A plane no field built

A projective plane is two axioms: every two points lie on exactly one line, and every two lines meet in exactly one point, with enough points around to rule out the degenerate cases. The seven-point plane builds one from the field with two elements, and the same recipe works for any field: points are the lines through the origin of three-dimensional space over the field, lines are the planes through it, and incidence is containment.

The recipe produces a plane of order $q$ for every prime power $q$, and for a while it looked as though it might produce all of them. The smallest orders are forced: every plane of order 2, 3, 4, 5, 7 or 8 is the one its field builds. Order nine is where that stops, and it stops in a specific, checkable way. There is a plane of order nine that obeys both axioms to the last incidence, and in which a theorem true in every field-built plane is false.

The theorem is Desargues’, and the way it fails turns out to say precisely which parts of the algebra of a field the plane has kept.

Two axioms that ask for half a distributive law #

A plane of order $q$ can be built from much less than a field, and seeing how much less is the start of the whole subject.

Take any set $R$ of $q$ symbols with an addition and a multiplication, and write $x \circ m$ for the product. Its points are the pairs $(x, y)$, plus one extra point for each “slope” $m$, plus one point at the end of the vertical direction. Its lines are the graphs $y = x \circ m + b$, each with its slope point added; the verticals $x = c$, each with the vertical point added; and one more line holding every slope point and the vertical point, the line at infinity.

When do two points lie on exactly one line? For two points $(x_1, y_1)$ and $(x_2, y_2)$ with different first coordinates, a line through both needs

$$y_1 - y_2 = x_1 \circ m - x_2 \circ m.$$

If the product distributes over addition from the right — $(a + b) \circ m = a \circ m + b \circ m$ — the right-hand side is $(x_1 - x_2) \circ m$, and the axiom asks only that the equation $a \circ m = c$ have exactly one solution $m$ for each non-zero $a$. Then $b$ follows. The other cases, involving the points at infinity, need similar solvability conditions and nothing more.

So the incidence axioms use addition, unique solvability, and one side of the distributive law. Commutativity is never asked for. Neither is the other side of distributivity, and associativity of the multiplication is not needed either, though the example below happens to have it.

Leonard Dickson had found, in 1905, finite number systems that keep almost everything a field has and lose exactly those things. The smallest interesting one has nine elements.

Eighty-one products, twenty-four of them changed #

Start with a finite field of order nine: the numbers $a + bi$ with $a, b$ in $\{0, 1, 2\}$, arithmetic modulo 3, and $i^2 = -1$. Half its eight non-zero elements are squares — every non-zero element is a power of one generator, and the squares are its even powers.

Dickson’s product keeps the field’s addition and changes the multiplication by one twist:

$$x \circ y = \begin{cases} x\,y & \text{if } y \text{ is a square,} \\ x^3\,y & \text{if } y \text{ is not.} \end{cases}$$

Cubing is an automorphism of this field — it is the Frobenius map, which in characteristic 3 respects addition — and it fixes exactly the three elements $0, 1, 2$.

Because cubing respects addition, $(a + b) \circ y = a \circ y + b \circ y$ in both cases, so right distributivity holds. Every equation $a \circ m = c$ with $a \neq 0$ has one solution, since each row of the table is a rearrangement of the nine elements. What fails is commutativity, and distributivity from the left: $i \circ (1 + i)$ is $1 + 2i$ where the field says $2 + i$.

Why the twist survives three factors #

Associativity is the property that looks as though it ought to break, and the reason it does not is a parity count.

Write $s(y) = 0$ when $y$ is a square and $s(y) = 1$ when it is not, so that $x \circ y = x^{3^{s(y)}}\,y$. Multiplying out both bracketings,

$$(x \circ y) \circ z = x^{3^{s(y) + s(z)}}\, y^{3^{s(z)}}\, z, \qquad x \circ (y \circ z) = x^{3^{s(y \circ z)}}\, y^{3^{s(z)}}\, z.$$

The two agree when $x$ is raised to the same power on both sides. Cubing twice gives $x^9$, and $x^9 = x$ for every element of a field of order nine, so only the parity of the exponent’s count matters. And $y \circ z$ is $y$ or $y^3$ times $z$; cubing sends squares to squares, and a product of two elements is a square exactly when an even number of them are not. So $s(y \circ z)$ has the parity of $s(y) + s(z)$, and the bracketings agree for all 729 triples — as the check behind the table confirms.

The same count explains the nonzero elements’ structure. Under $\circ$ they form a group of order eight that is not commutative, and it is the quaternion group: the eight units $\pm1, \pm i, \pm j, \pm k$ in a disguise of ordered pairs modulo 3.

A system with these properties — a group under addition, a group under multiplication away from zero, one distributive law — is a nearfield, and by the argument above it coordinatises a projective plane of order nine. Nothing about that plane has been assumed; it can be checked.

