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# 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 (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 $\sqrt[4]{2}$:

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

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

## Mathematics Subject Classification

12F05*no label found*12E05

*no label found*

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