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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 [[nth root|radicals]]. Specifically, if a cubic polynomial is [[irreducible polynomial|irreducible]] over the [[rational numbers]] and has three [[real numbers|real]] roots, then in order to express the roots with radicals, one must introduce [[complex numbers|complex]]-valued expressions, even though the resulting expressions are ultimately real-valued.
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| 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
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| :<math>ax^3+bx^2+cx+d=0.\,</math>
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| Then the discriminant ''D'' appearing in the algebraic solution is given by
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| : <math> D = 18abcd -4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2. \,</math>
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| * If ''D'' < 0, then the polynomial has two complex roots, so ''casus irreducibilis'' does not apply.
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| * 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.
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| * 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.
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| == Formal statement and proof ==
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| 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.
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| 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''<sub>3</sub>. Suppose that ''p''(''x'') = 0 can be solved by real radicals. Then ''p''(''x'') can be split by a tower of [[cyclic extension]]s (of prime degree)
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| :<math> F\sub F(\sqrt{\Delta})\sub F(\sqrt{\Delta}, \sqrt[p_1]{\alpha_1}) \sub\cdots \sub K\sub K(\sqrt[3]{\alpha})</math>
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| 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]].
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| 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, ω<sup>2</sup>, and 1 are all positive. But if ω<sup>2</sup>>ω, then cubing both sides gives 1>1, a contradiction; similarly if ω>ω<sup>2</sup>.
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| ==Solution in non-real radicals==
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| The equation <math>ax^3+bx^2+cx+d=0\,</math> can be depressed to a [[monic polynomial|monic]] [[trinomial]] by dividing by <math>a</math> and substituting <math>x = t-\frac{b}{3a}</math> (the [[Tschirnhaus transformation]]), giving the equation
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| :<math>t^3+pt+q=0 </math>
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| where
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| :<math> \begin{align}
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| p=&\frac{3ac-b^2}{3a^2}\\
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| q=&\frac{2b^3-9abc+27a^2d}{27a^3}.
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| \end{align}
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| </math>
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| Then regardless of the number of real roots, by [[Cubic function#Cardano's method|Cardano's solution]] the three roots are given by
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| :<math> t_k = \omega_k \sqrt[3]{-{q\over 2}+ \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}} + \omega_k^2 \sqrt[3]{-{q\over 2}- \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}}</math>
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| where <math> \omega_k</math> (''k''=1, 2, 3) is a cube root of 1: <math>\omega_1 = 1</math>, <math>\omega_2 = -\frac{1}{2} + \frac{\sqrt{3}}{2}i</math>, and <math>\omega_3 = -\frac{1}{2} - \frac{\sqrt{3}}{2}i</math>, where ''i'' is the [[imaginary unit]].
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| ''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 <math>{q^{2}\over 4}+{p^{3}\over 27} < 0 </math>, in which case Cardano's formula involves first taking the square root of a negative number, which is [[Imaginary number|imaginary]], and then taking the cube root of a complex number (which cannot itself be placed in the form <math>\alpha + \beta i</math> with specifically given expressions in real radicals for <math>\alpha</math> and <math>\beta</math>, 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.
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| ==Non-algebraic solution in terms of real quantities==
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| {{Main|Cubic function#Trigonometric (and hyperbolic) method}}
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| While ''casus irreducibilis'' cannot be [[Algebraic solution|solved in radicals]] in terms of real quantities, it ''can'' be solved [[Trigonometry|trigonometrically]] in terms of real quantities. Specifically, the depressed monic cubic equation <math>t^3+pt+q=0 </math> is solved by
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| :<math>t_k=2\sqrt{-\frac{p}{3}}\cos\left(\frac{1}{3}\arccos\left(\frac{3q}{2p}\sqrt{\frac{-3}{p}}\right)-k\frac{2\pi}{3}\right) \quad \text{for} \quad k=0,1,2 \,.</math>
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| These solutions are in terms of real quantities if and only if <math>{q^{2}\over 4}+{p^{3}\over 27} < 0</math> — i.e., if and only if there are three real roots.
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| ==Relation to angle trisection==
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| 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 [[Angle trisection|trisectible]] by the classical means of [[compass and straightedge constructions|compass and unmarked straightedge]]. If the cosine of an angle <math>\theta</math> 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
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| :<math> 4x^3 - 3x - \cos (\theta) = 0.</math>
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| Likewise, if the sine of <math>\theta</math> 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
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| :<math> -4y^3 + 3y - \sin (\theta) = 0.</math>
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| In either case, if the rational root test reveals a real 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 <math>\cos (\theta / 3)</math> or <math>\sin (\theta / 3)</math> is constructible and so is the associated angle <math>\theta / 3</math>. On the other hand, if the rational root test shows that there is no real root, then ''casus irreducibilis'' applies, <math>\cos (\theta / 3)</math> or <math>\sin (\theta / 3)</math> is not constructible, the angle <math>\theta / 3</math> is not constructible, and the angle <math>\theta</math> is not classically trisectible.
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| ==Generalization==
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| ''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 <math>K\subseteq R</math> which is an extension of ''F'' by radicals. Then the degree of ''p'' is a power of 2, and its splitting field is an iterated quadratic extension of ''F''.
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| ''Casus irreducibilis'' for [[Quintic function|quintic polynomials]] is discussed by Dummit.<ref>David S. Dummit [http://www.emba.uvm.edu/~dummit/quintics/solvable.pdf Solving Solvable Quintics] </ref>{{rp|p.17}}
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| == References == | |
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| {{reflist}}
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| * {{citation|first=Bartel Leendert|last=van der Waerden|authorlink=Bartel Leendert van der Waerden|title=Modern Algebra I|isbn=978-0-387-40624-4|publisher=Springer|year=2003|others= F. Blum, J.R. Schulenberg}}
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| ==External links==
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| *{{PlanetMath|urlname=CasusIrreducibilis|title=casus irreducibilis}}
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| [[Category:Abstract algebra]]
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| [[Category:Polynomials]]
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| [[de:Casus irreducibilis]]
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Alyson is what my husband enjoys to contact me but I don't like when people use my full name. It's not a common thing but what she likes doing is to perform domino but she doesn't have the time recently. Kentucky is exactly where I've always been residing. I am currently a journey agent.
Feel free to visit my page; tarot card readings (you can try these out)