How far from integrally closed can a finitely generated subring of a number field be? For example, is it possible to have a finitely generated subring of a number field that has infinite index (as additive group) in its integral closure? And is it true that every integrally closed finitely generated subring of a number field is a localization of a ring of integers?

  • $\begingroup$ You ask about subrings but then mention localization. Localization of a domain makes the ring bigger, not smaller. So perhaps when you ask about localization, you are asking about the localization of some other number field? $\endgroup$
    – RghtHndSd
    Jun 29, 2019 at 17:13
  • $\begingroup$ The integers is infinite index in any number ring of dimension at least 2. $\endgroup$ Jun 29, 2019 at 17:14
  • $\begingroup$ For $K$ a number field and a subring $A$ of $K$ then $A$ is finitely generated $\Bbb{Z}$-module iff $A \subset O_K$ (because $a \in K$ is algebraic integer iff $\Bbb{Z}[a]$ is finitely generated $\Bbb{Z}$-module) and $A$ is integrally closed iff it contains $O_K$ ($A$ contains $\Bbb{Z}$ so to be integrally closed it must contain $O_K$ so $A = O_K[ \bigcup a_j^{-1}]$ which is always integrally closed). A last problem is when $A=\sigma(A)$ and $A$ contains (the integral closure of) $N_{K/\Bbb{Q}}(A)$ $\endgroup$
    – reuns
    Jun 29, 2019 at 17:20
  • $\begingroup$ I am asking whether a finitely generated subring of a number field has finite additive index in its integral closure. And also whether the integral closure of such a subring is a localization of the ring of integers. Although I should have said of course that I naturally want to restrict attention to subrings whose field of fractions is the given number field. $\endgroup$ Jun 29, 2019 at 20:18


You must log in to answer this question.

Browse other questions tagged .