Three ways to pair four roots

The group that will not come apart explained why the general quintic has no formula in radicals: its symmetry group would have to descend through a chain of subgroups with commutative steps, and at degree five the chain gets stuck. In passing it noted that the quartic’s chain has steps of sizes 2, 3 and 4, and that Ferrari’s formula of 1540 has “a resolvent cubic solved first, which is the size-3 step, sandwiched between two square roots”.

That sentence names an object without showing it. The resolvent cubic is not a trick of algebra. It is the answer to a question about four points: in how many ways can four things be split into two pairs? Three — and the three pairings of a quartic’s roots are exactly what the cubic’s three roots are. This essay draws them, and follows what they say about the formula and about the quartic’s symmetries.

Three numbers the roots cannot hide #

Label the roots of a quartic $r_1, r_2, r_3, r_4$. Split them into two pairs and multiply within each pair: $r_1 r_2$ and $r_3 r_4$. Add the products. The result, $\theta_1 = r_1 r_2 + r_3 r_4$, depends only on the pairing, not on the order of the pairs or of the roots inside them. There are three pairings — $\{12, 34\}$, $\{13, 24\}$, $\{14, 23\}$ — and so three numbers:

$$\theta_1 = r_1 r_2 + r_3 r_4, \qquad \theta_2 = r_1 r_3 + r_2 r_4, \qquad \theta_3 = r_1 r_4 + r_2 r_3.$$

For $x^4 + x + 1$ the roots are two conjugate pairs, and the figure draws each pairing in its own panel. The values come out real — $-1.861$, $-0.254$ and $2.115$ — even though no root is.

Now permute the roots in any way. The three $\theta$’s are permuted among themselves, because a permutation of four things sends a pairing to a pairing. So any symmetric function of $\theta_1, \theta_2, \theta_3$ — their sum, the sum of their pairwise products, their product — is unchanged by every permutation of the roots, and what the coefficients already know proved that such a function is a polynomial in the quartic’s coefficients. So the cubic with roots $\theta_1, \theta_2, \theta_3$ can be written down without finding any root at all. For $x^4 + a x^3 + b x^2 + c x + d$ it is

$$y^3 - b\,y^2 + (ac - 4d)\,y - (a^2 d - 4bd + c^2).$$

For $x^4 + x + 1$, where $a = b = 0$ and $c = d = 1$, that is $y^3 - 4y - 1$. The figure computed the four roots numerically, formed the three pairings from them, and substituted each into this cubic; each gives zero to the accuracy of the roots. Two routes to the same three numbers, one through the roots and one that never touches them.

From a pairing back to the roots #

A formula for the quartic needs the roots, and a pairing gets there in two square roots. Suppose $\theta_1 = r_1 r_2 + r_3 r_4$ is known, from solving the cubic. The product of all four roots is $d$, so $r_1 r_2$ and $r_3 r_4$ are two numbers with known sum $\theta_1$ and known product $d$ — the roots of the quadratic $t^2 - \theta_1 t + d$. One square root separates the pair products.

The same pairing splits the roots’ sum. With $r_1 r_2$ and $r_3 r_4$ in hand, the quartic factors into two quadratics, $(x^2 - s x + r_1 r_2)(x^2 - s' x + r_3 r_4)$, where $s = r_1 + r_2$ and $s' = r_3 + r_4$; matching coefficients gives $s + s' = -a$ and a linear equation for the other combination. Each quadratic factor then yields its two roots by a second square root. Three stages: a cubic, a square root, a square root — the “sandwich” of the earlier essay, and Ferrari’s formula when it is all written out in one expression.

The order matters and the figure shows why. Nothing about the quartic distinguishes one pairing from another, so no formula in the coefficients can produce $\theta_1$ without also producing $\theta_2$ and $\theta_3$. The three have to be produced together, as the roots of one cubic, and only then can one be chosen. The cubic is the smallest equation the quartic’s symmetry allows.

Twenty-four symmetries seen as six #

The deeper reason the cubic exists is a fact about the group of permutations of four things. There are 24 permutations, and each one moves the three pairings somehow. Sorting the 24 by what they do to the pairings puts them in six columns, one for each of the six permutations of three things — and every column holds exactly four.

The first column holds the permutations that fix every pairing. There are four: doing nothing, and the three double swaps $(12)(34)$, $(13)(24)$, $(14)(23)$, each of which exchanges the two roots in both pairs of one pairing and so preserves all three. These four form a group of their own, Klein’s four-group $V_4$, and it is a normal subgroup: the permutations in any other column are one fixed permutation composed with the four in the first. That is the structure the blocks a subgroup cuts out builds — cosets of a normal subgroup forming a group of their own — and here the quotient $S_4/V_4$ has six elements and is the symmetric group $S_3$.

