\documentstyle[a4wide,11pt]{article} \title{ GROUP-ARCS OF PRIME POWER ORDER ON CUBIC CURVES} \author{J.W.P. Hirschfeld and J.F. Voloch} \begin{document} \maketitle \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{definition}[theorem]{Definition} \begin{abstract} This article continues the characterization of elliptic curves among sets in a finite plane which are met by lines in at most three points. The case treated here is that of sets of prime-power cardinality. \end{abstract} \section{Notation} \begin{tabular}{ll} $GF(q)$ & the finite field of $q$ elements\\ $PG(2,q)$ & the projective plane over $GF(q)$\\ $PG^{(1)}(2,q)$& the set of lines in $PG(2,q)$\\ ${\bf P}(X)$& the point of $PG(2,q)$ with coordinate vector $X$\\ $PQ$ & the line joining the points $P$ and $Q$\\ $\ell (P,Q)$& $PQ$\\ $\langle P\rangle$ & the group generated by $P$. \end{tabular} \section{Introduction} This article continues the work of \cite{HV} in considering sufficient conditions for a set of points in a finite plane to be embedded in a cubic curve. Similar results to those in \cite{HV} were obtained independently by Ghinelli, Melone and Ott \cite{GMO}. For completeness the main results in \cite{HV} need to be summarized. \begin{definition} A $(k;n)$-arc in $PG(2,q)$ is a set of $k$ points with at most $n$ points on any line of the plane. \end{definition} The fundamental problem is to decide when a $(k;n)$-arc ${\cal K}_{n}$ lies on an absolutely irreducible algebraic curve ${\cal C}_{n}$ of degree $n$. Here we consider the problem for $n = 3$. A crucial point is the number of points ${\cal K}_{3}$ contains and the number of rational points on ${\cal C}_{3}$. Let $m_{3}(2,q)$ be the maximum number of points on ${\cal K}_{3}$. Then \begin{equation} m_{3}(2,q) \leq 2q + 1\: for\: q > 3, \end{equation} \cite{T}, \cite[p.331]{H1}, and the exact values known are given in Table ~1. \begin{table} \begin{center} \begin{tabular}{c|c|c|c} $q$ & 2, 3 & 4, 5, 7 & 8, 9\\ \hline $m_{3}(2,q)$ & $2q + 3$ & $2q + 1$ & $2q - 1$ \end{tabular} \end{center} \caption{Values for $m_{3}(2,q)$} \end{table} For $q = 11$, $2q-1\leq m_{3}(2,q)\leq 2q+1$. For an elliptic curve, $N_{q}(1)$ is the maximum number of points it can contain. Its value, for $q = p^{h}$ with $p$ prime, is $$ N_{q}(1) =\left\{ \begin{array}{ll} q + [2\surd q] & {\rm when}\: h\: {\rm is\: odd}, h \geq 3\: {\rm and}\: p|[2\surd q]\\ q + 1 + [2\surd q] & {\rm otherwise}, \end{array} \right. $$ where $[t]$ denotes the integer part of $t$, \cite{W}, \cite[p. 273]{H2}. The precise values that are achieved by the number of points of an elliptic curve over $GF(q)$ are also known \cite{W}, as well as the number of isomorphism classes and the number of plane projective equivalence classes for a given value, \cite{S}. For such a value the possible structures of the abelian group the points form is also known, \cite{R}, \cite{V}. \section{Axioms} Now, we recall the axioms imposed on a $(k;3)$-arc in \cite{HV}, and then solve the main case unresolved there. For further motivation and details concerning the axioms, see \cite[section 2]{HV}. \\ Let $\cal K$ be a $(k;3)$-arc in $PG(2,q)$. Four axioms (E1) - (E4) are required. For each axiom, the property that it gives to $\cal K$ is mentioned in parentheses.