The quartics whose roots stay paired

The cubic census had only three outcomes to count, and the answer was simple: the full group almost always, reducible cubics about $15H^2$ times, and cyclic ones about $5H^{3/2}$ times. A quartic has more room. Its four roots can be permuted in twenty-four ways, and an irreducible quartic’s Galois group can be any of five transitive groups: all twenty-four, $S_4$; the twelve even permutations, $A_4$; the eight symmetries of a square, $D_4$; the four rotations of a square, $C_4$; or the four permutations that swap the roots in two pairs, Klein’s group $V_4$.

There is no reason in advance to expect these five to be equally common, and no obvious reason to expect any particular order among them. The natural guess is that larger groups are commoner, since a larger group imposes fewer coincidences on the roots. This essay sorts every monic quartic $x^4 + ax^3 + bx^2 + cx + d$ with coefficients up to 30 — $61^4 = 13{,}845{,}841$ of them — and the guess is wrong in a precise way.

How a resolvent decides five groups #

Three ways to pair four roots built the instrument. Split the four roots $\alpha_1, \ldots, \alpha_4$ into two pairs; there are three ways, and each gives a number such as $\alpha_1\alpha_2 + \alpha_3\alpha_4$. Those three numbers are the roots of a cubic whose coefficients are computable from the quartic’s own,

$$R(x) = x^3 - bx^2 + (ac - 4d)x - (a^2d - 4bd + c^2),$$

and the Galois group acts on them by acting on the pairings. So the group can be read from $R$. If $R$ has no rational root, the group moves all three pairings around, and it is $S_4$ or $A_4$, separated as in the cubic case by whether the discriminant is a square. If $R$ splits into three rational roots, every pairing is fixed and the group is $V_4$. If $R$ has exactly one rational root, one pairing is fixed and the group lies inside the symmetries of a square, $D_4$ or $C_4$.

Telling those last two apart needs one more test, due to Luise-Charlotte Kappe and Bette Warren in 1989. If $\theta$ is the rational root of $R$, the fixed pairing splits the quartic into two quadratics over a quadratic field, and the group is the cyclic $C_4$ exactly when both $x^2 - \theta x + d$ and $x^2 + ax + (b - \theta)$ split over $\mathbb{Q}(\sqrt D)$, where $D$ is the discriminant. That is a question about whether two whole numbers, each multiplied by $D$, are squares.

Before any of this the quartic must be irreducible, which is two tests rather than one. A rational root must be a whole-number divisor of $d$. And a split into two quadratics $(x^2 + px + q)(x^2 + rx + s)$ with whole coefficients needs $qs = d$; for each divisor $q$ the remaining equations determine $p$ and $r$ or rule them out. Each step is exact arithmetic on whole numbers, and every classical example — $x^4 + x + 1$ with $S_4$, $x^4 + 8x + 12$ with $A_4$, $x^4 - 2$ with $D_4$, $x^4 + 5x^2 + 5$ with $C_4$, $x^4 + 1$ with $V_4$ — comes out with its known group.

The plane with no full group #

The hero figure shows the census’s first surprise before any counting. On the plane of even quartics, $x^4 + bx^2 + d$, not one of the 1,681 quartics has the full group. There are 1,338 with $D_4$, 124 with $V_4$, 20 with $C_4$, and 199 that factor.

The reason is a symmetry the even quartics carry in their shape. If $\alpha$ is a root of $x^4 + bx^2 + d$, so is $-\alpha$, and the four roots come in two pairs $\pm\alpha, \pm\beta$. Any symmetry of the roots that respects the field’s arithmetic must send $-\alpha$ to the negative of wherever it sends $\alpha$, so it must carry pairs to pairs. That keeps one pairing fixed, and the permutations that keep a pairing fixed are exactly the eight symmetries of a square. So $D_4$ is the most an even quartic can have.

Within the plane there is a further rule, visible as rows in the figure. $V_4$ occupies the rows where $d$ is a perfect square, and $C_4$ the scattered points where $d(b^2 - 4d)$ is a perfect square. Everything else irreducible is $D_4$. The census checks this square-class rule on every irreducible quartic in the plane and finds no exception. Its content is short: $\alpha^2$ and $\beta^2$ are the roots of the quadratic $y^2 + by + d$, so their product is $d$, and when $d$ is a square, $\alpha\beta$ is rational and the extra symmetry that would swap $\alpha\beta$ with $-\alpha\beta$ is lost.

Add the single term $x$ and everything changes. On the right-hand plane, $x^4 + bx^2 + x + d$, the pairing symmetry is broken and 1,524 of the 1,681 quartics have the full group; the other 157 factor. Not one has any of the four intermediate groups. The smaller groups are not scattered evenly through the box. They are concentrated on special surfaces, of which the even quartics are the most visible.

Five groups, and not in order of size #

Now the whole box, $H$ from 2 to 30, with a running count for each group.