Read through the lattice that runs the other way, that is the resolvent in the language of fields. The subgroup $V_4$ fixes exactly the numbers that the double swaps cannot move, and the pairings $\theta_1, \theta_2, \theta_3$ are such numbers; the field they generate is the fixed field of $V_4$, and its symmetry group is the quotient $S_3$ — the symmetry group of a cubic. Solving the resolvent cubic is climbing from the bottom field to the fixed field of $V_4$. The remaining two square roots climb the rest of the way, through the two steps of size 2 in $V_4$ itself.

One quartic, taken all the way down #

The three stages are easiest to follow on a quartic whose answer is known in advance. $x^4 - 10x^2 + 1$ has the roots $\pm\sqrt 2 \pm \sqrt 3$, but suppose that is not known. Here $a = c = 0$, $b = -10$ and $d = 1$, so the resolvent is

$$y^3 + 10y^2 - 4y - 40 = (y + 10)(y - 2)(y + 2),$$

and all three pairings are rational: $-10$, $2$ and $-2$. Take $\theta = 2$. The two pair products have sum 2 and product $d = 1$, so they are the roots of $t^2 - 2t + 1$, and both equal 1. The quartic therefore splits as $(x^2 - s x + 1)(x^2 + s x + 1)$ — the second pair’s sum is $-s$, since all four roots add to $-a = 0$ — and multiplying out gives $x^4 + (2 - s^2)x^2 + 1$. Matching the $x^2$ coefficient, $2 - s^2 = -10$, so $s = 2\sqrt 3$: one square root. The factor $x^2 - 2\sqrt 3\,x + 1$ then has roots $\sqrt 3 \pm \sqrt 2$: a second square root.

Every step is a quadratic, because the cubic happened to factor completely — which is the $V_4$ verdict of the table below, arrived at by hand. The roots are the nested radicals every step is a square root builds towers of, and the tower here has two storeys of degree 2, as a tower whose degrees multiply requires of a field of degree 4 reached by square roots alone. For $x^4 + x + 1$ the cubic does not factor, so solving it needs a square root and then a genuine cube root, and the whole tower has degrees $2 \cdot 3 \cdot 2 \cdot 2 = 24$ — one storey for each step of the chain $S_4 \supset A_4 \supset V_4 \supset C_2 \supset 1$.

Where the idea came from #

Ferrari found the method in 1540, and his teacher Cardano published it in the Ars Magna of 1545, but neither saw it as a statement about pairings. That reading is Lagrange’s, from his long memoir of 1770–71, Réflexions sur la résolution algébrique des équations. Lagrange asked why the methods for degrees three and four worked, and found the same pattern in both: each formula first computes a function of the roots that takes fewer values than the roots do when the roots are permuted. For the cubic that function takes two values, and they are the roots of a quadratic; for the quartic, $r_1 r_2 + r_3 r_4$ takes three values under all 24 permutations, and they are the roots of a cubic.

He then looked for such a function for the quintic — a function of five roots taking fewer than five values but more than two — and could not find one. Ruffini and Abel turned the failure into a proof, and Galois turned the counting of values into the counting of cosets that the column figure above displays. The number of values a function of the roots takes is the index of the subgroup that fixes it. Three values means that the permutations fixing $r_1 r_2 + r_3 r_4$ form a subgroup of index 3 — eight permutations, the symmetries of a square — and the permutations fixing all three values at once form $V_4$, the normal subgroup of index 6 that the column figure isolates. Lagrange had found the subgroups; Galois saw that normality was the point.

Pairings that are real, and pairings that are not #

When all four roots are real, every pairing value is real too, and the three pairings are three ways of grouping four points on a line. The resolvent then has three real roots, and here is the historical irony: a cubic with three real roots is exactly the case in which Cardano’s formula passes through complex numbers — the casus irreducibilis — so Ferrari’s formula for a quartic with four real roots goes through the complex plane on its way from real coefficients to real roots. When the resolvent does not factor over the rationals, as here, there is no way round it: Hölder proved in 1891 that an irreducible cubic with three real roots cannot be solved by real radicals alone.

The contrast with the first example is worth stating exactly. There, the roots are complex and the pairings real; here, the roots are real and so are the pairings, but the route from one to the other is forced through numbers that are neither. The pairings of $x^4 + 8x + 12$ are all real although its roots are not, and they have a property that the pairings of $x^4 + x + 1$ lack: their resolvent’s discriminant is a perfect square. That single fact, invisible in the picture of the roots, cuts the quartic’s symmetry group in half.