\\ (E1) There exists $O$ in $\cal K$ such that $\ell\cap {\cal K} = \{ O\}$ for some line $\ell$. (INFLEXION)\\ (E2) There exists an injective map $\tau : {\cal K}\backslash\{ O\}\rightarrow PG^{(1)}(2,q) $ such that $P\in P\tau$ and $|P\tau\cap {\cal K}| \leq 2$, for all $P\in {\cal K}\backslash\{ O\}$. (TANGENT)\\ (E3) If $P, Q \in {\cal K}$ and $PQ \neq P\tau\: {\rm or}\: Q\tau$, then $|PQ \cap {\cal K}| = 3$. (FEW BISECANTS)\\ (E4) For$P \in {\cal K}$, define $\overline{P}$ to be the third point of $\cal K$ on $OP$. For $P, Q \in {\cal K}$, define $P + Q = \overline{R}$, where $R$ is the third point of $\cal K$ on $PQ$. Now, let $\cal K$ be an abelian group under the operation + with identity $O$ and $-P = \overline{P}$. (ABELIAN GROUP)\\ \begin{definition} A $(k;3)$-arc $\cal K$ satisfying (E1) - (E4) is called a group-arc or $k$-group-arc. \end{definition} It follows from the axioms that\\ (a) any subgroup of a group-arc is a group-arc,\\ (b) $P + Q + R = O$ if and only if $P, Q, R$ are collinear. \begin{definition} In $PG(2,q)$, the point set $\cal S$ is linearly determined by the set $\cal T$ of points and lines if every point of $\cal S$ is the intersection of two lines each of which is in $\cal T$ or is the join of two points of $\cal T$ or is the join of two points iteratively determined in this way. \end{definition} \begin{lemma} If $P$ is a point of an arbitrary group-arc, then the cyclic group $\langle P\rangle$ is linearly determined by $\{O, \pm P, \pm 2P, 3P, (-2P)\tau\}.$ \end{lemma} \begin{lemma} Let $P$ be a point of order at least six of a group-arc. Then $\langle P\rangle$ is a subgroup of a unique cubic curve with inflexion $O$. \end{lemma} \begin{lemma} Let ${\cal E}_{1}$ and ${\cal E}_{2}$ be cubic curves and $\cal K$ a $k$-group-arc which is a subgroup of both ${\cal E}_{1}$ and ${\cal E}_{2}$. If $k > 5$, then ${\cal E}_{1} = {\cal E}_{2}$. \end{lemma} \begin{lemma} Let $\cal K$ be a group-arc contained in a cubic curve $\cal E$ such that any cyclic subgroup of $\cal K$ is a subgroup of $\cal E$. Then $\cal K$ is a subgroup of $\cal E$. \end{lemma} \begin{theorem} Let $\cal K$ be a $k$-group-arc in $PG(2,q)$ such that one of the following hold:\\ {\rm (a)} $k = p_{1}p_{2}r$ where $p_{1}$ and $p_{2}$ are distinct primes $\geq 7$;\\ {\rm (b)} $k = 2^{a}3^{b}5^{c}p_{1}^{d}$, where $p_{1}$ is a prime $\geq 7$, $d \geq 1$ and $2^{a}3^{b}5^{c}\geq 6$.\\ Then $\cal K$ is a subgroup of the group of non-singular points of a cubic curve. \end{theorem} The theorem leaves the following values of $k$ to be considered:\\ (i) $k = 2^{a}3^{b}5^{c}$, with $a, b, c \geq 0$; (ii) $k = ep_{1}^{d}$ , with $p_{1}$ prime $\geq 7, d \geq 1, 1 \leq e \leq 5$.\\ In the next section we consider case (ii). \section{The main theorem} \begin{lemma} Suppose $P, Q$ are elements of a group-arc $\cal K$ both of prime order $p_{1}\neq 2, 3$ generating a subgroup $G$ of order $(p_{1})^{2}.