Casus irreducibilis

In algebra, casus irreducibilis is one of the cases that may arise in solving polynomials of degree 3 or higher with integer coefficients algebraically (as opposed to numerically), i.e., by obtaining roots that are expressed with radicals. It shows that many algebraic numbers are real-valued but cannot be expressed in radicals without introducing complex numbers. The most notable occurrence of casus irreducibilis is in the case of cubic polynomials that have three real roots, which was proven by Pierre Wantzel in 1843. One can see whether a given cubic polynomial is in the so-called casus irreducibilis by looking at the discriminant, via Cardano's formula.

The three cases of the discriminant
Let
 * $$ax^3+bx^2+cx+d=0$$

be a cubic equation with $$a\ne0$$. Then the discriminant is given by
 * $$D := \bigl((x_1-x_2)(x_1-x_3)(x_2-x_3)\bigr)^2 = 18abcd - 4ac^3 - 27a^2d^2 + b^2c^2 -4b^3d~.$$

It appears in the algebraic solution and is the square of the product
 * $$\Delta := \prod_{j<k}(x_j-x_k) = (x_1-x_2)(x_1-x_3)(x_2-x_3) \qquad \qquad \bigl(\!= \pm\sqrt{D}\bigr)$$

of the $$\tbinom32 = 3$$ differences of the 3 roots $$x_1,x_2,x_3$$.

Although there are cubic polynomials with negative discriminant which are irreducible in the modern sense, casus irreducibilis does not apply. All the cubic polynomials with zero discriminant are reducible. Because they require complex numbers (in the understanding of the time: cube roots from non-real numbers, i.e. from square roots from negative numbers) to express them in radicals, this case in the 16th century has been termed casus irreducibilis. James Pierpont in Annals of Mathematics 1900-1901 on p.&thinsp;42: „To Cardan and his contemporaries who had no idea how such cube roots could be found this case was highly paradoxical. Since that time mathematicians have attempted to present these real roots as sums of real radicals. As their efforts were unsuccessful, the case when D > 0 was known as the casus irreducibilis.“ Artur Ekert Complex and unpredictable Cardano takes Cardano’s example $$x^3-15x-4=0$$ having $q^2/4 + p^3/27 = (-4)^2/4 + (-15)^3/27 = -121 < 0 $ and writes on p.&thinsp;9: „Cardano knew that $$x = 4$$ was one of the solutions and yet it was a casus irreducibilis“. This shows that in the 16th century „irreducibilis“ must have meant something like „not reducible to real radicals“. On the other hand, Cardano’s example may be used to show how real roots can arise from cube roots of non-real numbers:
 * 1) If $D < 0$, then the polynomial has one real root and two complex non-real roots. $$\Delta\in i\R^\times$$ is purely imaginary.
 * 1) If $D = –31 < 0$, then $$\Delta=0$$ and there are three real roots; two of them are equal. Whether $D = 0$ can be found out by the Euclidean algorithm, and if so, the roots by the quadratic formula. Moreover, all roots are real and expressible by real radicals.
 * 1) If $D = 0$, then $$\Delta\in\R^\times$$ is non-zero and real, and there are three distinct real roots which are sums of two complex conjugates.

