The curve that no three points in line define

A conic in the ordinary plane is a curve, and a curve is something drawn. In a finite plane there is nothing to draw: the seven-point plane and its larger relatives over the fields of order $q$ are sets of points and sets of lines with an incidence relation, and a “conic” is only the set of points whose coordinates satisfy a quadratic equation.

What such a set has, and what can be stated without any equation, is one combinatorial property: no line meets a conic in more than two points. A quadratic restricted to a line is a quadratic in one variable, with at most two roots, and that survives the passage to any field.

The question this essay follows goes the other way. Take a set of points in a finite plane with no three on a line, as large as possible. Must it be a conic? In odd order the answer is yes, always — a purely combinatorial condition that forces an algebraic equation. In even order it is no, and the reason is a point that all the tangents pass through.

As many points as a plane allows #

A set of points with no three collinear is an arc, and in a plane of order $q$ there is a hard ceiling on its size.

Pick one point $P$ of the arc. There are exactly $q + 1$ lines through $P$, and each of them holds at most one other point of the arc — two others would make three in a line. So the arc has at most $1 + (q + 1) = q + 2$ points.

When $q$ is odd, it cannot reach $q + 2$. Suppose it did. Then every line through every point of the arc would hold exactly one more, so every line meets the arc in 0 or 2 points. Take a point $Q$ off the arc: the lines through $Q$ pair up the $q + 2$ arc points, so $q + 2$ is even — and $q + 2$ is odd. So in odd order an arc has at most $q + 1$ points, and in even order it may have $q + 2$.

An arc of $q + 1$ points is an oval; one of $q + 2$ points is a hyperoval. Both names borrow from the real plane, where an oval is a closed convex curve that any line meets at most twice.

A parabola with no three points in line #

The first oval to hand is the parabola.

A line $y = mx + b$ meets $y = x^2$ where $x^2 - mx - b = 0$, which has at most two roots in any field, and a vertical line $x = c$ meets it once and then continues to the vertical point at infinity. The line at infinity meets it only there. So the $q$ affine points and the one point at infinity make an arc of $q + 1$ points: an oval, in every plane over a field.

The drawing is a scatter, and no ordinary sense of curve is visible in it. The points $(2,4)$ and $(3,4)$ sit side by side, and the set wraps round because arithmetic mod 5 does. What makes it a conic is the equation, and what makes it an oval is the count of how it meets lines. The two properties coincide on this example. Whether they always do is the question, and in small planes it can be settled by listing every oval there is.

Two familiar distinctions dissolve on the way. Over the reals, one sign decides which curve a quadratic draws — ellipse, parabola or hyperbola. In a projective plane over a finite field there is only one non-degenerate conic up to symmetry, and the three affine types are the same curve meeting the line at infinity in one point, two, or none. And the number of points is never in doubt: a non-degenerate conic over a field of order $q$ has exactly $q + 1$, where a cubic curve’s count wanders by up to twice the square root of $q$. A conic is a projective line in disguise, parametrised by its slopes from any one of its points, and a line has no error term.

Every oval, counted #

In the planes of order 2 to 5 every arc can be found by a backtracking search — adding points one at a time and refusing any that would complete a line.

The count of non-degenerate conics has a clean derivation. The symmetries of the plane of order $q$ — invertible $3 \times 3$ matrices up to scaling — number $q^3(q^3 - 1)(q^2 - 1)$, and they carry any conic to any other. The ones fixing a given conic number $q(q^2 - 1)$, the symmetries of the $q + 1$ points of a line. Dividing gives $q^2(q^3 - 1) = q^5 - q^2$.

So in orders 3 and 5 the ovals are exactly as numerous as the conics, and every conic is an oval; since two distinct conics cannot share all their points, the two sets are the same set. The table has decided, for these planes, that no oval exists except a conic. In the plane of order 5 the direct check agrees: every oval’s six points satisfy a common quadratic, which is a determinant of their six monomials $x^2, y^2, z^2, xy, xz, yz$ vanishing.

Order 4 matches the count too, but the table’s fifth column says something else is going on there: 168 arcs of six points, which no conic can be.

Tangents in pairs, or none #

The route from counting to algebra goes through tangents, and the tangents are where odd and even order separate.

At each point of an oval exactly one of the $q + 1$ lines through it is a tangent — meeting the oval only there — because the other $q$ lines each pick up one of the other $q$ points. So an oval has $q + 1$ tangents. The question is how they sit relative to the points off the oval.

The parity argument predicts this. A point $Q$ off the oval sees the $q + 1$ oval points along its $q + 1$ lines, each line carrying 0, 1 or 2 of them. The number of tangents through $Q$ therefore has the parity of $q + 1$, which is even when $q$ is odd. Counting incidences pins it further: the points on two tangents number $q(q+1)/2$ and are called external, like points outside a circle; the rest, on none, are internal.

The whole distinction between inside and outside is here, with no notion of inside. A finite plane has no order and no continuity, and still its points off a conic split into two classes of sizes $15$ and $10$, exactly as a point in the real plane sees a circle with two tangents or none.

