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 $A$.


1 Answer 1


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 $c_A(\mathcal{O}_1)=c_A(\mathcal{O}_2)=XF(t)[[X]]$.

  • $\begingroup$ 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. $\endgroup$ Commented Feb 6, 2020 at 12:04
  • $\begingroup$ 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. $\endgroup$ Commented Feb 7, 2020 at 13:19

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