CONDITIONS ON POLYNOMIALS DESCRIBING AN OVAL IN PG(2, q)

Transcription

1 CONDITIONS ON POLYNOMIALS DESCRIBING AN OVAL IN PG(2, q) TIMOTHY L. VIS Abstract. An oval in a finite projective plane of order q is a set of q+1 points such that no three of the points lie on a common line. The classification of all ovals in the projective plane PG(2, q) with q even remains a significant and evidently difficult open problem. Ovals in PG (2, q) can be completely described by polynomials over GF (q) of degree at most q 1 with certain properties. In this paper, we explore the theory of polynomials over finite fields. In particular, we examine permutation polynomials and develop some of the theory characterizing o-polynomials, the name given to those polynomials which give rise to ovals. 1. Polynomials over a Field F Since polynomials play a very significant role in the classification of ovals, we first discuss some significant and important results relating to polynomials over finite fields. The results of greatest interest to us are those results that allow us to take arbitrary functions over a finite field and from them to construct polynomials acting on the field in the same way as the original function,while at the same time restricting the degree of the polynomial constructed. A critical and important result over finite fields allows us to construct such a polynomial. Theorem 1.1. (Lagrange Interpolation Formula) [7] For n 0, let a 0,..., a n be n + 1 distinct elements of F, and let b 0,..., b n be n + 1 arbitrary elements of F. Then there exists exactly one polynomial f F [x] of degree at most n such that f (a i ) = b i for all i and this polynomial is given by f (x) = n b i i=0 k=0 k i x a k a i a k Proof. For a given a j, notice that when i j, the appropriate term in the sum has (a j a j ) as a factor in the numerator of the fraction so that the only nonzero term occurs when i = j. But when i = j, the numerator and denominator of each factor in the product are identical, so that f (a j ) = b j as desired. On the other hand, if f and g are not equal and both satisfy the conditions, then h = f g has n + 1 zeros and at least one point that is not zero (and so is not constant). But since h has degree at most n, this is a contradiction, and f is the unique polynomial satisfying the desired conditions. Date: August 27,

2 2 TIMOTHY L. VIS While this result is true over arbitrary fields, it allows us in particular to determine a polynomial of degree less than q to represent any function over the finite field GF (q) by simply taking the elements of GF (q) as the a i and their images as the b i. An alternative form to the Lagrange interpolation polynomial where all elements of the field and their images are prescribed is given in [7]. Where φ(x) is an arbitrary function, the required polynomial is given by f (x) = ( φ(c) 1 (x c) q 1). c GF(q) If we wish a polynomial of degree less than q to represent a polynomial of arbitrary degree, the following result allows us to determine the appropriate small degree polynomial. Lemma 1.2. [7] For f, g GF (q)[x], f (c) = g (c) for all c GF (q) if and only if f (x) g (x) mod (x q x). Proof. Consider f g. By the division algorithm, f (x) g (x) = h (x) (x q x) + r (x), where r (x) has degree less than q. Then f (c) = g (c) for all c if and only if r (c) = 0, since c q = c. But this occurs if and only if r = 0, which is if and only if f (x) g (x) mod (x q x). We also have the following determination of the coefficients of such a polynomial. Lemma 1.3. [2] The polynomial of degree at most q 1 representing a function g (x) on a field has coefficients given by a 0 = g (0), a r = λ GF(q) g (λ) λ r, 1 r q 2, and a q 1 = λ GF(q) g (λ). These three results allow functions over a finite field to be greatly simplified by reduction to polynomials of degree less than the order of the field. As ovals and hyperovals are generally described in terms of functions giving their coordinates, these results are of fundamental importance in that they allow great simplification in the functions that must be examined. 