I was looking the valuation ring of dimension $2$. Then I found, it has two number of non-zero prime ideals and localization at prime is a valuation domain again. Moreover, there is a one-to-one correspondence between prime ideals and valuation overrings. Dimension 2 means, we have two non-zero prime ideals, then the valuation ring should have two valuation overrings different from quotient field. We will get one by localize at prime ideal of height 1 and another one by localized at maximal ideal. I have difficulty to figure out the overrvaluation ring with respect to maximal ideal. That is, let $V$ be a valuation rings and $0\subseteq p\subseteq m$ then $V\subseteq V_{m} \subseteq V_{(p)}\subseteq V_{(0)} = K$, quotient field. The second sequence forces me that $V$ has dimension $3$ instead of $2$. Where is my mistake?

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    $\begingroup$ The maximal chain $0 \subset p \subset m$ shows that $V$ has dimension $2$. That's all. $\endgroup$ – Harry Mar 18 '12 at 17:32
  • $\begingroup$ Thanks Harry. I think this is the case, the localization of local ring at its maximal ideal is equal to itself. $\endgroup$ – Rajnish Mar 18 '12 at 19:46

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