 6 t − 1, the polynomial ρijk (Y )
is either zero or a homogeneous polynomial of degree 2t − i − j − k which is zero on S.
Proof. For all a ∈ S, the polynomial fa (X) is the product of t linear forms whose kernels
contain the point a. Therefore, fa (X + a) = fa (X). By Theorem 13, for each a ∈ S,
F (X + a, a) − F (X, a) = fa (X + a) − fa (X) = fa (X) − fa (X) = 0
(mod Φ[X]).
However, F (X + a, a) − F (X, a) is a polynomial in X of degree at most t − 1, so this is in
fact zero. Hence, each coefficient of F (X + Y, Y ) − F (X, Y ), written as a polynomial in
X whose coefficients are polynomials in Y , is a (possibly zero) polynomial which vanishes
on S.
Example 5. Applying Lemma 14 to the arc of size 12 in Example 2, we see that S lies
on the intersection of the three quartic curves x43 = x21 x22 , x42 = x21 x23 and x41 = x23 x22 , as
observed in [1].
The following lemma will be used in the proof of Lemma 16.
PLANAR ARCS
15
Lemma 15. Let t be a positive integer and = blogp tc. Suppose that for all k ∈
{1, . . . , p } and for all i ∈ {0, . . . , t − k},
i+k X
d
t−d
gd = 0,
i
t−i−k
d=i
for some g0 , . . . , gt in a polynomial ring over Fq . Then
gt + (−1)t+1 g0 = 0.
Proof. Let t = t0 + t1 p + · · · + t p be the p-ary expansion of t.
First, we prove by induction that
t−i
s+1 i + s
gi + (−1)
gi+s = 0,
s
i
for all s ∈ {1, . . . , p } and for all i ∈ {0, . . . , t − s}.
For s = 1 this is the given equation with k = 1.
Assume the equation holds for j 6 s − 1. Then
i+s−1
s−1 X d t − d X
i+s
i+j
t−i−j
gi+s = −
gd = −
gi+j ,
i
i
t
−
i
−
s
i
t
−
i
−
s
j=0
d=i
which, by induction, is equal to
s−1
s−1
X
X
t−i−j
t−i
j t−i
j s
s t−i
−
(−1)
gi = −
(−1)
gi = (−1)
gi ,
j
t−i−s
j
s
s
j=0
j=0
as required.
For s = p and i = t − (k + 1)p , where k = 0, . . . , t − 1, this gives
(k + 1)gt−(k+1)p + (t − k)gt−kp = 0,
by Lucas’ theorem.
Combining these equations we have that
gt−t p = (−1)t gt .
For s = t − t p and i = 0 we obtain
g0 + (−1)t−t p
+1
gt−t p = 0.
Eliminating gt−t p from these two equations, the lemma follows.
16
SIMEON BALL AND MICHEL LAVRAUW
We say that a polynomial φ(X) is hyperbolic on an arc S if it has the property that if
the kernel of a linear form γ is a bisecant to S then φ modulo γ factorises into at most
two linear factors whose multiplicities sum to the degree of φ and which are zero at the
points of S on the bisecant.
Lemma 16. Let S be a planar arc of size q + 2 − t. If q is odd then one of the following
holds: (i) there are two co-prime polynomials of degree at most t + pblogp tc which are zero
on S; (ii) there is a non-zero homogeneous polynomial φ of degree at most t + pblogp tc
which is hyperbolic on S.
Proof. Let
W = {(w1 , w2 , w3 ) ∈ {0, . . . , t − 1}3 | t − p 6 w1 + w2 + w3 6 t − 1},
where = blogp tc.
Let φ(Y ) be the greatest common divisor of
{ρw (Y ) | w ∈ W } ∪ Φ[Y ].
Observe that the degree of φ is at most t + p . We do not yet discount the case that the
first set and the second set contain only the zero polynomial. In this case, which we shall
rule out, φ is the zero polynomial.
Let F (X, Y ) be a representative of the equivalence class of polynomials given by Theorem 13.
Let x and y be arbitrary points of S and let B be a basis, with respect to which, x =
(1, 0, 0) and y = (0, 1, 0). Let fa∗ (X) be the polynomial we obtain from fa (X) when
we change the basis from the canonical basis to B, and likewise let F ∗ (X, Y ) be the
polynomial we obtain from F (X, Y ), and let φ∗ be the polynomial we get from φ.
