Are the $n^\text{th}$ roots of an algebraic number algebraic? Suppose that $\alpha \in \mathbb{C}$ is algebraic over $\mathbb{Q}$, and that $n$ is a positive integer. Is it true that the roots of $x^n - \alpha \in \mathbb{C}[x]$ are algebraic over $\mathbb{Q}$?
This is certainly true when $\alpha \in \mathbb{Q}$. The case $n = 1$ is also trivial. Is the result true for arbitrary $\alpha$ and $n$? If so, how could one start a proof?
To say that $a$ is algebraic over $\Bbb Q$ is to say that there are numbers
$q_k, \ldots, q_2, q_1, q_0 \in \Bbb Q$ such that
$$
p(a) = q_k a^k + \ldots + q_1 a + q_0 = 0
$$
(where this evaluation takes place in $\Bbb C$ for the current problem).
Suppose that
$$
x^n - a = 0
$$
Then $x^n = a$. Consider the polynomial
$$
h(u) = q_k u^{nk} + \ldots + q_1 u^n + q_0.
$$
Note that all the coefficients of this polynomial are still rationals.
Let's evaluate $h(x)$ for a particular $x$ that satisfies $x^n - a = 0$:
\begin{align}
h(x) 
&amp;= q_k x^{nk} + \ldots + q_1 x^n + q_0 \\
&amp;= q_k (x^n)^k + \ldots + q_1 (x^n)^1 + q_0 \\
&amp;= q_k (a)^k + \ldots + q_1 (a)^1 + q_0 \\
&amp;= 0\end{align}
hence $x$, by satisfying a polynomial with rational coefficients, is also algebraic over $\Bbb Q$.
Concrete example: from
$$
x^2 - x - 1 = 0
$$
we know that
$$
a = \frac{1 + \sqrt{5}}{2}
$$
is algebraic. What about a cube root, $u$, of $a$? Well, look at the polynomial above, with $x$ replaced by $x^3$ in every term, i.e.,
$$
x^6 - x^3 - 1 = 0.
$$
When we plug in $u$, we get
$$
u^6 - u^3 - 1 = (u^3)^2 - (u^3) - 1 = a^2 - a - 1 = 0
$$
so $u$, also, satisfies a polynomial over the rationals.
The algebraic numbers form a so-called algebraically-closed field. That means every polynomial with algebraic coefficients has an algebraic root. It follows that all the roots of a polynomial with algebraic coefficients are algebraic.