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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?

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Me.
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    I wonder if you have to make the assumption that $I^r$ is not principal for $r<h$. For example, if $I=(y)$ is already principal, then $x=y^h$, and $X^h-y^h$ doesn't look irreducible. – Gerry Myerson Aug 01 '14 at 13:04
  • Interesting. Assume we do, i.e. $h > 0$ is minimal with the property that $I^h$ is principal. Does this mean, that $X^h - x$ is irreducible? I mean, we basically quotient out a minimal polynomial, why does it have to have the form $X^h - x$?

    I guess we can assume that for all non-trivial $j$-th roots of $x$, $j$ and $h$ are coprime, and the irreducibility should somehow follow... I find it a little too inelegant though.

     – Me. Aug 01 '14 at 13:53
  • Is it not enough to prove it for prime ideals first then take suitable compositums? – fretty Aug 02 '14 at 07:27
  • Related: https://math.stackexchange.com/questions/241348 – Watson Feb 17 '18 at 20:45

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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.)

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