Define homogeneous polynomials bd1 d2 d3 (Y ) of degree t by writing
X
F ∗ (X, Y ) =
bd1 d2 d3 (Y )X1d1 X2d2 X3d3 .
d1 +d2 +d3 =t
Then
F ∗ (X + Y, Y ) =
X
bd1 d2 d3 (Y )(X1 + Y1 )d1 (X2 + Y2 )d2 (X3 + Y3 )d3 ,
d1 +d2 +d3 =t
=
X
d1 +d2 +d3
d1
d2
d3
bd1 d2 d3 (Y )
X1i1 Y1d1 −i1 X2i2 Y2d2 −i2 X3i3 Y3d3 −i3 .
i1
i2
i3
=t
PLANAR ARCS
17
Let rijk (Y ) be the coefficient of X1i X2j X3k in F ∗ (X + Y, Y ) − F ∗ (X, Y ). Then rijk (Y ) is
a linear combination of the polynomials in the set {ρ∗w (Y ) | w1 + w2 + w3 = i + j + k},
where ρ∗w (Y ) is the polynomial we obtain from ρw (Y ), when we change the basis from the
canonical basis to B.
Since φ(Y ) is the greatest common divisor of
{ρw (Y ) | w ∈ W } ∪ Φ[Y ],
φ∗ (Y ) is a factor of all the polynomials in the set
{rw (Y ) | w ∈ W } ∪ Φ∗ [Y ],
where Φ∗ [Y ] is the subspace of homogeneous polynomials of degree t which are zero on
S, with respect to the basis B.
Let w = (i, t − i − k, 0), where k ∈ {1, . . . , p } and i ∈ {0, . . . , t − k}. Then w ∈ W and
X d1 d2 rw (Y ) =
Y1d1 −i Y2d2 −t+i+k Y3d3 bd1 d2 d3 (Y ).
t
t−i−k
d +d +d =t
1
2
3
The polynomial φ∗ is a factor of all these polynomials, so it is a factor of Y1i Y2t−i−k rw (Y )
and therefore,
i+k X
d
t−d
Y1d Y2t−d bd,t−d,0 (Y ) = 0 (mod Y3 , φ∗ ),
i
t
−
i
−
k
d=i
for all k ∈ {1, . . . , p } and i ∈ {0, . . . , t − k}.
By Lemma 15, with gd (Y ) = Y1d Y2t−d bd,t−d,0 (Y ),
Y1t bt,0,0 (Y ) + (−1)t+1 Y2t b0,t,0 (Y ) = 0
(mod Y3 , φ∗ ).
Note that if φ∗ = 0 then ρw = rw = 0 for all w ∈ W , so the expression is also zero in this
case.
By Theorem 13,
F (Y, x) = fx (Y )
(mod Φ[Y ]).
With respect to the basis B this gives,
F ∗ (Y, x) = fx∗ (Y )
(mod Φ∗ [Y ]).
Since fx∗ (Y ) is a polynomial in Y2 and Y3 ,
fx∗ (Y ) = fx∗ (y)Y2t
(mod Y3 ).
By Theorem 13,
F (X, Y ) = (−1)t+1 F (Y, X),
18
SIMEON BALL AND MICHEL LAVRAUW
so with respect to the basis B this gives,
F ∗ (X, Y ) = (−1)t+1 F ∗ (Y, X),
This implies that
bt,0,0 (Y ) = F ∗ (x, Y ) = (−1)t+1 F ∗ (Y, x) = (−1)t+1 fx∗ (y)Y2t
(mod Φ∗ [Y ], Y3 ).
Similarly
b0,t,0 (Y ) = (−1)t+1 fy∗ (x)Y1t
(mod Φ∗ [Y ], Y3 ).
Hence, we have that
Y1t Y2t (fx∗ (y) + (−1)t+1 fy∗ (x)) = 0 (mod Y3 , φ∗ ).
By Lemma 11 and the fact that fa and fa∗ define the same functions, this implies
2Y1t Y2t fx (y) = 0
(mod Y3 , φ∗ ).
By hypothesis q is odd, so the left-hand side is non-zero. Hence, this equation rules out
the possibility that φ∗ (and hence φ) is zero and we have proved that there is a curve of
degree at most t + p which contains S.
If the degree of φ is zero then there must be at least two co-prime polynomials of degree
at most t + p both of which are zero on S.
