If $c\in\mathbb C$ is an algebraic number then for every $k\in\mathbb Z$, $kc$ is also algebraic

Question

Prove that if $c\in\mathbb C$ is an algebric number then for every $k\in\mathbb Z$, $kc$ is also algebric

So if $c$ is happened to be a rational it's trivial.

How to solve the case of $c\in\mathbb C \setminus \mathbb Z$?

Answer 1

If $k=0$ the problem is trivial, and if $k \neq 0$, since $c$ is a solution to a rational coefficient polynomial, say $P(x)$, we have that $kc$ is a solution to $P\left( \dfrac{x}{k} \right)$, another rational coefficient polynomial (note $k$ is an integer). So $kc$ is a solution to a rational coefficient polynomial.

Answer 2

Hint: Use the definition of algebraic directly. Given a polynomial $p$ with rational coefficients such that $p(c) = 0$, and an integer $k$, can you make a polynomial $q$ such that $q(kc) = 0$?

Answer 3

Some authors define algebraic numbers as the roots of polynomials with integer coefficients (although of course, if we allow rational coefficients, the definition is equivalent).

If $p\in\Bbb Z[X]$, $\deg p=n$ and $p(c)=0$ then $q(X)=k^np(X/k)\in\Bbb Z[X]$ and $q(ck)=0$.