1
$\begingroup$

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 $A$.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – nowhere dense Feb 6 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. $\endgroup$ – Dario Spirito Feb 7 at 13:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.