Where Segre’s proof finds the equation #

Beniamino Segre proved in 1955 that in every plane over a field of odd order $q$, every oval is a conic. The small-plane counts are instances; the theorem is for all odd $q$, and its engine is a product that looks unrelated.

Put three points of the oval at the corners of the coordinate triangle. Every other oval point, and every tangent, then has coordinates whose ratios are non-zero field elements, and a line through a corner meeting the oval again determines one such ratio. As the line turns through all its positions, the ratio runs through all $q - 1$ non-zero elements of the field.

Segre multiplies them. The product of all the non-zero elements of a finite field is $-1$ — the field version of Wilson’s theorem, proved the same way, by pairing each element with its inverse so that only $1$ and $-1$ are left unpaired. Applied at each of the three corners, the three products force a relation between the tangents at the three chosen points: the triangle of three oval points and the triangle of their three tangents are in perspective from a point. That is Segre’s lemma of tangents.

The lemma holds for every triple of points, and from it a coordinate calculation produces a single quadratic equation satisfied by every oval point — which is to say, the oval is a conic. The real-plane analogue is Pascal’s theorem, whose degenerate cases are exactly statements of this kind about tangents and inscribed triangles.

The arithmetic fact is doing the geometry. Nothing in the lemma of tangents mentions multiplication, and nothing in Wilson’s theorem mentions lines. But the lemma needs the three tangents to form a triangle, and in characteristic 2 they do not: all three pass through a single point, as the next section shows, and there is no second triangle to be in perspective with the first.

An oval hidden in a list of numbers #

Ovals appear in places with no coordinates at all. A plane in a list of numbers builds the plane of order three from the residues $0, 1, 3, 9$ modulo 13, with its lines the thirteen shifts of that list. Negate the list: $0, 12, 10, 4$. No shift of the original contains three of those, so the negated list is an oval — four points, no three collinear, in a plane described by nothing but addition modulo 13.

The reason takes three lines. Suppose three negated members $-d_1, -d_2, -d_3$ lie on one shifted line $D + t$, so that $-d_i = e_i + t$ with each $e_i$ in $D$. Then $d_1 + e_1 = d_2 + e_2$, which says $d_1 - d_2 = e_2 - e_1$. Each non-zero difference occurs exactly once in $D$, so $e_2 = d_1$ and $e_1 = d_2$. The same argument for the first and third points gives $e_1 = d_3$ — and then $d_2 = d_3$, which is not three distinct points.

The same holds for the difference sets of orders 4, 5, 7 and 8, and it is a theorem for every cyclic plane: the negative of a planar difference set is always an oval. In the planes over fields, Segre’s theorem then says that in odd order this arithmetic set is a conic, though nothing in its definition is quadratic.

A code in every arc #

The same object has a third life, in coding theory. Write the coordinates of the points of an arc as the columns of a matrix with three rows. Any three columns are independent — that is exactly the statement that no three points are collinear — and the words the rows generate form a code in which two different words disagree in all but at most two positions. That is the most distance a code of dimension three can have, which makes it a maximum distance separable code, the best a code of its size can be.

The conic $y = x^2$ gives the columns $(1, x, x^2)$ and $(0, 0, 1)$: the values of every polynomial of degree at most two at every field element, plus its leading coefficient. That is a Reed–Solomon code, extended by one position. A hyperoval gives a code one symbol longer than any Reed–Solomon code of that dimension. The ceiling of $q + 1$ points on an arc in odd order and $q + 2$ in even order is the length limit of the best three-dimensional codes over that field, and the conjecture that no code of this kind is ever longer in other dimensions — the MDS conjecture — was proved by Simeon Ball in 2012 for fields of prime order and remains open for the rest.

In even order, the tangents all meet #

In characteristic 2 the parity count flips. When $q$ is even, $q + 1$ is odd, so every point off the oval lies on an odd number of tangents — at least one.

The argument that forces this is short. Take a line through two oval points: it holds $q - 1$ further points, each on at least one tangent. The tangents at the two oval points on that line meet it at those points, so only the other $q - 1$ tangents are available, and they must supply one each. Every point on a secant line lies on exactly one tangent. A point on no secant sees each oval point along a different line, all of them tangents — so it lies on all $q + 1$. Counting shows such a point exists, and exactly one does.

That point is the nucleus, and Qvist proved in 1952 that every oval in a plane of even order has one. For $y = x^2$ it is the horizontal point at infinity: the tangent at $(t, t^2)$ has slope $2t$, and in characteristic 2 that is $0$. Every tangent is horizontal.

One more point than any conic #

Adding the nucleus to the oval breaks no line — every line through the nucleus is a tangent, carrying one oval point — and gives an arc of $q + 2$ points.

The field of four elements makes the squaring map a permutation: $\omega^2$ squares to $\omega$, so the conic’s points $(\omega, \omega^2)$ and $(\omega^2, \omega)$ are mirror images. The table of all arcs counted 168 hyperovals in this plane and 1,008 ovals, and $168 \times 6 = 1{,}008$: every oval lies in exactly one hyperoval, the one its nucleus completes, and every hyperoval contains six ovals, one for each point removed.

