Are the $n^\text{th}$ roots of an algebraic number algebraic?

Question

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?

Answer 1

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) &= q_k x^{nk} + \ldots + q_1 x^n + q_0 \\ &= q_k (x^n)^k + \ldots + q_1 (x^n)^1 + q_0 \\ &= q_k (a)^k + \ldots + q_1 (a)^1 + q_0 \\ &= 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.

Answer 2

An algebraic extension of an algebraic extension is algebraic the roots of $X^n-\alpha$ are algebraic over $\mathbb{Q}(\alpha)$.

If $u$ is a root of $X^n-\alpha$, $[\mathbb{Q}(\alpha)(u):\mathbb{Q}]=[\mathbb{Q}(\alpha)(u):\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]$ is finite implies that there exists $m$ such that $1,u,...,u^m$ is not independednt over $\mathbb{Q}$ and $u$ is algebraic over $\mathbb{Q}$.

Answer 3

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.

Answer 4

This is a standard fact in the theory of field extensions.

Remember: $z\in\mathbb{C}$ is algebraic (over a subfield $F$) if and only if the extension $F\subset F(z)$ has finite degree $[F(z):F]$.

Here $F(z)$ denotes the smallest field containing $F$ and $z$.

Now suppose that $\alpha$ is algebraic over $\mathbb{Q}$ and $\beta$ is algebraic over $\mathbb{Q}(\alpha)$ (your situation is a special instance of this more general setting).

Note that we have $$ \mathbb{Q}\subset\mathbb{Q}(\alpha)\subset\mathbb{Q}(\alpha,\beta). $$

By multiplicativity in towers, the degree $$ [\mathbb{Q}(\alpha,\beta):\mathbb{Q}]= [\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)]\cdot[\mathbb{Q}(\alpha):\mathbb{Q}] $$ is finite because both factors are finite. Then also $[\mathbb{Q}(\beta):\mathbb{Q}]$ is finite since $\mathbb{Q}(\beta)\subset\mathbb{Q}(\alpha,\beta)$ and you are done.