on abelian cubic arcs

D. GHINELLI, N. MELONE* AND U. OTT*

O N A B E L I A N C U B I C A R C S

INTRODUCTION

The problem of characterizing algebraic varieties from their combinatorial properties dates back to Segre's famous theorems on normal rational curves in PG(r, q).

THEOREM I (B. Serge 1-17]). I f q is odd, then every (q + 1)-arc in PG(2, q) is a conic.

THEOREM II (B. Serge [18]). I f q >~ 5 is odd, then every (q + 1)-arc in PG( 3, q) is a normal rational curve (i.e. a twisted cubic).

Theorem I was extended to quadrics in PG(3, q) by Barlotti [1] and Panella [-13] and to quadrics in PG(r, q), r > 3, by Tallini [19], [20]. As for the extension of Theorem II to dimension r > 3: when q is odd, there are results by Thas [22] and by Glynn [6]; when q is even, there are results by Segre [18], Casse and Glynn [3], [41 and, recently, by Bruen et al. [2].

The more natural combinatorial objects to study algebraic curves of order n >~ 3 in PG(2, q) are (k, n)-arcs (see Definition 1.1). Obviously, considering an algebraic curve of order n, say C n, as a (k, n)-arc, we lose many essential properties. For instance, we lose the notion of intersection multiplicity of a line I and C n at a common point P (hence the notions of simple point, singular point and tangent line). Moreover, neither the maximum number of points of a (k, n)-arc nor the maximum number of points on an algebraic curve of order n >~ 3 is known, in general (see, for instance, Waterhouse [23], for elliptic curves). This yields the necessity to introduce a stronger notion than that of (k, n)-arc.

The first results in this direction are due to Tallini Scafati [2i], who introduced the notion of graphic curve. Further results are in Keedwell [10] and Raguso and Rella [-14], [15], [16]. These authors obtained characterization theorems for rational plane curves of order n >I 3 (i.e. curves of genus g = 0). Recently, Hirschfeld and Voloch [9] proved an embedding theorem in an irreducible plane cubic, for particular (k, 3)-arcs in PG(2, q) (called k-group arcs in [9]).

In this paper we use the notion of a curve as an element of the standard module of a projective plane (see [12]) in order to define abelian (cubic) arcs.

* Work partially supported by a NATO Grant.

Geometriae Dedicata 32: 31-52, 1989. © 1989 Kluwer Academic Publishers. Printed in the Netherlands.

32 D. G H I N E L L I ET AL.

We point out how this seems to be the more natural combinatorial object to study elliptic curves; in fact, many properties of elliptic curves can be generalized to abelian arcs. Finally we prove that, under an arithmetical condition, a finite abelian arc in a Pappian projective plane (not necessarily finite) can be embedded in a unique irreducible algebraic cubic, when the arc has no inflections (case not considered in [9]).

1. PRELIMINARIES

The source for the present research on abelian (cubic) arcs is given by the famous Segre theorem on (q + 1, 2)-arcs. We recall the following definition.

DEFINITION 1.1. Let P = (f~l, f~2, I) be a projective plane (if P is finite, we denote its order by q). A (k, n)-arc K in P is a set K of k = IKr points of P such that

(1) n is the maximum number of collinear points in K.

Conics in Desarguesian planes of order q are examples of (q + l, 2)-arcs. Conversely, the above mentioned result of Segre holds, namely:

THEOREM 1.2. In a Desarguesian projective plane of odd order q a (q + 1, 2)- arc is necessarily a conic.

In general the theory of (k, 2)-arcs is similar to the theory of conics. For algebraic curves of higher degree the situation is more complicated. For example, already for n = 3 the number of points on an algebraic cubic K in PG(2, q) is not fixed but (see Hasse [7], [8], Weil [24], [25] and Waterhouse

[233)

(2) q + 1 - 2 x / q ~< Igl ~< q + 1 + 2x/q

For an algebraic curve C n of order n, the notion of intersection multiplicity between C" and a line I at a common point P and the notion of multiplicity of a point play an important role. Another important notion is that of envelope of tangent lines to C". We describe similar notions for special kind of subsets of a projective plane not necessarily Desarguesian (see Definitions 1.3, 1.4 and 1.5 below). We note that these kind of ideas first appeared in Tallini Scafati [21].

Let F be the set of flags A = {A, a} of P:

(3) n: = {A = {A, a} I(A, a) ~ f~l x f~2, A I a}.

Let m i> 2 be an integer (when P is finite of order q, then 2 ~< m ~< q + 1).

O N A B E L I A N C U B I C A R C S 33

DEFINITION 1.3. A quasi-m-arc in P (see Ott [12]) is a function f : 0: --+ Z such that

(4)

(5)

f(A) e {0, 1 . . . . . m} for every A e ~c,

~ f (A) e {0, 1 . . . . . m - 2, m} for each fixed a e f~2 A~a

(the sum being over all flags A = {X, a} through the fixed line a). The set of points

(6) K = supp(f) = {Xe fl 1 If(A) # 0 for every A ~ a}

is called the support of the quasi-m-arc f . We also say that (f, K) is a quasi-m-arc, meaning that f : B: --. ~ is a function with support K, satisfying (4) and (5). Clearly, the value f(A) of f on the flag A = {A, a} gives the intersection multiplicity between the line a and K at the point A.

All the terminology is inspired from the theory of algebraic curves.

DEFINITION 1.4. A point A e supp(f) is said to be an s-point (or a point with multiplicity s, 1 ~< s < m - 1), if

(7) f(A) >~ s for each A ~ A,

(8) f(A)/> s + 1 for at most s flags containing A.

The lines of the flags in (8) are called tangent lines to K at the s-point A. An s-point with s ~> 2 is called singular (double if s = 2). If f (A) is a constant c for all flags containing a point A, then A is said to be an isolated c-point (c integer with 1 ~< c ~< m - 1). A simple point is a nonisolated 1-point or, equivalently, a nonsingular point A for which there exists exactly one line t, called the tangent line to K at A, such thatf(F) ~> 2 for F = {A, t}. Iff(F) >/3, the simple point A is called an inflection point.

