# rational algebraic integers

Proof.$1^{\circ}$.  Any rational integer $m$ has the minimal polynomial $x\!-\!m$, whence it is an algebraic integer.
$2^{\circ}$.  Let the rational number  $\alpha=\frac{m}{n}$  be an algebraic integer where $m,\,n$ are coprime integers and  $n>0$.  Then there is a polynomial

 $f(x)\;=\;x^{k}+a_{1}x^{k-1}+\ldots+a_{k}$

with  $a_{1},\,\ldots,\,a_{k}\in\mathbb{Z}$  such that

 $f(\alpha)\;=\;\left(\frac{m}{n}\right)^{k}+a_{1}\left(\frac{m}{n}\right)^{k-1}% +\ldots+a_{k}\;=\;0.$

Multiplying this equation termwise by $n^{k}$ implies

 $m^{k}\;=\;-a_{1}m^{k-1}n-\ldots-a_{k}n^{k},$

which says that  $n\mid m^{k}$ (see divisibility in rings).  Since $m$ and $n$ are coprime and $n$ positive, it follows that  $n=1$.  Therefore,  $\alpha=m\in\mathbb{Z}$.

Title rational algebraic integers RationalAlgebraicIntegers 2013-03-22 19:07:34 2013-03-22 19:07:34 pahio (2872) pahio (2872) 5 pahio (2872) Theorem msc 11R04 MultiplesOfAnAlgebraicNumber