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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. One can decide whether a given irreducible cubic polynomial is in ''casus irreducibilis'' using the discriminant ''D'', via Cardano's formula.〔, Theorem 1.3.1, p. 15.〕 Let the cubic equation be given by : Then the discriminant ''D'' appearing in the algebraic solution is given by : * If ''D'' < 0, then the polynomial has two complex 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 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''() 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 ''C''3. 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''(∛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. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Casus irreducibilis」の詳細全文を読む スポンサード リンク
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