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 \(k f, \frac{g}{k} \in \mathbb{Z}[X]\), 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.