A line of cubics that are all cyclic

The cubic census sorted every monic cubic with coefficients up to 150 by its Galois group and found the full group of six permutations in 98.7% of them. The three rotations, the cyclic group $A_3$, turned up in about $5H^{3/2}$ cubics of a box of side $H$ — a share that falls like one over the square root of the volume and tends to nothing. The quartic census found the same law one degree up, and ended on a question the boxes could frame and not answer: a cubic drawn at random from a box has the full group almost surely, but some families of cubics may have a smaller group always. Which families force the smaller group, and why?

For quartics the even ones $x^4 + bx^2 + d$ were such a family: none of them has the full group $S_4$, because their roots come in pairs $\pm r$. For cubics the answer is less visible and more beautiful. In 1974 Daniel Shanks wrote down the cubics

$$f_t(x) = x^3 - t\,x^2 - (t + 3)\,x - 1,$$

and every one of them, for every whole $t$ and indeed every rational $t$, has the three rotations as its group. They lie on a straight line in the space of coefficients, and that line runs through a region where the rotations are vanishingly rare.

The test the family passes #

The quintic essay gave the test that separates the two possible groups of an irreducible cubic: the group is the three rotations exactly when the discriminant is a perfect square. For the cubic $x^3 + ax^2 + bx + c$ the discriminant is

$$a^2b^2 - 4b^3 - 4a^3c - 27c^2 + 18abc,$$

and substituting $a = -t$, $b = -(t + 3)$, $c = -1$ and expanding gives, after the terms collapse,

$$t^4 + 6t^3 + 27t^2 + 54t + 81 = (t^2 + 3t + 9)^2.$$

The discriminant is not merely a square for many values of $t$; it is the square of a polynomial, so it is a square for every value at once. The hero checks it at the marked member: at $t = 2$ the cubic is $x^3 - 2x^2 - 5x - 1$, and its discriminant is $19^2 = 361$.

Irreducibility is just as uniform. A monic cubic with whole coefficients that factors has a whole-number root dividing its constant term, so the only candidates are 1 and $-1$. But $f_t(1) = -2t - 3$, which is odd and never zero, and $f_t(-1) = 1$ for every $t$. So no member of the family factors over the rationals, every member has a square discriminant, and every member has the cyclic group. The family is a line of exceptions.

The roots in the hero are real at every $t$, as they must be: a square discriminant is a positive one, and a cubic with positive discriminant has three real roots. For large $t$ one root runs off to infinity, close to $t + 1$, while the other two settle near $0$ and $-1$, at about $-1/t$ and $-1 - 1/t$. The picture’s left half mirrors its right half through $x \mapsto 1/x$: the member at $-3 - t$ has as its roots the reciprocals of the roots at $t$, which is the family’s one obvious symmetry and a hint that it has a less obvious one.

A map that turns the roots #

The less obvious one is in the hero’s list. At $t = 2$ the roots are $3.50702$, $-0.22188$ and $-1.28514$, and the map $x \mapsto -1/(1 + x)$ sends the first to the second, the second to the third and the third back to the first. This is not special to $t = 2$. Substitute $-1/(1 + x)$ into $f_t$ and clear denominators:

$$(1 + x)^3 f_t\!\left(\frac{-1}{1 + x}\right) = -f_t(x),$$

an identity in both $x$ and $t$. So whenever $\rho$ is a root, so is $-1/(1 + \rho)$, and so is the root after that. Applying the map three times returns every number to itself — as a matrix it is $\begin{pmatrix} 0 & -1 \\ 1 & 1 \end{pmatrix}$, whose cube is minus the identity — so it cycles the three roots of each member without fixing any.

This is the Galois group made visible, and visible in a strong form. For a general cyclic cubic the theorem says that some rational function of a root gives the next root, but says nothing about which function, and it changes from cubic to cubic. Here it is one function for the whole family, a single Möbius transformation with rational coefficients, and the family is precisely the cubics whose roots it permutes. That is the reason the discriminant is a square: the square root of the discriminant is $(\rho_1 - \rho_2)(\rho_2 - \rho_3)(\rho_3 - \rho_1)$, a quantity unchanged by cycling the roots, and since cycling is the only symmetry available it must be rational.

