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When solving Diophantine equations, often I pass to a number field $K$ and hope that the algebraic integers $O_K$ have unique factorization.

Suppose that $O_K$ is not a UFD. Is it possible that there exists a field extension $L \supset K$ such that $O_L$ has unique factorization? Does such an $L$ always exist? Is there a method to find such an $L$ when it exists? I have tried to construct examples, but the $L$'s always end up too large for me to compute the class number by hand.

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  • $\begingroup$ I don't know a lot about number theory, but so far as I know it's not even known if there exist infinitely many number fields with class number one... $\endgroup$ Mar 25, 2015 at 17:58
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    $\begingroup$ See en.wikipedia.org/wiki/Principal_ideal_theorem , en.wikipedia.org/wiki/Hilbert_class_field and en.wikipedia.org/wiki/Golod%E2%80%93Shafarevich_theorem (This is not a direct answer; I might come back.) $\endgroup$
    – quid
    Mar 25, 2015 at 18:04
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    $\begingroup$ Don't miss the article on Pincipalization (capitulation). With these keywords you should be able to locate much about this classical class field tower problem (Klassenkörper-turmproblem) $\endgroup$ Mar 25, 2015 at 18:28
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    $\begingroup$ There's an old MO question showing that doing this doesn't necessarily make your problem any easier; the price you pay is that $\mathcal{O}_L$ may have a more complicated group of units. I can't find it at the moment, though. $\endgroup$ Mar 25, 2015 at 19:16
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    $\begingroup$ Well, there's a much easier thing to do than passing to another $\mathcal{O}_L$, which is to invert a finite set of primes (say all primes lying over primes less than or equal to the Minkowski bound). You still run into the problem that you've enlarged the group of units this way. $\endgroup$ Mar 26, 2015 at 3:49

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