If the degree of φ is not zero then the above equation implies that
φ∗ (Y ) = cY1i Y2j
(mod Y3 ),
for some integers i, j such that i + j = deg φ∗ = deg φ and some c ∈ Fq . With respect to
the canonical basis this gives
φ(Y ) = α(Y )i β(Y )j
(mod γ(Y )),
where the kernel of the linear form γ is the line joining x and y and α and β are linear
forms with the property that α(y) = 0 and β(x) = 0. Thus, we have proved that if the
kernel of a linear form γ is a bisecant to S then φ modulo γ factorises into at most two
linear factors whose multiplicities sum to the degree of φ, i.e. φ is hyperbolic on S.
9. Proof of Theorem 1.
Lemma 17. Let S be a planar arc of size q + 2 − t > 8. If there is a homogeneous
polynomial φ of degree at most 21 (q − t + 1) which is hyperbolic on S, then all but at most
one point of S are contained in a conic and if q is odd then S is contained in a conic.
PLANAR ARCS
19
Proof. Let r be the degree of φ.
Suppose there is a point x in S which is not in the zero set of φ. If γ is a linear form such
that the kernel of γ is a bisecant joining x to a point y of S then, since φ is hyperbolic on
S, φ(X) = α(X)r modulo γ for some linear form α, where α(y) = 0. This implies that
φ(y) = 0, so φ is zero on all but at most one point of S.
Observe that r > 2, since S is not a line. Also, we can assume that φ is not a p-th power,
since we can replace φ by its p-th root and all the properties are preserved.
Choose a suitable basis so that (1, 0, 0), (0, 1, 0) and (0, 0, 1) are points of S \ {x}.
Suppose every term of φ(X) is of the form cijk X1i X2jp X3kp . Since φ(X) is hyperbolic on S,
φ(0, X2 , X3 ) = c0jk X2jp X3kp , for some j and k. Hence r = (j + k)p. Considering any term
with i > 0, it follows that p divides i. So φ is a p-th power, which it is not. Hence we can
assume, without loss of generality, that some exponent of X1 is not a multiple of p, some
exponent of X3 is not a multiple of p and that the degree of φ in X3 is at most the degree
of φ in X1 .
Let n be the degree of φ in X1 .
Write
φ(X) =
n
X
X1n−j cj (X2 , X3 ),
j=0
where cj is either zero or a homogeneous polynomial of degree r −n+j and by assumption
c0 (X2 , X3 ) 6= 0.
Let E = {e ∈ Fq | X3 − eX2 is a bisecant and c0 (X2 , eX2 ) 6= 0}. Since the degree of c0
is r − n we have that |E| > q − t − r + n. Since φ(X) is hyperbolic on S, for all e ∈ E,
there exists a d such that
φ(X1 , X2 , eX2 ) = (X1 + dX2 )n c0 (X2 , eX2 ).
The coefficient of X1n−j implies that for j = 1, . . . , n,
n j j
cj (X2 , eX2 ) =
d X2 c0 (X2 , eX2 ).
j
If p divides n and j is not a multiple of p then cj (X2 , eX2 ) = 0 for all e ∈ E. Since the
degree of cj is at most r and r 6 |E| − n + 1, by hypothesis, cj (X2 , X3 ) = 0. But then this
implies that each exponent of X1 in a term of φ(X) is a multiple of p, a contradiction.
Therefore, n is not a multiple of p.
20
SIMEON BALL AND MICHEL LAVRAUW
For e ∈ E and j = 1 we have that
c1 (X2 , eX2 ) = ndX2 c0 (X2 , eX2 ).
Thus, for j = 1, . . . , n, substituting for d we obtain
n
j−1
j
c0 (X2 , eX2 ) cj (X2 , eX2 )n =
c1 (X2 , eX2 )j .
j
Hence, hj (e) = 0, for all e ∈ E, where hj (Y ) is a polynomial in (Fq [X2 ])[Y ] defined by
n
j−1
j
hj (Y ) = c0 (X2 , Y X2 ) cj (X2 , Y X2 )n −
c1 (X2 , Y X2 )j .
j
Suppose n > 2. Let m be the degree of the polynomial h2 (Y ). Then m 6 2(r − n + 1)
since the degree of cj is r − n + j, and m 6 2n since the degree of cj (X2 , X3 ) in X3 is at
most n. To be able to conclude that h2 is identically zero, we need m 6 |E| − 1, which is
equivalent to min(3(r −n)+2, r +n) 6 q −t−1. If n 6 r −2 then r +n 6 2r −2 6 q −t−1
by hypothesis. If n = r − 1 then 3(r − n) + 2 = 5 6 q − t − 1, since |S| = q + 2 − t > 8.
