# Chain of closed irreducible sets on Zariski Riemann spaces

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?

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 $$c_A(\mathcal{O}_1)=c_A(\mathcal{O}_2)=XF(t)[[X]]$$.
• Thanks, that is really interesting! Do you have any idea of what will happen when $A$ is an algebra of finite type over a field? In this case by Abhyankar's inequality the rational rank, and hence the rank, is bounded by the krull dimension of $A$, so we can't have the same type of obstruction. – yamete kudasai Feb 6 at 12:04
• I'm not sure. For algebras of finite type probably one can try by induction (if $K=\mathrm{Frac}(A)=F[x_1,\ldots,x_n]$, restrict everything to $F[x_1,\ldots,x_{n-1}]$, then to \$F[x_1,\ldots.x_{n-2}], etc.: all dimensions should decrease by 1 each time), but I haven't worked it out. I also feel there should be an easier proof for arbitrary Noetherian rings. – Dario Spirito Feb 7 at 13:19