2. The Projective Plane P G(2, K) In order to discuss ovals and hyperovals in PG(2, K), we must first discuss what PG(2, K) is. We begin by defining a projective plane. Definition 2.1. A projective plane is a point-line incidence geometry satisfying the following four axioms: (1) Any two points are incident with exactly one line (2) Any two lines meet in exactly one point (3) Every line contains at least three points (4) There exists a set of four points such that no three are incident with a common line. The last axiom is a non-degeneracy axiom, excluding certain undesirable and uninteresting cases from consideration. Because we are restricting ourselves to projective planes, no classification is present. In fact, the classification of all projective planes remains perhaps one of the most outstanding and significant open problems in combinatorics. Higher dimensional projective spaces are classified and can all be constructed from vector spaces over skewfields, however the same cannot be said

3 o-polynomials 3 for planes. In this paper, we shall restrict our attention to those planes constructed from vector spaces over skewfields. Given a skewfield K (finite or infinite), we can construct a three dimensional vector space K 3 over that skewfield, where the vectors are considered as ordered triples of elements from K and the scalars as elements of K with scalar multiplication carried out on the left. (In the case where K is a field, this distinction is unnecessary). We define a geometry from this vector space in the following manner. Definition 2.2. The geometry P G(2, K) is a point-line incidence structure whose points are the one-dimensional vector subspaces of K 3, whose lines are the twodimensional vector subspaces of K 3, where a point is incident with a line if and only if the one-dimensional vector space corresponding to the point is a subspace of the two-dimensional vector space corresponding to the line. Where K is the finite field GF (q), we write simply PG(2, q). Theorem 2.3. The geometry PG(2, K) is a projective plane. Proof. We verify that the axioms of a projective plane hold. Let P and Q be two distinct points of PG(2, K). Then P = v and Q = w, with w / v. Thus the line l = v, w is a line incident with both v and w. But any two-dimensional vector space containing both v and w must contain l and, as such, l is the unique line incident with both v and w. Now let l = u, v and m = w, x be two distinct lines of PG(2, K) and consider l m. Since l and m both have rank two and are distinct, and since K 3 has rank three, it follows that l m has rank one. But then l m is a unique point incident with both l and m. Consider the line l = u, v. Then l contains the distinct points u, v, and u + v. Finally, since any skewfield contains both 0 and 1 distinctly, the points (1, 0, 0), (0, 1, 0), (0, 0, 1), and (1, 1, 1 ) form a set of four points with no three incident with a common line. Notice thus far that we have used a single vector to generate a one-dimensional subspace, or point. We will, throughout this paper, denote a point simply as an ordered triple, rather than as the subspace generated by that ordered triple, with the understanding that these coordinates are homogeneous and that two ordered triples differing only in a scalar multiple describe the same point. Thus, (a, b, c) will, as a point, be denoted simply as (a, b, c). It should be readily clear that the line through two points (x 0, x 1, x 2 ) and (x 0, x 1, x 2 ) contains precisely those points of the form (x 0, x 1, x 2 ) + α (x 0, x 1, x 2 ), where α ranges over the elements of K along with the point (x 0, x 1, x 2). Note that the use of homogeneous coordinates eliminates the need for any further scalar multiples of the point (x 0, x 1, x 2 ) to be used. Furthermore, in the finite case, we have described a total of K + 1 points on a given line. The following well known geometrical result is very useful. Theorem 2.4. For any finite projective plane, there exists some number n such that each line contains n + 1 points. In particular, every line of PG(2, q) contains exactly q + 1 points. Definition 2.5. The number n described in Theorem 2.4 is called the order of a plane.

