# Casus irreducibilis

In algebra, casus irreducibilis (Latin for "the irreducible case") is one of the cases that may arise in attempting to solve a cubic equation with integer coefficients, to obtain roots that are expressed with radicals. Specifically, if a cubic polynomial is irreducible (not factorable into lower-degree polynomials) over the rational numbers and has three real roots, then in order to express the roots with radicals, one must introduce complex-valued expressions, even though the resulting expressions are ultimately real-valued. This was proven by Pierre Wantzel in 1843.

One can decide whether a given irreducible cubic polynomial is in casus irreducibilis using the discriminant D, via Cardano's formula. Let the cubic equation be given by

$ax^{3}+bx^{2}+cx+d=0.\,$ with a≠0. Then the discriminant D appearing in the algebraic solution is given by

$D=18abcd-4b^{3}d+b^{2}c^{2}-4ac^{3}-27a^{2}d^{2}.\,$ • If D < 0, then the polynomial has two complex non-real roots, so casus irreducibilis does not apply.
• If D = 0, then there are three real roots, and two of them are equal and can be found by the Euclidean algorithm, and by the quadratic formula. All roots are real and expressible by real radicals. The polynomial is not irreducible.
• If D > 0, then there are three distinct real roots. Either a rational root exists and can be found using the rational root test, in which case the cubic polynomial can be factored into the product of a linear polynomial and a quadratic polynomial, the latter of which can be solved via the quadratic formula; or no such factorization can occur, so the polynomial is casus irreducibilis: all roots are real, but require complex numbers to express them in radicals.

## Formal statement and proof

More generally, suppose that F is a formally real field, and that p(x) ∈ F[x] is a cubic polynomial, irreducible over F, but having three real roots (roots in the real closure of F). Then casus irreducibilis states that it is impossible to find any solution of p(x) = 0 by real radicals.

To prove this, note that the discriminant D is positive. Form the field extension F(D). Since this is F or a quadratic extension of F (depending in whether or not D is a square in F), p(x) remains irreducible in it. Consequently, the Galois group of p(x) over F(D) is the cyclic group C3. Suppose that p(x) = 0 can be solved by real radicals. Then p(x) can be split by a tower of cyclic extensions

$F\subset F({\sqrt {D}})\subset F({\sqrt {D}},{\sqrt[{p_{1}}]{\alpha _{1}}})\subset \cdots \subset K\subset K({\sqrt[{3}]{\alpha }})$ At the final step of the tower, p(x) is irreducible in the penultimate field K, but splits in K(3α) for some α. But this is a cyclic field extension, and so must contain a primitive root of unity.

However, there are no primitive 3rd roots of unity in a real closed field. Suppose that ω is a primitive 3rd root of unity. Then, by the axioms defining an ordered field, ω, ω2, and 1 are all positive. But if ω2>ω, then cubing both sides gives 1>1, a contradiction; similarly if ω>ω2.

### Cardano's solution

The equation ax3 + bx2 + cx + d = 0 can be depressed to a monic trinomial by dividing by $a$ and substituting x = tb/3a (the Tschirnhaus transformation), giving the equation t3 + pt + q = 0 where

$p={\frac {3ac-b^{2}}{3a^{2}}}$ $q={\frac {2b^{3}-9abc+27a^{2}d}{27a^{3}}}.$ Then regardless of the number of real roots, by Cardano's solution the three roots are given by

$t_{k}=\omega _{k}{\sqrt[{3}]{-{q \over 2}+{\sqrt {{q^{2} \over 4}+{p^{3} \over 27}}}}}+\omega _{k}^{2}{\sqrt[{3}]{-{q \over 2}-{\sqrt {{q^{2} \over 4}+{p^{3} \over 27}}}}}$ where $\omega _{k}$ (k=1, 2, 3) is a cube root of 1 ($\omega _{1}=1$ , $\omega _{2}=-{\frac {1}{2}}+{\frac {\sqrt {3}}{2}}i$ , and $\omega _{3}=-{\frac {1}{2}}-{\frac {\sqrt {3}}{2}}i$ , where i is the imaginary unit). Here if the radicands under the cube roots are non-real, the cube roots expressed by radicals are defined to be any pair of complex conjugate cube roots, while if they are real these cube roots are defined to be the real cube roots.

Casus irreducibilis occurs when none of the roots is rational and when all three roots are distinct and real; the case of three distinct real roots occurs if and only if q2/4 + p3/27 < 0, in which case Cardano's formula involves first taking the square root of a negative number, which is imaginary, and then taking the cube root of a complex number (which cube root cannot itself be placed in the form α + βi with specifically given expressions in real radicals for α and β, since doing so would require independently solving the original cubic). Even in the reducible case in which one of three real roots is rational and hence can be factored out by polynomial long division, Cardano's formula (unnecessarily in this case) expresses that root (and the others) in terms of non-real radicals.