\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}}}$$ It may be noticed that $d := q^2/4 + p^3/27 = -121$ is not the discriminant $$D$$; it is $$D = -108 \, d = 13068 = 2^2 3^3 11^2 $$ with the sign inverted. Interestingly $\sqrt{d} = i\sqrt{D/108}$ occurs in Cardano’s formula (as well as the primitive 3rd roots of unity $$\omega_{2,3}$$ with their $i \sqrt{3}$), although $$\sqrt{D}~,$$ and not $$\sqrt{d}~,$$ is necessarily an element of the splitting field.
 * We have || $$p=-15, \; q=-4$$,
 * which yields      || $$d:={q^2\over 4}+{p^3\over 27}=-121$$,
 * from which ||        $$-{q\over 2} \pm \sqrt{{q^{2}\over 4}+{p^{3}\over 27}} = -{q\over 2} \pm \sqrt{d}=2 \pm 11\,i$$.
 * colspan=2| In the 16th century it was difficult („verè sophistica“) to find that
 * ||        $$\sqrt[3]{2 + 11^{\color{white};} i} = 2 + i =: u$$
 * and ||        $$\sqrt[3]{2 - 11^{\color{white};} i} = 2 - i =: v$$,
 * so that || $$t_k =
 * colspan=2| In the 16th century it was difficult („verè sophistica“) to find that
 * ||        $$\sqrt[3]{2 + 11^{\color{white};} i} = 2 + i =: u$$
 * and ||        $$\sqrt[3]{2 - 11^{\color{white};} i} = 2 - i =: v$$,
 * so that || $$t_k =
 * and ||        $$\sqrt[3]{2 - 11^{\color{white};} i} = 2 - i =: v$$,
 * so that || $$t_k =
 * so that || $$t_k =
 * ||    $$= \omega_k u + \omega_k^2 v $$.
 * colspan=2| This means in detail:
 * 1st root || $$t_1 = \omega_1 u + \omega_1^2 v $$
 * ||    $$= 1 \cdot (2 + i) + 1 \cdot (2 - i)$$
 * ||    $$= 4$$,
 * 2nd root || $$t_2 = \omega_2 u + \omega_2^2 v $$
 * ||    $$= \biggl(\!\!-\frac{1}{2} + \frac{\sqrt{3}}{2}i\biggr) \cdot (2 + i) + \biggl(\!\!-\frac{1}{2} - \frac{\sqrt{3}}{2}i\biggr) \cdot (2 - i)$$
 * ||    $$= -2 - \sqrt{3}$$,
 * 3rd root || $$t_3 = \omega_3 u + \omega_3^2 v $$
 * ||    $$= \biggl(\!\!-\frac{1}{2} - \frac{\sqrt{3}}{2}i\biggr) \cdot (2 + i) + \biggl(\!\!- \frac{1}{2} + \frac{\sqrt{3}}{2}i\biggr) \cdot (2 - i)$$
 * ||    $$= -2 + \sqrt{3}$$.
 * }
 * ||    $$= \biggl(\!\!-\frac{1}{2} + \frac{\sqrt{3}}{2}i\biggr) \cdot (2 + i) + \biggl(\!\!-\frac{1}{2} - \frac{\sqrt{3}}{2}i\biggr) \cdot (2 - i)$$
 * ||    $$= -2 - \sqrt{3}$$,
 * 3rd root || $$t_3 = \omega_3 u + \omega_3^2 v $$
 * ||    $$= \biggl(\!\!-\frac{1}{2} - \frac{\sqrt{3}}{2}i\biggr) \cdot (2 + i) + \biggl(\!\!- \frac{1}{2} + \frac{\sqrt{3}}{2}i\biggr) \cdot (2 - i)$$
 * ||    $$= -2 + \sqrt{3}$$.
 * }
 * ||    $$= \biggl(\!\!-\frac{1}{2} - \frac{\sqrt{3}}{2}i\biggr) \cdot (2 + i) + \biggl(\!\!- \frac{1}{2} + \frac{\sqrt{3}}{2}i\biggr) \cdot (2 - i)$$
 * ||    $$= -2 + \sqrt{3}$$.
 * }
 * ||    $$= -2 + \sqrt{3}$$.
 * }

Formal statement and proof
More generally, suppose that $D > 0$ is a formally real field, and that $F$ is a cubic polynomial, irreducible over $p(x) &isin; F[x]$, but having three real roots (roots in the real closure of $F$). Then casus irreducibilis states that it is impossible to express a solution of $F$ by radicals with radicands $p(x) = 0$.