4 4 TIMOTHY L. VIS Notice that if we exchange the roles of points and lines in the axioms of a projective plane, the axioms remain unchanged. This duality is an important property of a projective plane and implies that a statement is true if and only if its dual is also true. The dual of Theorem 2.4 thus implies that each point of PG(2, q) is incident with exactly q + 1 lines. Definition 2.6. The set of points (1, 0, 0), (0, 1, 0), (0, 0, 1) and (1, 1, 1) will be called the fundamental quadrangle. Definition 2.7. A bijective map π on the points and lines of a projective plane is called a collineation of the plane provided that whenever a point P is incident with a line l, the point π (P) is incident with the line π (l). It is immediate from the definition that the composition of two collineations will be a collineation and that any collineation has an inverse. As such, the set of all collineations of a projective plane forms a group, the full collineation group of the projective plane. In the case where the projective plane is PG(2, K), the structure of this group is known. Definition 2.8. The projective general linear group PGL (3, K) is the quotient group of all invertible 3 3 matrices over the skewfield K modulo the group of all nonzero scalar transformations of K 3. This group acts on vectors of K 3 as a linear map. Such a linear map on the vectors of K 3 is easily seen to produce a collineation of PG(2, K). Consider that subspace inclusion is preserved by the action of an invertible matrix. As such, the full collineation group of PG(2, K) must include all elements of PGL (3, K). In fact, we could argue that it must include all elements of GL (3, K); however, our use of homogeneous coordinates implies that nothing is added by loosening the restriction simply to invertible matrices, as any invertible matrix is obtained by some scalar transformation of an element of PGL (3, K). On the other hand, consider the action of a field automorphism σ on the vectors of K 3. The fact that σ preserves both multiplication and addition forces σ to preserve subspace inclusion in the vector space. As such, the full collineation group of PG(2, K) must include all elements of Aut (K). Definition 2.9. Given a matrix M PGL (3, K) and an automorphism σ Aut(K), the map on a vector P in K 3 determined by Mσ (P) is called a semilinear map. It is known (though not immediately obvious) that the composition of any two semilinear maps produces another semilinear map. As such, the set of all semilinear maps of K 3 forms a group. Definition The group PΓL (3, K) is the group of all semilinear maps on K 3. That is, PΓL (3, K) = PGL (3, K),Aut(K). Theorem P ΓL (3, K) is the full collineation group of P G(2, K). The proof of this theorem is somewhat lengthy and can be found in such sources as [1]. Theorem The full collineation group PΓL (3, K) of PG(2, K) acts transitively on quadrangles of PG(2, K).

5 o-polynomials 5 Proof. Any quadrangle consists of a set of four points, corresponding to four linearly dependent vectors, where any three are linearly independent. Thus, some vector v 0 = αv 1 + βv 2 + γv 3 for appropriate choices of α, β, γ. But the vectors αv 1 and v 1 represent the same point. Further, αv 1, βv 2, and γv 3 form a basis for the vector space such that v 0 = (1, 1, 1) under this basis. It is well known that any basis can be mapped to any other basis. Such a change of basis maps any quadrangle to any other quadrangle, so that the action is transitive. 3. Ovals and Hyperovals of PG(2, q) Of particular interest in projective planes are sets of points that are as independent as possible; that is, sets of points so that no three lie on a common line. Such sets of points are, in some sense, analogous to arcs in the Euclidean plane, and are named accordingly. Definition 3.1. A set of k points such that no three lie on a common line is called a k-arc. Lemma 3.2. In the projective plane PG(2, q), the number k of points on a k-arc is at most q + 2. Proof. Consider a point on a k-arc. There are q +1 lines through this point. These lines contain every point of the plane, and each line contains at most one other point of the k-arc. As such, there are at most q + 1 other points on the k-arc, for a total of q + 2 points on the k-arc. Definition 3.3. A nondegenerate conic in PG(2, q) is the set of points satisfying x x 1x 2 = 0. Lemma 3.4. [6] Every non-degenerate conic in P G(2, q) is a (q + 1)-arc. We have shown thus far that in PG(2, q) we can always find a (q + 1)-arc, and that we can never find a (q + 3)-arc. As such, we need only determine exactly when q + 2 is attainable and when q + 1 is the best possible. In a (q + 1)-arc, every point is on a common line with q other points, so that there is exactly one tangent line through each point of the arc that contains only one point of that arc. Lemma 3.5. [6] When q is even, the q + 1 tangent lines to a (q + 1)-arc meet in a common point. Theorem 3.6. [6] The size of the largest arc in PG(2, q) is either q + 1, when q is odd, or q + 2, when q is even. Proof. We have shown that an arc cannot exceed q + 2 points and that q + 1-arcs always exist. Lemma 3.5 implies the existence of a point that lies on the tangent lines to a (q + 1)-arc through every point where q is even. As such, any line from this point to a point on the arc contains only 1 point of the arc and this point can be appended to form a (q + 2)-arc. On the other hand, a (q + 2)-arc has no tangent lines, since every line through a point on the arc must contain a second point of the arc. Take some point not on the arc. The lines through such a point are a partition of the points of the arc into even numbers of points (either zero or two). But where q is odd, so is q + 2. Thus, no such arc can exist where q is odd.