Therefore, h2 (Y ) is identically zero. This implies that the polynomial c0 (X2 , Y X2 ) divides
c1 (X2 , Y X2 ) and so
c1 (X2 , Y X2 ) = (aX2 + bY X2 )c0 (X2 , Y X2 ),
for some a, b ∈ Fq . Hence,
j−1
hj (Y ) = c0 (X2 , Y X2 )
n
(cj (X2 , Y X2 )n −
c0 (X2 , Y X2 )(aX2 + bY X2 )j ).
j
j
But for each e ∈ E, the polynomial
n
cj (X2 , Y X2 )n −
c0 (X2 , Y X2 )(aX2 + bY X2 )j ,
j
j
is zero at Y = e. It has degree at most r − n + j 6 r 6 |E| + 1 − n in Y , so we conclude
that it is identically zero.
Substituting Y = X3 /X2 , we have that
n
j
cj (X2 , X3 )n =
c0 (X2 , X3 )(aX2 + bX3 )j ,
j
for j = 1, . . . , n.
Hence,
φ(X) =
n X
n
j=0
a
b
X1n−j c0 (X2 , X3 )( X2 + X3 )j ,
j
n
n
PLANAR ARCS
21
and therefore
φ(X) = c0 (X2 , X3 )(X1 +
a
b
X2 + X3 )n .
n
n
Since φ is zero at the points of S \ {x}, this implies that all but r − n points of S \ {x} are
contained in a line, which gives |S| − 1 6 r − n + 2 6 r and hence q + 1 − t 6 21 (q − t + 1),
an inequality which is not valid.
Hence, n = 1. Since the degree of φ in X3 is at most the degree of φ in X1 , this implies
that the degree of φ in X3 is one. Since φ is hyperbolic on S, φ(X1 , 0, X3 ) is a constant
times X1 X3 . Therefore, r = 2 and we have proved that φ is a quadratic form and that
S \ {x} is contained in a conic.
Now, let γ be a linear form whose kernel is a bisecant to S containing x. Then, as in
the first paragraph this bisecant to S is a tangent to φ. Since a point not in a conic is
on at most two tangents to a conic, when q is odd, and |S \ {x}| > 3 by assumption, we
conclude that no such point x exists and that S is contained in a conic if q is odd.
Proof. (of Theorem 1) Let d = t + pblogp tc .
Suppose that |S| 6 2d. Let a1 , . . . , ad be linear forms whose kernels give a set L of lines
which cover the points of S. Let b1 , . . . , bd be another d linear forms whose kernels give a
Q
set of lines, disjoint from L, but which also cover the points of S. Let a(X) = di=1 ai (X)
Q
and b(X) = di=1 bi (X). Then the zero sets of a(X) and b(X) both contain S and a(X)
and b(X) have no common factor, so we are done.
If |S| 6 7 then the theorem holds almost trivially, since we can cover the points with a
conic and a line in two ways (i.e. using different conics and different lines) and deduce
that S is in the intersection of two cubics, which do not share a common component.
Note that t > 2, since we are assuming that S is not contained in a conic.
Suppose that |S| > 2d + 1 and |S| > 8. Then q + 2 − t > 2(t + pblogp tc ) + 1 which implies
1
(q − t + 1) > t + pblogp tc and Lemma 16 applies.
2
If there is a non-zero homogeneous polynomial φ of degree at most t + pblogp tc which is
hyperbolic on S then, by Lemma 17, S is contained in a conic. This is ruled out by
hypothesis. Therefore, we have the other possibility given by Lemma 16, that there are
two co-prime polynomials of degree at most t + pblogp tc which are zero on S, i.e. there are
two curves of degree at most t + pblogp tc , which do not share a common component, both
containing S.