The full group takes 94.8% of the box at $H = 30$, and the reducible quartics 4.6%; together they leave a sliver of 0.67% for the four smaller groups. Within the sliver $D_4$ comes first and $C_4$ last in every box from $H = 8$ on. Between them, $A_4$ and $V_4$ change places five times between $H = 14$ and $H = 21$, and $A_4$ stays ahead from there to the end, and at $H = 30$ the order is $D_4$, $A_4$, $V_4$, $C_4$.

The order is not the order of size. $A_4$ has twelve elements and $D_4$ eight, yet at $H = 30$ there are thirty-five quartics with group $D_4$ for every one with $A_4$. $V_4$ and $C_4$ both have four elements, and $V_4$ is nearly three times as common.

The order follows the kind of coincidence each group demands rather than its size. $D_4$, $C_4$ and $V_4$ all need the resolvent cubic $R$ to have a rational root. That is the same kind of condition as a cubic being reducible, and the cubic census found reducibility to be the commonest coincidence there is, costing a share of only about $1/H$. $A_4$ instead needs the discriminant to be a perfect square while $R$ stays irreducible — a squareness condition on a number of size about $H^6$, which in the cubic census cost much more. $V_4$ needs $R$ to split completely, a stronger condition than one root, and $C_4$ needs one root and a square on top of it.

The exponents, box by box #

The counts suggest exponents, and the slopes between box sizes measure them.

The reducible quartics are the only clear line: an exponent of 2.90 at the smallest pair rising to 2.95 at the largest, approaching the 3 that van der Waerden’s bound allows and that the reducible count, like its cubic counterpart, is known to reach. Against a total growing like $H^4$, that is again a share falling like $1/H$.

The four smaller groups are less tidy. $D_4$ runs between 2.56 and 2.73 with no clear trend. $A_4$ falls steeply from 3.5 at the smallest boxes — where it has only a handful of members — to 2.3 at the largest. $V_4$ sits near 2.2 and $C_4$ drifts from 1.95 down to 1.71.

These numbers say two firm things. Every smaller group grows more slowly than the reducible quartics, so van der Waerden’s law holds for quartics with the same structure as for cubics: almost all exceptions are reducible. And $C_4$ grows more slowly than every other group, so its share of the sliver is falling. What the numbers do not say is the true exponent of any of the four, because a box of side 61 is far too small. An exponent that has moved by 0.2 between boxes of 15 and 30 can move by as much again before 1,000. Bounds have been proved for each of these groups, with exponents that differ from group to group, but which of them is sharp, and whether logarithmic factors accompany the powers, is not settled for all five.

Where the dihedral quartics come from #

The even quartics explain why no even quartic has the full group. They also suggest a reason for $D_4$’s abundance: every irreducible even quartic is $D_4$, $C_4$ or $V_4$, and an even quartic is a two-parameter family, so on its own it would supply about $H^2$ members to those groups.

A quartic is a translate of an even one when the substitution $x \to x - a/4$ removes both its cube term and its linear term, which happens exactly when $8c - 4ab + a^3 = 0$. Counting those along with the even quartics themselves accounts for 22% of the $D_4$ quartics in the box with $H = 20$, 42% of the $V_4$ and 17% of the $C_4$ — and, as the pairing argument requires, none of the $A_4$.

So the even quartics are a real source but not the main one. Most dihedral quartics are not translates of even ones. They have a rational root of the resolvent for other reasons, scattered through the box rather than on a plane. The plane is where the symmetry is forced. Elsewhere it is a coincidence among the coefficients, and those coincidences outnumber the forced cases by three and a half to one.

The same split shows why $A_4$ is so rare. It has no forcing family of comparable size: no linear condition on the coefficients makes the discriminant a square while keeping the resolvent irreducible. Every $A_4$ quartic in the box is a coincidence of the squareness kind, and those cost more than coincidences of the root kind.

The cyclic quartics and two squares #

The rarest group has a property no other group in the census shares, and it ties the census to one of the oldest results in number theory.

The group $C_4$ has exactly one subgroup of index two, so the field of a cyclic quartic contains exactly one quadratic field, and it is $\mathbb{Q}(\sqrt D)$. Remove every square factor from $D$ and what remains names that field. For the 406 cyclic quartics with coefficients up to 20, the commonest names are 5, then 2, 13 and 17. Every one is positive, and every one is a sum of two squares: $5 = 1^2 + 2^2$, $2 = 1^2 + 1^2$, $13 = 2^2 + 3^2$, $17 = 1^2 + 4^2$, and so on through 137. The census checks all 406 and finds no prime leaving remainder 3 on division by 4 in any squarefree part.