DEFINITION 1.5. An m-arcfwith support K is a quasi-m-arc 0 c, K) such that

(9) each point in K is an s-point (1 ~< s ~< m - 1),

(10) each 1-point is simple.

An m-arc (f, K) is said to be nonsingular when each point of K is simple. If an m-arc (f, K) has a 2-point (or double point) A, we say that A is either

a knot, or a cusp or an isolated double point, according as there are two, one or no tangent lines to K at A. Here, again, the terminology is inspired from the theory of algebraic curves.

From now on, (f, K) will be a 3-arc (or, as we also say, a cubic arc). By


definition, each point A in K is an s-point with 1 ~< s ~< 2 = m - 1 and each 1-point is not isolated.

LEMMA 1.6. A cubic arc has at most one singular point, necessarily double. I f P is finite of order q and (f, K) has a double point A, then

! if A is a knot

(11) IKI = + 1 if A is a cusp

+ 2 if A is an isolated 2-point.

Proof If there would be two distinct double points A and B in K, then for the line a joining the points A and B we would have ~,a~af(A) >1 4, in contradiction with (5) since m = 3. Hence there is at most a double point in K. Moreover, through a knot, a cusp or an isolated double point there pass exactly q - 1, q or q + 1 nontangent lines to K, respectively. The condition ~a~ f(A) ~ {0, 1, 3} implies that on each of these nontangent lines there is exactly one point of K distinct from A. Hence, we have (11).

Let A be the unique double point of a singular cubic arc (f , K) and set K' = K\{A} . The function f ' : D z --* 77 defined by

f f (A) if A does not contain A (12) if(A) lo otherwise

is a cubic arc with support K'. We say that (f ' , K') is obtained from (f, K) by removing the double point A. In other words:

LEMMA 1.7. The set K' of simple points of a cubic arc ( f , K) form the support of a nonsingular cubic arc (f', K').

EXAMPLE 1.8. Let P = PG(2, 8) and let P' be a subplane of order 2. Let X 1 , X 2 . . . . . X 7 be the points on P' and X a point of P not belonging to the seven lines ai of P which are 3-secant to P'. Moreover, let l be one of the two 0-secants of P' through X and t i the line joining X and Xi. The function f : D: ~ Z defined

by m

3 if A = { X , I }

2 if A = {X~, ti}

(13) f(A) = 1 if A = {Xi, a} a distinct from all the t~

1 if A = {X,a}, a distinct from l

0 elsewhere m

is a nonsingular cubic arc of P with exactly one inflection point X.

The above example can be generalized as follows.

ON ABELIAN CUBIC ARCS 35

EXAMPLE 1.9. Let C be the algebraic cubic in the affine plane AG(2, 2 a) represented by the equation y = x 3. The origin 0 is a simple point with tangent the line l: y = 0. Since f(A) = 3 with A = {O, l}, O is an inflection point. It can be shown that C\{0} form the projective space PG(a - 1, 2). For a = 3 we get Example 1.8 as a particular case.

2. ABELIAN (CUBIC) ARCS

In this section we give a brief description of some algebraic structures which occur, in a natural way, in the theory of cubic arcs. Further details on these arguments can be found in Manin [11].

From now on (f, K) will be a nonsingular cubic arc. If we count points with their intersection multiplicity, on each line there are exactly 0, 1 or 3 points. Counting multiplicities, for every two points X, Y in K, there is a unique third point Z ,= X o Y in K collinear with X and Y. We say

(1) X, Y, Z is a cycle ¢~ X, Y, Z are collinear ~ Z = X o Y.

We explicitly remark that X o X = Z means that the tangent line to K at X meets K also at Z and, in particular, that X o X = X means that X is an inflection point. Clearly, the three-place relation X o y = Z is invariant under all the permutations of X, Y, Z. This is equivalent, for all X, Y in K, to the following conditions:

(2) Xo Y = yoX,

(3) X o ( X o V ) = Y.

The above remarks may serve as geometrical background and motivation for the following definition (see Manin [11]).

D E F I N I T I O N 2.1. A set E with a binary composition law o : E x E ~ E is a symmetric quasigroup if the identities (2) and (3) hold for all X, Y in K.

Now, in order to be able to give the definition of an abelian quasigroup we need the following theorem, proved in [11]. We include the proof here to make the paper more or less self-contained.

T H E O R E M 2.2. Let (E, o) be a symmetric quasigroup. Then the following properties are equivalent:

(i) For some 0 ~ E the composition law

(4) X + Y: O o ( X o y )

turns E into an abelian group E o (with 0 as neutral element).

36 D. G H I N E L L I ET A L .

(ii) For any 0 ~ E the composition (4) turns E into an abelian group. Further-

more, the abelian groups {Eo}o~ are all isomorphic.

Proof (ii) = =~ (i) is trivial. (i) = =~ (ii). By (3) and (4), if Z = X o Y then

(5) (X + Y) + Z = O o {(X + V ) o Z} = O o {(O o Z) o Z} = O o O.

If we set

(6) U = O o O,

then, for every pair of points X, Y we have X o Y = U - (X + Y). F r o m this

and f rom (1) and (5) we easily see that

(7) X, Y, Z is a cycle ¢ : -X + Y + Z = U in Eo.

Now, let A be any fixed element in E; we have to show that the composi t ion law

(8) X + Y ..= A o ( X o If)

turns E into an abelian group E A and that E A is i somorphic to E o. By (8), the points A, X o Y and X ~ Y form a cycle; therefore, f rom (7) we have

A + X ~ Y + X o y = U, which implies that X @ Y = U - X o y _ A. Since

X, Y, X o y is a cycle, f rom (7) we have U - X ° Y = X + Y. Therefore

(9) X O Y = X + Y - A .

Since Eo is an abelian group, f rom (9) easily follows that (E, (~) = E A is an

abelian group with neutral element A. The one- to -one-map 7: Eo ~ EA

defined by 7(X) = X + A satisfies the relation 7(X + Y) = X + Y + A =

X + A + Y + A - A = 7(X) ~ 7(Y); hence, 7 is an i somorphism between E o

and E A for any fixed A in E, which complete the proof.

D E F I N I T I O N 2.3. A symmetr ic quas igroup is called abelian if it satisfies one

of the equivalent condit ions of Theorem 2.2.

