rational algebraic integers

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\textbf{Theorem.}\, A rational number is an algebraic integer iff it is a rational integer.\\

\emph{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
$$f(x) \;=\; x^k+a_1x^{k-1}+\ldots+a_k$$
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
$$m^k \;=\; -a_1m^{k-1}n-\ldots-a_kn^k,$$
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,\, $\alpha = m \in \mathbb{Z}$.

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