22
SIMEON BALL AND MICHEL LAVRAUW
10. Planar arcs in planes of order less than 32 of odd characteristic.
The entries in the following array are the number of complete arcs in planes of odd order
less than 32, for which |S| = q + 2 − t > 2d + 1, where d = t + pblogp tc . Again, we have used
[14, Table 9.4] and the articles [8], [9] and [10] to compile this data. Below these values
the trivial construction in the proof of Theorem 1 suffices to prove that S is contained in
the intersection of two curves of degree at most d, sharing no common component.
q
5 7 9 11 13 17 19 23 25 27 29 31
6 7 10 10 11 14 15 18 23 24 22 23
|S| >
# not contained in a conic 0 0 0 1 1 1 0 0 0 0 1 0
We consider each of the four complete arcs of size at least 2d + 1 in planes of odd characteristic of order at most 31. In each case, we construct two curves of degree d, sharing
no common component, containing the arc, whose existence follows from Theorem 1.
The unique arc S of size 10 in PG2 (F11 ) is Example 3. Since any five points lie on a conic,
we can cover the points of S by two conics C1 and C10 . By choosing four points on C1
and one point on C10 , we can cover at least these five points of S by another conic C2 and
therefore all the points of S by two conics C2 and C20 . This gives us two quartic curves
which share no common component, both of which contain S.
The unique arc of size 12 in PG2 (F13 ) is Example 2. As we already saw in Example 5 it is
contained in the intersection of the two quartic curves given by the equations x42 = x21 x23
and x41 = x23 x22 .
The unique arc S of size 14 in PG2 (F17 ) has 10 points on a conic C. Hence, we get one
curve of degree four covering S by choosing two lines covering the points of S \ C. Now,
let C1 be a conic through four points of S ∩ C and one point of S \ C and C2 be a conic
through four other points of S ∩ C and one other point of S \ C. We can cover the points
of S \ (C1 ∪ C2 ) with two lines and obtain a curve of degree 6 which contains S and does
not share a common component with the curve of degree 4 containing S.
The unique arc of size 24 in PG2 (F29 ) is Example 4. It is exactly the zero set of the
polynomial
g(x) = x31 x2 + x32 x3 + x33 x1 .
Let S1 and S2 be disjoint subsets of the arc of size 12 and choose two points a1 and a2
not in S = S1 ∪ S2 . Let fi be a quartic curve containing Si ∪ {ai }. Then f1 f2 defines a
PLANAR ARCS
23
degree 8 curve containing S and having no common component with the curve defined by
g (since g(ai ) 6= 0).
Appendix
The 136 term F (x, y) for the 24-arc in PG2 (F29 ) mentioned in Example 4 is given by
F (x, y) = x73 y27 +x73 y17 +12x2 x63 y26 y3 +28x2 x63 y1 y23 y33 +17x2 x63 y12 y35 +24x2 x63 y13 y24 +28x2 x63 y14 y2 y32
+19x22 x53 y25 y32 +28x22 x53 y1 y22 y34 +14x22 x53 y13 y23 y3 +18x22 x53 y14 y33 +3x22 x53 y16 y2 +27x32 x43 y24 y33 +20x32 x43 y1 y2 y35
+18x32 x43 y12 y25 +5x32 x43 y13 y22 y32 +13x32 x43 y16 y3 +27x42 x33 y23 y34 +3x42 x33 y1 y36 +27x42 x33 y12 y24 y3 +20x42 x33 y13 y2 y33