Go ing back to cubic arcs we have:

L E M M A 2.4. The support of a nonsingular cubic arc is, in a natural way,

a symmetric quasigroup.

D E F I N I T I O N 2.5. A nonsingular cubic arc (f , K) is called an abelian arc if its suppor t is an abelian symmetr ic quasigroup, i.e. it satisfies the equivalent condit ions of Theorem 2.2; now, for any O ~ K, the composi t ion law (4) turns

K into an abelian group Ko, with 0 as neutral element, and the abelian groups

{Ko}o~K are all isomorphic.


EXAMPLE 2.6. The cubic arc defined in 1.8 is abelian and Ko is EA(8), the elementary abelian group of order 8. More generally, the algebraic cubic C: y = x 3 in AG(2, 2 a) is an abelian arc with K o isomorphic to EA(2a).

Other examples can be obtained from the following proposition.

PROPOSITION 2.7. Let (f, K) be an abelian arc, 0 a fixed element in K and

U = 0 o O. Then any subgroup W o f K o containing U is the support of an abelian cubic arc. Furthermore, if the index [Ko: W] ~ 0 (mod 3) then each coset

X + W(X(EW) of W i n K o is a (IWI, 2)-arc. Proof. From (7) we have

(10) X + y + X o y = U = O o O i n K o ,

thus

(11 ) X o y = U - X - Y i s i n W f o r e v e r y X , Y e W

since Wis a subgroup containing U. Hence, Wis the support ofa subquasigroup of (K, o ), so (f, W) is a nonsingular cubic arc. Moreover, the composition law X + Y = O o (X o Y) turns W into an (abelian) subgroup of Ko; thus (f, W) is an abelian cubic arc. Now, let X be a point in K with X ¢ W, then the three points X + Y1, X + Yz, X + Y3 of X + W are collinear if and only if X + YI + X + Y2 + X + Y3 = U , that is 3X = U - ( Y t + Y2 + Y3)~ W , i.e. 3(X + W) = W. Thus the quotient group Ko/W has an element of order 3, a contradition with the hypothesis [Ko: W] ~ 0 (mod 3).

Two interesting problems arise.

PROBLEM 2.8. Under which conditions does the existence of an abelian arc imply that the projective plane is Desarguesian?

PROBLEM 2.9. Under which conditions is an abelian arc of a Pappian plane P embeddable in a unique algebraic cubic? An answer to Problem 2.9 will be given in Section 5.

3. P O I N T S OF I N F L E C T I O N

Let (f, K) be an abelian cubic arc. When K has an inflection point O it is easier to study the structure of the group K o. In this case O o O = O and from 2(7) we deduce the following

RESULT 3.1. I f ( f , K) is an abelian arc with an inflection point O, then for all X,


Y, Z i n K

(1) X, Y, Z is a cyclec~ X + Y + Z = 0 in K o

For every X e K, - X = Oo X

From the second of( l ) we have:

(2) 2X = O in K o ¢~ the line XO is tangent to K at X.

L E M M A 3.2. I f an abelian arc ( f , K) has a point of inflection O, then the number

of inflection points is a power of 3 (or it is infinite). Proof First we show that:

(i) The order o(X) of an inflection point X in Ko is 3.

Since X is an inflection then X o X = X; moreover, from the second of (1), we h a v e O o X = - X ; h e n c e X + X = 0 o ( X o X ) = 0 o X = - X , i . e . 3 X = O .

Now, i fF is the set of inflection points and U is the subgroup o f K o defined by

(3) U = { X ~ K o I 3X = 0} ~ Ko,

then F G U. Conversely, we shall show that U _ F. Let X be in U, X # O. If

X would not be an inflection point, on the tangent to K at X there would be another point Y = X o X # X. The third point on the line O Y is O o Y =

O o (X o X) = X + X = - X, as 3X = 0; thus the line O(X + X) contains the

points X and Y, which is a contradiction. Therefore, F = U and, since

obviously ]U] = 3 a, we have the result.

An easy consequence is the following property.

C O R O L L A R Y 3.3. I f X and Y are two distinct inflection points, then Y has order 3 in K x and the unique third point of K on the line X Y is again an inflec-

tion point. Proof The first part of the statement is contained in part (i) of the proof of

Lemma 3.2. Now, since F = U is a subgroup of K o and X, Y are in U, also

- ( X + Y)isin U. But - ( X + Y) + X + Y=0,i .e . - ( X + Y) ,X, Yis a cycle; hence X o Y = - (X + Y) ~ U = F as desired.

We remark that, by Corollary 3.3, the set F of inflection points of an abelian arc is a Steiner triple system (see Dembowski [5], for the definition).

L E M M A 3.4. I f (f, K) is an abelian arc with an inflection point O, then the subgroup U = F of K o is the support of an abelian arc.

Proof Since O o O = O, the proof easily follows from Proposition 2.7.

L E M M A 3.5. Let (f, K) be a finite cubic arc (not necessarily abelian), and let


F be the set of its inflection points. I f IK150 (rood 3), then

(4) [Vl = l(mod 3).

Proof Let M be the set {(X, Y) e K x KIX v a Y} and let s be the number of the 3-secants to K. On each 3-secant there are exactly six elements of M, while on each of the IKI - IF[ tangents to K at a noninflection point there are exactly two such elements. Hence, counting in two ways the order of M, we obtain

[KI(JKI - 1) = 6s + 2(IKI - IFI), and, modulo 3, we get I K [ 2 - IKI = - [K[ + iFI (rood 3), which gives IFI - IKI z (mod 3) and this implies (4), since IKI ~ 0

(mod 3).

COROLLARY 3.6. I f ( f K) is a finite abelian arc and [K[ @ 0 (mod 3), then there exists a unique inflection point.

Proof By (4) F ¢ Q, thus we may apply Lemma 3.2, i.e. [FI = 3 ~. Then (4) implies that s = O.

The result in Corollary 3.6 can be generalized to the infinite case as follows. For an infinite K, we say that K is a 3-divisible group if the map X - , 3X is a bijection.

