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 with roots that are expressed with radicals. Specifically, if a cubic polynomial is irreducible 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 proved by Pierre Wantzel in 1843.
Contents
- Formal statement and proof
- Cardanos solution
- Example
- Non algebraic solution in terms of real quantities
- Relation to angle trisection
- Generalization
- References
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
Then the discriminant D appearing in the algebraic solution is given by
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
At the final step of the tower, p(x) is irreducible in the penultimate field K, but splits in K(3√α) 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. Indeed, 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
Then regardless of the number of real roots, by Cardano's solution the three roots are given by
where
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). Note that 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
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
for k=1, 2, 3. This solution in radicals involves the imaginary number
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
These solutions are in terms of real quantities if and only if
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
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
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( θ⁄3) or sin( θ⁄3) is constructible and so is the associated angle θ⁄3. On the other hand, if the rational root test shows that there is no rational root, then casus irreducibilis applies, cos( θ⁄3) or sin( θ⁄3) is not constructible, the angle θ⁄3 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
Thus for any irreducible polynomial whose degree is not a power of 2 and which has all roots real, the roots cannot 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.