There is a converse with a long reach. Every cyclic cubic field — every field of degree three whose symmetries are the rotations — is generated by a root of $f_t$ for some rational value of $t$. The family is, in the language of Galois theory, a generic polynomial for the cyclic group of order three: specialise its one parameter and every cyclic cubic field appears, while every specialisation gives a cyclic field or nothing new. The whole of the cubic census’s rare category lives on this one line in disguise, each field reached at some fraction $t$ even when the box found it at quite different coefficients.

Equilateral triangles on a circle #

A Möbius transformation of order three has two fixed points, and here they are the roots of $x^2 + x + 1$, the complex cube roots of unity $\omega$ and $\bar\omega$ — the same numbers that, as chaos on the line between two roots found for Newton’s method, a change of variable can carry to $0$ and $\infty$. Make that change, $z = (x - \omega)/(x - \bar\omega)$. It sends the real line onto the unit circle, since a real $x$ is equally far from $\omega$ and from its conjugate, and it turns the map $x \mapsto -1/(1 + x)$ into multiplication by a cube root of unity, a plain turn of the circle by a third.

Now every member of the family is three points on the circle that a turn by a third permutes: an equilateral triangle. The figure places the roots of six members and checks the side lengths, which agree to a millionth of the radius. The member is decided by where one corner sits, and the table records that position modulo 120°: $67.3°$ at $t = -1$, $80.0°$ at $t = 0$, rising towards $120°$ as $t$ grows. Conversely any equilateral triangle on the circle with its corners at real points — which on this circle is every point except the image of infinity — is the root set of a cubic whose coefficients are fixed by the triangle, and that cubic is $f_t$ for the corresponding $t$. The family is the circle’s worth of triangles, one for each angle, and the parameter $t$ is an angle in disguise.

So a question about Galois groups has become a question about a regular polygon. A cyclic cubic, seen through the right coordinate, is a triangle that a rotation preserves, in exactly the way that the roots of $x^3 - 1$, which the polygon an equation forces drew, are a triangle that the same rotation preserves. The difference is that those roots are fixed and these slide round the circle with $t$.

A diagonal through the box #

To see how exceptional the line is, the figure takes the slice of the cubic box in which the constant term is $-1$ and colours every cubic $x^3 + ax^2 + bx - 1$ with $a$ and $b$ between $-40$ and $40$ by its group, deciding each exactly: it has a rational root if $1$ or $-1$ is a root, and otherwise its group is the rotations exactly when the discriminant, computed in whole numbers, is a perfect square.

Of the 6,561 cubics in the slice, 159 factor and 6,276 have the full group. The remaining 126 are cyclic, and 78 of them lie on one line, $b = a - 3$ — Shanks’s family with $t = -a$, filled along its whole length inside the square. The other 48 are scattered, without a pattern the eye can follow. They are cyclic for reasons of their own, each one a coincidence of whole numbers; and each of them, by the converse above, has a field that some member of the family also generates, at a value of $t$ that is a fraction rather than a whole number.

The slice also shows why the boxes never noticed the line. A box of side $H$ meets it in about $2H$ cubics, while the cyclic cubics in the box number about $5H^{3/2}$. So the family’s whole-number members are a share of the box’s cyclic cubics that itself falls like $1/\sqrt H$. The census counts coincidences scattered through the box, and the family is a structure that the census sees as a thin and shrinking sliver of them.

Why one family can and another cannot #

The real question is which families do this, and the discriminant turns it into geometry. A one-parameter family of irreducible cubics has a square discriminant at $t$ exactly when there is a number $y$ with $y^2 = \mathrm{disc}(t)$, which is to say when the curve $y^2 = \mathrm{disc}(t)$ has a point above $t$. For Shanks’s family that curve is $y^2 = (t^2 + 3t + 9)^2$, which falls apart into two parabolas $y = \pm(t^2 + 3t + 9)$, each with a point above every $t$ whatever. The family is cyclic everywhere because its discriminant curve contains a copy of the line.