To prove this, note that the discriminant $&isin; F$ is positive. Form the field extension $D$. Since this is $F(√D) = F(∆)$ or a quadratic extension of $F$ (depending in whether or not $F$ is a square in $D$), $F$ remains irreducible in it. Consequently, the Galois group of $p(x)$ over $p(x)$ is the cyclic group $F(√D)$. Suppose that $C_{3}$ can be solved by real radicals. Then $p(x) = 0$ can be split by a tower of cyclic extensions
 * $$ F\sub F(\sqrt{D})\sub F(\sqrt{D}, \sqrt[p_1]{\alpha_1}) \sub\cdots \sub K\sub K(\sqrt[3]{\alpha})$$

At the final step of the tower, $p(x)$ is irreducible in the penultimate field $p(x)$, but splits in $K$ for some $K(\sqrt{α$. But this is a cyclic field extension, and so must contain a conjugate of $α$ and therefore a primitive 3rd root of unity.

However, there are no primitive 3rd roots of unity in a real closed field. Suppose that &omega; is a primitive 3rd root of unity. Then, by the axioms defining an ordered field, &omega; and &omega;2 are both positive, because otherwise their cube (=1) would be negative. But if &omega;2>&omega;, then cubing both sides gives 1>1, a contradiction; similarly if &omega;>&omega;2.

Cardano's solution
The equation $\sqrt{α$ can be depressed to a monic trinomial by dividing by $$a$$ and substituting $ax^{3} + bx^{2} + cx + d = 0$ (the Tschirnhaus transformation), giving the equation $x = t − b⁄3a$ where
 * $$p=\frac{3ac-b^2}{3a^2}$$
 * $$q=\frac{2b^3-9abc+27a^2d}{27a^3}.$$

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 $t^{3} + pt + q = 0$ 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 are rational and when all three roots are distinct and real; the case of three distinct real roots occurs if and only if $i$, 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 (the cube root cannot itself be placed in the form $q^{2}⁄4 + p^{3}⁄27 < 0$ with specifically given expressions in real radicals for $α + βi$ and $α$, 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 cubic equation


 * $$2x^3-9x^2-6x+3=0$$

is irreducible, because if it could be factored there would be a linear factor giving a rational solution, while none of the possible roots given by the rational root test are actually roots. Since its discriminant is positive, it has three real roots, so it is an example of casus irreducibilis. These roots can be expressed as


 * $$t_k=\frac{3-\omega_k\sqrt[3]{39-26i}-\omega_k^2\sqrt[3]{39+26i}}{2}$$

for $$k\in\left\{1, 2, 3\right\}$$. The solutions are in radicals and involve the cube roots of complex conjugate numbers.

Trigonometric 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.

Although cosine and its inverse function (arccosine) are transcendental functions, this solution is algebraic in the sense that $$\cos\left[\arccos\left(x\right)/3\right]$$ is an algebraic function, equivalent to angle trisection.

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 is trisectible by the classical means of compass and unmarked straightedge. For any angle $β$, one-third of this angle has a cosine that is one of the three solutions to


 * $$4x^3-3x-\cos(\theta)=0.$$

Likewise, $θ$ has a sine that is one of the three real solutions to


 * $$4y^3-3y+\sin(\theta)=0.$$

In either case, if the rational root test reveals a rational solution, $x$ 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 $θ/3$ or $cos(θ/3)$ is constructible and so is the associated angle $sin(θ/3)$. On the other hand, if the rational root test shows that there is no rational root, then casus irreducibilis applies, $θ/3$ or $cos(θ/3)$ is not constructible, the angle $sin(θ/3)$ is not constructible, and the angle $θ/3$ is not classically trisectible.

As an example, while a 180° angle can be trisected into three 60° angles, a 60° angle cannot be trisected with only compass and straightedge. Using triple-angle formulae one can see that $θ$ where $cos π⁄3 = 4x^{3} − 3x$. Rearranging gives $x = cos(20°)$, which fails the rational root test as none of the rational numbers suggested by the theorem is actually a root. Therefore, the minimal polynomial of $8x^{3} − 6x − 1 = 0$ has degree 3, whereas the degree of the minimal polynomial of any constructible number must be a power of two.