Ninety-one points, and every incidence right #

Both planes of order nine — the one the field builds and the one the nearfield builds — can be tested against the axioms completely, because they are small.

The first four columns are the axioms, and the two planes are indistinguishable on them. Every count a plane of order nine could have, both have — the same numbers of points, lines, points per line and lines per point, and the same “exactly one” for every pair.

The last column is the difference. A Desargues configuration needs a centre, three lines through it, a pair of corresponding corners on each line, and then three meeting points of corresponding sides. In a plane over a field the three meets are always collinear: coordinates reduce the claim to a determinant identity, and in fact Desargues’ theorem holds in a plane exactly when its coordinates can be taken from a division ring, a field in which multiplication need not commute. In the nearfield plane, 895 of 1,200 random configurations fail — about three in four.

So the axioms do not imply Desargues’ theorem. That was first shown for infinite planes by Hilbert and for finite ones by Oswald Veblen and Joseph Wedderburn in 1907, with planes of order nine. The plane here is the one now called the Hall plane of order nine, and the complete classification of order nine — four planes, found by computer in 1991 — has it as one of the three that no division ring coordinatises.

One configuration that does not close #

A proportion is a summary, and a single counterexample is worth writing out, because it is the whole logical content of the claim.

Each entry is found by the plane’s own operations: two points determine a line by solving $y_1 - y_2 = (x_1 - x_2) \circ m$ for $m$, and two lines determine a point by solving a pair of such equations. Every step can be carried out by hand with the nine-by-nine table above, and the final check — does the third point satisfy the equation of the line through the first two — comes out no.

The same configuration written with the field’s product instead closes, because every one does. That is the practical meaning of “non-Desarguesian”: not a subtle defect of an exotic object, but a familiar construction that stops working on ordinary-looking coordinates, and stops working most of the time.

It also answers a question the Latin-square view of planes leaves hanging. A full set of orthogonal Latin squares of order nine can be read off either plane, and both sets satisfy every orthogonality condition. Desargues’ theorem is a property of the plane that the squares carry without stating, and it is what makes the two families genuinely different rather than relabellings of each other.

The one line from which every configuration closes #

Random configurations mix together every position a configuration can take. Choosing the axis and the centre first, and building the rest from them, separates the cases — and they do not behave alike.

The construction is direct. Fix a centre $O$ and an axis line. Choose three lines through $O$, corners $A$, $B$, $C$ on them, and a second corner $A'$ on the first line. Now $B'$ is forced: $AB$ meets the axis at some point $P$, and $B'$ has to be where the line $A'P$ crosses $OB$. $C'$ is forced the same way. Two of the three meets are on the axis by construction, and the theorem’s whole content is whether $BC$ and $B'C'$ meet there too.

One position never fails. The next figure asks whether that is about the line at infinity specifically, or about any line with the centre placed on it.

Exactly one line is special. From the line at infinity, with the centre on it, the plane looks Desarguesian; from every other line, it does not.

The reason is the right distributive law again. Shifting every point by a fixed $(a, b)$ sends the line $y = x \circ m + c$ to

$$y = (x - a) \circ m + c + b = x \circ m - a \circ m + c + b,$$

which is another line of the same slope, because $(x - a) \circ m$ splits. So every translation of the nearfield plane is a symmetry of it, and every translation fixes the line at infinity point by point. Reinhold Baer showed in the 1940s that a Desargues configuration with a given centre and axis always closes exactly when the plane has enough symmetries fixing that axis pointwise and that centre — and the translations are exactly those symmetries for the line at infinity and the slope points on it.

The failed configurations are the ones no symmetry reaches. Their lines have no translations of their own, because nothing in the nearfield distributes from the left.

Which law each theorem is #

The pattern generalises into one of the cleanest dictionaries in mathematics, and every entry pairs a configuration theorem with an algebraic law.

The little Desargues theorem on one line — centre on the axis, axis fixed — holds exactly when the coordinates can be taken from a quasifield: addition is a group, one distributive law holds, and equations are uniquely solvable. Those are translation planes, and the nearfield plane is one.

The little Desargues theorem on every line holds exactly when the coordinates are an alternative division ring, in which associativity survives for any product involving only two distinct elements. These are Ruth Moufang’s planes, from 1933.

The full Desargues theorem holds exactly when the coordinates form a division ring — associative, both distributive laws, but perhaps not commutative, as with the quaternions.

Pappus’s theorem — six points on two lines, and three meets on a third — holds exactly when the coordinates are a field.

Read downwards, each theorem is a stronger law. The surprise is what finiteness does to the list. Wedderburn proved in 1905 that every finite division ring is commutative, so in a finite plane Desargues implies Pappus. And Emil Artin and Max Zorn proved that every finite alternative division ring is associative, so in a finite plane Moufang’s condition implies Desargues. Three entries of the dictionary collapse into one when the plane is finite.