Try a family that looks just as innocent, $x^3 - tx - 1$. Its discriminant is $4t^3 - 27$, which is not the square of anything, and the curve $y^2 = 4t^3 - 27$ is an elliptic curve. Elliptic curves have only finitely many points with whole coordinates — Carl Ludwig Siegel’s theorem of 1929 — so the family can be cyclic only finitely often among whole $t$. The search runs through every $t$ up to a million and finds one: $t = 3$, where $4 \cdot 27 - 27 = 81 = 9^2$. The cubic $x^3 - 3x - 1$ is cyclic, its roots being $2\cos(\pi/9)$, $2\cos(7\pi/9)$ and $2\cos(13\pi/9)$, and it is the only cyclic member of its family in the range searched. The curve itself is an old acquaintance: multiplying through by 16 and writing $X = 4t$, $Y = 4y$ turns it into $Y^2 = X^3 - 432$, which is the elliptic curve hiding in Fermat’s equation $u^3 + v^3 = 1$ — the curve whose shortage of rational points is Fermat’s last theorem for cubes, the case Euler settled. A question about which cubics have the rotations as their group has landed on the exponent three of Fermat.

That is the general shape of the answer, and it is Hilbert’s irreducibility theorem read in reverse. A family’s group at a typical parameter is its group over the field of rational functions in $t$ — where the discriminant either is or is not a square of a polynomial — and Hilbert proved in 1892 that every whole-number specialisation outside a thin set keeps that group. So a cubic family is cyclic always when its discriminant is a square in $t$, as Shanks’s is; it is cyclic at a thin set of $t$ when the discriminant curve has a few points, as $x^3 - tx - 1$ is; and which kind a family is can be read off one polynomial identity.

A norm in the discriminant #

The square root of the discriminant has its own structure. With $\omega$ again a complex cube root of unity,

$$t^2 + 3t + 9 = (t - 3\omega)(t - 3\bar\omega),$$

the norm of the Eisenstein integer $t - 3\omega$ — a whole number of the form $a + b\omega$, in the system of integers which primes a form takes used for $x^2 + xy + y^2$. A prime that leaves remainder 2 on division by 3 stays prime among the Eisenstein integers, so it can divide a norm only to an even power. The table factorises $t^2 + 3t + 9$ for $t$ from $-1$ to 12, and every prime in it is 3 or leaves remainder 1; the check runs to $t = 2{,}000$ without exception.

Those are exactly the primes that can be ramified in a cyclic cubic field, the only primes that can divide its conductor. When $t^2 + 3t + 9$ is itself a prime $p$ — at $t = -1, 1, 2, 4, 7, 8, 10, 11$, giving $p = 7, 13, 19, 37, 79, 97, 139, 163$ — the field the member cuts out is the unique cubic field inside the $p$-th roots of unity, the field generated by the Gauss periods that the sums that obey a smaller equation built from sums of roots of unity. At $t = -1$ the cubic is $x^3 + x^2 - 2x - 1$, the minimal polynomial of $2\cos(2\pi/7)$, the period equation for the heptagon. Shanks’s family runs through the period cubics of a sequence of primes, and fills in between them with fields of composite conductor.

Units with nothing to spare #

The constant term of $f_t$ is $-1$, so the three roots multiply to 1, and each is an algebraic integer whose inverse is also an algebraic integer: each root is a unit of its field. Two of them, $\rho$ and its image $-1/(1 + \rho)$, are independent, and by Dirichlet’s theorem a cyclic cubic field, all three of whose embeddings are real, has a unit group of rank two. The regulator — the area of the parallelogram the units’ logarithms span — measures how large units must be. For these two it is computed in the figure for $t$ up to 200, and it tracks $(\ln t)^2$, within a few per cent by the end of the range.