P R O P O S I T I O N 3.7. I f ( f , K) is an abelian arc with a 3-divisible 9roup, then there exists a unique inflection point.

Proof Let O be a point in K such that U = OoO ¢ O. Since K o is 3-divisible, there exists a unique point A such that 3A -- U. First ot' all, notice that A o A ¢ O (otherwise A + A = O o ( A o A ) = U = 3 A , i.e. A = O ) and A + A ¢ O ( o t h e r w i s e U = A , s o O = A + A = U + U = O o ( U o U ) andthe line OU is tangent to K both at O and U, hence O = U). We shall show that A is an inflection point. If not, on the tangent to K at A there is another point A o A ~ A. Now, A, A + A, A o (A + A) are a cycle, thus by 2(7) A + (A + A) +

A o(A + A ) = U. Since 3A = U we have A o(A + A ) = O and this gives a contradiction, because the third point of K on the line O(A + A) is A o A ~ A. The contradiction proves that A is an inflection point. From Corollary 3.3 and since KA is 3-divisible, we derive that A is the unique inflection point.

Our purpose is to prove the following theorem.

T H E O R E M 3.8. Let (f, K) be a finite abelian arc with IK[ - 0 (rood p) for some prime p >~ 5, p ~ 37. Then one of the followin9 holds:

(i) there are no inflection points; (ii) there is exactly one inflection point;

(iii) there are exactly three collinear inflection points; (iv) there are exactly nine inflection points,formin9 an affine plane of order 3.


In order to prove this theorem we need a certain number of preliminary results. Thus, the very proof of it starts after Proposition 3.20.

By Lemma 3.2, we already know that, if IF] # 0, then IF] = 3 a. Moreover, by Corollary 3.3 we have that, if ]El < 9, then one of (i), (ii), (iii) holds. Thus we may assume, without loss of generality, that IF] >t 9.

REMARK 3.9. Let S be the set of secants containing three inflection points. The incidence structure G = (F, S, e) is a Steiner triple system.

LEMMA 3.10. Let X, Y, Z be three noncollinear inflection points. Then the span <X, Y, Z> in G is an affine plane of order 3. Furthermore, <X, Y, Z> is the subgroup of K z generated by X and Y.

Proof. We choose Z = O. By (1), we have in Ko the picture shown in Figure 1, which, obviously, can be completed.

X+Y "Y -X

X Y -X-Y

Fig. 1

To prove Theorem 3.8, it remain to show that, if IF[ = 3 a with a/> 3, we get a contradiction. From now on we suppose IF] ~> 27.

PROPOSITION 3.11. With the above notations, the incidence structure G = (F, S, e) is an affine space AG(n, 3) of dimension n >>. 3 over GF(3).

Proof. See Dembowski I-5, p.100].

Let p/> 5 be a prime divisor of [K]; since IF[ = 3 a with a ~> 3, we have IKI = 3apd with d ~> i. We may assume, without loss of generality, that d = 1 and that the 3-Sylow subgroup is elementary abelian. Indeed, we can always restrict ourselves to the cubic arc corresponding to the subgroup of K o generated by the elementary abelian 3-group F and by a cyclic p-subgroup.

Let N = K \ F be the set of noninflection points of K. Since IKr = 3"p and [El = 3% it follows that

(5) INI = 3a(p - 1).

PROPOSITION 3.12. For each 0 e F, there are exactly p - 1 points in N o/

order p in K o.

ON A B E L I A N CUBIC ARCS 41

Proof. I t is trivial, since K o is abelian. Conversely, we shall p rove that for each A in N there is exactly one O in

F such that A has order p in K o. To prove this, we need two prel iminary results.

L E M M A 3.13. Let I = GF(p), p >~ 5 a prime. I f T is a subset of the abelian 9roup

(~, +) such that

(i) X, Y ~ T==~ - X - Y e T,

(ii) 1 e T.

then T = ~, thus, in particular 0 ~ T.

Proof. Consider the binary compos i t ion law ~): ~ x ~ ~ ~ defined by

(6) X ~ ) Y , = X + Y - 1 .

Since (~, + ) is an abel ian group, it is s t ra ightforward that ~) turns ~ into an

abelian group with 1 as neutral element, the ' inverse' of X with respect to

bein9 0X = - X + 2. N o w X ~) Y = X + Y - I = - ( - X - Y ) - I;

thus if X and Y are in T, since by (i) and (ii) - X - Y and 1 are in T, applying (i)

to - X - Y and 1 we get that X ~) Y is in T. F r o m (i) and (ii), - 2 = - 1 - 1 ~ T;

hence, for each X in T, - X - ( - 2) = - X + 2 = OX is in T, by (i). Therefore

T is a subgroup of ([, ~)) which is not {1} because - 2 # 1. Since ~ has pr ime order p, it then follows that T = t.

C O R O L L A R Y 3.14. Let ( f K) be an abelian arc of prime order p v~ 2, 3. I f ( f ' , K') is a subarc (not necessarily abelian), then either K' = K or K' = {O},

where 0 is an inflection point of K.

Proof. By Corol la ry 3.6 and since I K[ = p # 2, 3 is a prime, ( f ,K) has exactly one inflection point, say O. Since ( f ' , K ' ) is a subarc, we have

- X - Y = X o y ~ K ' for every X, Y in K ' . Thus in the abelian group K o the subset T = K ' satisfies condi t ion (i) of L e m m a 3.13. If K ' ~ {O}, let A be

a point in K'\{O}. The m a p j: GF(p) ~ K o defined by j(O) = O,j(1) = A and

j(m) = mA if m ~ 0, m ~ GF(p), is a h o m o m o r p h i s m between the abel ian

groups GF(p) and K o of the same pr ime order p. Hence, we m a y identify K o with GF(p) and A with 1. Thus T also satisfies condi t ion (ii) of L e m m a 3.13, t h e n K ' = T = K .

We are now able to prove the following proposi t ion.

P R O P O S I T I O N 3.15. With the above notations, for each noniflection point

A ~ N = K \ F there is exactly one 0 in F such that A has order p in K o.