This is not a coincidence of the box. A quadratic field $\mathbb{Q}(\sqrt k)$ sits inside a cyclic extension of degree four exactly when $k$ is a sum of two rational squares — a classical theorem on embedding fields — which by Fermat’s two-squares theorem means that no prime $\equiv 3 \pmod 4$ divides $k$ to an odd power. So the cyclic quartics are confined twice. Their resolvent must have a rational root, which puts them among the dihedral-type coincidences. And their discriminant must have a squarefree part of a special arithmetic kind, a condition that excludes about half of all numbers at every size and which no other group in the census imposes.

The commonest field, $\mathbb{Q}(\sqrt 5)$, is the one inside the fifth roots of unity, and the quartic $x^4 + x^3 + x^2 + x + 1$ that defines them is cyclic. The second, $\mathbb{Q}(\sqrt 2)$, sits inside the sixteenth roots of unity, whose real part is the field of $x^4 - 4x^2 + 2$. Just as every cyclic cubic in the cubic census was secretly a sum of roots of unity, so is every cyclic quartic here, and the commonest ones come from the smallest such fields.

The same law in two degrees #

Set the two censuses side by side.

In both degrees, $H$ times the share of exceptions levels off at a constant: about 1.86 for cubics and 1.57 for quartics. That is van der Waerden’s law in its exact form — Bhargava’s theorem that the exceptions number at most a constant times $H^{n-1}$ among roughly $(2H)^n$ polynomials — and in both degrees nearly all the exceptions are reducible.

The two degrees differ in what lies beneath that line. A cubic has one smaller transitive group and a quartic four, and the quartic’s four arrive in an order set by the kind of coincidence each requires: root-type coincidences of the resolvent first, square-type ones after, and those that need both last. The order would be the same in any degree. A polynomial whose group fixes some structure among the roots — a pairing, a block, a partition — can be found where an auxiliary polynomial has a rational root, and that is cheap. A polynomial whose group lies inside the even permutations needs a square, and that is expensive.

What the box cannot show #

The census is a list of facts about the box of side 61, and the strongest statement it supports about larger boxes is the one already proved: exceptions are a share of order $1/H$, mostly reducible. $D_4$ first and $C_4$ last holds from $H = 8$ to $H = 30$, but an order among four counts growing at different, still-moving rates is the kind of fact that could change at a size no census can reach. $A_4$ and $V_4$ show how fragile it is: they swap places five times inside the box, are within 16% of each other at $H = 30$, and $A_4$’s exponent has fallen from 3.5 to 2.3 across the boxes while $V_4$’s has stayed near 2.2.

The figures also do not show the roots. The group is certified by arithmetic on the coefficients — the resolvent’s rational roots, two square tests and Kappe and Warren’s condition — and the census never computes a root. For any particular quartic the symmetries can be made visible, as three ways to pair four roots did by drawing the pairings, but the 768 cyclic quartics in the box are known to be cyclic without anyone having seen their rotation.

And the census is of monic quartics only. Allowing a leading coefficient other than 1 changes the counts and their constants. It also lets in polynomials with rational roots that are not whole numbers, which changes the reducible count’s structure, though not the exponent van der Waerden’s law predicts.

Still open: the true exponent for each group #

For each of the four intermediate groups, the box gives an exponent between 1.7 and 2.7 and a direction of drift, and theory gives bounds that do not yet meet in every case. Pinning down the exact order of growth of the $A_4$ quartics, or of the $C_4$ quartics, is a question about counting number fields with a prescribed group together with the polynomials that generate them. The quantity that has to be controlled is the same index that the integers a field contains measured: how much larger the discriminant of a polynomial is than that of its field. Polynomials with small coefficients tend to have small fields, but the relation between the two sizes is loose in exactly the way that makes the counts hard.

There is also a question this census frames and cannot answer. A random quartic, chosen from a large box, has the full group with probability tending to one. But a quartic chosen from a special surface — the even plane, or any other family defined by a polynomial condition on the coefficients — may have a smaller group always, as the even quartics do. Which families force a smaller group, and which merely make it more likely, is Hilbert’s irreducibility theorem asked in reverse. The answer is known for families of the simplest shapes and not in general.

A symmetry forced, and a symmetry found #

The quartics with a smaller group come in two kinds. Some have it because of their form: an even quartic cannot have more than eight symmetries, because its roots come in pairs that nothing can separate, and the census confirms that not one of the even quartics in the box escapes. Others have it by coincidence — a rational root of the resolvent where none was forced, or a square discriminant — and those are scattered through the box at rates set by how costly each coincidence is.

The forced kind is the more visible and the less numerous. It accounts for a fifth of the dihedral quartics and two-fifths of the Klein ones, and none of the quartics with group $A_4$. The coincidental kind is most of the sliver. In both kinds, what decides which group a quartic gets is not how large the group is but what the coefficients must satisfy for the roots to have it. That is why the eight symmetries of a square outnumber the twelve even permutations thirty-five to one, and why the lattice of fields that an even quartic generates always has the same shape.