In infinite planes they do not. The octonions are an alternative division algebra that is not associative — the property lost at dimension eight — and they coordinatise a real projective plane in which little Desargues holds on every line and full Desargues does not. That plane is Moufang and non-Desarguesian, and no finite plane can be both. The eight-dimensional algebra and the order-nine plane are two sides of one table, one showing two entries that can be separated and the other showing that they cannot, once the plane is finite.

Nearfields are rare, and every one is known #

The nine-element system is not an isolated accident, but nearly.

Dickson’s twist generalises: start from a field of order $q^n$, split its non-zero elements into $n$ classes by a multiplicative index, and multiply by a different power of the Frobenius map in each class. The result is a nearfield whenever every prime factor of $n$ divides $q - 1$, with one extra condition when $4$ divides $n$. Order nine is $q = 3$, $n = 2$, the smallest case. Hans Zassenhaus proved in 1936 that these Dickson nearfields, together with exactly seven exceptional nearfields — of orders $5^2$, $11^2$, $7^2$, $23^2$, $11^2$, $29^2$ and $59^2$ — are all the finite nearfields there are.

So every finite nearfield that is not a field has prime-power order, and each builds a plane that is not Desarguesian, of the same kind as this one: a translation plane, with one line from which it looks classical. A Desarguesian plane coordinatised by a nearfield would force that nearfield to be a division ring, and Wedderburn’s theorem below would then make it a field.

None of this reaches beyond prime powers. The nearfield construction needs an additive group of prime-power order to start, exactly as the field construction does, so it enlarges the list of planes without enlarging the list of orders.

What a sample of configurations cannot establish #

A test on random configurations is evidence, not proof, in one direction. The failures are genuine: each is a configuration written out, checkable by hand, and one is enough to show the nearfield plane is not Desarguesian. The successes are not. Zero failures in 1,600 configurations in the field’s plane is consistent with the theorem, and the theorem there is proved by algebra, not by the sample. The zero on the line at infinity is the same: Baer’s theorem and the translations make it certain, and the forty configurations per line only agree with it.

The counts per line do not show which positions on a failing line are safe. A line with twenty failures in forty is a line where half the tested configurations closed. Some centre-and-axis pairs close for reasons of their own — a symmetry fixing that centre and that line — and a random sample averages them away. A complete census of every configuration would separate them; the counts here do not attempt it.

The pictures of the plane are tables. The only drawn figure is the one in the ordinary plane, where Desargues holds. There is no way to draw the nearfield plane with straight lines that are lines of it, because a drawing in the real plane inherits the real plane’s theorems. A failure of Desargues is visible only as a row of coordinates that does not satisfy an equation.

And isomorphism is not tested. That the nearfield plane is the Hall plane, and that it is not isomorphic to the field’s plane, rest on the classification. The failure count settles the second — an isomorphism would carry Desargues configurations to Desargues configurations — but naming the plane is a citation, not a computation.

Still open: whether a prime order forces a field #

Every non-Desarguesian plane anybody has found has an order that is a prime power, and never a prime. Every plane of order 2, 3, 5 and 7 is Desarguesian, and those were settled by exhaustion. Order 11 has only one known plane, the field’s, and nobody has classified it; the search that took order nine to a computer is far out of reach at eleven.

Is every projective plane of prime order Desarguesian? It is widely conjectured, and the evidence is thin in a characteristic way. A prime number of symbols admits no field but the integers modulo that prime, and every construction of a nearfield or quasifield of that size collapses back into it — but a plane need not come from any construction. The difference-set searches in a plane in a list of numbers have found no cyclic plane that is not Desarguesian, and it is conjectured that none exists; those searches, though, see only planes with a cyclic symmetry, and most of the question is about planes with no symmetry at all.

The problem beneath it is the one Fisher’s inequality and every other counting argument cannot touch: whether the order of a plane must be a prime power at all. A non-Desarguesian plane of non-prime-power order would need a coordinate system with no field anywhere in it, and nobody has an example of that, or a proof that there is none.

A theorem that reads the algebra back out #

The habit here is to treat a theorem as a detector rather than as a fact.

In the ordinary plane, Desargues’ theorem is something to prove. Among projective planes in general, it is a question to ask of each one, and the answer tells what the plane’s coordinates can do. The nearfield plane answers it in a way that is almost too informative: yes, from one line, exactly where translations exist, and no everywhere else, because the algebra behind it distributes from one side only.

That reverses the usual order of explanation. Coordinates are normally chosen first and geometry follows. Here the plane comes first, with nothing but its incidences, and the configuration theorems recover the algebra: which laws hold, which fail, and on which lines. Ninety-one points and ninety-one lines carry, unannounced, a record of a multiplication that nobody wrote into them.

The same detector works on any plane anyone produces. Given only its lines as lists of points, the configuration tests say whether a field could have built it, and if not, from which lines it still looks as though one had — a question about algebra answered entirely with straight lines.