Three real roots and no real radicals

The quintics that have a formula sorted equations of degree five into those whose roots can be written with radicals and those whose roots cannot, and the sorting was done by a group. The cubic seems to need no such sorting: every cubic is solvable, and Cardano’s formula of 1545 solves it. But “solvable by radicals” hides a question about which radicals, and for the cubic the answer is strange enough that it took three and a half centuries to settle.

Take $x^3 - 3x + 1$. Its graph crosses the axis three times, near $-1.879$, $0.347$ and $1.532$. Its coefficients are whole numbers, its roots are real, and nothing about it involves the square root of a negative number. Yet Cardano’s formula, applied to it, asks for the cube root of $-\tfrac12 + \tfrac{\sqrt3}{2}i$, and every route to its roots that anyone found in the sixteenth century went through the complex plane and came back. Making a square out of it named this the casus irreducibilis and said that it cannot be avoided by any rearrangement. This essay draws the detour, proves that it is forced, and follows the proof to a much more general statement about which real numbers real radicals can reach.

The detour, drawn #

Cardano’s formula looks for a root of $x^3 + px + q$ in the form $x = u + v$. Substituting and asking that $3uv + p = 0$ reduces the cubic to a quadratic for $u^3$ and $v^3$: they are the two roots of $t^2 + qt - p^3/27$. For $x^3 - 3x + 1$, with $p = -3$ and $q = 1$, the quadratic is $t^2 + t + 1$, whose roots are $-\tfrac12 \pm \tfrac{\sqrt3}{2}i$ — the two non-real cube roots of unity. The quadratic’s discriminant is negative precisely because the cubic’s is positive, and a cubic with three real roots has positive discriminant. So the numbers under Cardano’s cube roots are not real whenever all three answers are.

The figure takes the formula literally. $u^3$ sits on the unit circle at $120^\circ$ and $v^3$ at $-120^\circ$. Each has three cube roots, at a third of its angle and at that angle plus a third and two thirds of a turn: $40^\circ$, $160^\circ$ and $280^\circ$ for $u$, and the mirror images for $v$. The condition $uv = 1$ pairs each $u$ with its conjugate, so each sum $u + v$ is twice a real part, $2\cos 40^\circ$, $2\cos 160^\circ$, $2\cos 280^\circ$. The imaginary parts are present at every stage and cancel exactly at the last one. The three real roots are reached, correctly, by a computation that leaves the real line and returns to it.

Nothing in the figure says that the departure was necessary. It would be natural to suspect an artefact of Cardano’s particular substitution $x = u + v$, and to look for a cleverer arrangement — a different substitution, a nested expression of real square roots and real cube roots — that stays on the line throughout. Mathematicians looked from the sixteenth century to the nineteenth. The answer, when it came, was that every arrangement fails, and the reason is a symmetry argument that the next two figures set up.

Bombelli’s lucky cube root #

The detour was first taken seriously in a case where it could be checked by hand. Rafael Bombelli, in his Algebra of 1572, applied Cardano’s rule to $x^3 = 15x + 4$. The rule gives

$$x = \sqrt[3]{2 + \sqrt{-121}} + \sqrt[3]{2 - \sqrt{-121}} = \sqrt[3]{2 + 11i} + \sqrt[3]{2 - 11i},$$

an expression that Cardano himself had called useless. But $x = 4$ is obviously a root: $64 = 60 + 4$. Bombelli guessed that the two cube roots might be conjugate numbers of the form $a + bi$ and $a - bi$ with whole $a$ and $b$, adding to $2a = 4$, so $a = 2$; and indeed $(2 + i)^3 = 8 + 12i - 6 - i = 2 + 11i$. The sum is $(2 + i) + (2 - i) = 4$. It was the first calculation in which numbers with $\sqrt{-1}$ in them were manipulated by fixed rules and produced a correct real answer, and it is the first figure’s picture with whole numbers in it.

