How do I prove the fact that for any valuation ring $V$ the ideals are totally ordered under inclusion?


closed as off-topic by user26857, Tianlalu, Brahadeesh, mrtaurho, KReiser Dec 19 '18 at 7:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user26857, Tianlalu, Brahadeesh, mrtaurho, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 4
    $\begingroup$ What is your definition of a valuation ring? $\endgroup$ – Bernard Nov 17 '18 at 20:55
  • 1
    $\begingroup$ @Bernard That V is a valuation ring if either x or x^-1 is contained in V where x is in its field of fractions F $\endgroup$ – Gentiana Nov 17 '18 at 20:56


  • Prove first the principal ideal in $V$ are totally ordered by inclusion: for this, let $a,b\in V$. Show that if $Va\not\subset Vb$, then $Vb\subset Va$.
  • Deduce that, if $\mathfrak a$ and $\mathfrak b$ are two ideals in $V$, if $\mathfrak a\not\subset \mathfrak b$, then $\mathfrak b\subset \mathfrak a$ (take $a\in\mathfrak a$, $\;a\notin\mathfrak b$. Show that, for any $b\in\mathfrak b$, $b\in Va$).

Not the answer you're looking for? Browse other questions tagged or ask your own question.