### Example

The depressed cubic equation

$x^{3}-3x+1=0$ is irreducible, because if it could be factored there would be a linear factor giving a rational solution, while by the rational root test there is no rational root. Since its discriminant is positive, it has three real roots, so it is an example of casus irreducibilis. Cardano's formula gives these three real roots as

$t_{k}=\omega _{k}{\sqrt[{3}]{-{1 \over 2}+{\sqrt {-3 \over 4}}}}+\omega _{k}^{2}{\sqrt[{3}]{-{1 \over 2}-{\sqrt {-3 \over 4}}}}$ for k=1, 2, 3. This solution in radicals involves the imaginary number ${\sqrt {-3 \over 4}}$ and hence involves the cube roots of complex conjugate numbers.

## Non-algebraic solution in terms of real quantities

While casus irreducibilis cannot be solved in radicals in terms of real quantities, it can be solved trigonometrically in terms of real quantities. Specifically, the depressed monic cubic equation $t^{3}+pt+q=0$ is solved by

$t_{k}=2{\sqrt {-{\frac {p}{3}}}}\cos \left({\frac {1}{3}}\arccos \left({\frac {3q}{2p}}{\sqrt {\frac {-3}{p}}}\right)-k{\frac {2\pi }{3}}\right)\quad {\text{for}}\quad k=0,1,2\,.$ These solutions are in terms of real quantities if and only if ${q^{2} \over 4}+{p^{3} \over 27}<0$ — i.e., if and only if there are three real roots. The formula involves starting with an angle whose cosine is known, trisecting the angle by multiplying it by 1/3, and taking the cosine of the resulting angle and adjusting for scale.

## Relation to angle trisection

The distinction between the reducible and irreducible cubic cases with three real roots is related to the issue of whether or not an angle with rational cosine or rational sine is trisectible by the classical means of compass and unmarked straightedge. If the cosine of an angle θ is known to have a particular rational value, then one third of this angle has a cosine that is one of the three real roots of the equation

$4x^{3}-3x-\cos(\theta )=0.$ Likewise, if the sine of θ is known to have a particular rational value, then one third of this angle has a sine that is one of the three real roots of the equation

$-4y^{3}+3y-\sin(\theta )=0.$ In either case, if the rational root test reveals a rational root of the equation, x or y minus that root can be factored out of the polynomial on the left side, leaving a quadratic that can be solved for the remaining two roots in terms of a square root; then all of these roots are classically constructible since they are expressible in no higher than square roots, so in particular cos( ​θ3) or sin( ​θ3) is constructible and so is the associated angle θ3. On the other hand, if the rational root test shows that there is no rational root, then casus irreducibilis applies, cos( ​θ3) or sin( ​θ3) is not constructible, the angle θ3 is not constructible, and the angle θ is not classically trisectible.

## Generalization

Casus irreducibilis can be generalized to higher degree polynomials as follows. Let p ∈ F[x] be an irreducible polynomial which splits in a formally real extension R of F (i.e., p has only real roots). Assume that p has a root in $K\subseteq R$ which is an extension of F by radicals. Then the degree of p is a power of 2, and its splitting field is an iterated quadratic extension of F.: pp. 571–572

Thus for any irreducible polynomial whose degree is not a power of 2 and which has all roots real, no root can be expressed purely in terms of real radicals. Moreover, if the polynomial degree is a power of 2 and the roots are all real, then if there is a root that can be expressed in real radicals it can be expressed in terms of square roots and no higher-degree roots, as can the other roots, and so the roots are classically constructible.

Casus irreducibilis for quintic polynomials is discussed by Dummit.:p.17

### Relation to angle pentasection

The distinction between the reducible and irreducible quintic cases with five real roots is related to the issue of whether or not an angle with rational cosine or rational sine is pentasectible (able to be split into five equal parts) by the classical means of compass and unmarked straightedge. If the cosine of an angle θ is known to have a particular rational value, then one fifth of this angle has a cosine that is one of the five real roots of the equation

$16x^{5}-20x^{3}+5x-\cos(\theta )=0.$ Likewise, if the sine of θ is known to have a particular rational value, then one fifth of this angle has a sine that is one of the five real roots of the equation

$16x^{5}-20x^{3}+5x-\sin(\theta )=0.$ In either case, if the rational root test yields a rational root x1, then the quintic is reducible since it can be written as a factor (x—x1) times a quartic polynomial. But if the test shows that there is no rational root, then the polynomial may be irreducible, in which case casus irreducibilis applies, cos( ​θ5) and sin( ​θ5) are not constructible, the angle θ5 is not constructible, and the angle θ is not classically pentasectible.

Pentasection of θ also implies pentasection of 450° - θ as: θ/5 + k ; k ∈ { 0°, 54°, 72°, 126°, 144°, 198°, 216°, 270°, 288°, 342° }.