Bombelli’s cubic is a misleading example in one respect, and the difference is the whole subject of this essay. $x^3 - 15x - 4$ factors: it is $(x - 4)(x^2 + 4x + 1)$, and its other two roots, $-2 \pm \sqrt3$, are real square-root expressions. So its roots can all be written with real radicals, and the complex cube roots in Cardano’s formula happened to be ones whose cube roots could be guessed. For $x^3 - 3x + 1$, which does not factor, the cube roots of $-\tfrac12 + \tfrac{\sqrt3}{2}i$ are $\cos 40^\circ + i\sin 40^\circ$ and its companions, and no guess will turn them into anything real-radical — because, as the proof below shows, there is nothing there to find. The lucky case taught the sixteenth century that the imaginary detour could be trusted; the unlucky one, which is the typical one, is why it could not be avoided.

The same three roots, as a trisection #

The angles in the first figure are not incidental. Write a root as $x = 2\cos\varphi$. The identity $\cos 3\varphi = 4\cos^3\varphi - 3\cos\varphi$ turns $x^3 - 3x$ into $2\cos 3\varphi$, so the equation $x^3 - 3x + 1 = 0$ becomes

$$\cos 3\varphi = -\tfrac12.$$

That is François Viète’s solution of 1591, and it says that solving this cubic is the same as dividing an angle of $120^\circ$ into three. The general cubic with three real roots behaves the same way after scaling: $x^3 + px + q$ with $p < 0$ is solved by a cosine of a third of an angle whose cosine is read off $p$ and $q$.

So the casus irreducibilis is the trisection problem in algebraic dress. The mark that changes what is reachable showed that a marked ruler trisects any angle because a sliding mark solves a cubic of exactly this kind, and the angle that will not divide by three showed that compass and straightedge cannot. Viète’s picture adds a third angle on the same fact: the roots are real numbers defined by a real geometric operation — take a third of an angle — and the only algebraic operation that performs that division is a cube root of a point on a circle, which is a cube root of a complex number.

The connection also explains why the trouble is confined to cubics with three real roots. When the cubic has one real root, $p^3/27 + q^2/4$ is positive, $u^3$ and $v^3$ are real, and Cardano’s formula takes two real cube roots and adds them. The detour appears exactly when the corresponding angle exists, which is exactly when $|{-q/2}|$ is small enough to be a cosine.

What a real cube root cannot do #

The obstruction is visible in one picture. Every number other than nought has three cube roots. For a real number $c$, one of them is real and the other two are that one turned by a third of a turn either way.

Adjoining $\sqrt[3]{2}$ to the rationals gives a perfectly good field of real numbers. But the field is lopsided: the equation $x^3 = 2$ that defines its new element has three roots, and the field holds only one of them. The lattice that runs the other way explained why lopsided fields are special in Galois theory. A field generated by roots of a polynomial is normal when it contains every root of every irreducible polynomial that has one root in it; normal extensions are the ones that correspond to normal subgroups, and they are what the correspondence handles cleanly. A real cube root step is never normal over a real field, because a normal one would have to contain $\omega\sqrt[3]{c}$, which is not real.

Hold that against a fact about the cubic. Suppose $x^3 - 3x + 1$ has a root $\alpha$ in some field $K$ that already contains $\sqrt{D}$, the square root of the cubic’s discriminant. For this cubic $D = 81$ and $\sqrt D = 9$ is rational, so any field will do. Then the field $K(\alpha)$ is normal over $K$: the cubic’s group is a cyclic group of three, and one root determines the others — here explicitly, since if $\alpha$ is a root then so are $\alpha^2 - 2$ and $2 - \alpha - \alpha^2$. Adjoining one root of the cubic adjoins all three, and they are all real. Adjoining one cube root of a real number adjoins one, and the others are not real. Hölder’s theorem is the observation that these two behaviours cannot be the same step.

Hölder’s proof, in three moves #

Suppose, for a contradiction, that some root of $x^3 - 3x + 1$ can be written using whole numbers, the four operations and real radicals. Then there is a chain of fields

$$\mathbb{Q} = K_0 \subset K_1 \subset \cdots \subset K_m \subset \mathbb{R},$$

each obtained from the one before by adjoining a real $n$-th root of one of its elements, with a root of the cubic in the last.

