Calculating the degree of $[\mathbb Q(\sqrt[n]{m}):\mathbb Q]$

Question

Let $m\in\mathbb Z$ with a prime factorization of the form $m=p\Pi p_i^{n_i}$, $p\neq p_i$.

How can I calculate $[\mathbb Q(\sqrt[n]{m}):\mathbb Q]$ for a natural number $n$?

Answer 1

For any natural number $n$, $[Q(\sqrt[n]{m}):Q]=n$. Note that $\alpha=\sqrt[n]{m}$ is a root of $f(x)=x^n-m$, since $m\in Q$, $f(x)$ is a polinomial in $Q[x]$. Note, for the Eisenstein criterion, $f(x)$ is irreducible over $Q$, since $p|m$, $p\not | 1$(the coefficient of $x^n$) and $p^2\not | m$, since $p\not= p_i$ for all $i$. So we can conclude that $f$ is the minimun polinomial for $\alpha$ in $Q$, so $[Q(\sqrt[n]{m}):Q]=n$.