# examples of minimal polynomials

Note that $\sqrt{2}$ is algebraic over the fields $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{2})$. The minimal polynomials for $\sqrt{2}$ 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 ($\sqrt{2}$ and $-\sqrt{2}$) 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 $\sqrt{2}$:

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

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 $\sqrt{2}+\sqrt{3}$.

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

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

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

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 $[\mathbb{Q}(\sqrt{2}+\sqrt{3})\!:\!\mathbb{Q}]=9$.) Thus, it is the minimal polynomial over $\mathbb{Q}$ for $\sqrt{2}+\sqrt{3}$.

Title examples of minimal polynomials ExamplesOfMinimalPolynomials 2013-03-22 16:55:18 2013-03-22 16:55:18 Wkbj79 (1863) Wkbj79 (1863) 13 Wkbj79 (1863) Example msc 12F05 msc 12E05 IrreduciblePolynomialsObtainedFromBiquadraticFields