That is about as small as possible. The roots are near $t$, near $-1/t$ and near $-1$, and no unit of a field of this size can be much smaller in logarithm than $\ln t$. Shanks’s reason for calling these the simplest cubic fields was what this forces. The class number formula ties the product of the class number and the regulator to the discriminant and an $L$-value, and that product grows roughly like the square root of the field’s discriminant, which for these fields is about $t^2$. With the regulator held down near $(\ln t)^2$, the class number must take up the rest, so the fields of this family have class numbers growing roughly like $t^2/(\ln t)^2$: they are a supply of cyclic cubic fields where unique factorisation fails badly, with units known in closed form. The same unit-for-unit trade appeared in the real fields that factorise uniquely with the roles reversed, where a large regulator left room for a class number of one.

What the pictures establish and what they do not #

The identities are exact. That the discriminant is $(t^2 + 3t + 9)^2$ is a polynomial identity, checked at every drawn member in whole-number arithmetic; that no member factors is a two-line argument; that the Möbius map permutes the roots is an identity, and the hero verifies the permutation numerically at the marked member. The equilateral triangles are an exact consequence of the map becoming a rotation, and the figure measures the sides to confirm the drawing rather than the fact.

The slice and the curve search are exact computations with finite reach. The slice decides each of its 6,561 cubics in whole numbers, so its counts are exact for that slice and say nothing about other slices beyond suggesting their shape. The search on $y^2 = 4t^3 - 27$ is exhaustive to a million and Siegel’s theorem guarantees finiteness, but no picture here proves that $t = 3$ is the only whole point on that curve; for this curve that is known by other methods, and the figure is evidence rather than proof. The regulator figure computes the regulator of the two units the roots supply, which is the regulator of the field only if those two generate the full unit group, and the comparison with $(\ln t)^2$ is a measurement over two hundred members, not an asymptotic theorem.

And the picture of a family as a line through the box shows one family. It does not show the other generic families for the cyclic cubic — there are many, related to Shanks’s by changes of parameter — or how their lines sit in the box together.

Still open: families of every shape #

For cubics the story is complete: the families that are always cyclic are the ones whose discriminant is a square in the parameter, and Shanks’s is the model for all of them. In higher degree the corresponding question is the inverse Galois problem with parameters. For a finite group $G$, is there a polynomial with coefficients in $\mathbb{Q}(t_1, \ldots, t_k)$ whose group is $G$ and whose specialisations give every field with group $G$ — a generic polynomial, as Shanks’s is for the cyclic group of order three? For many groups there is, including the symmetric groups and the cyclic groups of small order. For the cyclic group of order eight there provably is not over the rationals, a consequence of the failure of Grunwald’s theorem at the prime 2. For most groups nobody knows.

And for groups where any family at all is known, the best one is a further question. Shanks’s family is simple in a precise sense — one parameter, small coefficients, explicit units — and families of comparable simplicity for cyclic quartic, quintic and sextic fields were found by Marie-Nicole Gras and Emma Lehmer, with units as explicit and class numbers as large. Whether every abelian group has such a simplest family, and what simplest should mean when it does, is a question about how much of the structure of a whole class of fields can be packed into one polynomial.

A rotation written into the coefficients #

Every cubic $x^3 - tx^2 - (t + 3)x - 1$ is irreducible with discriminant $(t^2 + 3t + 9)^2$, so every one has the three rotations as its Galois group, and the rotation is the single map $x \mapsto -1/(1 + x)$, which on the right circle is a turn by a third that makes each member’s roots an equilateral triangle. The family is a straight line of cyclic cubics in a box where the full group fills nearly everything, and it reaches every cyclic cubic field at some rational $t$. Whether a family is cyclic always, sometimes or almost never is decided by the curve $y^2 = \mathrm{disc}(t)$: a parabola gives a point at every $t$, and the elliptic curve of $x^3 - tx - 1$ gives a point only at $t = 3$ among the first million.