Proof. Let 01 and O 2 be two distinct such points, and let H~ = ( A ) be the subgroup of order p generated by A in Ko, = 1, 2. By Propos i t ion 2.7, H i is

42 D. GHINELLI ET AL.

the support of an abelian arc. Since the H~ are subquasigroups of K, the subset

T = H1 n H 2 is a subquasigroup of Ha and hence a subarc in H1, so T contains Oa by Corollary 3.14. Then 01 is in H2 and this is a contradiction, since 01 is an inflection point and thus an element of order 3 in the subgroup H 2 of prime order p ¢ 3. Therefore, 01 = 02 and hence for the number nA of points O in F such that A has order p in K o we have 11A ~< 1. Counting twice the pairs (O,A)~ F x N, with A of order p in Ko, we get by Proposition 3.12

(7) 3a (p- 1)= 2 11.4" .4eN

Since na ~< 1, from (5) and (7) it follows that n A = 1 for each A s N.

LEMMA 3.16. For each pair O, O' of distinct points in K we have

(8) nX = O' in K o, ~ nX = nO' in K o.

Proof This is an immediate consequence of 2(9).

LEMMA 3.17. With the above notations, let A 1, A 2, A 3 be three noninflection

points, and let 0 i (i = 1, 2, 3) be the unique inflection such that pAi = O i in Ko,. I f

0 is a point in K \ ( 0 1 , 0 2 , 0 3 } , then

(9) p(A 1 + A 2 + A3) = p(O 1 + 02 + 03) in K o.

Proof This follows from (8).

LEMMA 3.18. Let A i and 0 i be as in Lema 3.17 (i = 1,2,3). I f A 1 , A z , A a are

collinear, then 01 ,02 , 0 a are also collinear. Proof If O e K \ { 0 1 , 0 2 , 03} is an inflection point, then the assumption

implies A 1 + A 2 + A 3 = O in K o and thus p(O 1 + 0 2 + O a ) = O in K o. Since the point O 1 + O 2 + 03 is an inflection and p/> 5, from p(01 + 02 +

0 3 ) = O i n K o w e d e d u c e O 1 + O 2 + O 3 = O .

LEMMA 3.19. Let 0 1 , 0 2 , O 3 be three eollinear inflection points and let

Aj( j = 1, 2) be noninflection points such that pA; = Oj in Koj. Then the point

A 3 = A 1 oA 2 is a noninflection point such that pA 3 = 03 in Ko3.

Proof IfO is an inflection point different from 01,02, 03, then A 1 + A 2 +

A 3 = 0 in Ko, hence pA3 = - (pAl + pA2). Since (p, 3) = 1, A 3 is a noninflection point if A 3 ¢ 03. Assume, on the contrary, A 3 = 03. Then, 03 =

A 3 = - A 1 - A 2 in Ko, so that

(i) pO 3 = - p ( A 1 + A2) in K o.

Moreover, from (8) and from the equalities p A i = O~ in Ko~, i = 1, 2, it follows

that pAi = pO i in Ko3. Hence p(A 1 + A 2 - 01 -- 02) = 03 in Ko3 and from


(8) we have

(ii) p(A 1 + A 2 - 01 - 0 2 ) = pO 3 in K o.

From (i) and (ii) it then follows that pO 3 = 0 in Ko, i.e. 03 = O, a contradiction. Thus, A 3 is a noninflection point. Since pAj = Oj in Koj, j = 1, 2, (8) yields

pAj = pO t in Ko, j = 1,2, so that pA 3 = - ( p A 1 + pA2) = - (p O 1 + p02) = - p ( O 1 + 02) = pO 3 (as O x + 02 + 03 = O in Ko, by the hypothesis of

collinearity for 01 ,02 , 03). Thus, from (8) it then follows that pA 3 = 03 in

Ko3 •

Lemmas 3.18 and 3.19 imply the following result.

P R O P O S I T I O N 3.20. For each triple 01, 02, 03 of collinear inflection points,

we have exactly (p - 1) 2triples A 1 , A z , A 3 of collinear noninflection points such that A~ has order p in Ko, (i = 1, 2, 3). Furthermore, all possible triples

A 1, A 2, A 3 of collinear noninflection points are obtained in this way from a triple 01 ,02 , 03 of collinear inflection points

Now we are able to prove Theorem 3.8. By contradiction, let [FI = 3 a (with a/> 3). If O is a fixed inflection point, let L be a subgroup of F of order 27 and let T be a subgroup of Ko, with IT[ = p. If G = L. T and R = G\L, then

[GI = 27p and [RI = 27(p - 1). The subgroup G is the support of an abelian arc with 27 inflection points, contained in K. Our purpose is to prove that this gives a contradiction.

We denote by Sec the set of secants to the abelian arc G. With the same argument as in Lemma 3.5, we get for G

(10)

hence

(11)

Ial(Ial - 1) = 61Secl + 2(161 - I L l ) ,

I Secl = 9/2(27p z - 3p + 2).

On the other hand, we have a partition of Sec in the following four subsets:

(a) Secx, the subset of secants of type XY, with the points X, Y in R. Thus the order of Sec 1 equals the number of tangents to G in the 27(p - 1) points of R.

(b) Sec2, the subset of secants of type X Y Z with the three points X, Y, Z in L. Since L is an affine space AG(3, 3) (see Proposition 3.11), the order of Sec2 is the number of lines in an AG(3, 3), that is 9.13.

(c) Sec3, the subset of secants of type X Y Z with only one point in L. Since through each point of L there are exactly 27(/9 - 1)/2 such secants, we have [Seca[ = 2 7 2 ( p - 1)/2.


(d) Sec 4, the subset of secants of type X Y Z with X,Y,Z in R. By Proposition 3.20, we get

[Sec4[ = [Sec2[( p - i) 2 = 9-13(p - 1) 2 .

Therefore:

272 [Secl = ~lSecil = 27(p - 1) + 9"13 + ~ - ( p - 1) + 9.13(p - 1) 2

i

= ~(26p z + 35p - 35),

which, together with (11), gives p2 _ 38p + 37 = 0, which is a contradiction, since p # 37. Hence IF[ = 9 and the proof of Theorem 3.8 is complete.

4 . T A N G E N T LINES

In this section we prove a result on the number of tangent lines to an abelian arc K through a point of K.

