- Let $u$ a algebraic number. Prove that exists a natural number $n\in \mathbb{Z}$ such that $nu$ is a algebraic integer
- If $u$ is algebraic integer and $n\in \mathbb{Z}$ then $u+n$ and $nu$ are algebraic integers.
I don't see how can I start.
Remember that: $u$ is algebraic integer if it is a root by a monic polynomial $f(x)\in \mathbb{Z}[x]$.
Best Answer
Let me give you a hint for the first part. Let's take a concrete example here. Suppose that $x$ satisfies the monic polynomial equation
$$ x^2 - \frac{3}{4}x + \frac{1}{5} = 0 $$
Then you want to prove that there exists an integer $n$ such that $nx$ satisfies a similar monic polynomial equation but with integer coefficients. Well the idea is that you can clear denominators. For example since $4$ and $5$ appear as denominators we can multiply by $20$ to get
$$ 20x^2 - 15x + 4 = 0 $$
But now the problem is that this is no longer monic. But then we can multiply again by 20 so that we get a factor $20^2$ in the leading coefficient which then can be absorbed by the square as follows
$$ 20^2 x^2 - 15\cdot 20x + 4\cdot20 = 0 \implies (\color{red}{20x})^2 -15 \cdot(\color{red}{20x}) + 4\times 20 = 0 $$
so you see that now $\alpha = 20x$ satisfies
$$ \alpha^2 - 15\alpha + 80 = 0 $$
Thus $\alpha = 20x$ is an algebraic integer. And this same idea works in general.