Let $A$ be a domain and $K=\mathrm{Frac}(A)$. 


The Zariski Riemann space $\mathrm{ZR}(K,A)$ is the set of all valuation rings of $K$ containing $A$. It comes with a natural center map
\begin{align}c\_A:\mathrm{ZR}(K,A)&\rightarrow \mathrm{Spec}\,A \\ (\mathcal{O},\mathfrak{m})&\mapsto\mathfrak{m}\cap A\end{align}



> 
> **Problem:** Let $\mathcal{O}\_1, \mathcal{O}\_2$ valuation rings in $\mathrm{ZR}(K,A)$ such that $\mathcal{O}\_1 \subsetneq \mathcal{O}\_2$. Is it true that $c\_A(\mathcal{O}\_1)\subsetneq c\_A(\mathcal{O}\_2)$?
> 
> 
> 


As $\mathcal{O}\_1 \subseteq \mathcal{O}\_2\iff\mathcal{O}\_1 \in \overline{\{\mathcal{O}\_2\}}$ with the Zariski topology of $\mathrm{ZR}(K,A)$, and as every irreducible closed subset is the closure of a point (because $\mathrm{ZR}(K,A)$ is a Spectral space). This can be used to understand how $c\_A$ relates the naive Krull dimension definition on $\mathrm{ZR}(K,A)$ to the Krull dimension of $\mathrm{Spec}(A)$.


Also notice that if $\mathcal{O}\_1 \subsetneq \mathcal{O}\_2$ then $\mathcal{O}\_2$ is the localization of $\mathcal{O}\_1$ on the prime $\mathfrak{m}\_2\in \mathrm{Spec}\,\mathcal{O}\_1$. Hence, it is equivalent to prove that the map $\mathrm{Spec}\,\mathcal{O}\_1\rightarrow \mathrm{Spec}(A\_{\mathfrak{m}\_1\cap A})$ is injective. So we can restate the problem as



> 
> **Problem':** Let $A$ be a local domain dominated by a valuation ring $\mathcal{O}$ of $\mathrm{Frac}(A)$. Does the extension of rings $A\subseteq \mathcal{O}$ satisfy the incomparability property?
> 
> 
> 


**Edit later:** Here is another more geometric counterexample on algebras of finite type. Let $X$ be the blow-up of $\mathbb{A}^2$ at $(0,0)$ and let $E$ be the exceptional divisor. By taking a point in $p\in E$ we can define a rank $2$ valuation in $K=K(X)$ that sends a rational function $f$ to $(n\_1,n\_2)$ where $n\_1$ is the order of vanishing at $E$ and $n\_2$ is the order of vanishing of $(f\pi^{-n\_1})\_{|E}$ at $p$ where $\pi$ is a local equation for $E$ around $p$. If $\mathcal{O}$ is the valuation ring of this valuation then it dominates $A=\mathcal{O}\_{\mathbb{A}^2,(0,0)}$ but the two primes ideals of it goes to the maximal ideal in $A$.



Chain of closed irreducible sets on Zariski Riemann spaces
The answer to the problem is no. One reason is that the dimension of a valuation ring $V\in\mathrm{ZR}(K,A)$ may be greater than the dimension of $A$. (The supremum of the dimension of the elements of the Zariski space of $A$ is called the *valuative dimension* of $A$.)


For example, let $F$ be a field, $t,X$ indeterminates over $F$ and define $A:=F+XF(t)[[X]]$ (that is, $A$ is the set of all power series over $F(t)$ such that the term of degree $0$ is in $F$). Then, $A$ is a one-dimensional local ring (the only prime ideals are $(0)$ and $XF(t)[[X]]$).


The elements of the Zariski space (aside from the quotient field) are those in the form $V+XF(t)[[X]]$, where $V$ is an extension of $F$ to $F(t)$, so they all have center $XF(t)[[X]]$. For example, one can take $\mathcal{O}\_1=F[t]\_{(t)}+XF(t)[[X]]$ and $\mathcal{O}\_2=F(t)[[X]]$: they are both valuation rings inside the quotient field, $\mathcal{O}\_1\subsetneq\mathcal{O}\_2$ but $c\_A(\mathcal{O}\_1)=c\_A(\mathcal{O}\_2)=XF(t)[[X]]$.


