Integral Multiple of an Algebraic Number

Theorem
Let $K$ be a number field and $\alpha\in K$. Then there exists a positive $n\in\Z$ such that $n\alpha \in O_K$.

Proof
If $\alpha$ is 0 then any integer works and we are done, so we assume $\alpha\neq 0$.

Let $f(x) = x^d + a_{d-1}x^{d-1} + \ldots + a_0$ be the minimal polynomial of $\alpha$ over $\Q$. Suppose that $a_i = \frac{b_i}{c_i}$ is a reduced fraction for each $i$ such that $a_i\neq 0$. Let $n$ be the least common multiple of the $c_i$, of which there must be at least one by our assumptions.

Consider the polynomial,

$g(x) = n^df\left(\frac{x}{n}\right) = n^d\left(\frac{x^d}{n^d} + a_{d-1}\frac{x^{d-1}}{n^{d-1}} + \ldots + a_0\right) = x^d + na_{d-1}x^{d-1} +\ldots + n^da_0.$

Note that $g$ is a monic polynomial with coefficients in $\Z$ by our choice of $n$. Furthermore, by construction, we see that $n\alpha$ is a root of $g$ and is therefore an algebraic integer.