Minimal Polynomial of Algebraic Integer

See classification of algebraic integers for other equivalent definitions of algebraic integers.

Proof

Suppose $\alpha$ is algebraic, that is, there exists a monic polynomial $h\in \mathbb{Z}[X]$ such that $h(\alpha )=0$. Since the minimal polynomial of $\alpha$, which we denote by $f(\alpha )$, divides $h$, we write

for some $g\in \mathbb{Q}[X]$. Since $f$ is monic by definition, and $h$ is monic by assumption $g$ must also be monic. Now, applying this form of Gauss' lemma, there exists a rational constant $k$ such that $kf,\frac{g}{k}\in \mathbb{Z}[X]$, and we may write

Since $f$ is already an integer polynomial, it follows that $k$ must be an integer. However, since $g$ is monic, $k$ must be an integer unit for the leading coefficient of $\frac{g}{k}$ to be an integer, that is $k\in \{\pm 1\}$. Therefore $f$ and $g$ must already have integer coefficients, and thus the minimal polynomial of $\alpha$, $f$, has integer coefficents.