How do I prove the fact that for any valuation ring $V$ the ideals are totally ordered under inclusion?
Asked
Active
Viewed 510 times
2
-
4What is your definition of a valuation ring? – Bernard Nov 17 '18 at 20:55
-
1@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 – Nov 17 '18 at 20:56
1 Answers
2
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 $b\in\mathfrak b$, $b\in Va$).
Bernard
- 179,256
-
damn it why did you get banned I have questions to ask you on this post :( – Clemens Bartholdy Jun 27 '22 at 00:10