multiples of an algebraic number

Theorem.  If $\alpha$ is an algebraic number, then there exists a non-zero multiple (http://planetmath.org/GeneralAssociativity) of $\alpha$ which is an algebraic integer.

Proof.  Let $\alpha$ be a of the equation

$$x^n+r_1{x}^{n-1}+r_2{x}^{n-2}+\mathrm{\cdots }+r_n=0,$$

where $r_1$, $r_2$, …, $r_n$ are rational numbers ($n>0$).  Let $l$ be the least common multiple of the denominators of the $r_j$’s.  Then we have

$$0=l^n({\alpha }^n+r_1{\alpha }^{n-1}+r_2{\alpha }^{n-2}+\mathrm{\cdots }+r_n)={(l\alpha )}^n+l{r}_1{(l\alpha )}^{n-1}+l^2{r}_2{(l\alpha )}^{n-2}+\mathrm{\cdots }+l^n{r}_n,$$

i.e. the the algebraic equation

$$x^n+l{r}_1{x}^{n-1}+l^2{r}_2{x}^{n-2}+\mathrm{\cdots }+l^n{r}_n=0$$

with rational integer coefficients.

According to the theorem, any algebraic number $\xi$ is aquotient (http://planetmath.org/Division) of an algebraic integer(of the field $\mathbb{Q}(\xi )$) and a rational integer.