DEF I NI TI ON 4.1 Let ( f K) be a cubic arc in a projective plane P and let A be a simple point of K. We call proper tangent to K at A every line a through A which is tangent to K at a point B # A.

In the classical case, when K is the set of nonsingular points of an algebraic cubic in P2(C), then for each A in K one of the following holds:

(i) IfA is an inflection point, then there are exactly three proper tangents to K through A;

(ii) if A is a noninflection point, then there are exactly four proper tangents to K through A.

In general, this is not true in the nonclassical case. For example, if the base field has characteristic 2, then all the tangent lines to the algebraic cubic C: y = x 3 contain the origin.

Anyway, if P is a Pappian anti-Fang plane (i.e. P does not admit a subplane PG(2, 2)), then we can prove the following theorem.

T H E O R E M 4.2. Let P be a Pappian anti-Fang plane and let ( f K) be an

abelian are with inflection points. Then there is an integer d ~ {1, 2, 4} such that:

(1) an inflection point is incident with exactly d - 1 proper tangent lines;

(ii) a noninfleetion point is incident either with zero or with exactly d proper tangent lines.


Proof. Fix an inflection point 0 and consider the subgroup of K o

(1) Ko, 2 = {X ~ KoJ2X = O}

of the elements of order 2 in K o. If X 4 0 is in Ko,z, we say that X is an involution. First we prove that no three noncollinear involutions exist. For, if

X, Y, Z ~ Ko,2\{O} are not collinear, then there is in P the subplane of order 2 (see Figure 2) as the following argument shows. Since O is an inflection point, by 3(1) we have

X, Y, X o Y is a cycle ¢~ X + Y + (X o y) = O in K o ¢~ X o Y

(2) = - ( X + Y).

Fig. 2.

X+¥

Now, X, Y~ Ko, 2 implies X + Y~ Ko,2, hence - ( X + Y) = X + Y and (2) yields that X o Y = X + Y, i.e. the third point on the line X Y is X + Y. Similarly, for the lines YZ, Z X , ( X + Y)Z, (Y + Z )X , (Z + X ) Y we deduce (from 3(1) and since Ko, 2 is a subgroup) that Y o Z = Y + Z, Z o X = z+x , (x+ v)oz=(x+ r)+z, (r+z)ox=(Y+Z)!+x,(z+x)o Y = (Z + X) + Y; moreover the last three points are the same point, since K o is an abelian group. Finally, since X , Y , Z e K o , 2 , we have that O = 2 ( X + Y + Z ) = ( X + Y ) + ( X + Z ) + ( Y + Z),i.e. b y 3 ( 1 ) X + Y , X +

Z, Y + Z are collinear. Therefore, in K o there are 0, 1 or 3 collinear involutions X ¢ O. Hence

(3) d = IKo,2l = 1,2,4.

By Theorem 2.2, the abelian groups K a a re all isomorphic; thus, KA, 2 = {X~ K I 2 X = O in KA) is isomorphic to Ko, 2, for all A~ K. Therefore

(4) IKA,zI=d=I,2,4 for any Ae K.

If A is an inflection point, then from 3(2) it follows that

(5) X ~ KA,z\ {A} ,~, the line X A is a proper tangent in A

thus we have (i). If A is a noninflection point and we suppose that there is

46 D. G H I N E L L I ET A L .

a proper tangent line I through A, then I is tangent to K at a point B ~ A. Clearly, B o B = A, thus by 2(6) and 2(7) we have

(6) X, Y, Z is a cycle in Kn ¢~ X + Y + Z = B o B = A.

In particular, for any X e K B , 2 = { Y e K n [ 2 Y = O in KB} , we get A = 2X + A = X + X + A and thus X, X, A is a cycle, which means that the line X A is tangent to K at X. Therefore, the number of proper tangent lines

through A is IKn,21 = d = 1,2,4 and (ii) is proved.

5. AN EMBEDDING THEOREM

In this section we give an answer to Problem 2.9, proving the following

theorem.

T HEOR EM 5.1. Let P be a Pappian plane (not necessarily finite) and let ( f K) be a finite abelian arc without inflection points. I f I KI is divisible by at least two different primes both greater than or equal to 5, then there is a unique irreducible algebraic cubic C containing K.

We remark that (see J. W. P. Hirschfeld and J. F. Voloch [9]), even when there are inflection points, an abelian arc can be embedded in an algebraic cubic. Furthermore, with the same kinds of arguments, it should be possible to prove a similar embedding theorem for a ([ K p, 3)-arc K with an abelian group

structure on it.

LEMMA 5.2. Let ( f K) be a finite abelian arc with no inflection points. For any fixed point 0 in K, let U = 0 o O. Then the subgroup ( U), 9enerated by U in Ko, is the support of an abelian arc ( f ' , ( U ) ) with no inflection points, thus

(1) I ( U ) I - 0(mod 3).

Proof Clearly, an inflection point o f ( f ' , ( U ) ) would be an inflection also for ( f K ) , so the proof follows immediately from Proposit ion 2.7 and

Corollary 3.6.

Of course, the above lemma still holds when the set F of inflection points is not empty, provided that ( U ) n F = Q.

CONSTRUCTION 5.3. Let K be as in Theorem 5.1. For a fixed point O in K, let U = O o O be the other point on the tangent to K at O(U ~ O, since F = Q); by Lemma 5.2, U is an element of order 3 h in K o. From the assumption on [ K [, there exists in K o an element Z ¢ O of order greater than or equal to 5 such that

(2) u ¢ ( z )


(for example, it suffices to choose Z either of prime order >~ 5 or of order

>/5 and prime with 3). We remind that for any two different points X, Y e K o,

the third point on the line X Y is

(3) X o r = U - X - Y.

Thus, the third point on the tangent to K at Z is U - 2Z; similarly, the third

points on the lines OZ, O ( U - 2Z) ,Z(2Z) are U - Z, 2Z, U - 3Z, respec-

tively. Therefore, we can draw the picture shown in Figure 3.

Z U-2Z

v U-3Z

U

Fig. 3.