$ Then $G$ is uniquely determined by $$ O, \pm P, \pm Q, P\pm Q, 2P.$$ \end{lemma} {\em Proof}: First,$$\begin{array}{rcl} -P-Q & = & \ell(P,Q) \cap \ell(O, P+Q),\\ -P+Q & = & \ell(P,-Q) \cap \ell(O, P-Q) .\end{array}$$ Now assume, by induction on $m < p_{1} - 1$, that we know $$ \pm(iP+Q), \pm iP$$ for $i = 0, \ldots, m$. This is true for $i = 1$. Now we determine these points for $i = m + 1$ as follows: $$ \begin{array}{rcl} - (m + 1)P - Q & = & \ell(P, mP + Q) \cap \ell(2P, (m - 1)P + Q),\\ (m + 1)P + Q & = & \ell(-P, -mP - Q) \cap \ell(O, -(m + 1)P - Q),\\ (m + 1) P & = & \ell(-P, -mP) \cap \ell(Q, -(m + 1)P - Q),\\ - (m + 1)P & = & \ell(P, mP) \cap \ell(O, (m + 1)P). \end{array} $$ The last equality works providing the two lines are distinct; that is, providing $(m + 1)P \neq O$ or $(2m + 2)P \neq O$. However, the first is true since otherwise the induction would have been finished at the previous step. In particular, $\langle P\rangle$ has been determined. Now $\langle P_{1}\rangle$, where $P_{1} = P + Q$, is found. From the previous step, $$ O, \pm P_{1} , \pm Q, P_{1}\pm Q, 2P_{1}$$ are required. Of these, the only ones lacking are $P_{1} + Q$ and $2P_{1}$. These are determined as follows: $$\begin{array}{rclcl} P_{1} + Q & = & P + 2Q & = & \ell(-P-Q, -Q) \cap \ell(-2P-Q, P-Q),\\ 2P_{1} & = & 2P + 2Q & = & \ell(-2P-Q, -Q) \cap \ell(-3P-Q, P-Q). \end{array}$$ Now, with $P_{1}$ instead of $P$, we can determine $\langle P_{2}\rangle$, where $P_{2} = P_{1} + Q = P + 2Q.$ Continuing this process, $\langle P + mQ\rangle$ for $m = 0, 1,..., p_{1}-1$ can be determined. To complete the proof, only $\langle Q\rangle$ needs to be found. By reversing the initial roles of $P$ and $Q$, we require $$ O, \pm P, \pm Q, Q \pm P, 2Q.$$ Of these, only $2Q$ is missing; this is given by $$ 2Q = \ell(P, -P-2Q) \cap \ell(-P, P-2Q).$$ \begin{corollary} A group-arc $\cal K$ isomorphic to $({\bf Z}_{p_{1}})^{2}, p_{1} \geq 5$, is a subgroup of a unique cubic curve. \end{corollary} {\em Proof}: Given $O, \pm P, \pm Q, P \pm Q, 2P$, where ${\cal K} = \langle P\rangle \oplus \langle Q\rangle$, the conditions that a cubic passes through these points and has an inflexion at $O$ are nine independent conditions and determine the cubic uniquely. \begin{theorem} Let $\cal K$ be a $k$-group-arc in $PG(2,q)$ such that $k$ is divisible by a prime $p_{1} \geq 7$. Then $\cal K$ is a subgroup of a unique cubic curve. \end{theorem} {\em Proof}: By Theorem 3.7, it suffices to consider the case that $k = ep_{1}^{d}$ with $1 \leq e \leq 5$. Consider first the case that the $p_{1}$-Sylow subgroup ${\cal P}_1$ of $\cal K$ is cyclic so that ${\cal P}_1 =\langle P_1\rangle$. Now, ${\cal K} = {\cal P}_1\oplus G$, where $|G| = e$ and $|{\cal P}_{1}| = p_{1}^{d}$. As ${\cal P}_1$ is cyclic it is contained in a cubic curve ${\cal E}_1$. For any point P in ${\cal P}_1$, the subgroup $\langle P\rangle$ is contained in a cubic curve $\cal E$, which coincides with ${\cal E}_1$ by