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Given a number field K show that there exists a number field extension L of K such that every ideal in K becomes a principal ideal in L.

Hoot
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user138710
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  • Is this not one of the deep results of class field theory? If so, then it is too broad for the site IMHO. As in, a full answer fills a book. You can ask for an explanation and/or insight for a specific step, or for Leitfaden. We have some users capable of providing those. – Jyrki Lahtonen Nov 22 '15 at 09:10
  • @JyrkiLahtonen It's a famous theorem of Furtwangler that the Hilbert class field will do (see here: https://en.wikipedia.org/wiki/Principal_ideal_theorem). But this question is, at least in theory, weaker, and can maybe be done with less technology. Since the class group is finite, it suffices to find a field extension killing just one ideal class (we can then take the compositum of all such fields over the class group). – hunter Nov 22 '15 at 09:17
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    Can some one prove just for integral ideal – user138710 Nov 22 '15 at 10:05
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    See this answer: http://math.stackexchange.com/questions/241348/every-ideal-of-an-algebraic-number-field-can-be-principal-in-a-suitable-finite-e – user219197 Nov 22 '15 at 17:10

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Assume for the sake of simplicity that the class group is cyclic and generated by the class $c$. Pick an integral ideal ${\mathfrak a}$ in $c$ and write ${\mathfrak a}^h = (\alpha)$. Set $L = K(\sqrt[h]{\alpha})$. The generalization to more than one generator is easy.