Casus_irreducibilis

In algebra, casus irreducibilis is a property of irreducible cubic polynomials with integer coefficients. If the roots of the polynomial are real numbers, then to express them as radicals one must introduce complex numbers, even though the resulting expressions are real. 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 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.

Formal statement and proof

More generally, suppose that F is a real field, and that p(x) ∈ F[x] is a cubic polynomial, irreducible over F, but having three real roots. 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 $F(\backslash sqrt\{D\})$. Since this is a quadratic extension, p(x) remains irreducible in it. Consequently, the Galois group of p(x) over $F(\backslash sqrt\{D\})$ 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)

$F\backslash sub\; F(\backslash sqrt\{\backslash Delta\})\backslash sub\; F(\backslash sqrt\{\backslash Delta\},\; \backslash sqrt[p\_1]\{\backslash alpha\_1\})\; \backslash sub\backslash cdots\; \backslash sub\; K\backslash sub\; K(\backslash sqrt[3]\{\backslash alpha\})$

At the final step of the tower, p(x) is irreducible in the penultimate field K, but splits in $K(\backslash sqrt[3]\{\backslash alpha\})$ for some α. But this is a cyclic field extension, and so must contain a primitive root of unity. There are no real primitive 3rd roots of unity, though.

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