Definition of valuation ring:
Let $R$ be an integral domain with $frac(R)=K$. Then $R$ is said to be a valuation ring if
(1) $R \neq K$
(2) $\forall x \in K, x \in R$ or $x^{-1} \in R$.
Now my problem is:
Let $A, B$ be two valuation rings with $K=frac(A)=frac(B)$. Now suppose $A⊂B$, can we conclude that $A=B$?
I can prove that this is not true if $A$ is a discrete valuation ring:
For suppose the maximal ideal in $A$ is generated by $t$, then the only non units in $A$ are of the form $ut^{n}$, with $u$ a unit in $A$, $n$ an integer $\geq 1$.
If $A \subsetneq B$, then $B$ must contains the inverse of some element of the above form, i. e. $u^{-1}t^{-n}=(ut^n)^{-1} \in B$, but then $ \frac 1 t = (u t^{n-1})(u^{-1}t^{-n}) \in B$. Hence $B\supset A[\frac 1 t ]= K$.
However, I can't think of anything useful in the general case that $A, B$ may not be discrete.
Any help will be appreciated.
If $A$ is a valuation ring with quotient field $K$, then any ring $B$ between $A$ and $K$ is a valuation ring.