First move: make every radical prime. A sixth root is a square root of a cube root, and in general an $n$-th root is a chain of prime roots, each real if the original was. So every step may be taken to be $K_{i+1} = K_i(\beta)$ with $\beta^p = c \in K_i$, $p$ prime, $\beta$ real. For a prime $p$, the polynomial $x^p - c$ is either irreducible over $K_i$ or has a root there, so each step has degree exactly $p$ or adds nothing.

Second move: find the step where the cubic splits. Let $K_j$ be the first field in the chain containing a root of the cubic. In $K_{j-1}$ the cubic has no root, and a cubic with no root cannot factor, so it is irreducible there and its root generates a step of degree 3. That degree divides $[K_j : K_{j-1}]$, which is a prime $p$ — so $p = 3$, and $K_j = K_{j-1}(\alpha)$. By the previous section, that field is normal over $K_{j-1}$.

Third move: the contradiction. $K_j$ is also $K_{j-1}(\beta)$ with $\beta^3 = c$, and $x^3 - c$ is irreducible over $K_{j-1}$ with the root $\beta$ in $K_j$. A normal extension containing one root of an irreducible polynomial contains all of them, so $\omega\beta \in K_j$. But every field in the chain lies inside the real numbers, and $\omega\beta$ is not real. The supposed chain cannot exist.

That is the whole proof, and it uses the Galois correspondence in exactly one place — the fact that a normal step is closed under all the roots of its polynomials. For a cubic whose group is the full six-element $S_3$ rather than the cyclic three, the only change is to adjoin $\sqrt D$ first; that is a real square root, since a cubic with three real roots has $D > 0$, and over $\mathbb{Q}(\sqrt D)$ the cubic’s group is cyclic again. Every irreducible cubic with three real roots has no real-radical expression for any of its roots, and the cyclic case drawn here is not special.

How common the trouble is #

The census below takes every cubic $x^3 + px + q$ with whole-number coefficients between $-30$ and $30$, removes those that factor, and colours the rest by whether all three roots are real.

The result is not a curiosity: 30% of the irreducible cubics in the box are afflicted. They fill the cusp-shaped region where $p$ is negative and $q$ is small relative to it, the same region where two roots run into each other draws as the set of cubics with three real roots. And the share grows with the box. The curve’s distance from the axis at height $q$ grows only like $q^{2/3}$, while the box grows like $q$, so in a large box the cusp swallows almost the whole left half: the share of afflicted cubics rises slowly towards one half. For every cubic in it, without exception, the formula must leave the real line.

The twenty ringed cubics are the cyclic ones, whose discriminants are perfect squares: $81 = 9^2$ for $x^3 - 3x \pm 1$, $49 = 7^2$ for $x^3 - 7x \pm 7$, $729 = 27^2$ for $x^3 - 9x \pm 9$, and so on up to $59{,}049 = 243^2$ for $x^3 - 27x \pm 27$. Their roots are the most symmetric real numbers a cubic can have — each is a polynomial in any other — and they are just as unreachable by real radicals as the rest. The census also shows the boundary curve passing between the two kinds with no cubic on it: a cubic on the curve has a repeated root and so factors.

The general rule: real radicals are square roots in disguise #

The argument did not use much about degree three. What it used was that the first step reaching a root had to have odd prime degree, and that a normal real step of odd prime degree cannot be made by a radical. Run the same reasoning on any irreducible polynomial whose roots are all real and the conclusion is stronger than Hölder’s: the polynomial can be solved by real radicals only if the degree of its splitting field is a power of two. I. Martin Isaacs gave this general form, with a short Galois-theoretic proof, in 1985.

A power of two means that the field is reached by a chain of steps of degree two, and a degree-two step is a square root. So for polynomials with all roots real, real radicals can reach exactly what real square roots can reach — and every step is a square root showed that real square roots reach exactly the lengths that ruler and compass construct. Among algebraic numbers whose conjugates are all real, the ones expressible by real radicals are exactly the constructible ones. A real cube root never helps unless the polynomial has non-real roots to receive its companions, as $x^3 - 2$ does.