6 6 TIMOTHY L. VIS Definition 3.7. A (q + 1)-arc will be called an oval. A (q + 2)-arc will be called an hyperoval. It should be clear that a conic is always an oval containing the fundamental quadrangle. But since the collineation group P ΓL (3, K) acts transitively on quadrangles and since every oval of PG(3, K) contains a quadrangle (except where K = GF (2), in which case every hyperoval is a quadrangle), we may always assume that an oval (or hyperoval in even characteristic) contains the fundamental quadrangle. When q is odd, the following result completes the classification of all ovals. Theorem 3.8. [10] Every oval of PG(2, q), q odd, is a conic. However, when q is even, the classification is far from complete and remains one of the most significant open problems in finite projective geometry. 4. Permutation Polynomials If we consider a hyperoval containing the fundamental quadrangle, we may notice several properties concerning the coordinates of the points. Since the points (0, 1, 0) and (0, 0, 1) are both contained in the hyperoval, no other point with a zero first coordinate may be in the hyperoval or it would be on the line spanned by these two points. We may therefore assume that the first coordinate of any other point on the hyperoval is one, dividing by the first coordinate as necessary from any other representation of such a point. Furthermore, suppose the points (1, a, b) and (1, a, c), with b c are both on the hyperoval. We have (1, a, c) = (1, a, b)+ (c b)(0, 0, 1), so that the three points lie on a common line. As such, we may further assume that the second coordinates of the remaining points of the hyperoval contain each of the q elements of the field exactly once. The same argument holds for the third coordinates of the points. Thus, we obtain the following lemma. Lemma 4.1. The q points of a hyperoval containing the fundamental quadrangle other than (0, 1, 0) and (0, 0, 1) can be represented so that the first coordinate of each is one and the second and third coordinates of each are permutations on the elements of GF ( 2 h). Under the conditions set down by this lemma, we may assume that the points of the hyperoval are { (1, t, f (t)) : t GF ( 2 h)} {(0, 1, 0), (0, 0, 1)}, where f (t) is a bijection from GF ( 2 h) to GF ( 2 h). By Theorem 1.1 we may assume that f is a polynomial. We further note that the inclusion of the remaining points of the fundamental quadrangle ensures that f (0) = 0 and f (1) = 1. Definition 4.2. A bijective polynomial on the field GF (q) is said to be a permutation polynomial. An extensive exposition on permutation polynomials can be found in chapter five of [2] and chapter seven of [7]. We recall some of the more important results within this paper, beginning with Hermite s Criterion, a necessary and sufficient condition for a polynomial to be a permutation polynomial. We first establish the following lemma which will be necessary in the proof of Hermite s Criterion. Lemma 4.3. [7] Let {a i } be elements of GF (q). Then the following are equivalent:

7 o-polynomials 7 (1) a i a j whenever i j. (2) q 1 i=0 ar i is equal to zero when r = 0, 1,...,q 2 and one when r = q 1. Proof. For some j with 0 j q 1, consider g j (x) = 1 q 1 k=0 aq 1 k j x k. This satisfies g j (a j ) = 1, since each term in the summation is one in this case and there are q such terms for a total of zero, as well as g j (b) = 0 for b a j. As such, the polynomial g (x) = q 1 j=0 g j (x) maps each element of GF (q) to one if and only if each element occurs as exactly one of the a i. On the other hand, since the degree of g (x) is less than q, g (x) maps