The cubic decides the symmetry #

A quartic’s roots rarely have all 24 symmetries. The Galois group — the permutations that preserve every polynomial relation among the roots with rational coefficients — can be smaller, and the resolvent is how to tell.

The rule follows from what the pairings are. A symmetry of the roots permutes the pairings; if a pairing’s value is rational, every symmetry must fix it, since a symmetry fixes every rational number. So the rational roots of the resolvent are the pairings the Galois group cannot move. No rational root: the group moves every pairing, and it is $S_4$ or $A_4$. One rational root: one pairing is fixed, and the group lies inside the eight permutations that preserve it — the symmetries of a square, $D_4$, or its rotation subgroup $C_4$. Three rational roots: every pairing is fixed, and the group lies inside $V_4$.

The table runs the rule on four examples. $x^4 + x + 1$ has an irreducible resolvent and a non-square discriminant, so its group is all of $S_4$. $x^4 + 8x + 12$ has an irreducible resolvent and a square discriminant, and a square discriminant means every symmetry is an even permutation — the crossings that will not come out even is where even and odd permutations are defined — so its group is $A_4$, of order 12. $x^4 - 2$ has resolvent $y^3 + 8y$ with the single rational root 0, and its group is $D_4$, the eight symmetries of a square, which is exactly the shape its roots make in the plane. $x^4 + 1$ has resolvent $y^3 - 4y$, all of whose roots are rational, and its group is $V_4$: its roots are the primitive eighth roots of unity, and the only symmetries are the double swaps.

For each row, the rational roots found by exact search in the resolvent were matched against the pairing values computed from the numerical roots — the ones that came out whole numbers. The two computations share nothing but the quartic, and they agree.

What the panels and the table cannot show #

The pictures show the roots, and the symmetry is invisible in them. The panels for $x^4 + x + 1$ and $x^4 + 8x + 12$ look alike — two conjugate pairs, real pairing values — and their Galois groups differ by a factor of two. What separates them is whether a single integer is a perfect square, and nothing in the drawing encodes that.

The dihedral row is incomplete. One rational pairing leaves two possibilities, $D_4$ or $C_4$, and telling them apart needs one more test — whether a certain quadratic splits over the field of the rational pairing — which the table does not run. For $x^4 - 2$ the answer is $D_4$, and the row names both.

And the numerical roots are approximations. The pairings were formed from roots found to about twelve places and checked against a cubic computed exactly; a pairing was called rational when it came within $10^{-7}$ of a whole number. That is a test that could in principle be fooled by an irrational number extraordinarily close to an integer, and the exact rational-root search is what makes the verdict certain.

Still open: which groups a polynomial can have #

Every group in the table occurs as the Galois group of some quartic over the rationals, and so do the others of degree four; for degree $n$, the symmetric group $S_n$ occurs, and most polynomials have it. The reverse question is the inverse Galois problem: is every finite group the Galois group of some polynomial with rational coefficients?

It is open. Shafarevich showed that every solvable group occurs, every symmetric and alternating group is known to occur, and so do all but one of the twenty-six sporadic simple groups — the exception being the Mathieu group $M_{23}$, whose status is unknown. For general finite groups there is no method, and Hilbert’s approach through irreducibility, which handles the symmetric groups, needs a special geometric input that most groups have not been shown to supply.

The question can be put on a single quartic’s scale. Given a group of order 8, such as $D_4$, it is easy to write down a quartic with that group; given an arbitrary finite group with a few hundred elements, nobody knows in general how to write down a polynomial of any degree whose roots have exactly that symmetry. The resolvent reads a group off a polynomial; the inverse problem asks for a polynomial given the group, and for most groups that direction has no algorithm at all.

A cubic made of pairs #

The resolvent cubic is not a device for making the quartic easier; it is the quartic’s own structure on display. Four roots have three pairings, the pairings’ values are a cubic’s roots, and the cubic can be written from the coefficients because permuting the roots only permutes the pairings. Ferrari’s formula is that cubic followed by two square roots, and the group-theoretic chain $S_4 \supset V_4 \supset \cdots$ is the same sequence of steps seen as symmetries: the six permutations of the pairings, then the four double swaps.

The same move explains why the method stops. A quintic’s 120 symmetries have no normal subgroup of the right kind — no function of five roots takes a handful of values the way the pairings do — so there is no resolvent of lower degree to solve first, and the chain the group that will not come apart could not find is the same resolvent this essay could.

When a set of roots has a symmetry, look for the combinations of them the symmetry only shuffles — the pairings here, the Gauss periods that the opening essay on roots of unity used for the seventeen-gon — because an equation for those combinations has fewer roots than the original, and its coefficients can be found without solving anything.