$k$ algebraically field, $A$ $k$ algebra and valuation ring of $K$ ($K$ field fraction of $A$) and we have the transcendence degree of $K$ over $k$ is one.


i want to ask if $A$ is noetherian ring? 



question about valuation ring
The answer is no via the following easy counterexample. Let $k$ be an algebraically closed field, and $K=\overline{k(t)}$. Let $v$ be an extension of the $t$-adic valuation to $K$, and let $A$ be the valuation ring of $v$. Then $A$ is a valuation ring of $K$ and $\operatorname{tr.deg}\_k(K)=1$, but since the value group of $A$ is $\mathbf Q$, $A$ is not noetherian.