each element of GF (q) to one if and only if g (x) = 1. But this is equivalent to q 1 j=0 aq 1 k j being zero except when k = q 1, when it must be 1, which is precisely the second condition, so that the equivalence holds. Theorem 4.4. (Hermite s Criterion) [7] Let GF (q) be a field of characteristic p. Then f GF (q) is a permutation polynomial of GF (q) if and only if the following two conditions hold: (1) [f (t)] r has degree at most q 2 modulo t q t for r with 1 r q 2, and r 0 mod p. (2) f has exactly one root in GF (q). Proof. Suppose that f is a permutation polynomial of GF (q). The second property then holds trivially. The reduction of [f (t)] r modulo t q q must be some polynomial of the form q 1 i=0 b(r) i t i, where b (r) q 1 = c GF(q) f (c)r, as a consequence of the alternative to the computation of the Lagrange interpolating polynomial. On the other hand, the preceding lemma requires that b (r) q 1 = 0 for all values of r with 1 r q 2, as desired. On the other hand, if both conditions are satisfied, the condition that exactly one root exists implies that c GF(q) f (c)q 1 = 1, as each non-zero element t GF (q) satisfies t q 1 = 1 and there are q 1 such elements being added together, which gives a sum of 1. Further, the first condition implies that c GF(q) f (c)r = 0 for the remaining terms. But since raising to a power of p (the characteristic of the field) is an automorphism, this must imply that c GF(q) f (c)r = 0 for 1 r q 2, and the identity is trivial when r = 0, since it simply states that the sum of q ones is zero. Thus, the preceding lemma implies that f is a permutation polynomial as desired. A consequence of this result is the special case where q = p for a prime p of the following result, which eliminates the condition that r 0 mod p. Theorem 4.5. [2] A function f is a permutation polynomial on GF (q) if and only if the following two conditions hold: (1) [f (t)] r has degree at most q 2 modulo t q t for r with 1 r q 2. (2) f has exactly one root in GF (q). This theorem has the following immediate corollary in the case that f (0) = 0, which is the case when the point (1, 0, 0) is to be included. Corollary 4.6. A function f with f (0) = 0 if and only if a = 0 is a permutation polynomial of GF (q) if and only if [f (t)] r has degree at most q 2 modulo t q t for r at most q 2.

8 8 TIMOTHY L. VIS As a final result related to permutation polynomials, we consider the special class of polynomials taking the form h k (x) = 1 + x+x x k. If such a polynomial is a permutation, we have the following result. Theorem 4.7. [8] Suppose h k (x) is a permutation polynomial over GF (q), where q = p r. If q is even, then (k + 1, q 1) = 1 and k mod p, and if q is odd, then (k + 1, q 1) = 2 and k mod p. Proof. Note that h k (x) = 1, so that h k (1) 1 and, in fact, h k (1) = k + 1 mod p, and k 0 mod p. If x 1, h k (x) = xk+1 1 x 1 so that the solutions of h k (x) = 1 are also solutions of x k=1 = x, where x 1. This gives a total of (k, q 1) solutions to h k (x) = 1. Since h k (x) is assumed to be a permutation, it follows that (k, q 1) = 1. If q is odd and k mod p, then h k (1) = 0 and there must be no solution other than one to x k+1 = 1. But then (k + 1, q 1) = 1, a contradiction, since q 1 is even, so that (k + 1, q 1) 1 when (k, q 1) = 1. But then k mod p, and h k (1) 0. Now 1 is a solution of x k+1 = 1, and since h k (x) is a permutation, there exists a unique other solution to x k+1 = 1 (as this polynomial is the numerator in the rational representation of h k (x)) and as such, (k + 1, q 1) = 2. If q is even, then k 0 mod p implies that k+1 0 mod p, so that h k (1) = k+ 1 = 0. So x k+1 = 1 must have 1 as the only solution so that (k + 1, q 1) = 1. Many examples of permutation polynomials exist. The simplest type of permutation polynomial over GF (q) is simply any linear polynomial. To see that such a polynomial is injective, ax 0 +b = ax 1 +b implies ax 0 = ax 1, which implies x 0 = x 1. To see that such a polynomial is surjective, ax + b = c has the solution x = c b a. Thus, this is a permutation polynomial. Another easy example in a finite field GF (q) is the monomial f (x) = x n where (n, q 1) = 1. Over a finite field, we need only prove that this is surjective. Let the elements of the field be expressed as α i, where α is a generator of the multiplicative group of the field. Then 0 = 0 n. Further, since (n, q 1) = 1, α n is also a generator of the multiplicative group of the field, every element is of the form α mn for some m. But then every element is of the form (α m ) n for some m and this function is surjective, and thus bijective. 5. o-polynomials We now turn our discussion to conditions which ensure that a given permutation polynomial is indeed an o-polynomial. As we have already discussed, a hyperoval of P G(2, q) containing the fundamental quadrangle can be described as the points D (f) = {(1, t, f (t)) : t GF (q)} {(0, 1, 0),(0, 0, 1)}. We now wish to place restrictions on the o-polynomial f (t). Theorem 5.1. [6] The polynomial f (t) is an o-polynomial over GF ( 2 h) if and only if f (t) and each of the f s (t) are permutation polynomials, where { 0 t = 0 f s (t) = f(t+s)+f(s) t t 0. Proof. Let H be a hyperoval. As established, the points are those points { (1, t, f (t)) : t GF ( 2 h )} {(0, 1, 0), (0, 0, 1)}.

9 o-polynomials 9 We have further established that f must be a permutation polynomial. Recall now that for three points to be collinear, one must be a linear combination of the other two. Further, to verify that no three points are collinear, we may ignore the points (0, 1, 0) and (0, 0, 1). No point of the form (0, a, b) other than these two exists, and no point of either of the forms (1, a + t, f (t)) or (1, t, a + f (t)) occurs, since f (t) is a permutation polynomial. Thus, we need only consider three points of the form (1, t, f (t)) being collinear. Suppose that no three such points (1, t 0, f (t 0 )), (1, t 1, f (t 1 )), and (1, t 2, f (t 2 )) are collinear. This can be true if and only if we have 1 t 0 f (t 0 ) 1 t 1 f (t 1 ) 1 t 2 f (t 2 ) 0 for all choices of t 0, t 1, and t 2. But this is true if and only if f (t 0 ) + f (t 1 ) f (t 0) + f (t 2 ) t 0 + t 1 t 0 + t 2 for all choices of t 0, t 1, and t 2. Performing the substitution t 0 = s and t = t i +s, we obtain the fact that the function f(t+s)+f(s) t is injective as a function of t. Further, the function cannot take the value of zero unless f (t + s) = f (s) and t = 0. Extending the function to f s (t), then, we obtain an injective function f s (t), which must therefore be a permutation polynomial. The converse holds in the same way, so that the theorem is proved. A further result that restricts the o-polynomial is due to Segre and Bartocci. Theorem 5.2. [11] The coefficient on any term of odd power in an o-polynomial is zero. However, perhaps the most significant result in the classification of hyperovals is the following result due to Glynn. Theorem 5.3. [4] The polynomial f (t) is an o-polynomial if and only if the coefficient of t a in [f (t)] b is congruent to zero modulo t q t for all a, b with 1 b a q 1, where b q 1 and where the binary expansion of a dominates the binary expansion of b. Currently, ten infinite families of hyperovals