If $a$ is an algebraic number, show that there exists a positive
integer $n$ such that $na$ is an algebraic integer.
I'm supposing that $a$ is algebraic over a field $F$, that is, there exists a polynomial $p(x)$ over $F$ such that $p(a) = 0$ . If we think about $F$ being $\mathbb{Q}$, then $a = p/q$. And if we think that $p(x) = x-a$, we have that $p(x) = x-p/q$ is our polynomial. But I don't know if it helps. I don't even know if we can suppose that $x-a$ is our polynomial
Best Answer
Let $f(a)=0$ be monic, degree $n$. If the polynomial is not already a polynomial over the integers, it has a term whose coefficient is rational, with denominator $q$ when fully reduced. Suppose that's the coefficient of $x_k$.
Let $g(x)=q^nf(x/q)$. Note that the coefficient of $x^k$ in $g$ is a whole number, $g$ is monic, and $g(qa)=0$.
If $g$ still has some non-integer rational coefficients, repeat until you do have a polynomial whose coefficients are all integers.