Lemma 3.5. If $Q$ is any point of ${\cal K}$, then $Q = P + R$ for some $P \in {\cal P}_1$ and $R\in G$. By Lemma 3.4, $\langle Q\rangle$ is contained in an cubic curve ${\cal E}'$; also, since the orders of $P$ and $R$ are coprime, $\langle Q\rangle$ contains both $\langle P\rangle$ and $\langle R\rangle$. Again, by Lemma 3.5, ${\cal E}'= {\cal E}_1$. Hence ${\cal K}\subset{\cal E}_1$. Now consider the non-cyclic case and let ${\cal K}_{1}\subset{\cal K}$ with ${\cal K}_{1}$ isomorphic to $({\bf Z}_{p_{1}})^{2}$. Then, by the previous corollary, ${\cal K}_{1}$ is contained in a cubic $\cal E$. As in the previous case, ${\cal K} = {\cal K}_{0}\oplus G$ where $|G| = e$ and $|{\cal K}_{0}| = p_{1}^{d}$. If $P$ in ${\cal K}_{0}\backslash {\cal K}_{1}$ has order $p_{1}^{\lambda}$, then $\langle P\rangle$ is contained in a cubic ${\cal E}'$ and, for $Q \in {\cal K}_{1}\backslash\{O\}$, the sum $\langle p_{1}^{\lambda - 1}P\rangle\oplus\langle Q\rangle$ is contained in a cubic ${\cal E}''$. Now ${\cal E}''\cap {\cal E} \supset \langle Q\rangle$, whence ${\cal E}'' = {\cal E}$ by Lemma 3.5. Also, ${\cal E}'\cap {\cal E}'' \supset \langle p_{1}^{\lambda - 1}P\rangle$ and so ${\cal E}' = {\cal E}''$. Hence ${\cal E} = {\cal E}'$ and therefore ${\cal K}_{0}\subset{\cal E}$. Now, let $R\in G$. Then there is a cubic ${\cal E}'''$ containing $\langle R + Q\rangle$. As $e(R+Q) = eQ \in {\cal E}$ and $eQ \neq O$, so $\langle eQ\rangle = \langle Q\rangle$ and ${\cal E}'''\cap {\cal E} \supset \langle Q\rangle$ . Therefore ${\cal E}''' = {\cal E}$ by Lemma 3.5 and $\langle R + Q\rangle\subset {\cal E}$, whence $p_{1}(R+Q) = p_{1}R \in {\cal E}$. So $R \in {\cal E}$. It has now been shown that both $G$ and ${\cal K}_0$ lie in $\cal E$, whence ${\cal K}\subset{\cal E}$. \newpage \section{Small cases} {\bf I. k = 8}\\ \begin{lemma} An $8$-group-arc $\cal K$ isomorphic to ${\bf Z}_{2}\times {\bf Z}_{2}\times {\bf Z}_{2}$ exists in $PG(2,q)$ if and only if $q = 2^{h}, h \geq 3$. Such a group-arc lies on a unique cuspidal cubic. \end{lemma} {\em Proof}: Let $O = {\bf P}(1,0,0), P = {\bf P}(0,1,0), Q = {\bf P}(0,0,1), R = {\bf P}(1,1,1)$ be points of $\cal K$. Then $$ \begin{array}{rcl} R + Q & = & {\bf P}(t,t,1), t \neq 0, 1;\\ P + R & = & {\bf P}(1,s,1), s \neq 1. \end{array}$$ Also $$\begin{array}{rclcl} P + Q & = & \ell(P,Q) \cap \ell(Q + R, P + R) & = &{\bf P}(0, t-ts, 1-t);\\ P + Q + R & = & \ell(P + R, Q) \cap \ell(P, Q + R) & = &{\bf P}(1, s, t^{-1}). \end{array}$$ Now, $P + Q + R \in \ell(P + Q, R) \Rightarrow$ $$ \left| \begin{array}{ccc} 1 & 1 & 1\\ 0 & t-ts & 1-t\\ 1 & s & t \end{array} \right| = 0 $$ $\Rightarrow 1 - s + 1 - t - (t-ts) - s(1-t) = 0$\\ $\Rightarrow 2(1-s)(1-t) = 0$\\ $\Rightarrow 2 = 0.