Since o(Z) >/5 and (2) holds, the set

(4) A = {O, U, Z, 2Z, U - Z, U - 2Z, U - 3Z},

consists of seven pairwise different points. Let V be an algebraic cubic passing through these seven points with

tangents in O and Z the lines OU and Z(U - 2Z), respectively. First we prove

that

(5) V is irreducible.

Proof. If V would be reducible, then Z and O cannot belong to a linear

component I of V For, if Z s l, the tangent line t z = Z(U - 2Z) to V at Z is l, so the remaining six points of A must lie on a conic C. If C would be reducible,

since t o = OU is the tangent to C at O, then C = t o r . Then, the three

points U - Z , 2 Z , U - 3Z must be on r, which is a contradiction, since

(U - Z) + 2Z + (U - 3Z) = 2(U - Z) ¢ U, as U # 2Z. Similarly, inter- changing the role of O and Z we get the contradiction that U - Z, 2Z, U - 3Z

must be on a line. Therefore, if V is reducible, then V splits in a line 1 and an irreducible conic C; furthermore, Z and O belong to C and the tangents to C at

Z and O are t z -- Z ( U - 2Z) and t o = OU, respectively. Hence, necessarily l = (U - 2Z(U). Now, the conic C cannot contain U - Z (otherwise the line OZ would be a component of C); thus U - Z s I and U - Z, U - 2Z, U is

a cycle, which is contradiction since (U - Z) + (U - 2Z) + U = 3(U - Z) ¢ U as, from U ¢ ( Z ) and o(U) = 3 h it follows that 2U ¢ 3Z. The contradiction proves that V is irreducible.


Coming back to the construction: for a fixed O in K, let U = O o O. I fZ is an element of order o(Z) ~> 5 such that U ¢ ( Z ) , then the unique algebraic cubic V, containing the set A defined in (4) (see Figure 3) with tangents in Z and O the lines t z = Z(U - 2Z) and to = OU, is irreducible. Therefore, the set V z of the nonsingular points of V is an abelian arc (see Lemma 1.7). In the abelian group Vz, o the relations between the elements of A are exactly those of Figure 2.

I fa point X of K belongs to Vz, then the same point will be denoted by {X}, when considered as a point of V z. For the points of A we have obviously in

Vz,o: {2Z} = 2{Z}, {V - mZ} = {V} - m{Z}, m = 1,2,3.

LEMMA 5.4. Let V = V z be the algebraic abelian arc in Construction 5.3. Then V contains all the points mZ and U - mZ (me 7/) of the cyclic subgroup

( Z } and of the subset U - ( Z } of K o. Furthermore, in V o

(6) m{Z} = {mZ}, {g} - m{Z} = {V - mZ}.

Thus, in particular, V is uniquely determined by the points of A.

Proof. First we prove that - Z is in Vand that { - Z} = - {Z} in V o. In K o

we have

(7) U + Z + ( - Z ) = U and U - Z + 2Z + ( - Z ) = U;

thus - Z is the intesection of the two lines UZ and (U - Z)(2Z). Since the same holds in Vo, we have that - Z is also in Vand { - Z } = - {Z} in V o. For m = 1, 2 the statement is true; by induction, we assume m > 2, hZ and U - hZ

in V, for all h ~< m, and we prove that (m + 1)Z and U - (m + 1)Z are in V. Furthermore, {(m + 1)Z} = (m + 1){Z} and {V} - (m + 1){Z} = {U - (m + 1)Z} in 11o. As before, we write (m + 1)Z as the intersection of two lines, whose other two points are already in V by induction. In K o we

have

U - mZ + ( - Z ) + (m + 1)Z = U,

U - ( m + l ) Z + ( - 2 Z ) + ( m + l ) Z = U ,

thus (m + 1)Z is the intersection of the two lines ( U - m Z ) ( - Z ) and (U - (m + 1)Z)(- 2Z). The previous ~irgument really works since, from (2) and from the induction assumption m > ~2, it easily follows that U - mZ ~ - Z

and U - (m + 1)Z ¢ - 2 Z . Since the same holds in Vo, by induction, we deduce that (m + 1)Z is in V, and in Vo we have {(m + 1)Z} = (m + 1){Z}. Similarly, in K o we can write

( U - ( m + 1)+mZ + Z = U,

( U - ( m + 1 ) Z ) ÷ ( m ÷ 1 ) Z + O = U.

ON A B E L I A N CUBIC ARCS 49

Thus U - (m + 1)Z, as intersection of the two lines (mZ)Z and ((m + 1)Z)O,

is in V o. Since the same identities hold in Vo, by induction, we also have {U - (m + 1)Z} = {U} - (m + 1){Z}. Once more, the argument works since from mZ = Z and (m + 1)Z = O it follows respectively that

U - ( m + I ) Z = U - 2 Z e A _ V o and U - ( m + I ) Z = U e A _ C V o . Now, another algebraic abelian arc V', satisfying the above condition on A, contains ( Z ) and U - ( Z ) , thus V' has in common with V at least the ten pairwise different points {O, U, mZ, U - mZ[m = 1, . . . ,4} . Since we proved that V and V' must be irreducible, we have V = V', i.e. V is uniquely determined by A.

LEMMA 5.5. Let • be the set

(8) ¢ = { x • Ko [o(X) >/5, v ¢ (x>} ,

of all elements X with order greater than or equal to 5 in K o, such that U 6 ( X ) . Then there is a unique algebraic abelian arc containing all points X • • and all points mX in ( X ) and U - mX • U - ( X ) (m • 7/). Furthermore, in the abelian group Vo:

(9) {mX} = re{X}, {U - reX} = {U} - re{X}.

Proof. By Construction 5.3 and Lemma 5.4, for any two fixed points Z i e O(i = 1, 2) we can construct two irreducible algebraic abelian arcs Vz,, with

(10) ( Z , ) , U - ( Z , ) C _ V z , ' i = 1 , 2

and Vz, uniquely determined by Z i. We remark that if

(11) I (Z1) c~ ( Z 2 ) I i> 5,

then there is an element X • • such that Vzl and Vz2 both contain the ten distinct elements mX and U - reX, m = 1 . . . . ,5. Since Vz~ and Vz2 are irreducible, we conclude that

(12) V z ,=Vz2 and ( Z i ) , U - ( Z ~ ) C _ V z =Vz: , i = 1 , 2 .