Expressing $cos(20°)$ in radicals results in


 * $$\cos\left(\frac{\pi}{9}\right)=\frac{\sqrt[3]{1-i\sqrt{3}}+\sqrt[3]{1+i\sqrt{3}}}{2\sqrt[3]{2}}$$

which involves taking the cube root of complex numbers. Note the similarity to $cos(20°)$ and $e^{iπ/3} = 1+i√3⁄2$.

The connection between rational roots and trisectability can also be extended to some cases where the sine and cosine of the given angle is irrational. Consider as an example the case where the given angle $e^{−iπ/3} = 1−i√3⁄2$ is a vertex angle of a regular pentagon, a polygon that can be constructed classically. For this angle $θ$ is 180°, and standard trigonometric identities then give


 * $$ \cos(\theta)+\cos(\theta/3) = 2\cos(\theta/3)\cos(2\theta/3)

=-2\cos(\theta/3)\cos(\theta)$$

thus


 * $$ \cos(\theta/3) = -\cos(\theta)/(1+2\cos(\theta)).$$

The cosine of the trisected angle is rendered as a rational expression in terms of the cosine of the given angle, so the vertex angle of a regular pentagon can be trisected (mechanically, by simply drawing a diagonal).

Generalization
Casus irreducibilis can be generalized to higher degree polynomials as follows. Let $5θ/3$ be an irreducible polynomial which splits in a formally real extension $p &isin; F[x]$ of $R$ (i.e., $F$ has only real roots). Assume that $p$ has a root in $$K\subseteq R$$ which is an extension of $p$ by radicals. Then the degree of $F$ is a power of 2, and its splitting field is an iterated quadratic extension of $p$.

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, i.e. it is a casus irreducibilis in the (16th century) sense of this article. 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.

Relation to angle pentasection (quintisection) and higher
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. For any angle $F$, 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, $θ$ has a sine that is one of the five real roots of the equation


 * $$16y^5-20y^3+5y-\sin(\theta)=0.$$

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, $θ⁄5$ and $cos(θ/5)$ are not constructible, the angle $sin(θ/5)$ is not constructible, and the angle $θ/5$ is not classically pentasectible. An example of this is when one attempts to construct a 25-gon (icosipentagon) with compass and straightedge. While a pentagon is relatively easy to construct, a 25-gon requires an angle pentasector as the minimal polynomial for $θ$ has degree 10:


 * $$\begin{align}

\cos\left(\frac{2\pi}{5}\right) &= \frac{\sqrt{5}-1}{4} \\ 16x^5-20x^3+5x+\frac{1-\sqrt{5}}{4} &= 0 \qquad\qquad x=\cos\left(\frac{2\pi}{25}\right) \\ 4\left(16x^5-20x^3+5x+\frac{1-\sqrt{5}}{4}\right)\left(16x^5-20x^3+5x+\frac{1+\sqrt{5}}{4}\right) &= 0 \\ 4\left(16x^5-20x^3+5x\right)^2+2\left(16x^5-20x^3+5x\right)-1 &= 0 \\ 1024x^{10}-2560x^8+2240 x^6+32x^5-800 x^4-40x^3+100x^2+10x-1 &= 0. \end{align}$$

Thus,


 * $$\begin{align}

e^{2\pi i/5} &= \frac{-1+\sqrt{5}}{4}+\frac{\sqrt{10+2\sqrt{5}}}{4}i \\ e^{-2\pi i/5} &= \frac{-1+\sqrt{5}}{4}-\frac{\sqrt{10+2\sqrt{5}}}{4}i \\ \cos\left(\frac{2\pi}{25}\right) &= \frac{\sqrt[5]{-1+\sqrt{5}-i\sqrt{10+2\sqrt{5}}}+\sqrt[5]{-1+\sqrt{5}+i\sqrt{10+2\sqrt{5}}}}{2\sqrt[5]{4}}. \end{align}$$