# 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 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:

\(\displaystyle \map g x\) | \(=\) | \(\displaystyle n^d \map f {\frac x n}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle n^d \paren {\frac {x^d} {n^d} + a_{d - 1} \frac {x^{d - 1} } {n^{d - 1} } + \dotsb + a_0}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x^d + n a_{d - 1} x^{d - 1} + \dotsb + n^d a_0\) |

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.

$\blacksquare$