The quartic in the figure is the positive case. Its four roots are $\pm\sqrt{2 \pm \sqrt 2}$, which are $2\cos$ of odd multiples of $22\tfrac12^\circ$, and every stage of the expression is real. The table below sets polynomials with all roots real beside their groups and the degrees of their fields.

The last row is the most striking. The quintic $x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1$, whose roots are $2\cos(2\pi k/11)$, has the commutative group of five rotations, and the quintics that have a formula placed it among the most easily solved quintics there are. Gauss showed that $\cos(2\pi/n)$ is expressible by radicals for every $n$. But its splitting field has degree five, so $\cos(2\pi/11)$ has no expression in real radicals: every formula for it passes through fifth roots of complex numbers. The same is true of $\cos 20^\circ$, $\cos 40^\circ$ and $\cos(2\pi/7)$. They are solvable, they are real, and the only radicals that reach them are complex ones.

What the drawings do not prove #

The figures are numerical, and the theorem is not. The first figure computes six cube roots in floating point and checks that their sums are roots to within $10^{-9}$; that shows the formula works, which nobody doubted. It cannot show that no other formula works, which is the content of Hölder’s theorem — that is an argument about every possible chain of fields, and no drawing enumerates those.

The census, similarly, checks discriminants exactly and irreducibility exactly, but its colouring by “afflicted” is an application of the theorem rather than evidence for it. And the table’s verdicts in the last column for the four rows marked “no” rest on the general theorem; the figure verifies only the inputs to it — that the roots are all real, by counting sign changes, and that the field degree is not a power of two. What the pictures can do, and do, is make the shape of the obstruction visible: the real line in the cube-root figure, holding one root of $x^3 = 2$ and missing the other two, is the proof’s third move.

It is also worth saying what “radicals” means in the statement, because the theorem is sharp about it. Trigonometric functions are not radicals, so Viète’s formula $2\cos\!\big(\tfrac13\arccos(-\tfrac12)\big)$ is a real expression for a root without contradicting anything. Hölder’s theorem forbids real radicals; it does not forbid real answers.

Still open: the real radicals of a field’s numbers #

Isaacs’s theorem settles the polynomials whose roots are all real. It leaves the mixed cases — a polynomial with some real roots and some complex ones — to a finer analysis, and there the question “which of its real roots can be written with real radicals” depends on how the complex roots sit in the group, not only on its size. For low degrees the answer can be read off a table of groups, as $x^3 - 2$ was. For general degree the question becomes one about the group together with the particular symmetry that complex conjugation induces on the roots, and deciding it for a given polynomial means first computing its Galois group, which for high degree is itself an expensive computation; a clean description of the answer, of the kind Isaacs’s theorem gives for the all-real case, is not available.

There is also a quantitative question with no general answer: when complex radicals are necessary, how short can the expression be? For $\cos(2\pi/11)$ the radicals needed are fifth roots of numbers in a field of degree four, and the classical expressions for it, worked out from Gauss’s periods, are long. How compactly such a number can be written — how many radicals of each kind, nested in the best arrangement — has no general theory, and the cases where a shortest expression is actually known are small.

Why the real line is not closed under its own equations #

The rational numbers are not closed under taking square roots, and the real numbers fix that. The real numbers are not closed under solving equations — $x^2 + 1 = 0$ has no real root — and the complex numbers fix that. Hölder’s theorem is a third kind of non-closure, subtler than either: the real numbers are closed under taking real radicals, and every root of $x^3 - 3x + 1$ is real, and still the radicals cannot get from the coefficients to the roots without leaving.

The reason is the asymmetry in the cube-root figure. A real field can hold one cube root of a real number and never all three, while a real field that holds one root of an irreducible cubic with real roots holds all three. A radical step and a root step have different shapes, and the Galois correspondence sees shape. That is why the sixteenth-century algebraists who wanted nothing to do with imaginary numbers had to accept them: not to solve equations with no real solutions, but to solve equations all of whose solutions were real.