Let O be an R-order for some Dedekind domain R, let F be the field of fractions of R and D be a division algebra over F. A fractional left ideal of O is an R-lattice I in D such that OI in I (I absorbs multiplication of O on the left).

How do I prove that in such a division algebra, every nonzero fractional (left) ideal of O is a complete R-lattice?

  • $\begingroup$ What's the problem? Every element of $D$ is gotten from an element of $O$ by dividing it with an element of $R$. The latter is in the center of $D$. $\endgroup$ – Jyrki Lahtonen Jun 19 '18 at 6:32
  • $\begingroup$ @JyrkiLahtonen But I need to be able to write every element in $D$ as an element in F multiplied by an element in my fractional left ideal of O, say I. What you said is just the definition of O being complete, no? $\endgroup$ – Jan Gerrit Jun 19 '18 at 6:41
  • $\begingroup$ I have always thought that for $O$ to be an $R$-order of $D$ it must, by definition, contain an $F$-basis of $D$. But, I'm not 100% what is your definition of a complete lattice. $\endgroup$ – Jyrki Lahtonen Jun 19 '18 at 6:47
  • $\begingroup$ @JyrkiLahtonen yes, sorry if that came out wrong. An order is a complete lattice by definition. But how about fractional (left) ideals of orders? $\endgroup$ – Jan Gerrit Jun 19 '18 at 6:48
  • $\begingroup$ @JyrkiLahtonen The definition i am working with says just that: for a lattice to be complete it must contain an F basis of D. $\endgroup$ – Jan Gerrit Jun 19 '18 at 6:49

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