are known [9]. These are listed here, along with restrictions on the fields GF ( 2 h) over which they occur. (1) Regular hyperovals: f (t) = t 2 for all h. (2) Translation hyperovals: f (t) = t 2i, (i, h) = 1, for all h. (3) Segre hyperovals: f (t) = t 6, h odd. (4) Two families of Glynn hyperovals: where σ, γ Aut ( GF ( 2 h)) with σ 2 = 2, γ 2 = σ, h odd. (a) Type I: f (t) = t σ+γ (b) Type II: f (t) = t 3σ+4. (5) Payne hyperovals: f (t) = t t t 5 6, h odd. (6) Two families of Subiaco hyperovals: Let d GF ( 2 h), where the absolute trace of d 1 is one and d 2 + d Let g (t) = t d 2 x 4 + d 2 ( 1 + d + d 2) x 3 + d 2 ( 1 + d + d 2) x 2 + d 2 x (x 2 + dx + 1) 2

10 10 TIMOTHY L. VIS h (x) = ( d 1 2 d 4 x 4 + d ) d + d 2 2 x 3 + d ( d 2) x d (1 + d + d 2 ) x1 2 + d (1 + d + d 2 )(x 2 + dx + 1) 2 (a) Type I: f (t) = g (t) + sh (x) + s 1 2x 1 2, s GF ( 2 h), for all h 2. (b) Type II: f (t) = h (t), for all h 2. (7) Cherowitzo hyperovals: f (t) = t σ + t σ+2 + t 3σ+4, σ = 2e, h = 2e 1. (8) Adelaide hyperovals: f (t) = T(βm )(t+1) T((βt+β T(β) + q ) m ) m 1 +t 2, 1 where T(β) t+t(β)t q = 2 k, β GF ( q 2), β q+1 = 1, β 1, T (t) = t + t q, m = q 1 3, h = 2k. In addition, the O Keefe-Penttila hyperoval, which does not currently lie in a known family, is given by f (t) = t 4 +t 16 +t 2 8+η 11 ( t 6 + t 10 + t 14 + t 18 + t 22 + t 26) + η 20 ( t 8 + t 20) + η 6 ( t 12 + t 24), η a primitive root of GF (32) satisfying η 5 = η Monomial Hyperovals Of the families of hyperovals in the preceding section, the first five are easily seen to have monomial o-polynomials. It is an interesting open problem to classify all hyperovals which have monomial o-polynomials, called monomial hyperovals. We earlier mentioned some results on the polynomials h k (x). Another result from [8] is the following. Theorem 6.1. f (t) = t k+1 is an o-polynomial if and only if the polynomial h k (x) is a permutation polynomial on GF ( 2 h), if and only if h k (x + 1) is a permutation polynomial on GF ( 2 h). This result easily demonstrates that both the regular and translation hyperovals are indeed hyperovals, as shown in [5]. Since h 1 (x) = 1+x is a linear polynomial, it is a permutation polynomial and f (t) = t 2 is an o-polynomial, so that the regular hyperovals are hyperovals. n = x x On the other hand h 2n 1 (x + 1) = (x+1)2n 1 x = x 2n 1, which was shown to be a permutation polynomial if and only if ( 2 n 1, 2 h 1 ) = 1. Let (n, h) = d. Then ( 2 n 1, 2 h 1 ) ( (2 d = 1 if and only if ) n d 1, ( 2 d) h d 1) = 1. Since 2 d 1 is a common divisor of these two elements, their greatest common divisor is one if and only if 2 d 1 = 1, which is if and only if d = 1, so that f (t) = t 2n is an o-polynomial if and only if (n, h) = 1, verifying the conditions placed on the translation hyperovals. However, the most powerful tool for classifying monomial hyperovals is the monomial case of Glynn s Criterion. We first state some important lemmas. Lemma 6.2. [6] f (t) = t k defines an o-polynomial if and only if (k, q 1) = 1, (k 1, q 1) = 1, and (t+1)k +1 t is a permutation of GF (q). Lemma 6.3. [3] The expansion of (1 + t) k modulo t q t over GF (q) is given by c k tc mod t q t, where 0 c q 1. Definition 6.4. [3] Define the partial ordering in the following manner. a b if and only if the binary expansion of b contains a one in every position in which the binary expansion of a contains a one. Theorem 6.5. [3] f (t) = t k is an o-polynomial over PG(2, q) if and only if d kd, for all 1 d q 2, where kd is reduced modulo q 1 under the convention that zero is reduced to zero and any other multiple of q 1 is reduced to q 1.

