Integral Multiple of an Algebraic Number
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Theorem
Let $K$ be a number field and $\alpha \in K$.
Then there exists a positive $n \in \Z$ such that $n \alpha \in \OO_K$.
In this context, $ \OO_K$ denotes the algebraic integers in $K$.
Proof
If $\alpha = 0$ then any integer works and the proof is finished.
Let $\alpha \ne 0$.
Let $\map f x = x^d + a_{d - 1} x^{d - 1} + \dotsb + a_0$ be the minimal polynomial of $\alpha$ over $\Q$.
Suppose that $a_i = \dfrac {b_i} {c_i}$ is a reduced fraction for each $i$ such that $a_i \ne 0$.
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:
\(\ds \map g x\) | \(=\) | \(\ds n^d \map f {\frac x n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n^d \paren {\frac {x^d} {n^d} + a_{d - 1} \frac {x^{d - 1} } {n^{d - 1} } + \dotsb + a_0}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x^d + n a_{d - 1} x^{d - 1} + \dotsb + n^d a_0\) |
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.
$\blacksquare$