This is where the correspondence between arcs and conics stops being forced. A hyperoval with one point removed is an oval, but it need not be a conic — the removed point was special only if it was the nucleus of the rest. In order 4 each hyperoval is a conic plus its nucleus in six different ways, because the plane is small. In larger even orders a hyperoval can be something no conic produces.

Powers of x that draw an oval #

A way to see the difference is to replace $x^2$ by other powers.

In odd order the column is a single cell, which is Segre’s theorem seen through one family: the graph of $x^k$ is an oval only if it is a conic, and for a monomial that means $k = 2$. In order 8 three exponents work, and all three are disguised parabolas — Segre showed in 1957 that every hyperoval of the plane of order 8 is a conic plus its nucleus.

The first hyperoval that is not appears in order 16. Lunelli and Sce found it in 1958 with one of the earliest computer searches in geometry. Beyond that, whole infinite families are known: the translation hyperovals from $x^{2^i}$, Segre’s $x^6$ in odd powers of two, families due to Glynn, Payne, Cherowitzo, and the Subiaco and Adelaide families from the 1990s and 2000s. They are sets that are as large as an arc can be and are not the zeros of any quadratic.

The odd column is short for a reason and the even column is long for one. In characteristic 2, squaring is additive, so $f(x) = x^{2^i}$ satisfies $f(x + s) = f(x) + f(s)$, and the condition that no three graph points are collinear stops depending on where the three points are. It reduces to one requirement: $f(x)/x = x^{2^i - 1}$ must take each non-zero value once, which happens exactly when $2^i - 1$ and $q - 1$ share no factor — when $i$ is coprime to $h$.

What exhaustive counts in small planes do not reach #

The counts stop at order 5. Every row of the arc table is complete for its plane, which settles “every oval is a conic” for orders 3 and 5 by exhaustion and says nothing about order 7. Segre’s theorem is what covers every odd order, and it is proved, not searched.

The monomial grid tests one family. A cell marked empty says that the graph of $x^k$ is not an arc; it does not say that the plane has no ovals besides the conic. In order 8 the grid shows three hyperovals of one shape, and the claim that every hyperoval of that plane is regular is Segre’s, not the grid’s.

The regularity check is a determinant. For the order-8 hyperovals, the check drops one point and asks whether the remaining nine satisfy one quadratic, by testing whether each sixth point lies on the conic through the first five. That is a complete test for those sets, but it is a test of those sets, not of the plane.

The grid pictures carry no geometry. A scatter of filled cells is the affine part of a plane with its points at infinity drawn off to the side, and neither the tangents nor the nucleus can be shown as lines — a line of the plane over GF(5) is five cells that obey an equation, and no straightedge joins them.

Still open: every hyperoval there is #

In odd order the question is closed: an oval is a conic. In even order it is wide open.

Hyperovals have been classified completely only in the smallest even orders, by computer, and each new order has taken far more computation than the last. Every known infinite family was found by guessing an exponent or a polynomial and proving that its graph is an arc, and no one knows whether the families found so far are all there are.

The sharpest special case is about monomials. For which exponents $k$ is the graph of $x^k$, completed at infinity, a hyperoval in the plane of order $2^h$? Up to the obvious equivalences — replacing $k$ by $1/k$ or $1 - k$ modulo $q - 1$ — the known exponents are $2$, the powers $2^i$ with $i$ coprime to $h$, $6$ when $h$ is odd, and two further families found by Glynn; it has been conjectured since the 1980s that there are no others, and it has been checked by computer for many $h$ without a proof appearing.

A more basic question underlies all of it. Segre’s theorem is about planes built from fields, where conics exist to be compared with. In a plane like the nearfield plane of order nine there is no quadratic equation to satisfy, and ovals there are sets with no algebraic description. Whether every finite projective plane contains an oval at all is not known: the axioms do not supply one, no plane without one is known, and nobody has a proof that one must exist.

A property of sets that turns out to be an equation #

The pattern worth carrying is a counting condition that secretly determines an algebraic object.

“No three points on a line” is a statement about subsets and nothing else — it can be checked with a list of lines and no arithmetic. Pushed to its maximum size in odd order, that condition leaves room for one kind of set only, and that kind is the zero set of a quadratic. The proof has to reach into the field’s multiplication to find the equation, and it does so through a product that has been known since the eighteenth century.

The failure in even order is as informative as the success. Characteristic 2 makes $-1$ equal to $1$, makes every tangent pass through a single point, lets one more point in, and lets the maximal sets escape the conics altogether. The same condition that forces a curve in one characteristic sets it free in the other, and the dividing line is a sign that the arithmetic can or cannot tell apart.

The counts in the small planes are where that becomes visible rather than merely stated. Two numbers agreeing — 3,100 ovals, 3,100 conics — is a proof for one plane, and a column of 168 six-point arcs is the first sign that the next plane of even order will not agree.