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 (see this entry 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 algebra'': 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 simple example is $\sqrt[4]{2}$ :

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

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 similar 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 $\sqrt[3]{2}+\sqrt[3]{3}$ .