examples of minimal polynomials


Note that 24 is algebraic over the fields and (2). The minimal polynomials for 24 over these fields are x4-2 and x2-2, respectively. Note that x4-2 is irreduciblePlanetmathPlanetmathPlanetmath over 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 (2) since it is a quadratic polynomial and neither of its roots (24 and -24) are in (2).

A common method for constructing minimal polynomials for numbers that are expressible over is “backwards ”: The number can be set equal to x, and the equation can be algebraically manipulated until a monic polynomialMathworldPlanetmath in [x] is equal to 0. Finally, if the monic polynomial is not irreducible, then it can be factored into irreducible polynomialsMathworldPlanetmath [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+522x=1+52x-1=5(2x-1)2=54x2-4x+1=54x2-4x-4=0x2-x-1=0

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

x=23+33x3=2+32233+32323+3x3-5=363(23+33)x3-5=363x(x3-5)3=276x3x9-35x6+325x3-125=162x3x9-15x6-87x3-125=0

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

On the other hand, x5-5x4+10x3-10x2+5x-2 factors over as (x-2)(x4-3x3+4x2-2x+1). Since 1+ω5 is not a root of x-2, it must be a root of x4-3x3+4x2-2x+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 . Since (1+ω5)=(ω5), degm(x)=[(1+ω5):]=[(ω5):]=φ(5)=4=deg(x4-3x3+4x2-2x+1). (Here φ denotes the Euler totient function.) Since m(x) divides x4-3x3+4x2-2x+1 and they have the same degree, it follows that m(x)=x4-3x3+4x2-2x+1.

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