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I think that what I ask must be true. I would need such lemma to finish the proof that for every ideal $I$ in $\mathcal{O}_K$ there exists a field extension $L$ of $K$ such that $I$ is a principal ideal in $\mathcal{O}_L$.

My proof goes like the first answer to this question: Every ideal of an algebraic number field can be principal in a suitable finite extension field

However, as I have not yet been introduced to fractional ideals, I do not see why, perhaps trivially, $I^m=J^m$ implies $I=J$, where $I$ and $J$ are both fractional ideals. But I have seen a proof of unique factorization into prime ideals for the ring of integers of a number field (I know that unique factorization into prime ideals holds for any Dedekind domain, despite not having yet learned the proof of this).

Can one prove what I ask in the question without needing fractional ideals? Or, if they are handy for this purpose, could you explain which properties of fractional ideals are used to deduce that $I^m=(a)^m$ implies $I=(a)$, for ideals in $\mathcal{O}_L$ ?

Federica Z.
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    As you say, any ideal in $\mathcal{O}{L}$ admits a unique factorization into a product of prime ideals of $\mathcal{O}{L}$. With that in mind, suppose you had the prove the analogous statement that two integers $a, b$ which satisfy $a^{m} = b^{m}$ for some $m \geqslant 1$ must be equal (up to sign). How could you prove this statement using unique factorization in $\mathbb{Z}$? Could you adapt your proof to the setting you're interested in? – Alex Wertheim Feb 17 '19 at 22:32
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    Yes, I can adapt the proof – I now see that my question was easy. Thanks very much! – Federica Z. Feb 17 '19 at 22:40
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    Glad I could help! – Alex Wertheim Feb 18 '19 at 06:59

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