Is the extension $\mathbb{Q}(\sqrt[m]{n})/\mathbb{Q}$ not Normal for $m > 1$ and $n$ square free?

Question

Is the extension $\mathbb{Q}(\sqrt[m]{n})/\mathbb{Q}$ not normal for $m > 2$ and $n>1$ square free?

My understanding is that $f(x) = x^m - n$ would be irreducible by Eisenstein's Criterion with respect to a prime factor of $n$, and so $f$ is the minimal polynomial of $\sqrt[m]{n}$ over $\mathbb{Q}$.

However, the extension does not contain the complex roots, and so is not normal. I am using the definition of normal that says if the extension contains a root of an irreducible polynomial, it must contain all the other roots.

Is my understanding and the statement in the title correct?

Answer 1

If $n>1$ (so that $n$ has some prime factor), then yes.