# Ideals in valuation rings [closed]

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, KReiserDec 19 '18 at 7:09

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• What is your definition of a valuation ring? – Bernard Nov 17 '18 at 20:55
• @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 – Gentiana Nov 17 '18 at 20:56

## 1 Answer

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 $$b\in\mathfrak b$$, $$b\in Va$$).