examples of minimal polynomials


Note that 24 is algebraic over the fields ℚ and ℚ⁢(2). The minimal polynomials for 24 over these fields are x4-2 and x2-2, respectively. Note that x4-2 is irreduciblePlanetmathPlanetmathPlanetmath over ℚ by using Eisenstein’s criterion and Gauss’s lemma (http://planetmath.org/GausssLemmaII) (see this entry (http://planetmath.org/AlternativeProofThatSqrt2IsIrrational) for more details), and x2-2 is irreducible over ℚ⁢(2) since it is a quadratic polynomial and neither of its roots (24 and -24) are in ℚ⁢(2).

A common method for constructing minimal polynomials for numbers that are expressible over ℚ is “backwards ”: The number can be set equal to x, and the equation can be algebraically manipulated until a monic polynomialMathworldPlanetmath in ℚ⁢[x] is equal to 0. Finally, if the monic polynomial is not irreducible, then it can be factored into irreducible polynomialsMathworldPlanetmath ℚ⁢[x], and the original number will be a root of one of these. A very example is 24:

x=24x4=2x4-2=0

This method will be further demonstrated with three more examples: One for 1+52, one for 1+ω5 where ω5 is a fifth root of unityMathworldPlanetmath, and one for 23+33.

x=1+522⁢x=1+52⁢x-1=5(2⁢x-1)2=54⁢x2-4⁢x+1=54⁢x2-4⁢x-4=0x2-x-1=0

x=1+ω5x-1=ω5(x-1)5=1x5-5⁢x4+10⁢x3-10⁢x2+5⁢x-1=1x5-5⁢x4+10⁢x3-10⁢x2+5⁢x-2=0

x=23+33x3=2+3⁢22⋅33+3⁢2⋅323+3x3-5=3⁢63⁢(23+33)x3-5=3⁢63⁢x(x3-5)3=27⋅6⁢x3x9-3⋅5⁢x6+3⋅25⁢x3-125=162⁢x3x9-15⁢x6-87⁢x3-125=0

Since x2-x-1 is a quadratic and has no roots in ℚ, it is irreducible over ℚ. Thus, it is the minimal polynomial over ℚ for 1+52.

On the other hand, x5-5⁢x4+10⁢x3-10⁢x2+5⁢x-2 factors over ℚ as (x-2)⁢(x4-3⁢x3+4⁢x2-2⁢x+1). Since 1+ω5 is not a root of x-2, it must be a root of x4-3⁢x3+4⁢x2-2⁢x+1. Moreover, this polynomialMathworldPlanetmathPlanetmathPlanetmath must be irreducible. This fact can be proven in the following manner: Let m⁢(x) be the minimal polynomial for 1+ω5 over ℚ. Since ℚ⁢(1+ω5)=ℚ⁢(ω5), degm(x)=[ℚ(1+ω5):ℚ]=[ℚ(ω5):ℚ]=φ(5)=4=deg(x4-3x3+4x2-2x+1). (Here φ denotes the Euler totient function.) Since m⁢(x) divides x4-3⁢x3+4⁢x2-2⁢x+1 and they have the same degree, it follows that m⁢(x)=x4-3⁢x3+4⁢x2-2⁢x+1.

It turns out that x9-15⁢x6-87⁢x3-125 is irreducible over ℚ. (This can be proven in a manner as above. Note that [ℚ(23+33):ℚ]=9.) Thus, it is the minimal polynomial over ℚ for 23+33.

Title examples of minimal polynomials
Canonical name ExamplesOfMinimalPolynomials
Date of creation 2013-03-22 16:55:18
Last modified on 2013-03-22 16:55:18
Owner Wkbj79 (1863)
Last modified by Wkbj79 (1863)
Numerical id 13
Author Wkbj79 (1863)
Entry type Example
Classification msc 12F05
Classification msc 12E05
Related topic IrreduciblePolynomialsObtainedFromBiquadraticFields