11 o-polynomials 11 Proof. Let k be given as desired. Then by Lemma 6.2, t k is an o-polynomial if and only if f k (t) = (t+1)k +1 t is a permutation. But f k (t) = (t+1)k +1 (t+1)+1, which simplifies to k 1 i=0 (t + 1)i. In particular, f k (0) is simply the sum of k ones and is zero or one as k is even or odd. For t 0, f k (t) = 0 if and only if (t + 1) k + 1 = 0 if and only if (t + 1) k = 1 if and only if t + 1 = 1 k 1 if and only if t = 0, a contradiction. So to be a permutation, we must have k even to obtain a root and [f k (t)] r modulo t q t having zero coefficient on t q 1 by Hermite s Criterion. By Lemma 1.3, this coefficient is ( ) (λ + 1) k r + 1 = 0 λ λ GF(q) [ r. for all 1 r q 2. Now let g (t) = (t + 1) k + 1] There is a unique polynomial representing this of degree at most q 1, so let this polynomial be f (t) having coefficient on t r of ( ) (λ+1) k +1 r, λ GF(q) λ so that the previous statement is [ r true if and only if the coefficient of t r in (t + 1) k + 1] is zero for all 1 r q 2. This occurs if and only if the expansion ( ) d r c kd tc, given by Lemma 6.3 has zero coefficient on t r for all 1 r q 2. But the terms with a t r are exactly d r r kd tr. Since a r = 0, we then have d r kd tr = 0 and conversely. With q even, this is equivalent to the statement that {d : d r kd} 0 mod 2 for all r. With this given, suppose that there exists some d such that d kd and such that d is minimal. Then r = d yields {d : d d kd} = {d }, a contradiction, so that this occurs if and only if d kd for all d. Now d kd for all d implies (k, q 1) = (k 1, q 1) = 1 in the following way. If (k, q 1) = l > 1, m = q 1 l satisfies mk q 1 mod q 1. Then m mk, a contradiction, so that (k, q 1) = 1. If (k 1, q 1) = n 1, p = q 1 n so p (k 1) 0 mod q 1, which implies that pk p mod q 1 giving them the same binary expansions, again a contradiction. So d kd for all d if and only if the three conditions of Lemma 6.2 are all satisfied. The structure of this result leads to a natural means of classification based on the number of ones appearing in the binary expansion of k. Use of this result has led to the classification of all monomial hyperovals whose o-polynomials are monomials with exponents having only one or two non-zero terms in their binary expansion. Several results are known which suggest that every monomial hyperoval is equivalent to one defined by a monomial having an exponent with at most three non-zero terms in its binary expansion. While this has yet to be proved, work on this conjecture, as well as on the classification of those monomial o-polynomials having three non-zero terms in the binary expansion of the exponent is underway, with some hope of completing the classification of monomial hyperovals. References [1] Albrecht Beutelspacher and Ute Rosenbaum. Projective Geometry. Cambridge University Press, New York, [2] L. E. Dickson. Linear Groups, with an Exposition of the Galois Field Theory. Dover, New York, 1958.

12 12 TIMOTHY L. VIS [3] David G. Glynn. Two new sequences of ovals in finite desarguesian planes of even order. In Combinatorial Mathematics X, Lecture Notes in Mathematics, volume 1036, pages Springer Verlag, [4] David G. Glynn. A condition for the existence of ovals in pg (2, q), q even. Geometriae Dedicata, 32: , [5] Roy G. Hadinata. Monomial hyperovals in desarguesian planes. Master s thesis, University of Adelaide, November [6] J. W. P. Hirschfeld. Projective Geometries over Finite Fields. Oxford University Press, New York, [7] Rudolf Lidl and Harald Niederreiter. Finite Fields, volume 20 of Encyclopedia of Mathematics and its Applications. Cambridge University Press, Cambridge, [8] Rex Matthews. Permutation properties of the polynomials 1 + x + + x k over a finite field. Proceedings of the American Mathematical Society, 120:47 51, [9] Tim Penttila. Configurations of ovals. Journal of Geometry, 76: , [10] B. Segre. Lectures on Modern Geometry. Casa Editrice Cremonese, Rome, [11] B. Segre and U. Bartocci. Ovali ed altre curve nei piani di galois di caratteristica due. Acta Arithmetica, 18: , 1971.