$\\ Since $O$ is on none of the lines $$\ell(Q, P+R), \ell(R+Q, P+R), \ell(P+Q, P+Q+R),$$ it follows that $s \neq 0, s \neq t, s \neq t^{-1}$; hence $q > 4$. Also the 7 points of ${\cal K} \backslash \{O\}$ form a $PG(2,2)$. The 8 points lie on the unique cubic $\cal C$ with equation $$ (s+1)x^{2}y + s(t+1)x^{2}z + (t+1)y^{2}z + t(s+1)yz^{2} = 0. $$ This is irreducible when $(s+t)(st+1) \neq 0$, which is satisfied in this case. It has a cusp at ${\bf P}(\sqrt{t}, \sqrt{st}, 1)$ and all tangents to $\cal C$ are concurrent at $O$. For more on cuspidal cubics, see \cite[section 11.3]{H1}. \begin{lemma} An $8$-group-arc $\cal K$ isomorphic to ${\bf Z}_{2}\times {\bf Z}_{4}$ exists in $PG(2,q)$ if and only if $q$ is odd with $q \geq 5$. Such a group-arc lies on a unique cubic curve, which is elliptic. \end{lemma} {\em Proof}: The eight points of $\cal K$ written as elements of ${\bf Z}_{2}\times {\bf Z}_{4}$ are \begin{displaymath} O = (0,0), P_{1} = (0,2), P_{2} = (1,0), P_{3} = (1,2), Q_{1} = (0,1), Q_{2} = (0,3), Q_{3} = (1,1), Q_{4} = (1,3) . \end{displaymath} Hence $$ 2P_{1} = 2P_{2} = 2P_{3} = O,\; P_{1} + P_{2} + P_{3} = O,\; 2Q_{1} = 2Q_{2} = 2Q_{3} = 2Q_{4} = P_{1} . $$ So $P_{1}, P_{2}, P_{3}$ are the points of contact of the tangents through $O$, and $Q_{1}, Q_{2}, Q_{3}, Q_{4}$ the points of contact of the tangents through $P_{1}$. Let $O = {\bf P}(0,0,1)$ with tangent $y = 0$. Let $P_{1} = {\bf P}(0,1,0), P_{2} = {\bf P}(1,1,1), P_{3} = {\bf P}(\alpha,1,\alpha)$ with respective tangents $x = 0, x = y, x = \alpha y$; so $\alpha\neq 0,1.$ Then, if $\cal K$ lies on the cubic curve $\cal E$, consider the intersection divisors in which $\cal E$ meets the two curves with equations $$y(x-z)^{2} = 0\; {\rm and}\; x(x-y)(x-\alpha y) = 0.$$ In both cases the divisor is $$ O\oplus O\oplus O\oplus P_{1}\oplus P_{1}\oplus P_{2}\oplus P_{2}\oplus P_{3}\oplus P_{3},$$ where $\oplus$ has been used to denote the formal sum to distinguish it from the sum on a cubic curve elsewhere in this paper. So $\cal E$ has equation \begin{equation} y(x-z)^{2} + \lambda x(x-y)(x-\alpha y) = 0. \end{equation} The common points of a line $z = tx$ through $P_{1}$ and $\cal C$ are determined by \begin{equation} (1-t)^{2} x^{2} y + \lambda x(x-y)(x-\alpha y) = 0; \end{equation} that is, apart from $P_{1}$ , the points defined by \begin{equation} \lambda x^{2}+ xy \{(1-t)^{2} - \lambda (1+\alpha)\} + \lambda \alpha y^{2} = 0. \end{equation} Since there are four tangents through $P_{1}$ , so $q$ is odd. For a tangent, the discriminant $\Delta = 0$. Here $$ \Delta = \{(1-t)^{2} - \lambda (1+\lambda)\}^{2} - 4\lambda^{2}\alpha = (1-t)^{4} - 2\lambda (1+\alpha)(1-t)^{2} + \lambda^{2}(1-\alpha^{2}).$$ Since $\Delta = 0$ has four solutions for $t$, so the discriminant $\Delta'$ of $\Delta$ considered as a quadratic in $(1-t)^{2}$ is a square. Now, $$ \Delta' = \lambda^{2}(1+\alpha)^{2} - \lambda^{2}(1-\alpha)^{2} 4\lambda^{2}\alpha. $$ Hence $\alpha = \beta^{2}$; this incidentally means that $GF(q)$ contains a square other than 0 and 1, whence $q\neq3$. Solving $\Delta = 0$ for $(1-t)^2$ gives $$ (1-t)^{2} = \lambda(1+ \beta^{2})\pm 2\lambda\beta = \lambda(1\pm \beta)^{2}. $$ Hence $\lambda = \gamma^{2}$ . Thus $$ 1 - t = \pm\gamma(1\pm \beta).$$ Therefore, (4) becomes $(x \pm \beta y)^{2} = 0.