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 $n \alpha \in \mathcal O_K$.

In this context, $ \mathcal O_K$ denotes the algebraic integers in $K$.

Proof

If $\alpha = 0$ then any integer works and we are done.

Let $\alpha \ne 0$.

Let $f \left({x}\right) = x^d + a_{d-1}x^{d-1} + \ldots + a_0$ 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 $a_i\neq 0$.

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle g(x)\)	\(=\)	\(\displaystyle n^d f\left({\frac x n}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle n^d \left({\frac{x^d}{n^d} + a_{d-1} \frac{x^{d-1} }{n^{d-1} } + \ldots + a_0}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x^d + n a_{d-1} x^{d-1} + \ldots + n^da_0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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.

$\blacksquare$