Casus irreducibilis

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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. Casus irreducibilis was the original reason for the introduction of the complex number system by Niccolò Fontana Tartaglia and Gerolamo Cardano in 1545. The term imaginary number referred then to a number which was imagined to exist in order to form an expression of the root.

One can decide whether a given irreducible cubic polynomial is in casus irreducibilis using the discriminant D, via Cardano's formula:

If D < 0, then the polynomial has two complex roots, so casus irreducibilis does not apply.
If D = 0, then two of the roots are equal and can be found by the Euclidean algorithm and the quadratic formula. All roots are real and expressible by real radicals. The polynomial is not irreducible.
If D > 0, then the polynomial is in casus irreducibilis. All roots are real, but require complex numbers to express them in radicals.

[edit] 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 $F(\sqrt{D})$ . Since this is a quadratic extension, p(x) remains irreducible in it. Consequently, the Galois group of p(x) over $F(\sqrt{D})$ 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 (of prime degree)

$F\sub F(\sqrt{\Delta})\sub F(\sqrt{\Delta}, \sqrt[p_1]{\alpha_1}) \sub\cdots \sub K\sub K(\sqrt[3]{\alpha})$

At the final step of the tower, p(x) is irreducible in the penultimate field K, but splits in $K(\sqrt[3]{\alpha})$ 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.