To prove the statement we consider the graph G -- (O, ~), with • as set of vertices and the adjacency relation defined by

(13) zl ~z2~.l<Z~>c~(z2)[>~5.

By the above remark, two adjacent vertices define a unique algebraic cubic arc satisfying (12). Clearly, any two vertices in the same connected component of G define a unique algebraic cubic arc satisfying (12). Hence, to prove the desired result, we have to prove that G is a connected graph. Let A be any fixed


vertex with prime order o(A) = p ~> 5. I fB is an element of,I, with order p' not divisible by p, then A ~ (A + B) and (A + B) ~ B, i.e. A and B are in the same connected component of G. If the order of B is divisible by p, we write B = Bp + Bp., with Bp ¢ O the component of B along the p-Sylow subgroup o f K o. By hypothesis, we can fix an element C of prime order q ¢ p and q/> 5. Then, A ,,~ (A + C) and (A + C) ,-~ C. Now, C and C + Bp are adjacent, since o(C) = q and o(Bp) are coprimes. For the same reason, C + Bp and B~ are adjacent. Since Bp ~ B, by definition, we have a path with vertices A, A + C, C, C + Bp, Bp, B between A and B, i.e. G is connected.

LEMMA 5.6. Let X be an arbitrary element of K o with U ¢ ( X ) . Then X belongs to the unique algebraic abelian arc V of Lemma 5.5. Furthermore,

V contains all the elements of ( X ) and of U - ( X ) and V o verifies (9). Proof By Lemma 5.5, this is true if o(X) >/5. Hence we may suppose

o(X) = 2, 3, 4. Let A be a fixed element of prime order o(A) = p/> 5. Since (p,o(X)) = 1, the element Z = X + A has order o(Z) = o(X).p. If U~ ( Z ) , 3 h = o(U) divides o(Z), i.e. o(U)/o(X) since (p, o(U)) = 1. It then follows that

o(U) = 3 = o(X) and ( U ) = (X) , for there is in ( Z ) a unique subgroup of order 3. Hence U e (X) , a contradiction with the assumption U¢ ( X ) . Therefore, we may assume U ¢ (Z ) . Now o(Z) > o(A) = p >/5 and U ¢ (Z) . By Lemma 5.5 ( Z ) and U - ( Z ) are in V. Since ( X ) _ ( Z ) , we deduce ( X ) and U - ( X ) are in V. Furthermore, in V o we have (9).

COROLLARY 5.7. For any X ~ K o the point 3X is in the unique algebraic

cubic arc V of Lemma 5.5. Furthermore, ( 3 X ) and U - ( 3 X ) are contained in V, and in V o we have

m{3X} = {m3X), {U} - m{3X} = {U - 3mX}.

Proof We note that U¢ (3X) , otherwise m3X = 3(reX)= U, i.e. mX would be an inflection point of K, a contradiction. Thus we can apply Lemma 5.6 and the proof is complete.

To prove Theorem 5.1, it remains to show that the set

(14) f~ = K - V

is empty.

LEMMA 5.8. I f X e fl, then there is a positive integer m = re(X) such that mX = U. Furthermore, such an integer satisfies

(15) m = 2(mod 3)"

Proof By Lemma 5.6 we have U e (X) ; thus there is an integer m/> 1 such that mX = U. First we prove that m ~ 0(mod 3). If not, m = 3t and


3(tX) = mX = U, i.e. tX is an inflection point, a contradiction. Suppose now m-= l(mod3). We note that U ¢ ( U - X ) . Otherwise s ( U - X ) = U, for some integer s >~ 1; thus m s U = m s X = m U and from m X = U we get m s U - sU = mU, which implies (ms - s - m)U = 0. Since o(U) = 3 h, we get that 3 divides ( m s - s - m), thus m s - s - m =-0 (mod 3), and, from m -= 1 (mod 3) we get s - s - 1 = 0 (rood 3), which is impossible. Hence U ¢ ( U - X ) and by Lemma 5.6 we would have X in V, against the assumption that X is in f~. Therefore m ~ 1 (mod 3) and the proof is complete.

LEMMA 5.9. I f X ~ f~, then o(X) = 1 (rood 3). P r o o f It follows from Lemma 5.6 that also - X e f~, since ( X ) = ( - X ) .

Let m be as in Lemma 5.8 a positive integer such that m X = U with m = 2 (mod 3). Since

(o(X) - m ) ( - X ) = o ( X ) ( - X ) + m X = 0 + U = U,

we have, by Lemma 5.8, that (o(X) - m) =- 2(mod 3); hence o(X) = 2 + m -

2 + 2 = 1 (rood 3). Now we are able to prove that fl is empty. By contradiction, let X be in f~;

since o(X) -= 1 (mod 3) (see Lemma 5.9), there is an element Ye ( X ) with X = 3 Y, and by Corollary 5.7, X must be a point of the algebraic cubic V, which gives the desired contradiction.

R E F E R E N C E S

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Sem. Geom. Comb. Dip. Mat. Univ. Roma La Sapienza 50 (1984).

52 D. GHINELLI ET AL.

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521-560. 24. Weil, A., Sur les courbes alg~briques et les vari~tds qui s'en d~duisent, Hermann 1948. 25. Weil, A., Varidt~s Abdliennes et courbes alg~briques, Hermann 1948.

Authors ' addresses:

D. Ghine l l i ,

D i p a r t i m e n t o di M a t e m a t i c a ,

Un ive r s i t~ degl i S tud i di R o m a ,

' L a Sap ienza ' ,

P i azza l e A l d o M o r o , 2,

00185 Roma,

I ta ly .

N. Me lo n e ,

D i p a r t i m e n t o di M a t e m a t i c a e App l i caz ion i ,

Via M e z z o c a n n o n e 8,

80134 Napol i ,

I ta ly .

U. Ot t ,

Univers i t / i t B raunschwe ig ,

I n s t i t u t f/Jr G e o m e t r i e ,

Pocke l s s t r aBe 14,

3300 Braunschweig,

F .R .G .

(Received November 21, 1988; revised version, March 14, 1989).

on abelian cubic arcs

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