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

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

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    $\begingroup$ What is your definition of a valuation ring? $\endgroup$ – Bernard Nov 17 '18 at 20:55
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    $\begingroup$ @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
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Hint:

  • 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 $b\in\mathfrak b$, $b\in Va$).
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