- Casus irreducibilis
In
algebra , "casus irreducibilis" (Latin for "the irreducible case") is one of the cases that may arise in attempting to solve acubic equation withinteger coefficients with roots that are expressed with radicals. Specifically, if a cubic polynomial is irreducible over therational 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 byNiccolò Fontana Tartaglia andGerolamo Cardano in 1545. The termimaginary 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", viaCardano'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 theEuclidean algorithm and thequadratic 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 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 aquadratic extension , "p"("x") remains irreducible in it. Consequently, theGalois group of "p"("x") over F(sqrt{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 ofcyclic extension s (of prime degree):Fsub F(sqrt{Delta})sub F(sqrt{Delta}, sqrt [p_1] {alpha_1}) subcdots sub Ksub 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.References
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