I want to show that, given an ideal $I \subseteq \mathcal O_K$ (where $K/\mathbb Q$ is an algebraic number field), there is a finite extension $K'/K$ such that, $I\mathcal O_{K'}$ becomes a principal ideal in $\mathcal O_{K'}$.
The example solution for this problem uses $K' = K[X]/(X^h - x)$ where $(x) = I^h$ and $h$ is the class number for the field $K$.
I see why $I\mathcal O_{K'}$ is a principal ideal, however what I fail to see is why $K'$ is a field. I know that $K'$ being a field is equivalent to $(X^h - x)$ being a maximal ideal, and this in turn is equivalent to $X^h - x$ being irreducible. Why is this the case?
We could just as easily write $K'=K(\sqrt[h]{x})$. One issue in writing $K'=K[X]/(X^h-x)$ is that the class group ${\rm Cl}(K)$ needn't be simple, so there can be elements with order properly dividing the class number $h$, in which case if $I^r=(y)$ with $r< h$ then $I^h=(y^{h/r})$ implying $X^h-y^{h/r}$ is reducible over $K$ since it's expressible as $(X^r)^{h/r}-y^{h/r}$, so indeed $K[X]/(X^h-x)$ wouldn't be a domain.
If $r$ is the order of the ideal class $[I]$ in ${\rm Cl}(K)$ and $I^r=(x)$ then $x\not\in K^\ell$ for any proper $\ell\mid r$, however this latter condition is not enough to imply $X^r-x$ is irreducible - see here; the extra necessary condition is that if $4\mid r$ then $x\not\in -4K^4$. (I am not sure of a counterexample for the ideal class condition implying the irreducibility of $X^r-x$ unfortunately, or if there can be one.)