examples of minimal polynomials

Note that $\sqrt[4]{2}$ is algebraic over the fields $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{2})$ . The minimal polynomials for $\sqrt[4]{2}$ over these fields are $x^{4}-2$ and $x^{2}-\sqrt{2}$ , respectively. Note that $x^{4}-2$ is irreducible over $\mathbb{Q}$ by using Eisenstein’s criterion and Gauss’s lemma (http://planetmath.org/GausssLemmaII) (see this entry (http://planetmath.org/AlternativeProofThatSqrt2IsIrrational) for more details), and $x^{2}-\sqrt{2}$ is irreducible over $\mathbb{Q}(\sqrt{2})$ since it is a quadratic polynomial and neither of its roots ( $\sqrt[4]{2}$ and $-\sqrt[4]{2}$ ) are in $\mathbb{Q}(\sqrt{2})$ .

A common method for constructing minimal polynomials for numbers that are expressible over $\mathbb{Q}$ is “backwards ”: The number can be set equal to $x$ , and the equation can be algebraically manipulated until a monic polynomial in $\mathbb{Q}[x]$ is equal to 0. Finally, if the monic polynomial is not irreducible, then it can be factored into irreducible polynomials $\mathbb{Q}[x]$ , and the original number will be a root of one of these. A very example is $\sqrt[4]{2}$ :

$\begin{array}[]{rl}x&=\sqrt[4]{2}\\ x^{4}&=2\\ x^{4}-2&=0\end{array}$

This method will be further demonstrated with three more examples: One for $\displaystyle\frac{1+\sqrt{5}}{2}$ , one for $1+\omega_{5}$ where $\omega_{5}$ is a fifth root of unity, and one for $\sqrt[3]{2}+\sqrt[3]{3}$ .

$\begin{array}[]{rl}x&=\displaystyle\frac{1+\sqrt{5}}{2}\\ 2x&=1+\sqrt{5}\\ 2x-1&=\sqrt{5}\\ (2x-1)^{2}&=5\\ 4x^{2}-4x+1&=5\\ 4x^{2}-4x-4&=0\\ x^{2}-x-1&=0\end{array}$

$\begin{array}[]{rl}x&=1+\omega_{5}\\ x-1&=\omega_{5}\\ (x-1)^{5}&=1\\ x^{5}-5x^{4}+10x^{3}-10x^{2}+5x-1&=1\\ x^{5}-5x^{4}+10x^{3}-10x^{2}+5x-2&=0\end{array}$

$\begin{array}[]{rl}x&=\sqrt[3]{2}+\sqrt[3]{3}\\ x^{3}&=2+3\sqrt[3]{2^{2}\cdot 3}+3\sqrt[3]{2\cdot 3^{2}}+3\\ x^{3}-5&=3\sqrt[3]{6}(\sqrt[3]{2}+\sqrt[3]{3})\\ x^{3}-5&=3\sqrt[3]{6}\,x\\ (x^{3}-5)^{3}&=27\cdot 6x^{3}\\ x^{9}-3\cdot 5x^{6}+3\cdot 25x^{3}-125&=162x^{3}\\ x^{9}-15x^{6}-87x^{3}-125&=0\end{array}$

Since $x^{2}-x-1$ is a quadratic and has no roots in $\mathbb{Q}$ , it is irreducible over $\mathbb{Q}$ . Thus, it is the minimal polynomial over $\mathbb{Q}$ for $\displaystyle\frac{1+\sqrt{5}}{2}$ .

On the other hand, $x^{5}-5x^{4}+10x^{3}-10x^{2}+5x-2$ factors over $\mathbb{Q}$ as $(x-2)(x^{4}-3x^{3}+4x^{2}-2x+1)$ . Since $1+\omega_{5}$ is not a root of $x-2$ , it must be a root of $x^{4}-3x^{3}+4x^{2}-2x+1$ . Moreover, this polynomial must be irreducible. This fact can be proven in the following manner: Let $m(x)$ be the minimal polynomial for $1+\omega_{5}$ over $\mathbb{Q}$ . Since $\mathbb{Q}(1+\omega_{5})=\mathbb{Q}(\omega_{5})$ , $\deg m(x)=[\mathbb{Q}(1+\omega_{5})\!:\!\mathbb{Q}]=[\mathbb{Q}(\omega_{5})\!:% \!\mathbb{Q}]=\varphi(5)=4=\deg(x^{4}-3x^{3}+4x^{2}-2x+1)$ . (Here $\varphi$ denotes the Euler totient function.) Since $m(x)$ divides $x^{4}-3x^{3}+4x^{2}-2x+1$ and they have the same degree, it follows that $m(x)=x^{4}-3x^{3}+4x^{2}-2x+1$ .

It turns out that $x^{9}-15x^{6}-87x^{3}-125$ is irreducible over $\mathbb{Q}$ . (This can be proven in a manner as above. Note that $[\mathbb{Q}(\sqrt[3]{2}+\sqrt[3]{3})\!:\!\mathbb{Q}]=9$ .) Thus, it is the minimal polynomial over $\mathbb{Q}$ for $\sqrt[3]{2}+\sqrt[3]{3}$ .

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