+14x42 x33 y15 y22 + 19x52 x23 y22 y35 + 16x52 x23 y1 y26 + 19x52 x23 y12 y23 y32 + 26x52 x23 y13 y34 + 9x52 x23 y15 y2 y3
+12x62 x3 y2 y36 + 16x62 x3 y1 y25 y3 + 7x62 x3 y12 y22 y33 + 5x62 x3 y14 y23 + 14x62 x3 y15 y32 + x72 y37 + x72 y17
+3x1 x63 y24 y33 +3x1 x63 y1 y2 y35 +3x1 x63 y12 y25 +2x1 x63 y13 y22 y32 +x1 x63 y16 y3 +20x1 x2 x53 y23 y34 +3x1 x2 x53 y1 y36
+12x1 x2 x53 y12 y24 y3 +4x1 x2 x53 y13 y2 y33 +10x1 x2 x53 y15 y22 +28x1 x22 x43 y22 y35 +27x1 x22 x43 y1 y26 +18x1 x22 x43 y12 y23 y32
+10x1 x22 x43 y13 y34 +11x1 x22 x43 y15 y2 y3 +28x1 x32 x33 y2 y36 +7x1 x32 x33 y1 y25 y3 +9x1 x32 x33 y12 y22 y33 +24x1 x32 x33 y14 y23
+26x1 x32 x33 y15 y32 +18x1 x42 x23 y12 y2 y34 +18x1 x42 x23 y14 y22 y3 +16x1 x52 x3 y26 y3 +7x1 x52 x3 y1 y23 y33 +4x1 x52 x3 y13 y24
+2x1 x52 x3 y14 y2 y32 +16x1 x62 y25 y32 +27x1 x62 y1 y22 y34 +16x1 x62 y13 y23 y3 +13x1 x62 y14 y33 +x1 x62 y16 y2 +17x21 x53 y2 y36
+4x21 x53 y12 y22 y33 +16x21 x53 y14 y23 +20x21 x53 y15 y32 +18x21 x2 x43 y1 y24 y32 +18x21 x2 x43 y14 y22 y3 +7x21 x22 x33 y26 y3
+9x21 x22 x33 y1 y23 y33 +4x21 x22 x33 y12 y35 +10x21 x22 x33 y13 y24 +3x21 x22 x33 y14 y2 y32 +19x21 x32 x23 y25 y32 +18x21 x32 x23 y1 y22 y34
+16x21 x32 x23 y13 y23 y3 +2x21 x32 x23 y14 y33 +28x21 x32 x23 y16 y2 +27x21 x42 x3 y24 y33 +12x21 x42 x3 y1 y2 y35 +12x21 x42 x3 y12 y25
+28x21 x42 x3 y13 y22 y32 + 11x21 x42 x3 y16 y3 + 18x21 x52 y23 y34 + 3x21 x52 y1 y36 + 12x21 x52 y12 y24 y3 + 27x21 x52 y13 y2 y33
+3x21 x52 y15 y22 + 26x31 x43 y25 y32 + 10x31 x43 y1 y22 y34 + 24x31 x43 y14 y33 + 20x31 x2 x33 y24 y33 + 4x31 x2 x33 y1 y2 y35
+27x31 x2 x33 y12 y25 +5x31 x22 x23 y23 y34 +2x31 x22 x23 y1 y36 +28x31 x22 x23 y12 y24 y3 +14x31 x32 x3 y22 y35 +16x31 x32 x3 y1 y26
+16x31 x32 x3 y12 y23 y32 + 24x31 x42 y2 y36 + 4x31 x42 y1 y25 y3 + 10x31 x42 y12 y22 y33 + 18x41 x33 y22 y35 + 13x41 x33 y1 y26
+2x41 x33 y12 y23 y32 +24x41 x33 y13 y34 +17x41 x33 y15 y2 y3 +28x41 x2 x23 y2 y36 +2x41 x2 x23 y1 y25 y3 +3x41 x2 x23 y12 y22 y33
+18x41 x22 x3 y1 y24 y32 + 18x41 x22 x3 y12 y2 y34 + 5x41 x32 y26 y3 + 24x41 x32 y1 y23 y33 + 16x41 x32 y12 y35 + 14x51 x23 y26 y3
+26x51 x23 y1 y23 y33 + 20x51 x23 y12 y35 + 9x51 x2 x3 y25 y32 + 11x51 x2 x3 y1 y22 y34 + 17x51 x2 x3 y14 y33 + 14x51 x22 y24 y33
+10x51 x22 y1 y2 y35 + 3x51 x22 y12 y25 + 13x61 x3 y23 y34 + x61 x3 y1 y36 + 11x61 x3 y12 y24 y3 + 3x61 x2 y22 y35
+x61 x2 y1 y26 + 28x61 x2 y12 y23 y32 + x71 y37 + x71 y27 .
24
SIMEON BALL AND MICHEL LAVRAUW
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PLANAR ARCS
Simeon Ball
Departament de Matemàtiques,
Universitat Politècnica de Catalunya,
Mòdul C3, Campus Nord,
c/ Jordi Girona 1-3,
08034 Barcelona, Spain
[email protected]
Michel Lavruaw
Università degli Studi di Padova,
Dipartimento di Tecnica e Gestione dei Sistemi Industriali,
Stradella S. Nicola, 3, I-36100 Vicenza, Italy
FENS 1015,
Faculty of Engineering and Natural Sciences,
Sabancı University, Turkey
[email protected]
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