rational algebraic integers

Theorem.  A rational number is an algebraic integer iff it is a rational integer.

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

$$f(x)=x^k+a_1{x}^{k-1}+\mathrm{\dots }+a_k$$

with  $a_1,\mathrm{\dots },a_k\in \mathbb{Z}$  such that

$$f(\alpha )={\left(\frac{m}{n}\right)}^k+a_1{\left(\frac{m}{n}\right)}^{k-1}+\mathrm{\dots }+a_k=\mathrm{\hspace{0.33em}0}.$$

Multiplying this equation termwise by $n^k$ implies

$$m^k=-a_1{m}^{k-1}n-\mathrm{\dots }-a_k{n}^k,$$

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
Canonical name	RationalAlgebraicIntegers
Date of creation	2013-03-22 19:07:34
Last modified on	2013-03-22 19:07:34
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	5
Author	pahio (2872)
Entry type	Theorem
Classification	msc 11R04
Related topic	MultiplesOfAnAlgebraicNumber