$ This gives for $Q_{1}, Q_{2}, Q_{3}, Q_{4}$ the points $$ {\bf P}(e\beta, 1, e\beta + f\beta\gamma - ef\beta^{2}\gamma) $$ where $e, f = \pm 1$. Also $\cal C$ has equation $$ y(x-z)^{2} + \gamma^{2} x(x-y)(x-\beta^{2} y) = 0, $$ which is elliptic. For the calculation of the equations of cubic curves with a precise number of points, see also \cite{HT}, \cite{K1}, \cite{K2}.\\ {\bf II. k = 25}\\ Each case not covered in this paper can be reduced to a finite calculation. An arbitrary group-arc $\cal K$ of a given order is given by a set of points, where some of the coordinates are elements of $GF(q)$ and some are indeterminates. The necessary collinearities are given by a set of polynomial equations in the indeterminates. An algebraic manipulation programme can then determine the consistency of these equations, and check whether or not $\cal K$ lies on a cubic curve. For example, A. Simis (personal communication) has verified that if $\cal K$ is isomorphic to $({\bf Z}_{5})^{2}$ , then this works, as one expects from Corollary 4.2. \begin{thebibliography}{999} \bibitem{GMO} Ghinelli, D., Melone, N. and Ott, U. (1989) On abelian cubic arcs, {\em Geom. Dedicata} {\bf 32}, 31-52. \bibitem{H1} Hirschfeld, J.W.P. (1979) {\em Projective geometries over finite fields}, Oxford University Press, Oxford. \bibitem{H2} Hirschfeld, J.W.P. (1985) {\em Finite projective spaces of three dimensions}, Oxford University Press, Oxford. \bibitem{HT} Hirschfeld, J.W.P. and Thas, J.A. (1990) Sets with more than one representation as an algebraic curve of degree three. {\em Finite Geometries and Combinatorial Designs}, American Mathematical Society, Providence, pp. 99-110. \bibitem{HV} Hirschfeld, J.W.P. and Voloch, J.F. (1988) The characterization of elliptic curves over finite fields, {\em J. Austral. Math. Soc. Ser. A} {\bf 45}, 275-286. \bibitem{K1} Keedwell, A. (1988) Simple constructions for elliptic cubic curves with specified small numbers of points, {\em European J. Combin.} {\bf 9}, 463-481. \bibitem{K2} Keedwell, A. (1991) More simple constructions for elliptic cubic curves with small numbers of points, preprint. \bibitem{R} R\"{u}ck, H.-G. (1987) A note on elliptic curves over finite fields, {\em Math. Comput.} {\bf 49}, 301-304. \bibitem{S} Schoof, R. (1987) Non-singular plane cubic curves over finite fields, {\em J. Combin. Theory Ser. A} {\bf 46}, 183-211. \bibitem{T} Thas, J.A. (1975) Some results concerning $\{(q+1)(n-1);n\}$-arcs and $\{(q+1)(n-1)+1;n\}$-arcs in finite projective planes of order $q$, {\em J. Combin. Theory Ser. A} {\bf 19}, 228-232. \bibitem{V} Voloch, J.F. (1988) A note on elliptic curves over finite fields, {\em Bull. Soc. Math. France} {\bf 116}, 455-458. \bibitem{W} Waterhouse, W.G. (1969) Abelian varieties over finite fields, {\em Ann. Sci. Ecole Norm. Sup.} {\bf 2}, 521-560. \end{thebibliography} \begin{flushleft} School of Mathematical and Physical Sciences\\ University of Sussex\\ Brighton BN1 9QH\\ United Kindom \end{flushleft} \begin{flushleft} Department of Mathematics\\ University of Texas at Austin\\ Austin\\ TX 78712\\ U S A \end{flushleft} \end{document}